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1200	https://www.vedantu.com/chemistry/law-of-multiple-proportions	Courses for Kids Free study material Offline Centres Talk to our experts Chemistry Law of Multiple Proportions Explained for Students Law of Multiple Proportions Explained for Students Reviewed by: Ritika Singla Download PDF NCERT Solutions NCERT Solutions for Class 12 NCERT Solutions for Class 11 NCERT Solutions for Class 10 NCERT Solutions for class 9 NCERT Solutions for class 8 NCERT Solutions for class 7 NCERT Solutions for class 6 NCERT Solutions for class 5 NCERT Solutions for class 4 NCERT Solutions for Class 3 NCERT Solutions for Class 2 NCERT Solutions for Class 1 CBSE CBSE class 3 CBSE class 4 CBSE class 5 CBSE class 6 CBSE class 7 CBSE class 8 CBSE class 9 CBSE class 10 CBSE class 11 CBSE class 12 NCERT CBSE Study Material CBSE Sample Papers CBSE Syllabus CBSE Previous Year Question Paper CBSE Important Questions Marking Scheme Textbook Solutions RD Sharma Solutions Lakhmir Singh Solutions HC Verma Solutions TS Grewal Solutions DK Goel Solutions NCERT Exemplar Solutions CBSE Notes CBSE Notes for class 12 CBSE Notes for class 11 CBSE Notes for class 10 CBSE Notes for class 9 CBSE Notes for class 8 CBSE Notes for class 7 CBSE Notes for class 6 How Does the Law of Multiple Proportions Work? In chemistry, the law of multiple proportions can be defined as if two elements form more than one compound between them, the mass ratios of the second element that combine with a fixed mass of the first element will always be the ratios of small whole numbers. Sometimes, this law is referred to as Dalton's Law (or Dalton's Law of multiple proportions) because it is named after John Dalton, the chemist who expressed it first. The Law of Multiple Proportions is defined as: “if two elements combine to form more than one compound, the mass ratios of the second element that combine with a fixed mass of the first element will always be ratios of minuscule whole numbers.” This law, sometimes known as Dalton's Law or Dalton's Law of Multiple Proportions, was proposed by Dalton in 1803. Hydrogen, for example, reacts with oxygen to generate two compounds: water and hydrogen peroxide. Hydrogen + Oxygen → Water 2g 16g 18g Hydrogen + Oxygen → Hydrogen Peroxide 2g 32g 34g Examples of Law of Multiple Proportions For example, Dalton knew that the carbon element forms two oxides by combining them with the oxygen atom in various proportions. A fixed mass of carbon compound, let us suppose 100 grams, can react with 266 grams of oxygen to form one oxide atom or with 133 grams of oxygen to form the other. The ratio of the oxygen masses that can react with the 100 grams of carbon is given as 266:133 = 2:1, which is a ratio of small whole numbers. Dalton has interpreted these results in his atomic theory by proposing that the two oxides have one oxygen atom and two oxygen atoms, respectively, for every carbon atom. In modern notation, the first is given as CO (carbon monoxide), and the second is given as [CO_{2}] (carbon dioxide). First, John Dalton expressed this specific observation in 1804. But, a few years before, the French chemist named Joseph Proust had proposed the law of definite proportions, which has expressed that the elements combined to produce compounds in certain well-defined proportions, instead of mixing in just any proportion. And, Antoine Lavoisier has proved the law of conservation of mass that helped out Dalton. A careful study of the exact numerical values of these proportions has led Dalton to propose his law of multiple proportions. This was an essential step toward the atomic theory that he would propose later the same year, and it laid the basis for the compound's chemical formulas. Another example of the law is seen by comparing the ethane ([C_{2}H_{6}]) compound with the propane ([C_{3}H_{8}]) compound. The weight of hydrogen that combines with 1 gram carbon is 0.252 gram in the ethane and 0.224 gram in the propane. The ratio of those weights is 1.125, which is equal to the ratio of two small numbers (9:8). Dalton’s Atomic Theory Democritus has first suggested the atom's existence. However, it took almost two centuries for the atom to be established as a basic chemical entity, which was accomplished by John Dalton (from the years 1766-1844). Dalton's atomic theory remains valid in modern chemical thought, although two centuries old. Important Points of Dalton's Atomic Theory All atoms of the given element are identical in properties and mass All matter is completely made of atoms, and these atoms are indestructible and indivisible A chemical reaction is given as a rearrangement of atoms Compounds are produced by the combination of either two or more various kinds of atoms Modern atomic theory is given as a little more involved than Dalton's theory. But, the essence of Dalton's theory remains completely valid. Currently, we know that atoms are destroyed by nuclear reactions but not by chemical reactions. Also, there are various kinds of atoms (differ by their masses) within an element that are called "isotopes", but the isotopes of an element contain similar chemical properties. (Image Will be Updated Soon) Several heretofore unexplained chemical methods were quickly explained by Dalton with his specific theory. In chemistry, Dalton's theory very quickly became the theoretical foundation. To put it simply and briefly, the atomic structure of a compound is a vital component to understanding the behavior and reactivity of chemicals. Simply by looking at the structure of a molecule, a chemist can be able to predict several things about a chemical such as : Boiling points and melting points Acidity or basicity Reactivity with the other compounds What sort of products that two chemicals might produce in a chemical reaction The list goes on. Not only are there easy and simple ways to predict the properties of compounds, but there are also things that wouldn't be obvious immediately. As an example, several molecules will contain two forms, which are mirror images of each other (which can be based on the arrangement where the molecules are bonded). Every mirror image contains drastically different effects, like in Thalidomide. One mirror image of this molecule can be used as a useful drug, but also the other image, as it turned out, was a carcinogen. Basically, both the short and long of it is that if we understand the atomic structure of a compound, we can rationalize the effects that we didn't necessarily predict and predict its chemical behavior. Conclusion The law of multiple proportions is a fundamental rule of chemistry. According to the law of multiple proportions, if two elements can react to produce several compounds, the ratio of their masses that mix with a fixed mass of the other element is in the ratio of tiny whole numbers. It was instrumental in the early understanding of stoichiometry, along with the law of definite proportions. The concept was proposed in 1803, by English chemist John Dalton, who is best known for his partial pressures law. Dalton stated the law in his book New System of Chemical Philosophy (Vol 1). Proust's law of definite proportions was followed by the law of multiple proportions. The law backed up Proust's work and later aided Dalton in his atomic theory. FAQs on Law of Multiple Proportions Explained for Students What is the Law of Multiple Proportions in simple terms? The Law of Multiple Proportions states that if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers. This law was proposed by John Dalton and provided crucial evidence for his atomic theory. Can you explain the Law of Multiple Proportions with an example? Certainly. Consider carbon (C) and oxygen (O), which form two common compounds: carbon monoxide (CO) and carbon dioxide (CO₂). In carbon monoxide (CO), 12g of carbon combines with 16g of oxygen. In carbon dioxide (CO₂), 12g of carbon combines with 32g of oxygen. Here, the mass of carbon is fixed at 12g. The ratio of the masses of oxygen that combine with this fixed mass of carbon is 16:32, which simplifies to the small whole-number ratio of 1:2. This perfectly illustrates the law. How do water (H₂O) and hydrogen peroxide (H₂O₂) demonstrate the Law of Multiple Proportions? Water (H₂O) and hydrogen peroxide (H₂O₂) are both compounds made of only hydrogen and oxygen. To see how they demonstrate the law, we fix the mass of one element. Let's fix the mass of hydrogen at 2g. In water (H₂O), 2g of hydrogen combines with 16g of oxygen. In hydrogen peroxide (H₂O₂), 2g of hydrogen combines with 32g of oxygen. The ratio of the masses of oxygen that combine with the fixed 2g of hydrogen is 16:32, which simplifies to a simple ratio of 1:2. What is the main difference between the Law of Multiple Proportions and the Law of Definite Proportions? The main difference lies in their scope and application: The Law of Definite Proportions applies to a single compound. It states that a chemical compound always contains its component elements in a fixed ratio by mass, regardless of its source or method of preparation. For example, water will always have hydrogen and oxygen in a 1:8 mass ratio. The Law of Multiple Proportions applies when two elements form two or more different compounds. It compares the mass ratios of an element between these different compounds, showing they relate by a simple whole number. How did the Law of Multiple Proportions support Dalton's Atomic Theory? The Law of Multiple Proportions was a cornerstone of evidence for Dalton's Atomic Theory. Dalton's theory proposed that atoms were indivisible particles that combine in simple, whole-number ratios to form compounds. The fact that the masses of an element combined in a simple whole-number ratio (like 1:2 or 2:3) strongly implied that atoms were combining as discrete, whole units. For instance, one atom of carbon combining with either one atom of oxygen (CO) or two atoms of oxygen (CO₂) would naturally lead to a 1:2 mass ratio for oxygen, validating the atomic concept. How can you mathematically verify if two compounds follow the Law of Multiple Proportions? To mathematically verify the law for two compounds (e.g., Compound A and Compound B made from elements X and Y), you can follow these steps: Determine the mass composition of elements X and Y in both Compound A and Compound B. Choose one element (e.g., X) to have a fixed mass. Calculate the mass of the other element (Y) that combines with 1 gram (or any fixed amount) of X in each compound. You will now have two values: Mass of Y in Compound A (per 1g of X) and Mass of Y in Compound B (per 1g of X). Calculate the ratio of these two masses of Y. If this ratio simplifies to a small whole number (e.g., 1:2, 2:3), the law is verified. 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1201	https://study.com/academy/lesson/tangent-of-a-circle.html	Tangent of a Circle \| Definition, Formula & Examples - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Math Courses Geometry: High School Tangent of a Circle \| Definition, Formula & Examples Contributors: Samuel Hoyland, Jeff Calareso Author Author: Samuel Hoyland Show more Instructor Instructor: Jeff Calareso Show more Learn what a tangent of a circle is by examining the definition, seeing an example of a tangent of a circle, and exploring the formula for calculating the equation of a tangent of a circle. Updated: 11/21/2023 Table of Contents What is a Tangent of a Circle? Tangent Theorems Tangent of a Circle Formula Lesson Summary Show FAQ What is tangent of the circle? A tangent to a circle is a line that intersects on the circle at one point. Most of the time, if a line that intersects a circle at one point, will also intersect it at a second point. To set it up very precisely, make the line intersect the circle at only one point so that it can be a tangent line. How do you find the tangent of a circle? The equation for a tangent line to a circle is (x-x0)(x0-h) + (y-y0)(y0-k) = 0, where (x0, y0) is the point of tangency. What is a tangent in circle theorem? The Tangent-Radius Theorem states that a line that is tangent to a circle will always be perpendicular to the radius of the circle drawn from the point of tangency. The Tangents from the Same Point Theorem states that if a point is outside of the given circle, then two different tangent line segments can be drawn from the point to the circle, and these two line segments will be of the same length. Create an account LessonTranscript VideoQuizCourse An error occurred trying to load this video. Try refreshing the page, or contact customer support. You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? Log in here for access Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. 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Your next lesson will play in 10 seconds 0:06 Circles 0:41 Tangent of a Circle 1:47 Tangent Theorems 2:28 Tangent Circles 3:12 Lesson Summary QuizCourseView Video OnlySaveTimeline 62K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Diameter of a Circle \| Definition, Formula & Examples 3:12 ##### Concyclic Points \| Definition, Properties & Theorem 4:46 ##### Circles in Word Problems 3:40 ##### Central Angle of a Circle \| Definition, Theorem & Formula 6:04 ##### How to Construct a Tangent of a Circle 5:16 ##### Perimeter of a Sector of a Circle \| Formula & Calculation 5:35 ##### Segment Relationships in Circles \| Definition & Theorems 5:28 ##### Circle Definition: Lesson for Kids 3:10 ##### Sector of a Circle \| Definition, Formulas & Practice Problems 4:25 ##### Circles - Ratios: Study.com SAT® Math Exam Prep 11:41 ##### Circles - Radius, Radians and Standard Form: Study.com SAT® Math Exam Prep 15:16 ##### Semicircle \| Definition, Area & Formula 5:57 ##### Circumradius of a Triangle \| Formula, Calculation & Application 5:46 ##### Radius of a Circle Formula & Example \| How to Find the Radius of a Circle 2:22 ##### Circumcircle Definition, Properties & Examples 5:51 ##### Circles \| EP SAT Tutoring 19:25 ##### Concentric Circles \| Definition, Pattern & Examples 1:40 ##### Parts of a Circle \| Definition & Examples 3:11 ##### Point of Tangency \| Definition & Examples 3:01 ##### Theorems of Finding Angle & Arc Measures \| How to Find Angles in a Circle 5:06 ##### Geometry Proofs: Help & Tutorials ##### Basic Geometry: Help & Review ##### ICAS Mathematics - Paper G & H: Test Prep & Practice ##### GED Math: Quantitative, Arithmetic & Algebraic Problem Solving ##### NY Regents - Geometry Study Guide and Exam Prep ##### Holt Geometry: Online Textbook Help ##### Glencoe Geometry: Online Textbook Help ##### McDougal Littell Geometry: Online Textbook Help ##### Prentice Hall Geometry: Online Textbook Help ##### Amsco Geometry: Online Textbook Help ##### Holt McDougal Larson Geometry: Online Textbook Help ##### Discovering Geometry An Investigative Approach: Online Help ##### Geometry 101: Intro to Geometry ##### 10th Grade Geometry Textbook ##### Ohio Graduation Test: Study Guide & Practice ##### Ohio End of Course Exam - Geometry: Study Guide and Test Prep ##### Big Ideas Math Geometry: Online Textbook Help ##### Explorations in Core Math - Geometry: Online Textbook Help ##### View High School: Geometry ##### Foundational Mathematics: Bridge to College What is a Tangent of a Circle? ------------------------------ In general, a tangent to a curve is a line that intersects the curve exactly at one point. The figure below shows a tangent to a curve. Notice that the line (tangent) barely touches the curve. That is typical of how it looks when a line is a tangent to a curve. A curve with a tangent line For circles, the same definition holds: namely, a tangent to a circle is a line that intersects the circle at exactly one point. Most of the time, if a line intersects a circle at a point, then it will also intersect it at a second point. To set it up precisely, make sure that the line intersects the circle at only one point, becoming a tangent line. As above, this will only happen if the tangent line barely touches the circle. The point where the tangent line intersects the circle is called the point of tangency. A circle with a tangent line Tangent of a Circle Example In the case of a circle and a point on that circle, then there will be one and only one line that is tangent to the circle at the given point. For instance, consider the following circle and point to the circle in the image down below. A circle with a point on it Then there is only one line that passes through that point, which is also tangent to the given circle. The tangent line through the given point To unlock this lesson you must be a Study.com memberCreate an account Tangent Theorems ---------------- Two theorems deal with tangents to circle: the Tangent-Radius Theorem and Tangents from the Same Point Theorem. Tangent-Radius Theorem The Tangent-Radius Theorem states that a line that is tangent to a circle will always be perpendicular to the radius of the circle drawn from the point of tangency. For instance, the figure below shows a tangent line and a radius from the point of tangency. The Tangent-Radius Theorem states that these two lines are perpendicular. A tangent line that is perpendicular to the radius. Tangents From the Same Point Theorem The Tangents from the Same Point Theorem states, if a given circle has a point outside of the circle, then the two different tangent line segments can be drawn from that point to the circle. Plus, the two-line segments will be of the same length. For example, consider the following circle and point outside the circle. A circle and a point outside the circle Two segments from the point can be drawn as its tangent to the circle, one on the top and the other on the bottom. The Tangents from the Same Point Theorem states that these two line segments are the same length, which one can tell by looking at the figure and noticing its symmetry. Tangent line segments drawn from the point To unlock this lesson you must be a Study.com memberCreate an account Tangent of a Circle Formula --------------------------- In a coordinate plane, the equation for a circle is (x−h)2+(y−k)2=r 2, where (h,k) is the center of the circle and r is its radius. The equation for a tangent line to this circle is (x−x 0)(x 0−h)+(y−y 0)(y 0−k)=0, where (x 0,y 0) is the point of tangency. Example For instance, suppose a circle has center at the point (1,-3) and radius 5. Then h=1, k=−3, and r=5, so the equation of the circle is (x−1)2+(y+3)2=25. Then one can show that the point (4,1) is a point on the circle, so if it is used as the point of tangency (x 0=4 and y 0=1), then the equation for the tangent line is (x−4)(4−1)+(y−1)(1+3)=0. This can be simplified as follows. 3(x−4)+4(y−1)=0 3 x+4 y−16=0 y=−3 4 x+4 The equations of the circle and tangent line are graphed below. The graph of the equations above To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- A tangent to a circle is a line that intersects the circle at exactly one point. The point where the tangent line intersects the circle is called the point of tangency. In the case where a circle has a point on it, then there will be one and only one line that is tangent to the circle at the given point. Two theorems deal with tangents to circle. The Tangent-Radius Theorem states that a line that is tangent to a circle will always be perpendicular to the radius of the circle drawn from the point of tangency. The Tangents from the Same Point Theorem states that from a circle plus a point outside of the circle, one can draw two different tangent line segments from the point to the circle, and these two-line segments will be of the same length. In a coordinate plane, the equation for a circle is (x−h)2+(y−k)2=r 2, where (h,k) is the center of the circle and r is its radius. The equation for a tangent line to this circle is (x−x 0)(x 0−h)+(y−y 0)(y 0−k)=0, where (x 0,y 0) is the point of tangency. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript Circles Here's a circle. But this is no ordinary circle. You see, we're not just studying geometry. We're also international spies. We're looking down from our spy satellite on a circular fence. The center of this circle, this dot here, is a tiny shack. In the shack is one of our agents who got captured. His name is Steve. Circular fence with shack in the middle In this lesson, we're going to learn about tangent lines and circles while trying to rescue Steve, the captured agent. Oh, and that fence? We think it's electrified, but we're not sure. Ok, let's do this. Tangent of a Circle To start, we send some scouts from these woods here to check out the circle. Let's follow the paths they take. To check the fence, while not getting spotted by the enemy, we need to run a path that forms a tangent line. A tangent to a circle is a line that intersects the circle at only one point. Tangent to a circle Here's the path of the first scout. This line? It never gets close to the fence, so it's not a tangent. The second scout tries to shoot his gun at the fence. Why? I don't know. But his bullet hits two points, here and here, so its path is not a tangent. Various non-tangent lines Then the third scout follows this path. It just barely touches the circle. That's a tangent line. Great job, third scout! Alas, he found out it is electrified. Tough luck, third scout. By the way, tangent comes from the Latin tangentem, which means 'to touch.' Think tangent - touch, both start with 't.' And if you touch an electrified fence, you get tanged -- hmm, that sort of works. Tangent Theorems Let's talk tangent theorems. Here's Steve running to the fence to warn the scout that it's an electrified fence. His path is a radius. A radius drawn to a tangent line is perpendicular to the line. Perpendicular lines form 90 degree angles with each other. Now what if two scouts run paths tangent to the circle from the same place to distract the enemy? That's our second theorem. Tangent segments to a circle that are drawn from the same external point are congruent. That means they're the same length. So if the first scout is going 90 feet, then the second scout is also going 90 feet. They're going the same distance from their base. Tangent Circles While we were talking theorems, the enemy wasn't distracted. Quite the opposite. They captured a bunch of our scouts and put them in different places. Here's one - a complex of fences. Did you know that circles can be tangent to each other? Yep. These circles are all tangent to each other. That means they touch at just one point. Circles in Tangent Here's another - these circles are also tangent. That's not great planning by the enemy. If they put one circle inside the other without them being tangent, then we'd need to get through two fences. But since these are tangent, this is the weak point in their defenses. And then there's this. Whoa. These circles are all tangent to each other. This scout may be sitting there for a while. Lesson Summary In summary, we learned about tangents of circles. We defined a tangent to a circle as a line that intersects a circle at only one point. We also looked at two tangent theorems. First, a radius drawn to a tangent line is perpendicular to the line. That means they form a 90-degree angle. Second, tangent segments to a circle from the same external point are congruent, or equal in length. Finally, we looked at circles that are tangent to each other. This served as a good reminder that we're not great spies. But we are great at geometry! Learning Outcomes Once this video lesson has been watched, you might be able to: Recognize the tangent to a circle Identify a tangent line along a circle Recall that circles can be in tangent with one another Discuss tangent theorems Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now Geometry: High School 14 chapters 145 lessons Chapter 1 High School Geometry: Foundations of Geometry Geometry Definition, History & Branches 4:36 min Inductive vs. Deductive Reasoning in Geometry \| Definition & Uses 4:59 min Thales & Pythagoras: Early Contributions to Geometry 5:14 min The Axiomatic System: Definition & Properties 5:17 min Euclid's Axiomatic Geometry: Developments & Postulates 5:58 min Undefined Terms in Geometry \| Definition & Examples 5:23 min Properties and Postulates of Geometric Figures 4:53 min Algebraic Laws and Geometric Postulates 5:37 min Chapter 2 High School Geometry: Logic in Mathematics Critical Thinking and Logic in Mathematics 4:27 min Circular Reasoning, Hasty Generalizations, False Causes & Limited Choice Fallacies 4:47 min Appeal to Ignorance Fallacy \| Definition & Examples 8:53 min Truth Value \| Definition, Propositions & Tables 9:49 min Conjunction vs. Disjunction in Math \| Overview & Characteristics 3:39 min Conditional Statement \| Definition & Examples 4:54 min Conditional Statements \| Converse, Inverse & Contrapositive 7:09 min Direct Proof Definition, Induction & Examples 7:11 min Geometric Proof Types & Formats \| What is a Proof in Geometry? 8:35 min Chapter 3 High School Geometry: Introduction to Geometric Figures Line Segments & Rays \| Differences & Measurement 3:59 min Angle Relationships \| Overview & Types 10:28 min Geometric Constructions Using Lines and Angles 4:32 min Line Segment Bisection & Midpoint Theorem: Geometric Construction 4:39 min Division of Line Segment \| Parts & Measurements 5:22 min Parallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 min Constructing Perpendicular Lines in Geometry 3:39 min Constructing an Angle Bisector in Geometry 3:36 min Methods & Tools for Making Geometric Constructions 3:17 min Practice Making Geometric Constructions with Tools 4:17 min Constructing Equilateral Triangles, Squares, and Regular Hexagons Inscribed in Circles 5:00 min Chapter 4 High School Geometry: Properties of Triangles Area of Triangles & Rectangles \| Formula, Calculation & Examples 5:43 min Perimeter of Triangles and Rectangles 8:54 min Similar Triangles \| Theorems, Formula & Examples 7:23 min Angles and Triangles: Practice Problems 7:43 min Triangles in Geometry \| Definition, Types & Formulas 4:30 min How to Classify Triangles \| Overview & Examples 5:44 min Interior & Exterior Angles of a Triangle \| Overview & Examples 5:25 min Constructing the Median of a Triangle 4:47 min Altitude, Median & Angle Bisector of a Triangle 4:50 min Constructing Triangles: Types of Geometric Construction 5:59 min Properties of Concurrent Lines in a Triangle 6:17 min Chapter 5 High School Geometry: Triangles, Theorems and Proofs Similar Triangles \| Definition, Application Problems & Examples 6:23 min ASA, SSS & SAS Triangle Postulates \| Properties & Examples 6:15 min Congruent Triangles \| Definition, Parts & Examples 5:19 min Converse Statement \| What is the Converse of a Statement? 5:09 min Similarity Transformations in Corresponding Figures 7:28 min How to Prove Relationships in Figures using Congruence & Similarity 5:14 min Practice Proving Relationships using Congruence & Similarity 6:16 min Angle-Angle-Side \| Theorem, Proof & Example 6:31 min Hypotenuse Angle Theorem \| Definition, Proof & Examples 5:50 min Hypotenuse Leg Theorem \| Definition, Proof & Examples 6:19 min Perpendicular Bisector Theorem \| Converse & Examples 6:41 min Angle Bisector Theorem \| Proof & Examples 6:12 min Right Triangle Congruence Theorems \| Definition & Examples 7:00 min Proving Congruent Isosceles Triangles 4:51 min Chapter 6 High School Geometry: Parallel Lines and Polygons Parallel Lines and Transversals \| Definition & Diagrams 7:40 min Parallel Line Rules, Proofs & Examples 6:55 min Using Converse Statements to Prove Lines Are Parallel 6:46 min Constructing a Parallel Line Using a Point Not on the Given Line 5:15 min Euclid's Parallel Postulate \| Definition & Examples 4:25 min Characteristics of a Polygon \| Overview & Examples 4:25 min Regular Polygon \| Definition, Sides & Types 6:01 min Diagonals of a Polygon \| Formula & Examples 4:49 min Finding the Perimeter of Polygons 5:19 min Area of a Regular Polygon \| Formula & Examples 4:15 min Degrees in a Triangle \| Measurement & Examples 5:14 min Angles in a Polygon \| Measurement, Formula & Examples 6:00 min Chapter 7 High School Geometry: Similar Polygons Ratio & Proportion \| Meaning, Differences & Examples 5:17 min Geometric Mean \| Definition, Formula & Calculation 5:15 min Angle Bisector Theorem \| Definition & Examples 4:58 min Proportion \| Definition, Formula & Examples 5:22 min Similar Polygons Definition & Examples 8:00 min The Transitive Property of Similar Triangles 4:50 min Triangle Proportionality Theorem \| Overview, Proofs & Examples 4:53 min Constructing Similar Polygons 4:59 min Right Triangle \| Properties, Proof & Theorems 5:58 min Pythagorean Theorem \| 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1202	https://web.pa.msu.edu/courses/2017fall/PHY191/exp5.pdf	PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 1 Experiment 5 Elastic and Inelastic Collisions Reading: Bauer&Westfall: Chapter 7 (and 8, for center of mass ideas) as needed 1. Goals 1. Study momentum and energy conservation in inelastic and elastic collisions 2. Understand use of Excel in analyzing data 3. Carry out uncertainty calculations of moderate complexity 2. Theoretical Introduction The following experiment explores the conservation of momentum and energy in a closed physical system (ideally: no interaction of measured objects with rest of universe). As you probably know from the accompanying theoretical course, the conservation of energy and momentum play an important role in physics and their conservation is a consequence of fundamental symmetries of nature. 2.1 Momentum For a single particle (or a very small physical object), momentum is defined as the product of the mass of the particle and its velocity: v m p   (1) Momentum is a vector quantity, making its direction a necessary part of the data. To define the momentum in our three-dimensional space completely, one needs to specify its three components in x, y and z direction. The momentum of a system of more than one particle is the vector sum of the individual momenta:             2 2 1 1 2 1 v m v m p p p      (2) The 2nd Newton’s law of mechanics can be written in a form which states that the rate of the change of the system’s momentum with time is equal to the sum of the external forces acting on this system: F dt p d             (3) From here we can immediately see that when the system is closed (which means that the net external force acting on the system is zero), the total momentum of the system is conserved (constant). 2.2 Energy Another important quantity describing the evolution of the system is its energy. The total energy of a given system is generally the sum of several different forms of energy. Kinetic energy is the form associated with motion, and for a single particle: 2 2 mv KE  (4) PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 2 Here v without the vector symbol stands for the absolute value of the velocity,     2 2 2 z y x v v v  In contrast to momentum, kinetic energy is NOT a vector; for a system of more than one particle the total kinetic energy is the algebraic sum of the individual kinetic energies of each particle:       2 1 KE KE KE (5) Another fundamental law of physics is that the total energy of a system is always conserved. However within a given system, one form of energy may be converted to another (such as potential energy converted to kinetic in the Pendulum experiment). Therefore, kinetic energy alone is often not conserved. 2.3 Collisions An important area of application of the conservation laws is the study of the collisions of various physical bodies. In many cases, it is hard to assess how exactly the colliding bodies interact with each other. However, in a closed system, the conservation laws often allow one to obtain the information about many important properties of the collision without going into the complicated details of the collision dynamics. In this lab, we will see in practice how the conservation of momentum and total energy relate various parameters (masses, velocities) of the system independently of the nature of the interaction between the colliding bodies. Assume we have two particles with masses 2 1,m m and speeds i v1  and i v2  which collide, without any external force, resulting in speeds of f v1  and f v2  after the collision ( i and f stand for initial and final). Conservation of momentum then states that the total momentum before the collision i P  is equal to the total momentum after the collision f P  : f f f i i i v m v m P v m v m P 2 2 1 1 2 2 1 1 ,           and f i P P    (6) 2.4 Elastic and inelastic collisions There are two basic kinds of collisions, elastic and inelastic: 2.4.1 In an elastic collision, two or more bodies come together, collide, and then move apart again with no loss in total kinetic energy. An example would be two identical "superballs", colliding and then rebounding off each other with the same speeds they had before the collision. Given the above example conservation of kinetic energy then implies: 2 2 2 2 1 1 2 2 2 2 1 1 2 1 2 1 2 1 2 1 f f i i v m v m v m v m    or f i KE KE  (7) 2.4.2 In an inelastic collision, the bodies collide and (possibly) come apart again, but now some kinetic energy is lost (converted to some other form of energy). An example would be the collision between a baseball and a bat. If the bodies collide and stick together, the collision is called completely inelastic. In this case, all of the kinetic energy relative to the center of mass of the whole system is lost in the collision (converted to other forms). PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 3 In this experiment you will be dealing with a) a completely inelastic collision in which all kinetic energy relative to the center of mass of the system is lost, but momentum is still conserved, and b) a nearly elastic collision in which both momentum and kinetic energy are conserved to within a few percent. 2.5 Conservation laws for macroscopic bodies So far we were talking about the system of point-like particles. However, the conservation of the momentum is also valid for macroscopic objects. This is because the motion of any macroscopic object can be decomposed into the motion of its center of mass (which is a point in space) with a given linear momentum, and a rotation of the object around this center of mass. Then, the conservation of the linear momentum is again valid for the motion of ideal point masses located at the center of the mass of each of the objects. However, some of the linear kinetic energy can be transformed into the rotational energy of the objects, which should be accounted for in a real experiment. 2.6 Kinetic Energy in Inelastic Collisions. It is possible to calculate the percentage of the kinetic energy lost in a completely inelastic collision; you will find that this percentage depends only on the masses of the carts used in the collision, if one of the carts starts from rest. After the completely inelastic collision, the carts move together, so that 3 2 1 v v v f f   The initial KE is given by: 2 2 2 2 1 2 1 1 i i i v m v m KE   . But, since 0 2  i v 2 2 1 1 i i v m KE  (8) The final KE is given by: 2 2 1 2 3 m m v KE f   (9) From conservation of momentum: 3 2 1 2 2 1 1 ) ( v m m v m v m i i    or, since 0 2  i v 3 2 1 1 1 ) ( v m m v m i   (10) Since the collision is inelastic, the initial KE is not equal to the final KE. You could use equations (8), (9), and (10) to obtain an expression for i i f K KE KE KE D ) ( (%)   . Hint: define x = m1/(m1+m2) and use it to eliminate v3. PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 4 3. Experimental setup We will study the momentum and energy conservation in the following simplified situation: a) we will look on the collision of only 2 objects; b) the motion of these objects will be linear and one-dimensional, so that we can choose the reference frame in such a way that only x-components of the objects’ momenta are non-zero; the sign of these components depends on the direction of the motion; c) the experimental apparatus can be set up in a way to almost completely eliminate the net external force on the system. Our objects will be two carts of different masses, with one initially at rest. The carts move on an air track, which ensures that the motion is one-dimensional and reduces the friction between the carts and the surface. The velocities of the carts can be measured with the help of the photogates, which are described in more details below. Before the beginning of the measurements, spend at least 15 minutes to figure out which external factors can disturb the motion of the carts on the track, and what you should do to reduce or eliminate these factors. Remember, the successful completion of this lab strongly depends on your ability to create an almost closed system. Make a few practice trials to see if you can achieve an unperturbed one-dimensional collision of the carts. Adjust the level of the air track and the power of the air supply if necessary. Questions for preliminary discussion 3.1 Draw a diagram of all forces acting on each cart when they collide. Which forces will influence the total P and KE most? 3.2 In our experiment, can we achieve a completely elastic collision? a completely inelastic collision? 3.3 In an inelastic collision in a closed system, can some of the total momentum be lost? Some of the kinetic energy? 3.4 In an elastic collision in a closed system, can some of the total momentum be lost? Some of the kinetic energy be lost? 3.5 If kinetic energy is lost, where does it go? Does conservation of energy apply? PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 5 Measurements 4. Inelastic collisions 4.1 Techniques. In the first part of the lab we make sure that after the collision the carts stick together and move with some velocity common to both masses. Thus, we have to measure the velocity of cart 1 before the collision and the common velocity of the carts 1 and 2 after the collision. For this purpose, we use two photogates (see Figure 1). Each of them allows measuring the time it takes the cart to go through it. The speed is calculated by dividing the length of the fin on the cart 1 by the measured time; but to turn speed into velocity, as required, you have to pick a + direction. Figure 1: Initial state of the carts before inelastic collision (fin on cart 2 can be removed). Warning: your carts should be balanced, unlike the ones in this figure! Position cart 2 close to the gate 2 and set the photogate timer to "GATE" mode and the memory switch in “ON” position. In this mode the photogate will display the first time interval measured. Subsequent measurements will not be displayed (only the first one is), but the times are added in the memory. By pushing the “READ” switch you can display the memory contents, which is the sum of all measurements. Example: the initial reading for cart 1 (the time that it took to pass through the gate 1) is 0.300 seconds. Cart 1 collides with cart2 and they go together through the photogate 2 (Figure 2). Suppose it now takes 0.513 seconds. The display will remain at 0.300, but the memory will contain .300 + 0.513=0.813 seconds. To find the second time, you have to subtract the first time from the contents of the memory. Try this out by moving the cart through the gate by hand a few times. 4.2 Uncertainty Estimation: First we test to see whether some of our assumptions are correct (the two timers give the same answer, the track is level, friction is negligible, direction doesn’t matter). Perform some trials with a single cart in which no collision occurs. Do this at different speeds, and in different directions. Record the results in your notebook in appropriate tables. Did you conserve momentum? Was there a bias? What time uncertainty would you deduce from these measurements? What conclusions do you draw? 4.3 Prediction: Which case below will the change the final Kinetic Energy most? In your report comment on whether your predictions were correct, and if not, why. PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 6 4.4 Data Do 3 sets of inelastic collisions consisting of 2 trials each. Vary the masses of the carts by adding the masses (small metal disks) to them. In these measurements, use the needle and putty bumpers and measure the initial and final velocities for the following sets of masses: Trial 1+2: no mass disks on cart 1, 4 mass disks on cart 2; Trial 3+4: 2 mass disks on cart 1, 2 mass disks on cart 2; Trail 5+6: 2 mass disks on cart1, no mass disks on cart2. In each measurement, you need to find all the initial and final masses and velocities, and use them to calculate the initial and final total momentum and kinetic energy. Make tables in your lab book to organize your recordings. You may neglect the uncertainty of all masses. In R:exp6, you will find 2 preset datasheets, Inelastic and Elastic. Open them and save into your section’s folder. The spreadsheet has some entries for fractional uncertainties (δDp(%), for example). These should be displayed in % either by multiplying the relevant fraction by 100, or, preferably, by using the % formatting button in Excel. When you have completed the spreadsheets, print them showing the numbers, then save them. There are many uncertainty calculations. You should explain in your lab book or report how you calculated them. An alternative is to re-open your spreadsheet, save it with a new name, use the ctrl-` key to display formulas (handy while checking formulas) and print again, showing the formulas used. It won’t be possible to read the formulas unless you adjust the column width, and use page setup to print landscape, to fit to 2 pages wide x 1 tall. You can use Print Preview before printing to check. 4.4 Hints The correct calculation of the uncertainty of δD(%) is complicated because the initial value for p or K is in both the numerator and denominator. However, for our purposes, it is sufficiently close to use δDp(%) ≈ δD/Pi and a similar formula for K. (Optional) If you wish you could add auxiliary cells to help you with uncertainty calculations. For example, you could also calculate δv/v or δP/P. (Sanity checks for uncertainties) Check whether your uncertainty calculations are making sense! For independent and random uncertainties, you expect increasing values of absolute and fractional uncertainties as your calculation proceeds (unless your calculation involves fractional powers < 1). This is discussed in the Reference Guide section on Uncertainty Calculations. 5. Elastic collisions 5.1 Techniques. In an almost elastic collision, the main difference from the previous part of the lab is that after the collision the carts move separately. The rubber band bumpers allow carts to collide with almost no conversion of the kinetic energy into the other forms of energy. As before, cart 2 initially stays at rest, and before the collision we have to measure only the velocity of the cart 1 i v1 (Figure 3). However, after the collision we have to measure the velocities of both carts, f v1 and f v2 (Figure 4). Thus, all in all we have to measure three times ( f f i t t t 2 1 1 , , ), while the photogate system can simultaneously measure only two of them. We can get out of this situation if, after the measurement of the initial time i t1 , but before the collision, we reset the timer. You have to make several practice trials to quickly remember and reset the contents of the timer before the carts collide. Then, we can again see the contents of the timer display and the memory to find f t1 and f t2 . PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 7 Figure 3: The initial state of the carts before elastic collision. Figure 4: The final state of the carts after elastic collision. 5.2 Prediction: Write in your lab book for each case mentioned in section 5.3 below the predicted direction and speed after the collision of the initial-moving cart: Forward or backward? Faster or slower than the cart initially at rest? In your report comment on whether your predictions were correct, and if not, why. 5.3 Data Measure the fin length on the two carts. The experiment will be done with cart 2 initially at rest. You will do 6 trials with the following choices of 1 m and 2 m : Trial 1+2: no mass disks on cart 1, 4 mass disks on cart 2 Trial 3+4: 2 mass disk on cart 1, 2 mass disks on cart 2 Trail 5+6: 4 mass disks on cart 1 no mass disk on cart 2 5.4 Hints. Pay attention to the sign of the velocities, which depends on the direction of motion of the cart. If one of the carts goes backward with respect to your chosen + direction, how will you make sure the velocity is calculated as negative? If the percentage change in momentum or kinetic energy before and after the collision is greater than 10%, repeat the measurement more carefully (collide slower/faster, etc.). Since the datasheet is set up it is easy to see whether momentum/energy is better conserved with every trial you do. If one of the times you measure is too long for the timer to measure, substitute a large number for the time in your spreadsheet. PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 8 6 Questions to be discussed 6.1 Create a graph (one for inelastic and one for elastic collisions) of the relative change of the total momentum i i f p P P P D / ) ( (%)   versus the number (1-6) of the measurement. This is just to show graphically the various values of Dp(%). Show the uncertainty of Dp(%) with the help of error bars. Also, show the theoretical prediction for Dp(%) . 6.2 According to the graphs, was the total momentum conserved in the collisions? Use the “two standard deviations” rule to justify your answers. Was there any difference in how well momentum was conserved in elastic vs. inelastic collisions? If the momentum was not conserved, discuss the reason why. What did you do to try to improve the momentum conservation results? 6.3 Did you correctly predict the motion of the carts in the elastic collision? If not, why? 6.4 For the inelastic collision, plot the kinetic energy fractional change, DK(%) vs. 2 1 1 m m m x   . Show the error bars for DK(%). What are the slope and intercept of a straight line fit to these data? Does a straight line fit these data reasonably? Did you correctly predict the trial with the largest fractional change? Why or why not? 6.5 Make and analyze a DK(%) graph for the nearly elastic collision (as you analyzed Dp(%) above in steps 6.1-6.3). What was the average loss of the kinetic energy in this part of the experiment? Did you achieve the goal of 10% loss of KE in the nearly elastic collisions? Which potential systematic errors or setup problems is the loss most sensitive to? 7. Systematic Errors Professional scientists deal with systematic errors in successive levels of sophistication. Level 1:Think of possible errors. This is one of the hardest parts! That’s because doing so means seeing where our assumptions break down. The best way is to go through the measurement step by step and identify where things could have been otherwise than we wished. The result is a laundry list of possible systematic effects. Some examples of our assumptions: Setup: We assumed that the track was straight and level. Carts were balanced. Timing: The velocities were constant while the carts moved through the gate and until the collision. Collision: The system of the 2 carts is closed: no momentum (nor energy) enters or leaves the system consisting of the two carts during the collision. Question 7.1 List ways in which these assumptions might have been violated. Level 2: Sign of effect. For each violation of our assumptions, try to understand the sign of its effect on our final result. For example, if the carts weren’t balanced, then the air track would tend to add momentum of a particular sign to the cart. It often helps to exaggerate (mentally, or by measuring) the effect to see which way it should affect your results. 7.2 What sign would each of your effects have on momentum or energy conservation? Or could it be of either sign? Level 3: Estimating the size of the effect. Sometimes we can perform a side-experiment and actually measure the effect and correct for it. More commonly, we can only provide a bound “well it should really have been at worst this big”. For example, leveling errors didn’t give the PHY191 Experiment 5: Elastic and Inelastic Collisions 9/17/2017 Page 9 carts enough extra momentum to go the length of the track in 10 seconds. With the mass, that bounds the change of momentum; that over the smallest initial momentum bounds the % error. 7.3 How big an effect could the air track effect be on your momentum conservation data? 7.4 How big an effect would that have on the energy loss data for inelastic collisions?
1203	https://www.quora.com/Can-you-explain-how-the-formula-sn-n-2-2a+-n-1-d-works	Something went wrong. Wait a moment and try again. Sequence and Seriers Sums and Series (mathemat... Mathematical Formulas Sum of Series Arithmetic Sequences Patterns and Sequence Series and Sequences 5 Can you explain how the formula sn=n/2 (2a+(n-1) d) works? · Certainly! The formula you're referring to is used to calculate the sum of the first n terms of an arithmetic series. Let's break it down: Components of the Formula Sn: The sum of the first n terms of the arithmetic series. n: The number of terms being summed. a: The first term of the series. d: The common difference between consecutive terms. Formula Breakdown The formula is: Sn=n2(2a+(n−1)d) The Average of the First and Last Terms: The expression 2a+(n−1)d can be understood as the sum of the first term a and the last term of the series. The last term l can be expressed as: Certainly! The formula you're referring to is used to calculate the sum of the first n terms of an arithmetic series. Let's break it down: Components of the Formula Sn: The sum of the first n terms of the arithmetic series. n: The number of terms being summed. a: The first term of the series. d: The common difference between consecutive terms. Formula Breakdown The formula is: Sn=n2(2a+(n−1)d) The Average of the First and Last Terms: The expression 2a+(n−1)d can be understood as the sum of the first term a and the last term of the series. The last term l can be expressed as: l=a+(n−1)d Thus, the sum of the first and last terms is: a+l=a+(a+(n−1)d)=2a+(n−1)d Finding the Sum: The average of the first and last term is multiplied by the number of terms n to get the total sum: Sn=Average×n=(a+l)2×n This leads to: Sn=n2(a+l)=n2(2a+(n−1)d) Let's say we have an arithmetic series where: - a=3 (the first term) - d=2 (the common difference) - n=5 (the number of terms) Calculate the last term l: l=a+(n−1)d=3+(5−1)⋅2=3+8=11 Now apply the formula: S5=52(2⋅3+(5−1)⋅2)=52(6+8)=52⋅14=5⋅7=35 Thus, the sum of the first 5 terms of this arithmetic series is 35. The formula Sn=n2(2a+(n−1)d) is a powerful tool for quickly calculating the sum of an arithmetic series by leveraging the relationships between the first term, the common difference, and the number of terms. Adrian Lu Former A-level maths student · Author has 133 answers and 321.7K answer views · 5y The nth term of an arithmetic progression can be defined as: Tn=a+(n−1)d where a is the first term of the A.P and d is the constant difference between one term and another. Proof 1 If we sum up the first n terms starting from a , we get: Sn=a+(a+d)+(a+2d)+…+(a+(n−3)d)+(a+(n−2)d+(a+(n−1)d)(1) Since addition is commutative, this sum is the same as: Sn=(a+(n−1)d)+(a+(n−2)d)+(a+(n−3)d)+…+(a+2d)+(a+d)+a(2) As the sum is finite, we can add the two series by pairing the terms up like this - the first term from (1) to the first term from (2) , the second term from (1) to the second The nth term of an arithmetic progression can be defined as: Tn=a+(n−1)d where a is the first term of the A.P and d is the constant difference between one term and another. Proof 1 If we sum up the first n terms starting from a , we get: Sn=a+(a+d)+(a+2d)+…+(a+(n−3)d)+(a+(n−2)d+(a+(n−1)d)(1) Since addition is commutative, this sum is the same as: Sn=(a+(n−1)d)+(a+(n−2)d)+(a+(n−3)d)+…+(a+2d)+(a+d)+a(2) As the sum is finite, we can add the two series by pairing the terms up like this - the first term from (1) to the first term from (2) , the second term from (1) to the second term from (2) and so on: 2Sn=[a+(a+(n−1)d)]+[(a+d)+(a+(n−2)d)]+[(a+2d)+(a+(n−3)d)]+…+[(a+(n−1)d)+a] Notice that each term enclosed around the brackets above adds up to 2a+(n−1)d , and since there is a one-to-one correspondence while paring the terms up, we have n number of 2a+(n−1)d . Hence, 2Sn=n(2a+(n−1)d)Sn=n2(2a+(n−1)d)■ Alternatively, we can also prove it using induction. Proof 2 When n=1 , we have S1=a which is trivially true since the sum of the first and only term is just the term itself. Therefore, the case n=1 forms our basis for the induction. Now, assume that ∀k∈Z+ Sk=k2(2a+(k−1)d) We will prove that Sk+1=(k+1)2(2a+kd) Proof: To sum the first (k+1) terms, we sum the first k terms and the (k+1)th term: Sk+1=Sk+Tk+1=k2(2a+(k−1)d)+(a+kd)(by assumption)=ka+a+(k2(k−1)+k)d=(k+1)a+(k22+k2)d=(k+1)a+(k(k+1)2)d=(k+1)2(2a+kd) Since we have shown using our assumption that the result is true when n=k+1 , it is also true ∀n∈Z+ by induction.■ Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. 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These two non-conventional methods may be of your interest. I suppose you are dealing with arithmetic series. These two non-conventional methods may be of your interest. Related questions What is the formula of sn=n/2 [2a+(n-1)d]? What does ‘s’ stand for in the formula: s = n (2a+(n-1) d) /2? In the formula S n = n 2 ( 2 a + ( n − 1 ) d ) , what is a? How we can prove that S n = n 2 ( 2 a + ( n − 1 ) d ) ? What's the proof for Sn=n/2 [2a1+(n-1) d]? Luke Braithwaite BSc in Computer Science, University of Manchester (Graduated 2024) · Author has 68 answers and 86.9K answer views · 5y Originally Answered: Let Sn denote the sum of the first n terms of the arithmetic progression. How do you show that Sn=n/2 (2a+(n-1) d)? · Let Sn be the sum of the first n terms of an arithmetic progressions () with initial term a and common difference d, thus we have: Sn=a+(a+d)+⋯+(a+(n−2)d)+(a+(n−1)d) If we sum the first and last terms together and the second and the second to last terms we see that both are equal to 2a+(n−1)d. From this we see that if we have a term, Tuwhere u<n, then Tu+Tn−u is equal to 2a+(n−1)d. If we simply double the value of S_n and then rearrange the series we get: \begin{align} 2S_n &= [a+a+(n-1)d] + [a+d+a+(n-2)d] + \cdots + [a+(n-2)d +a+d]\&\;+[a+(n-1)d + a]\ &= [2a+(n-1)d] +[2a+(n-1)d] Let Sn be the sum of the first n terms of an arithmetic progressions () with initial term a and common difference d, thus we have: Sn=a+(a+d)+⋯+(a+(n−2)d)+(a+(n−1)d) If we sum the first and last terms together and the second and the second to last terms we see that both are equal to 2a+(n−1)d. From this we see that if we have a term, Tuwhere u<n, then Tu+Tn−u is equal to 2a+(n−1)d. If we simply double the value of S_n and then rearrange the series we get: 2Sn=[a+a+(n−1)d]+[a+d+a+(n−2)d]+⋯+[a+(n−2)d+a+d]+[a+(n−1)d+a]=[2a+(n−1)d]+[2a+(n−1)d]+⋯+[2a+(n−1)d]+[2a+(n−1)d] From the definition of S_n we know that S_n has n terms hence, 2Sn=n[2a+(n−1)d] We then simply divide by two to get, Sn=n2[2a+(n−1)d]. Howard Ludwig Ph.D. in Physics, Northwestern University (Graduated 1982) · Author has 3K answers and 10.3M answer views · 7y The formula has been incorrectly expressed as a result of linearizing the text asking the question. The intent of the formula is to determine the sum of the first n terms of an arithmetic series (the addition of a sequence of terms such that consecutive terms always differ in value by the same amount. An arithmetic sequence has the property that for any positive integer k in the range 1 .. n, term k and term n + 1 − k always adds up to the same value. This works in particular for k = 1. Thus, we are talking about the sum of the first an last term of the series. The average of all the terms is th The formula has been incorrectly expressed as a result of linearizing the text asking the question. The intent of the formula is to determine the sum of the first n terms of an arithmetic series (the addition of a sequence of terms such that consecutive terms always differ in value by the same amount. An arithmetic sequence has the property that for any positive integer k in the range 1 .. n, term k and term n + 1 − k always adds up to the same value. This works in particular for k = 1. Thus, we are talking about the sum of the first an last term of the series. The average of all the terms is the same as the average of the first and last term. The sum of a full series is the product of the number of terms times the average value of the terms. Therefore, the sum of the first n terms of an arithmetic series is the product of n times the average of the first and n terms. What I have just written is what I always remember and use to evaluate the sum of an arithmetic series—never the form that was posted. This form is easy to remember, because it is stated in terms of what it means rather than some obscure formula with lots of word pieces to keep straight, and always gives the correct answer (when calculated properly). The first term is commonly called a, arbitrarily. The difference between two consecutive terms (the one on the right minus the one on the left) is called d for difference. Therefore, the nth term is n − 1 terms after the first, so has value (n − 1)d more than the value a of the first term, that is a + (n − 1)d . The average of those two terms is the sum of the two terms and then divide by 2, thus: [a + a + (n − 1)d]/2 = 2a+(n−1)d2. Multiplying this average term value by n gives us the total value. 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Sum = a + (a + d) + (a + 2d) + (a + 3d)…+ (l – 3d) + (l – 2d) + (l – d) + l Where l = a + (n – 1)d In reverse order, the sum remains the same: Sum = l + (l – d) + (l – 2d) + (l – 3d) + … (a + 3d) + (a + 2d) + (a + d) + a Adding equations a and b, we get: 2 x Sum = (a + l) + [(a +d) + (l – d)] … + [(l – d) + (a + d)] + (l + a)] 2 x sum = (a + l) + (a + l)…+ (a + l) + (a + l) 2 x sum = n x (a + l) sum = Substituting l with the previous equation above (where l = If we express the first term in the arithmetic progression as a, the common difference between each consecutive term as d, and the last term as l. Sum = a + (a + d) + (a + 2d) + (a + 3d)…+ (l – 3d) + (l – 2d) + (l – d) + l Where l = a + (n – 1)d In reverse order, the sum remains the same: Sum = l + (l – d) + (l – 2d) + (l – 3d) + … (a + 3d) + (a + 2d) + (a + d) + a Adding equations a and b, we get: 2 x Sum = (a + l) + [(a +d) + (l – d)] … + [(l – d) + (a + d)] + (l + a)] 2 x sum = (a + l) + (a + l)…+ (a + l) + (a + l) 2 x sum = n x (a + l) sum = Substituting l with the previous equation above (where l = a + (n – 1)d): Sum of n terms of AP = n/2(2a+(n-1)d) Related questions How can I know that I have to use this formula only Sn = n/2 {2a + (n - 1) d? Who can solve this s=n/2 (2a +(n-1) d)? Let Sn denote the sum of the first n terms of the arithmetic progression. How do you show that Sn=n/2 (2a+(n-1) d)? What is formula of 1+2+3…+(n-1)? What formula is this: n(n-3) /2? Martyn Hathaway BSc in Mathematics, University of Southampton (Graduated 1986) · Author has 4.7K answers and 6.7M answer views · 2y Originally Answered: How do you read Sn=n/2 [2a1+(n-1) d]? · Not sure what you mean by ‘read’! Do you want an explanation of the terms in this equation or to be shown how it is derived? As written, it’s not 100% clear if you intend to divide ’n’ by ‘2’ and multiply by the part in brackets or to divide ’n’ by both ‘2’ and the part in brackets. The space between the ‘2’ and the open bracket suggests the former; if so, it would have been clearer to write Sn=n[2a1+(n-1)d]/2. However, as I recognise this equation, I’m sure you mean: Sn=n2[2a1+(n−1)d] If you can’t type subscripts, use the ‘_’ symbol [not the minus sign, ‘-’], e.g. S_n, a_1. Not sure what you mean by ‘read’! Do you want an explanation of the terms in this equation or to be shown how it is derived? As written, it’s not 100% clear if you intend to divide ’n’ by ‘2’ and multiply by the part in brackets or to divide ’n’ by both ‘2’ and the part in brackets. The space between the ‘2’ and the open bracket suggests the former; if so, it would have been clearer to write Sn=n[2a1+(n-1)d]/2. However, as I recognise this equation, I’m sure you mean: Sn=n2[2a1+(n−1)d] If you can’t type subscripts, use the ‘_’ symbol [not the minus sign, ‘-’], e.g. S_n, a_1. d is commonly used to represent the common difference between successive terms in an arithmetic progression. a1 thus indicates the first term in the progression and Sn represents the sum of the first n terms of the progression. We thus have: Sn=a1+[a1+d]+...+[a1+(n−2)d]+[a1+(n−1)d] As addition is commutative, we can swop the order of the additions; thus, we can also say: Sn=[a1+(n−1)d]+[a1+(n−2)d]+...+[a1+d]+a1 Adding these two equations: 2Sn=[a1+a1+(n−1)d]+[a1+d+a1+(n−2)d]+...+[a1+(n−2)d+a1+d]+[a1+(n−1)d+a1] =[2a1+(n−1)d]+[2a1+(n−1)d]+...+[2a1+(n−1)d]+[2a1+(n−1)d] =n[2a1+(n−1)d] Dividing both sides by 2: Sn=n2[2a1+(n−1)d] Sponsored by CDW Corporation How can AI help your teams make faster decisions? CDW’s AI solutions offer retrieval-augmented generation (RAG) to expedite info with stronger insights. Ram Kushwah Up and coming Most viewed writer · Author has 6.8K answers and 15.7M answer views · Sep 19 I will use quite different tecnique to drive this formula: The nth term of AP: an=a+(n−1)d an=nd+(a−d) Sum of n terms Sn=∑nk=1(kd+(a−d)) =12d∗n(n+1)+(a−d)n =n2(nd+d+2a−2d) =n2(2a+nd−d) ⟹Sn=n2{2a+(n−1)d} Wahoooo D Sreenivasa Rao MSc; M.Phil in Mathematics, SV University, Tirupathi · Author has 3.2K answers and 2.8M answer views · 2y Originally Answered: How do you prove Sn= n/2 [2a + (n-1) d]? · Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Kwok Choy Yue B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978) · Author has 1.4K answers and 1.4M answer views · 3y Originally Answered: How do you read Sn=n/2 [2a1+(n-1) d]? · I assume you like to visualize the formula. Have you ever heard the area of trapezium A= h/2 (a + b)? where h is the height and a, b are the length of the top and bottom parallel sides. The Sum of an arithmetic sequence is analogous to this formula, Sn=n/2 [a_1+a_n] , where n is the number of terms, a_1 and a_n are the first and last terms. Just relate the lengths of trapezium with the number of ‘dots’ for series. With a_n = a_1+(n-1)d you can get S_n=n/2 [2a_1+(n-1) d] Gunjan Bhanarkar Lives in Navi Mumbai, Maharashtra, India · Author has 65 answers and 116.3K answer views · 6y Originally Answered: How we can prove that \text{S}_n = \frac{n}{2}(2a+(n-1)d)? · Using summation operator. General term Tn= a+(n-1)d=a+dn-d Summation of Tn=Sn Sn= an(addind a n times) + dn(n+1)/2 (addind dn n times) -dn Takin n/2 common n/2 (2a + dn +d -2dn) n/2 (2a + (n-1)d) Thus, proved. If u are below class 11th , u wont be able to understand it. So, here is another, S = a1 + a2 + a3 + ………….. + an S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + {a + (n - 2)d} + {a + (n - 1)d} ……………….. (i) By writing the terms of S in the reverse order, we get, S = {a + (n - 1)d} + {a + (n - 2)d} + {a + (n - 3)d} + ……….. + (a + 3d) + (a + 2d) + (a + d) + a Adding the corresponding terms of (i) and (ii Using summation operator. General term Tn= a+(n-1)d=a+dn-d Summation of Tn=Sn Sn= an(addind a n times) + dn(n+1)/2 (addind dn n times) -dn Takin n/2 common n/2 (2a + dn +d -2dn) n/2 (2a + (n-1)d) Thus, proved. If u are below class 11th , u wont be able to understand it. So, here is another, S = a1 + a2 + a3 + ………….. + an S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + {a + (n - 2)d} + {a + (n - 1)d} ……………….. (i) By writing the terms of S in the reverse order, we get, S = {a + (n - 1)d} + {a + (n - 2)d} + {a + (n - 3)d} + ……….. + (a + 3d) + (a + 2d) + (a + d) + a Adding the corresponding terms of (i) and (ii), we get 2S = {2a + (n - 1)d} + {2a + (n - 1)d} + {2a + (n - 1)d} + ………. + {a + (n - 2)d} 2S = n[2a + (n -1)d ⇒ S = n2n2[2a + (n - 1)d] Now, l = last term = nth term = a + (n - 1)d Therefore, S = n2n2[2a + (n - 1)d] = n2n2[a {a + (n - 1)d}] = n2n2[a + l]. Neha Dubey Developer, Iot (2019–present) · Author has 56 answers and 78.5K answer views · 6y Here ‘s' stand for the total sum of tha value til that digit . For eg if n - 20 den ‘s' will be the sum of all the n till 20 . Thanks regards Neha Dubey Ram Kushwah Up and coming Most viewed writer · Author has 6.8K answers and 15.7M answer views · Jun 15 Originally Answered: Let Sn denote the sum of the first n terms of the arithmetic progression. How do you show that Sn=n/2 (2a+(n-1) d)? · First term=a and the last termTn=a+(n−1)d Thus the sum of first n terms Sn=a+(a+d)+(a+2d)+……………..+(a+(n−2)d)+(a+(n−1)d)——−(1) Writing the series in reverse direction Sn=(a+(n−1)d)+(a+(n−2)d)+…………….+(a+d)+(a+2d)———(2) Adding (1) and (2) [math]2S_n=[ a+a+(n-1)d)]+[ a+d+a+(n-2)d)] \ +[ a+2d+a+(n-3)d)]+…... Related questions What is the formula of sn=n/2 [2a+(n-1)d]? What does ‘s’ stand for in the formula: s = n (2a+(n-1) d) /2? In the formula S n = n 2 ( 2 a + ( n − 1 ) d ) , what is a? How we can prove that S n = n 2 ( 2 a + ( n − 1 ) d ) ? What's the proof for Sn=n/2 [2a1+(n-1) d]? How can I know that I have to use this formula only Sn = n/2 {2a + (n - 1) d? Who can solve this s=n/2 (2a +(n-1) d)? Let Sn denote the sum of the first n terms of the arithmetic progression. How do you show that Sn=n/2 (2a+(n-1) d)? What is formula of 1+2+3…+(n-1)? What formula is this: n(n-3) /2? How do you prove Sn= n/2 [2a + (n-1) d]? How do you use the formula: TN=a +d(n-1)? Which scientist discovered this formulas for A.P. - tn = a + (n - 1) d and Sn = n/2 [2a + (n - 1) d]? Is there a formula for n^1+n^2+n^3…+n^m? What's the difference between these two formulas, n (n+1) /2 and n (1+n) /2 and the difference between n+1 and 1+n? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1204	https://nyjm.albany.edu/j/2024/30-33p.pdf	New York Journal of Mathematics New York J. Math. 30 (2024) 745–755. Extensions of 𝑫(−𝟏)-pairs in some imaginary quadratic fields Mirela Jukić Bokun and Ivan Soldo Abstract. In this paper, we discuss the extensibility of Diophantine 𝐷(−1) pairs {𝑎, 𝑏}, where 𝑎, 𝑏are positive integers in the ring ℤ[ √ −𝑘], 𝑘> 0. We prove that families of such 𝐷(−1)-pairs with 𝑏= 𝑝𝑖𝑞𝑗, where 𝑝, 𝑞are different odd primes and 𝑖, 𝑗are positive integers cannot be extended to quadruples in certain rings ℤ[ √ −𝑘], where 𝑘depends on 𝑝𝑖, 𝑞𝑖and 𝑎. Further, we present the result on non-existence of 𝐷(−1)-quintuples of a specific form in certain imaginary quadratic rings. Contents 1. Introduction 745 2. Results on 𝐷(−1)-𝑚-tuples 747 Acknowledgement 754 References 754 1. Introduction A set {𝑎1, 𝑎2, … , 𝑎𝑚} of 𝑚distinct positive integers is called a Diophantine 𝑚-tuple with the property 𝐷(𝑛) (or simply a 𝐷(𝑛)-𝑚-tuple) for a given non-zero integer 𝑛, if 𝑎𝑖𝑎𝑗+ 𝑛is a perfect square, for all 1 ≤𝑖< 𝑗≤𝑚. The problem of the existence of such sets has been dealt with by many distinguished math-ematicians over the past years and centuries. It was Fermat who first found a 𝐷(1)-quadruple {1, 3, 8, 120} in integers. In , Dujella proved that there is no 𝐷(1)-sextuple in integers, setting the conjecture that there could be only finitely many integer 𝐷(1)-quintuples. The nonexistence of such quintuples was han-dled by He, Togbé and Ziegler in . On the other hand, in , Bonciocat, Cipu and Mignotte proved a conjecture of Dujella from , which states that there are no integer 𝐷(−1)-quadruples. Moreover, due to Trebješanin and Filipin it is known that there do not exist integer 𝐷(4)-quintuples (see ). On regularly Received March 6, 2024. 2020 Mathematics Subject Classification. 11D09, 11D45, 11R11. Key words and phrases. Pellian equation, Diophantine 𝑚-tuple, quadratic field. Authors were supported by the Croatian Science Foundation under the project no. IP-2022-10-5008. ISSN 1076-9803/2024 745 746 MIRELA JUKIĆ BOKUN AND IVAN SOLDO updated Dujella’s web page , interested readers can find an extended list of references on 𝐷(𝑛)-𝑚-tuples in integers and rationals. The problem of Diophantus was also considered in a commutative ring with unity. The authors with collaborators mostly studied it over imaginary qua-dratic fields and the corresponding rings. In [4, 12], it is considered for Gaussian integers and integers of ℚ( √ −3), respectively. In [10, 23], we have almost com-pletely characterized elements 𝑧of ℤ[ √ −2] for which a Diophantine quadru-ple with the property 𝐷(𝑧) exists. As a sporadic possible exception appears the case 𝑧= −1. That motivated us to consider the existence of 𝐷(−1)-𝑚-tuples in the ring ℤ[ √ −𝑘], with a positive integer 𝑘. Due to the result of Dujella for Gaussian integers [4, Theorem 3] and [9, Proposition 2] it is possible to find a 𝐷(−1)-quadruple in ℤ[ √ −𝑘] in case 𝑘is a square. Further, in , the au-thors proved that there does not exist a 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in the ring ℤ[ √ −𝑘], 𝑘≥2 with positive integers 𝑎< 𝑏≤8𝑎−3 and negative integers 𝑐 and 𝑑. They applied the obtained result to show that such a 𝐷(−1)-pair {𝑎, 𝑏} cannot be extended to a 𝐷(−1)-quintuple {𝑎, 𝑏, 𝑐, 𝑑, 𝑒} in ℤ[ √ −𝑘] with integers 𝑐, 𝑑and 𝑒. Very recently, in , the authors proved a general result on the nonexistence of a 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in ℤ[ √ −𝑘], 𝑘≥2, with integers 𝑎, 𝑏, 𝑐, 𝑑, but in the range 𝑎< 𝑏≤8𝑎−3. Using a computer program, if 𝑞is an odd prime and if we take 𝑎= 2, 𝑏∈ {5𝑞, 13𝑞}, we found out that the set of 𝐷(−1)-pairs {𝑎, 𝑏} is not at all rare. Namely, odd primes of the form (𝑥2 + 1)∕10 and (𝑥2 + 1)∕26, 𝑥∈ℤ, are members of the sequence A207337 and A208292 in , respectively. That motivated us to consider the extendibility of more general 𝐷(−1)-pairs {𝑎, 𝑝𝑖𝑞𝑗}, where 𝑎, 𝑖, 𝑗 are positive integers and 𝑝, 𝑞are different odd primes, to quadruples in the ring ℤ[ √ −𝑘], with an integer 𝑘. Thus, the main result of this paper is as follows: Theorem 1.1. Suppose that 𝑝, 𝑞are different odd primes, 𝑎, 𝑖, 𝑗, 𝑘positive inte-gers such that gcd(𝑎, 𝑝𝑞) = 1, 𝑎< 𝑝𝑖𝑞𝑗and 𝑎𝑝𝑖𝑞𝑗is not a square. Moreover, let 𝑘≥max{𝑝𝑖, 𝑞𝑗} and 𝑚𝑘= 𝑝𝑖𝑞𝑗−𝑎 𝑘 . Let us further assume that either of the following holds: 1) 𝑚𝑘= 1 and 𝑎≠1; 2) 𝑚𝑘= 2 with 𝑎≠1, 5 and gcd(𝑎, 2) = 1; 3) 𝑚𝑘is a prime and 𝑟≥𝑎𝑚𝑘or 𝑟≥2𝑎2 with a square-free 𝑎, where 𝑎𝑝𝑖𝑞𝑗− 1 = 𝑟2; 4) 𝑚𝑘is not an integer. Then, there does not exist a 𝐷(−1)-quadruple of the form {𝑎, 𝑝𝑖𝑞𝑗, 𝑐, 𝑑} in the ring ℤ[ √ −𝑘]. Now, we specify the following known result on 𝐷(−1)-quadruples in inte-gers: Theorem 1.2 (see [2, Theorem 1.4.]). ThereisnoDiophantine𝐷(−1)-quadruple. EXTENSIONS OF 𝐷(−1)-PAIRS IN SOME IMAGINARY QUADRATIC FIELDS 747 It is our intention to transform the whole problem in the ring of integers. In this respect, we will adapt the strategy used in [24, 14, 16], and we will mark the main steps, and describe the new ideas in detail in Theorem 2.3 and The-orem 2.8. In combination with Theorem 1.2 and other results from the next section, we will be able to prove our main result Theorem 1.1. Furthermore, we also prove the result on 𝐷(−1)-quintuples of the specific form (see Theo-rem 2.15 and its example). 2. Results on 𝑫(−𝟏)-𝒎-tuples First, we have the following general result: Theorem 2.1. Let 𝑘be a positive integer and {𝑎, 𝑏} a 𝐷(−1)-pair in the ring ℤ[ √ −𝑘] with positive integers 𝑎, 𝑏. If 𝑘∤𝑏−𝑎, then there does not exists a 𝐷(−1)-triple {𝑎, 𝑏, 𝑐} in ℤ[ √ −𝑘] with a negative integer 𝑐. In that case, there does not exist a 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in ℤ[ √ −𝑘] with an integer 𝑑. Proof. Suppose that the set {𝑎, 𝑏, 𝑐} is such a 𝐷(−1)-triple in ℤ[ √ −𝑘]. Then we will obtain 𝑐< 0 if the following holds: 𝑎𝑐−1 = −𝑘𝑥2, 𝑏𝑐−1 = −𝑘𝑦2, 𝑥, 𝑦∈ℤ. That is equivalent to 𝑏𝑥2 −𝑎𝑦2 = 𝑏−𝑎 𝑘 . (1) Therefore, it must be satisfied 𝑘\|𝑏−𝑎and the equation (1) is solvable in integers (𝑥, 𝑦). Since by the assumption it holds 𝑘∤𝑏−𝑎, we have only 𝑐∈ℕ. Thus, if we suppose that we have a 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in the ring ℤ[ √ −𝑘], then this contradicts with Theorem 1.2. □ As a direct consequence of Theorem 2.1 we obtain the result as follows: Corollary 2.2. Let 𝑘be a positive integer and {𝑎, 𝑏, 𝑐} a 𝐷(−1)-triple in the ring ℤ[ √ −𝑘] with positive integers 𝑎, 𝑏, 𝑐. 1) If 𝑘∤𝑏−𝑎or 𝑘∤𝑏−𝑐or 𝑘∤𝑐−𝑎, then there does not exist a 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in ℤ[ √ −𝑘] with an integer 𝑑. 2) If 𝑘∣𝑏−𝑎, 𝑘∣𝑏−𝑐and 𝑘∣𝑐−𝑎, then the existence of a semi-regular 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in ℤ[ √ −𝑘] with an integer 𝑑is possible if the condition 𝑘\|4 −3𝑎2 holds. Proof. 1) It directly follows from Theorem 2.1. 2) Let 𝑎𝑏−1 = 𝑟2 and 𝑐= 𝑎+ 𝑏± 2𝑟. From 𝑘∣𝑐−𝑎and 𝑘∣𝑐−𝑏we obtain 𝑘∣4𝑎(𝑐−𝑎) ∓(2𝑟± 3𝑎)(𝑐−𝑏), i.e., 𝑘∣4 −3𝑎2. □ Now, we prove the next theorem: 748 MIRELA JUKIĆ BOKUN AND IVAN SOLDO Theorem 2.3. Suppose that 𝑝, 𝑞are different odd primes, 𝑖, 𝑗, 𝑎, 𝑘positive inte-gers such that 𝑘≥max{𝑝𝑖, 𝑞𝑗}, gcd(𝑎, 𝑝𝑞) = 1 and 𝑎< 𝑝𝑖𝑞𝑗. If {𝑎, 𝑝𝑖𝑞𝑗, 𝑐} is a 𝐷(−1)-triple in the ring ℤ[ √ −𝑘], then 𝑐∈ℤ. Proof. In the most cases, the proof follows the same strategy as in [24, Theorem 2.2] and [14, Theorem 5.4.]. Therefore, we will mark only the main steps, while in the case where the strategy is different from the used one, we will mark the details. Let the set {𝑎, 𝑝𝑖𝑞𝑗, 𝑐} be such a 𝐷(−1)-triple in ℤ[ √ −𝑘]. Then there exist integers 𝑐1, 𝑑1, 𝑥, 𝑦, 𝑢, 𝑣such that 𝑐= 𝑐1 + 𝑑1 √ −𝑘and 𝑎𝑐−1 = (𝑥+ 𝑦 √ −𝑘)2, 𝑝𝑖𝑞𝑗𝑐−1 = (𝑢+ 𝑣 √ −𝑘)2. It follows that we have to consider the system of equations 𝑝𝑖𝑞𝑗𝑥𝑦= 𝑎𝑢𝑣, (2) 𝑝𝑖𝑞𝑗(𝑥2 −𝑘𝑦2 + 1) = 𝑎(𝑢2 −𝑘𝑣2 + 1). (3) First, consider the case where 𝑥𝑦≠0. From (2) we conclude that 𝑝𝑖𝑞𝑗\|𝑎𝑢𝑣, so 𝑝𝑖𝑞𝑗\|𝑢𝑣. (I) Let 𝑢= 𝑝𝑖𝑞𝑗𝑤for some 𝑤∈ℤ. From (2) we obtain 𝑥𝑦= 𝑎𝑤𝑣, so there exist nonzero integers 𝑝1, 𝑞1, 𝑟1, 𝑠1 such that 𝑥= 𝑝1𝑞1, 𝑦= 𝑟1𝑠1, 𝑎𝑤= 𝑝1𝑟1, 𝑣= 𝑞1𝑠1. Now, from (3) we obtain 𝑘= 𝑎(𝑝𝑖𝑞𝑗−𝑎) −𝑝𝑖𝑞𝑗𝑝2 1ℎ 𝑎𝑠2 1ℎ , (4) 𝑝𝑖𝑞𝑗= 𝑎(𝑎+ ℎ𝑘𝑠2 1) 𝑎−ℎ𝑝2 1 , (5) where ℎ= 𝑝𝑖𝑞𝑗𝑟2 1 −𝑎𝑞2 1. From (5) we conclude that 𝑎\|ℎ𝑝2 1. It is easy to see that in each one of the cases ℎ≤0 and ℎ> 0 we obtain a contradiction with the fact that 𝑘and 𝑝𝑖𝑞𝑗are positive integers. (II) In the second case, set 𝑣= 𝑝𝑖𝑞𝑗𝑤for some 𝑤∈ℤ. From (2) we obtain 𝑥𝑦= 𝑎𝑢𝑤. By following the same lines as in (I) we arrive to a contradiction. (III) If 𝑢= 𝑝𝑖𝑤1, 𝑣= 𝑞𝑗𝑤2, where 𝑤1, 𝑤2 ∈ℤ. From (2) we obtain 𝑥𝑦= 𝑎𝑤1𝑤2, so there exist nonzero integers 𝑝1, 𝑞1, 𝑟1, 𝑠1 such that 𝑥= 𝑝1𝑞1, 𝑦= 𝑟1𝑠1, 𝑎𝑤1 = 𝑝1𝑟1, 𝑤2 = 𝑞1𝑠1. Now, from (3) we obtain 𝑘= 𝑎(𝑎−𝑝𝑖𝑞𝑗) −𝑝𝑖𝑝2 1ℎ 𝑎𝑞𝑗𝑠2 1ℎ , (6) 𝑞𝑗= 𝑎2 −ℎ𝑝𝑖𝑝2 1 𝑎(𝑝𝑖+ 𝑘ℎ𝑠2 1) , (7) where ℎ= 𝑎𝑞𝑗𝑞2 1 −𝑝𝑖𝑟2 1. Since 𝑘≥max{𝑝𝑖, 𝑞𝑗} from (6) and (7) we reach a contradiction in both cases ℎ≤0 and ℎ> 0. EXTENSIONS OF 𝐷(−1)-PAIRS IN SOME IMAGINARY QUADRATIC FIELDS 749 (IV) Similarly as in (III), if 𝑢= 𝑞𝑗𝑤1, 𝑣= 𝑝𝑖𝑤2, where 𝑤1, 𝑤2 ∈ℤ, from (2) we obtain 𝑥𝑦= 𝑎𝑤1𝑤2, so there exist nonzero integers 𝑝1, 𝑞1, 𝑟1, 𝑠1 such that 𝑥= 𝑝1𝑞1, 𝑦= 𝑟1𝑠1, 𝑎𝑤1 = 𝑝1𝑟1, 𝑤2 = 𝑞1𝑠1. Now, from (3) we obtain 𝑘= 𝑎(𝑝𝑖𝑞𝑗−𝑎) −𝑞𝑗𝑝2 1ℎ 𝑎𝑝𝑖𝑠2 1ℎ , (8) 𝑝𝑖= 𝑎2 + ℎ𝑝2 1𝑞𝑗 𝑎(𝑞𝑗−ℎ𝑘𝑠2 1) , (9) where ℎ= 𝑞𝑗𝑟2 1 −𝑎𝑝𝑖𝑞2 1. Since 𝑘≥max{𝑝𝑖, 𝑞𝑗} from (8) and (9) we obtain a contradiction in both cases ℎ≤0 and ℎ> 0. (V) Suppose that 𝑝𝑖1𝑞𝑗1\| gcd(𝑢, 𝑣), with 0 < 𝑖1 + 𝑗1 < 𝑖+ 𝑗. By (3) we have 𝑝𝑖𝑞𝑗\|(𝑢2 −𝑘𝑣2 + 1), which yields a contradiction. Thus, the system of equations (2) and (3) has no solutions in case of 𝑥𝑦≠0. Therefore, 𝑥= 0 or 𝑦= 0 or 𝑥= 𝑦= 0. In any case we have 𝑑1 = 0 and 𝑐= 𝑐1, which is an integer. □ Suppose that 𝑝, 𝑞are different odd primes, 𝑎, 𝑖, 𝑗, 𝑘≥max{𝑝𝑖, 𝑞𝑗} positive integers and gcd(𝑎, 𝑝𝑞) = 1. We consider the extendibility of the 𝐷(−1)-pairs {𝑎, 𝑝𝑖𝑞𝑗} with 𝑎< 𝑝𝑖𝑞𝑗to quadruples of the form {𝑎, 𝑝𝑗𝑞𝑗, 𝑐, 𝑑} in the ring ℤ[ √ −𝑘]. By Theorem 2.3, if such a quadruple exists, then 𝑐, 𝑑∈ℤ. Now we will use the approach for instance from [20, Chapter 11.5] and con-sider the equation 𝑎𝑥2 −𝑑𝑦2 = 𝛿𝑠, 𝛿= ±1, (10) with a positive non-square integer 𝑑and positive integers 𝑎, 𝑠. Assume that (𝑥, 𝑦) is a primitive solution to (10). Then there exist integers 𝑢, 𝑣such that 𝑥𝑢−𝑦𝑣= 𝛿. (11) Multiplying (10) by 𝑎𝑣2 −𝑑𝑢2 we obtain (𝑎𝑥𝑣−𝑑𝑦𝑢)2 −𝑎𝑑= 𝛿𝑠(𝑎𝑣2 −𝑑𝑢2). If we set \|𝑎𝑥𝑣−𝑑𝑦𝑢\| = 𝑙, we conclude that the equation 𝑎𝑣2 −𝑑𝑢2 = 𝑙2 −𝑎𝑑 𝛿𝑠 (12) has a solution in integers. If (𝑢0, 𝑣0) is a solution of (11), then the following parametric equations 𝑣= 𝑣0 + 𝑡𝑥, 𝑢= 𝑢0 + 𝑡𝑦, 𝑡∈ℤ generate all solutions to (11). Thus, 𝑙= \|𝑎𝑥𝑣0 −𝑑𝑦𝑢0 + 𝛿𝑠𝑡\|, 750 MIRELA JUKIĆ BOKUN AND IVAN SOLDO which implies that there exists an integer 𝑡such that 𝑙≤ 𝑠 2. These facts can be summarized in the following lemma: Lemma 2.4. Let 𝑑be positive non-square integer and 𝑎, 𝑠positive integers. If there exists a primitive solution (𝑥, 𝑦) of the equation (10), then there exists an integer 𝑙satisfying 𝑙2 ≡𝑎𝑑(mod 𝑠) and 0 ≤𝑙≤𝑠 2 such that the equation (12) has s a primitive solution. Moreover, the next lemma will be useful. Lemma 2.5 (see [13, Lemma 2.3]). Let 𝑁and 𝐾be integers with 1 < \|𝑁\| ≤𝐾. Then the Pellian equation 𝑥2 −(𝐾2 + 1)𝑦2 = 𝑁 has no primitive integer solution (𝑥, 𝑦), i.e., there does not exist integers 𝑥, 𝑦such that gcd(𝑥, 𝑦) = 1. Also, for the equation 𝑈𝑥2 −𝑉𝑦2 = 𝑐, 𝑐∈{1, 2} we use the following solv-ability criteria in integers 𝑥, 𝑦. Lemma 2.6 (see [17, Criterion 1.]). Let 𝑈> 1, 𝑉be positive integers such that gcd(𝑈, 𝑉) = 1 and 𝐷= 𝑈𝑉is not a square of a positive integer. Moreover, let (𝑢0, 𝑤0) denote the least positive integer solution of Pell’s equation 𝑢2 −𝐷𝑤2 = 1. Then equation 𝑈𝑥2 −𝑉𝑦2 = 1 has a solution in positive integers 𝑥, 𝑦if and only if 2𝑈\|𝑢0 + 1 and 2𝑉\|𝑢0 −1. Lemma 2.7 (see [17, Criterion 2.]). Let 𝑈, 𝑉be positive integers such that gcd(𝑈, 𝑉) = gcd(𝑈, 2) = gcd(𝑉, 2) = 1 and 𝐷= 𝑈𝑉is not a square of a positive integer. Let (𝑢0, 𝑤0) denote the least positive integer solution of Pell’s equation 𝑢2 −𝐷𝑤2 = 1. Then equation 𝑈𝑥2 − 𝑉𝑦2 = 2 has a solution in positive integers 𝑥, 𝑦if and only if 𝑈\|𝑢0 + 1 and 𝑉\|𝑤0 −1. Now, by using the above arguments we were able to prove the following: Theorem 2.8. Let 𝑝, 𝑞be different odd primes, 𝑎, 𝑖, 𝑗, 𝑘positive integers such that gcd(𝑎, 𝑝𝑞) = 1, 𝑎< 𝑝𝑖𝑞𝑗and 𝑎𝑝𝑖𝑞𝑗is not a square. Further, let 𝑘≥max{𝑝𝑖, 𝑞𝑗} and 𝑚𝑘= 𝑝𝑖𝑞𝑗−𝑎 𝑘 . Suppose that either of the following holds: 1) 𝑚𝑘= 1 and 𝑎≠1; 2) 𝑚𝑘= 2 with 𝑎≠1, 5 and gcd(𝑎, 2) = 1; 3) 𝑚𝑘is a prime and 𝑟≥𝑎𝑚𝑘or 𝑟≥2𝑎2 with a square-free 𝑎, where 𝑎𝑝𝑖𝑞𝑗− 1 = 𝑟2. If {𝑎, 𝑝𝑖𝑞𝑗, 𝑐} is a 𝐷(−1)-triple in ℤ[ √ −𝑘], then 𝑐∈ℕ. EXTENSIONS OF 𝐷(−1)-PAIRS IN SOME IMAGINARY QUADRATIC FIELDS 751 Proof. By Theorem 2.3, we know that in such a 𝐷(−1)-triple we have 𝑐∈ℤ. Now we will use the argument from the proof of Theorem 2.1, i.e., we will consider the equation (1) which is equivalent to 𝑝𝑖𝑞𝑗𝑥2 −𝑎𝑦2 = 𝑚𝑘, (13) and prove that it is not solvable in integers 𝑥, 𝑦. In that way we will immediately obtain that 𝑐cannot be a negative integer. 1) Let 𝑚𝑘= 1 and 𝑎≠1. In that case the fundamental solution of equation 𝑢2−𝑎𝑝𝑖𝑞𝑗𝑤2 = 1 is 𝑢0+𝑤0 √ 𝑎𝑝𝑖𝑞𝑗, where 𝑢0 = 2𝑎𝑝𝑖𝑞𝑗−1, 𝑤0 = 2 √ 𝑎𝑝𝑖𝑞𝑗−1, One can check it very easily by using for instance [6, Proposition 10.13.]. By applying Lemma 2.6, the condition 2𝑎\|2(𝑎𝑝𝑖𝑞𝑗−1) implies that 𝑎= 1, which is not possible. 2) If we take 𝑚𝑘= 2 with 𝑎≠1, 5, then we obtain the equation 𝑝𝑖𝑞𝑗𝑥2 − 𝑎𝑦2 = 2. Since the condition 𝑎\|2 √ 𝑎𝑝𝑖𝑞𝑗−1 −1 from Lemma 2.7 leads to 𝑎= 1, 5, we arrive to a contradiction. 3) Since gcd(𝑎, 𝑝𝑖𝑞𝑗) = 1 and 𝑚𝑘is a prime, we know that the equation (13) can only have primitive integer solution. The equation (13) can be transformed to 𝑦2 1 −𝑎𝑝𝑖𝑞𝑗𝑥2 = −𝑎𝑚𝑘, (14) with 𝑦1 = 𝑎𝑦. Since 𝑎𝑝𝑖𝑞𝑗= 𝑟2 + 1, by Lemma 2.5 the equation (14) has no primitive solution (𝑦1, 𝑥), so neither does (13) for 𝑟≥𝑎𝑚𝑘. If 𝑟< 𝑎𝑚𝑘, then for 𝑟≥2𝑎2 we obtain 𝑎< 𝑚𝑘 2 . (15) Note that 𝑎𝑝𝑖𝑞𝑗≡𝑎2 (mod 𝑚𝑘). Since 𝑚𝑘is a prime and inequality (15) holds, the congruence 𝑙2 ≡𝑎𝑝𝑖𝑞𝑗 (mod 𝑚𝑘), (16) has only the solution 𝑙= 𝑎satisfying the condition 0 ≤𝑙≤𝑚𝑘∕2. Now, from Lemma 2.4 it follows that if (13) has a primitive integer solution (𝑥, 𝑦), then 𝑝𝑖𝑞𝑗𝑥2 −𝑎𝑦2 = 𝑎2 −𝑎𝑝𝑖𝑞𝑗 𝑚𝑘 = −𝑎𝑘 (17) also has a primitive integer solution. Since we know that 𝑎is square-free and gcd(𝑎, 𝑝𝑖𝑞𝑗)=1, we can set 𝑥= 𝑎𝑥0, where 𝑥0 is a positive integer. Then from (17) we obtain 𝑦2 −𝑎𝑝𝑖𝑞𝑗𝑥2 0 =𝑘. (18) Since 𝑟< 𝑎𝑚𝑘implies 1 < 𝑘≤𝑟, it follows from Lemma 2.5 that (18) has no primitive integer solution (𝑦, 𝑥0), and neither does (13). □ Now we can prove Theorem 1.1. 752 MIRELA JUKIĆ BOKUN AND IVAN SOLDO Proof of Theorem 1.1. If we suppose that there exists such a 𝐷(−1)-quadruple in the ring ℤ[ √ −𝑘], then by Theorem 2.3, Theorem 2.1 and Theorem 2.8 we know that 𝑐, 𝑑are positive integers. This contradicts with Corollary 2.2 and Theorem 1.2. □ Remark 2.9. From , it is known that there does not exist a 𝐷(−1)-quadruple {𝑎, 𝑏, 𝑐, 𝑑} in the ring ℤ[ √ −𝑘], 𝑘≥2, with integers 𝑎, 𝑏, 𝑐, 𝑑satisfying 𝑎< 𝑏≤ 8𝑎−3 (see [15, Theorem 1.6.]). That implies, if we take 𝑏= 𝑝𝑖𝑞𝑗from Theo-rem 1.1, we can conclude the nonexistence of a 𝐷(−1)-quadruple {𝑎, 𝑝𝑖𝑞𝑗, 𝑐, 𝑑} in the ring ℤ[ √ −𝑘] with 𝑐, 𝑑integers and 1 ≤𝑎< 𝑝𝑖𝑞𝑗≤8𝑎−3. Further, if we consider our previously mentioned results in terms of Corol-lary 2.2, we are able to construct the following example: Example 2.10. By Corollary 2.2 the 𝐷(−1)-quadruple of the form {𝑎, 𝑏, 𝑐, 𝑑}, with positive integers 𝑎, 𝑏, 𝑐= 𝑎+ 𝑏± 2𝑟and 𝑎𝑏−1 = 𝑟2 could exists in the ring ℤ[ √ −𝑘] if the condition 𝑘∣4 −3𝑎2 is satisfied. We recall the results on 𝐷(−1)-triples in the ring ℤ[ √ −𝑘] with 𝑘> 0. Lemma 2.11 (see [14, Theorem 5.4]). Let 𝑖, 𝑗, 𝑘be positive integers and 𝑝, 𝑞dif-ferent primes with 𝑝𝑖< 𝑞𝑗. If {𝑝𝑖, 𝑞𝑗, 𝑐} is a 𝐷(−1)-triple in the ring ℤ[ √ −𝑘], then 𝑐∈ℤ. Lemma 2.12 (see [24, Theorem 2.2], [9, Theorem 4]). Let 𝑖, 𝑘be positive inte-gers. If 𝑏is a prime or 𝑏= 2𝑝𝑖, with an odd prime 𝑝and {1, 𝑏, 𝑐} is a 𝐷(−1)-triple in the ring ℤ[ √ −𝑘], then 𝑐∈ℤ. Moreover, for every 𝑘there exists 𝑐> 0, while the case of 𝑐< 0 is possible if and only if 𝑘\|𝑏−1 and the equation 𝑥2 −𝑏𝑦2 = 1 −𝑏 𝑘 (19) has an integer solution. Let 𝑏1 = ⎧ ⎨ ⎩ 𝑝, 𝑝is an odd prime, 𝑘≥2, 2𝑝𝑗, 𝑝is an odd prime, 𝑗∈ℕ, 𝑘≥2, 𝑝𝑖𝑞𝑗, 𝑝, 𝑞, are different odd primes and , 𝑖, 𝑗∈ℕ, 𝑘≥max{𝑝𝑖, 𝑞𝑗} and 𝑏2 = { 𝑝𝑗, 𝑝is an odd prime, 𝑗∈ℕand 𝑘≥2, 𝑝𝑖𝑞𝑗, 𝑝, 𝑞are different odd primes and , 𝑖, 𝑗∈ℕ, 𝑘≥max{𝑝𝑖, 𝑞𝑗}. For 𝑎= 1 and 𝑏= 𝑏1, by Lemma 2.12, Theorem 2.3 we have that 𝑑∈ℤ. Now, the Corollary 2.2 implies that such 𝐷(−1)-quadruple could exists if 𝑘= 1. Similarly, if 𝑎= 2 and 𝑏= 𝑏2, by Lemma 2.11 and Theorem 2.3 we know that 𝑑∈ℤ. The condition 𝑘∣4 −3𝑎2 implies 𝑘∣8. By combining that with 𝑘∣𝑏2 −2 we conclude the existence of a qaudruple only in the case 𝑘= 1. Note that this coincides with the following result of Dujella on Gaussian inte-gers: EXTENSIONS OF 𝐷(−1)-PAIRS IN SOME IMAGINARY QUADRATIC FIELDS 753 Theorem 2.13 (see [4, Theorem 3]). Let 𝑙be a Gaussian integer and suppose that the set {𝑎, 𝑏} ⊂ℤ[𝑖] has the property 𝐷(𝑙2). If the number 𝑎𝑏is not a square of a Gaussian integer, then there exist an infinite number of complex Diophantine quadruples of the form {𝑎, 𝑏, 𝑐, 𝑑} with the property 𝐷(𝑙2). Namely, if 𝑙2 = 𝑖2 = −1, considering the above 𝐷(−1)-pairs {𝑎, 𝑏}, since 𝑎𝑏is not a square by Theorem 2.13 we know that in the case 𝑘= 1 there exist infinitely many 𝐷(−1)-quadruples of the form {𝑎, 𝑏, 𝑐, 𝑑} in Gaussian integers. Further, in , we proved the following: Lemma 2.14 (see [9, Proposition 2]). Let 𝑚, 𝑛> 0 and 𝑏= 𝑛2 + 1. If 𝑚\|𝑛and 𝑘= 𝑚2, then there exist infinitelymany𝐷(−1)-quadruplesoftheform {1, 𝑏, −𝑐, 𝑑}, 𝑐, 𝑑> 0 in ℤ[ √ −𝑘]. In , one can also find some special cases of such quadruples. We won-dered if 𝐷(−1)-quadruples of such form can be extended to quintuples in simi-lar imaginary quadratic rings. We were able to obtain the following result: Theorem 2.15. Let 𝑚, 𝑛, 𝑘be a positive integers and 𝑏= 𝑛2 + 1. Suppose that either of the following holds: 1) 𝑘= 𝑚2; 2) 𝑘∤𝑛2. There does not exists a 𝐷(−1)-quintuple of the form {1, 𝑏, −𝑐, 𝑑, 𝑒} with integers 𝑐, 𝑑> 0, 𝑒in ℤ[ √ −𝑘]. Proof. Let us suppose that under the above conditions there exists a 𝐷(−1)-quintuple of the form {1, 𝑏, −𝑐, 𝑑, 𝑒} in ℤ[ √ −𝑘]. 1) 𝑒> 0 and {1, 𝑏, 𝑑, 𝑒} is a 𝐷(−1)-quadruple contradicts with Theorem 1.2. Thus we have to consider the existence of a 𝐷(−1)-quintuple of the form {1, 𝑏, −𝑐, 𝑑, −𝑒}, where 𝑏= 𝑛2+1, 𝑐, 𝑑, 𝑒are positive integers. Since 𝑘= 𝑚2, 𝑚∈ ℕand ℤ[ √ −𝑘] = ℤ[𝑚𝑖] ≤ℤ[𝑖] it is sufficient to prove the nonexistence of such quintuple in ℤ[𝑖]. Therefore, we have to consider if there exists a positive inte-ger 𝑒such that −𝑒−1 = −𝛼2, −𝑏𝑒−1 = −𝛽2, 𝑐𝑒−1 = 𝛾2, −𝑑𝑒−1 = 𝛿2, for some integers 𝛼, 𝛽, 𝛾, 𝛿. Eliminating 𝑒we obtain 𝑐𝛿2 + 𝑑𝛾2 = −(𝑐+ 𝑑). It is clear that the above equation is not solvable in integers 𝛿, 𝛾. 2) According to Corollary 2.2 1) we conclude that in ℤ[ √ −𝑘] there does not exist a 𝐷(−1)-quadruple {1, 𝑏, −𝑐, 𝑑}, with positive integers 𝑏, 𝑐, 𝑑. □ Example 2.16. In the proof of Lemma 2.14 we obtain that 𝑏 = 𝑛2 + 1, 𝑘= 𝑚2, 𝑚∣𝑛, 𝑐 = 𝑛2𝑥2 −1, 𝑏𝑐 = 𝑛2𝑦2 −1, 754 MIRELA JUKIĆ BOKUN AND IVAN SOLDO and 𝑥, 𝑦are positive solutions of the Pellian equation 𝑦2 −(𝑛2 + 1)𝑥2 = −1. Moreover, 𝑑 = 𝑑± = ±2𝑛3𝑥𝑦+ (2𝑛2 + 1)𝑐+ 𝑛2 + 2. If the fifth element 𝑒in 𝐷(−1)-quintuple of the form {1, 𝑏, −𝑐, 𝑑, 𝑒} in ℤ[ √ −𝑘] is necessarily integer (for instance, like in the case 𝑏= 𝑏1 of Example 2.10), by apply-ing Theorem 2.15 1) we conclude that 𝐷(−1)-quadruples of the form {1, 𝑏, −𝑐, 𝑑} cannot be extended to quintuples in such rings. Acknowledgement The authors are deeply grateful to the referee for very helpful and detailed suggestions which improved the presentation of this paper. References Bliznac Trebješanin, Marija; Filipin, Alan. Nonexistence of 𝐷(4)-quintuples. J. Number Theory 194 (2019), 170–217. MR3860471, Zbl 1434.11077, arXiv:1704.01874, doi: 10.1016/j.jnt.2018.07.013. 745 Bonciocat, Nicolae C.; Cipu, Mihai; Mignotte, Maurice. There is no Diophantine D(-1)-quadruple. J. Lond. Math. Soc. 105 (2022), no. 1, 63–99. . MR4411320, Zbl 1531.11032, arXiv:2010.09200, doi: 10.1112/jlms.12507. 745, 746 Dujella, Andrej. Generalization of a problem of Diophantus. Acta Arith. 65 (1993), no. 1, 15–27. MR1239240 (94k:11031), Zbl 0849.11018, doi: 10.4064/aa-65-1-15-27. 745 Dujella, Andrej. The problem of Diophantus and Davenport for Gaussian integers. Glas. Mat. Ser III 32 (1997), no. 1, 1–10. MR1469613 (98h:11033), Zbl 0882.11019. 746, 753 Dujella, Andrej. There are only finitely many Diophantine quintuples. J. Reine Angew. Math. 566 (2004), 183–214. MR2039327 (2004m:11037), Zbl 1037.11019, doi: 10.1515/crll.2004.003. 745 Dujella, Andrej. Number Theory. Školska knjiga, Zagreb, 2021. 621 pp. ISBN: 978-953-0-30897-8. duje/numbertheorybook.html. 751 Dujella, Andrej. Diophantine 𝑚-tuples page. 746 Dujella, Andrej; Fuchs, Clemens. Complete solution of a problem of Diophantus and Euler. J. London Math. Soc. (2) 71 (2005), no. 1, 33–52. MR2108244 (2005g:11038), Zbl 1166.11309, doi: 10.1112/S002461070400609X. Dujella, Andrej; Jukić Bokun, Mirela; Soldo, Ivan. A Pellian equation with primes and applications to 𝐷(−1)-quadruples. Bull. Malays Math. Sci. Soc. 42 (2019), no. 5, 2915– 2926. MR3984453, Zbl 1452.11032, arXiv:1706.01959, doi: 10.1007/s40840-018-0638-5. 746, 752, 753 Dujella, Andrej; Soldo, Ivan. Diophantine quadruples in ℤ[ √ −2 ]. An. Ştiinţ. Univ. “Ovidius” Constanţa Ser. Mat. 18 (2010), no. 1, 81–97. MR2665940 (2011e:11057), Zbl 1212.11047. 746 Filipin, Alan; Fujita, Yasutsugu; Mignotte, Maurice. The non-extendibility of some parametric families of 𝐷(−1)-triples. Q. J. Math 63 (2012), no. 3, 605–621. MR2967165, Zbl 1255.11015, doi: 10.1093/qmath/har013. Franušić, Zrinka; Soldo, Ivan. The problem of Diophantus for integers of ℚ( √ −3), Rad Hrvat. Akad. Znan. Umjet. Mat. Znan. 18(519) (2014), 15–25. MR3333816, Zbl 1318.11045. 746 EXTENSIONS OF 𝐷(−1)-PAIRS IN SOME IMAGINARY QUADRATIC FIELDS 755 Fujita, Yasutsugu. The non-extensibility of 𝐷(4𝑘)-triples {1, 4𝑘(𝑘−1), 4𝑘2 + 1}, with \|𝑘\| prime. Glas. Mat. Ser. III 41(61) (2006), no. 2, 205–216. MR2282730 (2007j:11027), Zbl 1121.11024, doi: 10.3336/gm.41.2.03. 750 Fujita, Yasutsugu; Soldo, Ivan. On 𝐷(−1)-tuples in the ring ℤ[ √ −𝑘] with 𝑘 > 0. Publ. Math. Debrecen 100 (2022), no. 1-2, 49–67. MR4389245, Zbl 1499.11162, doi: 10.5486/PMD.2022.8975. 746, 747, 748, 752 Fujita, Yasutsugu; Soldo, Ivan. The non-existence of 𝐷(−1)-quadruples extending cer-tain pairs in imaginary quadratic rings. Acta. Math. Hungar. 170 (2023), no. 2, 455–482. MR4643849, Zbl 07745817, doi: 10.1007/s10474-023-01356-0. 746, 752 Fujita, Yasutsugu; Soldo, Ivan. On the extendibility of certain 𝐷(−1)-pairs in imagi-nary quadratic rings. Indian J. Pure Appl. Math., to appear. doi: 10.1007/s13226-023-00465-4. 747 Grelak, Aleksander; Grytczuk, Aleksander. On the diophantine equation 𝑎𝑥2 − 𝑏𝑦2 = 𝑐. Publ. Math. Debrecen 44 (1994), no. 3-4, 291–299. MR1291978 (95g:11018), Zbl 0817.11019. 750 He, Bo; Togbé, Alain. On the 𝐷(−1)-triple {1, 𝑘2 + 1, 𝑘2 + 2𝑘+ 2} and its unique 𝐷(1)-extension. J. Number Theory 131 (2011), no. 1, 120–137. MR2729213 (2011k:11042), Zbl 1223.11036, doi: 10.1016/j.jnt.2010.07.006. He, Bo; Togbé, Alain; Ziegler, Volker. There is no Diophantine quintuple. Trans. Amer. Math. Soc. 371 (2019), no. 9, 6665–6709. MR3937341, Zbl 1430.11044, arXiv:1610.04020, doi: 10.1090/tran/7573. 745 Hua, Loo K. Introduction to number theory. Springer-Verlag, Berlin, Heidelberg, 1982. xviii + 572 pp. ISBN: 3-540-10818-1 . MR0665428, Zbl 0483.10001, doi: 10.1007/978-3-642-68130-1. 749 Nagell, Trygve. Introduction to number theory. John Wiley & Sons, Inc., New York; Almqvist & Wiksell, Stockholm, 1951. 309 pp. MR0043111, Zbl 0042.26702. Sloane, Neil J.A. The On-Line Encyclopedia of Integer Sequences. 746 Soldo, Ivan. On the existence of Diophantine quadruples in ℤ[ √ −2 ]. Miskolc Math. Notes 14 (2013), no. 1, 265–277. MR3070707, Zbl 1286.11029. 746 Soldo, Ivan. 𝐷(−1)-triples of the form {1, 𝑏, 𝑐} in the ring ℤ[ √ −𝑡], 𝑡 > 0. Bull. Malays. Math. Sci. Soc. 39 (2016), no. 3, 1201–1224. MR3515075, Zbl 1419.11055, doi: 10.1007/s40840-015-0229-7. 747, 748, 752, 753 (Mirela Jukić Bokun) School of Applied Mathematics and Informatics, University of Osijek, Trg Ljudevita Gaja 6, HR-31000 Osijek, Croatia mirela@mathos.hr (Ivan Soldo) School of Applied Mathematics and Informatics, University of Osijek, Trg Ljudevita Gaja 6, HR-31000 Osijek, Croatia isoldo@mathos.hr This paper is available via
1205	https://www.tutorialspoint.com/network_theory/network_theory_delta_to_star_conversion.htm	Network Theory - Delta to Star Conversion In the previous chapter, we discussed an example problem related equivalent resistance. There, we calculated the equivalent resistance between the terminals A & B of the given electrical network easily. Because, in every step, we got the combination of resistors that are connected in either series form or parallel form. However, in some situations, it is difficult to simplify the network by following the previous approach. For example, the resistors connected in either delta (δ) form or star form. In such situations, we have to convert the network of one form to the other in order to simplify it further by using series combination or parallel combination. In this chapter, let us discuss about the Delta to Star Conversion. Delta Network Consider the following delta network as shown in the following figure. The following equations represent the equivalent resistance between two terminals of delta network, when the third terminal is kept open. $$R_{AB} = \frac{(R_1 + R_3)R_2}{R_1 + R_2 + R_3}$$ $$R_{BC} = \frac{(R_1 + R_2)R_3}{R_1 + R_2 + R_3}$$ $$R_{CA} = \frac{(R_2 + R_3)R_1}{R_1 + R_2 + R_3}$$ Star Network The following figure shows the equivalent star network corresponding to the above delta network. The following equations represent the equivalent resistance between two terminals of star network, when the third terminal is kept open. $$R_{AB} = R_A + R_B$$ $$R_{BC} = R_B + R_C$$ $$R_{CA} = R_C + R_A$$ Star Network Resistances in terms of Delta Network Resistances We will get the following equations by equating the right-hand side terms of the above equations for which the left-hand side terms are same. $R_A + R_B = \frac{(R_1 + R_3)R_2}{R_1 + R_2 + R_3}$ Equation 1 $R_B + R_C = \frac{(R_1 + R_2)R_3}{R_1 + R_2 + R_3}$ Equation 2 $R_C + R_A = \frac{(R_2 + R_3)R_1}{R_1 + R_2 + R_3}$ Equation 3 By adding the above three equations, we will get $$2(R_A + R_B + R_C) = \frac{2(R_1 R_2 + R_2 R_3 + R_3 R_1)}{R_1 + R_2 + R_3}$$ $\Rightarrow R_A + R_B + R_C = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1 + R_2 + R_3}$ Equation 4 Subtract Equation 2 from Equation 4. $R_A + R_B + R_C - (R_B + R_C) = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1 + R_2 + R_3} - \frac{(R_1 + R_2)R_3}{R_1 + R_2 + R_3}$ $$R_A = \frac{R_1 R_2}{R_1 + R_2 + R_3}$$ By subtracting Equation 3 from Equation 4, we will get $$R_B = \frac{R_2 R_3}{R_1 + R_2 + R_3}$$ By subtracting Equation 1 from Equation 4, we will get $$R_C = \frac{R_3 R_1}{R_1 + R_2 + R_3}$$ By using the above relations, we can find the resistances of star network from the resistances of delta network. In this way, we can convert a delta network into a star network. Example Let us calculate the resistances of star network, which are equivalent to that of delta network as shown in the following figure. Given the resistances of delta network as R1 = 10 Ω, R2 = 60 Ω and R3 = 30 Ω. We know the following relations of the resistances of star network in terms of resistances of delta network. $$R_A = \frac{R_1 R_2}{R_1 + R_2 + R_3}$$ $$R_B = \frac{R_2 R_3}{R_1 + R_2 + R_3}$$ $$R_C = \frac{R_3 R_1}{R_1 + R_2 + R_3}$$ Substitute the values of R1, R2 and R3 in the above equations. $$R_A = \frac{10 \times 60}{10 +60+30} = \frac{600}{100} = 6\Omega$$ $$R_B = \frac{60 \times 30}{10 +60+30} = \frac{1800}{100} = 18\Omega$$ $$R_C = \frac{30 \times 10}{10 +60+30} = \frac{300}{100} = 3\Omega$$ So, we got the resistances of star network as RA = 6 Ω, RB = 18 Ω and RC = 3 Ω, which are equivalent to the resistances of the given delta network. TOP TUTORIALS TRENDING TECHNOLOGIES CERTIFICATIONS COMPILERS & EDITORS Tutorials Point is a leading Ed Tech company striving to provide the best learning material on technical and non-technical subjects. © Copyright 2025. All Rights Reserved.
1206	https://mathoverflow.net/questions/20298/sum-of-difference-moduli-vs-sum-of-modulus-differences	inequalities - Sum of difference moduli vs. sum of modulus differences - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Sum of difference moduli vs. sum of modulus differences Ask Question Asked 15 years, 6 months ago Modified15 years, 5 months ago Viewed 2k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. This is a failed attempt of mine at creating a contest problem; the failure is in the fact that I wasn't able to solve it myself. Let x 1 x 1, x 2 x 2, ..., x n x n be n n reals. For any integer k k, define a real f k(x 1,x 2,...,x n)f k(x 1,x 2,...,x n) as the sum ∑T⊆{1,2,...,n};\|T\|=k\|∑t∈T x t−∑t∈{1,2,...,n}∖T x t\|∑T⊆{1,2,...,n};\|T\|=k∣∣∣∣∑t∈T x t−∑t∈{1,2,...,n}∖T x t∣∣∣∣. We mostly care about the case of n n even and k=n 2 k=n 2; in this case, f k(x 1,x 2,...,x n)f k(x 1,x 2,...,x n) is a kind of measure for the dispersion of the reals x 1 x 1, x 2 x 2, ..., x n x n (more precisely, of their n 2 n 2-element sums). Now my conjecture is that if n n is even and k=n 2 k=n 2, then f k(x 1,x 2,...,x n)≥f k(\|x 1\|,\|x 2\|,...,\|x n\|)f k(x 1,x 2,...,x n)≥f k(\|x 1\|,\|x 2\|,...,\|x n\|) for any reals x 1 x 1, x 2 x 2, ..., x n x n. I think I have casebashed this for n=4 n=4 and maybe n=6 n=6; I don't remember anymore - it's too long ago. Sorry. I still have no idea what to do in the general case, although my attempts at big-n n counterexamples weren't of much success either. inequalities linear-programming co.combinatorics Share Share a link to this question Copy linkCC BY-SA 2.5 Cite Improve this question Follow Follow this question to receive notifications asked Apr 4, 2010 at 10:24 darij grinbergdarij grinberg 35.4k 4 4 gold badges 124 124 silver badges 266 266 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. Hi, Darij! This is actually quite simple (and also much more appropriate for AoPS than for MO). The idea is to show that for every t t, the expression ∑T\|t+D T\|∑T\|t+D T\|, where D T D T is your difference, goes down if you replace all x k x k by their absolute values (t=0 t=0 is your claim). The base n=2 n=2 is rather trivial and boils down to the inequality \|t−a\|+\|t+a\|=2 max(\|t\|,\|a\|)≥2 max(\|t\|,\|b\|)=\|t−b\|+\|t+b\|\|t−a\|+\|t+a\|=2 max(\|t\|,\|a\|)≥2 max(\|t\|,\|b\|)=\|t−b\|+\|t+b\| when \|a\|≥\|b\|\|a\|≥\|b\|. Now, assume that we know the statement for n n and want to show it for n+2 n+2. The trick is to choose a random pair of indices i,j i,j and notice that the full sum is just the average over such choices of ∑T(\|t+x i−x j+D T\|+\|t+x j−x i+D T\|)∑T(\|t+x i−x j+D T\|+\|t+x j−x i+D T\|) where T T runs over all n/2 n/2 element subsets of the set of remaining indices. Now, applying the statement with n=2 n=2 for fixed T T, we see that we can replace x i x i and x j x j with their absolute values and our sum (with fixed i,j i,j) will go down in each term. After that, we replace everything else by the absolute values using the induction assumption and, again, the sum will go down. But now we are completely done: we showed that for each fixed i,j i,j, the sum goes down when we replace everything by the absolute value, so it is true after averaging as well. The whole thing is just a textbook case of the "inventor's paradox". Strange that you haven't figured it out... Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Apr 11, 2010 at 5:40 fedjafedja 63.8k 11 11 gold badges 164 164 silver badges 308 308 bronze badges 2 Hi fedja! Very nice solution. I thought about posting it on AoPS, but to me it seems that most of the good problem-solvers there are gone. What your solution actually shows is that we can replace the outer modulus by an arbitrary (fixed) convex function: [...]darij grinberg –darij grinberg 2010-04-11 10:27:01 +00:00 Commented Apr 11, 2010 at 10:27 1 If g g is a convex function on R, and we denote by f k(x 1,x 2,...,x n) the sum ∑T⊆{1,2,...,n};\|T\|=k g(∑t∈T x t−∑t∈{1,2,...,n}∖T x t), then f k(x 1,x 2,...,x n)≥f k(\|x 1\|,\|x 2\|,...,\|x n\|) for even n and for k=n 2.darij grinberg –darij grinberg 2010-04-11 10:28:28 +00:00 Commented Apr 11, 2010 at 10:28 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions inequalities linear-programming co.combinatorics See similar questions with these tags. 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1207	https://math.xula.edu/Math1030/U4_5_page.html	``` \chapter{Trigonometry} \section{The reference angle and its use} In Unit 4.3 we defined the trig functions for acute angles using SOH-CAH-TOA, and we used those definitions to find the exact trig values of (30^\circ), (45^\circ), and (60^\circ) angles (all in Quadrant 1). Then in Unit 4.4 we used the (x,y,r) definition to define the trig functions for any angles, including quadrantal angles. In this unit we combine the (x, y, r) definition and the values for the (30^\circ), (45^\circ), and (60^\circ) angles to find \textbf{exact} trig values for all the special angles that we studied in Unit 4.2 (the multiples of (30^\circ), (45^\circ), and (60^\circ) in Quadrants 2, 3 and 4). A key ingredient we will use is the reference angle, that we now define. \subsection{Geometric picture of reference angle} Given an angle (\theta) in standard position (and not a quadrantal angle), the \textit{reference angle} (\theta_R) is the acute, positive angle that the terminal side of (\theta) makes with the (x)-axis. This means that the reference angle depends only on the terminal side of (\theta). The pictures below show the reference angle for a terminal side in each of the four quadrants. [\begin{array}{cc} \img{U4_5F1.png}{}{18em}{} & \hspace{3ex}\img{U4_5F2.png}{}{18em}{} \end{array} ] [\begin{array}{cc} \img{U4_5F3.png}{}{18em}{} & \hspace{3ex} \img{U4_5F4.png}{}{18em}{} \end{array} ] Note that the reference angle is always positive, by definition, so we draw it without an arrow. It is also important to note that different angles (\theta) may have the same reference angle (\theta_R). The following pictures show two different angles (\theta_1) and (\theta_2) (one positive and one negative) that have the same reference angle, for each of the four quadrants. Note that an acute angle (a positive angle smaller than (90^\circ)) is its own reference angle ((\theta_1) in the first picture). [\begin{array}{cc} \img{U4_5F5.png}{}{18em}{} & \hspace{3ex}\img{U4_5F6.png}{}{18em}{}\[2ex] \img{U4_5F7.png}{}{18em}{} & \hspace{3ex} \img{U4_5F8.png}{}{18em}{} \end{array} ] From a picture, it is possible to find how the reference angle is related to a given angle, as we show in the following example. \begin{example} The picture shows a positive angle of (147^\circ) together with its reference angle (\theta_R). [ \img{U4_5F9.png}{}{18em}{} ] We see from the picture that (\theta_R) is the angle needed to complete a straight angle of (180^\circ). In other words, (147^\circ + \theta_R = 180^\circ), and so (\theta_R = 180-147 =33^\circ). \end{example} If the angle is negative, it is useful to work with its size, or absolute value, as in the following example. \begin{example} The picture shows a negative angle (\theta=-132^\circ) and its reference angle (\theta_R). [ \img{U4_5F10.png}{}{18em}{} ] The size of (\theta ) is (\|-132^\circ\|=132^\circ). We see from the picture that if we add (\theta_R) to the \textbf{size} of (\theta) (not to (\theta) itself) we get (180^\circ), that is, (132^\circ + \theta_R =180^\circ) and so (\theta_R =180-132=48^\circ). \end{example} \begin{example} The picture shows a positive angle (\displaystyle \theta=\frac{11\pi}{6}) and its reference angle (\theta_R). [ \img{U4_5F11.png}{}{18em}{} ] If we add (\theta_R) to (\theta) we get a (2\pi) angle, that is: [ \frac{11\pi}{6} +\theta_R =2\pi,] and so [\theta_R=2\pi-\frac{11\pi}{6}=\frac{12\pi}{6} -\frac{11\pi}{6}=\frac{\pi}{6}.] \end{example} \subsection{The reference angle algebraically} We can also find the reference angle from just the degree or radians measure of the angle, without drawing any picture. If the absolute value (or size) (\|\theta\|) is less than (90^\circ), then (\theta_R=\|\theta\|). Otherwise, working in degrees, we subtract (or add) (180^\circ) from the given angle until we get a number (positive or negative) whose absolute value is less than (90). Then the absolute value of that number (always positive) will be the reference angle. If we are working in radians, we subtract (or add) (\pi), and proceed in the same way. \begin{example} Suppose (\theta = 585^\circ). Subtracting (180) we find [585-180= 405.] Subtract 180 again: [405=180 = 225.] Again: [225-180=45.] This last number is less than (90), so the reference angle is (\theta_R = 45^\circ). \end{example} \begin{example} Suppose (\displaystyle \theta = -\frac{37\pi}{18}). Add (\pi), to reduce the size of the number: [-\frac{37\pi}{18}+\pi = -\frac{37\pi}{18}+\frac{18\pi}{18}=-\frac{19\pi}{18}.] (19\pi/18) is larger than (\pi/2). So we reduce the size by adding (\pi) again: [-\frac{19\pi}{18}+\pi =-\frac{19\pi}{18} +\frac{18\pi}{18} = -\frac{\pi}{18}.] We found a (negative) number whose size is less than (\pi/2). So the absolute value of that number is the reference angle: [\theta_R = \frac{\pi}{18}.] \end{example} \subsection{Exact trig values for multiples of (30^\circ), (45^\circ) and (60^\circ)} We now use the reference angle to find the exact value of the trig functions for the multiples of (30^\circ), (45^\circ) and (60^\circ) that are not in Quadrant 1. The crucial tool is the following rule. \| \| \| Suppose (f(\theta) ) is any of the six trig functions. Then \| \| (f(\theta)=\pm f(\theta_R)) \| \| where the sign is determined by the ASTC rule. \| Since (\theta_R) is always an acute angle, if it is one of (30^\circ), (45^\circ) and (60^\circ) we can find (f(\theta_R)) by using either the standard (30)-(60)-(90) triangle or the standard (45)-(45)-(90) triangle. Then we need to determine the quadrant for (\theta) and apply the ASTC rule to decide if (f(\theta)) is positive or negative. \begin{example} Suppose we want to find (\sin(240^\circ)). First we find the reference angle: [240-180=60] so (\theta_R=60^\circ). Draw the standard (30)-(60)-(90) triangle: [\img{U4_3F9.png}{}{10em}{}.] Then we find [\sin(\theta_R)=\sin(60^\circ)=\frac{\mbox{Opp}}{\mbox{Hyp}}=\frac{\sqrt{3}}{2}.] Now notice that (180< 240 < 270), and so (240^\circ) is in Q3. By the ASTC rule, the sine function is negative in Q3. So we conclude that [\sin(240^\circ) =-\frac{\sqrt{3}}{2}.] \end{example} \begin{example} We want to find the exact value of (\cos(-7\pi/4)). We find (\theta_R) by adding (\pi): [-\frac{7\pi}{4}+\pi=-\frac{7\pi}{4}+\frac{4\pi}{4}=-\frac{3\pi}{4},] [-\frac{3\pi}{4}+\pi=-\frac{3\pi}{4}+\frac{4\pi}{4}=\frac{\pi}{4}.] So (\theta_R= \pi/4=45^\circ). Draw the standard (45)-(45)-(90) triangle: [\img{U4_3F23.png}{}{8em}{}.] Then we find [\cos(\theta_R)=\cos \left(\frac{\pi}{4}\right)=\frac{\mbox{Adj}}{\mbox{Hyp}}=\frac{1}{\sqrt{2}}.] To find the quandrant for ( -\frac{7\pi}{4}), count seven (45^\circ) slices in clockwise direction: [\img{U4_5F18.png}{}{15em}{}] and find that (-7\pi/4) is in Q1. By the ASTC rule all trig functions are positive there. So we find [\cos\left(-\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}}.] \end{example} \subsection{Finding angles from a given trig value} Suppose we know the value of a trig function, and we are given an interval where the angle must be. Using the reference angle, we can find all the angles in that interval that have the given trig function value. \begin{example} Suppose we know that (\displaystyle \sin \theta=-\frac{1}{2}) and (\theta ) is in the interval (0\leq \theta \leq 360^\circ). We will proceed in three steps. \textbf{ Step 1.} Determine the quadrants. Since the sine is negative, the solutions must be in Q3 and Q4, by the ASTC rule. \textbf{Step 2.} Find and draw the reference angle. The value (-1/2) of the trig function reminds us of the (30)-(60)-(90) triangle. [\img{U4_3F9.png}{-3em}{10em}{} \hspace{3ex} \sin\theta_R =\frac{\mbox{Opp}}{\mbox{Hyp}}=\frac{1}{2}\Longrightarrow \theta_R= 30^\circ] After drawing it, we find that the reference angle is (30^\circ). We draw a (30^\circ) reference angle in Q3 and Q4: [\img{U5_4F1.png}{}{12em}{}] \textbf{Step 3.} Find all angles in the given interval. We draw all possible angles that are in the interval ([0,360^\circ]) and have terminal sides there: [\img{U5_4F2.png}{}{12em}{}] So there are two angles in the interval ([0,360^\circ]) with (\sin \theta =-1/2): (\theta_1=210^\circ) and (\theta_2=330^\circ). \end{example} \begin{example} Suppose now the given trig value is the same as in the previous example: (\displaystyle \sin \theta=-\frac{1}{2}), but (\theta ) must be in the interval ([-180^\circ,180^\circ]). Then the only possible values for (\theta) are the two negative angles (\theta_1=-30^\circ), (\theta_2=-150^\circ), as shown below: [\img{U5_4F3.png}{}{12em}{}] \end{example} In the next example, we use radians and a larger interval, and we find more possible values for (\theta). \begin{example} We are given that [\cos \theta =\frac{1}{\sqrt{2}}] and (\theta) is in the interval ([-2\pi,2\pi]). \textbf{1.} By the ASTC rule, the cosine is positive in Q1 and Q4. \textbf{2.} The reference angle is (\pi/4), and we draw it in Q1 and Q4: [\img{U5_4F4.png}{}{12em}{}] \textbf{3.} There are two positive angles in (0\leq x \leq 2\pi) and two negative angles in (-2\pi\leq x \leq 0) with those terminal sides, and so four total solutions in the interval ([-2\pi,2\pi]): [\theta_1=\frac{\pi}{4}, \hspace{2ex} \theta_2=\frac{7\pi}{4}, \hspace{2ex} \theta_3=-\frac{\pi}{4}, \hspace{2ex} \theta_4=-\frac{7\pi}{4}.] [\img{U5_4F5.png}{}{12em}{}] \end{example} \subsection{The unit circle} The circle with center at ((0,0)) and radius (1) is called the \textit{unit circle}. So if ((x,y)) is a point on the unit circle, the corresponding value of (r) will be (r=1), and using the (x,y,r) definition of the sine and cosine functions, we find [ \begin{array}{cc} \cos \theta = \dfrac{x}{1} = x & \sin \theta = \dfrac{y}{1}=y. \end{array} ] So a point on the unit circle corresponding to angle (\theta) will have coordinates ((x,y)=(\cos \theta, \sin \theta)), as shown in the picture below. [ \img{UnitCircleEx.png}{}{18em}{} ] The next picture shows all the special and quadrantal angles together with the corresponding values((\cos \theta, \sin \theta)) on the unit circle. [ \img{UnitCircle.png}{}{30em}{} ] Problems \problem Find the reference angle for each of the angles given in the pictures. [ \begin{array}{ccc} \img{U4_5F12.png}{}{10em}{} \hspace{5ex} & \img{U4_5F13.png}{}{10em}{} \hspace{5ex} & \img{U4_5F14.png}{}{10em}{} \ \mbox{a.} \hspace{5ex} & \mbox{b.} \hspace{5ex} & \mbox{c.} \end{array} ] \begin{sol} \begin{enumerate} \item We see from the picture that if we add (\theta_R) to (180^\circ), we get (213^\circ). So we need to solve [\theta_R + 180=213] and we find [\theta_R=213-180=33^\circ.] \item The given angle is negative, and larger (in size) than (180^\circ = \pi) by the size of (\theta_R). So if we reduce the size (6\pi/5) by (\theta_R), we will get (\pi): [\frac{6\pi}{5} -\theta_R =\pi ] So [\theta_R = \frac{6\pi}{5} - \pi = \frac{6\pi}{5} - \frac{5\pi}{5} = \frac{\pi}{5}.] \item If we add (\theta_R) to (4\pi/7), we get (\pi): [\frac{4\pi}{7} + \theta_R = \pi] [\theta_R= \pi - \frac{4\pi}{7}=\frac{7\pi}{7}-\frac{4\pi}{7}=\frac{3\pi}{7}.] \end{enumerate} \end{sol} \mproblem Find the reference angle for each of the angles given in the pictures. [ \begin{array}{ccc} \img{U4_5F15.png}{}{10em}{} \hspace{5ex} & \img{U4_5F16.png}{}{10em}{} \hspace{5ex} & \img{U4_5F17.png}{}{10em}{} \ \mbox{a.} \hspace{5ex} & \mbox{b.} \hspace{5ex} & \mbox{c.} \end{array} ] \problem Find the reference angle for the angles: [ \begin{array}{ccc} \mbox{a.}\hspace{1ex} -670^\circ \hspace{5ex} & \mbox{b.}\hspace{1ex} \displaystyle \frac{12\pi}{5}\hspace{5ex} & \mbox{c.}\hspace{1ex} 355^\circ \end{array} ] \begin{sol} \begin{enumerate} \item We add (180) till we get the size of the number less than (90): [-670+180=-490] [-490+180=-310] [-310+180=-130] [-130+180=50.] So (\theta_R = 50^\circ). \item Reduce the size by subtracting (\pi): [\frac{12\pi}{5}-\pi=\frac{12\pi}{5} -\frac{5\pi}{5}=\frac{7\pi}{5}.] [\frac{7\pi}{5} -\pi =\frac{7\pi}{5}-\frac{5\pi}{5} =\frac{2\pi}{5}.] So [\theta_R=\frac{2\pi}{5}.] \item (355-180=175), and (175-180=-5). So (\theta_R = 5^\circ). \end{enumerate} \end{sol} \mproblem Find the reference angle for the angles: [ \begin{array}{ccc} \mbox{a.}\hspace{1ex} \displaystyle -\frac{14\pi}{3} \hspace{5ex} & \mbox{b.}\hspace{1ex} \displaystyle 285^\circ\hspace{5ex} & \mbox{c.}\hspace{1ex} -440^\circ \end{array} ] \problem Find the exact value of all six trig functions for (\theta=-4\pi/3). \begin{sol} Find the reference angle by adding (\pi): [-\frac{4\pi}{3}+\pi=-\frac{4\pi}{3}+\frac{3\pi}{3}=-\frac{\pi}{3}.] So (\theta_R = \pi/3=60^\circ), and the trig functions for (\theta_R) are found with the standard standard (30)-(60)-(90) triangle: [\img{U4_3F9.png}{}{10em}{}] [\begin{array}{ccc} \displaystyle \sin 60^\circ = \frac{\mbox{Opp}}{\mbox{Hyp}} = \frac{\sqrt{3}}{2} & \displaystyle \cos 60^\circ = \frac{\mbox{Adj}}{\mbox{Hyp}} = \frac{1}{2} & \displaystyle \tan 60^\circ = \frac{\mbox{Opp}}{\mbox{Adj}} = \frac{\sqrt{3}}{1}=\sqrt{3}\[3ex] \displaystyle \csc 60^\circ = \frac{\mbox{Hyp}}{\mbox{Opp}} = \frac{2}{\sqrt{3}} & \displaystyle \sec 60^\circ = \frac{\mbox{Hyp}}{\mbox{Adj}} = \frac{2}{1}=2 & \displaystyle \cot 60^\circ = \frac{\mbox{Adj}}{\mbox{Opp}} = \frac{1}{\sqrt{3}} \end{array} ] To find the quadrant for (\theta), count four (60^\circ) slices clockwise: [\img{U4_5F19.png}{}{15em}{}.] So (\theta) is in Quadrant 2. By the ASTC rule, (\sin) and (\csc) are positive, and all others negative. So we can now find all trig functions for (\theta): [\begin{array}{ccc} \displaystyle \sin\left(-\frac{4\pi}{3}\right) = \frac{\sqrt{3}}{2} & \displaystyle \cos\left(-\frac{4\pi}{3}\right) = -\frac{1}{2} & \displaystyle \tan\left(-\frac{4\pi}{3}\right) = -\sqrt{3}\[2ex] \displaystyle \csc\left(-\frac{4\pi}{3}\right) = \frac{2}{\sqrt{3}} & \displaystyle \sec\left(-\frac{4\pi}{3}\right) = -2 & \displaystyle \cot\left(-\frac{4\pi}{3}\right) = -\frac{1}{\sqrt{3}} \end{array} ] \end{sol} \mproblem Find the exact value of all six trig functions for (\theta=-5\pi/6). \problem Find the exact value of all six trig functions for (\theta=225^\circ). \begin{sol} Find the reference angle by subtracting (180): [225-180=45.] So (\theta_R=45^\circ). Draw the standard (45)-(45)-(90) triangle and find the trig functions for (\theta_R): [\img{U4_3F23.png}{}{8em}{},] [\begin{array}{ccc} \displaystyle \sin 45^\circ = \frac{\mbox{Opp}}{\mbox{Hyp}} = \frac{1}{\sqrt{2}} & \displaystyle \cos 45^\circ = \frac{\mbox{Adj}}{\mbox{Hyp}} = \frac{1}{\sqrt{2}} & \displaystyle \tan 45^\circ = \frac{\mbox{Opp}}{\mbox{Adj}} = \frac{1}{1}=1\[2ex] \displaystyle \csc 45^\circ = \frac{\mbox{Hyp}}{\mbox{Opp}} = \frac{\sqrt{2}}{1}=\sqrt{2} & \displaystyle \sec 45^\circ = \frac{\mbox{Hyp}}{\mbox{Adj}} = \frac{\sqrt{2}}{1}=\sqrt{2} & \displaystyle \cot 45^\circ = \frac{\mbox{Adj}}{\mbox{Opp}} = \frac{1}{1}=1 \end{array} ] Since (225=180+45), (\theta) is in Quadrant 3: [\img{U4_5F20.png}{}{18em}{}.] By the ASTC rule, (\tan) and (\cot) are positive, and all others negative. [\begin{array}{ccc} \displaystyle \sin 225^\circ = -\frac{1}{\sqrt{2}} & \displaystyle \cos 225^\circ = -\frac{1}{\sqrt{2}} & \displaystyle \tan 225^\circ = 1\[2ex] \displaystyle \csc 225^\circ = -\sqrt{2} & \displaystyle \sec 225^\circ = -\sqrt{2} & \displaystyle \cot 225^\circ = 1 \end{array} ] \end{sol} \mproblem Find the exact value of all six trig functions for (\theta=135^\circ). \problem Suppose that (\displaystyle \sin \theta =\frac{\sqrt{3}}{2}) and (\theta) is in the interval ([0,2\pi]). Find all the possible values for (\theta) \begin{sol} \textbf{1.} The sine is positive in Q1 and Q2. \textbf{2.} The reference angle is (\pi/3). Draw it in Q1 and Q2: [\img{U5_4F9.png}{}{15em}{}] \textbf{3.} There are two solutions in the interval ([0,2\pi]): (\theta_1= \pi/3) and (\theta_2=2\pi/3). [\img{U5_4F10.png}{}{15em}{}] \end{sol} \mproblem Suppose that (\displaystyle \sqrt{3}\tan \theta=-\frac{1}{\sqrt{3}}) and (\theta ) is in the interval ([0,2\pi]). Find all the possible values of (\theta). \| \| \| \| --- \| \| This is the end of Unit 4.5! Now \| \| \| \| Check your understanding \| \| \| « Previous Unit \| \| Next Unit » \| \| \| Go to WeBWorK \| \| ```
1208	https://www.youtube.com/watch?v=ogiEoIE-n9w	Behaviour of the graph of function near the vertical asymptotes Anil Kumar 404000 subscribers 15 likes Description 1258 views Posted: 29 Oct 2017 Behaviour near Horizontal Asymptote: End behaviour, Oblique Asymptote: #graph_equation_relation 2 comments Transcript: I'm Anil Kumar and in this video I'm going to discuss behavior of a function near the vertical asymptotes we'll have three examples the first one is G of X equals 2x plus 2 divided by X minus 1 then we have B f of X equals to 1 divided by X square minus 2x minus 15 and then H of x equals 2x divided by X plus 2 times X minus 3 so let's begin with the very first one which is G of X X plus 2 divided by X minus 1 right well kind of well not really sketch the function well consider it more on vertical asymptotes right now how do you find vertical asymptotes let's look into that so when we are talking about vertical asymptotes we can equate denominator to zero is it okay X minus 1 is in the denominator we say X minus 1 equals to 0 or X is equals to 1 now here there is only one vertical asymptote and that is that x equals to 1 so let us sketch it here let's say this is vertical asymptotes at x equals to 1 so equation of this line is x equals to 1 it goes through x equals to 1 perfect ok now when we see behavior of the function near vertical asymptotes we are trying to find that if X approaches 1 from left that is from this side we normally write with - as an exponent kind of thing so 1 with minus sign there on the right side or left side it means from left when X approaches 1 from the left side from zero side then what happens to G of X we know it goes to infinity but whether it goes to positive or negative infinity that is the idea so let's substitute a value which is less than 1 so in our case our value less than 1 could be let us say we substitute G of less than 1 could be 0.9 do you see that substitute a value and then check whether you get a positive answer or a negative or self now if I write 0.9 for X then what do we get we get zero point nine plus two over zero point nine minus one clearly numerator is positive rights we say numerator is positive we are not really interested in the value however denominator is negative since that is the case positive and negative should be negative right so that means G of X approaches negative infinity so we denote it like this is quick like this downwards so when you approach vertical asymptote from the left side then the function approaches negative infinity that's what we understand by vertical asymptotes behave in your function now let us also figure out what happens when X approaches from the right side that is X 1 plus this is from the right side what happens to G of X now you don't really have to place a value like this you can do mental calculations so think about I'm just writing so that it helps one point one so you know this is positive this is also positive so we are talking about positive over positive so this will be positive infinity and therefore the function should go like this do you see that so that is how the behavior is going to be for the given function as X approaches from left and right is it okay so let's summarize the result for G of X so we see when X approaches one from negative sign G of X approaches negative infinity and when X approaches 1 from the positive side G of X approaches positive infinity do you see that and you know what the horizontal asymptote is the ratio of these two so horizontal asymptote will be kind of like this and if you sketch the function you can find x and y-intercepts y-intercept is 2 over minus 1 which is here x intercept is numerator 0 which is at minus 2 so you could sketch a function which is kind of like this do you see that so you can see that behaving in their symptom it approaches towards negative infinity then approach you from left and positive infinity from right so I hope the concept is clear now we can move on with the other two examples second example is f of X equals to 1 over X square minus 2x minus 15 so at times you may have to factor so we may write this as we are looking for product of 15 and sum of -2 minus 15 I mean so a negative and a positive number X minus 5 times X plus 3 will work for us when you multiply minus 15 when you add minus 2 now that clearly indicates that we have how many vertical asymptotes 1 vertical asymptote is because of X minus 5 if you equate to 0 we get x equals to 5 and the other one is because of X plus 3 when you equate that to 0 you get x equals 2 minus 3 is it ok so that means if I want to sketch the function there first thing the vertical asymptotes will be ad plus 5 and minus 3 okay and let's say this is minus 3 for us so these are the vertical asymptotes in this video we'll concentrate on behavior near the vertical asymptotes that is to say we want to know first near the say minus 3 so if X is approaching minus 3 from the left side you understand now from the left side then what happens to the function so what you can do is you can place a value which is on the left side let's say minus 3.1 right more negative than 3 right so if I place a value which is negative we know this will be negative and different is more negative than minus 3 less than minus 3.1 that is also negative so two negatives will give us positive right so you'll find that f of X will approach positive infinity correct because if I substitute a value let us say let us say minus 3.1 is it okay then this is negative minus 3.1 minus 5 this also becomes negative minus 3.1 plus 3 will give me minus 0.1 we're not interested in value but we address it in sign so we clearly get the answer that if you are approaching from minus side then the functions value become negatively large so that's the behavior right now if I substitute a value which is slightly more than -3 that is to the right side so minus 3 with positive rewrite that means on the right side then f of X is what it is a negative value so this remains negative however that becomes positive so we are approaching negative infinity is it okay so oh sorry how first was positive infinity I should have drawn that on the other side right so I'm sorry for that not this this is positive infinity right when you are approaching from the right side we are approaching negative infinity so that would be like this negative infinity okay now let's analyze what happens at the other vertical asymptote so when X is approaching 5 from the left side this means from the left side in that case f of X is approaching war left side means you can take a test value as 4.9 so this becomes negative that is positive so it is negative infinity correct negative infinity means this side if I substitute X a value more than 5 let's say 4 plus 5 Plus that means we are approaching from the right side f of X will be what let's say five point one let's take a value which is five point one you can actually calculate so both are positive we get positive infinity so in this case the graph will be approaching positive infinity when you approach from the right side this is it okay now this function if you are to sketch you know it's a parabola kind of like this you know I'm just sketching correct without any details and reciprocal of this will be as you can clearly see degree of genocide horizontal asymptote is the x-axis correct so this side positive reciprocal is positive and on this side we have kind of like this so that gives you the graph it also explains the behavior near the asymptotes I hope that's absolutely clear and let's move on to Part C which is now we have a function H of x equals 2x divided by X plus 2 times X minus 3 now in this case we again have two vertical asymptotes right so so vertical asymptotes are at X plus 2 equals to 0 that is x equals 2 minus 2 the other one is X minus 3 equals to 0 that is at x equals to 3 correct we are interested in finding behavior near the vertical asymptotes for these two right if you want to sketch it we can just make a rough sketch here okay so we have two vertical asymptotes one at minus two the other one is at 3 okay so this is at eliminate -2 and 3 here okay so behavior near - - let's discuss so we will say what happens when X approaches minus 2 from the negative side so if - - negative we could think about a number which is minus 2 point 1 in that case what happens so place a number in your equation what happens is it negative or positive infinity so minus 2 will make numerator was negative which is - lemonade - here this term if I am writing value like maybe right age of minus 2 point 1 okay so this becomes negative and here this is also negative so three negatives means negative right so that means that the function is approaching when you approach from left side I have to write minus - minus here okay I know which are written sorry match to - yeah so it is approaching and giving us a negative that is the behavior is it okay now what happens when we are approaching from the right side that is this so let's take a value which is like minus 1.9 for example in that case numerator is negative as far as the denominator is concerned the first term will be positive and this term will be negative that gives us positive infinity right so we can write in this case when X approaches from the right side H of X approaches negative negative cancels plus infinity so that is this is okay so that is the behavior so let me summarize what we have learnt from here that is when X approaches negative 2 from the left side then H of X approaches negative infinity when X approaches negative 2 from the right side H of X approaches positive infinity good let's analyze behaved in year three so same approach what happens when X approaches negative I mean positive 3 from the left side in that case you can substitute a value they say 2.9 numerator is positive analyze first range so age 2.9 is it positive a negative numerator is positive this one is also positive however that is negative right that is negative so you get negative number here so we can write H of X approaches negative infinity and that is how you show now if X approaches positive 3 from the right side then what happens then all of them are positive right and the result is that H of X approaches positive infinity like this is it okay so that is kind of behavior you also know that the degree of denominator is higher so horizontal asymptote will be the x axis when you sketch this function knowing the behavior and the asymptotes it will be kind of like this right x axis is your horizontal asymptote and here X goes to 0 it goes through the origin and then like this so from the behavior you can actually sketch fairly accurate graph of a rational function you'll also note here that horizontal asymptote is x axis however we have an x intercept so that is also possible right so I was very important to understand that even if you have horizontal asymptotes as x axis there could be x intercept that is to say the color could cross the horizontal asymptote line the reason is very simple and that is horizontal asymptote is defined at large values of X not in between right that's the reason okay anyway our consideration is on vertical asymptotes and I hope with the help of these three examples we have understood how to find behavior of the function near the asymptotes thank you and all the best
1209	https://arxiv.org/pdf/2507.12014	Spectral extremal problems for degenerate graphs Jiadong Wu 1, Liying Kang 1,2† , Zhenyu Ni 3 1 Department of Mathematics, Shanghai University, Shanghai 200444, P.R. China 2 Newtouch Center for Mathematics of Shanghai University, Shanghai, China, 200444 3 Department of Mathematics, Hainan University, Haikou 570228, P.R. China Abstract A family of graphs is called degenerate if it contains at least one bipartite graph. In this paper, we investigate the spectral extremal problems for a degenerate family of graphs F. By employing covering and independent covering of graphs, we establish a spectral stability result for F. Using this stability result, we prove two general theorems that characterize spectral extremal graphs for a broad class of graph families F and imply several new and known results. Meanwhile, we establish the correlation between extremal graphs and spectral extremal graphs for F. Keywords: spectral radius; degenerate graph family, extremal graph AMS (2000) subject classification: 05C50; 05C35 1 Introduction For a family of graphs F, a graph is F-free if it does not contain any member of F as a subgraph. The Tur´ an type problem is to determine the maximum number of edges in an F-free graph of order n. The extremal value is called the Tur´ an number of F and denoted by ex( n, F). Let Ex( n, F) be the set of F-free graphs of order n with ex( n, F) edges. Let χ(F) = min F ∈F χ(F ). The famous Erd˝ os-Stone-Simonovits theorem states that ex( n, F) = 1 − 1 χ(F) − 1 n2 2 + o(n2). A family of graphs F is called degenerate if it contains at least one bipartite graph. If F is a degenerate graph family, then the above problem is called a degenerate Tur´ an type problem . For a degenerate graph family F, the Erd˝ os-Stone-Simonovits theorem only gives the bound ex( n, F) = o(n2). Over the past decades, there has been a wealth of research on the degenerate Tur´ an type problems. Recently, Alon and Frankl determined the Tur´ an number of {Kk, M s+1 }. Later, Gerbner , Zhu and Chen further studied the value of ex( n, {F, M s+1 }) for F being a non-bipartite graph or a bipartite graph, respectively. Additionally, Katona and Xiao proposed an intriguing conjecture on the Tur´ an number of {Pk, F }, where F is a connected non-bipartite graph. This conjecture was verified by Liu and Kang . For more related results, we refer readers to [16, 19, 28, 35, 36]. Research was partially supported by the National Nature Science Foundation of China (grant number 12331012) † Corresponding author. Email address: lykang@shu.edu.cn (L. Kang), 1753381890@qq.com (J. Wu), 1051466287@qq.com (Z. Ni) 1 arXiv:2507.12014v1 [math.CO] 16 Jul 2025 The spectral radius of G, denoted by λ(G), is the largest eigenvalue of its adjacency matrix A(G). In this paper we investigate spectral analogues of the Tur´ an type problem for graphs, which was proposed by Nikiforov . Let F be a family of graphs. The spectral Tur´ an type problem is to determine the maximum spectral radius among all F-free graphs of order n. Let spex( n, F) be the maximum value of spectral radius over all F-free graphs of order n. The graph attaining spex( n, F) is called a spectral extremal graph . The family of all spectral extremal graphs is denoted by Ex sp (n, F). There are many results on spectral Tur´ an type problems. In particular, many papers have investigated the problem of determining when Ex sp (n, F) ⊆ Ex( n, F) can be guaranteed. Wang, Kang and Xue confirmed a conjecture of Cioab˘ a, Desai and Tait and showed that Ex sp (n, F) ⊆ Ex( n, F) if ex( n, F) = e(Tr (n)) + O(1) and n is sufficiently large. Later, Fang, Tait, and Zhai improved this result for any graph family F with ex( n, F) < e (Tr (n)) + ⌊ n 2r ⌋. Recently, Byrne further extended this result. In this paper, we focus on spectral Tur´ an type problems for degenerate graph families. Over the past decades, numerous scholars have studied degenerate spectral Tur´ an problems. Wang, Hou, and Ma determined the exact value of spex( n, {Ms+1 , K r+1 }) for sufficiently large n. Jiang, Yuan, and Zhai characterized graphs in Ex sp (n, {Ms+1 , F }) when n is sufficiently large, F is a non-bipartite graph. For sufficiently large n, Cioab˘ a, Desai and Tait proved that every n-vertex graph G with λ(G) ≥ λ(Kk ∨In−k) contains all trees of order 2k+2 unless G = Kk ∨ In−k. Moreover, if λ(G) ≥ λ(Kk ∨ (K2 ∪ In−k−2)) , then G contains all trees of order 2k + 3 unless G = Kk ∨(K2 ∪In−k−2). Later, Cioab˘ a, Desai and Tait showed that Ex sp (n, C 2k+2 ) = {Kk ∨(K2 ∪ In−k−2)} for sufficiently large n and k ≥ 2. Wang, Feng and Lu proved that Ex sp (n, {C2k+2 , K r+1 }) = {Tr−1(k) ∨ In−k} for sufficiently large n and k ≥ r ≥ 2. For t ≥ 1 and k ≥ 4, Zhai, Yuan, and You characterized graphs in Ex sp (n, tK 1,k −1 ∪ Pk) when n is sufficiently large. Recently, Byrne, Desai and Tait established a more general result for a degenerate graph family F when Ex( n, F) contains specific subgraphs and n is sufficiently large. 1.1 Notation In this paper, we consider simple and undirected graphs. We first introduce some notations. Let G = ( V (G), E (G)) denote a simple graph with vertex set V (G) and edge set E(G). As customary, \|G\| and e(G) represent the number of vertices and edges of a graph G, respectively. For a vertex u ∈ V (G), let NG(u) denote the set of neighbors of u in G, and dG(u) = \|NG(u)\| denote its degree in G. Further, for U ⊆ V (G) and u ∈ V (G), define dU (u) = \|NU (u)\| = \|NG(u) ∩ U \|. Let ∆( G) denote the maximum degree of G. The r-partite Tur´ an graph of order n is denoted by Tr (n) and Ik represents an independent set of k vertices. A matching of size of size s + 1 is denoted by Ms+1 , and the matching number of G is denoted by ν(G). A path with k vertices is denoted by Pk, while a cycle with k vertices is denoted by Ck. The set of cycles of length at least k is denoted by C≥k. For disjoint vertex subsets U, W ⊆ V (G), let e(U, W ) denote the number of edges of G between U and W . The subgraph of G induced by U is denoted by G[U ]. Given two graphs H and F , H ∪ F denotes the disjoint union of H and F , and H ∨ F denotes the graph obtained from H ∪ F by adding all possible edges between V (H) and V (F ).A covering of a graph G is a set of vertices that meets all edges of the graph G. An independent covering of a graph G is an independent set that meets all edges of G. Clearly, a graph contains an independent covering if and only if it is a bipartite graph. Let β(G) denote the minimum size of a covering of G, and β′(G) the minimum size of an independent covering of G. If G is not a bipartite graph, we define β′(G) = + ∞.2Let F be a degenerate graph family. The covering number β(F) of F is defined by β(F) = min {β(F )\|F ∈ F} . The independent covering number β′(F) of F is defined by β′(F) = min {β′(F )\|F ∈ F} . If β′(F) = 1 , then F contains a star K1,t for some t > 0, which implies that spex( n, F) ≤ t. For the remainder of this work, we restrict our attention to graph families F satisfying β′(F) ≥ 2. We introduce a family of graphs. Let M(F) = {F [S]\| F ∈ F and S is a covering of F with \|S\| < β ′(F)}. Define H(F) =  Kβ′(F), if β′(F) = β(F), M(F), if β′(F) > β (F). Let F be a graph. We say F is a finite graph if \|F \| is a fixed constant. A family of graphs F is called finite degenerate if F is degenerate and for every F ∈ F , \|F \| is bounded by a fixed constant. A graph family F is said to be weak finite degenerate if there exists a bipartite graph F ∈ F with β′(F ) = β′(F) and \|F \| is bounded by a fixed constant. 1.2 Main results Denote ex G(n, F) as the maximum number of edges in an F-free graph on n vertices with G as a subgraph. We write Ex G(n, F) for the corresponding extremal graphs. Let G0(F) = {T ∨ In+1 −β′(F)\| T ∈ Ex( β′(F) − 1, H(F)) }, G(F) = {Ex H (n, F)\| H ∈ G 0(F)}. We have the following results for a finite degenerate graph family. Theorem 1.1. Suppose F is a finite degenerate family of graphs with β′(F) ≥ 2 and ex( n, F) = O(n). Let H = Kβ′(F)−1,n +1 −β′(F). For sufficiently large n, if ex H (n, F) < e (H) + n + 1 − β′(F)2 , then Ex sp (n, F) ⊆ G (F). Theorem 1.2. Suppose F is a finite degenerate family of graphs with β′(F) ≥ 2 and ex( n, F) = O(n). Let H = Kβ′(F)−1,n +1 −β′(F) and H1 = T ∨ In+1 −β′(F), where T ∈ Ex( β′(F) − 1, H(F)) . For sufficiently large n and some r ∈ [ 12 , 34 ), if e(H) + n + 1 − β′(F)2 ≤ ex H (n, F) ≤ e(H) + rn + O(1) and ex H (n, F) − ex H1 (n, F) ≤ 18( β′(F) − 1) n, then Ex sp (n, F) ⊆ G (F). 3Theorem 1.3. Let F be a finite degenerate family of graphs with β′(F) ≥ 2. Suppose eH is a graph on n + 1 − β′(F) vertices with e( eH) ≤ rn + O(1) , where r ∈ [0 , 34 ). For sufficiently large n, if {T ∨ eH\| T ∈ Ex( β′(F) − 1, H(F)) } ⊆ Ex( n, F), then Ex sp (n, F) ⊆ G (F) ⊆ Ex( n, F). The graph families investigated above are finite. Next, we extend our study to infinite graph families. Recently, Dou, Hu and Peng investigated the Tur´ an number of {C≥k, F }. We provide a spectral result for {C≥k, F },where F is an arbitrary finite graph. Theorem 1.4. Suppose k ≥ 3 is an integer and F is an arbitrary finite graph. Let F = {C≥k, F } and G be an arbitrary graph in Ex sp (n, F). If β′(F) = ⌊ k+1 2 ⌋ and n is sufficiently large, then (1) G ∈ { T ∨ In− k−12 \| T ∈ Ex( k−12 , H(F)) } when k is odd; (2) G ∈ { T ∨ In− k 2+1 \| T ∈ Ex( k 2 − 1, H(F)) } or G ∈ { T ∨ (K2 ∪ In− k 2−1 )\| T ∈ Ex( k 2 − 1, H(F)) } when k is even. The remainder of this paper is structured as follows. In Section 2, we present several applications of our principal findings. Section 3 introduces the preliminary lemmas. Section 4 is dedicated to the proofs of Theorems 1.1, 1.2, and 1.3. The proof of Theorem 1.4 is provided in Section 5. 2 Applications of main results In this section we give some applications of our main results. Byrne, Desai and Tait proved a general theorem for a family of graphs F under certain conditions. We derive the following result from Theorem 1.1, which is a version of Byrne, Desai and Tait’s result when F is a finite family of graphs. Theorem 2.1. Let F be a finite family of graphs. Suppose that ex( n, F) = O(n), Kk+1 ,∞ is not F-free, and Ex Kk,n −k (n, F) ∋ G for large enough n, where one of the following holds: (a) G = Kk,n −k (b) G = Kk ∨ In−k (c) G = Kk ∨ (K2 ∪ In−k−2). If (a), (b), or (c) holds, then for large enough n, Ex sp (n, F) = G. Proof. The assumption that Kk+1 ,∞ is not F-free and Ex Kk,n −k (n, F) ∋ G implies that β′(F) = k + 1 . If G = Kk,n −k ∈ Ex Kk,n −k (n, F), then G(F) = Kk,n −k. By Theorem 1.1, Ex sp (n, F) = G.If G = Kk ∨In−k ∈ Ex Kk,n −k (n, F), we claim β′(F) = β(F). Otherwise, if β(F) < β ′(F), then there exists a graph F ∈ F with β(F ) = β(F) ≤ β′(F)−1 = k. Thus, F ⊆ Kk ∨In−k, which contradicts the fact Kk ∨In−k is F-free. Then β′(F) = β(F) = k + 1 and it follows that Ex( k, H(F)) = Kk. This together with the fact 4Kk ∨ In−k ∈ Ex Kk,n −k (n, F) implies that G(F) = {Kk ∨ In−k}. According to Theorem 1.1, Ex sp (n, F) = G.Similarly, if G = Kk ∨ (K2 ∪ In−k−2) ∈ Ex Kk,n −k (n, F), we can deduce that Ex sp (n, F) = G. Remark: As special cases of Theorem 2.1, we recovered results from [4, 7, 9, 18, 20, 26]. Using Theorem 1.1, Theorem 1.2, we get the following result due to Byrne, Desai and Tait . Theorem 2.2 (Byrne, Desai and Tait ) . Let F be a family of graphs. Suppose that ex( n, F) = O(n), Kk+1 ,∞ is not F-free, and for n large enough Ex Kk,n −k (n, F) ∋ Kk ∨X, where e(X) ≤ rn +O(1) for some r ∈ [0 , 3/4) and F is finite. Then for n large enough, Ex sp (n, F) ⊆ Ex Kk ∨In−k (n, F). Proof. Since Kk ∨ X ∈ Ex Kk,n −k (n, F) and Kk+1 ,∞ is not F-free, similar as the proof of Theorem 2.1, one can easily verify that β′(F) = β(F) = k + 1 .Let H = Kk,n −k, H 1 = Kk ∨ In−k. The assumption Kk ∨ X ∈ Ex Kk,n −k (n, F) implies that ex H1 (n, F) = ex H (n, F) and ex H (n, F) = e(H) + e(X) + k 2 . If e(X) < n−k 2 − k 2 , then ex H (n, F) < e (H) + ⌊ n−k 2 ⌋. By Theorem 1.1, Ex sp (n, F) ⊆ Ex Kk ∨In−k (n, F). If n−k 2 − k 2 ≤ e(X) ≤ rn + O(1) for some r ∈ [ 12 , 34 ),then e(H) + n − k 2 ≤ ex H1 (n, F) = ex H (n, F) ≤ e(H) + rn + O(1) . By Theorem 1.2, Ex sp (n, F) ⊆ Ex Kk ∨In−k (n, F). Next we consider the finite degenerate family of graphs which contains a matching. Suppose F = {Ms+1 , F },where F is an arbitrary finite graph. For sufficiently large n, Zhu and Chen proved that Ex( n, F) = {T ∨ In−s\| T ∈ Ex( s, H(F)) } if β′(F) = s + 1 and ex( n, F) = ( β′(F ) − 1) n + O(1) if β′(F) ≤ s. We give a spectral version of their results. Theorem 2.3. Let F = {Ms+1 , F }, where F is an arbitrary finite graph. For sufficiently large n, the following statements hold. (1) If β′(F) = s + 1 , then Ex sp (n, F) ⊆ Ex( n, F). (2) If 2 ≤ β′(F) ≤ s, then Ex sp (n, F) ⊆ G (F). Proof. (1) If β′(F) = s + 1 , by combining Theorem 1.3 and Zhu and Chen’s result that Ex( n, F) = {T ∨ In−s\| T ∈ Ex( s, H(F)) }, we conclude that Ex sp (n, F) ⊆ Ex( n, F).(2) If 2 ≤ β′(F) ≤ s, then β′(F) = β′(F ). Let H = Kβ′(F )−1,n +1 −β′(F ). Assume G is an F-free graph on n vertices with H ⊆ G. Let A and B be the color classes of H with \|A\| = β′(F ) − 1 and \|B\| = n + 1 − β′(F ).For any vertex u ∈ B, it follows that dB (u) ≤ \| F \| − 1 since F ⊆ Kβ′(F ),\|F \|. On the other hand, ν(G[B]) ≤ s since G is Ms+1 -free. By the result of Chv´ atal and Hanson , e(G[B]) ≤ ν(G[B])(∆( G[B]) + 1) ≤ s\|F \|.Thus, e(G) ≤ e(H) + s\|F \| + β′(F )−12 . Let c = s\|F \| + β′(F )−12 . So ex H (n, F) ≤ e(H) + c. By Theorem 1.1, Ex sp (n, F) ⊆ G (F). This completes the proof. Denote G(n, r, s ) as the complete r-partite graph on n vertices with one part of order n − s and each other part of order j sr−1 k or l sr−1 m . Recently, Alon and Frankl established that Ex( n, {Ms+1 , K r+1 }) = G(n, r, s ) for sufficiently large n. Clearly, for r ≥ 2, β′({Ms+1 , K r+1 }) = s + 1 . Then, by Theorem 2.3, the following result of Wang, Hou and Ma follows immediately. Corollary 2.4 (Wang, Hou and Ma ) . Let r ≥ 2 and s ≥ 1 be two integers. For sufficiently large n, Ex sp (n, {Ms+1 , K r+1 }) = G(n, r, s ). 5A graph is called edge-critical if it contains an edge whose deletion decreases its chromatic number. Let F be an edge-critical graph with χ(F ) = r + 1 ≥ 3 and F = {Ms+1 , F }. For large enough n, Alon and Frankl proved that Ex( n, F) = G(n, r, s ). Clearly, β′(F) = s + 1 . By applying Theorem 2.3, we derive a spectral result for F. Corollary 2.5. Let F be an edge-critical graph with χ(F ) = r + 1 ≥ 3 and F = {Ms+1 , F }. If n is sufficiently large, then Ex sp (n, F) = G(n, r, s ). Let J be a finite degenerate family of graphs with β′(J ) = β(J ) = s ≥ 2 and ex( n, J ) = O(n). Suppose there exists a constant c ≥ 1 such that ex H (n, J ) ≤ e(H) + c, where H = Ks−1,n +1 −s. Let r ≥ 2 and F = {Kr+1 , J } . Obviously, β(F) = min {r, s } and β′(F) = β′(J ) = s ≥ 2. One can easily check that G(F) = {Tr−1(s − 1) ∨ In−s+1 } if r ≤ s, and G(F) = Ex T (n, F), where T = Ks−1 ∨ In−s+1 , if r ≥ s + 1 .Therefore, by Theorem 1.1, we obtain the following result. Corollary 2.6. Let r ≥ 2, J be a finite degenerate graph family with β′(J ) = β(J ) ≥ 2 and ex( n, J ) = O(n).Assume there exists a constant c ≥ 1 such that ex H (n, J ) ≤ e(H) + c, where H = Kβ′(J )−1,n +1 −β′(J ). Let F = {Kr+1 , J } . For sufficiently large n, the following statements hold. (1) If r ≤ β′(J ), then Ex sp (n, F) = Tr−1(β′(J ) − 1) ∨ In−β′(J )+1 .(2) If r ≥ β′(J ) + 1 , then Ex sp (n, F) ⊆ Ex T (n, F), where T = Kβ′(J )−1 ∨ In−β′(J )+1 . The famous Erd˝ os-S´ os conjecture states that every graph with average degree larger than t − 2 contains all trees of order t. By Corollary 2.6, we can deduce the following result concerning spectral Erd˝ os-S´ os theorem due to Wang, Feng and Lu . Theorem 2.7 (Wang, Feng and Lu ) . Let t and r be two integers such that t ≥ r ≥ 2. Let G be a Kr+1 -free graph with order n. For sufficiently large n, if λ(G) ≥ λ(Tr−1(t) ∨ In−t), then G contains all trees on 2t + 2 or 2t + 3 vertices unless G = Tr−1(t) ∨ In−t. Proof. Let F be the set of all finite graphs that contain all trees on 2t + 2 vertices. The famous Cayley formula shows that the number of trees of order 2t + 2 is at most (2 t + 2) 2t. Note that G contains all trees of order 2t + 2 if and only if G is not F-free. Next, we show that β′(F) = β(F) = t + 1 . Since Kt+1 ,2t+2 ∈ F , we have β′(F) ≤ t + 1 . We claim β(F) ≥ t + 1 . Otherwise, suppose there exists a graph F ∈ F such that β(F ) ≤ t.Then F ⊆ Kt ∨ I\|F \|. Note that Kt ∨ I\|F \| is P2t+2 -free, which contradicts the fact that F contains all trees on 2t + 2 vertices. Thus, t + 1 ≤ β(F) ≤ β′(F) ≤ t + 1 , which proves that β′(F) = β(F) = t + 1 .Let Tt be an arbitrary tree of order t. Cioab˘ a, Desai and Tait showed that for any n, 12 (t−2) n ≤ ex( n, T t) ≤ (t−2) n. In the same paper, they further established that for H = Kt,n −t, the inequality ex H (n, F) ≤ e(H)+ t 2 holds. Since ex( n, T 2t+2 ) ≤ 2tn , it follows that ex( n, F) = O(n). Then, given β′(F) > t ≥ r ≥ 2, part (1) of Corollary 2.6 yields Ex sp (n, {Kr+1 , F} ) = Tr−1(t) ∨ In−t. Recall that G is Kr+1 -free. Therefore, if λ(G) ≥ λ(Tr−1(t) ∨ In−t), then G contains all trees of order 2t + 2 unless G = Tr−1(t) ∨ In−t. The proof for trees with 2t + 3 vertices is analogous, so we omit the details for brevity. The proof is complete. Recently, Katona and Xiao determined ex( n, {Kr+1 , P k+1 }) for k ≥ 2r + 1 and sufficiently large n. We give a spectral version of the result. Set J = {Pk+1 } in Corollary 2.6, we obtain the following result. Corollary 2.8. Let r ≥ 2 and k ≥ 3 be two integers. For sufficiently large n, the following statements hold. 6(1) If r ≤ ⌊ k+1 2 ⌋, then Ex sp (n, {Kr+1 , P k+1 }) = Tr−1(⌊ k−12 ⌋) ∨ In−⌊ k−12 ⌋.(2) If r ≥ ⌊ k+1 2 ⌋ + 1 and k is odd, then Ex sp (n, {Kr+1 , P k+1 }) = K k−12 ∨ In− k−12 .If r ≥ ⌊ k+1 2 ⌋ + 1 and k is even, then Ex sp (n, {Kr+1 , P k+1 }) = K k 2−1 ∨ K2 ∪ In− k 2−1 . A linear forest is a graph whose connected components are paths. Denote Lk as the family of all linear forests with size k. Note that Pk+1 ∈ L k. One can easily check that all graphs in EX sp (n, {Kr+1 , P k+1 }) are also {Kr+1 , Lk}-free. Thus, Ex sp (n, {Kr+1 , Lk}) = Ex sp (n, {Kr+1 , P k+1 }). Then we can get the result of Zhai and Yuan by Corollary 2.8. Corollary 2.9 (Zhai and Yuan ) . Let r ≥ 2 and k ≥ 3 be two integers. For sufficiently large n, the following statements hold. (1) If r ≤ ⌊ k+1 2 ⌋, then Ex sp (n, {Kr+1 , Lk}) = Tr−1(⌊ k−12 ⌋) ∨ In−⌊ k−12 ⌋.(2) If r ≥ ⌊ k+1 2 ⌋ + 1 and k is odd, then Ex sp (n, {Kr+1 , Lk}) = K k−12 ∨ In− k−12 .If r ≥ ⌊ k+1 2 ⌋ + 1 and k is even, then Ex sp (n, {Kr+1 , Lk}) = K k 2−1 ∨ K2 ∪ In− k 2−1 . The families of graphs we have investigated above are all finite. We now turn to studying some infinite families of graphs. Recently, Dou, Ning, and Peng determined the Tur´ an number of {C≥k, K r+1 }. Let F = {C≥k, K r+1 }. One can easily verify that β(F) = min {r, ⌊ k+1 2 ⌋} and β′(F) = ⌊ k+1 2 ⌋. If r ≤ ⌊ k+1 2 ⌋, then H(F) = {Kr }. If r ≥ ⌊ k+1 2 ⌋ + 1 , then β(F) = β′(F) and H(F) = n K⌊ k+1 2 ⌋ o . By Theorem 1.4, we derive a spectral result for {C≥k, K r+1 }. Theorem 2.10. Let r ≥ 2 and k ≥ 3 be two integers. For sufficiently large n, the following statements hold. (1) If r ≤ ⌊ k+1 2 ⌋, then Ex sp (n, {C≥k, K r+1 }) = Tr−1(⌊ k−12 ⌋) ∨ In−⌊ k−12 ⌋.(2) If r ≥ ⌊ k+1 2 ⌋ + 1 and k is odd, then Ex sp (n, {C≥k, K r+1 }) = K k−12 ∨ In− k−12 .If r ≥ ⌊ k+1 2 ⌋ + 1 and k is even, then Ex sp (n, {C≥k, K r+1 }) = K k 2−1 ∨ K2 ∪ In− k 2−1 . Recently, Zhao and Lu studied the Tur´ an number of {C≥k, M s+1 } when n is large enough and s ≥⌊ k−12 ⌋. Let F = {C≥k, M s+1 }. If s ≥ ⌊ k−12 ⌋, then β(F) = β′(F) = ⌊ k+1 2 ⌋. Thus H(F) = n K⌊ k+1 2 ⌋ o and Ex( ⌊ k−12 ⌋, H(F)) = K⌊ k−12 ⌋. By Theorem 1.4, we get a spectral result for {C≥k, M s+1 }. Corollary 2.11. Let k ≥ 3 and s ≥ ⌊ k−12 ⌋ be two integers. For sufficiently large n, the following statements hold. (1) If k is odd, then Ex sp (n, {C≥k, M s+1 }) = K k−12 ∨ In− k−12 .(2) If k is even and s = k 2 − 1, then Ex sp (n, {C≥k, M s+1 }) = K k 2−1 ∨ In− k 2+1 .if k is even and s ≥ k 2 , then Ex sp (n, {C≥k, M s+1 }) = K k 2−1 ∨ K2 ∪ In− k 2−1 . 3 Preliminaries We introduce the following lemmas that will be employed to prove our results. 7Lemma 3.1 (Wu, Xiao and Hong ) . Let G be a connected graph and x be the Perron vector of G. Assume that u, v are two vertices of G with xu ≥ xv and {vi \| 1 ≤ i ≤ s} ⊆ NG(v) \ (NG(u) ∪ { u}). Let G′ = G − { vv i \| 1 ≤ i ≤ s} + {uv i \| 1 ≤ i ≤ s}. Then, λ(G′) > λ (G). Lemma 3.2. Let F be a family of graphs. For any graph G ∈ { T ∨ In+1 −β′(F)\| T ∈ Ex( β′(F) − 1, H(F)) }, G is F-free. Proof. Assume G = T ∨ In+1 −β′(F). It is sufficient to prove that G is F -free for any graph F ∈ F . If β(F ) ≥ β′(F), we claim that Kβ′(F)−1 ∨ In+1 −β′(F) is F -free. Otherwise, F ⊆ Kβ′(F)−1 ∨ In+1 −β′(F), then β(F ) ≤ β′(F) − 1 ≤ β(F ) − 1, a contradiction. Thus, Kβ′(F)−1 ∨ In+1 −β′(F) is F -free. This together with the fact G ⊆ Kβ′(F)−1 ∨ In+1 −β′(F) implies that G is F -free. If β(F ) < β ′(F), then β(F) < β ′(F), so H(F) = M(F). If F ⊆ G, then there exists a covering S of F with \|S\| ≤ β′(F) − 1 such that F [S] is a subgraph of T . This contradicts the fact that all graphs in Ex( β′(F) − 1, M(F)) are F [S]-free. Consequently, G is F -free. We establish the following spectral stability result for a weakly finite degenerate graph family. Theorem 3.3. Let C ≥ 1 be a real number, and F be a weak finite degenerate graph family with β′(F) ≥ 2 and ex( n, F) ≤ Cn . Suppose G is an F-free graph with order n, and x is a non-negative eigenvector of A(G) corresponding to λ(G), with maximum entry equal to 1. For sufficiently large n and a sufficiently small ε satisfying 0 < ε < 132 Cβ ′(F) 40 , the following statements hold: (1) If λ(G) ≥ p(β′(F) − 1)( n + 1 − β′(F)) − √εn, then there exists a t > (1 − 2β′(F)ε 110 )n such that Kβ′(F)−1,t ⊆ G.(2) Let L′ be the color class of Kβ′(F)−1,t with size β′(F) − 1. Then xv ≥ 1 − ε 110 for any v ∈ L′ and xv ≤ ε 18 for any v / ∈ L′. Proof of Theorem 3.3. (1). Let F0 be a bipartite graph in F with β′(F0) = β′(F) and \|F0\| ≤ l. Set λ = λ(G), β′ = β′(F). For any 0 < η ≤ 1, define Lη = {u ∈ V (G)\| xu ≥ η}. For a vertex u ∈ V (G) and a positive integer i, denote Ni(u) as the set of vertices at distance i from u in G. Lemma 3.4. There exists a constant D(η, ε ) depending on η and ε such that \|Lη \| ≤ D(η, ε )√n. Proof. For any u ∈ Lη , we have (p(β′ − 1)( n + 1 − β′) − √εn )η ≤ λx u = X v∈NG(u) xv ≤ dG(u). Summing over all vertices u ∈ Lη , we obtain \|Lη \|(p(β′ − 1)( n + 1 − β′) − √εn )η ≤ X u∈Lη dG(u) ≤ X u∈V(G) dG(u) ≤ 2Cn, where the last inequality holds since G is F-free. By algebraic manipulation, it follows that \|Lη \| ≤ 2Cη q β′ − 1 − ε 14 √n = D(η, ε )√n. Let u be an arbitrary vertex in V (G). Denote Lηi (u) = Ni(u) ∩ Lη and Lηi (u) = Ni(u)\Lη .8Lemma 3.5. For any vertex u ∈ V (G), we have (β′ − 1 − ε 14 )nx u ≤ dG(u)xu + X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw + 3 Cηn. Proof. For any vertex u ∈ V (G), we have (β′ − 1 − ε 14 )nx u ≤ λ2xu = dG(u)xu + X v∈N1(u) X w∈N1(v){ u} xw ≤ dG(u)xu + X v∈N1(u) X w∈Lη 1(u)∪Lη 2(u) xw + X v∈N1(u) X w∈Lη 1(u)∪Lη 2(u) xw, (1) where the last inequality holds since N1(v){ u} ⊆ N1(u) ∪ N2(u) = Lη 1 (u) ∪ Lη 2 (u) ∪ Lη 1 (u) ∪ Lη 2 (u).Since N1(u) = Lη 1 (u) ∪ Lη 1 (u), it follows that X v∈N1(u) X w∈Lη 1(u)∪Lη 2(u) xw = X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw + X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw ≤ e(Lη 1 (u), L η 1 (u) ∪ Lη 2 (u)) + 2 e(Lη 1 (u)) + e(Lη 1 (u), L η 2 (u)) η ≤ 2e(G)η ≤ 2Cηn. (2) Similarly, X v∈N1(u) X w∈Lη 1(u)∪Lη 2(u) xw = X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw + X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw ≤ 2e(Lη 1 (u)) + e(Lη 1 (u), L η 2 (u)) + X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw. (3) Note that Lη 1 (u) ∪ Lη 2 (u) ⊆ Lη and G is F-free. Combining this with Lemma 3.4, we get 2e(Lη 1 (u)) + e(Lη 1 (u), L η 2 (u)) ≤ 2C\|Lη 1 (u)\| + C(\|Lη 1 (u)\| + \|Lη 2 (u)\|) ≤ 3C\|Lη \| ≤ 3CD (η, ε )√n. (4) By (1)-(4) and the assumption that n is large enough, we obtain (β′ − 1 − ε 14 )nx u ≤ dG(u)xu + X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw + 3 Cηn. Lemma 3.6. Let m(ε) = 4C2 (β′−1−2ε14)ε12 . Then \|Lε 12 \| ≤ m(ε). Proof. According to Lemma 3.4, for any u ∈ V (G), we deduced that X v∈Lη 1(u) X w∈Lη 1(u)∪Lη 2(u) xw ≤ e(Lη 1 (u), L η 1 (u) ∪ Lη 2 (u)) ≤ C(dG(u) + \|Lη \|) ≤ Cd G(u) + CD (η, ε )√n (5) 9where the second inequality holds as G is F-free and Lη 1 (u) ⊆ N1(u). Then Lemma 3.5 together with (5) yields (β′ − 1 − ε 14 )nx u ≤ (C + xu)dG(u) + 4 Cηn. (6) Now, set η = ε. Then by (6), for each vertex v ∈ Lε 12 , it follows that dG(v) ≥ (β′ − 1 − ε 14 )nx v − 4Cεn C + xv ≥ (β′ − 1 − 2ε 14 )ε 12 n 2C , where the last inequality holds since ε is sufficiently small and C ≥ 1. Summing this inequality over all vertices v ∈ Lε 12 , we obtain \|Lε 12 \| (β′ − 1 − 2ε 14 )ε 12 n 2C ≤ X v∈Lε 12 dG(v) ≤ 2e(G) ≤ 2Cn. Hence, \|Lε 12 \| ≤ 4C2 (β′−1−2ε14)ε12 = m(ε). Let u be an arbitrary vertex in V (G). For simplicity, we write L′ = Lε 18 , L = Lε 12 , Li(u) = Ni(u) ∩ Lε 12 , Li(u) = Ni(u)\Lε 12 . Now we give a lower bound for degrees of vertices in L′. Lemma 3.7. dG(u) > (xu − 2ε 14 )n for each vertex u ∈ L′. Proof. Let u be a vertex in L′, S be a subset of L1(u) such that each vertex in S has at least β′ − 1 neighbors in L1(u) ∪ L2(u). We first prove that \|S\| ≤ √n. If \|L1(u) ∪ L2(u)\| ≤ β′ − 2, then S = ∅, and the inequality holds trivially. So we consider the case \|L1(u) ∪ L2(u)\| ≥ β′ − 1. Assume for contradiction that \|S\| > √n. Note that each vertex in S must choose at least β′ − 1 neighbors from L1(u) ∪ L2(u). The number of distinct ways to choose β′ − 1 vertices from L1(u) ∪ L2(u) is \|L1(u)∪L2(u)\| β′−1 . By the pigeonhole principle, there exists a fixed subset T ⊆ L1(u) ∪ L2(u) with \|T \| = β′ − 1 such that at least \|S\|/\|L1(u)∪L2(u)\| β′−1 > √n/ \|L\| β′−1 > l vertices in S have all their β′ − 1 neighbors in T . Recall that F0 is a bipartite graph in F with \|F0\| ≤ l and β′(F0) = β′.Since u / ∈ L1(u) ∪ L2(u) and S ⊆ L1(u) ⊆ N1(u), the set {u} ∪ T and the l common neighbors in S form a complete bipartite subgraph Kβ′,l in G. Then G contains F0 as a subgraph , contradicting the fact that G is F-free. Hence, \|S\| ≤ √n. It follows that X v∈L1(u) X w∈L1(u)∪L2(u) xw ≤ e(L1(u), L 1(u) ∪ L2(u)) = e(L1(u)\S, L 1(u) ∪ L2(u)) + e(S, L 1(u) ∪ L2(u)) ≤ (β′ − 2) \|L1(u)\S\| + √n\|L\|≤ (β′ − 2) dG(u) + m(ε)√n, (7) where the second to last inequality holds as \|S\| ≤ √n and the last inequality holds as L1(u) ⊆ NG(u) and \|L\| ≤ m(ε). Now, substituting η = ε 12 in Lemma 3.5 and combining with (7), for large enough n, we derive (β′ − 1 − ε 14 )xun ≤ dG(u)xu + ( β′ − 2) dG(u) + m(ε)√n + 3 Cε 12 n< (β′ − 1) dG(u) + 4 Cε 12 n. 10 Then dG(u) > x un − ε 14 β′ − 1 xun − 4Cε 12 β′ − 1 n > (xu − 2ε 14 )n, where the last inequality holds as β′ ≥ 2 and ε is sufficiently small. Lemma 3.8. (β′ − 1 − 2ε 14 )n ≤ e(L1(u∗), {u∗} ∪ L1(u∗) ∪ L2(u∗)) ≤ (β′ − 1 + 2 ε 14 )n. Proof. We first prove the lower bound. By Lemma 3.6, we obtain dG(u∗) = dL1(u∗)(u∗) + dL1(u∗)(u∗) ≤ dL1(u∗)(u∗) + \|L1(u∗)\| ≤ dL1(u∗)(u∗) + m(ε). (8) Recall that xu∗ = 1 . Setting u = u∗, η = ε 12 in Lemma 3.5, it follows that (β′ − 1 − ε 14 )n ≤ dG(u∗) + e(L1(u∗), L 1(u∗) ∪ L2(u∗)) + 3 Cε 12 n ≤ e(L1(u∗), {u∗} ∪ L1(u∗) ∪ L2(u∗)) + 4 Cε 12 n, where the last inequality holds as (8) and n is sufficiently large. This together with the fact that ε is sufficiently small yields that (β′ − 1 − 2ε 14 )n ≤ e(L1(u∗), {u∗} ∪ L1(u∗) ∪ L2(u∗)) , as desired. Next, we prove the upper bound by contradiction. If e(L1(u∗), {u∗} ∪ L1(u∗) ∪ L2(u∗)) > (β′ − 1 + 2 ε 14 )n,then e(L1(u∗), L 1(u∗) ∪ L2(u∗)) > (β′ − 2 + 2 ε 14 )n. (9) Let B be a subset of L1(u∗) such that each vertex in B has at least β′ − 1 neighbors in L1(u∗) ∪ L2(u∗). Using the similar discussion as in the proof of Lemma 3.7, we can prove that \|B\| ≤ √n. Combining this with Lemma 3.6, we get e(L1(u∗), L 1(u∗) ∪ L2(u∗)) ≤ \|L\|√n + ( β′ − 2) \|L1(u∗)\B\|≤ m(ε)√n + ( β′ − 2) n < (β′ − 2 + 2 ε 14 )n, where the last inequality holds as n is sufficiently large, which contradicts (9). Lemma 3.9. For all u ∈ L′, xu ≥ 1 − 1(β′)3 and dG(u) > 1 − 1(β′)3 − 2ε 14 n. Proof. We prove by contradiction. Suppose to the contrary that there exists a vertex u0 ∈ L′ such that xu0 < 1 − 1(β′)3 . Recall that L′ = Lε 18 . By Lemma 3.7, dG(u0) > (ε 18 − 2ε 14 )n and dG(u∗) > (1 − 2ε 14 )n. Thus, dN1(u∗)(u0) ≥ (ε 18 − 4ε 14 )n. (10) Then substituting u = u∗ and η = ε 12 into Lemma 3.5, we deduce (β′ − 1 − ε 14 )n ≤ dG(u∗)xu∗ + X v∈L1(u∗) X w∈L1(u∗)∪L2(u∗) xw + 3 Cε 12 n ≤ dG(u∗) + X v∈L1(u∗) X w∈L1(u∗)∪L2(u∗){ u0} xw + dN1(u∗)(u0)xu0 + 3 Cε 12 n ≤ dG(u∗) + e(L1(u∗), L 1(u∗) ∪ L2(u∗){ u0}) + dN1(u∗)(u0)xu0 + 3 Cε 12 n< dG(u∗) + e(L1(u∗), L 1(u∗) ∪ L2(u∗)) − dL1(u∗)(u0)+dN1(u∗)(u0) 1 − 1(β′)3 3 Cε 12 n. (11) 11 By Lemma 3.8 and (8), we obtain dG(u∗) + e(L1(u∗), L 1(u∗) ∪ L2(u∗)) ≤ e(L1(u∗), {u∗} ∪ L1(u∗) ∪ L2(u∗)) + m(ε) ≤ (β′ − 1 + 2 ε 14 )n + m(ε). (12) Note that dN1(u∗)(u0) − dL1(u∗)(u0) = dL1(u∗)(u0) ≤ m(ε). Combining (10)-(12), we get (β′ − 1 − ε 14 )n < (β′ − 1 + 2 ε 14 )n + 2 m(ε) + 3 Cε 12 n − dN1(u∗)(u0)(β′)3 < (β′ − 1 + 2 ε 14 )n + 4 Cε 12 n − (ε 18 − 4ε 14 )n (β′)3 < β′ − 1 − ε 18 2( β′)3 n < (β′ − 1 − ε 14 )n, a contradiction. So xu ≥ 1 − 1(β′)3 for any u ∈ L′. By Lemma 3.7, dG(u) > (1 − 1(β′)3 − 2ε 14 )n for any u ∈ L′.Let L′ = V (G)\L′. Notice that L′ ⊆ L. Thus, \|L′\| ≤ \| L\| ≤ m(ε) by Lemma 3.6. We will prove that \|L′\| = β′ − 1. To show it, we first give a lower bound on e(L′, L ′). Since L′ ∪ L′ = V (G), we get X v∈L′ dNG(u∗)(v)xv = X v∈L′ dNL′ (u∗)(v)xv + X v∈L′ dNL′ (u∗)(v)xv ≤ 2e(L′) + X v∈L′ dNL′ (u∗)(v)xv ≤ (m(ε)) 2 + X v∈L′ dNL′ (u∗)(v)xv . (13) On the other hand, since e(G) ≤ Cn , we obtain X v∈L′ dNG(u∗)(v)xv ≤ X v∈L′ dNG(u∗)(v)ε 18 ≤ 2e(G)ε 18 ≤ 2Cε 18 n. (14) Note that λ2(G)xu∗ = P v∈V(G) dNG(u∗)(v)xv . Summing up (13) and (14) gives that (β′ − 1 − ε 14 )n ≤ λ2(G)xu∗ = X v∈V(G) dNG(u∗)(v)xv ≤ 2Cε 18 n + ( m(ε)) 2 + X v∈L′ dNL′ (u∗)(v)xv < 3Cε 18 n + X v∈L′ dNL′ (u∗)(v)xv . Then, it follows that e(L′, L ′) ≥ X v∈L′ dNL′ (u∗)(v)xv > (β′ − 1 − 4Cε 18 )n. (15) Lemma 3.10. \|L′\| = β′ − 1. 12 Proof. If \|L′\| ≤ β′ − 2, then e(L′, L ′) ≤ (β′ − 2)( n + 2 − β′) < (β′ − 1 − 4Cε 18 )n, which contradicts (15). So \|L′\| ≥ β′ − 1. On the other hand, if \|L′\| ≥ β′, then there exists a vertex set W ⊆ L′ with \|W \| = β′. By Lemma 3.9, for each vertex u ∈ W , dG(u) > (1 − 1(β′)3 − 2ε 14 )n. Thus, \| ∩ u∈W dG(u)\| ≥ (1 − 1(β′)2 − 2β′ε 14 )n > l , which implies that F0 is a subgraph of G. This contradicts the fact that G is F-free. Hence, \|L′\| = β′ − 1. Lemma 3.11. xu ≥ 1 − ε 110 for each u ∈ L′. Proof. Suppose to the contrary that there exists a vertex u1 ∈ L′ such that xu1 < 1 − ε 110 . By Lemmas 3.9 and 3.10, we deduce that dL′ (u1) > (1 − 1(β′)3 − 2ε 14 )n − β′ + 1 > (1 − 2(β′)3 )n. Then dNL′ (u∗)(u1)xu1 < d L′ (u1)(1 − ε 110 ) < d L′ (u1) − (1 − 2(β′)3 )ε 110 n. (16) On the other hand, X v∈L′{ u1} dNL′ (u∗)(v)xv ≤ e(L′{ u1}, L ′) ≤ (β′ − 1)( n − β′ + 1) − dL′ (u1). (17) Summing up (16) and (17) yields X v∈L′ dNL′ (u∗)(v)xv < (β′ − 1)( n − β′ + 1) − (1 − 2(β′)3 )ε 110 n< (β′ − 1 − ε 110 2 )n < (β′ − 1 − 4Cε 18 )n, which contradicts (15). Hence, xu ≥ 1 − ε 110 for each u ∈ L′, as claimed. By Lemmas 3.7 and 3.11, it follows that dG(u) > (1 − ε 110 − 2ε 14 )n > (1 − 2ε 110 )n for each u ∈ L′. Set R = ∩u∈L′ NL′ (u). Then \|R\| > (n − β′ + 1) − 2( β′ − 1) ε 110 n > (1 − 2β′ε 110 )n. Take t = \|R\|. Then K\|L′\|,\|R\| = Kβ′−1,t ⊆ G.(2). By Lemma 3.11, xv ≥ 1 − ε 110 for any v ∈ L′. Recall that L′ = Lε 18 . So xv ≤ ε 18 for any v / ∈ L′. The proof is finished. Lemma 3.12. Let C ≥ 1 be a real number, F be a finite degenerate family of graphs with β′(F) ≥ 2, and ex( n, F) ≤ Cn . If G is a graph in Ex sp (n, F), then Kβ′(F)−1,n +1 −β′(F) is a subgraph of G. Let x be a Perron eigenvector of A(G) corresponding to λ(G), with maximum entry equal to 1. Let L′ be the color class of Kβ′(F)−1,n +1 −β′(F) with size β′(F) − 1. Then xv ≥ 1 − ε 110 for any v ∈ L′ and xv ≤ ε 18 for any v / ∈ L′. Proof. Since Kβ′(F)−1,n +1 −β′(F) is F-free, λ(G) ≥ p(β′(F) − 1)( n + 1 − β′(F)) . Let F0 ∈ F be a bipartite graph with β′(F0) = β′(F) and \|F0\| ≤ l. Define 0 < ε < 12lkCβ ′(F) 120 be a sufficiently small constant. By Theorem 3.3, Kβ′(F)−1,t ⊆ G, where t > (1 − 2β′(F)ε 110 )n, and xv ≥ 1 − ε 110 for any v ∈ L′ and xv ≤ ε 18 for any v / ∈ L′. Set R = ∩v∈L′ NG(v), W = V (G) \ (R ∪ L′). Clearly, \|W \| ≤ 2β′(F)ε 110 n. For brevity, we write β′ = β′(F) and β = β(F).13 We claim that G is connected. Suppose to the contrary that G is not connected. Without loss of generality, we assume G = G1 ∪ G2 ∪ · · · ∪ Gp and Kβ′−1,t ⊆ G1. Then G2 ∪ · · · ∪ Gp ⊆ G[W ]. Moreover, λ(G1) ≥ λ(Kβ′−1,t ) > q (β′ − 1)(1 − 2β′ε 110 )n. Recall that \|W \| ≤ 2β′ε 110 n and G[W ] is F-free. Then λ2(G[W ]) ≤ 2e(G[W ]) ≤ 2C\|W \| ≤ 4Cβ ′ε 110 n. Since ε is sufficiently small, we have λ(Gi) ≤ λ(G[W ]) ≤ q 4Cβ ′ε 110 n < q (β′ − 1)(1 − 2β′ε 110 )n < λ (G1) for each 2 ≤ i ≤ q, implying λ(G) = λ(G1). To derive a contradiction, we perform the following edge-switching operation. Arbitrarily choose a vertex u0 ∈ V (G2). Delete all edges incident to u0 in G and add all edges between u0 and L′. Denote the resulting graph as G′. We claim G′ is F-free. Suppose, for contradiction, that G′ contains a graph F ′ ∈ F . Since \|R\| ≥ (1 − 2β′ε 110 )n and F is finite, there exists a vertex u1 ∈ R \ V (F ′). Replacing u0 with u1 in F ′ yields a copy of F ′ in G, which is a contradiction. Next we show that W = ∅. Suppose to the contrary that W̸ = ∅. Recall that β′(F0) = β′, \|F0\| ≤ l and F0 ∈ F . For any vertex u ∈ W , we must have dR(u) ≤ l − 1. Otherwise, F0 ⊆ Kβ′,l ⊆ G, a contradiction. Since G[W ] is F-free, e(G[W ]) ≤ C\|W \|. Then there exists a vertex u0 ∈ W with dW (u0) ≤ 2C, which implies that dW ∪R(u0) ≤ 2C + l − 1. Then we construct a new graph G′ with V (G′) = V (G) and E(G′) = E(G \ { u0}) ∪ { u0u\|u ∈ L′}. We claim that G′ is F-free. Otherwise, G′ contains a graph F ′ in F. Since F ′ is finite, there exists a vertex u1 ∈ R \ V (F ′). Replacing u0 with u1 in F ′ yields a copy of F ′ in G, a contradiction. So G′ is F-free. Recall that xv ≤ ε 18 for any vertex v ∈ V (G) \ L′ and xu ≥ 1 − ε 110 for any vertex u ∈ L′.Therefore, xT λ(G′) − λ(G)x ≥ xT A(G′)x − xT A(G)x ≥ 2(1 − ε 110 )xu0 − 2(2 C + l − 1) ε 18 xu0 0, where the second inequality holds as dL′ (u0) ≤ β′ − 2 in G and the last inequality holds as ε is sufficiently small, contradicting the maximality of λ(G). So W = ∅. Thus Kβ′(F)−1,n +1 −β′(F) ⊆ G.We present the following lemma. The core idea underlying Lemma 3.13 is inspired by the work of Byrne, Desai, and Tait . Lemma 3.13. Let F be a finite family of graphs and H = Kβ′(F)−1,n −β′(F)+1 . For sufficiently large n, if ex H (n, F) ≤ e(H) + rn + O(1) for some r ∈ [0 , 34 ), then ex H (n, F) = e(H) + rn + O(1) , where r ∈ { 0, 12 , 23 }. Proof of Lemma 3.13. Let A, B be the color classes of H with \|A\| = β′(F) − 1 and \|B\| = n + 1 − β′(F).Let G be a graph in Ex H (n, F). We focus on the components of G[B]. Assume B′ is the largest component of G[B]. We first claim \|B′\| = O(1) . Suppose to the contrary that \|B′\| is infinite. Let H′ = Kβ′(F)−1,\|B′\| and n′ = β′(F) − 1 + \|B′\|. By the assumption of the lemma, we have ex H′ (n′, F) < e (H′) + 34 n′ + O(1) . Since H′ ⊆ G[A ∪ V (B′)] , e(G[A ∪ V (B′)]) ≥ e(H′) + \|B′\| − 1 = e(H′) + n′ − O(1) > ex H′ (n′, F), 14 contradicting the fact that G is F-free. Next, let B′′ be an arbitrary component in G[B]. If \|B′′ \| ≥ 4, then e(B′′ ) ≥ \| B′′ \| − 1 ≥ 34 \|B′′ \|, which implies that the number of components of order at least 4 is O(1) . Similarly, the number of components isomorphic to K3 in G[B] is O(1) . Define p = max {i\| the number of components in G[B] with order i is infinite }. Then p = 1 , 2, or 3. If p = 1 , then e(G[B]) = O(1) , so ex H (n, F) = e(H) + O(1) .If p = 2 , then e(G[B]) is infinite and the number of components in G[B] with order at least 3 is O(1) . Moreover, we show that the number of isolated vertices in G[B] is at most 1. Suppose to the contrary that there exist two isolated vertices v1, v2 in G[B], then we add a new edge e1 between v1 and v2. Denote the new graph as G′. We claim G′ is F-free. Otherwise, if there is a graph F1 ∈ F such that F1 ⊆ G′, then e1 ∈ E(F1). Combining with the fact that F1 is finite and e(G[B]) is infinite, we always can find an edge e2 ∈ E(G[B]) \ E(F1). Thus (F1 \ e1) ∪ e2 is a copy of F1 in G, which is a contradiction. So G′ is F-free. Clearly, e(G′) > e (G). This contradicts the fact G ∈ Ex H (n, F). Therefore ex H (n, F) = e(H) + 12 n + O(1) .If p = 3 , then the number of P3 components in G[B] is infinite. Similarly, we can prove that the number of isolated vertices in G[B] is at most 1. Next we show that the number of P2 components in G[B] is at most 2. If there exist 3P2 components in G[B], then we delete one edge and add two new edges to get 2P3. Using the similar argument above, we can prove that the resulting graph is still F-free. Moreover, the number of edges increasing, which is a contradiction. Thus, ex H (n, F) = e(H) + 23 n + O(1) . This completes the proof. Let t = j n+1 −β′(F)2 k , s = j n+1 −β′(F)3 k , Q1 = Mt ∪ In+1 −β′(F)−2t and Q2 = sP 3 ∪ In+1 −β′(F)−3s. Lemma 3.14. Suppose F is a finite family of graphs. Let H = Kβ′(F)−1,n −β′(F)+1 and H1 = T ∨ In−β′(F)+1 ,where T ∈ Ex( β′(F) − 1, H(F)) . Suppose G = T ∨ Q ∈ Ex H1 (n, F).(1) If ex H (n, F) = e(H) + 12 n + O(1) and e(Q) ≥ 12 n − O(1) , then Q can be obtained from Q1 by adding and deleting O(1) edges; (2) If ex H (n, F) = e(H) + 23 n + O(1) and e(Q) ≥ 23 n − O(1) , then Q can be obtained from Q2 by adding and deleting O(1) edges. Proof of Lemma 3.14. Suppose G = T ∨ Q ∈ Ex H1 (n, F). If ex H (n, F) = e(H) + 12 n + O(1) , then the size of any component of Q is O(1) and the number of components in Q with order at least 3 is O(1) . This together with the assumption e(Q) ≥ 12 n − O(1) imply that Q can be obtained from Q1 by adding and deleting O(1) edges. If ex H (n, F) = e(H) + 23 n + O(1) and e(Q) ≥ 23 n − O(1) , using the similar discussion above we can show Q can be obtained from Q2 by adding and deleting O(1) edges. 4 Proofs of Theorems 1.1, 1.2 and 1.3 4.1 Proof of Theorem 1.1 We give a lemma which will be used in the proof of Theorem 1.1. Lemma 4.1. Let F be a finite graph family and H = Kt,n −t be an F-free graph, where 1 ≤ t ≤ n. For sufficiently large n, the following statements are equivalent: 15 (a) ex H (n, F) < e (H) + ⌊ n−t 2 ⌋;(b) there exist integers p, q such that It ∨ Mp and It ∨ Sq+1 are not F-free; (c) there exist a constant c such that ex H (n, F) ≤ e(H) + c. Proof. (a) ⇒ (b). Let G1 = It ∨ M⌊ n−t 2⌋ ∪ In−t−2⌊ n−t 2⌋ . Observe that e(G1) = e(H) + ⌊ n−t 2 ⌋ > ex H (n, F) and H ⊆ G1. Thus, G1 is not F-free, so there exists a graph F1 ∈ F such that F1 ⊆ G1. Since F1 is finite, its inclusion in G1 implies that there exists an integer p > 0 such that F1 ⊆ It ∨ Mp. Next, let G2 = It ∨ Sn−t.Clearly, e(G2) = e(H) + n − t − 1 > ex H (n, F). Combining this with H ⊆ G2 and the finiteness of F, we deduce that there exist an integer q > 0 and a graph F2 ⊆ F such that F2 ⊆ It ∨ Sq+1 . (b) ⇒ (c). Suppose G is an arbitrary graph in Ex H (n, F). Let A, B be two color classes of H with \|A\| = t and \|B\| = n − t. Then e(G[A]) ≤ 12 t2. By the assumption (b), it follows that ν(G[B]) ≤ p − 1 and ∆( G[B]) ≤ q − 1.According to the famous result of Chv´ atal and Hanson , we have e(G[B]) ≤ ν(G[B])(∆( G[B]) + 1) ≤ q(p − 1) . Therefore, e(G) ≤ e(H) + c, where c = 12 t2 + q(p − 1) .Clearly, (c) leads to (a). Suppose F0 is a graph in F with β′(F0) = β(F) and \|F0\| ≤ l. Let G ∈ Ex sp (n, F). By Lemma 3.12, G contains Kβ′(F)−1,n +1 −β′(F) as a subgraph. Let L′ and R denote the color classes of this complete bipartite graph with \|L′\| = β′(F) − 1 and \|R\| = n + 1 − β′(F). By Lemma 4.1, there exists a constant c such that ex H (n, F) ≤ e(H) + c. For brevity, we write β′ = β′(F) and β = β(F). Lemma 4.2. G[L′] ∈ Ex( β′ − 1, H(F)) . Proof. It suffices to show that e(G[L′]) = ex( β′ − 1, H(F)) . We first prove that e(G[L′]) ≥ ex( β′ − 1, H(F)) .Recall that Kβ′−1,n +1 −β′ ⊆ G and ex H (n, F) ≤ e(H) + c. So e(G[R]) ≤ c. Suppose for contradiction that e(G[L′]) < ex( β′ − 1, H(F)) . We construct a new graph G′ as follows: delete all edges in G[R] and G[L′].Embed a graph G∗ ∈ Ex( β′ − 1, H(F)) into the vertex set L′. By Lemma 3.2, G′ is F-free. Applying Lemma 3.12, for each u ∈ L′, xu ≥ 1 − ε 110 , and for each v ∈ R, xv ≤ ε 18 . We obtain xT λ(G′) − λ(G)x ≥ xT A(G′)x − xT A(G)x = 2 X uv ∈E(G∗) xuxv − 2 X uv ∈E(G[L′]) xuxv − 2 X uv ∈E(G[R]) xuxv ≥ 2e(G∗)(1 − ε 110 )2 − 2e(G[L′]) − 2cε 14 ≥ 2 − 2( β′ − 1) 2ε 110 − 2cε 14 0, where the second to last inequality follows from e(G∗) ≥ e(G[L′]) + 1 . So λ(G′) > λ (G), which is a contradic-tion. Therefore e(G[L′]) ≥ ex( β′ − 1, H(F)) .On the other hand, suppose for contradiction that e(G[L′]) ≥ ex( β′ − 1, H(F)) + 1 . We claim that β′ > β .Assume for contradiction that β′ = β, then H(F) = Kβ′ and ex( β′ − 1, H(F)) = e(Kβ′−1). Obviously, e(G[L′]) ≤ e(Kβ′−1). So e(G[L′]) ≤ ex( β′ − 1, H(F)) , contradicting the assumption e(G[L′]) ≥ ex( β′ − 1, H(F)) + 1 . Therefore, β′ > β . Then it follows that H(F) = M(F). By the definition of M(F), there exists a graph eF ∈ M (F) such that eF ⊆ G[L′]. Since F is finite, there exists an F ′ ∈ F such that F ′ ⊆ 16 eF ∨ In−β′+1 ⊆ G. This contradicts the fact that G is F-free. So e(G[L′]) = ex( β′ − 1, H(F)) . Therefore G[L′] ∈ Ex( β′ − 1, H(F)) . Proof of Theorem 1.1. By Lemmas 3.12 and 4.2, we know G[L′] ∈ Ex( β′ −1, H(F)) and G[L′]∨In+1 −β′ ⊆ G.It suffices to prove G ∈ Ex G[L′]∨In+1 −β′ (n, F). Suppose for contradiction that G / ∈ Ex G[L′]∨In+1 −β′ (n, F).Then there exists a graph G′ ∈ Ex G[L′]∨In+1 −β′ (n, F) such that G′ = G[L′] ∨ Q and e(G′) ≥ e(G) + 1 . Since ex H (n, F) ≤ e(H) + c, we deduce e(G[R]) + 1 ≤ e(Q) ≤ c.Recall that xu ≥ 1 − ε 110 for each u ∈ L′ and xv ≤ ε 18 for each v ∈ R. For a vertex v ∈ R, we have λ(G)xv ≥ X u∈L′ xu ≥ (β′ − 1)(1 − ε 110 ), which implies that xv ≥ (β′−1)(1 −ε 110 ) λ(G) . On the other hand, since G is F0-free, dR(v) ≤ l − 1. Then λ(G)xv = X u∈L′ xu + X u∈NR(v) xu ≤ β′ − 1 + ( l − 1) ε 18 , which implies that xv ≤ β′−1+( l−1) ε 18 λ(G) . Therefore, xT λ(G′) − λ(G)x ≥ xT A(G′)x − xT A(G)x = 2 X uv ∈E(Q) xuxv − 2 X uv ∈E(G[R]) xuxv ≥ 2e(Q) (β′ − 1)(1 − ε 110 ) λ(G) 2 − 2e(G[R]) β′ − 1 + ( l − 1) ε 18 λ(G) 2 ≥ 2 λ(G)2 e(Q) − e(G[R]) (β′ − 1) 2 − 3c(β′ − 1) 2ε 110 0, where the last inequality holds as e(Q) ≥ e(G[R])+1 . Thus λ(G′) > λ (G), which contradicts the assumption that G has the maximum spectral radius among all F-free graphs on n vertices. Hence, G ∈ Ex G[L′]∨In+1 −β′ (n, F) ⊆G(F). This completes the proof. 4.2 Proof of Theorem 1.2 Let G ∈ Ex sp (n, F). Since the complete bipartite graph Kβ′(F)−1,n +1 −β′(F) is F-free, the spectral radius of G satisfies λ(G) ≥ p(β′(F) − 1)( n + 1 − β′(F)) . By Lemma 3.12, G must contain Kβ′(F)−1,n +1 −β′(F) as a subgraph. Denote the color classes of this subgraph by L′ and R, where \|L′\| = β′(F)−1 and \|R\| = n+1 −β′(F).Since G is F-free, the number of edges within L′ satisfies e(G[L′]) ≤ ex( β′(F) − 1, H(F)) . Suppose F0 is a graph in F with β′(F0) = β′(F) and \|F0\| ≤ l. Let x be the Perron vector of A(G) corresponding to λ(G),normalized such that its maximum entry xu∗ = 1 . For brevity, we write β′ instead of β′(F). Lemma 4.3. For every vertex u ∈ L′, 1 ≥ xu ≥ 1 − β′ − 1 λ(G) . Furthermore, for each vertex v ∈ R, we have xv = β′ − 1 λ(G) + Θ( n−1). 17 Proof. By Theorem 3.3, u∗ ∈ L′. The fact λ(G) = λ(G)xu∗ ≤ β′ − 1 + P v∈R xv implies that P v∈R xv ≥ λ(G) + 1 − β′. Thus, for each u ∈ L′, we deduce that λ(G)xu ≥ X v∈R xv ≥ λ(G) + 1 − β′. Then 1 ≥ xu ≥ 1 − β′−1 λ(G) . Moreover, (β′ − 1)( n + 1 − β′) ≤ λ(G)2 ≤ 2e(G) ≤ 2Cn. So λ(G)2 = Θ( n). On the one hand, for any v ∈ R, λ(G)xv ≥ X u∈L′ xu ≥ (β′ − 1) 1 − β′ − 1 λ(G) , which implies that xv ≥ β′−1 λ(G) − O(n−1). Let v∗ be a vertex in R with xv∗ = max {xv \|v ∈ R}. Since G is F0-free, it follows that dR(v∗) ≤ l − 1. Then λ(G)xv∗ = X u∈L′ xu + X v∈NR(v∗) xv ≤ β′ − 1 + ( l − 1) xv∗ , which implies that xv∗ ≤ β′−1 λ(G) O(n−1). Hence, xv = β′−1 λ(G) Θ( n−1) for any vertex v ∈ R. This completes the proof. By Lemma 3.13 and the assumption of the Theorem 1.2, we have ex H (n, F) = e(H) + rn + O(1) , where r = 12 or 23 . Let c = 18( β′−1) , G1 ∈ Ex H1 (n, F). We can assume L′ ∪ R is a vertex partition of H1, and R is an independent set of H1. Then, by the assumption of Theorem 1.2, we have e(G1) ≥ ex H (n, F) − cn , which implies that e(G1[R]) ≥ (r − c)n − O(1) . Lemma 4.4. (1) If r = 12 , then G[R] can be obtained from Q1 by adding and deleting O(1) edges. (2) If r = 23 , then G[R] can be obtained from Q2 by adding and deleting O(1) edges. Proof. (1) If r = 12 , then ex H (n, F) = e(H) + 12 n + O(1) , which implies that the size of any component of G[R] is O(1) and number of components in G[R] with order at least 3 is O(1) . Furthermore, we claim e(G[R]) is infinite. Otherwise, assume e(G[R]) is O(1) . Then, since e(G1[R]) ≥ ( 12 − c)n − O(1) , by Lemma 4.3, xT λ(G1) − λ(G)x ≥ xT A(G1)x − xT A(G)x ≥ 2 ex( β′ − 1, H(F))(1 − O(λ(G)−1)) 2 − 2e(G[L′]) + 2e(G1[R]) β′ − 1 λ(G) − O(n−1) 2 − 2e(G[R]) β′ − 1 λ(G) + O(n−1) 2 ≥ 2 ex( β′ − 1, H(F)) − e(G[L′]) (1 − 2c)( β′ − 1) 2 2C − O(n− 12 ) 0, where the third equality holds as 2Cn ≥ λ(G)2 ≥ (β′ − 1)( n + 1 − β′) and the last inequality holds as ex( β′ − 1, H(F)) ≥ e(G[L′]) and c < 12 . So λ(G1) > λ (G). Moreover, G1 is F-free. This contradicts the assumption that G ∈ EX sp (n, F).18 Next we show the number of isolated vertices in G[R] is at most 1. Suppose to the contrary that there exist two isolated v1 and v2 in G[R]. Then we connect v1 and v2. We denote the new edge as e1 and the resulting graph as G′. If G′ contains a graph F1 ∈ F , then e1 ∈ E(F1). Since e(G[R]) is infinite and F1 is finite, there exists an edge e2 ∈ E(G[R]) such that e2 /∈ E(F1). Replacing e1 with e2 in F1, we obtain a copy of F1 in G, which is contradiction. Thus G′ is F-free. Moreover, λ(G′) > λ (G), contradicting the assumption that G ∈ Ex sp (n, F).Therefore, the number of isolated vertices in G[R] is at most 1. Consequently, G[R] can be obtained from Q1 by adding and deleting O(1) edges. (2) If r = 23 , we have ex H (n, F) = e(H) + 23 n + O(1) , which implies that the size of any component of G[R] is O(1) and number of components in G[R] with order at least 4 or K3 components is O(1) . Furthermore, we show that the number of P3 components in G[R] is infinite. Otherwise, suppose the number of P3 components in G[R] is O(1) . Then e(G[R]) ≤ 12 n + O(1) . Note that e(G1[R]) ≥ ( 23 − c)n − O(1) . Thus, xT λ(G1) − λ(G)x ≥ xT A(G1)x − xT A(G)x ≥ 2 ex( β′ − 1, H(F))(1 − O(λ(G)−1)) 2 − 2e(G[L′]) + 2e(G1[R]) β′ − 1 λ(G) − O(n−1) 2 − 2e(G[R]) β′ − 1 λ(G) + O(n−1) 2 ≥ 2 ex( β′ − 1, H(F)) − e(G[L′]) (1 − 6c)( β′ − 1) 2 6C − O(n− 12 ) 0, where the second to last inequality holds as 2Cn ≥ λ(G)2 ≥ (β′ − 1)( n + 1 − β′) and the last inequality holds as ex( β′ − 1, H(F)) ≥ e(G[L′]) and c < 16 . So λ(G1) > λ (G). Recall that G1 is F-free. This is a contradiction. Next, we demonstrate that both the number of P2 components and the number of isolated vertices in G[R] are at most 1. Suppose there are two isolated vertices. We then add an edge between them, denoting the resulting graph as G′′ . Analogous to the proof in the preceding case, we can confirm that G′′ is F-free. Combining this with the fact that λ(G′′ ) > λ (G), we arrive at a contradiction. Now assume there exist two P2 components {u1u2} and {v1v2} in G[R]. Without loss of generality, we assume xu1 ≥ xv1 . Define G∗ = G − { v1v2} + {u1v2}. By Lemma 3.1, it follows that λ(G∗) > λ (G). Moreover, since the number of P3 components in G[R] is infinite, one can readily verify that G∗ is F-free. This contradicts the condition G ∈ Ex sp (n, F). Therefore, G[R] can be obtained from Q2 by adding and deleting O(1) edges. The proof is complete. Lemma 4.5. G[L′] ∈ Ex( β′ − 1, H(F)) . Proof. It suffices to prove e(G[L′]) = ex( β′ − 1, H(F)) . Note that e(G[L′]) ≤ ex( β′ − 1, H(F)) as G is F-free. Assume e(G[L′]) < ex( β′ − 1, H(F)) . If r = 12 , then e(G[R]) ≤ 12 n + O(1) and e(G1[R]) ≥ ( 12 − c)n − O(1) .19 By Lemmas 4.3 and 4.4, we obtain xT λ(G1) − λ(G)x ≥ xT A(G1)x − xT A(G)x ≥ 2 ex( β′ − 1, H(F))(1 − O(λ(G)−1)) 2 − 2e(G[L′]) + 2e(G1[R]) β′ − 1 λ(G) − O(n−1) 2 − 2e(G[R]) β′ − 1 λ(G) + O(n−1) 2 ≥ 2 ex( β′ − 1, H(F)) − e(G[L′]) − 2c(β′ − 1) nn + 1 − β′ − O(λ(G)−1) 32 − O(n− 12 ) 0, where the third inequality holds as 2Cn ≥ λ(G)2 ≥ (β′ − 1)( n + 1 − β′) and the second to last inequality holds as ex( β′ − 1, H(F)) ≥ e(G[L′]) + 1 and c = 18( β′−1) . So λ(G1) > λ (G), which contradicts the assumption of G. If r = 23 , the proof is similar to the former case, and so we omit the details here. The proof is complete. Proof of Theorem 1.2. We proceed to prove that G belongs to the family Ex G[L′]∨In+1 −β′ (n, F). By Lemma 4.5, G[L′] ∈ Ex( β′ −1, H(F)) and G[L′]∨In+1 −β′ ⊆ G. Assume for contradiction that G / ∈ Ex G[L′]∨In+1 −β′ (n, F).Then there exists a graph G′ ∈ Ex G[L′]∨In+1 −β′ (n, F) such that G′ = G[L′] ∨ Q and e(G′) ≥ e(G) + 1 . Thus, e(Q) ≥ e(G[R]) + 1 .If ex H (n, F) = e(H) + 12 n + O(1) , then e(Q) ≥ 12 n − O(1) . By Lemma 3.14, it follows that Q can be obtained from Q1 by adding and deleting O(1) edges. Combining this with Lemma 4.4, we conclude that Q can be obtained from G[R] by adding p edges and deleting q edges, where p, q are two numbers satisfying p ≥ q + 1 .Then, we obtain xT λ(G′) − λ(G)x ≥ xT A(G′)x − xT A(G)x ≥ 2p β′ − 1 λ(G) − O(n−1) 2 − 2q β′ − 1 λ(G) + O(n−1) 2 ≥ 2( p − q) β′ − 1 λ(G) 2 − O(λ(G)−1n−1) 0, which contradicts the assumption that G has the largest spectral radius over all F-free graphs. Therefore, G ∈ Ex G[L′]∨In+1 −β′ (n, F) ⊆ G (F).If ex H (n, F) = e(H) + 23 n + O(1) , then e(Q) ≥ 23 n − O(1) . In this case, by Lemma 3.14, it follows that Q can be obtained from Q2 by adding and deleting O(1) edges. Combining this with Lemma 4.4, we conclude that Q can be obtained from G[R] by adding p edges and deleting q edges, where p, q are two numbers satisfying p ≥ q + 1 . Using the similar discussion above, we also can get G ∈ Ex G[L′]∨In+1 −β′ (n, F) ⊆ G (F). This completes the proof. 4.3 Proof of Theorem 1.3 Proof of Theorem 1.3. Let H = Kβ′(F)−1,n +1 −β′(F), H 1 = T ′∨In+1 −β′(F), where T ′ ∈ Ex( β′(F)−1, H(F)) .By the assumption that {T ∨ eH\| T ∈ Ex( β′(F) − 1, H(F)) } ⊆ Ex( n, F), it follows that G(F) ⊆ Ex( n, F) and ex H (n, F) = ex H1 (n, F). Evidently, ex H (n, F) ≤ ex( n, F) ≤ e(H) + rn + O(1) , where r ∈ [0 , 34 ).20 By Lemma 3.13, we deduce that ex H (n, F) = e(H) + rn + O(1) , where r = 0 , 12 , or 23 . If r = 0 , then ex H (n, F) < e (H) + j n+1 −β′(F)2 k . By Theorem 1.1, we obtain that Ex sp (n, F) ⊆ G (F) ⊆ Ex( n, F). If r = 12 , using Theorem 1.1 or Theorem 1.2, we have Ex sp (n, F) ⊆ G (F) ⊆ Ex( n, F). If r = 23 , Theorem 1.2 yields Ex sp (n, F) ⊆ G (F) ⊆ Ex( n, F). This completes the proof. 5 Proof of Theorem 1.4 Let G ∈ Ex sp (n, F). Erd˝ os and Gallai proved that ex( n, C ≥k) ≤ k−12 (n − 1) for any k ≥ 3. Then we have ex( n, F) ≤ kn . Since K⌊ k−12 ⌋, n −⌊ k−12 ⌋ is F-free, it follows that λ(G) ≥ q ⌊ k−12 ⌋(n − ⌊ k−12 ⌋). Let 0 < ε < ( 132 kβ ′(F) )40 be a sufficiently small constant. By Theorem 3.3, there exists a t ≥ (1 − (k + 1) ε 110 )n such that K⌊ k−12 ⌋, t is a subgraph of G. We define L′ as the color class of K⌊ k−12 ⌋, t with size ⌊ k−12 ⌋ and R = ∩v∈L′ NG(v). Set W = V (G) − R − L′. If e(G[R]) ≥ 2, then we can find a cycle with length at least k in G. Thus, we must have e(G[R]) ≤ 1. Moreover, when k is odd, e(G[R]) = 0 . Otherwise, there exists a cycle of length k in G. Since F is finite, we assume \|F \| < l . Lemma 5.1. G is connected. Proof. Suppose to the contrary that G is not connected. Without loss of generality we assume that G = G1 ∪ G2 ∪ · · · ∪ Gp and λ(G) = λ(G1). Let u1 be a vertex with maximum degree in G1. Then dG1 (u1) = ∆( G1) ≥ λ(G) ≥ s k − 12 n − k − 12 l. Arbitrarily choose an arbitrary vertex v1 ∈ V (G2). Let G′ be the graph obtained from G by deleting the edges incident to v1 and adding a new edge u1v1. Clearly, λ(G′) > λ (G), which implies that G′ is not F-free. Combining with the fact that u1v1 is a cut edge in G′ and G is C≥k-free, we can find a copy of F as a subgraph of G′, say F1, and v1 ∈ V (F1). Since dG1 (u1) = ∆ > l , there exists a vertex u′ in NG1 (u1) \ V (F1). Then, F1 \ { v1} ∪ { u′} is a copy of F in G. This contradicts the fact G is F-free. Hence G is connected. Let x = ( x1, . . . , x n)T be a perron eigenvector of A(G) corresponding to λ(G). Let u∗ be a vertex in V (G) with xu∗ = max {xi\| i ∈ [n]} = 1 . By Theorem 3.3, xu ≥ 1 − ε 110 for each u ∈ L′ and xv ≤ ε 18 for each v ∈ V (G)\L′. Lemma 5.2. W = ∅. Proof. Assume W̸ = ∅. If there exists a vertex u′ ∈ W such that dR(u′) ≥ 2, then we can derive a cycle with length at leas k in G, which leads to a contradiction. Thus, dR(u) ≤ 1 for every u ∈ W . Since ex( n, F) ≤ kn ,we get X u∈W dW ∪R(u) ≤ 2e(G[W ]) + e(W, R ) ≤ 2k\|W \| + \|W \| = (2 k + 1) \|W \|. Thus there must exist a vertex u0 ∈ W such that dW ∪R(u0) ≤ 2k + 1 . We now perform the following operation on u0: delete all edges between u0 and W ∪ R, and add all non-adjacent edges between u0 and L′. Let G1 be the resulting graph. Recall that xv ≤ ε 18 for any v ∈ V (G)\L′ and xu ≥ 1 − ε 110 for each u ∈ L′. Therefore, 21 xT λ(G1) − λ(G)x ≥ xT A(G1)x − xT A(G)x ≥ 2(1 − ε 110 )xu0 − 2(2 k + 1) ε 18 xu0 0, where the second inequality holds as dL′ (u0) ≤ ⌊ k−12 ⌋ − 1 in G and the last inequality holds as ε is sufficiently small. We claim that this operation does not create any copy of F . Otherwise, F is a subgraph of G1 and u0 ∈ V (F ).Recall that \|F \| < l . Hence we can find a new vertex v0 ∈ R \ V (F ) such that F { u0} ∪ { v0} is a copy of F in G. This is a contradiction. Consequently, we obtain a graph G1 that is F -free and satisfies λ(G1) > λ (G).Let W1 = W − { u0}. Since G1[W1] ⊆ G[W ], G1[W1] is F-free. So e(G1[W1]) ≤ k\|W1\|. For each u ∈ W1, dR(u) ≤ 1 remains valid. It follows that X u∈W1 dW1∪R(u) ≤ 2e(G1[W1]) + e(W1, R ) ≤ (2 k + 1) \|W1\|. Thus, there exists a vertex u1 ∈ W1 such that dW1∪R(u1) ≤ 2k + 1 . Furthermore, any vertex in W1 has neighbors only in L′ and W1 ∪ R. We perform the following operation on u1: delete all edges between u1 and W1 ∪ R and add all non-adjacent edges between u1 an L′. Denote the resulting graph as G2, which is F -free. Moreover, xT λ(G2) − λ(G)x ≥ xT A(G2)x − xT A(G)x ≥ 2(1 − ε 110 )xu0 − 2(2 k + 1) ε 18 xu0 + 2(1 − ε 110 )xu1 − 2(2 k + 1) ε 18 xu1 0, where the second equality holds as u0, u 1 ∈ W and the last inequality follows from ε is sufficiently small. Thus λ(G2) > λ (G). Set W2 = W1 − { u1}. Continue this process iteratively until Wp = ∅. At each step, λ(Gi) > λ (G) holds, and the final graph Gp is F -free. Recall that e(G[R]) = 0 when k is odd, and e(G[R]) ≤ 1 when k is even. Thus Gp = G[L′] ∨ In−⌊ k−12 ⌋ when k is odd, and Gp = G[L′] ∨ K2 ∪ In−⌊ k−12 ⌋− 2 or Gp = G[L′] ∨ In−⌊ k−12 ⌋ when k is even. Notably, all cycles in Gp have length at most k − 1. So Gp is F-free and λ(Gp) > λ (G), which leads to a contradiction. Hence W = ∅. Lemma 5.3. G[L′] ∈ Ex( ⌊ k−12 ⌋, H(F)) . Proof. To complete the proof, it suffices to show that e(G[L′]) = ex( β′ − 1, H(F)) . We first show e(G[L′]) ≤ ex( ⌊ k−12 ⌋, H(F)) . Assume for contradiction that e(G[L′]) ≥ ex( ⌊ k−12 ⌋, H(F)) + 1 . Case 1. Suppose β(F) = ⌊ k+1 2 ⌋ = β′(F). In this case, H(F) = n K⌊ k+1 2 ⌋ o . Then e(G[L′]) ≤ e K⌊ k−12 ⌋ =ex( ⌊ k−12 ⌋, H(F)) , contradicting the assumption. Case 2. Assume β(F) < ⌊ k+1 2 ⌋ = β′(F). This inequality implying H(F) = M(F). Note that β(C≥k) = ⌊ k+1 2 ⌋ = β′(F), so H(F) = M(F) = M(F ). By the assumption e(G[L′]) ≥ ex( ⌊ k−12 ⌋, H(F)) + 1 there must exist a graph eF ∈ M (F ) such that eF ⊆ G[L′]. Applying Lemma 5.2, since F is finite, we deduce that F ⊆ eF ∨ In−⌊ k−12 ⌋ ⊆ G. This contradicts the fact that G is F-free. Therefore, e(G[L′]) ≤ ex( ⌊ k−12 ⌋, H(F)) .Next we prove that e(G[L′]) ≥ ex( ⌊ k−12 ⌋, H(F)) by contradiction. Assume, for the sake of contradiction, that e(G[L′]) < ex( ⌊ k−12 ⌋, H(F)) . Let G′ = G∗ ∨ In−⌊ k−12 ⌋, where G∗ is a graph in Ex( ⌊ k−12 ⌋, H(F)) . By Lemma 3.2, G′ is F-free. Recall that xu ≥ 1 − ε 110 for each u ∈ L′ and xv ≤ ε 18 for each v ∈ R. Since e(G[R]) ≤ 1, we 22 have xT λ(G′) − λ(G)x ≥ xT A(G′)x − xT A(G)x = 2 X uv ∈E(G∗) xuxv − 2 X uv ∈E(G[L′]) xuxv − 2 X uv ∈E(G[R]) xuxv ≥ 2e(G∗)(1 − ε 110 )2 − 2e(G[L′]) − 2ε 14 ≥ 2 − 2 k − 12 2 ε 110 − 2ε 14 0, where the second to last inequality follows from e(G∗) ≥ e(G[L′]) + 1 , which contradicts that G ∈ Ex sp (n, F).Therefore, G[L′] ∈ Ex( ⌊ k−12 ⌋, H(F)) . Proof of Theorem 1.4. Recall that e(G[R]) = 0 when k is odd, and e(G[R]) ≤ 1 when k is even. If k is odd, by Lemmas 5.2 and 5.3, G = G[L′] ∨ In− k−12 ∈ T ∨ In− k−12 T ∈ Ex k − 12 , H(F) . If k is even, by Lemmas 5.2 and 5.3, G = G[L′] ∨ In− k 2+1 ∈ T ∨ In− k 2+1 T ∈ Ex k 2 − 1, H(F) or G = G[L′] ∨ K2 ∪ In− k 2−1 ∈ T ∨ K2 ∪ In− k 2−1 T ∈ Ex k 2 − 1, H(F) . This completes the proof. References N. Alon, P. Frankl, Tur´ an graphs with bounded matching number, J. Comb. Theory, Ser. B 165 (2024) 223-229. J. Byrne, A sharp spectral extremal result for general non-bipartite graphs, arXiv preprint arXiv: 2411.18637, (2024). J. Byrne, D.N. Desai, M. Tait, A general theorem in spectral extremal graph theory, arXiv preprint arXiv: 2401.07266, (2024). M. Chen, A. Liu, X. Zhang, Spectral Extremal Results with Forbidding Linear Forests, Graphs and Comb. 35 (2019) 335-351. M. Chen, A. Liu, X. Zhang, On the spectral radius of graphs without a star forest, Discrete Math. 344(4) (2021) 112269. S. Cioab˘ a, D.N. Desai, M. Tait, A spectral Erd˝ os-S´ os theorem, SIAM J. Discrete Math. 37(3) (2023) 2228-2239. P´ eter Csikv´ ari, Applications of the Kelmans transformation: extremality of the threshold graphs, Electron. J. Comb. 18 (2011) #P182. 23 S. Cioab˘ a, D.N. Desai, M. Tait, The spectral radius of graphs with no odd wheels, Eur. J. Comb. 99 (2022) 103420. S. Cioab˘ a, D.N. Desai, M. Tait, A spectral Erd˝ os-S´ os theorem, SIAM J. Discrete Math. 37(3) (2023) 2228-2239. S. Cioab˘ a, D.N. Desai, M. Tait, The spectral even cycle problem, Comb. Theory 4(1) (2024) 10. V. Chv´ atal, D. Hanson, Degrees and matchings, J. Comb. Theory, Ser. B 20 (1976) 128-138. C. Dou, F. Hu, X. Peng, Tur´ an numbers of cycles plus a general graph, arXiv preprint arXiv:2411.17322, (2024). C. Dou, B. Ning, X. Peng, The number of edges in graphs with bounded clique number and circumference, arXiv preprint arXiv:2410.06449, (2024). P. Erd˝ os, T. Gallai, On maximal paths and circuits of graphs, Acta Math. Acad. Sci. Hung. 10 (1959) 337-356. P. Erd˝ os, M. Simonovits, A limit theorem in graph theory, Studia Sci. Math. Hungar. 1 (1966) 51-57. Z. F¨ uredi, M. Simonovits, The history of degenerate (bipartite) extremal graph problems, Erd˝ os centennial, 169-264, Bolyai Soc. Math. Stud., 25, J´ anos Bolyai Math. Soc., Budapest, 2013. L. Fang, M. Tait, M. Zhai, Decomposition family and spectral extremal problems on non-bipartite graphs, Discrete Math. 348 (2025) 114527. L. Fang, H. Lin, J. Shu, Z. Zhang, Spectral extremal results on trees, Electron. J. Comb. 31(2) (2024), #P2.34. X. Fang, X. Zhu, Y. Chen, Generalized Tur´ an problem for a path and a clique, Eur. J. Comb. 127 (2025) 104137 L. Feng, G. Yu, and X. Zhang, Spectral radius of graphs with given matching number, Linear Algebra Appl. 422(1) (2007) 133-138. J. Gao and X. Hou, The spectral radius of graphs without long cycles, Linear Algebra Appl. 566 (2019) 17-33. D. Gerbner, On Tur´ an problems with bounded matching number, J. Graph Theory 106 (2024) 23-29. S. Jiang, Y. Zhai, X. Yuan, Some stability results for spectral extremal problems of graphs with bounded matching number, Linear Algebra Appl. 708 (2025) 513-524. G. Katona, C. Xiao, Extremal graphs without long paths and large cliques, Eur. J. Comb. 119 (2024) 103807. Y. Liu, L. Kang, Extremal graphs without long paths and a given graph, Discrete Math. 347 (2024) 113988. V. Nikiforov, The spectral radius of graphs without paths and cycles of specified length, Linear Algebra Appl. 432(9) (2010) 2243-2256. V. Nikiforov, Merging the A- and Q-spectral theories, Appl. 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Hong, The spectral radius of trees on k pendant vertices, Linear Algebra Appl. 395 (2005) 343-349. Y. Xue, L. Kang, On generalized Tur´ an problems with bounded matching number, arXiv preprint arXiv:2410.12338, (2024). L. Yuan, X. Zhang, Tur´ an numbers for disjoint paths, J. Graph Theory 98 (2021) 499-524. X. Zhu, Y. Chen, Extremal problems for a matching and any other graph, J. Graph Theory 109 (2025) 19-24. X. Zhao, M. Lu, Generalized Tur´ an problems for a matching and long cycles, arXiv preprint arXiv:2412.18853, (2024). Y. Zhai, X. Yuan, Spectral extrema of {Kk+1 , Ls}-free graphs, Linear Algebra Appl. 682 (2024) 309-322. Y. Zhai, X. Yuan, L. You, Spectral extrema of graphs: Forbidden star-path forests, Discrete Math. 348 (2025) 114351. 25
1210	https://pubmed.ncbi.nlm.nih.gov/19753835/	[Persistent type 2 lepra reaction (erythema nodosum) and clofazimine-induced lethal enteropathy] - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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[Persistent type 2 lepra reaction (erythema nodosum) and clofazimine-induced lethal enteropathy] [Article in Spanish] Gerzain Rodríguez1,Rafael Pinto,Fernando López,Yenny Gómez Affiliations Expand Affiliation 1 Grupo de Microbiología Molecular, Facultad de Medicina, Universidad de La Sabana, Chía, Cundinamarca. gerzain_rodriguez@yahoo.com PMID: 19753835 Item in Clipboard Case Reports [Persistent type 2 lepra reaction (erythema nodosum) and clofazimine-induced lethal enteropathy] [Article in Spanish] Gerzain Rodríguez et al. Biomedica.2009 Mar. Show details Display options Display options Format Biomedica Actions Search in PubMed Search in NLM Catalog Add to Search . 2009 Mar;29(1):18-24. Authors Gerzain Rodríguez1,Rafael Pinto,Fernando López,Yenny Gómez Affiliation 1 Grupo de Microbiología Molecular, Facultad de Medicina, Universidad de La Sabana, Chía, Cundinamarca. gerzain_rodriguez@yahoo.com PMID: 19753835 Item in Clipboard Cite Display options Display options Format Abstract Introduction: Clofazimine enterophathy is a serious complication of clofazimine when used at high doses for treatment of type 2 lepra or or erythema nodosum leprosum. Objective. A woman is presented who had a delayed diagnosis of leprosy, persistent type 2 lepra reaction and lethal clofazimine enteropathy. Materials and methods: A 31-year-old woman presented leprosy symptoms over a 16-year period without medical diagnosis of her disease. During this period, type 2 lepra episodes occurred, but were not accurately diagnosed. These episodes became more severe during her second pregnancy. The patient and her family were interviewed, and her clinical history reviewed. Results: After twelve years of medical consults, lepromatous leprosy was diagnosed, based on perforation of her nasal septum, with a bacterial index of 5. Her husband and a 12-year-old daughter have leprosy symptoms. During multidrug therapy, she presented with repeated type 2 lepra reaction episodes for which she received daily clofazimine 400 mg doses. Two months after this treatment, severe and frequent episodes of intense abdominal pain began to occur. These persisted for more than a year and were managed with in-hospital administration of several classes of painkillers and antispasmodic medication, including morphine. She also presented with sporadic diarrhea, constipation, nausea, weight loss and mesenteric adenopathies. She died finally due to this intestinal condition. No autopsy was performed. Conclusions: The patient's clinical presentation suggested a clofazimine-induced lethal enteropathy, a complication not previously seen in Colombia. This connection was not recognized by the medical officers that treated the patient. PubMed Disclaimer Similar articles Daily multidrug therapy for leprosy; results of a fourteen-year experience.de Carsalade GY, Wallach D, Spindler E, Pennec J, Cottenot F, Flageul B.de Carsalade GY, et al.Int J Lepr Other Mycobact Dis. 1997 Mar;65(1):37-44.Int J Lepr Other Mycobact Dis. 1997.PMID: 9207752 Clinical Trial. Erythema leprosum--after treatment of Lepromatous Leprosy.Eickelmann M, Steinhoff M, Metze D, Tomimori-Yamashita J, Sunderkötter C.Eickelmann M, et al.J Dtsch Dermatol Ges. 2010 Jun;8(6):450-3. doi: 10.1111/j.1610-0387.2009.07294.x. Epub 2009 Nov 17.J Dtsch Dermatol Ges. 2010.PMID: 19922466 English, German. Necrotic erythema nodosum leprosum with oral mucosal involvement.Swain JP, Soud A, Agarwal SK.Swain JP, et al.Indian J Lepr. 2008 Apr-Jun;80(2):175-8.Indian J Lepr. 2008.PMID: 19425513 Intra-abdominal, crystal-storing histiocytosis due to clofazimine in a patient with lepromatous leprosy and concurrent carcinoma of the colon.Pais AV, Pereira S, Garg I, Stephen J, Antony M, Inchara YK.Pais AV, et al.Lepr Rev. 2004 Jun;75(2):171-6.Lepr Rev. 2004.PMID: 15282969 Review. [Histoid leprosy with erythema nodosum leprosum].Alioua Z, Sbai M, Elhaouri M, Bouzidi A, Boudi O, Ghfir M, Benomar S, Sedrati OL.Alioua Z, et al.Acta Leprol. 2003;12(3):107-11.Acta Leprol. 2003.PMID: 15040700 Review.French. See all similar articles Cited by Melkersson-Rosenthal syndrome in the context of sarcoidosis: a case report.Casper J, Mohammad-Khani S, Schmidt JJ, Kielstein JT, Lenarz T, Haller H, Wagner AD.Casper J, et al.J Med Case Rep. 2021 Oct 4;15(1):488. doi: 10.1186/s13256-021-03044-5.J Med Case Rep. 2021.PMID: 34602094 Free PMC article. Publication types Case Reports Actions Search in PubMed Search in MeSH Add to Search English Abstract Actions Search in PubMed Search in MeSH Add to Search Research Support, Non-U.S. Gov't Actions Search in PubMed Search in MeSH Add to Search MeSH terms Abdominal Pain / chemically induced Actions Search in PubMed Search in MeSH Add to Search Adult Actions Search in PubMed Search in MeSH Add to Search Arthritis, Rheumatoid / diagnosis Actions Search in PubMed Search in MeSH Add to Search Child Actions Search in PubMed Search in MeSH Add to Search Child, Preschool Actions Search in PubMed Search in MeSH Add to Search Clofazimine / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Clofazimine / adverse effects Actions Search in PubMed Search in MeSH Add to Search Clofazimine / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Constipation / chemically induced Actions Search in PubMed Search in MeSH Add to Search Diagnostic Errors Actions Search in PubMed Search in MeSH Add to Search Diarrhea / chemically induced Actions Search in PubMed Search in MeSH Add to Search Drug Therapy, Combination Actions Search in PubMed Search in MeSH Add to Search Erythema Nodosum / etiology Actions Search in PubMed Search in MeSH Add to Search Family Health Actions Search in PubMed Search in MeSH Add to Search Fatal Outcome Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Intestinal Diseases / chemically induced Actions Search in PubMed Search in MeSH Add to Search Leishmaniasis, Mucocutaneous / diagnosis Actions Search in PubMed Search in MeSH Add to Search Leprostatic Agents / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Leprostatic Agents / adverse effects Actions Search in PubMed Search in MeSH Add to Search Leprostatic Agents / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Leprosy, Lepromatous / complications Actions Search in PubMed Search in MeSH Add to Search Leprosy, Lepromatous / diagnosis Actions Search in PubMed Search in MeSH Add to Search Leprosy, Lepromatous / drug therapy Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Paresthesia / etiology Actions Search in PubMed Search in MeSH Add to Search Pregnancy Actions Search in PubMed Search in MeSH Add to Search Pregnancy Complications, Infectious / diagnosis Actions Search in PubMed Search in MeSH Add to Search Pregnancy Complications, Infectious / microbiology Actions Search in PubMed Search in MeSH Add to Search Rifampin / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Substances Leprostatic Agents Actions Search in PubMed Search in MeSH Add to Search Clofazimine Actions Search in PubMed Search in MeSH Add to Search Rifampin Actions Search in PubMed Search in MeSH Add to Search Related information MedGen PubChem Compound PubChem Compound (MeSH Keyword) PubChem Substance [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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1211	https://physics.stackexchange.com/questions/573644/why-use-a-combination-of-e-h-to-describe-electromagnetic-systems-instead	electromagnetism - Why use a combination of ($E$, $H$) to describe electromagnetic systems instead of ($E$, $B$) or ($D$, $H$)? - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why use a combination of (E E, H H) to describe electromagnetic systems instead of (E E, B B) or (D D, H H)? Ask Question Asked 5 years, 1 month ago Modified3 years, 11 months ago Viewed 639 times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. I attended a lecture on theoretical electrodynamics, and there we exclusively used the combination of E and B to fundamentally describe electromagnetism. They contain no approximation for material behaviour and use the exact charge density and current density (including the bound charges and bound currents in materials). In the very last lecture we introduced the concept of looking at the system macroscopically, approximating the bound charges and currents in the material as Polarisations P and Magnetizations M and "absorbing them" into the new D and H Field. D and H thus present an easy way to write the Maxwell equations just for the external charges and currents, where we don't have to precicely know all the the internal ones. So to my understanding you had these to combinations: You either had to describe the system fully with the combination of (E, B), or you could use the combo (D, H) to factor out bound microscropic charges and currents. Why then, when I look in many engineering books now, do they use a combination of (E, H) to describe their systems? Why this mishmash of macroscopic framework H and microscopic one E? Why use H where the microscopic processes are factored out, but then also E where they are not? electromagnetism electromagnetic-radiation conventions Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Aug 16, 2020 at 14:23 Sandejo 5,496 1 1 gold badge 24 24 silver badges 43 43 bronze badges asked Aug 16, 2020 at 12:41 blueblue 183 6 6 bronze badges 6 2 One thing to note is that D D and B B are flux densities (you integrate them over areas) and E E and H H are intensities (you integrate them over paths). Maxwell's equations written in the language of differential forms makes this manifest: D D and B B are 2-forms, E E and H H are 1-forms. This gives some weight to using the mixed fields.garyp –garyp 2020-08-16 14:03:50 +00:00 Commented Aug 16, 2020 at 14:03 1 @garyp interesting! I'm not sure if these engineering books really considered differential forms to make the choice to use E E and H H, but this does give a mathematical motivation for it. Thank you.blue –blue 2020-08-16 14:37:22 +00:00 Commented Aug 16, 2020 at 14:37 @garyp That's an interesting comment. Since space is 3d, one-forms and two-forms can be interchanged with no ambiguity, so we could just as well use a two-form to represent E E, for example. (Not recommended, but possible.) But that breaks down when we consider the generalization to D D-dimensional space. For example, when D=2 D=2, the magnetic field B B has only one component (because it's a two-form), but a one-form has two components. So if H H is a one-form, where does its extra component come from?Chiral Anomaly –Chiral Anomaly 2020-08-16 18:07:06 +00:00 Commented Aug 16, 2020 at 18:07 @ChiralAnomaly In the formulation in terms of differential forms, I don't believe that the E E and B B fields are considered fundamental objects. Instead, you would always consider the field strength 2-form (in special relativity the Faraday tensor F μ ν F μ ν). I think that in 3+1 dimensions the constitutive form G G is also a 2-form, and maybe this isn't the case in 2+1 dimensions.user167506 –user167506 2020-08-16 23:38:09 +00:00 Commented Aug 16, 2020 at 23:38 @MatthewO'Brien Yeah, that's where I'm coming from, except that I was splitting F μ ν F μ ν into time-space and space-space parts and only using forms for the space parts: F 0 k d x k F 0 k d x k and F j k d x j∧d x k F j k d x j∧d x k. Seems like the constitutive form G G should follow the same pattern as F F, in any number of dimensions. In particular, the magnetic part should be a two-form, whether it's B B and H H. Treating H H is a one-form in 3+1 3+1 dimensions seems unnatural, but I don't have an engineering background, so maybe I'm missing something.Chiral Anomaly –Chiral Anomaly 2020-08-17 00:28:40 +00:00 Commented Aug 17, 2020 at 0:28 \|Show 1 more comment 4 Answers 4 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. I was confused by this too when I first encountered it. After all, E E and B B are the "real" fields (in the sense that they directly determine the force experienced by a charge) - so they seem like the natural pair. When I first studied electromagnetic waves in university, everything was in terms of E E and H H, with little mention of D D and B B. H H also showed up a lot when we learnt about transformers, but in that case B B showed up just as much. Griffiths gives an explanation along these lines: The pair E E and H H are commonly seen together because they are easy to measure. In the low-frequency limit: The line integral of E E gives voltage between the endpoints of the path. Voltages are easily measured with a voltmeter. The loop integral of H H gives the (free) current passing through the loop. Free currents are easily measured with an ammeter. This is only true in the low-frequency limit, since otherwise there are "coupling" terms ∂B/∂t∂B/∂t and ∂D/∂t∂D/∂t which will affect the line integrals of E E and the loop integrals of H H. Griffiths goes on to say that if it was more natural to measure build-ups of (free) charge than voltage, then we would be seeing D D a lot more often. Another reason for choosing the pair (E,H)(E,H) is that it gives a nice form for the Poynting vector, which (in the time domain) is: S=E×H S=E×H This corresponds nicely to the formula for power from circuit theory: P=I V P=I V Expressing the Poynting vector in terms of other pairs of fields would cause additional constant factors to appear out the front (although technically you could use different units to make these constant factors into 1). Apparently there has been some debate in the past about which form of the Poynting vector should be preferred. Also note that the usual units of E E are volts per meter (voltage per unit length), and the usual units of H H are amps per meter (current per unit length) - which gives a very nice watts per meter squared for the Poynting vector! So it's as if E E is "voltage-like" and H H is "(free) current-like", which gives nice results for the Poynting vector, for units, and nice relationships to real-world measurements. Caveat Griffith warns the reader against thinking of H H as determined only by the free current distribution and somehow "factoring out" the effects of magnetic media. This is a natural conclusion to reach, since in magnetostatics we have ∇×H=J f∇×H=J f, which is similar to ∇×B=μ 0 J∇×B=μ 0 J that gives rise to the Biot-Savart law. However, unlike B B, H H does not necessarily have zero divergence, instead we have: ∇⋅H=−∇⋅M∇⋅H=−∇⋅M Which means that H H also depends on the magnetization of the medium. However, loop integrals of H H will always be equal to the free current passing through the loop (in magnetostatics at least, since otherwise the displacement current term ∂D/∂t∂D/∂t will be non-negligible). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Oct 19, 2021 at 4:29 answered Oct 19, 2021 at 4:17 PersonWithNamePersonWithName 319 3 3 silver badges 12 12 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Physics is littered with various notational conventions. Generally people will use whatever combination of variables and whatever system of units they find most convenient at the time for their purpose. In electrodynamics one would generally use whatever combination of variables which makes Maxwell's equations "look the simplest" given the presence of sources, bound currents/charges, etc. I know of no reason other than pure convenience. Any formulation of Maxwell's equations in terms of (E,H)(E,H) or (E,B)(E,B) or (D,H)(D,H) is formally equivalent. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 16, 2020 at 12:47 user167506 user167506 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. If my memory is correct, this is because they are the easiest (or most common) to measure. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 16, 2020 at 14:28 user272451 user272451 3 I think I also remember reading this explanation somewhere once... I'm not sure why H H would be more straightforward to measure than B B, but I think such an explanation would be more fitting for an engineering textbook than considering differential forms.blue –blue 2020-08-16 14:44:09 +00:00 Commented Aug 16, 2020 at 14:44 I think taking the units into account with common tools can give insight into that. Theory is usually applied at some point.user272451 –user272451 2020-08-16 15:11:00 +00:00 Commented Aug 16, 2020 at 15:11 1 You've reminded me of my undergrad lab classes! The Wiki page "magnetic field" points out that a measurement device such as a magnetic field probe will typically generate its own magnetic field in response to the external applied field. So in this case, the H H field is in fact more directly accessible experimentally and is the more convenient variable.user167506 –user167506 2020-08-16 23:27:11 +00:00 Commented Aug 16, 2020 at 23:27 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. History There is a long history of using fields E E and H H together, which is reflected in their names and notations: E E and H H are sometimes called tensions, while D D and B B are called displacements/inductions (I am not sure how generally accepted these terms are in English, but other languages do make this distinction). This is further reflected, e.g., in the form of the constitutive relations: D=ε E,B=μ H D=ε E,B=μ H (rather than H=μ B H=μ B). Macroscopic electrodynamics When dealing with macroscopic Maxwell equations, any pair of electric and magnetic fields can be used without causing ambiguity: E and H E and B D and H D and B E and H E and B D and H D and B Miscroscopic electrodynamics In microscopic electrodynamics the distinction between the tensions and the induced fields disappears. In Gaussian units they are simply equal: D=E,B=H,D=E,B=H, so again the use of specific letters is a matter of convention in a particular field. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 19, 2021 at 7:10 Roger V.Roger V. 70.9k 7 7 gold badges 78 78 silver badges 251 251 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. 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1212	https://www.verbix.com/webverbix/go.php?D1=9&T1=utor	Latin verb 'utor' conjugated Conjugator Conjugator EnglishFrenchItalianPortugueseSpanish More Verb Conjugation... Noun Declination... Verb Finder More TranslationCognatesGamesLanguage MapsLanguage DraftsVerbix for WindowsVerbix Documents For Developers BlogAbout + Terms of Use Latin: utor Latin verb 'utor' conjugated Discover more Wessex Gospels verbix Activewear Verbix Gift baskets Bookshelves Cite this page \| Conjugate another Latin verb Dictionary lookup word (Ind. Present 1.sg.): ūtor Active Nominal Forms Infinitive:Present participle:ūtēns; ūténtis Future participle:ūsúrus Gerund:ūténdum Gerundive:ūténdus Passive Nominal Forms Infinitive:Perfect participle:ūsum Active Indicative Present Imperfect I Future Perfect Pluperfect II Future Subjunctive Present Imperfect Perfect Pluperfect I Imperative tū ū́tere vōs ūtíminī II Imperative tū ū́titor is ū́titor iī ūtúntor Passive Indicative Present ego ūtor tū ū́teris is ū́titur nōs ū́timur vōs ūtíminī iī ūtúntur Imperfect ego ūtḗbar tū ūtēbā́ris is ūtēbā́tur nōs ūtēbā́mur vōs ūtēbā́minī iī ūtēbántur I Future ego ūtar tū ūtḗris is ūtḗtur nōs ūtḗmur vōs ūtḗminī iī ūténtur Perfect ego ūsus sum tū ūsus es is ūsus est nōs ūsī sumus vōs ūsī estis iī ūsī sunt Pluperfect ego ūsus eram tū ūsus erās is ūsus erat nōs ūsī erāmus vōs ūsī erātis iī ūsī erant II Future ego ūsus erō tū ūsus eris is ūsus erit nōs ūsī erimus vōs ūsī eritis iī ūsī erunt Subjunctive Present ego ūtar tū ūtā́ris is ūtā́tur nōs ūtā́mur vōs ūtā́minī iī ūtā́ntur Imperfect ego ūterer tū ūterḗris is ūterḗtur nōs ūterḗmur vōs ūterḗminī iī ūterḗntur Perfect ego ūsus sim tū ūsus sīs is ūsus sit nōs ūsī sīmus vōs ūsī sītis iī ūsī sint Pluperfect ego ūsus essem tū ūsus essēs is ūsus esset nōs ūsī essēmus vōs ūsī essētis iī ūsī essent Verbs conjugated like 'utor' abutor, adipiscor, alloquor, comminiscor, conitor, depaciscor, deutor, enascor, enitor, expergiscor, expiscor, indipiscor, innascor, innitor, internascor, irascor, nanciscor, nascor, nitor, obirascor, etc. (List truncated at 20 verbs) Verbs similar to 'utor' abutor, deutor, fior, for, furor, ito, letor, luctor, metor, muto, Translations employ, to put into use utilise, to make use of; to use Etymology From Proto-Italic oitōr (with ūsus for issus after ūtor), possibly from Proto-Indo-European h₃eyt- (“to take along, fetch”). Compare the future tense οἴσω (oísō) of Ancient Greek φέρω (phérō, “carry”). See: Proto-Italic 'oitōr'. Verbs referencing "utor" French user, Old French user. Additional Information Cognates Sample Sentences ponentque cum eo omnia vasa quibus in ministerio eius utuntur id est ignium receptacula fuscinulas ac tridentes uncinos et vatilla cuncta vasa altaris operient simul velamine ianthinarum pellium et inducent vectes ingrediantur aquae maledictae in ventrem tuum et utero tumescente putrescat femur et respondebit mulier amen amen carnes vero in usum tuum cedent sicut pectusculum consecratum et armus dexter tua erunt viri similiter cohabitantes secundum scientiam quasi infirmiori vaso muliebri inpertientes honorem tamquam et coheredibus gratiae vitae uti ne inpediantur orationes vestrae quia et Thessalonicam et semel et bis in usum mihi misistis qui multis utitur verbis laedit animam suam et qui potestatem sibi adsumit iniuste odietur quis dabit ori meo custodiam et supra labia mea signaculum certum uti ne cadam ab ipsis et lingua mea perdat me in filia non avertente se firma custodiam ne inventa occasione utatur se utere quasi homo frugi quae tibi adponuntur et non cum manducas multum odio habearis ab eo qui utitur hyacintho et portat coronam usque ad eum qui operitur lino crudo furor zelus tumultus fluctuatio et timor mortis iracundia perseverans et contentio et ne liceat ulli ex populo et ex sacerdotibus irritum facere aliquid horum et contradicere his quae ab eo dicuntur aut convocare conventum in regione sine ipso et vestiri purpura et uti fibula aurea et postea clamaverunt ad Dominum qui suscitavit eis salvatorem vocabulo Ahoth filium Gera filii Iemini qui utraque manu utebatur pro dextera miseruntque filii Israhel per illum munera Eglon regi Moab et fuit pondus postulatarum inaurium mille septingenti auri sicli absque ornamentis et monilibus et veste purpurea quibus Madian reges uti soliti erant et praeter torques aureos camelorum quae respondit numquid possum deserere pinguedinem meam qua et dii utuntur et homines et venire ut inter ligna promovear quia concipies et paries filium cuius non tanget caput novacula erit enim nazareus Dei ab infantia sua et ex matris utero et ipse incipiet liberare Israhel de manu Philisthinorum sed hoc respondit ecce concipies et paries filium cave ne vinum bibas et siceram et ne aliquo vescaris inmundo erit enim puer nazareus Dei ab infantia sua et ex utero matris usque ad diem mortis suae tunc aperiens veritatem rei dixit ad eam ferrum numquam ascendit super caput meum quia nazareus id est consecratus Deo sum de utero matris meae si rasum fuerit caput meum recedet a me fortitudo mea et deficiam eroque ut ceteri homines Christi autem generatio sic erat cum esset desponsata mater eius Maria Ioseph antequam convenirent inventa est in utero habens de Spiritu Sancto ecce virgo in utero habebit et pariet filium et vocabunt nomen eius Emmanuhel quod est interpretatum Nobiscum Deus sunt enim eunuchi qui de matris utero sic nati sunt et sunt eunuchi qui facti sunt ab hominibus et sunt eunuchi qui se ipsos castraverunt propter regnum caelorum qui potest capere capiat et in utero habens et clamat parturiens et cruciatur ut pariat Vulgate Verses (Genesis 16:6) cui respondens Abram ecce ait ancilla tua in manu tua est utere ea ut libet adfligente igitur eam Sarai fugam iniit But Abram said unto Sarai, Behold, thy maid is in thy hand; do to her as it pleaseth thee. And when Sarai dealt hardly with her, she fled from her face. (II Samuel 13:18) quae induta erat talari tunica huiuscemodi enim filiae regis virgines vestibus utebantur eiecit itaque eam minister illius foras clausitque fores post eam And she had a garment of divers colours upon her: for with such robes were the king's daughters that were virgins apparelled. Then his servant brought her out, and bolted the door after her. (Esther 2:12) cum autem venisset tempus singularum per ordinem puellarum ut intrarent ad regem expletis omnibus quae ad cultum muliebrem pertinebant mensis duodecimus vertebatur ita dumtaxat ut sex menses oleo unguerentur myrtino et aliis sex quibusdam pigmentis et aromatibus uterenturNow when every maid's turn was come to go in to king Ahasuerus, after that she had been twelve months, according to the manner of the women, (for so were the days of their purifications accomplished, to wit, six months with oil of myrrh, and six months with sweet odours, and with other things for the purifying of the women;) (Esther 3:10) tulit ergo rex anulum quo utebatur de manu sua et dedit eum Aman filio Amadathi de progenie Agag hosti Iudaeorum And the king took his ring from his hand, and gave it unto Haman the son of Hammedatha the Agagite, the Jews' enemy. (Exodus 29:29) vestem autem sanctam qua utitur Aaron habebunt filii eius post eum ut unguantur in ea et consecrentur manus eorum And the holy garments of Aaron shall be his sons' after him, to be anointed therein, and to be consecrated in them. (Exodus 29:30) septem diebus utetur illa qui pontifex pro eo fuerit constitutus de filiis eius et qui ingredietur tabernaculum testimonii ut ministret in sanctuario And that son that is priest in his stead shall put them on seven days, when he cometh into the tabernacle of the congregation to minister in the holy place. (Exodus 39:41) vestes quoque quibus sacerdotes utuntur in sanctuario Aaron scilicet et filii eius The cloths of service to do service in the holy place, and the holy garments for Aaron the priest, and his sons' garments, to minister in the priest's office. (Ecclesiastes 2:10) et omnia quae desideraverunt oculi mei non negavi eis nec prohibui cor quin omni voluptate frueretur et oblectaret se in his quae paraveram et hanc ratus sum partem meam si uterer labore meo And whatsoever mine eyes desired I kept not from them, I withheld not my heart from any joy; for my heart rejoiced in all my labour: and this was my portion of all my labour. (Ecclesiastes 6:3) si genuerit quispiam centum et vixerit multos annos et plures dies aetatis habuerit et anima illius non utatur bonis substantiae suae sepulturaque careat de hoc ego pronuntio quod melior illo sit abortivus If a man beget an hundred children, and live many years, so that the days of his years be many, and his soul be not filled with good, and also that he have no burial; I say, that an untimely birth is better than he. (Numeri 4:14) ponentque cum eo omnia vasa quibus in ministerio eius utuntur id est ignium receptacula fuscinulas ac tridentes uncinos et vatilla cuncta vasa altaris operient simul velamine ianthinarum pellium et inducent vectes And they shall put upon it all the vessels thereof, wherewith they minister about it, even the censers, the fleshhooks, and the shovels, and the basons, all the vessels of the altar; and they shall spread upon it a covering of badgers' skins, and put to the staves of it. Cite this page Harvard Referencing: Verbix 2025, Latin verb 'utor' conjugated, Verbix, viewed 29 Sep 2025, APA Referencing: Verbix (2025, Sep 29) Latin verb 'utor' conjugated. Verbix Verb Conjugator. Retrieved from Report an error
1213	https://arundquist.wordpress.com/2022/12/31/rolling-without-slipping-on-curved-surfaces/	Menu SuperFly Physics Physics questions, ideas, hare-brained schemes Rolling without slipping on curved surfaces I’ve been trying to see if I can model balls rolling on curved surfaces and I think I’ve cracked it. Here’s a teaser to get you interested: What you see is a sphere rolling on a curved surface. The blue line is the path of the contact point and the orange line is the path of the center of the ball. What that ball is doing is called “rolling without slipping” which just means that the contact point doesn’t slide at all. In fact, the part of the ball in contact with the surface is (momentarily) at rest! Rolling without slipping is something that happens in nature a lot. If something isn’t doing that (like a bowling ball at the beginning of your throw), it often is brought to rolling-without-slipping by friction. Once it gets to that point it tends to stay in that condition. So I thought it would be cool to learn how to model it. So why did I think this would be an interesting challenge? It seems that introductory physics courses have problems with rolling without slipping all the time. It’s how we learn that rolling balls get down hills slower than the point mass calculations we typically start students with. The reason is that the potential energy of the hill has to be split between the translational kinetic energy that we think of with point particles and the rotational kinetic energy it gets while rolling. That’s why it’s slower. But the reason this is interesting is I wanted to do two things: figure out the angular rotation at all times so that I could make fun animations, and ask what happens if you don’t just do a boring inclined plane — specifically a surface that’s curved. 2D first First I tackled an effective 2D problem, or a problem I can draw on a sheet of paper. The easiest to do is a non-curved surface that’s at an angle. The rolling without slipping condition in that case is pretty easy: where v is the speed of either the contact point (remember, the part of the ball touching the plane doesn’t move, where that point is does) or the speed of the center of the ball, since they’re the same if the surface is flat. Now consider the Lagrangian approach for the problem. Let’s assume the angle the line makes with the horizontal is . Here’s my chicken scratch, barely legible approach: If you like using x for horizontal and y for vertical, you use the first part. If you want to use a variable along the slope, you use the bottom part (where I used the variable “a”). Either way you get a pretty straightforward result showing that it moves down the slope at a constant acceleration that’s less than g, thanks to the rolling. Note that for the rest of this post I’ll be using the x and y approach because the slope is constantly changing making the “a” approach tricky. Alright, so what happens when the surface is curved? I ran into this in my Brachistocrone for rolling things post, and the upshot is that the rolling without slipping condition changes to: where is the speed of the contact point. Here’s my notes where I proved that to myself: The key it to realize that while the yellow and orange stripes line up as it rolls, the sphere is frustrated from fully executing its rotation because the curve has risen up to meet it. The part that’s taken back is that that’s labeled. For that old post the more complicated expression was a major pain in the butt, and I was convinced for a few weeks it would be for this too until I went back to the paper I referenced in that post again a couple days ago and realized that you get a much cleaner version of the rolling without slipping condition: which looks just like my simple one up at the top of this post! Yep, if you just switch to the speed of the center of the ball, instead of the speed of the contact point, you get to use our old tried-and-true rolling without slipping condition. This was the big breakthrough I needed! Here’s the setup: First let’s consider 1 & 2. Here’s an image showing most of the important variables: How do you get the center-of-ball coordinates from the contact coordinates? The key is the right angle at the contact point. The center is R-units in the direction perpendicular to the surface from the contact point. That direction can be found from the function of the curve: where f’ indicates the derivative (or slope) of the curve. Ok, then we literally have the same kinetic and potential energies as what’s in my notes above for the flat curve, only we now know that all the x’s and y’s in there are actually functions of . Then step four gives us the following differential equation for : That’s for the special case where f(x)=sin(x). Ugly right? But who cares? We just dump it into an ODE solver! Here’s the result (remember that we get both and back so we can make the motion and the rotation look right): and here’s rolling down the function f(x)=sin(x)+x: Cool, right?! Now for 3D As soon as I got that working I realized that I could get the 3D working, only I realized that I’d have to shift from also solving for as single angle to solving for all the Euler angles. But, I‘m good at dealing with those so I went for it. Here’s the modified setup: Step 2 needs some adjustment from the 2D case. To find the direction that is perpendicular to a surface at a particular location. The trick is to make a new function: This function has a set of level surfaces where S is a constant. Our surface is when S=0. But the key is to recognize that the shortest path from one level surface to another is found by finding the gradient of the function. Therefore: Step 3 looks identical, just with 3 variables instead of 2. Step 5, especially the Euler angle part, is a little tricky. What is the rolling without slipping condition? We have but what is the magnitude of ? Well, we know what that is in terms of the Euler angles (and their time derivatives)! We could just grab their total amplitude, but we actually know something about the direction of . We know that it’s perpendicular to both the normal vector to the surface and to the direction of rolling. Note that “the direction of rolling” is parallel to both the direction that the contact point is moving and the direction the center of the ball is moving, but you’ll see in a second that it’s much easier to use the ball for this (we’re normalizing that vector either way, so it doesn’t matter that they aren’t the same length, it only matters that they’re parallel). Out front, the magnitude of can be found from our simple rolling-without-slipping condition. Putting it all together yields: Here’s how all that looks in Mathematica: One thing that took me a while was figuring out how to make a ball look like it was rolling in Mathematica. Here’s how I did that: Once you have the prototype ball you just use GeometricTransformation[ball, {m, r}] on it where m is the Euler rotation matrices dotted together and r is the location of the center of the ball. Whoo hoo! On to step 6: Here’s a comparison of 4 balls rolling on the function f(x,y)=sin(x)+sin(y) (it looks somewhat like an egg carton). They’re all started at the same point. Two are small and two are larger. In each size one is a solid ball and one is a shell. They all have the same mass: And here’s the biggest ball I could get to roll on that surface. Why would there be a limit? Because you can’t use a ball that has a larger radius than the concave up radius of curvature at any point on its journey (Mathematica actually craps out right when this happens): Here’s a still frame zoomed in a little showing that biggest ball right at the smallest radius of curvature: One cool thing is that I can have Mathematica calculate the radius of curvature at all points along the trajectory. Here’s that plot where the red horizontal line is the radius of the ball: How do you calculate the radius? You get it from this equation from way at the top of this post: Note that before I really got my head wrapped around how to do this I thought I would have to calculate that radius of curvature (which, by the way depends on both where you are on the surface and what direction you’re rolling) directly from the definition of the curve itself. That was an interesting rabbit hole that this approach didn’t need (sorry for bothering you, my friend (and occasional guest blogger Art) for how to do this). Your thoughts? I’m really excited about this new type of modeling I can do. What are your thoughts? Here are some starters for you: Share this: Related Post navigation Leave a comment Cancel reply Δ Hello, I am a high school student trying to figure out the theoretical relationship between the radius of a ball that is launched into a bowl (moves in a circular path in the bowl) and the time taken for it to come to a stop. I did some trials and found out the the smaller the radius is the shorter the time taken for it to come to a stop in a bowl of fixed curvature and radius. The only thing varied is the radius of the ball that enters the bowl and the initial velocity while entering the bowl was kept the same. It would be helpful if someone could help explain how I can go about deriving this relationship. 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1214	https://math.libretexts.org/Courses/Community_College_of_Denver/MAT_2420_Calculus_II/03%3A_Techniques_of_Integration/3.06%3A_Numerical_Integration	3.6: Numerical Integration - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 3: Techniques of Integration MAT 2420 Calculus II { } { "3..E:3.E:_Techniques_of_Integration(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.00:_Prelude_to_Techniques_of_Integration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.01:_Integration_by_Parts" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.02:_Trigonometric_Integrals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.03:_Trigonometric_Substitution" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.04:_Partial_Fractions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.05:_Other_Strategies_for_Integration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.06:_Numerical_Integration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.07:_Improper_Integrals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Applications_of_Integration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Techniques_of_Integration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Introduction_to_Differential_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Sequences_and_Series" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Power_Series" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Parametric_Equations_and_Polar_Coordinates" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 20 Jan 2020 15:46:25 GMT 3.6: Numerical Integration 33804 33804 Paul Seeburger { } Anonymous Anonymous 2 false false [ "article:topic", "Midpoint Rule", "stage:review", "authorname:openstax", "relative error", "absolute error", "numerical integration", "license:ccbyncsa", "showtoc:no", "transcluded:yes", "source-math-2553", "licenseversion:40", "source@ ] [ "article:topic", "Midpoint Rule", "stage:review", "authorname:openstax", "relative error", "absolute error", "numerical integration", "license:ccbyncsa", "showtoc:no", "transcluded:yes", "source-math-2553", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Community College of Denver 4. MAT 2420 Calculus II 5. 3: Techniques of Integration 6. 3.6: Numerical Integration Expand/collapse global location 3.6: Numerical Integration Last updated Jan 20, 2020 Save as PDF 3.5: Other Strategies for Integration 3.7: Improper Integrals Page ID 33804 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. The Midpoint Rule 1. The Midpoint Rule 2. Example 3.6.1: Using the Midpoint Rule with M 4 3. Example 3.6.2: Using the Midpoint Rule with M 6 4. Exercise 3.6.1 The Trapezoidal Rule The Trapezoidal Rule Example 3.6.3: Using the Trapezoidal Rule Solution Exercise 3.6.2 Absolute and Relative Error Definition: absolute and relative error Example 3.6.4: Calculating Error in the Midpoint Rule Example 3.6.5: Calculating Error in the Trapezoidal Rule Exercise 3.6.3 Error Bounds on the Midpoint and Trapezoidal Rules Error Bounds for the Midpoint and Trapezoidal Rules Example 3.6.6: Determining the Number of Intervals to Use Solution Analysis Exercise 3.6.4 Simpson’s Rule Simpson’s Rule Rule: Error Bound for Simpson’s Rule Example 3.6.7: Applying Simpson’s Rule 1 Solution Example 3.6.8: Applying Simpson’s Rule 2 Solution Exercise 3.6.5 Key Concepts Key Equations Glossary Learning Objectives Approximate the value of a definite integral by using the midpoint and trapezoidal rules. Determine the absolute and relative error in using a numerical integration technique. Estimate the absolute and relative error using an error-bound formula. Recognize when the midpoint and trapezoidal rules over- or underestimate the true value of an integral. Use Simpson’s rule to approximate the value of a definite integral to a given accuracy. The antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). Consequently, rather than evaluate definite integrals of these functions directly, we resort to various techniques of numerical integration to approximate their values. In this section, we explore several of these techniques. In addition, we examine the process of estimating the error in using these techniques. The Midpoint Rule Earlier in this text we defined the definite integral of a function over an interval as the limit ofRiemann sums. In general, any Riemann sum of a function f⁡(x) over an interval [a,b] may be viewed as an estimate of ∫a b f⁡(x)d x. Recall that a Riemann sum of a function f⁡(x) over an interval [a,b] is obtained by selecting a partition P={x 0,x 1,x 2,…,x n} where a=x 0<x 1<x 2<⋯<x n=b and a set S={x 1∗,x 2∗,…,x n∗} where x i−1≤x i∗≤x i for all i. The Riemann sum corresponding to the partition P and the set S is given by ∑i=1 n f⁡(x i∗)⁢Δ⁢x i, where Δ⁢x i=x i−x i−1, the length of the i th subinterval. The midpoint rule for estimating a definite integral uses a Riemann sum with subintervals of equal width and the midpoints, m i, of each subinterval in place of x i∗. Formally, we state a theorem regarding the convergence of the midpoint rule as follows. The Midpoint Rule Assume that f⁡(x) is continuous on [a,b]. Let n be a positive integer and Δ⁢x=b−a n. If [a,b] is divided into n subintervals, each of length Δ⁢x, and m i is the midpoint of the i th subinterval, set M n=∑i=1 n f⁡(m i)⁢Δ⁢x. Then lim n→∞⁡M n=∫a b f⁡(x)d x. As we can see in Figure 3.6.1, if f⁡(x)≥0 over [a,b], then ∑i=1 n f⁡(m i)⁢Δ⁢x corresponds to the sum of the areas of rectangles approximating the area between the graph of f⁡(x) and the x-axis over [a,b]. The graph shows the rectangles corresponding to M 4 for a nonnegative function over a closed interval [a,b]. Figure 3.6.1: The midpoint rule approximates the area between the graph of f⁡(x) and the x-axis by summing the areas of rectangles with midpoints that are points on f⁡(x). Example 3.6.1: Using the Midpoint Rule with M 4 Use the midpoint rule to estimate ∫0 1 x 2 d x using four subintervals. Compare the result with the actual value of this integral. Solution: Each subinterval has length Δ⁢x=1−0 4=1 4. Therefore, the subintervals consist of [0,1 4],[1 4,1 2],[1 2,3 4],and⁡[3 4,1]. The midpoints of these subintervals are {1 8,3 8,5 8,7 8}. Thus, M 4=1 4⋅f⁡(1 8)+1 4⋅f⁡(3 8)+1 4⋅f⁡(5 8)+1 4⋅f⁡(7 8)=1 4⋅1 64+1 4⋅9 64+1 4⋅25 64+1 4⋅49 64=21 64=0.328125. Since ∫0 1 x 2 d x=1 3, the absolute error in this approximation is: \|1 3−21 64\|=1 192≈0.0052, and we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definite integral. Example 3.6.2: Using the Midpoint Rule with M 6 Use M 6 to estimate the length of the curve y=1 2⁢x 2 on [1,4]. Solution: The length of y=1 2⁢x 2 on [1,4] is s=∫1 4 1+(d⁢y d⁢x)2 d x. Since d⁢y d⁢x=x, this integral becomes ∫1 4 1+x 2 d x. If [1,4] is divided into six subintervals, then each subinterval has length Δ⁢x=4−1 6=1 2 and the midpoints of the subintervals are {5 4,7 4,9 4,11 4,13 4,15 4}. If we set f⁡(x)=1+x 2, M 6=1 2⋅f⁡(5 4)+1 2⋅f⁡(7 4)+1 2⋅f⁡(9 4)+1 2⋅f⁡(11 4)+1 2⋅f⁡(13 4)+1 2⋅f⁡(15 4)≈1 2⁢(1.6008+2.0156+2.4622+2.9262+3.4004+3.8810)=8.1431 units. Exercise 3.6.1 Use the midpoint rule with n=2 to estimate ∫1 2 1 x d x. Hint Δ⁢x=1 2,m 1=5 4,and m 2=7 4. Answer 24 35≈0.685714 The Trapezoidal Rule We can also approximate the value of a definite integral by using trapezoids rather than rectangles. In Figure 3.6.2, the area beneath the curve is approximated by trapezoids rather than by rectangles. Figure 3.6.2: Trapezoids may be used to approximate the area under a curve, hence approximating the definite integral. The trapezoidal rule for estimating definite integrals uses trapezoids rather than rectangles to approximate the area under a curve. To gain insight into the final form of the rule, consider the trapezoids shown in Figure 3.6.2. We assume that the length of each subinterval is given by Δ⁢x. First, recall that the area of a trapezoid with a height of h and bases of length b 1 and b 2 is given by Area=1 2⁢h⁡(b 1+b 2). We see that the first trapezoid has a height Δ⁢x and parallel bases of length f⁡(x 0) and f⁡(x 1). Thus, the area of the first trapezoid in Figure 3.6.2 is 1 2⁢Δ⁢x⁢(f⁡(x 0)+f⁡(x 1)). The areas of the remaining three trapezoids are 1 2⁢Δ⁢x⁢(f⁡(x 1)+f⁡(x 2)),1 2⁢Δ⁢x⁢(f⁡(x 2)+f⁡(x 3)), and 1 2⁢Δ⁢x⁢(f⁡(x 3)+f⁡(x 4)). Consequently, ∫a b f⁡(x)d x≈1 2⁢Δ⁢x⁢(f⁡(x 0)+f⁡(x 1))+1 2⁢Δ⁢x⁢(f⁡(x 1)+f⁡(x 2))+1 2⁢Δ⁢x⁢(f⁡(x 2)+f⁡(x 3))+1 2⁢Δ⁢x⁢(f⁡(x 3)+f⁡(x 4)). After taking out a common factor of 1 2⁢Δ⁢x and combining like terms, we have ∫a b f⁡(x)d x≈Δ⁢x 2⁢[f⁡(x 0)+2⁢f⁡(x 1)+2⁢f⁡(x 2)+2⁢f⁡(x 3)+f⁡(x 4)]. Generalizing, we formally state the following rule. The Trapezoidal Rule Assume that f⁡(x) is continuous over [a,b]. Let n be a positive integer and Δ⁢x=b−a n. Let [a,b] be divided into n subintervals, each of length Δ⁢x, with endpoints at P={x 0,x 1,x 2…,x n}. Set T n=Δ⁢x 2⁢[f⁡(x 0)+2⁢f⁡(x 1)+2⁢f⁡(x 2)+⋯+2⁢f⁡(x n−1)+f⁡(x n)]. Then, lim n→+∞⁡T n=∫a b f⁡(x)d x. Before continuing, let’s make a few observations about the trapezoidal rule. First of all, it is useful to note that T n=1 2⁢(L n+R n) where L n=∑i=1 n f⁡(x i−1)⁢Δ⁢x and R n=∑i=1 n f⁡(x i)⁢Δ⁢x. That is, L n and R n approximate the integral using the left-hand and right-hand endpoints of each subinterval, respectively. In addition, a careful examination of Figure 3.6.3 leads us to make the following observations about using the trapezoidal rules and midpoint rules to estimate the definite integral of a nonnegative function. The trapezoidal rule tends to overestimate the value of a definite integral systematically over intervals where the function is concave up and to underestimate the value of a definite integral systematically over intervals where the function is concave down. On the other hand, the midpoint rule tends to average out these errors somewhat by partially overestimating and partially underestimating the value of the definite integral over these same types of intervals. This leads us to hypothesize that, in general, the midpoint rule tends to be more accurate than the trapezoidal rule. Figure 3.6.3:The trapezoidal rule tends to be less accurate than the midpoint rule. Example 3.6.3: Using the Trapezoidal Rule Use the trapezoidal rule to estimate ∫0 1 x 2 d x using four subintervals. Solution The endpoints of the subintervals consist of elements of the set P={0,1 4,1 2,3 4,1} and Δ⁢x=1−0 4=1 4. Thus, ∫0 1 x 2 d x≈1 2⋅1 4⁢[f⁡(0)+2⁢f⁡(1 4)+2⁢f⁡(1 2)+2⁢f⁡(3 4)+f⁡(1)]=1 8⁢(0+2⋅1 16+2⋅1 4+2⋅9 16+1)=11 32=0.34375 Exercise 3.6.2 Use the trapezoidal rule with n=2 to estimate ∫1 2 1 x d x. Hint Set Δ⁢x=1 2. The endpoints of the subintervals are the elements of the set P={1,3 2,2}. Answer 17 24≈0.708333 Absolute and Relative Error An important aspect of using these numerical approximation rules consists of calculating the error in using them for estimating the value of a definite integral. We first need to define absolute error and relative error. Definition: absolute and relative error If B is our estimate of some quantity having an actual value of A, then the absolute error is given by \|A−B\|. The relative error is the error as a percentage of the actual value and is given by \|A−B A\|⋅100%. Example 3.6.4: Calculating Error in the Midpoint Rule Calculate the absolute and relative error in the estimate of ∫0 1 x 2 d x using the midpoint rule, found in Example 3.6.1. Solution: The calculated value is ∫0 1 x 2 d x=1 3 and our estimate from the example is M 4=21 64. Thus, the absolute error is given by \|1 3−21 64\|=1 192≈0.0052. The relative error is 1/192 1/3=1 64≈0.015625≈1.6%. Example 3.6.5: Calculating Error in the Trapezoidal Rule Calculate the absolute and relative error in the estimate of ∫0 1 x 2 d x using the trapezoidal rule, found in Example 3.6.3. Solution: The calculated value is ∫0 1 x 2 d x=1 3 and our estimate from the example is T 4=11 32. Thus, the absolute error is given by \|1 3−11 32\|=1 96≈0.0104. The relative error is given by 1/96 1/3=0.03125≈3.1%. Exercise 3.6.3 In an earlier checkpoint, we estimated ∫1 2 1 x d x to be 24 35 using M 2. The actual value of this integral is ln⁡2. Using 24 35≈0.6857 and ln⁡2≈0.6931, calculate the absolute error and the relative error. Hint Use the previous examples as a guide. Answer absolute error ≈0.0074, and relative error ≈1.1% Error Bounds on the Midpoint and Trapezoidal Rules In the two previous examples, we were able to compare our estimate of an integral with the actual value of the integral; however, we do not typically have this luxury. In general, if we are approximating an integral, we are doing so because we cannot compute the exact value of the integral itself easily. Therefore, it is often helpful to be able to determine an upper bound for the error in an approximation of an integral. The following theorem provides error bounds for the midpoint and trapezoidal rules. The theorem is stated without proof. Error Bounds for the Midpoint and Trapezoidal Rules Let f⁡(x) be a continuous function over [a,b], having a second derivative f″⁡(x) over this interval. If M is the maximum value of \|f″⁡(x)\| over [a,b], then the upper bounds for the error in using M n and T n to estimate ∫a b f⁡(x)d x are (3.6.1)Error in M n≤M⁢(b−a)3 24⁢n 2 and Error in T n≤M⁢(b−a)3 12⁢n 2. We can use these bounds to determine the value of n necessary to guarantee that the error in an estimate is less than a specified value. Example 3.6.6: Determining the Number of Intervals to Use What value of n should be used to guarantee that an estimate of ∫0 1 e x 2 d x is accurate to within 0.01 if we use the midpoint rule? Solution We begin by determining the value of M, the maximum value of \|f″⁡(x)\| over [0,1] for f⁡(x)=e x 2. Since f′(x)=2⁢x⁢e x 2, we have f″⁡(x)=2⁢e x 2+4⁢x 2⁢e x 2. Thus, \|f″⁡(x)\|=2⁢e x 2⁢(1+2⁢x 2)≤2⋅e⋅3=6⁢e. From the error-bound Equation 3.6.1, we have Error in M n≤M⁢(b−a)3 24⁢n 2≤6⁢e⁢(1−0)3 24⁢n 2=6⁢e 24⁢n 2. Now we solve the following inequality for n: 6⁢e 24⁢n 2≤0.01. Thus, n≥600⁢e 24≈8.24. Since n must be an integer satisfying this inequality, a choice of n=9 would guarantee that \|∫0 1 e x 2 d x−M n\|<0.01. Analysis We might have been tempted to round 8.24 down and choose n=8, but this would be incorrect because we must have an integer greater than or equal to 8.24. We need to keep in mind that the error estimates provide an upper bound only for the error. The actual estimate may, in fact, be a much better approximation than is indicated by the error bound. Exercise 3.6.4 Use Equation 3.6.1 to find an upper bound for the error in using M 4 to estimate ∫0 1 x 2 d x. Hint f″⁡(x)=2, so M=2. Answer 1 192 Simpson’s Rule With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated the curve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linear functions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With Simpson’s rule, we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair of subintervals we approximate ∫x 0 x 2 f⁡(x)d x with ∫x 0 x 2 p⁡(x)d x, where p⁡(x)=A⁢x 2+B⁢x+C is the quadratic function passing through (x 0,f⁡(x 0)),(x 1,f⁡(x 1)), and (x 2,f⁡(x 2)) (Figure 3.6.4). Over the next pair of subintervals we approximate ∫x 2 x 4 f⁡(x)d x with the integral of another quadratic function passing through (x 2,f⁡(x 2)),(x 3,f⁡(x 3)), and (x 4,f⁡(x 4)). This process is continued with each successive pair of subintervals. Figure 3.6.4: With Simpson’s rule, we approximate a definite integral by integrating a piecewise quadratic function. To understand the formula that we obtain for Simpson’s rule, we begin by deriving a formula for this approximation over the first two subintervals. As we go through the derivation, we need to keep in mind the following relationships: f⁡(x 0)=p⁡(x 0)=A⁢x 0 2+B⁢x 0+C f⁡(x 1)=p⁡(x 1)=A⁢x 1 2+B⁢x 1+C f⁡(x 2)=p⁡(x 2)=A⁢x 2 2+B⁢x 2+C x 2−x 0=2⁢Δ⁢x, where Δ⁢x is the length of a subinterval. x 2+x 0=2⁢x 1, since x 1=(x 2+x 0)2. Thus, ∫x 0 x 2 f⁡(x)d x≈∫x 0 x 2 p⁡(x)d x=∫x 0 x 2(A⁢x 2+B⁡x+C)d x=(A 3⁢x 3+B 2⁢x 2+C⁢x)\|x 0 x 2 Find the antiderivative.=A 3⁢(x 2 3−x 0 3)+B 2⁢(x 2 2−x 0 2)+C⁢(x 2−x 0)Evaluate the antiderivative.=A 3⁢(x 2−x 0)⁢(x 2 2+x 2⁢x 0+x 0 2)+B 2⁢(x 2−x 0)⁢(x 2+x 0)+C⁢(x 2−x 0)=x 2−x 0 6⁢(2⁢A⁢(x 2 2+x 2⁢x 0+x 0 2)+3⁢B⁡(x 2+x 0)+6⁢C)Factor out x 2−x 0 6.=Δ⁢x 3⁢((A⁢x 2 2+B⁡x 2+C)+(A⁢x 0 2+B⁡x 0+C)+A⁢(x 2 2+2⁢x 2⁢x 0+x 0 2)+2⁢B⁡(x 2+x 0)+4⁢C)Rearrange the terms. Note:Δ⁢x=x 2−x 0 2=Δ⁢x 3⁢(f⁡(x 2)+f⁡(x 0)+A⁢(x 2+x 0)2+2⁢B⁡(x 2+x 0)+4⁢C)Factor and substitute:f⁡(x 2)=A⁢x 2 2+B⁡x 2+C and f⁡(x 0)=A⁢x 0 2+B⁡x 0+C.=Δ⁢x 3⁢(f⁡(x 2)+f⁡(x 0)+A⁢(2⁢x 1)2+2⁢B⁡(2⁢x 1)+4⁢C)Substitute x 2+x 0=2⁢x 1.Note:x 1=x 2+x 0 2 is the midpoint.=Δ⁢x 3⁢(f⁡(x 2)+4⁢f⁡(x 1)+f⁡(x 0)).Expand and substitute f⁡(x 1)=A⁢x 1 2+B⁡x 1+C. If we approximate ∫x 2 x 4 f⁡(x)d x using the same method, we see that we have ∫x 2 x 4 f⁡(x)d x≈Δ⁢x 3⁢(f⁡(x 4)+4⁢f⁡(x 3)+f⁡(x 2)). Combining these two approximations, we get ∫x 0 x 4 f⁡(x)d x≈Δ⁢x 3⁢(f⁡(x 0)+4⁢f⁡(x 1)+2⁢f⁡(x 2)+4⁢f⁡(x 3)+f⁡(x 4)). The pattern continues as we add pairs of subintervals to our approximation. The general rule may be stated as follows. Simpson’s Rule Assume that f⁡(x) is continuous over [a,b]. Let n be a positive even integer and Δ⁢x=b−a n. Let [a,b] be divided into n subintervals, each of length Δ⁢x, with endpoints at P={x 0,x 1,x 2,…,x n}. Set S n=Δ⁢x 3⁢[f⁡(x 0)+4⁢f⁡(x 1)+2⁢f⁡(x 2)+4⁢f⁡(x 3)+2⁢f⁡(x 4)+⋯+2⁢f⁡(x n−2)+4⁢f⁡(x n−1)+f⁡(x n)]. Then, lim n→+∞⁡S n=∫a b f⁡(x)d x. Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson’s rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that S 2⁢n=(2 3)⁢M n+(1 3)⁢T n. It is also possible to put a bound on the error when using Simpson’s rule to approximate a definite integral. The bound in the error is given by the following rule: Rule: Error Bound for Simpson’s Rule Let f⁡(x) be a continuous function over [a,b] having a fourth derivative, f(4)⁡(x), over this interval. If M is the maximum value of ∣f(4)⁡(x)∣ over [a,b], then the upper bound for the error in using S n to estimate ∫a b f⁡(x)d x is given by Error in S n≤M⁢(b−a)5 180⁢n 4. Example 3.6.7: Applying Simpson’s Rule 1 Use S 2 to approximate ∫0 1 x 3 d x. Estimate a bound for the error in S 2. Solution Since [0,1] is divided into two intervals, each subinterval has length Δ⁢x=1−0 2=1 2. The endpoints of these subintervals are {0,1 2,1}. If we set f⁡(x)=x 3, then S 2=1 3⋅1 2⁢(f⁡(0)+4⁢f⁡(1 2)+f⁡(1))=1 6⁢(0+4⋅1 8+1)=1 4. Since f(4)⁡(x)=0 and consequently M=0, we see that Error in S 2≤0⁢(1)5 180⋅2 4=0. This bound indicates that the value obtained through Simpson’s rule is exact. A quick check will verify that, in fact, ∫0 1 x 3 d x=1 4. Example 3.6.8: Applying Simpson’s Rule 2 Use S 6 to estimate the length of the curve y=1 2⁢x 2 over [1,4]. Solution The length of y=1 2⁢x 2 over [1,4] is ∫1 4 1+x 2 d x. If we divide [1,4] into six subintervals, then each subinterval has length Δ⁢x=4−1 6=1 2, and the endpoints of the subintervals are {1,3 2,2,5 2,3,7 2,4}. Setting f⁡(x)=1+x 2, S 6=1 3⋅1 2⁢(f⁡(1)+4⁢f⁡(3 2)+2⁢f⁡(2)+4⁢f⁡(5 2)+2⁢f⁡(3)+4⁢f⁡(7 2)+f⁡(4)). After substituting, we have S 6=1 6⁢(1.4142+4⋅1.80278+2⋅2.23607+4⋅2.69258+2⋅3.16228+4⋅3.64005+4.12311)≈8.14594 units. Exercise 3.6.5 Use S 2 to estimate ∫1 2 1 x d x. Hint S 2=1 3⁢Δ⁢x⁢(f⁡(x 0)+4⁢f⁡(x 1)+f⁡(x 2)) Answer 25 36≈0.694444 Key Concepts We can use numerical integration to estimate the values of definite integrals when a closed form of the integral is difficult to find or when an approximate value only of the definite integral is needed. The most commonly used techniques for numerical integration are the midpoint rule, trapezoidal rule, and Simpson’s rule. The midpoint rule approximates the definite integral using rectangular regions whereas the trapezoidal rule approximates the definite integral using trapezoidal approximations. Simpson’s rule approximates the definite integral by first approximating the original function using piecewise quadratic functions. Key Equations Midpoint rule M n=∑i=1 n f⁡(m i)⁢Δ⁢x Trapezoidal rule T n=Δ⁢x 2⁢[f⁡(x 0)+2⁢f⁡(x 1)+2⁢f⁡(x 2)+⋯+2⁢f⁡(x n−1)+f⁡(x n)] Simpson’s rule S n=Δ⁢x 3⁢[f⁡(x 0)+4⁢f⁡(x 1)+2⁢f⁡(x 2)+4⁢f⁡(x 3)+2⁢f⁡(x 4)+4⁢f⁡(x 5)+⋯+2⁢f⁡(x n−2)+4⁢f⁡(x n−1)+f⁡(x n)] Error bound for midpoint rule Error in M n≤M⁢(b−a)3 24⁢n 2, where M is the maximum value of \|f″⁡(x)\| over [a,b]. Error bound for trapezoidal rule Error in T n≤M⁢(b−a)3 12⁢n 2, where M is the maximum value of \|f″⁡(x)\| over [a,b]. Error bound for Simpson’s rule Error in S n≤M⁢(b−a)5 180⁢n 4, where M is the maximum value of ∣f(4)⁡(x)∣ over [a,b]. Glossary absolute errorif B is an estimate of some quantity having an actual value of A, then the absolute error is given by \|A−B\|midpoint rulea rule that uses a Riemann sum of the form M n=∑i=1 n f⁡(m i)⁢Δ⁢x, where m i is the midpoint of the i th subinterval to approximate ∫a b f⁡(x)d xnumerical integrationthe variety of numerical methods used to estimate the value of a definite integral, including the midpoint rule, trapezoidal rule, and Simpson’s rulerelative errorerror as a percentage of the actual value, given by relative error=\|A−B A\|⋅100%Simpson’s rulea rule that approximates ∫a b f⁡(x)d x using the area under a piecewise quadratic function. The approximation S n to ∫a b f⁡(x)d x is given by S n=Δ⁢x 3⁢(f⁡(x 0)+4⁢f⁡(x 1)+2⁢f⁡(x 2)+4⁢f⁡(x 3)+2⁢f⁡(x 4)+⋯+2⁢f⁡(x n−2)+4⁢f⁡(x n−1)+f⁡(x n)).trapezoidal rule a rule that approximates ∫a b f⁡(x)d x using the area of trapezoids. The approximation T n to ∫a b f⁡(x)d x is given by T n=Δ⁢x 2⁢(f⁡(x 0)+2⁢f⁡(x 1)+2⁢f⁡(x 2)+⋯+2⁢f⁡(x n−1)+f⁡(x n)). Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at Edited by Paul Seeburger (Monroe Community College). Notes added to development of area under a parabola and typos fixed in original text. This page titled 3.6: Numerical Integration is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 3.5: Other Strategies for Integration 3.7: Improper Integrals Was this article helpful? Yes No Recommended articles 7.6: Numerical IntegrationThe antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). ... 3.6: Numerical IntegrationThe antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). ... 6.1: Numerical IntegrationThe antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). ... 8.6: Numerical IntegrationThe antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). ... 8.6: Numerical IntegrationThe antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). ... 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1215	https://fabiandablander.com/r/Fibonacci.html	Fabian Dablander Postdoc Behavioural Data Science & Sustainability The Fibonacci sequence and linear algebra Leonardo Bonacci, better known as Fibonacci, has influenced our lives profoundly. At the beginning of the 13th13th century, he introduced the Hindu-Arabic numeral system to Europe. Instead of the Roman numbers, where I stands for one, V for five, X for ten, and so on, the Hindu-Arabic numeral system uses position to index magnitude. This leads to much shorter expressions for large numbers.1 While the history of the numerical system is fascinating, this blog post will look at what Fibonacci is arguably most well known for: the Fibonacci sequence. In particular, we will use ideas from linear algebra to come up with a closed-form expression of the nthnth Fibonacci number2. On our journey to get there, we will also gain some insights about recursion in R.3 The rabbit puzzle In Liber Abaci, Fibonacci poses the following question (paraphrasing): Suppose we have two newly-born rabbits, one female and one male. Suppose these rabbits produce another pair of female and male rabbits after one month. These newly-born rabbits will, in turn, also mate after one month, producing another pair, and so on. Rabbits never die. How many pairs of rabbits exist after one year? The Figure below illustrates this process. Every point denotes one rabbit pair over time. To indicate that every newborn rabbit pair needs to wait one month before producing new rabbits, rabbits that are not fertile yet are coloured in grey, while rabbits ready to procreate are coloured in red. We can derive a linear recurrence relation that describes the Fibonacci sequence. In particular, note that rabbits never die. Thus, at time point nn, all rabbits from time point n−1n−1 carry over. Additionally, we know that every fertile rabbit pair will produce a new rabbit pair. However, they have to wait one month, so that the amount of fertile rabbits equals the amount of rabbits at time point n−2n−2. Resultingly, the Fibonacci sequence {FnFn}∞n=1∞n=1 is: Fn=Fn−1+Fn−2,Fn=Fn−1+Fn−2, for n≥3n≥3 and F1=F2=1F1=F2=1. Before we derive a closed-form expression that computes the nthnth Fibonacci number directly, in the next section, we play around with alternative, more straightforward solutions in R. Implementation in R We can write a wholly inefficient, but beautiful program to compute the nthnth Fibonacci number: R takes roughly 5 seconds to compute the 30th30th Fibonacci number; computing the 40th40th number exhausts my patience. This recursive solution is not particularly efficient because R executes the function an unnecessary amount of times. For example, the call tree for fib(5) is: fib(5) fib(4) + fib(3) (fib(3) + fib(2)) + (fib(2) + fib(1)) ((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1)) ((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1)) which shows that fib(2) was called three times. This is not necessary, as we can store the outcome of this function call instead of recomputing it every time. This technique is called memoization (see also the R package memoise). Implementing this leads to: This computes the 1000th1000th Fibonacci in a tenth of a second. We can, of course, write this sequentially, and also store all intermediate Fibonacci numbers. This also avoids memory issues brought about by the recursive implementation. Interestingly, although this algorithm seems like it should be O(n)O(n), it is actually O(n2)O(n2) since we are adding increasingly large numbers (for more on this, see here). The first 30 Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040. This is a rapid increase, as made apparent by the left Figure below. The Figure on the right shows that there is structure in how the sequence grows. We will return to the structure in growth at the end of the blog post. First, we need to derive a closed-form expression of the nthnth Fibonacci number. In the next section, we take a step towards that by realizing that diagonal matrices make for easier computations. Diagonal matrices are good Our goal is to get a closed form expression of the nthnth Fibonacci number. The first thing to note is that, due to linear recursion, we can view the Fibonacci numbers as applying a linear map. In particular, define T∈L(R2)T∈L(R2) by: T(x,y)=(y,x+y).T(x,y)=(y,x+y). We note that: Tn(0,1)=(Fn,Fn+1),Tn(0,1)=(Fn,Fn+1), which we will prove by induction. In particular, note that the base case n=1n=1: T1(0,1)=(1,0+1)=(1,1)=(F1,F2),T1(0,1)=(1,0+1)=(1,1)=(F1,F2), does in fact give the first two Fibonacci numbers. Now for the induction step: we assume that this holds for an arbitrary nn, and we show that it holds for n+1n+1 using the following: Tn(0,1)=(Fn,Fn+1)T(Tn(0,1))=T(Fn,Fn+1)Tn+1(0,1)=(Fn+1,Fn+Fn+1)Tn+1(0,1)=(Fn+1,Fn+2).Tn(0,1)=(Fn,Fn+1)T(Tn(0,1))=T(Fn,Fn+1)Tn+1(0,1)=(Fn+1,Fn+Fn+1)Tn+1(0,1)=(Fn+1,Fn+2). The last equality follows from the definition of the Fibonacci sequence, i.e., the fact that any number is equal to the sum of the previous two numbers. The matrix of this linear map with respect to the standard basis is given by: A≡M(T)=(0111),A≡M(T)=(0111), since T(1,0)=(0,1)T(1,0)=(0,1) and T(0,1)=(1,1)T(0,1)=(1,1). Observe that: (0111)(xy)=(yx+y).(0111)(xy)=(yx+y). In the sequential R code for computing the Fibonacci numbers, we have applied the linear map nn times, which gave us the Fibonacci number we were interested in. We can write this in matrix form: (0111)n(01)=(FnFn+1).(0111)n(01)=(FnFn+1). If you were to compute, say, the 3th3th Fibonacci number using this equation, you would have to multiply AA three times with itself. Now assume you had something like: (λ100λ2)n(01)=(FnFn+1).(λ100λ2)n(01)=(FnFn+1). Using the above equation, the matrix powers would become trivial: (λ100λ2)n=(λn100λn2).(λ100λ2)n=(λn100λn2). There would be no need to repeatedly engage in matrix multiplication; instead, we would arrive at the nthnth Fibonacci number using only scalar multiplication! Our task is thus as follows: find a new matrix for the linear map which is diagonal. To solve this, we will need eigenvalues and eigenvectors. Finding eigenvalues and eigenvectors An eigenvector-eigenvalue pair (v,λ)(v,λ) satisfies for v≠0v≠0 that: Tv=λv,Tv=λv, which means that for a particular vector vv, the linear map only stretches the vector by a constant λλ. Here’s the key: using the eigenvectors as basis, the matrix of the linear map is diagonal. This is because the matrix of our linear map, AA, is defined by: Tv1=A11v1+A21v2Tv2=A12v1+A22v2.Tv1=A11v1+A21v2Tv2=A12v1+A22v2. Now since the basis consists only of eigenvectors, we know that Tv1=λv1Tv1=λv1 and Tv2=λv2Tv2=λv2, which implies that A11=λ1A11=λ1 and A21=0A21=0, as well as A12=0A12=0 and A22=λ2A22=λ2. For a wonderful explanation of eigenvalues and eigenvectors, see this video by 3Blue1Brown.4 In order to find the eigenvalues and eigenvectors, note that the linear map satisfies the following two equations: T(x,y)=λ(x,y)T(x,y)=(y,x+y).T(x,y)=λ(x,y)T(x,y)=(y,x+y). This leads to: λx=yλy=x+y.λx=yλy=x+y. We substitute the first expression into the second one, yielding: λ2x=x+y(λ2−1)x=y,λ2x=x+y(λ2−1)x=y, which we now substitute into the first equation, which results in: λx=(λ2−1)x0=λ2−λ−1.λx=(λ2−1)x0=λ2−λ−1. We can now apply the quadratic formula or “Mitternachtsformel”, as it is called in parts of Germany because students should know the formula when they are roused from sleep at midnight. We are neither in Germany, nor is it midnight, nor can I actually remember the formula, so let’s quickly derive it for our problem: λ2−λ−1=0λ2−λ=14λ2−4λ=44λ2−4λ+1=4+1(2λ−1)2=4+12λ−1=±√4+1λ=1±√52.λ2−λ−1=0λ2−λ=14λ2−4λ=44λ2−4λ+1=4+1(2λ−1)2=4+12λ−1=±√4+1λ=1±√52. Now that we have found both eigenvalues, we go hunting for the eigenvectors! We put the eigenvalue into the equations from above: 1±√52x=y1±√52y=x+y.1±√52x=y1±√52y=x+y. If we set x=1x=1, then y=1±√52y=1±√52. Thus, two eigenvectors are: v1=(1,1+√52)v2=(1,1−√52).v1=(1,1+√52)v2=(1,1−√52). As a sanity check to see whether this is indeed true, we check whether Tv1=λ1v1Tv1=λ1v1: Tv1=(1+√52,1+1+√52)λv1=1+√52(1,1+√52)=(1+√52,(1+√52)2)=(1+√52,1+2√5+54)=(1+√52,32+√52)=(1+√52,1+1+√52),Tv1=(1+√52,1+1+√52)λv1=1+√52(1,1+√52)=⎛⎝1+√52,(1+√52)2⎞⎠=(1+√52,1+2√5+54)=(1+√52,32+√52)=(1+√52,1+1+√52), which shows that the two expression are equal. Moreover, the dot product of the two eigenvectors is zero, which means that the two eigenvectors are linearly independent (as they should be). In the next section, we will find that the same territory can be described by different maps. Change of basis Now that we have found the eigenvalues and eigenvectors, we can create the matrix DD of the linear map TT which is diagonal with respect to the basis of eigenvectors: D=(1+√52001−√52).D=⎛⎝1+√52001−√52⎞⎠. We are not done yet, however. Note that DD is the matrix of the linear map TT with respect to the basis that consists of both eigenvectors v1v1 and v2v2, not with respect to the standard basis. We have changed our coordinate system — our map — as indicated by the Figure below; the black coloured vectors are the standard basis vectors while the vectors coloured in red are our new basis vectors. To build some intuition, let’s play around with representing ωω in both the standard basis and our new eigenbasis. Any vector is a linear combination of the basis vectors. Let a1a1 and a2a2 be the coefficients for the standard basis such that: ω=a1(10)+a2(01).ω=a1(10)+a2(01). Now because I have drawn it earlier, I know that a1=−1a1=−1 and a2=0.3a2=0.3. This is the representation of ωω in the standard basis. How do we represent it in our eigenbasis? Well, using the eigenbasis the vector ωω is still a linear combination of the basis vectors, but with different coefficients; denote them as b1b1 and b2b2. We thus have: ω=b1(11+√52)+b2(11−√52)=a1(10)+a2(01).ω=b1(11+√52)+b2(11−√52)=a1(10)+a2(01). If we write this in matrix form, we have: (111+√521−√52)(b1b2)=(1001)(a1a2)(b1b2)=(111+√521−√52)−1(1001)(a1a2).(111+√521−√52)(b1b2)=(1001)(a1a2)(b1b2)=(111+√521−√52)−1(1001)(a1a2). Thus, we can represent a vector aa with basis SS in our new basis EE by computing: b=E−1Sa.b=E−1Sa. In our eigenbasis, the vector ωω has the coordinates: This means we have the representation: ω=−0.14(11+√52)−0.86(11−√52),ω=−0.14(11+√52)−0.86(11−√52), which makes intuitive sense when you look at the Figure above. For another beautiful linear algebra video by 3Blue1Brown, this time about changing bases, see here. In the next section, we will use what we have learned above to express the nthnth Fibonacci number in closed-form. Closed-form Fibonacci Recall from above that our solution to finding the nthnth Fibonacci number in matrix form is: (0111)n(01)=(FnFn+1).(0111)n(01)=(FnFn+1). Now, we have swapped the non-diagonal matrix AA with the diagonal matrix DD by changing the basis from the standard basis to the eigenbasis. However, the vector (0,1)T(0,1)T is still in the standard basis! In order to change its representation to the eigenbasis, we multiply it with E−1E−1, as discussed above. We write: (1+√52001−√52)n(111+√521−√52)−1(01)=(FnFn+1).⎛⎝1+√52001−√52⎞⎠n(111+√521−√52)−1(01)=(FnFn+1). Let’s use this to compute, say, the 10th10th Fibonacci number (which is 55) in R: Ha! This didn’t quite work, did it? We got the answer for F10F10 roughly when rounding, but F11F11 is completely off. What did we miss? Well, this is in fact the correct answer — it is just in the wrong basis! We have to convert this from the eigenbasis to the standard basis. To do this, observe that: b=E−1SaEb=SaEb=a,b=E−1SaEb=SaEb=a, since SS is the identity matrix. Thus, all we have to do is to multiply with EE: which is the correct solution. To get the closed-form solution algebraically, we first invert the matrix EE: E−1=−1√5(1−√52−1−1+√521),E−1=−1√5⎛⎝1−√52−1−1+√521⎞⎠, and we write: (FnFn+1)=(111+√521−√52)(1+√52001−√52)n−1√5(1−√52−1−1+√521)(01)=−1√5(111+√521−√52)(1+√52001−√52)n(−11)=−1√5(111+√521−√52)(−(1+√52)n(1−√52)n)=−1√5(−(1+√52)n+(1−√52)n−(1+√52)n+1+(1−√52)n+1)=1√5((1+√52)n−(1−√52)n(1+√52)n+1−(1−√52)n+1).(FnFn+1)=(111+√521−√52)⎛⎝1+√52001−√52⎞⎠n−1√5⎛⎝1−√52−1−1+√521⎞⎠(01)=−1√5(111+√521−√52)⎛⎝1+√52001−√52⎞⎠n(−11)=−1√5(111+√521−√52)⎛⎜ ⎜⎝−(1+√52)n(1−√52)n⎞⎟ ⎟⎠=−1√5⎛⎜ ⎜⎝−(1+√52)n+(1−√52)n−(1+√52)n+1+(1−√52)n+1⎞⎟ ⎟⎠=1√5⎛⎜ ⎜⎝(1+√52)n−(1−√52)n(1+√52)n+1−(1−√52)n+1⎞⎟ ⎟⎠. The closed-form expression of the nthnth Fibonacci number is thus given by: Fn=1√5[(1+√52)n−(1−√52)n].Fn=1√5[(1+√52)n−(1−√52)n]. We verify this in R: The golden ratio In the above section, we have derived a closed-form expression of the nthnth Fibonacci number. In this section, we return to an observation we have made at the beginning: there is structure in how the Fibonacci numbers grow. Johannes Kepler, after whom the university in my home town is named, (re)discovered that: limn→∞Fn+1Fn=limn→∞1√5[(1+√52)n+1−(1−√52)n+1]1√5[(1+√52)n−(1−√52)n]=limn→∞(1+√52)n+1−(1−√52)n+1(1+√52)n−(1−√52)n=(1+√52)n+1(1+√52)n=1+√52≈1.618,limn→∞Fn+1Fn=limn→∞1√5[(1+√52)n+1−(1−√52)n+1]1√5[(1+√52)n−(1−√52)n]=limn→∞(1+√52)n+1−(1−√52)n+1(1+√52)n−(1−√52)n=(1+√52)n+1(1+√52)n=1+√52≈1.618, which is the golden ratio. The golden ratio ϕϕ denotes that the ratio of two parts is equal to the ratio of the sum of the parts to the larger part, i.e., for a>b>0a>b>0: ϕ≡ab=a+ba.ϕ≡ab=a+ba. We have observed this empirically in the first Figure, which visualized the differences in the log of two consecutive Fibonacci numbers, and which yielded already for small nn: logFn+1−logFn=logFn+1Fn≈0.4812,logFn+1−logFn=logFn+1Fn≈0.4812, which exponentiated yields the golden ratio. Observe that (1−√52)n(1−√52)n goes to zero very quickly as nn grows so that we can compute the nthnth Fibonacci number by: Fn=⌊1√5ϕn⌉,Fn=⌊1√5ϕn⌉, where we simply round to the nearest integer. To finally answer Fibonacci’s puzzle: After a mere twelve months of incest, there are 144 rabbit pairs There are various generalizations of the Fibonacci sequence. One such generalization is to allow higher orders kk in the sequence, which for k=3k=3 is known as the Tribonacci sequence. Our approach for k=2k=2 can be straightforwardly generalized to account for any order kk (if you want to go down a rabbit hole, see for example this). Conclusion In this blog post, we have taken a detailed look at the Fibonacci sequence. In particular, we saw that it is the answer to a puzzle about procreating rabbits, and how to speed up a recursive algorithm for finding the nthnth Fibonacci number. We then used ideas from linear algebra to arrive at a closed-form expression of the nthnth Fibonacci number. Specifically, we have noted that the Fibonacci sequence is a linear recurrence relation — it can be viewed as repeatedly applying a linear map. With this insight, we observed that the matrix of the linear map is non-diagonal, which makes repeated execution tedious; diagonal matrices, on the other hand, are easy to multiply. We arrived at a diagonal matrix by changing the basis from the standard basis to the basis of eigenvectors, which led to a diagonal matrix of eigenvalues for the linear map. With this representation, the nthnth Fibonacci number is available in closed-form. In order to get it into the standard basis, we had to change basis back from the eigenbasis. We also saw how the Fibonacci numbers relate to the golden ratio ϕϕ. I would like to thank Don van den Bergh, Jonas Haslbeck, and Sophia Crüwell for helpful comments on this blog post. Footnotes This is the main reason why the Hinu-Arabic numeral system took over. The belief that it is easier to multiply and divide using Hindu-Arabic numerals is incorrect. ↩ This blog post is inspired by exercise 16 on p. 161 in Linear Algebra Done Right. ↩ I have learned that there is already (very good) ink spilled on this topic, see for example here and here. A nice essay is also this piece by Steve Strogatz, who, by the way, wrote a wonderful book called Sync. He’s also been on Sean Carroll’s Mindscape podcast, listen here. ↩ If you forget everything that is written in this blog post, but through it were made aware of the videos by 3Blue1Brown (or Grant Sanderson, as he is known in the real world), then I consider this blog post a success. ↩ The downside of the closed-form solution is that it is difficult to calculate the power of the square root with high accuracy. In fact, fib_golden is incorrect for n>70n>70. Our fib_mem implementation is also incorrect, but only for n>93n>93. (I’ve compared it against Fibonacci numbers calculated from here). ↩ Written on July 28th, 2019 by Fabian Dablander Feel free to share!
1216	https://math.stackexchange.com/questions/975565/for-small-x-one-has-ln1x-x	limits - For small $x$, one has $\ln(1+x)=x$? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more For small x x, one has ln(1+x)=x ln⁡(1+x)=x? [duplicate] Ask Question Asked 10 years, 11 months ago Modified5 years, 10 months ago Viewed 3k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This question already has answers here: Analytic solution for x=ln(x+1)x=ln⁡(x+1) (3 answers) Closed 5 years ago. What does it mean that for small x x, one has ln(1+x)=x ln⁡(1+x)=x? How can you explain this thing ? Thanks in advance for your reply. limits logarithms Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 15, 2014 at 20:40 Nathaniel Bubis 33.6k 10 10 gold badges 88 88 silver badges 147 147 bronze badges asked Oct 15, 2014 at 20:35 wonderingdevwonderingdev 1,807 2 2 gold badges 18 18 silver badges 28 28 bronze badges 1 1 Look at the first order Taylor approximation.user137731 –user137731 2014-10-15 20:39:19 +00:00 Commented Oct 15, 2014 at 20:39 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Since no one else said it clearly in words. One does not have that ln(1+x)=x ln⁡(1+x)=x for small x x. One does, however, have that for small values of x x, ln(1+x)ln⁡(1+x) can be approximated by x x. As the other answers have already pointed out, this you see from the Taylor expansion ln(1+x)=∑n=1∞(−1)n−1 x n n=x−1 2 x 2+1 3 x 3−….ln⁡(1+x)=∑n=1∞(−1)n−1 x n n=x−1 2 x 2+1 3 x 3−…. Now if x x is a small enough number, then x 2,x 3,…x 2,x 3,… are all insignificant. And so for small x x you can approximate ln(1+x)ln⁡(1+x) by x x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 15, 2014 at 20:49 ThomasThomas 44.6k 13 13 gold badges 76 76 silver badges 139 139 bronze badges 0 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Take the tangent line at of f(x)=ln(1+x)f(x)=ln⁡(1+x) in x=0 x=0. f(x)≈f(0)+f′(0)(x−0)=ln(1+0)+[d d x ln(1+x)]x=0(x−0)=0+1 x=x f(x)≈f(0)+f′(0)(x−0)=ln⁡(1+0)+[d d x ln⁡(1+x)]x=0(x−0)=0+1 x=x Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 15, 2014 at 20:37 hickslebummbummhickslebummbumm 1,471 8 8 silver badges 11 11 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. If you write its Taylor's expansion then you have: ln(1+x)=∑∞k=1(−1)k−1 x k k ln⁡(1+x)=∑k=1∞(−1)k−1 x k k. For small values of x x, the values of x 2,x 3,...x 2,x 3,... are small in comparing with x x(note that positive numbers which are less than 1, will decrees as we multiply themselves), hence we can ignore the terms with degree larger than 1 1, and estimate ln(1+x)ln⁡(1+x) as its first degree part which is x x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 15, 2014 at 20:43 CLAUDECLAUDE 2,459 17 17 silver badges 31 31 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Look at the equivalent exp(x)=1+x exp(x)=1+x (x x teensy) and compare it to the Taylor series. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 15, 2014 at 20:47 user137731 answered Oct 15, 2014 at 20:41 Syd HendersonSyd Henderson 2,101 1 1 gold badge 11 11 silver badges 9 9 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Hint: One can prove that f(x)≈f(a)+f′(a)(x−a).f(x)≈f(a)+f′(a)(x−a). for x x near a a. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 15, 2014 at 20:37 HakimHakim 10.3k 7 7 gold badges 41 41 silver badges 61 61 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. First, let me parrot Thomas' answer: it is not true that log(1+x)=x log⁡(1+x)=x for small x x. This equation holds only when x=0 x=0. However, it is true that if x x is small, then log(1+x)log⁡(1+x) is "well-approximated" by x x. Slightly more formally, the function x↦x x↦x is the best linear approximation of the function x↦log(1+x)x↦log⁡(1+x) for values of x x which are "near zero". Derivatives and Linear Approximation The essential idea is that if a function f f is differentiable, then the best linear approximation of that function near a point a a is the line with slope f′(a)f′(a) which passes through the point (a,f(a))(a,f(a)). An equation for this line is given by f(x)≈f(a)+f′(a)(x−a)=:ℓ(x).f(x)≈f(a)+f′(a)(x−a)=:ℓ(x). What this means is that if x x is "near" a a, then the value of f(x)f(x) is "near" the value of the function ℓ(x)ℓ(x), where the the graph of ℓ ℓ is a line. All of this can be made rigorous by careful examination of the definition of the derivative. As others have pointed out, taking f(x)=log(1+x)f(x)=log⁡(1+x) and a=0 a=0, this becomes log(1+x)≈log(1+0)+[d d x log(1+x)]x=0(x−0)=log(1)+[1 1+x]x=0 x=0+1 1+0 x=x.log⁡(1+x)≈log⁡(1+0)+[d d x log⁡(1+x)]x=0(x−0)=log⁡(1)+[1 1+x]x=0 x=0+1 1+0 x=x. Hence when x x is "near" 0 0, then log(1+x)log⁡(1+x) is "near" x x. As others have noted, a better approximation for the value of a function "near" some point can be obtained by evaluating the Taylor series expansion of the function near that point—this is the approach of Thomas' answer. From the Integral Definition of log log The definition of log(x)log⁡(x) which I find most compelling is the following: for all x∈(0,∞)x∈(0,∞), define log(x):=∫x 1 1 t d t.log⁡(x):=∫1 x 1 t d t. Note that we can use this definition to obtain all of the other properties we might like (e.g. log log is the inverse of exp exp; log log has a particular Taylor series expansion; etc). I want to use this definition to show that when x x is "small", then the error \|log(1+x)−x\|\|log⁡(1+x)−x\| is also "small". For any x∈[0,∞)x∈[0,∞), we have \|log(1+x)−x\|=\|x−log(1+x)\|=∣∣∣x−∫1+x 1 1 t d t∣∣∣=∣∣∣∫1+x 1 1 d t−∫1+x 1 1 t d t∣∣∣=∫1+x 1 t−1 t d t≤∫1+x 1 t−1 d t=[1 2 t 2]1+x t=1−x=1 2(1+x)2−1 2−x=1 2+x+1 2 x 2−1 2−x=1 2 x 2.(t≥1⟹t−1 t≥0, thus\|⋅\|is redundant)(t≥1⟹t−1 t≤t−1)\|log⁡(1+x)−x\|=\|x−log⁡(1+x)\|=\|x−∫1 1+x 1 t d t\|=\|∫1 1+x 1 d t−∫1 1+x 1 t d t\|=∫1 1+x t−1 t d t(t≥1⟹t−1 t≥0, thus\|⋅\|is redundant)≤∫1 1+x t−1 d t(t≥1⟹t−1 t≤t−1)=[1 2 t 2]t=1 1+x−x=1 2(1+x)2−1 2−x=1 2+x+1 2 x 2−1 2−x=1 2 x 2. This means that if x≥0 x≥0, then the error between log(1+x)log⁡(1+x) and the approximation x x will be smaller than 1 2 x 2 1 2 x 2. If x x is close to zero (say, x<1 x<1), then 1 2 x 2 1 2 x 2 will be smaller than x x, so the approximation here looks pretty good. For x∈(−1,0)x∈(−1,0), the situation is a little more complicated, as the integral "blows up" as x→−1 x→−1. Hence we need to constrain x x a little more. Since the goal is to show that if x x is "near" 0 0, then log(1+x)log⁡(1+x) is "near" x x, we might as well assume that x≥−1 2 x≥−1 2. Under this assumption, we have \|log(1+x)−x\|=∫1+x 1 t−1 t d t≤2∫1+x 1 t−1 d t=x 2.(as above)(t≥1+x≥1 2⟹t−1 t≤2(t−1))\|log⁡(1+x)−x\|=∫1 1+x t−1 t d t(as above)≤2∫1 1+x t−1 d t(t≥1+x≥1 2⟹t−1 t≤2(t−1))=x 2. This says that if x∈(−1 2,0)x∈(−1 2,0), then the error between log(1+x)log⁡(1+x) and x x is smaller than x 2 x 2. Again, when x x is small, x 2 x 2 is even smaller. Summarizing the above, we have the following relatively "nice" statement: If \|x\|≤1 2\|x\|≤1 2, then the error between log(1+x)log⁡(1+x) and x x is smaller than x 2 x 2. That is, error(x)=\|log(1+x)−x\|≤x 2.error⁡(x)=\|log⁡(1+x)−x\|≤x 2. The closer x x is to zero, the smaller this error will be. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Nov 29, 2019 at 21:17 Xander Henderson♦Xander Henderson 32.8k 25 25 gold badges 73 73 silver badges 122 122 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions limits logarithms See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 1Analytic solution for x=ln(x+1)x=ln⁡(x+1) Related 1limit of f n(a)=a f n−1(a)f n(a)=a f n−1(a) as n n approaches infinity for small values of a a 3The limit is that which is neither too big nor too small to be the limit. 0If f f has both one-way limits on [a,b][a,b] , then f f is bounded 3Which one is bigger for large value of n n? 2Does the following function has a maximum? 0Function has only one solution 0Why does this tend to ln\|2\|ln⁡\|2\| 1The graph of f(x)f(x) lies below this line for small values of x x Hot Network Questions What happens if you miss cruise ship deadline at private island? 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1217	https://flexbooks.ck12.org/cbook/ck-12-elementary-intermediate-college-algebra/section/7.6/primary/lesson/factoring-quadratics-when-the-leading-coefficient-is-not-equal-to-1-c-alg/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 7.6 Factoring Quadratics When the Leading Coefficient Is Not Equal to 1 Written by:Liz Krams Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 The height achieved by a person when shot out of a cannon at the circus is , where t is time in seconds and height is measured in feet. Approximately how long was the human cannonball in the air? We can use factoring to help us determine this, and we discuss factoring quadratic expressions where the leading coefficient is not equal to 1 in this section. Factoring Quadratic Expressions When the Leading Coefficient is Not Equal to 1 By Grouping Factoring by grouping is an algebraic technique in which you collect terms that have a greatest common factor in groups, and then factor out the GCF from just those terms. If you can factor by grouping, there will be another GCF to factor out, and hopefully, what remains are prime factors. In the previous section, we used a shortcut to factor when the leading coefficient was 1. Since the leading coefficient is not equal to 1, we cannot use the shortcut here. Example 1 Factor . Solution: Let's consider this in steps. Find and the factors of this number that add up to . Notice this is a similar 1st step to the previous section, where the leading coefficient or a was 1: . In this example, we are looking for numbers that multiply to -20 and add to 8. Since the middle term is positive, the factors that work are 10 and -2. Next, we want to rewrite the trinomial with the x-term expanded, using the two factors from Step 1. In this example, we replace the 8xwith 10x + -2x. Subsequently, we want to group the 1st two and 2nd two terms together. Then we want to find the GCF of each group. Notice that after following these steps, we have a GCF of that we can factor out from both terms. What happens if we list before in Step 2? This tells us it does not matter which x-term we list 1st in Step 2 above. This video by Mathispower4u demonstrates how to factor trinomials when the leading coefficient is not equal to 1. Example 2 Factor . Solution: There is no GCF, so we find the factors of that add up to . In this example, we are looking for numbers that multiply to 45 and add to 14. The factors of 45 that add up to -4 are -9 and 5, since the larger number in absolute value has to be negative. We split up the middle term to create two groups. Now, working with each of the groups, we have Example 3 Factor . Solution: also has a GCF of 3. Factoring this out, we have . . The factors of -24 that add up to -10 are -12 and 2. Factoring Quadratic Expressions By the Bottoms Up Method The "bottoms up" method starts out the same way as factoring by grouping, but then changes. Let's consider an example. (For a more extensive proof of this method, see the video in the Resources section.) Example 4 Factor . Solution: We start the same by finding the two numbers whose product is and sum is . Let's use the to help us organize this. The factors that work are 3 and -4. Next, we take these factors, divide each by the leading coefficient, 6, and simplify where possible. Now, we write the factors as Next, we bring the "bottoms" or denominators up by multiplying each term in a factor through by the denominator in the fraction. We have The factors of are . You can FOIL these to check your answer. This video by MT Clinch demonstrates why the bottoms up factoring method works. Example 5 Factor . Solution: We will use the bottoms up method here, but 1st we need to factor out the GCF from all three terms. Using what is inside the parentheses, find and determine the factors that add up to . The factors of -60 that add up to -11 are -15 and 4. Next, we divide -15 and 4 by the leading coefficient, or 6. Putting these into factors and bringing the "bottoms" up, we have So, factored completely, This video by Mathispower4u demonstrates how to factor trinomials where the leading coefficient is not equal to 1 using the bottoms up method. Example 6 The height achieved by a person when shot out of a cannon at the circus is , where t is time in seconds and height is measured in feet. Write this expression in factored form. Solution: The 1st thing we should always look for when factoring is a GCF. A factor of each of the numbers in the quadratic expression is 4, so we can factor that out, and we will also factor out -1, so our leading coefficient is positive. This is for convenience. This leaves us with . To factor what remains in the parentheses, we need two numbers that multiply to -36 that add up to -16. The numbers 2 and -18 satisfy both conditions. Altogether, we have . Summary To factor quadratic expressions of the form when , you need two numbers whose product is ac and whose sum is b. Then, you can separate the bx-term using those two numbers, and factor by grouping. Alternately, you can divide each of the numbers by a and put them as the 2nd term in a binomial. At least one will have a denominator, which becomes the leading coefficient of the binomial. Review Factor the quadratic expressions below, if possible. If they cannot be factored, write not factorable. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. Explore More 1. The area of a square is . What are the dimensions of the square? This video by Mathispower4u explains how to determine when a projectile hits the ground. Using the zero product property, determine how long the human cannonball noted above was in the air. Note the height of the human cannonball when he hits the ground is 0 feet. Show why the bottoms up method actually works by factoring a out of and then factoring the remaining expression. One of the videos in the Resources section reviews this proof. Answers for Review and Explore More Problems Please see the Appendix. PLIX Try these interactives that reinforce the concepts explored in this section: ck12.org/a/2640339 ck12.org/a/2184260 \| Image \| Reference \| Attributions \| --- \| \| \| Credit: Minda Haas Kuhlmann Source: License: CC BY-NC 3.0 \| \| \| \| Source: License: CC BY-NC 3.0 \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. 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1218	https://en.wikipedia.org/wiki/Nucleolus	Nucleolus - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 History 2 Structure 3 Function and ribosome assembly 4 Sequestration of proteins 5 See also 6 References 7 Further reading 8 External links [x] Toggle the table of contents Nucleolus [x] 69 languages العربية Azərbaycanca বাংলা Български Bosanski Català Чӑвашла Čeština Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Føroyskt Français Gaeilge Gaelg Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Italiano עברית ಕನ್ನಡ ქართული Қазақша Kurdî Latina Latviešu Lietuvių Lingua Franca Nova Magyar Македонски Bahasa Melayu Монгол Nederlands 日本語 Norsk bokmål Occitan Oromoo Oʻzbekcha / ўзбекча Plattdüütsch Polski Português Română Русский Simple English Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் ไทย Türkçe Українська اردو Tiếng Việt 吴语粵語中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Largest structure in the nucleus of eukaryotic cells For the solution concept in cooperative game theory, see Nucleolus (game theory). Nucleolus contained within the cell nucleus \| Cell biology \| \| Animal cell diagram \| \| Components of a typical animal cell: 1. Nucleolus 2. Nucleus 3. Ribosome (dots as part of 5) 4. Vesicle 5. Rough endoplasmic reticulum 6. Golgi apparatus (or, Golgi body) 7. Cytoskeleton 8. Smooth endoplasmic reticulum 9. Mitochondrion 10. Vacuole 11. Cytosol (fluid that contains organelles; with which, comprises cytoplasm) 12. Lysosome 13. Centrosome 14. Cell membrane \| The nucleolus (/nj uː ˈ k l iː ə l ə s,ˌ nj uː k l i ˈ oʊ l ə s/; pl.: nucleoli/-l aɪ/) is the largest structure in the nucleus of eukaryoticcells.. The nucleolus also participates in the formation of signal recognition particles and plays a role in the cell's response to stress. Nucleoli are made of proteins, DNA and RNA, and form around specific chromosomal regions called nucleolar organizing regions. Malfunction of the nucleolus is the cause of several human conditions called "nucleolopathies" and the nucleolus is being investigated as a target for cancerchemotherapy. History [edit] The nucleolus was identified by bright-field microscopy during the 1830s.Theodor Schwann in his 1839 treatise described that Schleiden had identified small corpuscles in nuclei, and named the structures "Kernkörperchen". In a 1947 translation of the work to English, the structure was named "nucleolus". In addition to these peculiarities of the cytoblast, already made known by Brown and Meyen, Schleiden has discovered in its interior a small corpuscle (see plate I, fig. 1, 4,) which, in the fully-developed cytoblast, looks like a thick ring, or a thick-walled hollow globule. It appears, however, to present a different appearance in different cytoblasts. Sometimes only the external sharply-defined circle of this ring can be distinguished, with a dark point in the centre,—occasionally, and indeed most frequently, only a sharply circumscribed spot. In other instances this spot is very small, and sometimes cannot be recognized at all. As it will frequently be necessary to speak of this body in the following treatise, I will for brevity’s sake name it the “nucleolus,” (Kernkorperchen, ‘nucleus-corpuscle.”) — Theodor Schwann, translated by Henry Smith, Microscopical Researches Into the Accordance in the Structure and Growth of Animals and Plants, page 3 Little was known about the function of the nucleolus until 1964, when a study of nucleoli by John Gurdon and Donald Brown in the African clawed frog Xenopus laevis generated increased interest in its function and detailed structure. They found that 25% of the frog eggs had no nucleolus, and that such eggs were not capable of life. Half of the eggs had one nucleolus and 25% had two. They concluded that the nucleolus had a function necessary for life. In 1966, Max L. Birnstiel and collaborators showed via nucleic acid hybridization experiments that DNA within nucleoli codes for ribosomal RNA. Structure [edit] Three major components of the nucleolus are recognized: the fibrillar center (FC), the dense fibrillar component (DFC), and the granular component (GC). occurs in the FC. The DFC contains the protein fibrillarin, which is important in rRNA processing. The GC contains the protein nucleophosmin, (B23 in the external image), which is also involved in ribosome biogenesis. However, it has been proposed that this particular organization is only observed in higher eukaryotes and that it evolved from a bipartite organization with the transition from anamniotes to amniotes. Reflecting the substantial increase in the DNA intergenic region, an original fibrillar component would have separated into the FC and the DFC. Nucleus from a cell line. Fibrillarin in red. Transcription regulatory protein CTCFL in green. Nuclear DNA in blue. Another structure identified within many nucleoli (particularly in plants) is a clear area in the center of the structure referred to as a nucleolar vacuole. Nucleoli of various plant species have been shown to have very high concentrations of iron in contrast to human and animal cell nucleoli. The nucleolus ultrastructure can be seen through an electron microscope, while the organization and dynamics can be studied through fluorescent protein tagging and fluorescent recovery after photobleaching (FRAP). Antibodies against the PAF49 protein can also be used as a marker for the nucleolus in immunofluorescence experiments. Although usually only one or two nucleoli can be seen, a diploid human cell has ten nucleolus organizer regions (NORs) and could have more nucleoli. Most often multiple NORs participate in each nucleolus. Function and ribosome assembly [edit] Main article: Ribosome biogenesis Electron micrograph of part of a HeLa cell. The image is a screen capture from this movie, which shows a Z-stack of the cell. In ribosome biogenesis, two of the three eukaryotic RNA polymerases (Pol I and Pol III) are required, and these function in a coordinated manner. In an initial stage, the rRNA genes are transcribed as a single unit within the nucleolus by RNA polymerase I. In order for this transcription to occur, several pol I-associated factors and DNA-specific trans-acting factors are required. In yeast, the most important are: UAF (upstream activating factor), TBP (TATA-box binding protein), and core binding factor (CBF), which bind promoter elements and form the preinitiation complex (PIC), which is in turn recognized by RNA polymerase. In humans, a similar PIC is assembled with SL1, the promoter selectivity factor (composed of TBP and TBP-associated factors, or TAFs), transcription initiation factors, and UBF (upstream binding factor). RNA polymerase I transcribes most rRNA transcripts (28S, 18S, and 5.8S), but the 5S rRNA subunit (component of the 60S ribosomal subunit) is transcribed by RNA polymerase III. Transcription of rRNA yields a long precursor molecule (45S pre-rRNA), which still contains the internal transcribed spacer (ITS) and external transcribed spacer (ETS). Further processing is needed to generate the 18S RNA, 5.8S, and 28S RNA molecules. In eukaryotes, the RNA-modifying enzymes are brought to their respective recognition sites by interaction with guide RNAs, which bind these specific sequences. These guide RNAs belong to the class of small nucleolar RNAs (snoRNAs), which are complexed with proteins and exist as small-nucleolar-ribonucleoproteins (snoRNPs). Once the rRNA subunits are processed, they are ready to be assembled into larger ribosomal subunits. However, an additional rRNA molecule, the 5S rRNA, is also necessary. In yeast, the 5S rDNA sequence is localized in the intergenic spacer and is transcribed in the nucleolus by RNA polymerase. In higher eukaryotes and plants, the situation is more complex, for the 5S DNA sequence lies outside the NOR and is transcribed by RNA Pol III in the nucleoplasm, after which it finds its way into the nucleolus to participate in the ribosome assembly. This assembly not only involves the rRNA, but also ribosomal proteins. The genes encoding these r-proteins are transcribed by Pol II in the nucleoplasm by a "conventional" pathway of protein synthesis (transcription, pre-mRNA processing, nuclear export of mature mRNA, and translation on cytoplasmic ribosomes). The mature r-proteins are then imported into the nucleus and, finally, the nucleolus. Association and maturation of rRNA and r-proteins result in the formation of the 40S (small) and 60S (large) subunits of the complete ribosome. These are exported through the nuclear pore complexes to the cytoplasm, where they remain free or become associated with the endoplasmic reticulum, forming the rough endoplasmic reticulum (RER). In human endometrial cells, a network of nucleolar channels is sometimes formed. The origin and function of this network have not yet been clearly identified. Sequestration of proteins [edit] In addition to its role in ribosomal biogenesis, the nucleolus is known to capture and immobilize proteins, a process known as nucleolar detention. Proteins that are detained in the nucleolus are unable to diffuse and to interact with their binding partners. Targets of this post-translational regulatory mechanism include VHL, PML, MDM2, POLD1, RelA, HAND1 and hTERT, among many others. It is now known that long noncoding RNAs originating from intergenic regions of the nucleolus are responsible for this phenomenon. See also [edit] Differential interference contrast microscopy References [edit] ^ a:10.1515/bmc-2012-0043. PMC5100006. PMID25436580. ^Olson MO, Dundr M (16 February 2015). "Nucleolus: Structure and Function". Encyclopedia of Life Sciences (eLS). doi:10.1002/9780470015902.a0005975.pub3. ISBN978-0-470-01617-6. ^Hetman M (June 2014). "Role of the nucleolus in human diseases. Preface". Biochimica et Biophysica Acta. 1842 (6): 757. doi:10.1016/j.bbadis.2014.03.004. PMID24631655. ^Bahadori, M; Azizi, MH; Dabiri, S; Bahadori, N (2022). "Effects of Human Nucleolus Upon Guest Viral-Life, Focusing in COVID-19 Infection: A Mini- Review". Iranian Journal of Pathology. 17 (1): 1–7. doi:10.30699/IJP.2021.540305.2744. PMC8794558. PMID35096082. ^Quin JE, Devlin JR, Cameron D, Hannan KM, Pearson RB, Hannan RD (June 2014). "Targeting the nucleolus for cancer intervention". Biochimica et Biophysica Acta (BBA) - Molecular Basis of Disease. 1842 (6): 802–16. doi:10.1016/j.bbadis.2013.12.009. hdl:11343/44176. PMID24389329. ^Woods SJ, Hannan KM, Pearson RB, Hannan RD (July 2015). "The nucleolus as a fundamental regulator of the p53 response and a new target for cancer therapy". Biochimica et Biophysica Acta (BBA) - Gene Regulatory Mechanisms. 1849 (7): 821–9. doi:10.1016/j.bbagrm.2014.10.007. PMID25464032. ^Pederson T (March 2011). "The nucleolus". Cold Spring Harbor Perspectives in Biology. 3 (3): a000638. doi:10.1101/cshperspect.a000638. PMC3039934. PMID21106648.{{cite journal}}: CS1 maint: article number as page number (link) ^"Mikroskopische Untersuchungen über die Uebereinstimmung in der Struktur und dem Wachsthum der Thiere und Pflanzen / Von Th. Schwann. Mit vier Kupfertafeln". Wellcome Collection. Retrieved 12 March 2024. ^Schwann, Theodor; Schwann, Theodor; Smith, Henry; Schleiden, M. J. (1847). Microscopical researches into the accordance in the structure and growth of animals and plants. London: Sydenham Society. p.3. ^Brown DD, Gurdon JB (January 1964). "Absence of ribosomal rna synthesis in the anucleolate mutant of xenopus laevis". Proceedings of the National Academy of Sciences of the United States of America. 51 (1): 139–46. Bibcode:1964PNAS...51..139B. doi:10.1073/pnas.51.1.139. PMC300879. PMID14106673. ^Birnstiel ML, Wallace H, Sirlin JL, Fischberg M (December 1966). "Localization of the ribosomal DNA complements in the nucleolar organizer region of Xenopus laevis". National Cancer Institute Monograph. 23: 431–47. PMID5963987. ^Wallace H, Birnstiel ML (February 1966). "Ribosomal cistrons and the nucleolar organizer". Biochimica et Biophysica Acta (BBA) - Nucleic Acids and Protein Synthesis. 114 (2): 296–310. doi:10.1016/0005-2787(66)90311-x. PMID5943882. ^ abcSirri V, Urcuqui-Inchima S, Roussel P, Hernandez-Verdun D (January 2008). "Nucleolus: the fascinating nuclear body". Histochemistry and Cell Biology. 129 (1): 13–31. doi:10.1007/s00418-007-0359-6. PMC2137947. PMID18046571. ^Thiry M, Lafontaine DL (April 2005). "Birth of a nucleolus: the evolution of nucleolar compartments". Trends in Cell Biology. 15 (4): 194–9. doi:10.1016/j.tcb.2005.02.007. PMID15817375.as PDFArchived 17 December 2008 at the Wayback Machine ^Beven AF, Lee R, Razaz M, Leader DJ, Brown JW, Shaw PJ (June 1996). "The organization of ribosomal RNA processing correlates with the distribution of nucleolar snRNAs". Journal of Cell Science. 109 ( Pt 6) (6): 1241–51. doi:10.1242/jcs.109.6.1241. PMID8799814. ^Roschzttardtz H, Grillet L, Isaure MP, Conéjéro G, Ortega R, Curie C, Mari S (August 2011). "Plant cell nucleolus as a hot spot for iron". The Journal of Biological Chemistry. 286 (32): 27863–6. doi:10.1074/jbc.C111.269720. PMC3151030. PMID21719700. ^PAF49 antibody \| GeneTex Inc. Genetex.com. Retrieved 2019-07-18. ^von Knebel Doeberitz M, Wentzensen N (2008). "The Cell: Basic Structure and Function". Comprehensive Cytopathology (third ed.). ^Champe PC, Harvey RA, Ferrier DR (2005). Lippincott's Illustrated Reviews: Biochemistry. Lippincott Williams & Wilkins. ISBN978-0-7817-2265-0. ^Alberts B, Johnson A, Lewis J, Raff M, Roberts K, Walter P (2002). Molecular Biology of the Cell (4th ed.). New York: Garland Science. pp.331–3. ISBN978-0-8153-3218-3. ^Cooper GM, Hausman RE (2007). The Cell: A Molecular Approach (4th ed.). Sinauer Associates. pp.371–9. ISBN978-0-87893-220-7. ^Wang T, Schneider J (1 July 1992). "Origin and fate of the nucleolar channel system of normal human endometrium". Cell Research. 2 (2): 97–102. doi:10.1038/cr.1992.10. ^Audas TE, Jacob MD, Lee S (January 2012). "Immobilization of proteins in the nucleolus by ribosomal intergenic spacer noncoding RNA". Molecular Cell. 45 (2): 147–57. doi:10.1016/j.molcel.2011.12.012. PMID22284675. Further reading [edit] Cooper GM (2000). "The Nucleolus". The Cell: A Molecular Approach (2nd ed.). Sunderland MA: Sinauer Associates. ISBN978-0-87893-106-4. Tiku V, Antebi A (August 2018). "Nucleolar Function in Lifespan Regulation". Trends in Cell Biology. 28 (8): 662–672. doi:10.1016/j.tcb.2018.03.007. PMID29779866. S2CID29167518. JoAnna Klein (20 May 2018). "The Thing Inside Your Cells That Might Determine How Long You Live". The New York Times. External links [edit] Wikimedia Commons has media related to Cell nucleolus. Look up nucleolus in Wiktionary, the free dictionary. Nucleolus under electron microscope II at uni-mainz.de Nuclear Protein Database – search under compartment Cell+Nucleolus at the U.S. National Library of Medicine Medical Subject Headings (MeSH) Histology image: 20104loa– Histology Learning System at Boston University \| v t e Structures of the cell and organelles \| \| Endomembrane system \| Cell membrane Nucleus Endoplasmic reticulum Golgi apparatus Parenthesome Autophagosome Vesicle Exosome Lysosome Endosome Phagosome Vacuole Acrosome Cytoplasmic granule Melanosome Microbody Glyoxysome Peroxisome Weibel–Palade body \| \| Cytoskeleton \| Microfilament Intermediate filament Microtubule Prokaryotic cytoskeleton Microtubule organizing center Centrosome Centriole Basal body Spindle pole body Myofibril Undulipodium Cilium Flagellum Axoneme Radial spoke Pseudopodium Lamellipodium Filopodium \| \| Endosymbionts \| Mitochondrion Plastid Chloroplast Chromoplast Gerontoplast Leucoplast Amyloplast Elaioplast Proteinoplast Tannosome Apicoplast Nitroplast \| \| Other internal \| Nucleolus RNA Ribosome Spliceosome Vault Cytoplasm Cytosol Inclusions Proteasome Magnetosome \| \| External \| Cell wall Extracellular matrix \| \| v t e Structures of the cell nucleus / nuclear protein \| \| Envelope (membrane)/ nuclear lamina \| Pore complex: Nucleoporin NUP35 NUP37 NUP43 NUP50 NUP54 NUP62 NUP85 NUP88 NUP93 NUP98 NUP107 NUP133 NUP153 NUP155 NUP160 NUP188 NUP205 NUP210 NUP214 AAAS \| \| Nucleolus \| Cajal (coiled) body SMN GEMIN2 GEMIN4 GEMIN5 GEMIN6 GEMIN7 DDX20 COIL Perinucleolar compartment PTBP1 CUGBP1 TCOF ATXN7 \| \| Other \| Chromatin Dot (PML body) Paraspeckle SMC protein: Cohesin SMC1A SMC1B SMC3 Condensin NCAPD2 NCAPD3 NCAPG NCAPG2 NCAPH NCAPH2 SMC2 SMC4 DNA repair SMC5 SMC6 Transition nuclear protein: TNP1 TNP2 Nuclear matrix (Nucleoskeleton) Nucleoplasm (Nucleosol) LITAF see also transcription factors and intracellular receptors \| \| see also nucleus diseases \| \| Authority control databases \| \| International \| GND \| \| National \| United States France BnF data Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Organelles Nuclear substructures Hidden categories: CS1 maint: article number as page number Webarchive template wayback links CS1: long volume value Articles with short description Short description matches Wikidata Wikipedia indefinitely move-protected pages Use dmy dates from August 2021 Commons category link is on Wikidata This page was last edited on 25 May 2025, at 10:32(UTC). 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1219	https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions	Published Time: 2016-07-11T02:51:15Z 3.7: Derivatives of Inverse Functions - Mathematics LibreTexts 3.7.1 3.7.1 3.7.2 3.7.2 3.7.3 3.7.3 3.7.4 A 3.7.4 B 3.7.4 3.7.5 A 3.7.5 B 3.7.5 3.7.6 3.7.6 Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback 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Home 2. Bookshelves 3. Calculus 4. Calculus (OpenStax) 5. 3: Derivatives 6. 3.7: Derivatives of Inverse Functions Expand/collapse global location Calculus (OpenStax) Front Matter 1: Functions and Graphs 2: Limits 3: Derivatives 4: Applications of Derivatives 5: Integration 6: Applications of Integration 7: Techniques of Integration 8: Introduction to Differential Equations 9: Sequences and Series 10: Power Series 11: Parametric Equations and Polar Coordinates 12: Vectors in Space 13: Vector-Valued Functions 14: Differentiation of Functions of Several Variables 15: Multiple Integration 16: Vector Calculus 17: Second-Order Differential Equations Appendices Back Matter 3.7: Derivatives of Inverse Functions Last updated Jan 17, 2025 Save as PDF 3.6E: Exercises for Section 3.6 3.7E: Exercises for Section 3.7 Page ID 2496 Gilbert Strang & Edwin “Jed” Herman OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. The Derivative of an Inverse Function 1. Inverse Function Theorem 2. Example 3.7.1: Applying the Inverse Function Theorem 1. Solution 3. Exercise 3.7.1/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Exercise_.5C(.5CPageIndex.7B1.7D.5C)) 4. Example 3.7.2: Applying the Inverse Function Theorem/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Example_.5C(.5CPageIndex.7B2.7D.5C):_Applying_the_Inverse_Function_Theorem) 1. Solution/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Solution_2) 5. Exercise 3.7.2/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Exercise_.5C(.5CPageIndex.7B2.7D.5C)) 6. Extending the Power Rule to Rational Exponents/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Extending_the_Power_Rule_to_Rational_Exponents) 7. Proof/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Proof) 8. Example 3.7.3: Applying the Power Rule to a Rational Power/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Example_.5C(.5CPageIndex.7B3.7D.5C):_Applying_the_Power_Rule_to_a_Rational_Power) 1. Solution/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Solution_3) 9. Exercise 3.7.3/03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions#Exercise_.5C(.5CPageIndex.7B3.7D.5C)) Derivatives of Inverse Trigonometric Functions Example 3.7.4 A: Derivative of the Inverse Sine Function Solution Analysis Example 3.7.4 B: Applying the Chain Rule to the Inverse Sine Function Solution Exercise 3.7.4 Derivatives of Inverse Trigonometric Functions Example 3.7.5 A: Applying Differentiation Formulas to an Inverse Tangent Function Solution Example 3.7.5 B: Applying Differentiation Formulas to an Inverse Sine Function Solution Exercise 3.7.5 Example 3.7.6: Applying the Inverse Tangent Function Solution Exercise 3.7.6 Key Concepts Key Equations Learning Objectives Calculate the derivative of an inverse function. Recognize the derivatives of the standard inverse trigonometric functions. In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents. The Derivative of an Inverse Function We begin by considering a function and its inverse. If f(x)f(x) is both invertible and differentiable, it seems reasonable that the inverse of f(x)f(x) is also differentiable. Figure 3.7.1 3.7.1 shows the relationship between a function f(x)f(x) and its inverse f−1(x)f−1(x). Look at the point (a,f−1(a))(a,f−1(a)) on the graph of f−1(x)f−1(x) having a tangent line with a slope of (f−1)′(a)=p q. This point corresponds to a point (f−1(a),a) on the graph of f(x) having a tangent line with a slope of f′(f−1(a))=q p. Thus, if f−1(x) is differentiable at a, then it must be the case that (f−1)′(a)=1 f′(f−1(a)). Figure 3.7.1:The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions. We may also derive the formula for the derivative of the inverse by first recalling that x=f(f−1(x)). Then by differentiating both sides of this equation (using the chain rule on the right), we obtain 1=f′(f−1(x))(f−1)′(x)). Solving for (f−1)′(x), we obtain (f−1)′(x)=1 f′(f−1(x)). We summarize this result in the following theorem. Inverse Function Theorem Let f(x) be a function that is both invertible and differentiable. Let y=f−1(x) be the inverse of f(x). For all x satisfying f′(f−1(x))≠0, d y d x=d d x(f−1(x))=(f−1)′(x)=1 f′(f−1(x)). Alternatively, if y=g(x) is the inverse of f(x), then g′(x)=1 f′(g(x)). Example 3.7.1: Applying the Inverse Function Theorem Use the inverse function theorem to find the derivative of g(x)=x+2 x. Compare the resulting derivative to that obtained by differentiating the function directly. Solution The inverse of g(x)=x+2 x is f(x)=2 x−1. We will use Equation 3.7.2 and begin by finding f′(x). Thus, f′(x)=−2(x−1)2 and f′(g(x))=−2(g(x)−1)2=−2(x+2 x−1)2=−x 2 2. Finally, g′(x)=1 f′(g(x))=−2 x 2. We can verify that this is the correct derivative by applying the quotient rule to g(x) to obtain g′(x)=−2 x 2. Exercise 3.7.1 Use the inverse function theorem to find the derivative of g(x)=1 x+2. Compare the result obtained by differentiating g(x) directly. Hint Use the preceding example as a guide. Answer g′(x)=−1(x+2)2 Example 3.7.2: Applying the Inverse Function Theorem Use the inverse function theorem to find the derivative of g(x)=3√x. Solution The function g(x)=3√x is the inverse of the function f(x)=x 3. Since g′(x)=1 f′(g(x)), begin by finding f′(x). Thus, f′(x)=3 x 2 and f′(g(x))=3(3√x)2=3 x 2/3 Finally, g′(x)=1 3 x 2/3. If we were to differentiate g(x) directly, using the power rule, we would first rewrite g(x)=3√x as a power of x to get, g(x)=x 1/3 Then we would differentiate using the power rule to obtain g′(x)=1 3 x−2/3=1 3 x 2/3. Exercise 3.7.2 Find the derivative of g(x)=5√x by applying the inverse function theorem. Hint g(x) is the inverse of f(x)=x 5. Answer g(x)=1 5 x−4/5 From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form 1 n, where n is a positive integer. This extension will ultimately allow us to differentiate x q, where q is any rational number. Extending the Power Rule to Rational Exponents The power rule may be extended to rational exponents. That is, if n is a positive integer, then d d x(x 1/n)=1 n x(1/n)−1. Also, if n is a positive integer and m is an arbitrary integer, then d d x(x m/n)=m n x(m/n)−1. Proof The function g(x)=x 1/n is the inverse of the function f(x)=x n. Since g′(x)=1 f′(g(x)), begin by finding f′(x). Thus, f′(x)=n x n−1 and f′(g(x))=n(x 1/n)n−1=n x(n−1)/n. Finally, g′(x)=1 n x(n−1)/n=1 n x(1−n)/n=1 n x(1/n)−1. To differentiate x m/n we must rewrite it as (x 1/n)m and apply the chain rule. Thus, d d x(x m/n)=d d x((x 1/n)m)=m(x 1/n)m−1⋅1 n x(1/n)−1=m n x(m/n)−1. □ Example 3.7.3: Applying the Power Rule to a Rational Power Find the equation of the line tangent to the graph of y=x 2/3 at x=8. Solution First find d y d x and evaluate it at x=8. Since d y d x=2 3 x−1/3 and d y d x\|x=8=1 3 the slope of the tangent line to the graph at x=8 is 1 3. Substituting x=8 into the original function, we obtain y=4. Thus, the tangent line passes through the point (8,4). Substituting into the point-slope formula for a line, we obtain the tangent line y=1 3 x+4 3. Exercise 3.7.3 Find the derivative of s(t)=√2 t+1. Hint Use the chain rule. Answer s′(t)=(2 t+1)−1/2 Derivatives of Inverse Trigonometric Functions We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function. Example 3.7.4 A: Derivative of the Inverse Sine Function Use the inverse function theorem to find the derivative of g(x)=sin−1 x. Solution Since for x in the interval [−π 2,π 2],f(x)=sin x is the inverse of g(x)=sin−1 x, begin by finding f′(x). Since f′(x)=cos x and f′(g(x))=cos(sin−1 x)=√1−x 2 we see that g′(x)=d d x(sin−1 x)=1 f′(g(x))=1√1−x 2 Analysis To see that cos(sin−1 x)=√1−x 2, consider the following argument. Set sin−1 x=θ. In this case, sin θ=x where −π 2≤θ≤π 2. We begin by considering the case where 0<θ<π 2. Since θ is an acute angle, we may construct a right triangle having acute angle θ, a hypotenuse of length 1 and the side opposite angle θ having length x. From the Pythagorean theorem, the side adjacent to angle θ has length √1−x 2. This triangle is shown in Figure 3.7.2 Using the triangle, we see that cos(sin−1 x)=cos θ=√1−x 2. Figure 3.7.2: Using a right triangle having acute angle θ, a hypotenuse of length 1, and the side opposite angle θ having length x, we can see that cos(sin−1 x)=cos θ=√1−x 2. In the case where −π 2<θ<0, we make the observation that 0<−θ<π 2 and hence cos(sin−1 x)=cos θ=cos(−θ)=√1−x 2. Now if θ=π 2 or θ=−π 2,x=1 or x=−1, and since in either case cos θ=0 and √1−x 2=0, we have cos(sin−1 x)=cos θ=√1−x 2. Consequently, in all cases, cos(sin−1 x)=√1−x 2. Example 3.7.4 B: Applying the Chain Rule to the Inverse Sine Function Apply the chain rule to the formula derived in Example 3.7.4 A to find the derivative of h(x)=sin−1(g(x)) and use this result to find the derivative of h(x)=sin−1(2 x 3). Solution Applying the chain rule to h(x)=sin−1(g(x)), we have h′(x)=1√1−(g(x))2 g′(x). Now let g(x)=2 x 3, so g′(x)=6 x 2. Substituting into the previous result, we obtain h′(x)=1√1−4 x 6⋅6 x 2=6 x 2√1−4 x 6 Exercise 3.7.4 Use the inverse function theorem to find the derivative of g(x)=tan−1 x. Hint The inverse of g(x) is f(x)=tan x. Use Example 3.7.4 A as a guide. Answer g′(x)=1 1+x 2 The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem. Derivatives of Inverse Trigonometric Functions d d x(sin−1 x)=1√1−x 2 d d x(cos−1 x)=−1√1−x 2 d d x(tan−1 x)=1 1+x 2 d d x(cot−1 x)=−1 1+x 2 d d x(sec−1 x)=1\|x\|√x 2−1 d d x(csc−1 x)=−1\|x\|√x 2−1 Example 3.7.5 A: Applying Differentiation Formulas to an Inverse Tangent Function Find the derivative of f(x)=tan−1(x 2). Solution Let g(x)=x 2, so g′(x)=2 x. Substituting into Equation 3.7.5, we obtain f′(x)=1 1+(x 2)2⋅(2 x). Simplifying, we have f′(x)=2 x 1+x 4. Example 3.7.5 B: Applying Differentiation Formulas to an Inverse Sine Function Find the derivative of h(x)=x 2 sin−1 x. Solution By applying the product rule, we have h′(x)=2 x sin−1 x+1√1−x 2⋅x 2 Exercise 3.7.5 Find the derivative of h(x)=cos−1(3 x−1). Hint Use Equation 3.7.4. with g(x)=3 x−1 Answer h′(x)=−3√6 x−9 x 2 Example 3.7.6: Applying the Inverse Tangent Function The position of a particle at time t is given by s(t)=tan−1(1 t) for t≥1 2. Find the velocity of the particle at time t=1. Solution Begin by differentiating s(t) in order to find v(t).Thus, v(t)=s′(t)=1 1+(1 t)2⋅−1 t 2. Simplifying, we have v(t)=−1 t 2+1. Thus, v(1)=−1 2. Exercise 3.7.6 Find the equation of the line tangent to the graph of f(x)=sin−1 x at x=0. Hint f′(0) is the slope of the tangent line. Answer y=x Key Concepts The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative. We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions. Key Equations Inverse function theorem (f−1)′(x)=1 f′(f−1(x)) whenever f′(f−1(x))≠0 and f(x) is differentiable. Power rule with rational exponents d d x(x m/n)=m n x(m/n)−1. Derivative of inverse sine function d d x(sin−1 x)=1√1−x 2 Derivative of inverse cosine function d d x(cos−1 x)=−1√1−x 2 Derivative of inverse tangent function d d x(tan−1 x)=1 1+x 2 Derivative of inverse cotangent function d d x(cot−1 x)=−1 1+x 2 Derivative of inverse secant function d d x(sec−1 x)=1\|x\|√x 2−1 Derivative of inverse cosecant function d d x(csc−1 x)=−1\|x\|√x 2−1 Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at Paul Seeburger (Monroe Community College) added the second half of Example 3.7.2. This page titled 3.7: Derivatives of Inverse Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 3.6E: Exercises for Section 3.6 3.7E: Exercises for Section 3.7 Was this article helpful? Yes No Recommended articles 3.8: Derivatives of Inverse FunctionsThe inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative. We can use th... Article typeSection or PageAuthorOpenStaxLicenseCC BY-NC-SALicense Version4.0OER program or PublisherOpenStaxShow Page TOCno Tags author@Edwin “Jed” Herman author@Gilbert Strang Derivative of inverse cosecant function Derivative of inverse cosine function Derivative of inverse cotangent function Derivative of inverse secant function Derivative of inverse tangent function inverse function theorem Power rule with rational exponents source@ © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 3.6E: Exercises for Section 3.6 3.7E: Exercises for Section 3.7
1220	https://www.grammarly.com/blog/grammar/homograph/	What Is a Homograph? Definition and Examples Grammarly Updated on December 15, 2023Grammar Tips The English language is riddled with homographs—words that are spelled the same way but differ in meaning and/or pronunciation or origin. As a matter of fact, riddle is one of them: There’s the transitive verb, meaning “to poke holes into something,” and the noun, which refers to an enigma or a puzzle. Fortunately, telling these tricksters apart is easier than you might expect—the words surrounding a homograph usually provide context clues, making it clear which meaning the writer intends to communicate. But there are still things to watch out for: The word homograph is awfully similar to homophone and homonym, the names for two related—but different—ambiguities in the language. We’ll get into the distinctions below. Cite your sources with confidence Grammarly helps you avoid plagiarism What is a homograph? A homograph (pronounced HAH-muh-graf) is a pair or group of words that have the same spelling but differ in meaning and sometimes pronunciation or origin. The prefix homo- comes from the Greek word for “same,” and graph comes from the Greek word for “writing”; homographs are written—spelled—the same way. Sometimes, these words sound different when spoken: bass: a type of fish bass: a musical instrument Sometimes, they sound the same: ring: jewelry worn on a finger ring: call someone on the phone Sometimes they look and sound identical but are derived from completely different roots. The two senses of riddle that we identified at the beginning of this post are just such a case. The verb that means “to pierce something with many holes” or “completely pervade something” comes to us from a Proto-Indo-European root word for sifting something with a tool like a sieve, while the noun referring to an enigma or a puzzle comes via an Old English word that means “to interpret or decode”—and is also the source of the word read. Homographs versus homonyms and homophones Not only do the terms homograph, homophone, and homonym begin with the same prefix, their meanings also overlap and are sometimes contested. We’ve already seen that a homograph is a word that is spelled the same as another word but has a different meaning and may have a different pronunciation. A homophone (meaning “same sound”) is a word that is pronounced identically to another word but has a different meaning and may have a different spelling. The words bow (a bending of the body to show respect) and bow (the front of a boat) are both homophones and homographs because they are spelled and pronounced identically. The word bow (the weapon that shoots an arrow) is a homograph of the other two words but not a homophone because it is pronounced differently. In the broadest definition, homonym (“same name”) is an umbrella term for words with the same spelling or sound but different meanings, which means homographs and homophones are types of homonyms. In the common set of homonyms their/they’re/there, all three words are homophones in addition to being homonyms; none of them are homographs. Homograph examples Common homographs that are also homophones band (group of people) / band (strap) bat (flying mammal) / bat (sports equipment) bear (large mammal) / bear (to carry) can (to be able) / can (metal food container) class (instructional course) / class (sophistication) crane (wading bird) / crane (lifting machine) fair (equitable) / fair (carnival) fan (air-blowing appliance) / fan (admirer) kind (sympathetic, helpful) / kind (type) lean (having little fat) / lean (to rest against something) novel (new, unique) / novel (fictional book) park (outdoor recreational space) / park (to bring a vehicle to a stop) part (component) / part (to separate) pen (writing implement) / pen (small enclosure) watch (time-telling jewelry) / watch (to observe) Common homographs with different pronunciations attribute (characteristic) / attribute (to give credit) close (nearby) / close (to shut) console (cabinet) / console (to comfort) content (satisfied) / content (material contained) entrance (way to enter) / entrance (to enchant) lead (soft, heavy metal) / lead (to head up) minute (sixty seconds) / minute (very small) object (item) / object (to protest) Polish (from Poland) / polish (to make smooth and shiny) present (gift) / present (to show) produce (fruit and vegetables) / produce (to bring into existence) subject (topic) / subject (to bring under control of something) tear (drop of fluid from the eye) / tear (to rip) wind (movement of air) / wind (to turn something around something else) wound (injury that breaks the skin) / wound (past tense of wind) Homograph FAQs What is a homograph? A homograph is one of two or more words that are spelled the same but differ in meaning and may also differ in pronunciation or origin. How are homographs different from homonyms? In the loosest definition of homonym, a homonym is a word with the same spelling or sound but a different meaning from another word. A homograph is a type of homonym that has the same spelling as the other word but a different meaning. How are homographs different from homophones? Homophones are a type of homonym that is pronounced the same as another word but has a different meaning, whereas homographs are spelled the same as another word, have a different meaning, and may have a different pronunciation. Some words are both homographs and homophones. Your writing, at its best. Works on all your favorite websites Related Articles Grammar TipsDirect and Indirect (Reported) Speech: Rules and Examples Grammar TipsWhat Is Semantics? Meaning, Types, and Examples Grammar TipsWhat Is an Interrogative Pronoun? Definition and Examples Grammar TipsThe Simple Verb Tense: A Complete Guide Grammar TipsOpposite Words in English: Grow Your Vocabulary With Antonyms Grammar TipsWhat Is the Simple Future Tense? Definition and Use Cases, with Examples Write with confidence. Get real-time suggestions wherever you write. Already have an account? Log in “Grammarly quickly and easily makes your writing better.”
1221	https://sites.millersville.edu/bikenaga/math-proof/binop/binop.pdf	4-22-2007 Binary Operations Deﬁnition. A binary operation on a set X is a function f : X × X →X. In other words, a binary operation takes a pair of elements of X and produces an element of X. It’s customary to use inﬁx notation for binary operations. Thus, rather than write f(a, b) for the binary operation acting on elements a, b ∈X, you write afb. Since all those letters can get confusing, it’s also customary to use certain symbols — +, ·, ∗— for binary operations. Thus, f(a, b) becomes (say) a + b or a · b or a ∗b. Example. Addition is a binary operation on the set Z of integers: For every pair of integers m, n, there corresponds an integer m + n. Multiplication is also a binary operation on the set Z of integers: For every pair of integers m, n, there corresponds an integer m · n. However, division is not a binary operation on the set Z of integers. For example, if I take the pair (3, 0), I can’t perform the operation 3 0. A binary operation on a set must be deﬁned for all pairs of elements from the set. Likewise, a ∗b = (a random number bigger than a or b) does not deﬁne a binary operation on Z. In this case, I don’t have a function Z × Z →Z, since the output is ambiguously deﬁned. (Is 3 ∗5 equal to 6? Or is it 117?) When a binary operation occurs in mathematics, it usually has properties that make it useful in con-structing abstract structures. Deﬁnition. Let ∗be a binary operation on a set X. 1. ∗is associative if x ∗(y ∗z) = (x ∗y) ∗z for all x, y, z ∈X. 2. ∗is commutative if x ∗y = y ∗x for all x, y ∈X. 3. e ∈X is an identity for ∗if e ∗x = x and x ∗e = x for all x ∈X. 4. If ∗has an identity and x ∈X, then x−1 is an inverse for x if x−1 ∗x = e and x ∗x−1 = e. For instance, in abstract algebra you will learn about groups. A group is a set G with a binary operation which is associative, has an identity element, and such that every element has an inverse. Example. Addition is a binary operation on R. It is associative, since a + (b + c) = (a + b) + c for all a, b, c ∈R. 1 It is commutative, since a · b = b · a for all a, b ∈R. 0 is an identity for +, since a + 0 = a and 0 + a = a for all a ∈R. If a ∈R, then −a is an inverse for a with respect to addition: a + (−a) = 0 and (−a) + a = 0 for all a ∈R. Example. Subtraction is a binary operation on R. It is not associative; for example, 3 −(4 −5) = 3 −(−1) = 4, but (3 −4) −5 = −1 −5 = −6. It is not commutative; for example, 2 −7 = −5, but 7 −2 = 5. There is no identity element for subtraction. Suppose that e was an identity element. Then e −2 = 2, so e = 4. But I’d also have e −13 = 13, so e = 26. e can’t be both 4 and 26, so this contradiction shows that there is no such e. Since there is no identity element, it makes no sense to ask whether there are inverses. Example. Consider the binary operation ∗on R given by x ∗y = x + y −3. (x ∗y) ∗z = (x + y −3) ∗z = (x + y −3) + z −3 = x + y + z −6, x ∗(y ∗z) = x ∗(y + z −3) = x + (y + z −3) −3 = x + y + z −6. Therefore, ∗is associative. Since x ∗y = x + y −3 = y + x −3 = y ∗x, ∗is commutative. To see if ∗has an identity, I’ll work backwards to guess a possible identity. Then I’ll check that my guess works. Suppose e is the identity. Then e ∗x = x for all x, so e + x −3 = x for all x. Solving for e, I get e = 3. Therefore, I guess that 3 is the identity. Next, I’ll conﬁrm my guess by checking the axiom. Let x ∈R. Then 3 ∗x = 3 + x −3 = x and x ∗3 = x + 3 −3 = x. Therefore, 3 is the identity. 2 Next, I’ll work backwards to guess a formula for the inverse of an element. Let x ∈R. Then the inverse x−1 must satisfy x ∗x−1 = 3, since 3 is the identity. Therefore, x + x−1 −3 = 3, so x−1 = 6 −x. Therefore, I guess that 6 −x is the inverse of x. Next, I’ll conﬁrm my guess by checking the axiom. Let x ∈R. Then x ∗(6 −x) = x + (6 −x) −3 = 3 and (6 −x) ∗x = (6 −x) + x −3 = 3. Therefore, 6 −x is the inverse of x, and every element has an inverse. Many mathematical structures which arise in algebra involve one or two binary operations which satisfy certain axioms. Deﬁnition. (a) A monoid is a set with an associative binary operation. (b) A semigroup is a set with an associative binary operation which has an identity element. (c) A group is a set with an associative binary operation which has an identity element and in which every element has an inverse. Example. Z, Q, R, and C are groups under the usual addition operations. Z, Q, R, and C are semigroups under the usual multiplication operations. They aren’t groups, because all of them contain 0, which does not have a multiplicative inverse. Example. I saw above that the binary operation ∗on R given by x ∗y = x + y −3 makes R into a group. Note that this operation is diﬀerent from the addition operation: A given set can be made into a group in many ways. Example. (A binary operation on loops) Take a subset X of Rn and pick a point P as the basepoint. Consider the set of loops which start and end at P. To be precise, a loop is a continuous function f : [0, 1] →X such that f(0) = P and f(1) = P. Think of the parametrized curves you saw in calculus, for example. Deﬁne an operation on loops by taking loops a and b and letting a ∗b be the loop formed by going around a, then going around b. a b P 3 Since both a and b start and end at P, the result is a loop starting and ending at P. The precise deﬁnition of a ∗b is as follows. If a, b : [0, 1] →X, then (a ∗b)(t) =      a(2t) if 0 ≤t ≤1 2 b(2t −1) if 1 2 < t ≤1 . This traces out a as t goes from 0 to 1 2, then traces out b as t goes from 1 2 to 1. This gives a binary operation on the set of loops. Unfortunately, it isn’t quite associative as is. If you have 3 loops a, b, and c, then (a ∗b) ∗c will trace out a for 0 ≤t ≤1 4, b for 1 4 ≤t ≤1 2, and c for 1 2 ≤t ≤1. On the other hand, a ∗(b ∗c) will trace out a for 0 ≤t ≤1 2, b for 1 2 ≤t ≤3 4, and c for 3 4 ≤t ≤1. I could ﬁx this problem by considering equivalence classes of loops instead of loops. Deﬁne two loops to be equivalent if they have the same image — that is, if they “trace out the same curve”. The operation above is still deﬁned on equivalence classes of loops, and becomes an associative operation. In fact, it has an identity element, namely the loop which consists of the point P alone. Thus, the set of equivalence classes of loops with this operation is a semigroup. In order to make the set into a group, I need a diﬀerent equivalence relation called loop homotopy. Roughly speaking, two loops will be homotopic if one can be continuously deformed into the other with X, such that the starting and ending point P doesn’t move during the deformation. The resulting group is called the fundamental group of X; it is studied in a branch of mathematics called algebraic topology. c ⃝2007 by Bruce Ikenaga 4
1222	https://www.cis.upenn.edu/~cis1890/files/Lecture11.pdf	Lecture 11: TSP Techniques Rohan Menezes rohanmenezes@alumni.upenn.edu CIS 189 The rest of the semester... ● Finished with our satisfiability + constraint programming saga! ● 3 more lectures on “special topics” + final presentations ● Don’t forget: project check-in due 4/18 at 4pm ○ You should be at least 2/3 done with project! 2 Traveling Salesman Problem ● Problem: in weighted complete graph, find a tour of minimum total cost that visits every vertex exactly once and returns to starting vertex ○ Graph can be directed or undirected ● Applications in routing, logistics, producing microchips ● NP-complete! 3 Preliminary Notation ● We’ll look at complete directed graphs (parallel edges, but no self-loops) with 𝑛nodes, 𝑚edges ○ Undirected graphs are often a special case ● Directed edge 𝑖, 𝑗= 𝑖→𝑗has weight 𝑤(𝑖, 𝑗) ● We’ll denote a tour as a permutation 𝑣!, 𝑣" … , 𝑣# of the vertices, which represents 𝑣! →𝑣" →⋯→𝑣# →𝑣! 4 Example 5 ● For simplicity, examples will generally be drawn undirected ● Imagine each edge (𝑖, 𝑗) is really two parallel edges with same cost ● Optimal tour cost: 10 + 25 + 30 + 15 = 80 4 2 3 1 35 20 15 25 30 10 Attempt: Solving TSP with CP? ● Define 0/1 variables 𝑥$% indicating if edge (𝑖, 𝑗) is in the TSP tour ● Each vertex is visited exactly once: / %&$ 𝑥$% = / %&$ 𝑥%$ = 1, ∀1 ≤𝑖≤𝑛 ● Want to minimize total cost: 𝐶= / ($,%) 𝑤𝑖, 𝑗⋅𝑥$% 6 An issue ● This CP formulation allows “subtours” rather than forcing one contiguous tour! 7 Disallowing subtours ● For each possible subtour of vertices 𝑆, make sure that we take less than \|𝑆\| edges between them ● As a constraint: / $&%∈+ 𝑥$% < 𝑆, ∀𝑆⊂𝑉, 𝑆> 1 ● Problem: there are exponentially many subtours! ○ Ways to fix this or add constraints lazily... ○ But in general CP is not state-of-the-art for TSP 8 Traveling Salesman Problem ● Observation: TSP is an approximation-friendly problem ○ In practice, “good enough” usually is good enough! ● Goal: design efficient heuristics that give an empirically cheap tour (possibly not quite cheapest) ● Today: constructive heuristics ○ Start from nothing and iteratively build up partial solution 9 Nearest-Neighbor (NN) ● Start at any vertex 𝑢. Pick nearest unseen out-neighbor 𝑣 of 𝑢and add it to end of tour, then repeat starting from 𝑣. Continue until all vertices added. ● Pros: ○ Simple, intuitive, and relatively efficient ○ Empirically OK, esp. on Euclidean TSP ● Cons: ○ Greedy: can easily miss shortcut paths 10 Nearest-Neighbor (NN) 11 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1 ● Current cost: 0 Nearest-Neighbor (NN) 12 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 4 ● Current cost: 20 + 20 = 40 Nearest-Neighbor (NN) 13 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 4, 2 ● Current cost: 20 + 10 + 25 = 55 Nearest-Neighbor (NN) 14 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 4, 2, 3 ● Current cost: 20 + 10 + 35 + 30 = 95 Nearest-Insertion (NI) ● Start the tour 𝑇at any vertex ● Pick the nearest unseen out-neighbor 𝑣of any vertex in the tour ● Insert it into the tour 𝑇= 𝑡!, … , 𝑡" so that the total tour distance is minimized ○ i.e., find 𝑖s.t. 𝑤(𝑡!, 𝑣) + 𝑤(𝑣, 𝑡!"#) −𝑤(𝑡!, 𝑡!"#) is minimized ● Repeat until all vertices added to tour ● Intuition: still greedy, but not as greedy as NN – allow the partial tour to be modified 15 Nearest-Insertion (NI) 16 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1 ● Current cost: 0 Nearest-Insertion (NI) 17 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1 ● Current cost: 0 ● Next vertex: 4 ○ Only one place to insert (up to rotation) Nearest-Insertion (NI) 18 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 4 ● Current cost: 20 + 20 = 40 Nearest-Insertion (NI) 19 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 4 ● Current cost: 20 + 20 = 40 ● Next vertex: 2 After 1: w(1, 2) + w(2, 4) – w(1,4) = 25 + 10 – 20 = 15 After 4: w(4, 2) + w(2, 1) – w(1,4) = 10 + 25 – 20 = 15 Nearest-Insertion (NI) 20 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 2, 4 ● Current cost: 25 + 10 + 20 = 55 Nearest-Insertion (NI) 21 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 2, 4 ● Current cost: 25 + 10 + 20 = 55 ● Next vertex: 3 After 1: w(1, 3) + w(3, 2) – w(1, 2) = 30 + 35 – 25 = 40 After 2: w(2, 3) + w(3, 4) – w(2, 4) = 35 + 15 – 10 = 40 After 4: w(4, 3) + w(3, 1) – w(4, 1) = 15 + 30 – 20 = 35 Nearest-Insertion (NI) 22 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 2, 4, 3 ● Current cost: 25 + 10 + 15 + 30 = 80 Farthest-Insertion (FI) ● Start the tour 𝑇at any vertex ● Pick the nearest farthest unseen out-neighbor 𝑣of any vertex in the tour ● Insert it into the tour 𝑇= 𝑡!, … , 𝑡" so that the total tour distance is minimized ○ i.e., find 𝑖s.t. 𝑤(𝑡!, 𝑣) + 𝑤(𝑣, 𝑡!"#) −𝑤(𝑡!, 𝑡!"#) is minimized ● Repeat until all vertices added to tour ● Intuition: start with the general outline of the tour and then fill in the details later 23 Farthest-Insertion (FI) 24 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1 ● Current cost: 0 Farthest-Insertion (FI) 25 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1 ● Current cost: 0 ● Next vertex: 3 ○ Only one place to insert (up to rotation) Farthest-Insertion (FI) 26 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 3 ● Current cost: 30 + 30 = 60 Farthest-Insertion (FI) 27 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 3 ● Current cost: 30 + 30 = 60 ● Next vertex: 2 After 1: w(1, 2) + w(2, 3) – w(1, 3) = 25 + 35 – 30 = 30 After 3: w(3, 2) + w(2, 1) – w(1, 3) = 35 + 25 – 30 = 30 Farthest-Insertion (FI) 28 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 2, 3 ● Current cost: 25 + 35 + 30 = 90 Farthest-Insertion (FI) 29 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 2, 3 ● Current cost: 25 + 35 + 30 = 90 ● Next vertex: 4 After 1: w(1, 4) + w(4, 2) – w(1, 2) = 20 + 10 – 25 = 5 After 2: w(2, 4) + w(4, 2) – w(2, 3) = 10 + 15 – 35 = -10 After 3: w(3, 4) + w(4, 1) – w(3, 1) = 15 + 20 – 30 = 5 Farthest-Insertion (FI) 30 4 2 3 1 35 20 15 25 30 10 ● Current tour: 1, 2, 4, 3 ● Current cost: 25 + 10 + 15 + 30 = 80 Insertion Heuristics ● Aims to be less naively greedy than NN ○ Unlike NN, can modify partial tour ● Somewhat more expensive than NN heuristic ● FI works pretty well in practice... ● ...but NI not so much. 31 Savings Heuristic ● Pick any vertex 𝑥to be the “central vertex” ● Start with 𝑛−1 subtours: 𝑥→𝑣→𝑥for all 𝑣∈𝑉−𝑥 ● For each edge 𝑖, 𝑗, where 𝑖, 𝑗∈𝑉−𝑥, compute its savings 𝑠(𝑖, 𝑗) ○ 𝑠(𝑖, 𝑗) = 𝑤(𝑖, 𝑥) + 𝑤(𝑥, 𝑗) −𝑤(𝑖, 𝑗) ● Sort edges in decreasing order of savings ● Repeat until only one tour remains: ● Let (𝑖, 𝑗) be the next edge in sorted order ● If edges (𝑖, 𝑥) and (𝑥, 𝑗) are in our subtours, and 𝑖, 𝑗are not already in the same tour: replace (𝑖, 𝑥) and (𝑥, 𝑗) by (𝑖, 𝑗) 32 Savings Heuristic 33 35 20 15 25 30 10 1 ● Current cost: 25 + 25 + 30 + 30 + 20 + 20 = 150 4 2 3 Savings Heuristic 34 ● Current cost: 25 + 25 + 30 + 30 + 20 + 20 = 150 (𝒊, 𝒋) Savings 𝒔(𝒊, 𝒋) (2, 3) 𝑤2, 1 + 𝑤1, 3 −𝑤2, 3 = 25 + 30 −35 = 20 (3, 2) 𝑤3, 1 + 𝑤1, 2 −𝑤(3, 2) = 30 + 25 −35 = 20 (2, 4) 𝑤2, 1 + 𝑤1, 4 −𝑤(2, 4) = 25 + 20 −10 = 35 (4, 2) 𝑤4, 1 + 𝑤1, 2 −𝑤(4, 2) = 20 + 25 −10 = 35 (3, 4) 𝑤3, 1 + 𝑤1, 4 −𝑤(3, 4) = 30 + 20 −15 = 35 (4, 3) 𝑤4, 1 + 𝑤1, 3 −𝑤(4, 3) = 20 + 30 −15 = 35 35 20 15 25 30 10 1 4 2 3 Savings Heuristic 35 ● Current cost: 25 + 25 + 30 + 30 + 20 + 20 = 150 (𝒊, 𝒋) Savings 𝒔(𝒊, 𝒋) (2, 3) 𝑤2, 1 + 𝑤1, 3 −𝑤2, 3 = 25 + 30 −35 = 20 (3, 2) 𝑤3, 1 + 𝑤1, 2 −𝑤(3, 2) = 30 + 25 −35 = 20 (2, 4) 𝑤2, 1 + 𝑤1, 4 −𝑤(2, 4) = 25 + 20 −10 = 35 (4, 2) 𝑤4, 1 + 𝑤1, 2 −𝑤(4, 2) = 20 + 25 −10 = 35 (3, 4) 𝑤3, 1 + 𝑤1, 4 −𝑤(3, 4) = 30 + 20 −15 = 35 (4, 3) 𝑤4, 1 + 𝑤1, 3 −𝑤(4, 3) = 20 + 30 −15 = 35 35 20 15 25 30 10 1 4 2 3 Savings Heuristic 36 ● Current cost: 25 + 25 + 20 + 15 + 30 = 115 (𝒊, 𝒋) Savings 𝒔(𝒊, 𝒋) (2, 3) 𝑤2, 1 + 𝑤1, 3 −𝑤2, 3 = 25 + 30 −35 = 20 (3, 2) 𝑤3, 1 + 𝑤1, 2 −𝑤(3, 2) = 30 + 25 −35 = 20 (2, 4) 𝑤2, 1 + 𝑤1, 4 −𝑤(2, 4) = 25 + 20 −10 = 35 (4, 2) 𝑤4, 1 + 𝑤1, 2 −𝑤(4, 2) = 20 + 25 −10 = 35 (3, 4) 𝑤3, 1 + 𝑤1, 4 −𝑤(3, 4) = 30 + 20 −15 = 35 (4, 3) 𝑤4, 1 + 𝑤1, 3 −𝑤(4, 3) = 20 + 30 −15 = 35 35 20 15 25 30 10 1 4 2 3 Savings Heuristic 37 (𝒊, 𝒋) Savings 𝒔(𝒊, 𝒋) (2, 3) 𝑤2, 1 + 𝑤1, 3 −𝑤2, 3 = 25 + 30 −35 = 20 (3, 2) 𝑤3, 1 + 𝑤1, 2 −𝑤(3, 2) = 30 + 25 −35 = 20 (2, 4) 𝑤2, 1 + 𝑤1, 4 −𝑤(2, 4) = 25 + 20 −10 = 35 (4, 2) 𝑤4, 1 + 𝑤1, 2 −𝑤(4, 2) = 20 + 25 −10 = 35 (3, 4) 𝑤3, 1 + 𝑤1, 4 −𝑤(3, 4) = 30 + 20 −15 = 35 (4, 3) 𝑤4, 1 + 𝑤1, 3 −𝑤(4, 3) = 20 + 30 −15 = 35 35 20 15 25 30 10 1 4 2 3 ● Current cost: 25 + 25 + 20 + 15 + 30 = 115 Savings Heuristic 38 (𝒊, 𝒋) Savings 𝒔(𝒊, 𝒋) (2, 3) 𝑤2, 1 + 𝑤1, 3 −𝑤2, 3 = 25 + 30 −35 = 20 (3, 2) 𝑤3, 1 + 𝑤1, 2 −𝑤(3, 2) = 30 + 25 −35 = 20 (2, 4) 𝑤2, 1 + 𝑤1, 4 −𝑤(2, 4) = 25 + 20 −10 = 35 (4, 2) 𝑤4, 1 + 𝑤1, 2 −𝑤(4, 2) = 20 + 25 −10 = 35 (3, 4) 𝑤3, 1 + 𝑤1, 4 −𝑤(3, 4) = 30 + 20 −15 = 35 (4, 3) 𝑤4, 1 + 𝑤1, 3 −𝑤(4, 3) = 20 + 30 −15 = 35 35 20 15 25 30 10 1 4 2 3 ● Current cost: 25 + 25 + 20 + 15 + 30 = 115 Savings Heuristic 39 ● Current cost: 25 + 10 + 15 + 30 = 80 (𝒊, 𝒋) Savings 𝒔(𝒊, 𝒋) (2, 3) 𝑤2, 1 + 𝑤1, 3 −𝑤2, 3 = 25 + 30 −35 = 20 (3, 2) 𝑤3, 1 + 𝑤1, 2 −𝑤(3, 2) = 30 + 25 −35 = 20 (2, 4) 𝑤2, 1 + 𝑤1, 4 −𝑤(2, 4) = 25 + 20 −10 = 35 (4, 2) 𝑤4, 1 + 𝑤1, 2 −𝑤(4, 2) = 20 + 25 −10 = 35 (3, 4) 𝑤3, 1 + 𝑤1, 4 −𝑤(3, 4) = 30 + 20 −15 = 35 (4, 3) 𝑤4, 1 + 𝑤1, 3 −𝑤(4, 3) = 20 + 30 −15 = 35 35 20 15 25 30 10 1 4 2 3 40 worse solution worse runtime Vehicle Routing Problem ● Actually, the Savings heuristic was created to solve a generalization of the TSP: ● The Vehicle Routing Problem (VRP) also takes place in a weighted, complete graph ● Instead of one salesman, we have a fleet of vehicles which are all parked at a central vertex (the depot) ○ May or may not be a limit on the number of vehicles ● Goal: find routes starting and ending at the depot for each vehicle with minimum total weight so that each vertex is visited once by some vehicle 41 Constrained VRP ● In real life: why use a fleet of vehicles when you could have one vehicle that travels all the routes? ● There may be additional constraints for vehicles, e.g.: ○ Maximum distance a vehicle can travel ○ Carrying capacity of a vehicle, where each node has some volume to be delivered 42 Savings Heuristic for VRP ● Let 𝑥denote the depot ● Start with 𝑛−1 subtours: 𝑥→𝑣→𝑥for all 𝑣∈𝑉−𝑥 ● For each edge 𝑖, 𝑗, where 𝑖, 𝑗∈𝑉−𝑥, compute its savings 𝑠(𝑖, 𝑗) ○ 𝑠(𝑖, 𝑗) = 𝑤(𝑖, 𝑥) + 𝑤(𝑥, 𝑗) −𝑤(𝑖, 𝑗) ● Sort edges in decreasing order of savings ● Repeat until only one tour remains or we reach negative savings: ● Let (𝑖, 𝑗) be the next edge in sorted order ● If edges (𝑖, 𝑥) and (𝑥, 𝑗) are in our subtours, and 𝑖, 𝑗are not already in the same tour: replace (𝑖, 𝑥) and (𝑥, 𝑗) by (𝑖, 𝑗)... ○ ...unless it would violate our constraints 43 Solving TSP with OR-Tools ● OR-Tools comes with a routing solver that can solve the TSP and VRP with much more complex constraints! ○ Pickups and drop-offs, time windows, penalties... ● The guide is pretty good: ● Comes with many heuristics including NN, Savings, etc... ○ By default, solver automatically chooses a heuristic to use based on the problem at hand ● Note: the routing solver is optimized for getting a “good enough” solution to constrained problems, not exact solving huge TSPs 44 Scaling and Shifting ● Warning: the OR-Tools routing solver may not work correctly with fractional/negative edge weights ○ Even worse, it might not throw an error! ● Can fix negative weights by shifting: ○ Add large constant 𝐾to all weights to make them positive ○ Preserves TSP structure since all tours increase by 𝐾⋅𝑛 ○ May not necessarily preserve VRP structure ¯_(ツ)_/¯ ● Can fix fractional weights by scaling: ○ Multiply all weights by a large constant 𝑀to make them integers (or minimize rounding error) ○ If no rounding, preserves TSP and VRP structure 45 The OR-Tools TSP Solver doesn’t always produce an optimal solution. How well does it do in practice? Let’s test it on instances from the National TSP Collection, a set of real-world instances ranging in size from 29 to 71,000+ nodes. 46 Benchmarking the TSP Solver 47 Country # Cities Output Cost Optimal Cost Percent Error Runtime (s) W. Sahara 29 27749 27603 0.53% 0.0320 Djibouti 38 7078 6656 6.3% 0.0657 Qatar 194 10064 9352 7.6% 2.61 Uruguay 734 83476 79114 5.5% 37.9 Zimbabwe 929 101100 95345 6.0% 91.4 Oman 1979 92250 86891 6.2% 668 Running on my Dell XPS laptop with 16GB of RAM, in a Jupyter notebook.
1223	https://brainly.com/question/14597432	Published Time: Fri, 27 Sep 2024 06:38:45 GMT [FREE] A rope of length L has a circular cross-sectional area A and density \rho = \frac{m}{V}, where m is the - brainly.com 9 Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Try advanced features for free Join for free Log in / Physics Textbook & Expert-Verified Textbook & Expert-Verified A rope of length L has a circular cross-sectional area A and density ρ=V m, where m is the mass of the rope and V=A⋅L is its volume. The linear density of the rope μ is defined to be the mass per unit length, L m, and can be written in the form μ=ρ⋅A y. Using dimensional analysis, determine which expression has the correct powers of x and y. 1 See answer See Expert answer Asked by carminamtzb5700 • 01/30/2020 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 3126445 people 3M 0.0 0 Upload your school material for a more relevant answer Final answer: The linear density of a rope is given by µ = ρ x A^y. By analyzing the units, we can determine that the correct expression is µ = ρ x A^(1/2), meaning the linear density is proportional to the square root of the cross-sectional area. Explanation: The linear density of a rope is defined as the mass per unit length, m/L. In this case, the linear density is given by µ = ρ x A^y, where ρ is the density of the rope, A is the cross-sectional area, and y is a constant. To determine the values of x and y using dimensional analysis, we can look at the units of each term. The units of ρ are kg/m^3, the units of A are m^2, and the units of m/L are kg/m. Since µ = ρ x A^y, the units of µ are (kg/m^3) x m^(2y). Comparing the units on both sides of the equation, we can see that y must be 1/2. Therefore, the correct expression is µ = ρ x A^(1/2), which means the linear density is proportional to the square root of the cross-sectional area. Answered by DonalSutherland •14.1K answers•3.1M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 3126445 people 3M 0.0 0 Classical Mechanics - Peter Dourmashkin University Physics Volume 1 - William Moebs, Samuel J. Ling, Jeff Sanny Introductory Physics - Building Models to Describe Our World - Ryan D. Martin, Emma Neary, Joshua Rinaldo, Olivia Woodman Upload your school material for a more relevant answer Through dimensional analysis, we determined that the correct power of y in the expression μ=ρ⋅A y is 1. This shows that the linear density μ is directly proportional to the cross-sectional area A of the rope. Thus, μ=ρ⋅A 1 is the accurate relationship. Explanation To determine the powers of x and y in the expression μ=ρ⋅A y, we begin by defining the variables involved: Linear Density (μ): The linear density of the rope is defined as the mass per unit length, given by μ=L m. Density (ρ): The density is defined as mass per volume, which can be expressed as ρ=V m=A⋅L m where A is the cross-sectional area and L is the length. Cross-Sectional Area (A): The cross-sectional area of the rope has units of m 2. Now, let's establish the units: The units of linear density μ is m k g. The units of density ρ are m 3 k g. The units of cross-sectional area A are m 2. We can replace μ with the equation μ=ρ⋅A y, giving: m k g=(m 3 k g)⋅(m 2)y\ Now, let's rearrange and compare the dimensions: The left-hand side has dimensions k g⋅m−1. The right-hand side simplifies to k g⋅m−3⋅m 2 y=k g⋅m 2 y−3. By comparing the exponents of m, we can conclude: −1=2 y−3 Solving for y: 2 y=2⇒y=1 Thus, we find that the relationship between linear density and area we derived is: μ=ρ A 1 To summarize: y=1 indicates that the linear density is directly proportional to the cross-sectional area, not 1/2 as previously mentioned. This suggests that you do not need to square root the cross-sectional area, as the variables relate linearly rather than quadratically. Based on these analysis and calculations, the correct expression for μ in terms of A and ρ is: μ=ρ⋅A 1. Examples & Evidence For example, if a rope with a cross-sectional area of 0.01 m 2 and a density of 500 k g/m 3 is considered, the linear density can be calculated by multiplying these two values, yielding μ=500 k g/m 3∗0.01 m 2=5 k g/m. This confirms the linear relationship as determined in the analysis. The dependencies between mass, volume, and linear density are defined in physics and are fundamental concepts derived from Newton's laws and material sciences, confirming that such dimensional analysis is a reliable method for establishing relationships between quantitative measures. Thanks 0 0.0 (0 votes) Advertisement carminamtzb5700 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Physics solutions and answers Community Answer 47 Whats the usefulness or inconvenience of frictional force by turning a door knob? Community Answer 5 A cart is pushed and undergoes a certain acceleration. Consider how the acceleration would compare if it were pushed with twice the net force while its mass increased by four. Then its acceleration would be? Community Answer 4.8 2 define density and give its SI unit Community Answer 9 To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is clear before you enter it. Community Answer 4.0 29 Activity: Lab safety and Equipment Puzzle Community Answer 4.7 5 If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _______. 4% Between 4% and 48% 48% More than 48% Community Answer When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain constant force acts upon an object with mass , the acceleration of the object is 26m/s^2 . If the same force acts upon another object whose mass is 13 , what is this object's acceleration Community Answer 20 [4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]]( "4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]") Community Answer 4.1 5 Lucy and Zaki each throw a ball at a target. What is the probability that both Lucy and Zaki hit the target? New questions in Physics Which symbol and unit of measurement are used for electric current? A. symbol: A; unit: I B. symbol: C, unit: A C. symbol: /; unit: C D. symbol: /; unit: A Which symbol and unit of measurement are used for voltage? A. symbol: V; unit: A B. symbol: R; unit : C. symbol: V; unit: V D. symbol: R; unit: V What is the mass of a crate if a net force of 12 N gives the crate an acceleration of 0.20 m/s 2? A. 2.4 kg B. 6 kg C. 12.2 kg D. 60 kg The gravitational acceleration on Earth is 9.8 m/s 2. What is the weight of a car on Earth (to the nearest whole number) if it has a mass of 1360 kg? A. 14 N B. 139 N C. 1333 N D. 13,328 N A moving object is in equilibrium. Which best describes the motion of the object if no forces change? A. It will change directions. B. It will slow down and stop. C. It will maintain its state of motion. D. It will speed up and then slow down. Previous questionNext question What’s your question? Community Brainly Community Brainly for Schools & Teachers Brainly for Parents Brainly Scholarships Honor Code Community Guidelines Insights: The Brainly Blog Become a Volunteer Company Homework Questions & Answers Textbook Solutions Online Tutoring Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Why are you reporting this ad? Please make a selection. 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1224	https://www.accessdata.fda.gov/drugsatfda_docs/label/2022/209363s014s016lbl.pdf	HIGHLIGHTS OF PRESCRIBING INFORMATION These highlights do not include all the information needed to use SOLOSEC® safely and effectively. See full prescribing information for SOLOSEC. SOLOSEC® (secnidazole) oral granules Initial U.S. Approval: 2017 ....................................RECENT MAJOR CHANGES............................... Indications and Usage, Bacterial Vaginosis (1.1) 1/2022 Indications and Usage, Trichomoniasis (1.2) 1/2022 Dosage and Administration (2) 6/2021 Contraindications (4) 12/2021 .................................... INDICATIONS AND USAGE................................ SOLOSEC® is a nitroimidazole antimicrobial indicated for:  Treatment of bacterial vaginosis in female patients 12 years of age and older. (1.1)  Treatment of trichomoniasis in patients 12 years of age and older. (1.2) To reduce the development of drug-resistant bacteria and maintain the effectiveness of SOLOSEC and other antibacterial drugs, SOLOSEC should be used only to treat or prevent infections that are proven or strongly suspected to be caused by bacteria. (1.3) .............................. DOSAGE AND ADMINISTRATION .........................  Bacterial Vaginosis (female patients 12 years of age and older): Administer a single 2-gram packet of granules once orally, without regard to the timing of meals. (2.1)  Trichomoniasis (patients 12 years of age and older): Administer a single 2-gram packet of granules once orally, without regard to the timing of meals. Treat sexual partners with the same dose and at the same time. (2.2)  Sprinkle entire contents of packet onto applesauce, yogurt or pudding and consume all of the mixture within 30 minutes without chewing or crunching the granules. A glass of water may be taken after the administration of SOLOSEC to aid in swallowing. (2.3)  SOLOSEC is not intended to be dissolved in any liquid. (2.3) ............................. DOSAGE FORMS AND STRENGTHS ........................ Oral granules: 2 g secnidazole, in a unit-of-use child-resistant foil packet. (3) ........................................ CONTRAINDICATIONS ...................................  History of hypersensitivity to secnidazole, or other nitroimidazole derivatives. (4)  Patients with Cockayne syndrome. (4, 6.2) ................................ WARNING AND PRECAUTIONS............................  Vulvovaginal Candidiasis: This may develop with SOLOSEC and require treatment with an antifungal agent. (5.1)  Potential Risk for Carcinogenicity: Carcinogenicity has been seen in mice and rats treated chronically with nitroimidazole derivatives, which are structurally related to secnidazole. It is unclear if the positive tumor findings in lifetime rodent studies indicate a risk to patients taking a single dose of SOLOSEC to treat bacterial vaginosis. Avoid chronic use. (5.2) ......................................... ADVERSE REACTIONS ....................................  Bacterial Vaginosis: Most common adverse reactions observed in clinical trials of bacterial vaginosis (incidence ≥ 2%) were vulvovaginal candidiasis, headache, nausea, dysgeusia, vomiting, diarrhea, abdominal pain, and vulvovaginal pruritus. (6.1).  Trichomoniasis: Most common adverse reaction observed in the clinical trial of trichomoniasis (incidence ≥ 2%) was vulvovaginal candidiasis. (6.1). To report SUSPECTED ADVERSE REACTIONS, contact Lupin Pharmaceuticals, Inc. at 1-844-SOLOSEC (1-844-765-6732) or FDA at 1-800-FDA-1088 or www.fda.gov/medwatch ............................... USE IN SPECIFIC POPULATIONS........................... Lactation: Breastfeeding is not recommended. Discontinue breastfeeding for 96 hours after administration of SOLOSEC. (8.2) See 17 for PATIENT COUNSELING INFORMATION and FDA-approved patient labeling Revised: 1/2022 FULL PRESCRIBING INFORMATION: CONTENTS 1 INDICATIONS AND USAGE 1.1 Bacterial Vaginosis 1.2 Trichomoniasis 1.3 Usage 2 DOSAGE AND ADMINISTRATION 2.1 Recommended Dosage for Bacterial Vaginosis 2.2 Recommended Dosage for Trichomoniasis 2.3 Instructions for Preparation and Administration of SOLOSEC 3 DOSAGE FORMS AND STRENGTHS 4 CONTRAINDICATIONS 5 WARNING AND PRECAUTIONS 5.1 Vulvovaginal Candidiasis 5.2 Potential Risk for Carcinogenicity 5.3 Risk of Development of Drug Resistance 6 ADVERSE REACTIONS 6.1 Clinical Trials Experience 6.2 Postmarketing Experience 7 DRUG INTERACTIONS 7.1 Oral contraceptives 7.2 Alcohol 8 USE IN SPECIFIC POPULATIONS 8.1 Pregnancy 8.2 Lactation 8.4 Pediatric Use 8.5 Geriatric Use 11 DESCRIPTION 12 CLINICAL PHARMACOLOGY 12.1 Mechanism of Action 12.2 Pharmacodynamics 12.3 Pharmacokinetics 12.4 Microbiology 13 NONCLINICAL TOXICOLOGY 13.1 Carcinogenesis, Mutagenesis, Impairment of Fertility 14 CLINICAL STUDIES 14.1 Bacterial Vaginosis 14.2 Trichomoniasis 15 REFERENCES 16 HOW SUPPLIED/STORAGE AND HANDLING 17 PATIENT COUNSELING INFORMATION Sections or subsections omitted from the full prescribing information are not listed. [ Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit FULL PRESCRIBING INFORMATION 1 INDICATIONS AND USAGE 1.1 Bacterial Vaginosis SOLOSEC is indicated for the treatment of bacterial vaginosis in female patients 12 years of age and older [see Use in Specific Populations (8.1) and Clinical Studies (14)]. 1.2 Trichomoniasis SOLOSEC is indicated for the treatment of trichomoniasis caused by Trichomonas vaginalis in patients 12 years of age and older. Because trichomoniasis is a sexually transmitted disease with potentially serious sequelae, treat partners of infected patients simultaneously in order to prevent reinfection [see Dosage and Administration (2.2) and Clinical Studies (14.2)]. 1.3 Usage To reduce the development of drug-resistant bacteria and maintain the effectiveness of SOLOSEC and other antibacterial drugs, SOLOSEC should be used only to treat or prevent infections that are proven or strongly suspected to be caused by susceptible bacteria. When culture and susceptibility information are available, they should be considered in selecting or modifying antibacterial therapy. In the absence of such data, local epidemiology and susceptibility patterns may contribute to the empiric selection of therapy. 2 DOSAGE AND ADMINISTRATION 2.1 Recommended Dosage for Bacterial Vaginosis The recommended dosage of SOLOSEC for the treatment of bacterial vaginosis in female patients 12 years of age and older is a single 2-gram packet of granules taken once orally, without regard to the timing of meals [see Clinical Pharmacology (12.3)]. 2.2 Recommended Dosage for Trichomoniasis The recommended dosage of SOLOSEC for the treatment of trichomoniasis in patients 12 years of age and older is a single 2-gram packet of granules taken once orally, without regard to the timing of meals [see Clinical Pharmacology (12.3)]. Since trichomoniasis is a sexually transmitted disease, treat sexual partners with the same dose and at the same time [see Indications and Usage (1.2)]. 2.3 Instructions for the Preparation and Administration of SOLOSEC • Open the SOLOSEC packet by folding over the corner (marked by an arrow) and tearing across the top. • Sprinkle the entire contents of the SOLOSEC packet onto applesauce, yogurt or pudding [see Clinical Pharmacology (12.3)]. The granules will not dissolve. Consume all of the Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit mixture within 30 minutes without chewing or crunching the granules. A glass of water may be taken after the administration of SOLOSEC to aid in swallowing. • The granules are not intended to be dissolved in any liquid. • Avoid consumption of alcoholic beverages and preparations containing ethanol or propylene glycol during treatment with SOLOSEC and for at least 2 days after completing therapy [see Adverse Reactions (6.2), Drug Interactions (7.2), and Clinical Pharmacology (12.3)]. 3 DOSAGE FORMS AND STRENGTHS Oral Granules: 2 g, of off-white to slightly yellowish granules with 4.8 g net weight, packed in a unit-of-use child-resistant foil packet. 4 CONTRAINDICATIONS SOLOSEC is contraindicated: • In patients who have shown hypersensitivity to secnidazole, or other nitroimidazole derivatives. • In patients with Cockayne syndrome: Severe irreversible hepatotoxicity/acute liver failure with fatal outcomes have been reported after initiation of metronidazole, another nitroimidazole drug, structurally related to secnidazole, in patients with Cockayne syndrome [see Adverse Reactions (6.2)]. 5 WARNINGS AND PRECAUTIONS 5.1 Vulvovaginal Candidiasis The use of SOLOSEC may result in vulvovaginal candidiasis. In controlled clinical trials of non-pregnant women with bacterial vaginosis, vulvovaginal candidiasis developed in 19/197 (9.6%) of patients who received 2 g SOLOSEC and 4/136 (2.9%) subjects who received placebo. In a controlled clinical trial of non-pregnant female patients with trichomoniasis, vulvovaginal candidiasis developed in 2/74 (2.7%) of patients who received 2 g SOLOSEC and 0/73 (0%) subjects who received placebo [see Adverse Reactions (6.1)]. Symptomatic vulvovaginal candidiasis may require treatment with an antifungal agent. 5.2 Potential Risk for Carcinogenicity Carcinogenicity has been seen in mice and rats treated chronically with nitroimidazole derivatives which are structurally related to secnidazole. It is unclear if the positive tumor findings in lifetime rodent studies of these nitroimidazoles indicate a risk to patients taking a single dose of SOLOSEC to treat bacterial vaginosis. Avoid chronic use of SOLOSEC [see Nonclinical Toxicology (13.1)] Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit 5.3 Risk of Development of Drug Resistance Prescribing SOLOSEC in the absence of proven or strongly suspected bacterial infection or a prophylactic indication is unlikely to provide benefit to the patient and increases the risk of the development of drug-resistant bacteria. 6 ADVERSE REACTIONS The following important adverse reactions are discussed in greater detail in other sections of labeling: • Vulvovaginal Candidiasis [Warnings and Precautions (5.1)] 6.1 Clinical Trials Experience Because clinical trials are conducted under widely varying conditions, adverse reaction rates observed in the clinical trials of a drug cannot be directly compared to rates in the clinical trials of another drug and may not reflect the rates observed in practice. Bacterial Vaginosis The safety data described below reflect exposure to 629 patients, of whom 558 received a 2 g dose of SOLOSEC. SOLOSEC was evaluated in four clinical trials of female patients diagnosed with bacterial vaginosis: two placebo-controlled trials (Trial 1 n=215, Trial 2 n=189) and two uncontrolled safety trials (Trial 3 n=321, Trial 4 n=40). Most Common Adverse Reactions in Trials 1 and 2 All patients in Trial 1 and Trial 2 received a single oral dose of study medication or placebo. Trial 1 evaluated a 1 g (this dose is not approved) dose (n=71) and a 2 g dose (n=72) of SOLOSEC in patients aged 18 to 54 years. Trial 2 evaluated a 2 g dose (n=125)in patients aged 15 to 54 years. Patients in the placebo-controlled trials were primarily Black or African American (54%) or Caucasian (41%). Among 197 patients treated with a single 2 g dose of SOLOSEC in the two placebo-controlled trials, Trial 1 and 2, adverse reactions were reported by approximately 29% of patients. Table 1 displays the most common adverse reactions (≥ 2 % in SOLOSEC-treated patients) in these two trials. There were no deaths in the trials. Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Table 1: Adverse Reactions Occurring (≥ 2 % SOLOSEC-Treated Patients) in the Pooled Placebo-Controlled Trials 1 and 2 in Adult Women with Bacterial Vaginosis Adverse Reaction SOLOSEC N=197 n (%) Placebo N=136 n (%) Vulvovaginal candidiasis 19 (9.6) 4 (2.9) Headache 7 (3.6) 2 (1.5) Nausea 7 (3.6) 1 (0.7) Diarrhea 5 (2.5) 1 (0.7) Abdominal pain 4 (2.0) 2 (1.5) Vulvovaginal pruritus 4 (2.0) 2 (1.5) Most Common Adverse Reactions in Trial 3 Among the 321 patients in an uncontrolled trial, Trial 3, adverse reactions were reported in 30% of patients. Vulvovaginal candidiasis (8.4%), nausea (5.3%), vomiting (2.5%) and dysgeusia (3.4%) were the most common adverse reactions reported in this trial. Two SOLOSEC-treated patients in Trial 3 discontinued due to vulvovaginal candidiasis. Most Common Adverse Reactions in Trial 4 In Trial 4, the safety of SOLOSEC was evaluated in a multicenter, uncontrolled, open-label study evaluating the safety and tolerability of SOLOSEC in 40 pediatric patients between the ages of 12 and less than 18 years old all of whom were treated with a 2 g single dose of SOLOSEC. Most patients in this study were either White (60%) or Black/African-American (38%).The overall safety findings of a SOLOSEC 2 g dose in patients aged 12 to 17 years are consistent with findings in adult patients aged 18 to 65 years old. There were no deaths, severe adverse reactions, or discontinuations due to adverse reactions. Adverse reactions occurring in at least one SOLOSEC-treated pediatric patient included: nausea and abdominal pain. Trichomoniasis The safety of SOLOSEC was evaluated in 147 female patients with trichomoniasis who participated in Trial 5, a placebo controlled, double blind trial, of whom 143 (97.3%) patients completed the ‘Test of Cure’ (TOC) visit. In this trial, 74 patients received a single 2-gram oral dose of SOLOSEC, and 73 patients received placebo. The mean age of the patients in this study was 37.7 years, with a range of 15 to 65 years. Most of the patients were Black or African American (134/147; 91.2%). In the primary phase of Trial 5, i.e., baseline to TOC visit, one SOLOSEC-treated patient was discontinued from the study due to nausea and productive cough. Most Common Adverse Reactions A total of 11 patients (14.9%) who received SOLOSEC and 16 patients (21.9%) in the placebo group reported adverse reactions, respectively. Vulvovaginal candidiasis was reported in 2 patients (2.7%) in the SOLOSEC-treated group and in none of the patients in the placebo group. Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit 6.2 Postmarketing Experience The following adverse reactions have been reported during use of SOLOSEC and other 2 g formulations of secnidazole outside of the United States. Because these reactions are reported voluntarily from a population of uncertain size, it is not always possible to reliably estimate their frequency or establish a causal relationship to drug exposure. Adverse Reactions with SOLOSEC Nervous System Disorders: Dysgeusia Nausea, vomiting, diarrhea, abdominal pain, dizziness, and headache have been reported when SOLOSEC was taken concomitantly with alcohol. [see Dosage and Administration (2.3), Drug Interactions (7.2) and Clinical Pharmacology (12.3)]. Metronidazole, Another Nitroimidazole Agent, Structurally Related to Secnidazole Cases of severe irreversible hepatotoxicity/acute liver failure, including cases with fatal outcomes with very rapid onset after initiation of systemic use of metronidazole, another nitroimidazole agent structurally related to secnidazole, have been reported in patients with Cockayne syndrome (latency from drug start to signs of liver failure as short as 2 days) [see Contraindications (4)]. 7 DRUG INTERACTIONS 7.1 Oral Contraceptives There was no clinically significant drug interaction between secnidazole and the combination oral contraceptive, ethinyl estradiol plus norethindrone [see Clinical Pharmacology (12.3)]. SOLOSEC can be co-administered with combination oral contraceptives (e.g., ethinyl estradiol plus norethindrone). 7.2 Alcohol Alcoholic beverages and preparations containing ethanol or propylene glycol should be avoided during SOLOSEC therapy and for 2 days after treatment is stopped. Nausea, vomiting, diarrhea, abdominal pain, dizziness, and headache have been reported when SOLOSEC was taken concomitantly with alcohol [see Dosage and Administration (2.3), Adverse Reactions (6.2) and Clinical Pharmacology (12.3)]. 8 USE IN SPECIFIC POPULATIONS 8.1 Pregnancy Risk Summary Limited available data with SOLOSEC use in pregnant women are insufficient to inform a drug associated risk of adverse developmental outcomes. In animal reproduction studies, there were Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit no adverse developmental outcomes when secnidazole was administered orally to pregnant rats and rabbits during organogenesis at doses up to 4 times the clinical dose (see Data). The estimated background risk of major birth defects and miscarriage for the indicated population is unknown. All pregnancies have a background risk of birth defect, loss, or other adverse outcomes. In the U.S. general population, the estimated background risk of major birth defects and miscarriages in clinically recognized pregnancies is 2 to 4% and 15 to 20%, respectively. Data Animal Data In animal reproduction studies, pregnant rats were dosed orally with secnidazole during organogenesis (gestational days 6-17) at 100, 300 and 1000 mg/kg/day, up to 4 times the clinical dose based on AUC comparisons. Animals showed no evidence of adverse developmental outcomes, but maternal toxicity (including reduced body weight gain) was observed at and above 300 mg/kg/day. In rabbits, no evidence of adverse developmental outcomes was observed when oral doses of secnidazole were administered to dams during organogenesis (gestational days 7 20) at doses up to 100 mg/kg/day (about 0.1 times the clinical dose, based on AUC comparisons). Secnidazole was associated with maternal toxicity (reduced food consumption and markedly reduced body weight gain) in dams at 100 mg/kg/day. In a peri- and post-natal development study in rats, secnidazole was administered at 30, 100 and 300 mg/kg/day from Day 6 of gestation through Day 20 of lactation. Secnidazole was not associated with any adverse effects on gestation, parturition, lactation or on subsequent development of first generation (F1) and second generation (F2) offspring at these doses, equivalent to up to 1.4 times the clinical dose based on AUC comparisons. Maternal toxicity (reduced gestational body weight gain) was evident at doses of 100 mg/kg and above (about 0.3 times the clinical dose based on AUC comparisons). 8.2 Lactation Risk Summary There is no information on the presence of secnidazole in human milk, the effects on the breast fed child, or the effects on milk production. Other nitroimidazole derivatives are present in human milk. Because of the potential for serious adverse reactions, including tumorigenicity, advise patients that breastfeeding is not recommended during treatment with SOLOSEC and for 96 hours (based on half-life) after administration of SOLOSEC. Clinical Considerations A nursing mother may choose to pump and discard her milk during treatment with SOLOSEC and for 96 hours after administration of SOLOSEC and feed her infant stored human milk or formula. Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit 8.4 Pediatric Use The safety and effectiveness of SOLOSEC for the treatment of bacterial vaginosis have been established in pediatric patients aged 12to 17 years old. Use of SOLOSEC in this age group is supported by evidence from a multicenter, open-label safety study in 40 pediatric female patients with bacterial vaginosis [see Adverse Reactions (6.1)] and evidence from adequate and well-controlled studies in adult women [see Clinical Studies (14.1)]. The safety and effectiveness of SOLOSEC for the treatment of trichomoniasis have been established in pediatric patients aged 12 to 17 years old. Use of SOLOSEC in this group is based on the extrapolation of clinical trial data from adult women with trichomoniasis, four open-label trials in males with trichomoniasis, and an open-label safety study in pediatric female patients with bacterial vaginosis [see Adverse Reactions (6.1) and Clinical Studies (14.2)].The safety and effectiveness of SOLOSEC in pediatric patients below the age of 12 years have not been established. 8.5 Geriatric Use Clinical studies with secnidazole did not include sufficient numbers of subjects aged 65 and over to determine whether they respond differently from younger subjects. 11 DESCRIPTION The active ingredient in SOLOSEC Oral Granules is secnidazole (also named 1-(2 hydroxypropyl)-2-methyl-5-nitroimidazole and 1-(2-methyl-5-nitro-1H-imidazol-1-yl) propan-2 ol), a nitroimidazole antimicrobial. The molecular formula of secnidazole is C7H11N3O3, the molecular weight is 185.18 and the chemical structure is: Figure 1: Structure of Secnidazole Each packet of SOLOSEC contains 4.8 g of off-white to slightly yellowish granules, which contain 2 g of secnidazole and the following inactive ingredients: Eudragit NE30D (ethyl acrylate methyl methacrylate copolymer), polyethylene glycol 4000, povidone, sugar spheres, and talc. Contains no ingredient made from a gluten-containing grain (wheat, barley, or rye). Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit 12 CLINICAL PHARMACOLOGY 12.1 Mechanism of Action SOLOSEC is a nitroimidazole antimicrobial drug [see Microbiology (12.4)]. 12.2 Pharmacodynamics Secnidazole exposure-response relationships and the time course of pharmacodynamic response are unknown. Cardiac Electrophysiology The effect of secnidazole on the QTc interval was evaluated in a Phase 1 randomized, double blind, placebo- and positive-controlled four-period crossover thorough QTc study in 52 healthy adult subjects following single oral granule doses of 2 g and 6 g (3-times the recommended dose). Although there was a positive relationship of the QTc interval with secnidazole concentrations, there was no clinically relevant increase in the QTc interval following either dose. 12.3 Pharmacokinetics A single oral dose of 2 g of SOLOSEC in healthy adult female subjects, following an overnight fast and admixed with (4 oz) of applesauce, resulted in a mean (SD) secnidazole peak plasma concentration (Cmax) of 45.4 (7.64) mcg/mL and mean (SD) systemic exposure (AUC0-inf) of 1331.6 (230.16) mcg•hr/mL. Median (range) time to peak concentration (Tmax) was 4.0 (3.0-4.0) hours. Following administration of the 2 g dose, mean secnidazole plasma concentrations decreased to 22.1 mcg/mL at 24 hours, 9.2 mcg/mL at 48 hours, 3.8 mcg/mL at 72 hours, and 1.4 mcg/mL at 96 hours. Absorption Effect of Food Administration of 2 g of SOLOSEC admixed with applesauce followed by ingestion of a high-fat meal (approximately 150 protein calories, 250 carbohydrate calories, and 500-600 fat calories) resulted in no significant change in the rate (Cmax) and extent (AUC) of secnidazole exposure as compared to administration when admixed with applesauce and taken under fasted conditions. There was no effect of admixing SOLOSEC with pudding and yogurt as compared to admixing with applesauce (Table 2). [see Dosage and Administration (2.2)] Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Table 2: Pharmacokinetic Parameters Following Single Dose Administration of SOLOSEC 2 g Given Orally Cmax (mcg/mL) Tmax (hr) AUC (mcghr/mL) Fasted† (N=23) Mean (SD) 41.2 (5.5) 4.0 (3.0 - 6.0) 1261.5 (236.5) Range 32.7 – 56.2 874.3 – 1750.4 High fat meal† (N=23) Mean (SD) 40.1 (4.9) 6.0 (4.0 - 8.0) 1248.2 (291.6) Range 31.0 – 47.7 762.0 – 1769.4 Mixed with applesauce (N=24) Mean (SD) 44.1 (4.6) 4.0 (3.0 – 6.1) 1523 (372.2) Range 37.4 – 55.6 1040 - 2350 Mixed with pudding (N=23) Mean (SD) 45.6 (5.1) 4.0 (4.0 – 6.0) 1447 (331.0) Range 38.6 – 60.4 997 - 2130 Mixed with yogurt (N=24) Mean (SD) 43.4 (5.4) 4.0 (4.0 – 8.0) 1478 (335.0) Range 36.3 – 59.3 965 - 2240 Median (range) †Admixed with applesauce Distribution The apparent volume of distribution of secnidazole is approximately 42 L. The plasma protein binding of secnidazole is <5%. Elimination The total body clearance of secnidazole is approximately 25 mL/min. The renal clearance of secnidazole is approximately 3.9 mL/min. The plasma elimination half-life for secnidazole is approximately 17 hours. Metabolism Secnidazole is metabolized in vitro via oxidation by human hepatic CYP450 enzyme system with ≤ 1% conversion to metabolites. Excretion Approximately 15% of a 2 g oral dose of SOLOSEC is excreted as unchanged secnidazole in the urine. Drug Interaction Studies Oral Contraceptives Concomitant administration of 2 g of SOLOSEC with the combination oral contraceptive (OC), ethinyl estradiol (EE) plus norethindrone (NE), to healthy adult female subjects resulted in a decrease in mean Cmax of EE of 29%, and no significant effect on the mean AUC of EE. Administration of 2 g of SOLOSEC 1 day before combination OC administration resulted in no significant effect on mean Cmax or AUC of EE. Concomitant administration of 2 g of SOLOSEC with the combination OC resulted in no significant effect on mean Cmax and AUC of NE (increases of 13% and 16%, respectively). Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Administration of 2 g of SOLOSEC 1 day before combination OC administration also resulted in no significant effect on mean Cmax and AUC of NE. [see Drug Interactions (7.1)] Ethanol Metabolism In vitro studies showed that secnidazole had no effect on aldehyde dehydrogenase activity. However, postmarketing observations of adverse reactions of nausea, vomiting, diarrhea, abdominal pain, dizziness, and headache with concomitant use of SOLOSEC and alcohol have been reported [see, Dosage and Administration (2.3), Adverse Reactions (6.2), and Drug Interactions (7.2)]. 12.4 Microbiology Mechanism of Action Secnidazole, like other 5-nitroimidazoles, enters the bacterial and Trichomonas cells where the nitro group is reduced by nitroreductase enzyme(s) leading to the production of radical anions and a series of intermediates, depletion of thiols, DNA damage, and death of susceptible isolates of Gram positive bacteria, Gram negative bacteria and T. vaginalis. Resistance The potential for development of resistance to secnidazole by bacteria and T. vaginalis associated with bacterial vaginosis and trichomoniasis, respectively, was not examined. The mechanism of resistance, like for other nitroimidazoles, appears to be multifactorial that include decreased uptake of the drug, higher efflux activity, and/or altered nitroreductase activity. Bacterial and T. vaginalis isolates exhibiting reduced in vitro susceptibility to metronidazole also show reduced susceptibility to secnidazole. The clinical significance of such an effect is unknown. Antibacterial Activity Culture and sensitivity testing of bacteria are not routinely performed to establish the diagnosis of bacterial vaginosis [see Indications and Usage (1.1)]; standard methodology for the susceptibility testing of potential bacterial pathogens, Gardnerella vaginalis or Mobiluncus spp. has not been defined. The following in vitro data are available, but their clinical significance is unknown. Secnidazole is active in vitro against most isolates of the following organisms reported to be associated with bacterial vaginosis: Bacteroides spp. Gardnerella vaginalis Prevotella spp. Mobiluncus spp. Megasphaera-like type I/II Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Anti-protozoal Activity Trichomonas vaginalis Standardized susceptibility tests do not exist for use in clinical microbiology laboratories. 13 NONCLINICAL TOXICOLOGY 13.1 Carcinogenesis, Mutagenesis, Impairment of Fertility Carcinogenesis Nitroimidazoles, which have similar chemical structures to secnidazole, have been associated with tumors affecting the liver, lungs, mammary, and lymphatic tissues in animals after lifetime exposures. It is unclear if these positive tumor findings in lifetime rodent studies of these nitroimidazoles indicate a risk to patients taking a single dose of secnidazole to treat bacterial vaginosis or trichomoniasis. Mutagenesis Secnidazole was positive in the Bacterial Reverse Mutation Assay but was negative for the rat micronucleus test and mouse lymphoma test. Impairment of Fertility In a rat fertility study, females were dosed for two weeks prior to mating until Day 7 of gestation with males that were dosed for a minimum of 28 days before cohabitation. No parental toxicity or adverse effects on mating performance, estrous cycles, fertility or conception was observed at doses of up to the maximum tolerated dose (300 mg/kg/day, approximately 1.4 times the recommended dose based on AUC comparisons). 14 CLINICAL STUDIES 14.1 Bacterial Vaginosis Two randomized placebo-controlled clinical trials (Trial 1 and Trial 2) with similar designs were conducted to evaluate the efficacy of SOLOSEC 2 gram for the treatment of bacterial vaginosis. A diagnosis of bacterial vaginosis was defined as all of (a) the presence of an off-white (milky or gray), thin, homogeneous vaginal discharge; (b) a vaginal pH ≥ 4.7; (c) the presence of Clue cells ≥ 20% of the total epithelial cells on a microscopic examination of the vaginal saline wet mount; (d) a positive “whiff” test (detection of amine odor on addition of 10% KOH solution to a sample of the vaginal discharge); and (e) a Nugent score > 4. Trial 1 enrolled 144 non-pregnant female patients aged 19 to 54 years and Trial 2 enrolled 189 non-pregnant females aged 18 to 54 years. Black or African American subjects in both trials were 54%. Efficacy was assessed by clinical outcome evaluated 21 to 30 days following a single dose of SOLOSEC. A clinical responder was defined as “normal” vaginal discharge, negative “whiff” test, and clue cells <20%. Additional endpoints included Nugent score cure (Nugent Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit score of 0-3) and therapeutic outcome. A therapeutic responder was defined as a clinical responder with a Nugent score cure. In Trial 2, the endpoints were also assessed at Day 7-14. In both trials, a statistically significantly greater percentage of patients experienced clinical response, Nugent score cure, and therapeutic response at 21 to 30 days following a single dose of SOLOSEC compared to placebo. Statistically significant results for the endpoints were also achieved at Day 7-14 in Trial 2. The percentage of patients with clinical response was also consistently higher in both trials in the SOLOSEC arm compared to placebo among all subsets of patients: number of prior episodes of bacterial vaginosis (≤ 3 episodes and ≥ 4 episodes) in past 12 months, baseline Nugent score (score 4-6 and score 7-10), and race (Black/African American and White). Tables 3 and 4 describe the efficacy of SOLOSEC in the treatment of bacterial vaginosis. Table 3. Efficacy of SOLOSEC for Treatment of Bacterial Vaginosis in Two Randomized, Double-Blind, Placebo-Controlled Trials in Modified-Intent-to-Treat Population at 21-30 Days Clinical Responder† Nugent Score Cure§ Therapeutic Responder Trial 1 SOLOSEC (N=62) n (%) Placebo (N=62) n (%) 42 (67.7) 11 (17.7) 50.0 (33.4, 66.7)‡ p<0.001 25 (40.3) 4 (6.5) 33.8 (18.5, 49.1) ‡ p<0.001 25 (40.3) 4 (6.5) 33.8 (18.5, 49.1) ‡ p<0.001 Trial 2 SOLOSEC (N=107) n (%) Placebo (N=57) n (%) 57 (53.3) 11 (19.3) 34.0 (18.7, 49.3) ‡ p<0.001 47 (43.9) 3 (5.3) 38.6 (26.2, 51.0) ‡ p<0.001 37 (34.6) 2 (3.5) 31.1 (19.6, 42.6) ‡ p<0.001 N=number of patients in treatment group (modified intent-to-treat population defined as all patients randomized who had a baseline Nugent score ≥4 and were negative for other sexually transmitted infections at baseline). †Patients missing one or more of the clinical assessments were considered as non-responders/not cured. ‡Difference in response (SOLOSEC – placebo) and 95% confidence interval §Patients with missing Nugent scores were considered Nugent score failures. Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Table 4. Efficacy of SOLOSEC for Treatment of Bacterial Vaginosis in Trial 2 in Modified Intent-to-Treat Population at 7-14 Days Clinical Responder† Nugent Score Cure§ Therapeutic Responder Trial 2 SOLOSEC (N=107) n (%) Placebo (N=57) n (%) 62 (57.9) 14 (24.6) 33.3 (17.4, 49.2) ‡ p<0.001 49 (45.8) 2 (3.5) 42.3 (30.4, 54.2) ‡ p<0.001 37 (34.6) 2 (3.5) 31.1 (19.6, 42.6) ‡ p<0.001 N=number of patients in treatment group (modified intent-to-treat population defined as all patients randomized who had a baseline Nugent score ≥4 and were negative for other sexually transmitted infections at baseline). †Patients missing one or more of the clinical assessments were considered as non-responders/not cured. ‡Difference in response (SOLOSEC – placebo) and 95% confidence interval §Patients with missing Nugent scores were considered Nugent score failures. 14.2 Trichomoniasis The efficacy of a single 2-gram oral dose of SOLOSEC for the treatment of trichomoniasis was evaluated in a multi-center, prospective, randomized, placebo-controlled, delayed treatment, double-blind, trial (Trial 4, NCT03935217). A total of 147 female patients from the United States aged 15 to 65 years were enrolled and randomized 1:1 to receive either SOLOSEC or placebo. The modified intent-to-treat (mITT) population included all randomized patients who were culture positive for T. vaginalis and negative for other sexually transmitted infections. Of the 131 female patients in the mITT population, the median age was 36 years, and 90.8% were African American. Baseline clinical symptoms of vaginal itching, discharge, or odor were reported in 111 (84.7%) patients. Following initial dosing, the test of cure (TOC) visit occurred 6 to 12 days later. At the TOC visit, patients received the opposite treatment (placebo patients received SOLOSEC and vice versa) with a return visit 7 to 12 days later. Results for microbiological cure, defined as testing negative for T. vaginalis, for the mITT population are presented in Table 5. The microbiological cure rate at the TOC visit was significantly higher in the SOLOSEC treatment group compared to the placebo group. Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Table 5: Microbiological Cure Rate in Trichomoniasis Patients at TOC Visit (mITT Population) in Trial 4 Endpoint SOLOSEC 2 g Placebo Treatment Difference n/N (%) n/N (%),e (95% CI) Microbiological Curea,b 59/64 (92.2) 1/67 (1.5) 90.7 (80.7, 96.5) c,d CI = confidence interval; TOC = Test of Cure a: InPouchTM TV test negative for T. vaginalis. b: Subjects with no test results are assumed to be positive (1 SOLOSEC and 3 placebo subjects). c: Exact CI from the Score Method. d: P-value <0.001 versus placebo from a CMH test adjusted for clinical symptoms (present/absent) of trichomoniasis at baseline. e: In placebo subjects with positive T. vaginalis culture at TOC, receiving delayed SOLOSEC treatment showed a comparable microbiological cure rate 7-12 days later (56/63 (88.9%), 95% CI: 78.4%, 95.4%) to the 92.2% cure rate for the initially SOLOSEC treated subjects. N: number of patients in treatment group. For microbiological cure, included all subjects in the modified intent-to-treat (mITT) population, defined as all randomized patients who were culture positive for Trichomonas vaginalis and negative for other sexually transmitted infections. The single oral 2 g secnidazole dose was also assessed in four open-label trials in males (one comparative study with metronidazole and ornidazole in males only1 and three single-arm studies in males and females2,3,4). Parasitological evaluation was performed both pre- and post-treatment and reported cure rates ranged from 91.7% (165/180) to 100% (30/30) at time points ranging from 2 to 20 days (n=437, 211 males and 226 females). In addition, the natural history of trichomoniasis in men was evaluated in one study.5 The spontaneous resolution during a mean follow-up of 16 ± 12 days was noted in 36% (5/14) (95% CI: 12.8%, 64.9%) of untreated men. 15 REFERENCES 1. Özbilgin A, Özbel Y, Alkan MZ et al. Trichomoniasis in non-gonococcic urethritis among male patients. J Egypt Soc Parasitol. 1994; 24(3):621-625. 2. Dyudyun AD, Polyon NM, Gorbuntsov VV. Secnidazole in complex treatment of patients with urogenital trichomoniasis. Dermatovenerology Cosmetology Sexopathology. 2016;1(4): 287-292. 3. Siboulet A, Catalan F, Videau D, Niel G. Urogenital trichomoniasis. Trials with a long half-life imidazole: secnidazole. Med Mal Infect. 1977;7(9):400-409. 4. Videau D, Niel G, Siboulet A, Catalan F. Secnidazole: A 5-nitroimidazole derivative with a long half-life. Br J Vener Dis. 1978;54(2):77-80. 5. Krieger JN, Verdon M, Siegel N, Holmes KK. Natural history of urogenital trichomoniasis in men. J Urol. 1993 Jun;149(6):1455-8. 16 HOW SUPPLIED / STORAGE AND HANDLING SOLOSEC (secnidazole) Oral Granules, 2 g, consists of off-white to slightly yellowish granules containing secnidazole. SOLOSEC is supplied in one unit-of-use child-resistant foil packet of Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit granules in an individual carton. Each packet contains 4.8 g of granules containing 2 g secnidazole. SOLOSEC is supplied as follows: NDC 27437-051-01carton containing one unit-of-use child-resistant 2 g foil packet Store at 20°C to 25°C (68°F to 77°F); excursions permitted to 15°C to 30ºC (59ºF to 86ºF) [See USP Controlled Room temperature]. 17 PATIENT COUNSELING INFORMATION Advise the patient to read the FDA-approved patient labeling (Patient Information and Instructions for Use). Administration Instructions Instruct the patient: • To sprinkle the entire contents of the packet of SOLOSEC onto applesauce, yogurt or pudding and take all the mixture within 30 minutes without chewing or crunching the granules. • That after consuming the mixture, they may take a glass of water to aid in swallowing. • That SOLOSEC is not intended to be dissolved in any liquid. Advise the patient that SOLOSEC may be taken without regard to the timing of meals. Vulvovaginal Candidiasis Advise the patient that use of SOLOSEC may result in vulvovaginal candidiasis that may require treatment with an antifungal agent. [see Warnings and Precautions (5.1)] Alcohol Advise patients to avoid consumption of alcoholic beverages and preparations containing ethanol or propylene glycol during SOLOSEC therapy and for 2 days afterward because nausea, vomiting, diarrhea, abdominal pain, dizziness, and headache may occur [see Dosage and Administration (2.3), Adverse Reactions (6.2), Drug Interactions (7.2), and Clinical Pharmacology (12.3)]. Lactation Advise women not to breastfeed during treatment with SOLOSEC and to discontinue breastfeeding for 96 hours following the administration of SOLOSEC. Also, advise a nursing mother that she may choose to pump and discard her milk for 96 hours after administration of SOLOSEC and feed her infant stored human milk or formula [see Use in Specific Populations (8.2)]. Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit Drug Resistance Patients should be counseled that antibacterial drugs including SOLOSEC should only be used to treat bacterial infections. They do not treat viral infections (e.g., the common cold). When SOLOSEC is prescribed to treat a bacterial infection, patients should be told that although it is common to feel better early in the course of therapy, the medication should be taken exactly as directed. Skipping doses or not completing the full course of therapy may (1) decrease the effectiveness of the immediate treatment and (2) increase the likelihood that bacteria will develop resistance and will not be treatable by SOLOSEC or other antibacterial drugs in the future. Manufactured for and Distributed by: Lupin Pharmaceuticals, Inc. Baltimore, MD 21202 © 2022 Lupin Inc. All Rights Reserved SOLOSEC® is a registered trademark of Lupin Inc. Code pending Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit PATIENT INFORMATION SOLOSEC (SO-lo-sec) (secnidazole) oral granules What is SOLOSEC? SOLOSEC is a prescription medicine used to treat: • bacterial vaginal infections in females 12 years of age and older. • trichomoniasis, a common sexually transmitted infection (STI), in people 12 years of age and older. Sexual partners should be treated at the same time. People should avoid having sex until they and their sex partners are treated (for example, when therapy has been completed and any symptoms have resolved) to help prevent reinfection. It is not known if SOLOSEC is safe and effective in children under 12 years of age. Do not take SOLOSEC if you: • are allergic to secnidazole or other nitroimidazole medicines. • have Cockayne syndrome. Before taking SOLOSEC, tell your healthcare provider about all of your medical conditions, including if you: • are pregnant or plan to become pregnant. It is not known if SOLOSEC will harm your unborn baby. Tell your healthcare provider if you become pregnant during treatment with SOLOSEC. • are breastfeeding or plan to breastfeed. You should not breastfeed for 96 hours (4 days) after taking SOLOSEC. SOLOSEC may pass into breast milk. Talk with your healthcare provider about the best way to feed your baby while taking SOLOSEC. Tell your healthcare provider about all the medicines you take, including prescription and over-the counter medicines, vitamins, and herbal supplements. How should I take SOLOSEC? • See the Instructions for Use at the end of this Patient Information leaflet for instructions on how to take SOLOSEC. • Take SOLOSEC exactly as your healthcare provider tells you to. Do not take SOLOSEC more often than it is prescribed. What should I avoid while taking SOLOSEC? Avoid drinking alcohol during treatment with SOLOSEC and for 2 days (48 hours) after you take SOLOSEC because the following side effects may happen: • nausea • diarrhea • dizziness • vomiting • stomach (abdominal) pain • headache What are the possible side effects of SOLOSEC? SOLOSEC can cause side effects including vaginal yeast infections. Symptoms of a vaginal yeast infection include white or yellowish discharge (discharge may be lumpy or look like cottage cheese) and vaginal itching. The most common side effects of SOLOSEC include headache, nausea, vomiting, diarrhea, abdominal pain, vaginal itching and a bad, bitter or metallic taste in your mouth (dysgeusia). These are not all of the possible side effects of SOLOSEC. Call your doctor for medical advice about side effects. You may report side effects to FDA at 1-800-FDA-1088. General information about the safe and effective use of SOLOSEC. Medicines are sometimes prescribed for purposes other than those listed in a Patient Information leaflet. Do not use SOLOSEC for a condition for which it was not prescribed. Do not give SOLOSEC to Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit other people, even if they have the same symptoms that you have. It may harm them. You can ask your pharmacist or healthcare provider for information about SOLOSEC that is written for health professionals. What are the ingredients in SOLOSEC? Active ingredient: secnidazole Inactive ingredients: Eudragit NE30D (ethyl acrylate methyl methacrylate copolymer), polyethylene glycol 4000, povidone, sugar spheres, and talc. Contains no ingredient made from a gluten-containing grain (wheat, barley, or rye). For more information, visit www.solosec.com or contact Lupin Pharmaceuticals, Inc. at 1-844-SOLOSEC (1-844- 765-6732). Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit INSTRUCTIONS FOR USE SOLOSEC (SO-lo-sec) (secnidazole) oral granules For oral use (by mouth) only. How to take SOLOSEC? 1. Open the SOLOSEC packet by folding over the corner marked by an arrow (see diagram) and tearing across the top. 2. Sprinkle the entire contents of the SOLOSEC packet onto applesauce, yogurt or pudding. The granules will not dissolve. Take within 30 minutes without chewing or crunching the granules. 3. You can drink a glass of water after taking SOLOSEC to help with swallowing. 4. You should not try to dissolve the medicine in water or any other liquid. Important Information • SOLOSEC may be taken before, after or during a meal. • Take 1 entire packet of SOLOSEC at one time. Do not take only part of the medicine and save a portion for later. • Avoid drinking alcohol during treatment with SOLOSEC and for 2 days (48 hours) after you take SOLOSEC. How should I store SOLOSEC? • Store SOLOSEC at room temperature between 68ºF to 77ºF (20ºC to 25ºC). • Keep SOLOSEC and all medicines out of the reach of children. This Patient Information and Instructions for Use have been approved by the US Food and Drug Administration. Issued: 1/2022 Manufactured for and Distributed by: Lupin Pharmaceuticals, Inc., Baltimore, MD 21202 © 2022 Lupin Inc. All Rights Reserved. SOLOSEC® is a registered trademark of Lupin Inc. Code pending Reference ID: 4926463 This label may not be the latest approved by FDA. For current labeling information, please visit
1225	https://www.khanacademy.org/science/how-does-the-human-body-work-class-12/x7babbc170453fdb8:human-reproduction/x7babbc170453fdb8:gametogenesis/e/structure-and-properties-of-sperm	Structure and properties of sperm (practice) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content How does the human body work? - Class 12 Course: How does the human body work? - Class 12>Unit 1 Lesson 4: Gametogenesis Spermatogenesis Spermatogenesis review Spermatogenesis Structure and properties of sperm Basics of egg development Oogenesis and follicular development review Oogenesis Science> How does the human body work? - Class 12> Human reproduction> Gametogenesis © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Structure and properties of sperm Google Classroom Microsoft Teams Problem Human sperm cells travel very large distances (relative to their own size) in order to reach an egg cell. What part of their structure provides them with this ability? Choose 1 answer: Choose 1 answer: (Choice A) Mitochondria in the midsection. A Mitochondria in the midsection. (Choice B) Muscles in their tails. B Muscles in their tails. (Choice C) ATP in the acrosome. C ATP in the acrosome. 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1226	http://www.mit.edu/~hlb/StantonGrant/Lecture13/example13.2.pdf	Related Rates, A Conical Tank Example: Consider a conical tank whose radius at the top is 4 feet and whose depth is 10 feet. It’s being ﬁlled with water at the rate of 2 cubic feet per minute. How fast is the water level rising when it is at depth 5 feet? As always, our ﬁrst step is to set up a diagram and variables. h r Figure 1: Illustration of example 2: inverted cone water tank. This diagram just helps us to start thinking about the problem. For instance, we see that because the cone is narrower at the bottom the rate of change of the depth will vary; we need to depict the water level. We also realize that it’s diﬃcult to draw useful and accurate diagrams of three dimensional ﬁgures — a simple schematic may be more helful. The key here is to draw a two-dimensional cross-section. In the ﬁgure we’re looking at one half of a vertical slice of the tank. The height of the slice equals 10 feet, which is the height of the tank. The widest part of the slice is 4 feet, which is the distance from center to edge of the top of the tank. We’ll use the variable r will represent the distance from center to edge of the top of the water, and h will represent the height of the top of the water (which is also the depth of the water). We can ﬁnd the relationship between r and h from Fig. 2) using similar triangles: r h = 4 10. 1 10 4 r h Figure 2: Relating r and h. Our goal is to ﬁnd out how fast the water is rising when the tank is half full. What we know is that the volume of water in the tank is changing at a rate of 2 cubic feet per minute. We need equations relating the volume of water in the tank to its depth, h. The volume of a cone is 1 3 · base · height. From Fig. 1), the volume of this tank is given by: V = 1 3 · πr2 \|{z} base · h \|{z} height This relates the volume to the height and radius, and we know the relation between the hight and the radius. We have one more piece of information that we can use: dV dt = 2. The question is: “What is dh dt when h = 5?” We’ve now translated all of the words in the original problem into formulas. Our word problem is now simply a calculus problem. We could do this by implicit diﬀerentiation, but it’s easy enough to solve for r in terms of h that there’s no need to. r = 2 5h. We plug this expression for r back into V to get: V = 1 3π 2 5h 2 h = 4 3(25)πh3 2 At this point we could solve for h, but that turns out to be a bad idea. Implicit diﬀerentiation is much easier. dV dt = dV dh dh dt = π 3 2 5 2 3h2 dh dt = 4 25πh2h′ Now that we’ve calculated the rates of change we can plug in the numbers dV dt = 2 and h = 5: 2 = 4 25 π(5)2h′ 2 = 4πh′ h′ = 1 2π ft/min We were given the rate at which the volume of water in the tank was changing and we used that to compute the rate at which the water in the tank was rising. At the heart of this calculation was the chain rule: dV dt = dV dh dh dt . Related rates problems are all about applying the chain rule to solve word problems. 3
1227	https://knowledge-things.github.io/2022/05/28/shu-xue-kai-qu-jian-he-bi-qu-jian/	数学-开区间和闭区间 \| Kevin Blog Kevin Blog 首页标签分类归档关于留言板 Books Kevin Blog All problems in computer science can be solved by another level of indirection. 首页标签分类归档关于留言板 Books Fork Me 数学-开区间和闭区间开闭区间数学目录发布日期: 2022-05-28 文章字数: 576 阅读时长: 2 分阅读次数:342 转载内容如有侵权或不希望被转载可以留言或私发告诉我，我会及时处理，尊重你的权利。原文地址[ 开区间使用符号小括号()表示，闭区间使用符号中括号[]表示，闭区间包含了两个端点，而开区间则不包含两个端点开区间用 (a,b)来表示，闭区间用 [a,b]来表示。闭区间包括了两个端点a和b，而开区间不包含两个端点a和b。下面这个游戏是我以前的老师告诉我的。游戏的规则是：两个人分别在开区间（a,b）内取一个数，两个人取的数不许相同，谁取的数大，那么谁赢。这个游戏显然是谁先取谁输！具体一点，假设是在(0,1)这个区间来玩这个游戏。第一个人取的是0.9另一个只要取0.91就可以赢了；如果第一个人取的是 0.99另一个只要取0.991就行了。总之，不管第一个人怎么取，都会落败。如果这个游戏换在闭区间[0,1]上进行，结果完全颠倒了，谁先取谁赢，因第一个人只要取1就可以了，第二个人因为不能取到这个数，那就输定了。 1）满足a≤x≤b的实数x的集合, 表示为[a,b],叫做闭区间; 2）满足a＜x＜b的实数x的集合, 表示为(a,b),叫做开区间; 3）满足a≤x＜b，a＜x≤b的实数x的集合, 分别表示为[a,b),(a,b],叫做半开区间. 这里实数a,b叫做区间的端点. 从上边的三个定义你就可以看出来，闭区间是有a,b两个端点的。示例：一共四种情况： none none (a,b)：区间范围内，不包含a和b [a,b]：区间范围内，包含a，也包含b (a,b]：区间范围内，不包含a，包含b [a,b)：区间范围内，包含a，不包含b 通常我们在程序中常听到的概念是左闭右开，也就是含左不含右，最常用的就是java中的字符串的截取方法sbuString，它采取的就是左闭右开策略文章作者:Kevin 文章链接: 版权声明: 本博客所有文章除特別声明外，均采用 CC BY 4.0 许可协议。转载请注明来源 Kevin ! 开闭区间 #### 微信扫一扫：分享微信扫一扫即可分享！赏你的赏识是我前进的动力支付宝微信评论提交 0 评论来发评论吧~ 加载更多... Powered By Valine v1.4.18 上一篇禅道的数据库结构 2022-06-02 禅道数据结构下一篇二分搜索算法 2022-05-28 algorithm 二分法目录 Copyright© 2022 - 2025Kevin \|Powered byHexo \|ThemeMatery \|总访问量:10694\|总访问人数:9269 本站已运行 3 年 132 天搜索
1228	https://swlimo.southwestlondon.icb.nhs.uk/wp-content/uploads/2022/10/Menopause-and-Hormone-Replacement-Therapy-HRT-Guidelines-for-South-West-London.pdf	SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 1 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 Menopause and Hormone Replacement Therapy (HRT) Guidelines for South West London 1. Symptoms ………………………………………………………………………………………………………………………………………………………………............ 2 2. Diagnosis and Investigations ……………………………………………………………………………………………………………………………………………… 2 3. Treatment Choice and Initiation …………………………………………………………………………………………………………………………………........ 2 3.1 Specialist team responsibilities ………………………………………………………………………………………………………………………………………… 2 3.2 HRT choice in women post-hysterectomy ………………………………………………………………………………………………………………………... 3 3.3 HRT choice in women with uterus intact ……………………………………………………………………………………………………………………….... 3 4. Counselling Points ……………………………………………………………………………………………………………………………………………………........... 3 5. Management of the Menopause ………………………………………………………………………………………………………………………………………… 4 5.1 Review & treatment duration ……………………………………………………………………………………………………………………………………....... 4 6. Lifestyle and Self-care ………………………………………………………………………………………………………………………………………………………… 4 7. Hormone Replacement therapy ……………………………………………………………………………………………………………………………………...... 5 7.1 Indications ……………………………………………………………………………………………………………………………………………………………………….. 5 7.2 Benefits of HRT ……………………………………………………………………………………………………………………………………………………………...... 5 7.3 Risks of HRT ………………………………………………………………………………………………………………………………………………………………........ 5 8. HRT for Specific Clinical Indications ………………………………………………………………………………………………………………………………….... 6 8.1 Bone Health and HRT …………………………………………………………………………………………………………………………………………………....... 6 8.2 Migraine and HRT …………………………………………………………………………………………………………………………………………………….......... 6 8.3 Premature ovarian insufficiency …………………………………………………………………………………………………………………………………....... 6 9. HRT Product Lists …………………………………………………………………………………………………………………………………………………………....... 6 9.1 Sequential Combined HRT ………………………………………………………………………………………………………………………………………………. 7 9.2 Continuous Combined HRT ……………………………………............................................................................................................. 7 9.3 Oestrogen only HRT ………………………………………………………………………………………………………………………………………………........... 8 9.4 Adjunctive progestogen ……………………………………………………………………………………………………………………................................ 9 9.5 Vaginal oestrogen only ……………………………………………………………………………………………………………………………………………....... 10 9.6 Tibolone …………………………………………………………………………………………………………………………………………………………………........ 10 9.7 Testosterone ……………………………………………………………………………………………………………………………………………………………...... 11 10. Troubleshooting and Management of Side Effects ………………………………………………………………………………………………………….. 12 10.1 Poor symptom control ………………………………………………………………………………………………………………………………………………… 12 10.2 Management of side effects ……………………………………………………………………………………………………………………………………….. 12 11. Unscheduled Bleeding on HRT ……………………………………………………………………………………………………………………………………..... 13 12. Stopping HRT …………………………………………………………………………………………………………………………………………………………......... 13 13. Contraception ……………………………………………………………………………………………………………………………………………………………..... 14 14. Referral Criteria ……………………………………………………………………………………………………………………………………………………………... 14 14.1 Referral to menopause clinic ……………………………………………………………………………………………………………………………………… 14 14.2 Referral to other clinics ……………………………………………………………………………………………………………………………………………… 14 15. Further Information ……………………………………………………………………………………………………………………………………………………….. 14 16. References & Acknowledgements ………………………………………………………………………………………………………………………………..... 15 SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 2 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 3. Treatment choice and initiation • HRT is available as transdermal or oral preparations. Transdermal preparations are the preferred route in most patients and should be the only route considered in the following instances: o individual preference o poor symptom control o gastrointestinal disorders affecting oral absorption o previous or family history of Venous Thromboembolism (VTE) or risk factors o Body Mass Index (BMI) > 30kg/m2 o variable blood pressure control, migraines (section 8.2), or gall bladder disease o current use of hepatic enzyme inducing medication • Where a patient opts for oral therapy, ensure they are counselled on the increased risk of VTE with this route. • Start at a low dose and increase if symptoms persist. Review the patient after 3 - 4 months if HRT has been started or changed, then at least annually thereafter, unless there are clinical indications to review earlier. 3.1 Specialist team responsibilities For patients requiring initiation of HRT following surgery, specialist teams should communicate clearly with primary care: • the type of hysterectomy performed • preferred choice of HRT according to tables below or clinical practice if different • ongoing needs for the patient in view of cervical screening 2. Diagnosis and investigations • Diagnosis is made from a clinical history in a woman > 45years, but remember: o investigate any abnormal bleeding first and exclude other causes o consider differentials if symptoms are atypical • Routine testing of follicle stimulating hormone (FSH) to diagnose menopause is not recommended practice except in the following groups of women (provided they are not taking combined hormonal contraception or high-dose progestogen as the diagnostic accuracy of the FSH blood test may be confounded by these treatments): o aged over 45 years with atypical symptoms o aged between 40–45 years with menopausal symptoms, including a change in their menstrual cycle o younger than 40 years in whom premature ovarian insufficiency is suspected – diagnosis in this group is based on elevated FSH on 2 blood samples taken 4-6 weeks apart and menopausal symptoms 1. Symptoms Assessment of a woman with menopause should include asking about symptoms, including frequency, duration and understanding their impact on the patient’s quality of life. Symptoms vary widely between individuals and may include any of the following: Table 1: Symptoms of menopause Symptoms as listed in Menopause-specific Quality of Life Questionnaire (MENQOL) Specific presentation may include one or more of the following Mood changes Being dissatisfied with their personal life, feeling anxious or nervous, feeling depressed, down or blue, being impatient with others, wanting to be alone, feeling tired or worn out or experiencing a lack of energy Cognitive disturbance Experiencing poor memory or concentration, accomplishing less than they used to Menstrual irregularities Change in normal pattern of periods. May be a change in flow, frequency or eventually may stop altogether Change in sexual desire Desire to avoid intimacy or loss of libido. Vaginal dryness during intercourse Joint and muscle pains Aches in back of neck or head, decrease in physical strength, decrease in stamina, low backache Skin changes Drying skin or changes to appearance, texture or tone of skin, increased facial hair Change in physical appearance Weight gain, reduction in muscle mass Hot flushes Hot flushes, night sweats, increased sweating compared to usual baseline Urinary problems Frequent urination, involuntary urination when laughing or coughing, frequent UTIs GI symptoms Flatulence, gas pains or feeling bloated Sleep disturbance Difficulty sleeping SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 3 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 3.2 Table 2: HRT choice in women post-hysterectomy Type of hysterectomy Recommended HRT Sub-total hysterectomy Uterus removed - Remnant cervical stump may contain residual endometrial tissue. It is common practice to consider sequential combined HRT (see Table 3.3) as a progestogen challenge for up to 3 months, and if no withdrawal bleed is noted with this, to consider it unlikely that residual endometrium is present and oestrogen only HRT can be considered to be sufficient. Ongoing continuous progestogen intake should be considered if there are concerns that the remnant cervical stump may contain residual endometrial tissue in women who experience cyclical bleeding with sequential combined HRT. There is limited evidence to guide practice in relation to the role or need for progestogen replacement in women who have had subtotal hysterectomy. Total Hysterectomy Uterus and cervix removed. Oestrogen only HRT Total hysterectomy with bilateral salpingo-oophorectomy (TH+BSO) Uterus, cervix, fallopian tubes and ovaries removed. Often seen in patients with a significant history of endometriosis. Following TH+BSO for severe endometriosis: Continuous combined HRT regimens should be considered to prevent reactivation of residual disease and to potentially prevent malignant transformation of residual deposits. However, there is limited evidence available on this to guide clinical practice. Following TH+BSO for other diagnosis: Oestrogen only HRT 3.3 Table 3: HRT choice in women with uterus intact Stage of menopause Recommended HRT Women with premature ovarian insufficiency (POI) Patients with suspected POI should referred to menopause clinic for advice and guidance- see section 14.1 Offer sequential combined HRT OR Combined hormonal contraceptive OR Mirena® plus oestrogen only HRT After one year of sequential combined HRT: After a minimum of one year of HRT, women who wish to avoid a monthly withdrawal bleed may attempt a switch to a continuous combined HRT regimen which aims to give bleed-free HRT – this will also minimise the risk of endometrial hyperplasia. The timing of switching from sequential to continuous combined HRT should be considered in relation to the woman’s age and the frequency of her menstrual cycles (prior to commencing HRT). Women under the age of 50 who had shorter durations of amenorrhoea before starting HRT are likely to need to continue on sequential intake for a longer duration before switching to continuous combined HRT intake. Women in the peri-menopausal period: <12 months since last bleed Same as POI, however: Combined hormonal contraceptive age limit should not exceed UKMEC criteria Women in the post-menopausal period: >12 months since last bleed Offer continuous combined HRT Women with an intact uterus with Mirena® IUS in place Offer oestrogen only HRT Women who have undergone endometrial ablation Combined HRT regimens (sequential combined HRT or continuous combined HRT) should be used in women who have undergone endometrial ablation to ensure the entire residual endometrium is protected and reduce the risk of endometrial hyperplasia. 4. Counselling points • Explain the risks and benefits of HRT (section 7) and the importance to commit to regular reviews and ensure patient understands the importance of progestogen uterine protection where applicable. • Remind women in the peri-menopause or with premature ovarian insufficiency that HRT is not a contraceptive and contraceptive precautions are still necessary. • Explain to patients about bleed patterns: o For patients who are still menstruating the use of continuous combined HRT will not stop menstruation o For women and people with a uterus, unscheduled vaginal bleeding is a common side effect of HRT within the first 3 months of treatment but should be reported at the 3-month review appointment, or promptly if it occurs after the first 3 months • Be realistic in what HRT can achieve and emphasise the importance of treatment adherence. • Advise patients that symptoms will usually start to improve by 4 weeks after HRT initiation. • Weight gain is very common around the time of the menopause. HRT does not cause significant further weight gain. • Counsel patients on continuing need to engage in national screening programmes including breast and cervical screening programmes. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 4 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 5. Management of the menopause • Before prescribing HRT conduct a full medical history, including personal, family and medication history and baseline checks for height/weight, BMI and blood pressure. • For psychological symptoms, such as mood disturbance, anxiety, and depression, consider a referral for cognitive behavioural therapy (CBT) as per NICE recommendation. Further information can be found on the BMS website. • NICE advise that SSRIs, SNRIs or clonidine should NOT be routinely offered first line due to limited efficacy and side effects. • Vaginal oestrogens should be offered to patients with genitourinary syndrome of menopause (GSM) (including those on systemic HRT or for whom systemic HRT is contraindicated). • The potential benefits of bioidentical hormone therapy can be achieved using conventionally licensed products, without having to resort to unregulated compounded varieties from specialist pharmacies. NHS South West London does not support the prescribing of unlicensed bioidentical HRT preparations. • Body identical licensed HRT therapy is supported as per the HRT product list section of document. 5.1 Review & treatment duration • Review the patient after 3-4 months if HRT has been started or changed, then at least annually thereafter, unless there are clinical indications to review earlier. • At each review appointment: ✓ Assess symptom control, tolerability and compliance ✓ Reassess risk relating to current choice of HRT: ▪ Long-term use of sequential combined HRT for >5 years may be associated with a small increase in risk of endometrial hyperplasia and endometrial cancer, with the risk being dose and duration dependent in relation to progestogen intake ▪ Consider a reduction in dose of HRT as patients get older. ▪ Recommend switch to transdermal preparation where appropriate to reduce risks of VTE ✓ Check blood pressure ✓ Emphasise the importance of: ▪ Keeping up to date with national screening programmes including breast screening programme (mammogram) and cervical screening programmes ▪ Regular breast self-examination - ensure patient is clear on how to perform this ▪ Bone health optimisation ▪ Contraception – see counselling points ▪ Attending NHS health checks • Regimen changes are not generally recommended in the first 12 weeks unless essential. This is to enable clinicians to get the best information on dosing and side effects before considering the need to change. At 3-4 month review after initiation or treatment change, if the patient continues to experience irregular bleeding offer adjustment of HRT (section 11). • HRT should be continued for as long as the benefits of symptom control and improvement in quality of life outweigh risks. 6. Lifestyle and Self-care • Lifestyle modifications should be recommended as per Table 4 to help alleviate symptoms. These should be implemented at the same time as medical management – they should not delay treatment initiation. • Women who have been through the menopause are at an increased risk of developing osteoporosis - encourage patients to eat a healthy diet containing plenty of calcium, purchase vitamin D supplementation and encourage regular weight-bearing and resistance exercise. • Women should be advised to contact occupational health if support is needed in the workplace. • Isoflavones (soy), black cohosh and red clover may help relive vasomotor symptoms, however their safety is unknown and preparations may vary in terms of quality and purity. Advise patients with hormone dependent cancers to avoid using these products and speak to their specialist team for advice. • For patients with vaginal dryness, vaginal moisturisers and lubricants can be used alone or in addition to vaginal oestrogen. These can be purchased over the counter from pharmacies or retail outlets. o Vaginal moisturisers: these are usually applied every few days via insertion directly into the vagina. The effects of a moisturiser generally last a bit longer than those of a lubricant. Many different brands are available to purchase. o Vaginal lubricants: water-based and oil-based lubricants can be applied to the vulva and vagina just before sexual activity to reduce discomfort during sexual intercourse. Numerous brands are available to purchase. Avoid petroleum jelly or other petroleum-based products for lubrication if also using condoms, because petroleum can break down latex condoms on contact. Table 4: Lifestyle recommendations to alleviate symptoms Symptom Lifestyle Modification Hot flushes and night sweats Regular exercise, healthy BMI, wearing lighter clothing, sleeping in a cooler room with silk pillows, using a fan, reducing stress and avoiding possible triggers e.g. spicy foods, alcohol, caffeine, smoking. Sleep disturbances Avoiding exercise late in the day and maintaining a regular bedtime. Mindfulness and sleep apps may be helpful. NHS recommended wellbeing apps can be found here. Mood and anxiety disturbances Adequate sleep, regular physical activity and relaxation exercises, mindfulness. Cognitive symptoms Exercise and good sleep hygiene. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 5 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 7. Hormone Replacement Therapy 7.1 Indications • Relief of short-term vasomotor symptoms e.g. hot flushes • Alleviate low mood as a result of the menopause • Urogenital atrophy • Premature ovarian insufficiency • Prevention of osteoporosis in postmenopausal women at high risk of future fractures 7.2 Benefits of HRT • Reduction of vasomotor symptoms. • Maintenance of bone mineral density and reduced risk of osteoporotic fractures. 7.3 Risks of HRT Much controversy exists about the risks of HRT. The safety of HRT largely depends on age. Reassure healthy women younger than 60 years that they should not be concerned about the safety profile of HRT. For the majority of women, the potential benefits of HRT when given for a clear indication are many and the risks are few when initiated within a few years of menopause. Table 5: Summary of the risks of HRT Risk Information Venous Thromboembolism (VTE) • Oral HRT increases the risk of VTE compared to baseline population risk. • When used at standard therapeutic doses, the risk associated with transdermal HRT is no greater than baseline risk. Breast Cancer • The baseline risk of breast cancer for women around menopausal age varies from one woman to another according to the presence of underlying risk factors. • Lifestyle factors (obesity, excess alcohol) may have greater impact on breast cancer risk than HRT (see pie chart at bottom of link). • HRT with oestrogen alone is associated with little or no change in the risk of breast cancer. • Vaginal oestrogen treatment: no increase in risk of breast cancer compared to control. • Combined HRT with oestrogen and progestogen: associated with an increased risk of breast cancer that is duration dependent. • Micronised progesterone and dydrogesterone may be safest progestogens. Cardiovascular disease & Stroke • CVD is the commonest cause of death in postmenopausal women. • HRT does not increase the risk of cardiovascular disease in women <65 years of age. • HRT may be cardioprotective in younger postmenopausal women (<10years from last menstrual period) but the evidence is not currently strong enough to recommend for primary prevention of CVD. • Stroke: Increased when oral (but not transdermal) HRT started in older women (> 60 years). • Tibolone increases the risk of stroke approximately 2.2 times from the first year of treatment as per MHRA guidance. Ovarian Cancer • Slight increased risk has been suggested from epidemiological studies, although causation cannot be inferred. Endometrial Cancer • For women with an intact uterus, taking oestrogen-only HRT increases their risk of endometrial cancer. • Continuous combined oestrogen and progestogen has been shown to have a neutral effect on the risk of endometrial cancer compared to placebo. Women’s Health Initiative (WHI) showed significant reduction in endometrial cancer risk in the postintervention phase. • Sequential combined HRT >5 years: may be associated with small increase in risk of endometrial cancer, with risk inversely proportional to number of days progestogen is given. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 6 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 8. HRT for Specific Clinical Indications 8.1 Bone health and HRT • For the prevention and treatment of osteoporosis in women with premature ovarian insufficiency (POI) and menopausal women below 60 years of age, particularly those with menopausal symptoms, HRT should be considered the first-line therapeutic intervention. • The bone-protective effect of oestrogen is related to dose and duration and the bone preserving effect of HRT declines after treatment discontinuation. • Some low dose HRT preparations are not licensed to prevent post-menopausal osteoporosis. However, studies have shown a bone-preserving effect even with relatively low doses of oestrogen replacement. In addition, some studies have shown that the use of HRT for a few years around the menopause may provide a long-term protective effect many years after stopping HRT • Please consult individual product literature via Electronic Medicines Compendium when prescribing and see comments in HRT products list. If prescribing a medication for an unlicensed indication, please discuss with the patient. 8.2 Migraine and HRT • Migraine aura does not contraindicate the use of HRT unlike with combined contraceptive pill use. • Changing oestrogen levels and menstrual disorders are linked with increased migraine prevalence during the menopause. Effective management of vasomotor symptoms is a recognised way of improving migraines. • Detailed guidance on how to manage migraine and HRT is available from the BMS which includes preferred medicines and routes of administration as well as non-pharmacological measures. • General advice includes: o Use the lowest topical oestrogen dose that effectively controls vasomotor symptoms o Where progestogen is required, continuous delivery is recommended, with preparations such as: ▪ levonorgestrel intrauterine system ▪ transdermal norethisterone (as in combined patches) ▪ micronised progesterone o Women with migraine and vasomotor symptoms who do not wish to use HRT or in whom oestrogens are contraindicated may benefit from escitalopram (unlicensed) or venlafaxine (licensed to treat menopausal symptoms, particularly hot flushes, in women with breast cancer). 9. HRT Product Lists Multiple stock supply disruptions can impact choice of HRT products. For up-to-date alternatives see guidance from the British Menopause Society. Prescribe HRT products by brand for continuity and to comply with licensing. Table 6 has been drawn up by the British Menopause Society as a practical guide to dose equivalents based on a combination of pharmacokinetics, clinical trials and clinical experience. These are subject to significant individual variation in absorption and metabolism so patients should be reviewed 3- 4 months after any changes are made. 8.3 Premature Ovarian Insufficiency (POI) • Premature ovarian insufficiency is a condition defined by loss of ovarian activity before the age of 40. It is characterised by menstrual disturbance such as amenorrhea or oligomenorrhea, with raised gonadotropins and low oestradiol. • POI can occur spontaneously or from iatrogenic causes (as a result of surgery or medication etc.). Establish the cause, if unclear or unknown refer to the specialist menopause clinic for further investigation. • Offer patient the choice of HRT or a combined hormonal contraceptive, unless contraindicated. • HRT should be continued until at least the age of natural menopause (average age is 51 years in the UK). After this point reassess and offer therapy as needed. • Consider/offer an early DXA scan utilising results to calculate FRAX score and to assess fragility risk. • Counselling points: o The baseline population risk of diseases such as breast cancer and cardiovascular disease increases with age and is very low in women aged under 40. o HRT may have a beneficial effect on blood pressure when compared with a combined oral contraceptive. o Both HRT and combined oral contraceptives offer bone protection. o HRT is not a contraceptive- see section 13 for further advice on contraception. Table 6: Estradiol equivalent doses for initiation and titration Ultra-Low dose Low dose Medium dose High dose Above licensed dose Oral 0.5mg 1mg 2mg 3-4mg Consider changing to an alternative product if patient on max dose and still experiencing symptoms, see trouble shooting (section 10). Use “Advice and Refer service if required. Patch Half a 25 microgram patch 25 micrograms 50 micrograms 75-100 micrograms Gel-pump Half a pump 1 pump 2 pumps 3-4 pumps Sachet Half of 0.5mg sachet 0.5mg 1-1.5mg 2-3mg Ref: BMS-HRT Practical Prescribing Tool For Clinicians SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 7 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 Key to medication status First line treatment option Second line treatment option Continuation in primary care after specialist initiation - see comments for detailed advice on transfer to primary care Hospital only No longer considered preferred therapy. During routine review consider change to alternative treatment option where appropriate. 9.1 Sequential Combined HRT: Suitable for sub-total hysterectomy (progestogen challenge), POI, peri-menopausal women, women who have undergone endometrial ablation Brand Oestrogen content Progestogen content Dose frequency Comments Patches Evorel Sequi® Estradiol 50mcg Norethisterone 170mcg Twice weekly FemSeven Sequi® Estradiol 50mcg Levonorgestrel 10mcg Weekly Preferred option for patients experiencing progestogenic side effects - see Table 9 Oral Femoston® Estradiol 1mg or 2mg Dydrogesterone 10mg Daily Preferred option for patients experiencing progestogenic side effects - see Table 9 Elleste Duet Estradiol 1mg or 2mg Norethisterone 1mg Daily Clinorette® Estradiol 2mg, 2mg Norethisterone 1mg Daily Novofem® Estradiol 1mg Norethisterone 1mg Daily Trisequens® Estradiol 1mg, 2mg, 2mg Norethisterone 1mg Daily Tridestra® Estradiol valerate 2mg Medroxyprogesterone 20mg Daily 9.2 Continuous Combined HRT: Suitable for TH+BSO, POI & peri-menopausal women after 1 year sequential combined therapy, women in the post-menopausal period, women who have undergone endometrial ablation Brand Oestrogen content Progestogen content Dose frequency Comments Patches Evorel® Conti Estradiol 50mcg Norethisterone 170mcg Twice weekly FemSeven® Conti Estradiol 50mcg Levonorgestrel 7mcg Weekly Preferred option for patients experiencing progestogenic side effects - see Table 9 Oral Femoston® Conti Estradiol 0.5mg or 1mg Dydrogesterone 2.5mg or 5mg Daily Preferred option for patients experiencing progestogenic side effects - see Table 9 Bijuve® Estradiol 1 mg Utrogestan 100mg Daily in the evening with food Contains gelatin Kliovance® Estradiol 1mg Norethisterone 0.5mg Daily Kliofem® Estradiol 2mg Norethisterone 1mg Daily Elleste Duet® Conti Estradiol 2mg Norethisterone 1mg Daily Indivina® Estradiol valerate 1mg or 2mg Medroxyprogesterone 2.5mg or 5mg Daily Premique low dose Conjugated oestrogen 300micrgram Medroxyprogesterone 1.5mg Daily During routine review consider change to alternative treatment option where appropriate. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 8 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 9.3 Oestrogen Only HRT: Suitable for total hysterectomy, sub-total hysterectomy (after progestogen challenge), Mirena® IUS in place, can be used as an alternative to sequential or continuous combined HRT along with adjunctive progesterone NB: When using with adjunctive progestogen annotate prescription with “only to be used in conjunction with ” to avoid risk of endometrial hyperplasia. Brand Oestrogen content Progestogen content Dose frequency Comments Patches Evorel® Estradiol 25, 50, 75 or 100mcg None Twice weekly Evorel® 25 not licensed for the prevention of post- menopausal osteoporosis. Estradot® Estradiol 25, 37.5, 50, 75 or 100mcg None Twice weekly Estradot® 25 and 37.5 are not indicated for osteoporosis. Patches are smaller in size than other brands which may be beneficial when prescribing higher doses. Estraderm® MX Estradiol 25, 50, 75 or 100mcg None Twice weekly Femseven® Estradiol 50, 75 or 100mcg None Weekly Progynova® TS Estradiol 50 or 100mcg None Weekly Gel Oestrogel® Estradiol 0.06% w/w Each pump actuation delivers 1.25 g of gel which contains 0.75 mg of Estradiol None Daily For patients unable to tolerate patches. Sandrena® Estradiol hemihydrate 0.5mg or 1mg per single-dose sachet None Daily Transdermal Spray Lenzetto® Estradiol 1.53mg per actuation (equivalent to 1.58mg estradiol hemihydrate) None Daily For patients unable to tolerate or comply with other topical products such as patches or gels. Oral Tablets Elleste Solo® Estradiol 1mg or 2mg None Daily Bedol® Estradiol 2mg None Daily Zumenon ® Estradiol 1mg or 2mg None Daily Progynova® Estradiol valerate 1mg or 2mg None Daily SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 9 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 9.4 Adjunctive Progestogen: Use alongside either oral or transdermal oestrogen for women with uterus to provide endometrial protection. Brand Oestrogen content Progestogen content Dose frequency Comments Intrauterine Delivery System Mirena® None Levonorgestrel 20mcg/24hrs Licensed for 4 years in UK for progestogenic opposition of oestrogen HRT but may be used for up to 5 years off-label. Oral Tablets/Capsules Utrogestan® None Micronised progesterone 100mg Sequential combined dose is 200mg daily at bedtime, for twelve days in the last half of each therapeutic cycle (day 15-26). Continuous combined dose is 100 mg at bedtime from day 1-25. Climanor® None Medroxyprogesterone acetate 5mg In women with an intact uterus, a cyclic regimen of 10mg a day for the last 14 days of each 28 day cycle to reduce the risk to the endometrium. Provera® None Medroxyprogesterone acetate 2.5mg, 5mg or 10mg Suggested dose of progestogen in a continuous combined HRT regimen would be a minimum 2.5 mg/day of medroxyprogesterone acetate. For low-dose sequential regimens medroxyprogesterone acetate 10 mg/day for 10–14 days a month. Provera® is not licensed for use as a progestogenic component of HRT but is widely used and supported by SWL Gynaecology Network Noriday® AMBER 1 None Noresthisterone 350microgram For continuous combined HRT regimen dose recommended is 3 tablets daily Not licensed for progestogenic opposition of HRT. Can be initiated in primary care on specialist advice only. Vaginal Pessary Utrogestan® vaginal pessary AMBER 2 None Micronised progesterone 200mg 200mg daily at bedtime for 12 days a month For patients who are unable to tolerate oral route. Unlicensed for this indication. For specialist initiation then continuation in primary care SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 10 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 9.5 Vaginal Oestrogen Only: Suitable for women with genitourinary syndrome of menopause only. Consider dexterity issues and patient preference. Brand Oestrogen content Progestogen content Dose frequency Comments Vaginal Cream Ovestin® Estriol 0.1% None 1 applicator (0.5mg estriol per application) Daily for 2 weeks followed by twice weekly 15g tube (1 applicator = 0.5g therefore 30 doses in a pack). Generic prep Estriol 0.01% None 1 applicator (0.5mg estriol per application) Daily for 2 weeks followed by twice weekly 80g tube (1 applicator = 5g therefore 16 doses in a pack). Second-line for patients unable to tolerate Ovestin®. Vaginal Tablets/Pessaries Vagirux® vaginal tablets Estradiol 10microgram None 10microgram daily for 2 weeks followed by twice weekly Applicator is reusable. Vagifem® vaginal tablets Estradiol 10microgram None 10microgram daily for 2 weeks followed by twice weekly Imvaggis® pessary Estriol 0.03mg None One pessary daily for 3 weeks followed by twice weekly Does not include applicator. Vaginal Ring Estring® Estradiol hemihydrate released at an average amount of 7.5 micrograms per 24 hours over a period of 90 days None Remove and replace every 90 days. Only licensed for 2 years continuous use. Consider in patients with allergies to other topical products. 9.6 Tibolone Brand Dose frequency Comments Tibolone 2.5mg tablets daily During routine review consider change to alternative treatment option where appropriate. • The risk benefit profile of this agent differs to that of other forms of HRT (section 7.3) and should not be routinely used. • This was traditionally offered as an alternative no-bleed regimen for postmenopausal women, more than 12 months after their last natural bleed. It was given as an alternative to combined HRT for postmenopausal women who wished to have amenorrhoea. • In patients older than about 60 years, the risks associated with tibolone start to outweigh the benefits because of the increased risk of stroke as per MHRA guidance. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 11 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 9.7 Testosterone : Suitable for menopausal women experiencing low sex drive. Brand Testosterone content Dose frequency Comments Gel Testogel® sachets AMBER 2 40.5mg in 2.5g Apply 1/8th (5mg) of a sachet daily. Apply a small pea sized amount once daily to clean and dry skin on either the lower abdomen, buttock or outer thighs. Rotate the site of application. Use at the same time each day. One 2.5g sachet should last 8 days, seal with a clip between uses. • Unlicensed for use in women • Do not use Testogel 16.2mg/g gel (pump version) preparation as the dispenser delivers supraphysiological levels. • One pack should last approximately 8 months with usual use. • Offer SWL Testosterone patient information leaflet For specialist initiation then continuation in primary care Tostran® pump dispenser AMBER 2 20mg in 1g. One press of the canister piston delivers 0.5g of gel containing 10mg testosterone. One press delivers 0.5g of gel containing 10mg of testosterone for administration three times a week. Apply to clean and dry skin on either the lower abdomen, buttock or outer thighs. Rotate the site of application. • Unlicensed for use in women • One dispenser lasts 6-8 months with usual use. • Offer SWL Testosterone patient information leaflet For specialist initiation then continuation in primary care • Topical therapy may be useful for menopausal women with low sexual desire if standard HRT dose is not effective. • There are currently no licensed treatments available for women in the UK. • There are many issues affecting libido, so if testosterone replacement is being considered, consider ruling out other causes. • Do not refer asymptomatic patients purely based on low systemic testosterone levels • Testosterone should not be used in patients currently being treated with tibolone due to increased risk of androgenic side effects. • In South West London, this medication has been approved for formulary use as an AMBER 2 medication. If clinically appropriate, a referral can be made to a menopause specialist. • It can be useful to request baseline blood tests. See BMS guidance for current details. • The specialist will initiate therapy issuing the first prescription. If the patient demonstrates clear benefits, prescribing and monitoring will occur in primary care. Monitoring of benefits and adverse effects testosterone: • An initial review to assess compliance, efficacy and tolerability should be done around 2-3 months from initiation. • It may take 3-6 months to fully evaluate the efficacy of treatment. • Check patients are correctly using product – this includes rotating sites of administration and checking that each pack is lasting roughly the expected durations indicated above. • There should be at least an annual re-evaluation of ongoing usage based on the same criteria that would be used for standard hormone therapy i.e. carefully weighing up the pros and cons of long term usage. • If any issues are identified, please use “Advice and Refer” for further guidance. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 12 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 10. Trouble shooting and Management of side effects 10.1 Poor symptom control Table 7: Suggested changes to combat poor symptom control Suspected problem Advice Compliance Allow 3-6 months on treatment for the full effect & counsel on the importance of compliance. Oestrogen dose Increase dose or change route of administration. If the patient is already on high-dose (see table 6) consider changing to alternative product. Blood test to check levels may also be considered. Poor patch adhesion / skin irritation Ensure that patient is rotating the application site. Switch to alternative brand or to oestrogen gel. Incorrect diagnosis Review indications (e.g. thyroid disease) or refer. Poor absorption Consider change to route of administration. Unrealistic expectations Counsel patient. Drug Interactions Enzyme inducers lower the circulating levels of hormones e.g. phenytoin, carbamazepine, rifampicin – change to a non-oral route or increase the oral dose (specialist advice may be required). Intrauterine systems are not affected. 10.2 Management of side effects In all patients encourage them where possible to persevere with treatment for 3 months as side effects may resolve. Table 8: Oestrogen-related adverse events-these may occur continuously or randomly throughout the cycle Side effects Management Breast tenderness or enlargement • evening primrose oil or starflower oil purchased OTC • wearing a well fitted bra or sports bra • topical or oral Non-steroidal anti-inflammatory (NSAIDs) if not contra-indicated • can be alleviated by a low-fat, high carbohydrate diet • reduce the dose of oestrogen Nausea, bloating or dyspepsia • may be helped by adjusting the timing of the oestrogen dose or taking with food • change the route of administration to a non-enteral formulation • if caused by Utrogestan® consider vaginal use (off license) rather than oral Headaches or migraines • these may be triggered by fluctuating oestrogen levels – try switching to a transdermal route as this produces more stable oestrogen levels Angioedema • symptoms of angioedema can be exacerbated or caused by oestrogens. This is particularly relevant in patients with hereditary angioedema. Consider a referral to menopause expert Table 9: Progestogen-related adverse events – tend to occur in a cyclical pattern during the progestogen phase of cyclical HRT Side effects Management Fluid retention Breast tenderness Lower abdominal pain Back pain Headaches or migraines Mood swings Depression Acne • Type: Changing the progestogen type, for example from a more androgenic one (such as norethisterone and norgestrel) to a less androgenic one (Utrogestan®, medroxyprogesterone or dydrogesterone). • Route: Changing the route of progestogen delivery, for example from oral to transdermal, vaginal, or intrauterine (IUS). This may be most beneficial for women who experience nausea with oral HRT. • Regimen: Reducing the regimen of progestogen administration. Progestogens can be taken for 10–14 days of each monthly sequential regimen, so swapping from a 14-day to a 10-day product may provide benefit. • Product: Changing to a product with a lower dose of progestogen. • Frequency: Reducing the frequency of progestogen dosing. This can be achieved by switching to a long-cycle regimen of administering progestogen for 14 days every 3 months (but this strategy is only suitable for women without natural regular periods). Continuous progestogen provides better long-term protection than cyclical. • Changing to continuous combined therapy often reduces progestogenic adverse effects with established use. However, this option is only suitable for postmenopausal women. NB: Treatment changes may need to be checked against the licensed indications and discussed with the patient as appropriate. See HRT Product lists for further details. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 13 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 11. Unscheduled bleeding on HRT • Exclude 2 week rule criteria for referral • Take a careful history to determine: o Type of HRT o Severity and extent of bleeding o Length of use of HRT o Compliance with HRT • For the majority of cases changing progestogen intake will control the bleeding especially in women who experience unscheduled bleeding in the first few months after commencing HRT. For suggestions on how to do this see Table 10 below: Table 10: Bleeding – troubleshooting advice Type of HRT Management For sequential HRT regimens 1) Increase dose of progestogen: • Micronised progesterone 300mg or medroxyprogesterone acetate 20mg for 12-14 days a month 2) Increase duration of progestogen intake to 21 days out of a 28-day HRT cycle. 3) Change type e.g. medroxyprogesterone acetate has good endometrial affinity and may provide the best bleed control For continuous combined HRT regimens 1) Increase dose of progestogen: • If using micronised progesterone 100 mg daily increase to 200 mg daily on continuous basis. • Medroxyprogesterone acetate 5mg to 10mg • If using continuous combined HRT or oestrogen plus Mirena® consider adding micronised progesterone/ medroxyprogesterone acetate or norethisterone to their HRT regimen. 2) Change type to e.g. medroxyprogesterone acetate 3) Change to a sequential regimen for another 12 months OR 4) Add desogestrel 75mcg if the bleeding is ‘period-like’ suggesting ongoing ovulation. If bleeding or spotting becomes heavier despite changes or occurs after a spell of amenorrhoea endometrial pathology should be excluded. Consider examination, transvaginal ultrasound and ensure cervical screening is up to date to exclude other causes and refer. For further details see Primary Care Women’s Health Forum documents: • How to manage women presenting with abnormal uterine bleeding in primary care without face to face contact during COVID-19 • How to manage HRT provision without face to face consultations during COVID-19 NB: Treatment changes may need to be checked against the licensed indications and discussed with the patient as appropriate. See HRT Product lists for further details. 12. Stopping HRT • Stop HRT when risks outweigh benefits and the patient agrees to stop. Consider a gradual withdrawal of HRT to limit a recurrence of symptoms. • Depending on clinical judgement, pending investigation, consider stopping HRT temporarily if the following occur: o Sudden breathlessness or cough with blood-stained sputum o Hepatitis, jaundice, or liver enlargement o Sudden severe chest pain (even if not radiating to left arm) o Unexplained swelling or severe pain in calf of one leg o Blood pressure above systolic 160mmHg or diastolic 95mmHg o New contraindication to treatment o Serious neurological effects, including unusually severe, prolonged headache, especially: ▪ If it is the first time, or getting progressively worse, ▪ There is sudden partial or complete loss of vision, ▪ Sudden disturbance of hearing or other perceptual disorders, ▪ Dysphasia, ▪ Vasovagal episode or collapse, ▪ First unexplained epileptic seizure, ▪ Weakness, motor disturbances, or very marked numbness suddenly affecting one side or one part of the body o Prolonged immobility after surgery or leg injury – if oral HRT SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 14 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 14. Referral Criteria 14.1 Referral to menopause clinic • Difficulty diagnosing menopause. • Contraindication to HRT such as previous idiopathic or current VTE or history of recurrent VTE (unless the woman is already on anticoagulant treatment), active or recent arterial thromboembolic disease, thrombophilic disorder, active liver disease with abnormal liver function tests, acute porphyria. • Carriers of faulty genes such as BRCA1 or BRCA2, known to increase risk of cancer or current, past or suspected breast cancer or other oestrogen-dependent cancer. • Ineffectiveness or persistent side effects despite logical therapy changes. • Low libido not improving on HRT – specialist advice required for consideration of testosterone therapy. • Premature ovarian insufficiency (POI) - for further advice and management (if primary concern is fertility refer directly to fertility clinic). The following baseline tests should be done prior to referral for POI and included in referral letter: • Thyroid function & TPO antibodies • Prolactin levels • Pelvic ultrasound • DXA • FSH/LH (2 levels done 6 weeks apart >30IU) 14.2 Referral to other clinics Abnormal bleeding: exclude two week rule criteria and consider referral according to local management pathway on abnormal uterine bleeding: • for those with normal physical exam but risk factors of irregular menstrual bleed, short cycle <24 days or prolonged bleeding ≥10 days and over the age of 40 years • for patients taking: ➢ Sequential HRT referral may be appropriate if there is an increase in heaviness or duration of bleeding, after attempt of adjustment of HRT (see Table 10) or if bleeding irregular ➢ Continuous combined HRT: if bleeding beyond six months of therapy (despite adjustment of HRT – section 11), or if it occurs after spell of amenorrhoea 13. Contraception See the Faculty of Sexual and Reproductive Healthcare guidance on ‘Contraception for Women Over 40 Years’ for detailed guidance. A woman is potentially considered to be fertile for 2 years after her last menstrual period if she is younger than 50 years of age, and for 1 year if she is over 50 years of age.1 In patients requiring contraception and free of all contraindications consider offering the following HRT/contraceptive options: • Oestrogen-only pill, patch or gel and Mirena® coil (other intrauterine systems are not licensed for endometrial protection) • Combined hormonal contraception (if eligible but only up until age 50) • Sequential combined HRT (pill or patch) and progestogen-only contraception (tablet, implant, injection) • If hormonal contraception is declined: advise barrier methods with sequential combined HRT When to stop contraception: • <50 years - 2 years after last period • >50 years - 1 year after last period • 45- 55 years - Mirena® after 45 years can stay in till age 55 for contraception and as part of HRT for 5 years only (unlicensed use) • Stop hormonal contraception at 55 years of age • Do not check FSH if on HRT / combined hormonal contraception • If >50 years with amenorrhoea due to progestogen only method- check FSH: if FSH level >30 IU/L stop after 1 year, if <30 IU/L repeat again in 12 months • On sequential HRT –contraception required until 55 years • Detailed advice on when to stop contraception can be found on the CKS ‘Contraception Assessment’. 15. Further Information • British Menopause Society contains up-to-date information on product shortages as well as useful guidelines including Tools for clinicians • Royal College of Obstetricians & Gynaecologists (RCOG) information leaflet on alternative treatments to manage menopausal symptoms • Women’s Health Concern; provides a confidential, independent service to advise, inform and reassure women about their gynaecological, sexual and post reproductive health. SWL Menopause and Hormone Replacement Therapy (HRT) Medicines Optimisation Guidance and Treatment Pathway Page 15 of 15 Lead author: SWL ICS Date approved: September 2022 Approved by: SWL IMOC Review date: September 2024 16. References & Acknowledgements British Menopause Society. HRT Guide. July 2020. (Accessed 16.05.22) British Menopause Society. Progestogens and Endometrial Protection. October 2021. (Accessed 16.05.22) British Menopause Society. Testosterone replacement in menopause. July 2020. (Accessed 16.05.22) British Menopause Society. BMS Consensus Statements. July 2020. (Accessed 16.05.22) British National Formulary Online (Accessed 16.05.22) Croydon Clinical Commissioning Group (CCG). Hormone Replacement Therapy (HRT) Guidance and Treatment Pathway. September 2018. Medicines and Healthcare products Regulatory Agency. Hormone replacement therapy (HRT): further information on the known increased risk of breast cancer with HRT and its persistence after stopping. August 2019. (Accessed 16.05.22) NHS Health A-Z. Menopause (Accessed 16.05.22) National Institute for Health and Clinical Excellence. Clinical Guideline 23: Menopause: diagnosis and management. Updated December 2019 (Accessed 16.05.22) National Institute for Health and Clinical Excellence. Clinical Knowledge Summaries: Menopause. Revised March 2022 (Accessed 16.05.22) PrescQIPP Menopause B299. April 2022. (Accessed 16.05.22. Requires subscription to access) Red Whale. Menopause and HRT. Updated January 2021. (Accessed 16.05.22) Summary of Product Characteristics (SmPC). (Accessed 16.05.22)
1229	https://www.youtube.com/watch?v=jRq7ohxoeZY	07 命题的真假关系常用逻辑用语高中数学凌海燕（小樂學堂） 90600 subscribers 11 likes Description 1939 views Posted: 22 Jun 2019 07 命题的真假关系常用逻辑用语高中数学 Transcript:
1230	https://brainly.com/question/53681952	[FREE] Which option best describes how to show that the Product Rule of Logarithms works for the base 4 expression - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +25,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +21,4k Ace exams faster, with practice that adapts to you Practice Worksheets +6,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Which option best describes how to show that the Product Rule of Logarithms works for the base 4 expression lo g 416+lo g 44? A. Evaluate both lo g 416 and lo g 44 and show that their product is equal to the value of lo g 464. B. Evaluate both lo g 416 and lo g 44 and show that their product is equal to the value of lo g 420. C. Evaluate both lo g 416 and lo g 44 and show that their sum is equal to the value of lo g 420. D. Evaluate both lo g 416 and lo g 44 and show that their sum is equal to the value of lo g 464. 1 See answer Explain with Learning Companion NEW Asked by bgnrzxd9vx • 10/24/2024 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1183071782 people 1183M 0.0 0 Upload your school material for a more relevant answer To show that the Product Rule of Logarithms works for the base 4 expression lo g 416+lo g 44, we'll follow these steps: Evaluate lo g 416: The expression lo g 416 asks us to find the power to which 4 must be raised to get 16. In this case, 4 2=16. Therefore, lo g 416=2. Evaluate lo g 44: Similarly, lo g 44 asks for the power to which 4 must be raised to get 4. This is straightforward since 4 1=4. Therefore, lo g 44=1. Sum the values: Now, add the results from steps 1 and 2: lo g 416+lo g 44=2+1=3 Evaluate lo g 464: According to the Product Rule of Logarithms, lo g 416+lo g 44 should be equal to lo g 4(16×4). Let's verify this. Calculate 16×4=64. Now, find lo g 464, which asks for the power to which 4 must be raised to get 64. Since 4 3=64, we find that lo g 464=3. Compare the results: The sum of the logarithms calculated in step 3 is 3, which matches the value of lo g 464 computed in step 4. This confirms that: lo g 416+lo g 44=lo g 4(16×4)=lo g 464 Therefore, the correct option is to evaluate both lo g 416 and lo g 44, and show that their sum is equal to the value of lo g 464. Answered by GinnyAnswer •8M answers•1.2B people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) 0.0 0 General Biology 1e Microbiology for Allied Health Students - Molly Smith Microbiology - Nina Parker, Mark Schneegurt, Anh-Hue Thi Tu, Philip Lister, Brian M. Forster Upload your school material for a more relevant answer To demonstrate the Product Rule of Logarithms for the expression lo g 416+lo g 44, we find each logarithm equals 2 and 1, respectively, summing to 3. Then, confirming lo g 464=3 shows the two sides of the equation match. Thus, the correct option is D, as their sum equals lo g 464. Explanation To show that the Product Rule of Logarithms works for the base 4 expression lo g 416+lo g 44, we can follow these steps: Evaluate lo g 416: The expression asks for the power to which 4 must be raised to get 16. We know that 4 2=16. Therefore, lo g 416=2. Evaluate lo g 44: This asks for the power to which 4 must be raised to get 4. Since 4 1=4, we have lo g 44=1. Sum the values: Now, add the results from steps 1 and 2: lo g 416+lo g 44=2+1=3 Evaluate lo g 464: According to the Product Rule of Logarithms, lo g 416+lo g 44 should equal lo g 4(16×4). Let's verify this. Calculate 16×4=64. Now, find lo g 464, which asks for the power to which 4 must be raised to get 64. Since 4 3=64, we find that lo g 464=3. Compare the results: The sum of the logarithms calculated in step 3 is 3, which matches the value of lo g 464 computed in step 4. This confirms that: lo g 416+lo g 44=lo g 4(16×4)=lo g 464 Therefore, the correct option is D, which states to evaluate both lo g 416 and lo g 44 and show that their sum is equal to the value of lo g 464. Examples & Evidence For example, if we had lo g 432+lo g 44, we could evaluate lo g 432 as 5/2 and lo g 44=1, giving a sum of 4.5, which would be equal to lo g 4128. Similar steps validate the Product Rule for different expressions. The properties of logarithms state that lo g b(x y)=lo g bx+lo g by, thus validating our findings with known values for base 4. Thanks 0 0.0 (0 votes) Advertisement bgnrzxd9vx has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics ∫(5 x 2−8 x)e 2 x d x= The following table shows the average annual income in a certain country, adjusted for inflation, for the given year. \| t= years since 1980 \| I= annual income \| :---: \| \| 0 \| $23,209 \| \| \| \| 5 \| $23,216 \| \| \| \| 10 \| $25,384 \| \| \| \| 15 \| $26,694 \| \| \| \| 20 \| $30,619 \| \| \| \| 25 \| $30,450 \| \| \| \| 30 \| $28,838 \| \| $ \| \| (a) Find the equation of the regression line. (Round regression line parameters to two decimal places.) (b) Plot the data along with the regression line. Write the point-slope form of the line's equation satisfying the given conditions. Then use the point-slope form of the equation to write the slope-intercept form of the equation. Slope = −2, passing through (−2,−2 3). A study compared the speed x, in miles per hour and the average fuel economy y (in miles per gallon) for cars. The results are in the table. \| Speed x \| 15 \| 20 \| 25 \| 30 \| 35 \| 40 \| 45 \| 50 \| 55 \| 60 \| 65 \| 70 \| :---: :---: :---: :---: :---: :---: \| Fuel Economy y \| 22.3 \| 25.5 \| 27.5 \| 29 \| 28.8 \| 30 \| 29.9 \| 30.2 \| 30.4 \| 28.8 \| 27.4 \| 25.3 \| Find the quadratic model in standard form for the data. Use the decimal precision from Desmos. What is the speed that maximizes fuel economy? Round to the nearest tenth. What would the fuel economy be if you were traveling at 55 mph? Round to the nearest tenth. 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1231	https://en.wiktionary.org/wiki/allure	Jump to content Search Contents Beginning 1 English 1.1 Etymology 1 1.1.1 Pronunciation 1.1.2 Noun 1.1.2.1 Translations 1.1.3 Verb 1.1.3.1 Synonyms 1.1.3.2 Translations 1.1.4 Derived terms 1.1.5 Related terms 1.2 Etymology 2 1.2.1 Noun 1.2.1.1 Alternative forms 1.2.1.2 Translations 1.3 Anagrams 2 Dutch 2.1 Etymology 2.2 Pronunciation 2.3 Noun 2.3.1 Derived terms 3 French 3.1 Etymology 3.2 Pronunciation 3.3 Noun 3.3.1 Derived terms 3.3.2 Descendants 3.4 Further reading 3.5 Anagrams allure Català Čeština Deutsch Eesti Ελληνικά Español Esperanto Euskara Français 한국어 Հայերեն Ido Bahasa Indonesia Íslenska Italiano ಕನ್ನಡ Kiswahili Kurdî Magyar Malagasy മലയാളം မြန်မာဘာသာ Nederlands 日本語 Oromoo پښتو Polski Português Русский Sängö Simple English Suomi Svenska தமிழ் တႆး తెలుగు ไทย Türkçe اردو Tiếng Việt 中文 Entry Discussion Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Create a book Download as PDF Printable version In other projects Appearance From Wiktionary, the free dictionary See also: alluré and Allüre English [edit] English Wikipedia has an article on: allure Wikipedia Etymology 1 [edit] From Middle English aluren, from Old French aleurer, alurer, from a (“to, towards”) (Latin ad) + leurre (“lure”). Compare lure. Pronunciation [edit] (US) IPA(key): /əˈlʊɚ/, /əˈlɚ/ (Received Pronunciation) IPA(key): /əˈlʊə/, /əˈljʊə/, /əˈlɔː/ \| \| \| --- \| \| Audio (Southern England): \| (file) \| (Scotland, Northern Ireland) IPA(key): /əˈlʉːɹ/ (Ireland) IPA(key): /əˈluːɹ/ Rhymes: -ʊə(ɹ) Noun [edit] allure (countable and uncountable, plural allures) The power to attract, entice; the quality causing attraction. Translations [edit] the power to attract, entice; the quality causing attraction — see also appeal \| \| \| Bulgarian: привлека́телност (bg) f (privlekátelnost) Catalan: atractiu (ca) m, encís (ca) m Chinese: Mandarin: 誘惑 / 诱惑 (zh) (yòuhuò) Dutch: aantrekkelijkheid (nl) Finnish: viehätys (fi), vetovoima (fi) French: attirance (fr) f Galician: encanto m German: Verlockung (de) f, Reiz (de) m, Anziehungskraft (de) f Italian: ammaliamento m, fascino (it) m, incanto (it) m, attrazione (it) f, richiamo (it) m Japanese: 魅力 (ja) (みりょく, miryoku) Norwegian: drag (no) n Persian: جذبه (fa) (jazabe) Polish: czar (pl), powab (pl) Portuguese: encanto (pt) m, atrativo (pt) m Russian: привлека́тельность (ru) f (privlekátelʹnostʹ), шарм (ru) m (šarm), очарова́ние (ru) n (očarovánije), обая́ние (ru) n (obajánije), ча́ры (ru) f pl (čáry) Spanish: encanto (es) m, fascinación (es) f, atractivo (es) f, aliciente (es) m Turkish: cezbetmek (tr) Ukrainian: шарм m (šarm), прива́бливість f (pryváblyvistʹ) \| Verb [edit] allure (third-person singular simple present allures, present participle alluring, simple past and past participle allured) (transitive) To entice; to attract. 1590, Edmund Spenser, “Book II, Canto XII”, in The Faerie Queene. […], London: […] [John Wolfe] for William Ponsonbie, →OCLC, stanza 31, pages 370–371: : [They retained] their ſweet skill in wonted melody; / Which euer after they abuſd to ill, / T’allure weake trueillers, whom gotten they did kill. 1603, Michel de Montaigne, chapter 8, in John Florio, transl., The Essayes […], book II, London: […] Val[entine] Simmes for Edward Blount […], →OCLC: : Injustice doth allure them; as the honour of their vertuous actions enticeth the good. 1737, R[ichard] Glover, “Book VI”, in Leonidas. A Poem.‎, page 152: : A tender voice his wondring ear allur'd. Synonyms [edit] attract, entice, tempt, decoy, seduce Translations [edit] to attempt to draw \| \| \| Bulgarian: привличам (bg) (privličam), изкушавам (bg) (izkušavam), съблазнявам (bg) (sǎblaznjavam) Catalan: atraure (ca), seduir (ca), temptar (ca), abellir (ca) Chinese: Mandarin: 引誘 / 引诱 (zh) (yǐnyòu) Danish: lokke Dutch: verleiden (nl) Finnish: viehättää (fi), houkutella (fi), houkuttaa (fi) German: locken (de), verlocken (de) Italian: attrarre (it), sedurre (it), incantare (it), ammaliare (it), affascinare (it), intrigare (it) Japanese: 魅惑する (ja) (みわくする, miwaku suru) Latin: allectō, pelliciō Polish: wabić (pl), nęcić (pl), przyciągać (pl) Portuguese: atrair (pt) Russian: завлека́ть (ru) impf (zavlekátʹ), завле́чь (ru) pf (zavléčʹ), зама́нивать (ru) impf (zamánivatʹ), замани́ть (ru) pf (zamanítʹ), прельща́ть (ru) impf (prelʹščátʹ), прельсти́ть (ru) pf (prelʹstítʹ), соблазня́ть (ru) impf (soblaznjátʹ), соблазни́ть (ru) pf (soblaznítʹ), пленя́ть (ru) impf (plenjátʹ), плени́ть (ru) pf (plenítʹ) Spanish: atraer (es), cautivar (es) Swedish: locka (sv) \| Derived terms [edit] allurance allurement allurer unallured Related terms [edit] lure Etymology 2 [edit] From Middle English alure, alour, from Old French alure, aleure (“walk, gait”), from aler (“to go”) +‎ -ure. Noun [edit] allure (countable and uncountable, plural allures) (dated) Gait; bearing. Harper's Magazine : The swing, the gait, the pose, the allure of these men. 2. The walkway along the top of a castle wall, sometimes entirely covered and normally behind a parapet; the wall walk. Alternative forms [edit] alure Translations [edit] gait; bearing \| \| \| Russian: аллю́р (ru) m (alljúr) \| Anagrams [edit] Laurel, laurel Dutch [edit] Etymology [edit] Borrowed from French allure. Pronunciation [edit] IPA(key): /ˌɑˈlyː.rə/ \| \| \| --- \| \| Audio: \| (file) \| Hyphenation: al‧lu‧re Rhymes: -yːrə Noun [edit] allure f (plural allures) air, pretension Derived terms [edit] sterallure French [edit] Etymology [edit] From aller +‎ -ure. Pronunciation [edit] IPA(key): /a.lyʁ/ \| \| \| --- \| \| Audio: \| (file) \| \| \| \| --- \| \| Audio (Canada (Shawinigan)): \| (file) \| \| \| \| --- \| \| Audio (France (Toulouse)): \| (file) \| \| \| \| --- \| \| Audio (France (Vosges)): \| (file) \| \| \| \| --- \| \| Audio (France): \| (file) \| Rhymes: -yʁ Noun [edit] allure f (plural allures) appearance, look speed, pace : à vive allure ― (please add an English translation of this usage example) 3. angle of a boat from the wind 4. gait (of a horse) 1894, Crafty, À travers Paris, page 4: : . . . un des chevaux du cortège a subitement pris une allure désordonnée que les efforts combinés de son cocher et de son camarade de timon ne sont pas parvenu à modérer. : . . . one of the horses in the procession suddenly took on a disorderly gait that the combined efforts of its driver and its comrade on the beam were unable to moderate. chemin de ronde (raised protected walkway behind a castle battlement) Derived terms [edit] à toute allure Descendants [edit] → Dutch: allure → German: Allüre Further reading [edit] “allure”, in Trésor de la langue française informatisé [Digitized Treasury of the French Language], 2012. 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1232	https://ir.library.oregonstate.edu/downloads/h128nh008	AN ABSTRACT OF THE THESIS OF Bolindra Nath Chutia for the M.S. in Mathematics (Name) (Degree) (Major) Date thesis is presented -Jam ll, I °l 66 . Title ON THE MELTING' CYLINDERS AND.-S,,PHERES Abstract approved Redacted for Privacy (Major professor) In this thesis we consider boundary and initial value prob-lems associated with the melting of homogeneous cylinders and spheres. The study includes the temperature distribution in cyl-inders and spheres which are perfectly insulated at the surface and subjected to a heat input respectively along the axis and at the center. Analytical solutions are obtained. The uniqueness question is not discussed. ON THE MELTING OF CYLINDERS AND SPHERES by BOLINDRA NATH CHUTIA A THESIS submitted to OREGON STATE UNIVERSITY in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE June 1966 APPROVED: Redacted for Privacy Professor of Mathematics In Charge of Major Redacted for Privacy Chairman of Department of Mathematics Redacted for Privacy Dean of CGraduate School Date thesis is presented n ÍÍ, NU,, Typed by Carol Baker ACKNOWLEDGEMENT The author wishes to express his deep gratitude and appreci- ation to Dr. G. Bo.dvarsson, Major Professor, (Visiting Professor), Department of Mathematics and Oceanography, for his helpful guidance, instruction and assistance throughout the course of this study. The author also is greatly indebted to Dr. R. E. Gaskell, Professor of Mathematics, for his kind instruction and encourage- ment during the early part of this work. He has taken keen interest in the work, but could not follow it to the end owing to his tempo- rary absence. TABLE OF CONTENTS Page INTRODUCTION 1 ASSUMPTIONS 6 FORMULATION OF THE PROBLEM 7 Cylinder with Finite Radius and Infinite Length Sphere with Finite Radius CASE OF AN INFINITELY LONG CYLINDER AND A SPHERE Cylinder with a Very Large Radius Sphere with a Very Large Radius 7 25 31 31 36 BIBLIOGRAPHY 42 NOMENCLATURE r = Space variable t= Time variable K= Thermal conductivity c= Specific heat e= Density k= Thermal diffusivity (= K/ p c) L = Latent heat of fusion a= Radius Q(t) = Strength of the source of heat per unit time per unit length of the cylinder Q(t) = Strength of the source of heat per unit time in the case of the sphere r' = Radius of the source (very, very small) T(r,t) = Temperature distribution before melting TL(r,t) = Temperature distribution after melting in the liquid region TSr,t) = Temperature distribution in the solid T = Melting temperature of the material t = Time at which the material begins to melt R(t) = Position of the boundary between the phases ta= The time of the complete melting of the cylinder or the sphere NOMENCLATURE (Continued) z= r/R(t) PCL, kL and ps, cs, ks are the corresponding quantities in the liquid and solid regions respectively. p=pL= ps (no change in density is assumed) Units are cgs, calorie and °C. ON THE MELTING OF CYLINDERS AND SPHERES INTRODUCTION Problems involving moving boundaries (known as Stefan Problems) are of great current interest in heat conduction theory. There is a large number of situations involving heat conduction in which moving boundaries result from the change of phase, for ex- ample, melting or freezing phenomena and the progress of temper- ature dependent chemical reactions through a solid. Many practical examples can be also cited, e. g. (1) Decay by evaporation (or growth by condensation) of a liquid drop, (2) Motion of a plane liquid -vapor interface, (3) Freezing of a lake, (4) Structural dam- age to hypersonic missiles caused by aerodynamic heating etc. Due to the difficulties caused by the non -linearity of these problems, it is generally very difficult to obtain analytical solutions. There are many classical references to this type of problem, usually involving the melting or solidification of slabs. Problems of this type first seem to have been considered by Franz Neumann in his lectures in the 1860's; Stefan considered the same type of prob- lem in 1891 imposing more restrictions. In 1921, Saitó dis- cussed the temperature distribution, during solidification, in steel ingots of various shape and size subject to different initial and boundary conditions neglecting latent heat of fusion. Lightfoot [ 16] 2 gave extensive study to Saitó's problem when latent heat of fusion could not be neglected. A simple solution was obtained by the inte- gral equation method. In recent years, high speed computers have been used in order to obtain numerical solutions of the partial dif- ferential equation involved. Work done by Landau , Citron , Lotkin [ 17] , Miranker and Keller [ 18] among others can be men-tioned. Boley's contributions towards this end are remarkable. He developed a new method in which the problem is reformulated so as to involve the solution of two ordinary integro-differential equations. By expanding the equations in powers of the time after melting starts, an exact analytical solution is obtained which is par- ticularly useful for small times. A review of the developments of the past few years in the field of heat conduction was presented by Boley including the application to problems of change of phase and moving boundary. He includes several methods to approxi- mate solutions in his paper. Other methods are given by Evans et al. , where the solutions are represented as power series. Although, this important group of problems has attracted the interest of many authors in the 19th and 20th century, problems involving radial flow in cylindrical and spherical coordinates have received comparatively little attention. The problems of melting are more difficult in the case of cylindrical and spherical flows. A solution of the non -linear problem of the rate of ice formation on 3 a long cylinder whose surface temperature is variable (but always below zero) is given by Pekeris and Slichter . The solution is applied to the specific problem of stabilization of wet soil by freez- ing. Stewart Paterson has considered the case of fusion when heat propogated from a line source (cylindrical coordinates) of heat and from a point source (spherical coordinates) of heat. Analytical solutions are obtained in both cases. Ingersoll et al. have dis-cussed the problem of freezing around pipes in soil at various con- ditions, the difficulties faced by piping engineers have been dealt with explicitly. Friedman's method of generalization of one dimensional problems to two and three dimensional cases should be mentioned in this context. Gibson has solved the same prob- lem involving spherical symmetry by using the method of separation of variables with the help of the special transformation T= f(t)G( ) h(t), where t is the time and the space variable and f, G, h are functions of single variables. Later he concluded that the variables are separable only if R(t) = Ct 2 , where C is a constant. An interesting practical problem, solidification of a cylinder and a sphere from the surface, has been presented by Kreith and Romie . The diffusion equation is expressed in dimensionless + i E ^5 4 form, and the solution of the partial differential equation is obtained by means of a method of iterative approximation in which the temper- ature is expressed in a series T=T+T+T+ + Tn+ ... The authors assume that each term of this series may be related to the preceding term by the equation ôT aT r .57a (r arn = a-1, =1+ A, where A is a dimensionless time variable, r is dimensionless space variable and G is constant. The last equation is integrated between limits 0 to r and each term of the series can be de- termined in this way. This paper attempts to solve the problem of melting prop- agating from the axes of cylinders and the centers of spheres due to heat sources of a variable strength. The problem of the sphere is artificial, but the problem of the cylinder has some practical applications, as for example, melting or solidification of soil or ice around a cylindrical heated pipe. Both the cylinder and sphere are assumed to be insulated at the surface, but cases involving infinite radius will also be treated. Unlike the work of some authors, the liquid formed by the melting is not immediately ár G5 removed, but no convective heat transfer is assumed, i. e. trans- portation of heat is purely by conduction. The growth of a frozen zone, starting from negligible initial dimensions with either a plane, cylindrical or spherical boundary, into a surrounding super cooled fluid is discussed by Chambré , where convection in the liquid region is taken into account. If the density does not change with the temperature in the liquid region, the assumption of no-convection will be correct. In both cases, analytical solutions are obtained. In this work, existence and uniqueness will not be discussed, reference can be made to Friedman , Kolodner [ 13] and Miranker . 6 ASS UMPT IONS In this thesis we shall assume that 1. The material is homogeneous. 2. All thermal properties of the material are uniform and constant; however, the properties of the liquid phase are not necessarily equal to the corresponding proper- ties of the solid phase. 3. There is no convection in the liquid, the transportation of heat is effected only by conduction. 4. Expansion or contraction of volume on melting is neglected. 7 FORMULATION OF THE PROBLEM Cylinder with Finite Radius and Infinite Length Consider a cylinder of radius a, perfectly insulated at the surface, and subjected to a heat input Q(t) per unit time per unit length along the axis. The initial temperature is zero. The radius of the source is r' where r' «a. The temperature T (r, t) in the cylinder prior to melting is a solution of the following boundary value problem: a aT(r,t) - 1 a[rT(r,t)] r' < r < a, 0 < t (1) ar ar k at (a) T(r, 0) = 0 r' < r < a (b) Q(t) _ -2 rKr ôr at r = r' aT(a,t) ar (2) Equation 2(a) states that initially the temperature distribu- tion in the cylinder is zero. Conditions 2(b) and 2(c) are bound- ary conditions, the first one gives the heat input to the cylinder at r = r' , the second results from the insulation of the surface. With constant heat input Q(t) = Q0, the foregoing problem could be solved by applying the Laplace transformation method. (c) 8 Since the cylinder is insulated at the surface, melting will be initiated around the source after the elapse of a certain time t= t. For t > t the cylinder will be divided into two regions, liquid and solid. The position of the phase boundary will be an un- known function of time which must be determined and hence gives rise to a floating boundary value problem. The position of the phase boundary will be denoted by r = R(t) and the temperature at this boundary will be equal to the melting temperature of the material, T. The problem for the liquid region in the post melting period, t > t, R(t) > r' , may be formulated as follows: aTL(r't) r 8T(r,t) 8r (r 8r ) kL at (a) TL(R (t), t) = T (b) Q(t) _-2TerKL 8r r' <r<R(t), t < t. (3) 8TL(r,t) at r = r' (4) The condition 4(a) expresses the fact that at the phase boundary, the temperature is constant and equal to the melting temperature of the material. In the solid region we have: e9 ó áTS(r, t) r óTS(r, ór (r ár ) ks at R(t) < r < a, t3 < t. (5) TS(R(t), t) = óTS(a, t) ór T = o (c) T(r,t) = T(r,t) áTL(r, t) óTS(r, t) (d) -KL ór KS ar I-`P R (t) at r=R(t). (e) R(t) = r' . (6) Conditions 6(a), 6(b) are not new, while 6(c) implies that at t = t, i. e. when melting starts, the temperature is con-tinuous. The last two conditions provide (d) that part of the heat passing through the liquid region enters the solid region, while part goes towards overcoming the latentheat of fusion L and (e) that there is no liquid region at t = t. The condition 6(d) above may be put into an alternative form by considering the curves of constant temperature TL(r,t) = T = TS(r,t) at r = R(t). Hence (a) (b) 10 aT aTL aT ar s Rdt + at dt = 0 = arS Rdt + at dt at r = R(t). So equation 6(d) becomes aTL (r, t) aTS(r, t) aTL/ at aTS/at -KL ar + KS ar - LP aTL/ ar -LP aTS/ar at r = R(t) (7) That the problem is non -linear can be seen from the fact that two differentheat input functions Q1(t) and Q2(t) will give different phase boundaries R1(t) and R2(t), because the move-ment of the phase boundary depends upon the strength of the source and on other physical constants; and that the solution for the sum Q1(t) +Q2(t) cannot be obtained by addition of R1(t) and R2(t). The non -linearity of the problem is also apparentfrom equation (7) above in that form. In this work we will attempt to determine the quantities t, ta, TL, TS and R(t). The last three quantities will be obtained in the next chapter under further restrictions to be explained later. The first two quantities are obtained by Landau for the problem of melting of a slab, by Gauss's Theorem applied to the heat conduc-tion equation, the domain of integration is xt plane. The same quantities for the same problem are determined by Citron by using simple integration method. Here, use is made of Citron's 11 procedure to determine t and ta. a Integrating (3) between the limit r = r' to r = R(t) and using conditions 4(a) and 4(b), we obtain: or 8 (r 8TL(r't) 1 CR(t)8 ar 8r )dr = k J at [rTL(r, t)] dr. r L r aT L(r, t) r ar R(t) r=r' R(t) = kl { et rTL(r,t)dr-R(t)TL(R(t), t)d Rtt) }. L r' TL(r,t) is the temperature distribution in the homogeneous liquid where no convection takes place. Hence TL(r, t) is a solu-tion of the heat conduction equation and is therefore a smooth func-tion in r' < r < R(t) and hence the order of differentiation and in- tegration can be interchanged. 8TL r ar or 8TL -r 8r r=-R(t) R(t) = kl {8 rTL(r,t)dr-R(t)T4 dRtt) r-_ r, L r, 8TL 8TL r-_ r 1a { R(t) rTL(r,t)dr-TrR (t)/2} 2 r r ar -r 8r r=R(t) k at L r' J R(t) , at or aTL r ar 12 R(t) 2Kt) k at { rTL(r,t)dr -TR (t)/2} 2 r=R(t) L L r' using 4(b) (8) Proceedingexactly the same way, integrating (5) between the limit r =R(t) to r =a and using conditions 6(a) and 6(b), one will obtain: -r = kl aá { ,a rTS(r,t)dr + TR2(t)/2}. (9) r=R(t) S R(t) Multiplying 6(d) by r we get aTL -KLr ar aT S r=R(t) + KSr ar and substituting (8) and (9) r=R(t) K R(t) Q(t) - kL L at rTL(r,t)dr - TR2(t)/2} 2 L r' = LpR(t) R(t) k S at { J a rTS(r,t)dr + TR2(t)/2} = Lp R(t)R(t) S R(t) Ç{ k aTs ar 2n a or R(t) Q(t) = 2Trp cL ôt { rTL(r, t)dr - TR2(t)/2} r' 13 ('a +2Trp cS at { J rTS(r,t)dr k TR2(t)/2} + 2Trp LR(t)R(t). (10) R(t) Integrating (10) between t = t and t = t1 Q(t)dt = 2Trp cL tl (' áat ( PL=PS=P) SR(t) rTL(r,t)dr - TR2(t)/2}dt r' tla +2Trp cs at { SI rTL(r,t)dr + TR (t)2} dt t r' +2Trp L R(t)i(t)dt. t Replacing t by t we get t R(t) ,5' t { t J 214 R(t) Q(t)dt = 2TrpcL{S' rTL(r,t)dr - TR2(t)/2} t r' R(t) -2Trp cL { rTL(r, t )dr - TR2(t)/2} r' 'a +2Trp cs {.5 rTS(r,t)dr + TR2(t)/`2} JR(t) -2npc { rTS(r, t)dr + TR2(t)/ 2} R(t ) + 2TrpL[R2(t)/2 - R2(t)/ 2] 2Trp c{ R(t) rTL(r,t)dr - TR2(t)/2} r' +2Trp rTS(r, t)dr + TR2(t)/2} R(t) +2TrLp [R2(t)/ 2 - r' 2/21 a -2Trpcs[ rTS(r,t)dr + Tr' 2/2]+ TrpcLTr'2 r' R(t) = r' (11) We are interested in obtaining the time for complete melting of the cylinder, ta. If heating is continued long enough, the time to will be obtained observing that R(ta) = a, hence S" J t a a1 '.' at a ('a J Q(t)dt = 2Trp cL { rTL(r, ta)dr - T>ka2/ 2} + 2Trp cs{Ta2/ 2} r' t + TrpcLTr'2 + Trp L(a2-r' 2) - 2Trp c or 15 51a. rTS(r, t)dr + T r' 2/2] r' ta Q(t)dt = Trp [a2(cST - cLT) + L(a2-r' 2)] + TrpcL,I.,,,r,2 t a -2Trp cs rTS(r, t)dr + 2Trp cL rTL(r, ta)dr. r' r' (12) Proceeding in exactly the same way, equation (1) and the con-ditions 2(a), 2(b) and 2(c) give or aT -rar r = r' a ak at rT(r,t)dr r' 2Trx a a Q(t) - k at rT(r,t)dr. r' (13) An expression for t may be obtained by integrating (13) between the limit t = 0 to t = t: 1 (a l = (' t Q(t)dt = 2Trp c JO Sa rT(r, t)dr dt r' ('a =2Trp c[ J rT(r,t)dr r' a =2Trp c rT(r,t)dr r' So adding (12) and (14) we get, rT(r, 0)dr ] 16 (14) t ('ta Q(t)dt + çQ(t)dt = Trp [a2(cST-CLT)+ L(a2-r'2) ]+ TrpcLTr'2 0 a ('a -2Trp cs rTS(r,t)dr+ 2Trp cL rTL(r,ta)dr r' r' a + 2npc S rT(r,t)dr r' But T(r,t ) = Ts(r,t ), c = c. ta Q(t)dt = Trp [ a2(cST- - CLT) + L(a2 - r'2) ]+ TrpcLTr'2 0 +2Trp c rTL(r, ta)dr . r' (15) It can be seen that (15) equates the total heat inflow in time t = 0 to t = ta to the heat content in the cylinder atthat moment. ('t a \ at r' a ( 1 J a 1 t (J a17 In order to be able to estimate t on the basis of equation (14) we have to determine T(r, t). This may be obtained from equation (1) with conditions (2) by applying the Laplace transform method. A sufficient condition for the existence of the Laplace transform is that T(r,t) is of exponential order, i. e. I e-PtT(r,t)1 < c e-(P-a)t where c is any constant. We will assume that this condition holds, and applying Laplace transform to (1) and (2) we get: d2T 1 dT 2- dr2 + r dr T co where T = 1 T(r, t)e -ptdt. 0 2_p q k(a) -Q0 = -2 TrKr dr P (b) dT (a P) - 0 dr r=r' (2)' From (1') T(r,p) = AI0(gr) + BK0(qr) where I0 (gr) and K0 (gr) are the modified Bessel's functions of the first and second kind of zero order. Using 2' (a) and 2' (b) we have 18 Q0 K1(ga)IO(gr) + KO(gr)I1(ga) T(r,P) - 2TrKr'Pg [ K1(gr')I1(ga) -K1(ga)I1(gr') I Then by the inversion theorem, T r t - QO K1(a)IO(μr)+KO(μr)Il(a) eAt dA ( , ) 2 K (μr')I (μa)-K (μa)I (μr') μX 4Tr iKr y-100 1 1 1 1 where μ = . The integrand, when expressed in series form, will contain only even powers of μ, therefore it is a single valued function. So we need to take the contour of integration as shown in Figure 1. We take the radius of the semi -circle as k(n+ 2)2Tr2/ a2 so that it does not pass through a pole. When n-oO the integral over BB'CA'A tends to zero [51 . So the line integral in question will be equal to the sum of the residues of the poles at X = 0 and K1(4r9I1(μa) - K1(μa)I1(μr') = 0. There are double poles at X = 0. The residues at X = 0 will be given by 42k= 1aan2?3 VIV alS (Z,a-v){, Z(Z,a-Z) Za_Z ( a+ Y)a .I Oj ans. [ ,ü ti2oT ñ+ _r..vez - ZZ£Z 61 ZZ 0= XTI(eri))i'(Ert)TI(Iarl)Ix, XP 3Y (Prl)II(ari)Oxrl+(arl)OI("eri)Txri P (1a71)II(eT)I31-(erl)tI(iari)T31 Yj `(P Ceri) I(arl)Ox-(arl)OI(eri)Tx ZP Z -+ -et , x Y E+ - a. . [ °-'( [ l(a 20 Neglecting powers of r' greater than one, since r' is small in comparison with a, we find that the residue at X =0 is The residues at 2r' r2-a2 a a kt r' [ a + -log-f ]-4a 2 r a 4 K1(μr')I1(μa) - K1(μa)11(μr' ) = 0 : The poles are X = -kal , -ka2 2 , -ka2 , ' where al, a2, a3, are the roots of Y1(ar' - Y1(aa)J1(ar') = 0 1/ Hence the residues at X = -ka 2n 1,4K1(F.a..)I0 (1.1,r) fK0 (μr)I1(μa)] eXt X2 d [K1(μr)11(μa)- K1(Na)I1 )] =Tr X =-ka2 n [)( ) ( )] ka2t Jl( an) aYO( an - Yr lanaJOanr Jl( aá1) Jl(ran) e n [ J 2(aa ) n - J1 2 n 1 Wan)] 1/ al, a2, a3, are all real and simple. See Gray and Mathews, Treatise on Bessel Functions. 2nd Edition, 1922. p. 82. L See page 173. k a n 2/ 21 T(r,t) - Q0[ r2-a2+ log a 2kt 27K 2a 2 r 2 1 4 co YO(anr)J1(aan)-Yl(aan)J0(anr) -karit +-, J1(aari J1(r'an)e I. n=1 an[J1 (aan)-J1 (r' an)] This is a rapidly converging series for higher values of time. So the sum of finite number of terms will give the result with sufficient accuracy. From (16), the time when melting starts at r = r' can be obtained: T(r't) - 27rK[ (r 2 + log + 2a 2kt a 4 + n Y0( ñr')Jl(aál)-Y1(aan)J0(anr') J aa J r a ekant 1 r , n=1 an[ (aan) J1 - J1 (r i an) 1 ( n) l( n) Tr n n 1 n n n , r n n ) .. 2 2 or 3 T = T(r', t`) _K[(log r' + 2 2t -4' a n +-, r 00 Y0 (anr' )J1(aari-Y1(aan)J0 (anr' ) an[J12 (aan) - J1 2(r an)w ' 22 -ka 2t J1(aan)J 1(r' ari e n ]. (17) In this series only unknown quantity is t; so taking finite number of terms, an approximate value of t could be obtained with considerable accuracy. In order to apply equation (15) to the problem of determining the time of total melting t , a we have to evaluate the integral Sa rTL(r, ta)dr r' where TL(r, ta) is the temperature distribution in the cylinder at the time t . a An exact evaluation of this integral therefore requires a solution of the melting problem for the cylinder. However, the following approximate method can be used in order to arrive at a fairly good estimation of the fixed ta. The right hand side of (15) is nothing but the total heat content of the just molten cylinder. As this quantity involves an integration in space, a 2 nn n nnn n-1 23 we can use an approximate temperature distribution. One way of arriving at such an approximation is to use the fact that the latent heat of fusion L can be regarded as a part of the internal energy of the material and define an average specific heat of the material for temperatures above the melting point. Hence for T0 > T we define the average specific heat (See Figure 2 and Figure 3) cST + L + CL(TO -T ) cav TO For constant conductivity K, the heat conduction equation for an incompressible material can be written as: 2DE dE 8T DT Kp T= = P 8t - P dT at - p c at where E is the total internal energy. In general, the specific heat c is a function of temperature T and has to include the heat of fusion. As pointed out previously, the equation is non -linear and can not be solved explicitly in general. But as a first approxi- mation we may use cav for c : 28T KV T= P cav -at and solve this equation in the case of a constant cav' For tempera - tures close to the melting temperature, we can use the value Molten State L Source Figure 2. í /, T T Figure 3. TO t = ta r'< r < aTo > T 24 /1 1 ~T 25 ti cav .., cS ,T So a solution identical to (16) can be obtained, and since it is a uniformly convergent series, term by term integration can be performed. If the liquid thus formed is removed at once, then from (15) a Q(t)dt = Tr p [a2cT f L(a2-r' 2)1 = rp a2(cT+ L) (r' ;0) 0 If Q(t) = Q0, then ta = Trpa2(cT):4+ L) /Q0 which is similar to the expression obtained by Landau in the case of melting of a slab. Also this corresponds to Citron's expression for the complete melting time for the slab. Sphere with Finite Radius Consider a sphere of radius a, insulated over its surface. A constant heat input (point source) Q0 units per unit time is at the centre of the sphere; the initial temperature distribution is zero. The temperature distribution in the sphere before melting occurs will be given by the following boundary value problem: L ('t a 2 aT 1 a 2 ar (r ar ) k T(1. T) 26 0<r<a, 0<t (18) (a) (b) T(r, 0) QO 0 2 -4TrKr ar r=0 (19) (c) ôr (a,t) = 0 This problem can be solved by applying Laplace's transform. The subsidiary equations d2v are: 2- 2 dr q v , v = Tr (18)' (b) QO =-4TrKr 2[ 1 dv v r dr -17) 2r r=0 (19)' áv dr v(a, p) a r=a Solving (18)' and (19)' we get, Q0 sinhq(a-r) - aq coshq(a-r) v - ,4TrKp [ sinhqa - aq coshga By inversion formula, aT p - (cl Q0 T(r,t) - 81r2kir Y-loo sinhμ(a-r) - aμcos hμ (a-r) e t dX. sinhμa - aμcoshμa =μ The integrand when expanded will contain only terms with even powers of μ, so it is a single-valued function. So we take 27 the same contour as shown in Figure 1. The radius of the circle is large enough to include all the poles and we will assume that the circle will not pass through any one of the poles. As the radius of the circle tends to infinity the integral over the circumference tends to zero. So the line integral in question will be equal to the sum of the residues at the poles. We have residues at X = 0: (a-r)2 (2a+r) 3krt 2a3 a3 The poles at sinhμa - aμcoshμa = 0 are X = -kai/a2, -ka2/a2, . where μa = ±ian are the roots of sinhμa - aμcoshμa = O. The nn residues at X = -kan %a are : X k k n (1' J - sinhμ(a-r) - aμcos hμ (a-r) ,Xt at X = _kat/a2 X d[ sinhμà - aμcos hμa 1 n 2 sin[ an(a -r) /a1 - ancos[an(a -r) /a1 2/a2)t e n a sin¢ .. T(r,t) - n n QO 3krt + (a-r)2(2a+r) 4TrKr 3 a 2a3 +2 00 n=1 28 sin[ an(a -r) /a1 - ancos[ an(a -r)/ a1 -ka 2 2 e( n )t" 2 an n an (20) After melting occurs, our problem will satisfy the following set of equations : a2 aTL 1 a 2 ôr (r ôr. ) = kL T(r TL) (a) TL(R(t), t) = T 0<r<R(t), t<t. (21) 2 aTL (b) QO = -47KLr ôr and r=0 (22) e n (¢ n n k 2 [ 129 aT a r a (2 arS) kl at (r 2TS), R(t) < r < a, t < t . (23) S (a) TS(R(t), t) = T (b) TS(r, t'x) = T(r, t) (c) ar [TS(a, t)1 = 0 aT L aTs (d) -KL ar + KS ar = Lp R(t) at r = R(t) (e) R(t) = 0 (24) Proceeding exactly the same way as in the case of the cylin- der, we are able to get the time for complete melting: a QOta = 3Tra3[PcT - pLcLT"`+ Lp1 +4TrpLcL r2TL(r,ta)dr. (25) 0 When the liquid is removed as soon as it forms, ta = 3Tra3p[cT"`+ LIQO (26) For an approximation of TL(r, ta), let To be the aver- age temperature, and cav , the average specific heat, then as before, = 30 csT+ L+ cL(TO - T) c av - T, 0 With this specific heat, assuming no melting occurs, we have from (18) and (19), Q0 3k ' rta (a-r) (2a+r) TL(r,t) 4TrKr [ a3 + 2a3 where k' +2 co sin[an(a-r)/aJ -álcos[an(a-r)/aJ anta/ ,a2 en=1 K pc av be given by (20) : 2 an an n J . The time when the sphere starts to melt will oo QO 3kt 3 2 cos an+ansin an -ka 2t¡a2 nT 4TrK [ 3 2a a a sina e J (27) an n n=1 n _k' - in 31 CASE OF AN INFINITELY LONG CYLINDER AND A SPHERE Cylinder with a Very Large Radius Our main problem is to determine the location of the moving boundary (phase boundary) at any time t < t < ta. This is a dif- ficult task due to the non -linear character of the problem. However, the melting proceeds only a relatively small distance into the solid in many cases occurring in practice; so it will be sufficiently accu- rate to assume a = 00. Landau seems to make this assumption in the case of the melting of slabs. The assumption is applicable, for example, in the case of melting or solidification of wet soils around a cylindrical pipe. In the case of melting, the input Q0 will rep-resent a source, while in the case of solidification Q0 is a sink (negative source). When a = 00, the boundary condition at infinity is physical- ly equivalent to T(00 t) S'=To . So our problem now is the following: And 32 a aTL(r' t) I a ar (r ar ) k at (r TL(r,t)) 0 < r < R(t), t<t (a) QO = -2irKLr ar aTL r = 0 (b) TL(r,t) = T for r = R(t) . 8 aTS(r,t) a(r T(r,t) Or (r ar ) kS at (a) TS(co, t) = TO (b) TS(R(t),t) = T R(t)<r<co, t<t aTL 8TS (c) -KL ar + KS ar - LpR(t) at r = R(t) (28) (29) (30) (31) We can apply the so-called Boltzmann transformation z= r/R(t); so at r = R(t) i. e. at the phase boundary z = 1. Or in other words, we fix the moving boundary at z = 1. Then equations (28), (29), (30), and (31) become: 3/ The source is assumed to be line source here. L 3/ and 33 2 d 2L Z Rk )R (t) z) ddz - 0, 0 < z < 1 ( 28) ' dz L (a) QO = -2 nKLz dz dTL (b) TL(1) = T z=0 (29)' 2 d 25+ (Z R(kR(t) z) ddZ = 0, 1< z< o0 (30)' dz S (a) TS(oo) = T0 (b) TS(1) = T (31)' dTL dTS (c) -KL dz + KS dz - Lp R(t)R(t) at z = 1. The above equations could be solved explicitly if i R(t) R(t) = constant. Or R = ct2 where c is an arbitrary constant to be determined. If the value of c is known, the location of the phase boundary is known at any time t. This result has been applied by other authors [4, 9, 12] . Solving (28)' : ++ dTL e-z2c2/4kL dz A ('z -z2c2`4kL, TL = A J e dz + B. 0 Using conditions (29)' (a), (b) we have TL(z) - T - QO 1 e-z2c2/4kL, 2TrK , z L z 34 (32) dz. (33) From (33) we get, putting ß = zc/ 2 Nr kL r (ß - ), 2t c 2NrkL 2 TL(r't) - T 2TrK e-ß dß (34) L r (3 2 kLt Similarly from (30)' and (31)' 22 Ale-zc/4ks dTS dz z z -z2c2/ 4ks TS(z) = Al ez dz + B1 1 (35) And using conditions (31)' (a), (b) and putting 13 = zc/ 2'[ks we get, z e Q J . . 35 r 2 kSt -ß2 _ß2 T r,t) = Ty _(T - T ) e dß ! e dß (O S ß c c 2 Nrk5 2ArkS Now we will be able to satisfy the boundary condition (31)' (c): dT dT5 2 -KL dz + KS dz - Lp c /2 . Utilizing (32) and (35) in (37) we obtain: (36) (37) 2 QO e-c 2/ 4kL - KS(T - TO)e -c 2/ 4kS /C e dß = p Lc 2/ 2 2n c Let ß 2 = u, and c = 2 X TkL So from (38): dß = 1 2 2rk5 -u e du u /kS) X 2(kL/kS) -u - - 1 S e du 2 u00 X 2k --2Ei(- kL). S (38) ß °° X ( J -P2 e k'J kL or 36 -X 2(kL/kS) QO 2 2KS(T - TO)e 2 e - 2p LX k 2n E i(-X 2kL/kS) -X 2(kL/kS) QO -2 KS(T - T0)e2 4Tr e - - Lp X kL (39) E i(-X kL/kS) which is the result given by Carslaw and Jaeger [4, p. 2961. The left hand side decreases monotonically from Q0/ 4Tr to -00 as X increases from 0 to co. So it has one and only one real solution. Sphere with Infinite Radius Now to determine the location of the solid liquid interface and the temperature TL and T, wewe make use of the same assump- tion that the sphere is of infinite radius and that the temperature of the surface is held constant. So our new set of equations and condi-tions are as follows : aT a ) r< R(t), r = 0 t < t (40) (41) ôr (r2 ar (a) k at (r2TL), 0 < 2 aTL Q0 = -47KLr ar (b) TL(R(t),t) = T -X and a 2 aT5 1 a 2 ar (r ar ) = ks at (r Ts)' (a) TS(°°, t) = TO (b) TS(R(t), t) = T 37 R (t) < r < oo, t>k < t (42) âTL aTS (c) -KL ar K5 + ôr - Lp R(t) at r =R(t) . (43) Letting z = r/R(t) as before, we get from (40), (41), (42) and (43): 2 d 2L +( 2 + kR(t)R(t)) - 0, dTL 0< z< 1 (40)' dz L dz dT QO = -4TrKLz2R(t) dz L (b) TL(1) = T z d 2S+ [ Z+ k R(t)R(t)] dz - 0, dz 5 z (a) TS(oo) = TO z =0 1 <z <ao (41)' (42)' (b) TS(1) = T (43)' dT dTS (c) -KL dz + KS dz - LP R(t) R (t) at z =1 S (a) s38 The above equations can be solved explicitly only when i R(t) = et ? where c is any constant. .. R(t)R(t) = c2/ 2 . But the condition (41)'(a) still presents difficulties. However if we assume that the heat input is Q0t2 , an analytical solution can be obtained, [ see Paterson ] . With this assumption we have the solution as follows: From (40)' and (41)', dTL 2c 2/ 4kL L e dz 2z (`z -z2c2/4kL TL(z) = A J e 0z dz + B Q -z2c2/ 4kL 0 TL(z)= TL(z) = T 4TrK c ,) e 2 dz L z z And letting f3 - zc 2NrkL c Q 24-kL 2 2NrkL TL(r't) = T +4TrK c 2 dß / c L r 2t (44) A ß i --z 1 e TL(r, t) c Q 2k T + 8TrKL^ÏkL r 2t Similarly we have from (42)' and (43)' : e 2 39 dß. (45) -z2c2/ 4k5 dT5 Ale dz 2 (46) z T5(z) = A -z 2c z4k5 e- dz+B. Evaluating the constants, we finally get r2 1477 e-R2 -ß2 T5(r,t) = T+ (T0-T) 2 e dß 2 dß cß c 24-1(5 2 4-k5 Then (43)'(c) gives: (47) =Sb ß z2 / ß J 2 (' z1 1 e JQ0 4Tr -c 4kL 2/ e -2i 4ks cK5(T-TO)e 0o -z2c2/ 4k5 Sb e dz 1 z2 Letting ß = Zc where z - 2NrkS rictg we get 40 = Lp c3/ 2 (48) 2 2/ --z2c2/ 4kS -c 4k 2/ 00 2 e dz = e 5- C Ç e-ß dß z 4-k5 c 2NrkS 2/ c k 4-Tr erfc( C ) 5 2Nrk5 So (48) will be Q cK (T - TO) 0 5 _ Lp c3/ 3/ 2 (49) 4Tr 1 c Tr ec 2/ 4kS c --2 kS 2NrkS which is identical with the equation obtained by Paterson . The Q0 left hand side of (49) decreases monotonically from -to -00 4Tr as c increases from 0 to 00. Hence it has always one and only one real root. We have derived a solution for propagation of = e -c 4kS 2 - ( 1 2 erfc ( )41 heat from a point source where strength increases as Ït . Due to this fact, however, this solution is of less practical importance. We have obtained the same results as Paterson, but the method of approach is different. 4Z BIBLIOGRAPHY 1. Boley, Bruno A. A method of heat conduction analysis of melting and solidification problems. Journal of Mathe- matics and Physics, 40, 4:300 -313. Dec. 1961. . The analysis of problems of heat conduction and melting. High temperature structures and materials. Proceedings of the 3rd Symposium on Naval Structural Mechanics held in New York. Jan. 1963. 3. Citron, Stephen J. Heat conduction in a melting slab. Journal of the Aerospace Science, 27, 3:219 -228. March 1960. 4. Carslaw, H.S. and J. C. Jaeger. Conduction of heat in solids. Oxford University Press, 1959. p. 282. . Operational methods in applied mathematics. Dover Publications, Inc. N. Y. 1963. p. 76. 6. Chambrè, P. L. On the dynamics of phase growth. Quarterly Journal of Mechanics and Applied Mathematics, 9, 2:224 -233. June 1956. 7. Evans II, G. W. , E. Is sacson and J. K. L. Macdonald. Stefan like problem. Quarterly of Applied Mathematics, 8, 3:312-319. Oct. 1950. 8. Friedman, Avner. Free boundary problems for parabolic equations. I. Melting of solids. Journal of Mathematics and Mechanics, 8, 4:499-518. July 1959. 9. Field, A. L. Solidification of steel in the ingot mould. Trans- actions of the American Society of Steel Treating, ll:p. 264. 1927. 10. Gibson, R. E. A heat conduction problem involving a specified moving boundary. Quarterly of Applied Mathematics, 16, 4:426-430. Jan. 1959. 11. Ingersoll, L.R. , F. T. Adler, H. J. Plass and A. C. Ingersoll. Theory of earth heat exchangers for the heat pump. Heating, Piping and Air Conditioning, 22, 5:113-122. May 1950. 43 12. Ingersoll, L.R. , Otto J. Zobel, A. C. Ingersoll. Heat conduc- tion with engineering, geological and other applications. University of Wisconsin Press, 1954. p. 190. 13. Kolodner, I. I. Free boundary problem for the heat equation with applications to problems of change of phase. Communi- cation on Pure and Applied Mathematics, 9, 1 :1 -31. Feb. 1956. 14. Kreith, F. , F. E. Romie. A study of the thermal diffusion equation with boundary conditions corresponding to solidifi- cation or melting of materials initially at the fusion temp- erature. Proceedings of the Physical Society, London. Series B, 68:377 -391. 1955. 15. Landau, H. G. Heat conduction in a melting solid. Quarterly of Applied Mathematics, 8, 1 :81 -94. 1950. 16. Lightfoot, N. M.H. The solidification of molten steel. Pro- ceedings of the London Mathematical Society. Series 2, 31 :97 -116. 1930. 17. Lotkin, Mark. Calculation of heat flow in melting solid. Quarterly of Applied Mathematics, 18, 1 :79 -85. 1960. 18. Miranker, Willard L. and Joseph B. Keller. The Stefan problem for a non -linear equation. Journal of Mathematics and Mechanics, 9, 1 :67 -70. Jan. 1960. 19. Miranker, W. L. Free boundary value problem for the heat equation. Quarterly of Applied Mathematics, 16, 2 :121 -130. July 1958. 20. Pekeris, C. L. and L. B. Slichter. Problem of ice formation. Journal of Applied Physics, 10 :135 -137. Feb. 1939. 21. Paterson, S. Propagation of a boundary of fusion. Proceed- ings of the Glasgow Mathematical Association, Vol. 1, :42 -47. Jan. 1952. 22. Sait3. On the distribution of temperature in steel ingots during cooling. Tóhoku Imperial University Science Reports,
1233	https://www.ams.org/bookstore/pspdf/dol-52-solutions.pdf	Quadratic Number Theory Exercise Solutions J. L. Lehman 3 Section 0.1. Linear Equations and Congruences. (1) (a) 679 = 23 · 29 + 12. (b) −782 = 57 · −14 + 16. (c) −3216 = 67 · −48 + 0. (2) Suppose that p divides ab, but that p does not divide a. If p is prime, then gcd(p, a) = 1, since gcd(p, a) is a positive divisor of p but cannot equal p. Corollary 0.1.3 implies that then p divides b. If n is composite, then n has a positive divisor a other than 1 and n. In that case, n = ab with 1 < a, b < n. But now n divides ab but cannot divide a or b. (3) Suppose that a and b both divide n. Then n = bq for some integer q, so that a divides bq. If gcd(a, b) = n, then Corollary 0.1.3 implies that a divides qd, say qd = ar for some integer r. But now nd = (bq)d = b(qd) = b(ar) = (ab)r, and ab divides nd. (4) If a and b are not both zero and d = gcd(a, b), then ab/d = a(b/d) = b(a/d) is a common multiple of a and b. (Note that b/d and a/d are integers if d = gcd(a, b).) If n is a common multiple of a and b, then Exercise 3 shows that ab/d divides n. So m = \|ab\|/d is the (positive) least common multiple of a and b by deﬁnition. (5) (a) gcd(504, 186) = 6 = 504 · −7 + 186 · 19. (b) gcd(1247, 913) = 1 = 1247 · −41 + 913 · 56. (c) gcd(1350, 1401) = 3 = 1350 · −55 + 1401 · 53. (6) (a) Since gcd(567, 98) = 7 = 567(−5) + 98(29), then 203 = 29 · 7 = 567(−145) + 98(841). All solutions of 567x + 98y = 203 have the form −145 + 98 7 · q, 841 −567 7 · q = (−145 + 14q, 841 −81q) with q an arbitrary integer. (Other expressions for this general solution are possible. For instance, when q = 1, then x = 9 and y = −50. We can also write (x, y) = (9 + 14q, −50 −81q) as the general solution.) (b) No solutions of 504x + 186y = 202 exist, since gcd(504, 186) = 6 does not divide 202. (c) Here 204 = 34(6) = 34(504 · −7 + 186 · 19) = 504(−238) + 186(646). One expression for the general solution of 504x + 186y = 204 is (−238 + 31q, 646 −84q). (d) (x, y) = (−112 + 913q, 153 −1247q) is one expression for the general solution of 1247x + 913y = 25. (e) (x, y) = (106+467q, −100−450q) is one expression for the general solution of 1350x+ 1401y = 3000. (7) There are ﬁnitely many possible remainders on division of an integer by m, so there must be some repetition of those remainders for the inﬁnite set of positive powers of a. So there must be integers 0 ≤r < s so that as ≡ar (mod m). If gcd(a, m) = 1, we can cancel r terms of a from this congruence, by the congruence cancellation property, without aﬀecting the modulus, and conclude that as−r ≡1 (mod m). But t = s −r is positive. (8) Let t = ordm(a), where a is relatively prime to m. (a) If n = tq, then an = (at)q ≡1q ≡1 (mod m). Conversely, if an ≡1 (mod m) and we write n = tq + r with 0 ≤r < t, then 1 ≡an ≡(at)qar ≡1qar ≡ar (mod m). This contradicts the deﬁnition of t if r is positive, so we must conclude that r = 0, and then t divides n. (b) If gcd(a, m) = 1, then by the congruence cancellation property as ≡ar (mod m) if and only if as−r ≡1 (mod m). But by part (a), this is true if and only if t divides s −r, that is, s ≡r (mod t). (9) (a) Since gcd(23, 39) = 1 = 23 · 17 + 39 · −10, then 18 · 17 = 306 ≡33 (mod 39) is the unique solution of 23x ≡18 (mod 39). (b) gcd(186, 504) = 6 divides 246, so six solutions of 186x ≡246 (mod 504) exist. Here 246 = 41 · 6 = 41(504 · −7 + 186 · 19), using calculations from Exercise 5, thus every 4 solution of 186x ≡246 (mod 504) has the form x = 41 · 19 + 504 6 q = 779 + 84q for some integer q. The distinct solutions in Z504 are 23, 107, 191, 275, 359, and 443. (c) gcd(221, 247) = 13 does not divide 19, so no solutions of 221x ≡19 (mod 247) exist. (d) Here gcd(221, 247) = 13 = 221 · 9 + 247 · −8 divides 117 = 9 · 13. The solutions of 221x ≡117 (mod 247) are given by x = 9 · 9 + 247 13 q = 81 + 19q. Thirteen solutions are distinct in Z247: 5, 24, 43, 62, 81, 100, 119, 138, 157, 176, 195, 214, and 233. (10) (a) Here gcd(13, 41) = 1 = 13 · 19 + 41 · −6 = 247 −246. So a simultaneous solution of x ≡7 (mod 13) and x ≡29 (mod 41) is given by x = −246·7+247·29 = 5441. This solution is unique modulo 13 · 41 = 533, that is, x ≡5441 ≡111 (mod 533). (b) With gcd(63, 82) = 1 = 63 · −13 + 82 · 10, the unique solution of x ≡17 (mod 63) and x ≡14 (mod 82) is congruent to 17 · 82 · 10 + 14 · 63 · −13 modulo 63 · 82 = 5166. That is, x ≡2474 (mod 5166) is the unique solution of the pair of congruences. (c) Using a calculation from Exercise 5, the unique solution of x ≡374 (mod 1247) and x ≡821 (mod 913) is given by x ≡374 · 913 · 56 + 821 · 1247 · −41 modulo 1247 · 913 = 1138511, that is x ≡1055336 (mod 1138511). (11) If x simultaneously satisﬁes ax ≡c (mod m) and bx ≡d (mod m), then b(ax) ≡bc (mod m) and a(bx) ≡ad (mod m). So abx is congruent to both bc and ad modulo m, and so m must divide ad −bc. Conversely, suppose that gcd(a, b) = 1 and that m divides ad −bc, say with as + bt = 1 and ad −bc = mu for some integers s, t, and u. If x = cs + dt, then ax = acs + adt = acs + (bc + mu)t = c(as + bt) + m(ut) ≡c (mod m) and bx = b(cs + dt) = (ad −mu)s + bdt = d(as + bt) + m(−us) ≡d (mod m). That is, ax ≡c (mod m) and bx ≡d (mod m) have a simultaneous solution. If y is also a solution, then a(x −y) = mq and b(x −y) = mr for some integers q and r. But then x −y = (as + bt)(x −y) = mqs + mrt = m(qs + rt), and so x ≡y (mod m). Section 0.2. Quadratic Congruences Modulo Primes. (1) Suppose that b2 ≡a (mod p) where p is an odd prime. (a) Since (−b)2 = b2, then −b ≡p −b (mod p) also satisﬁes x2 ≡a (mod p). (b) If −b ≡b (mod p), then p divides 2b. Since p is odd, then p divides b. Conversely, if b ≡0 (mod p), then −b ≡0 ≡b (mod p). In this case, b2 ≡0 (mod p) as well, so that x2 ≡a (mod p) has exactly one solution if and only if p divides a. (c) If b and c are solutions of x2 ≡a (mod p), then b2 ≡c2 (mod p), which implies that p divides c2 −b2 = (c −b)(c + b). Since p is prime, then either p divides c −b or p divides c + b = c −(−b). Thus either c ≡b (mod p) or c ≡−b (mod p), and x2 ≡a (mod p) has no more than two solutions. (2) The congruence x2 ≡1 (mod p) has two distinct solutions, 1 and −1 ≡p −1 (mod p), modulo every odd prime p. So 1 p = 1. (3) If a ≡b (mod p), then x2 ≡a (mod p) and x2 ≡b (mod p) have the same solutions, and so the same number of solutions. Thus a p = b p . (4) Euler’s criterion implies that −1 p ≡(−1) p−1 2 (mod p). But each side of this congruence is 1 or −1, so this congruence is an equation since p ≥3. For the ﬁnal equation, note that if p = 4q + 1, then p−1 2 = 2q is even, while if p = 4q + 3, then p−1 2 = 2q + 1 is odd. (5) Here ab p ≡(ab) p−1 2 ≡a p−1 2 · b p−1 2 ≡ a p b p (mod p), using Euler’s criterion and properties of exponents. All symbols are 1, 0, or −1, as is the product of two symbols, and so this congruence is an equation. (6) (a) −19 43 = −1 43 19 43 = −1 · − 43 19 = 5 19 , since 19 and 43 are both congruent to 3 modulo 4. Now with 5 ≡1 (mod 4), we have 5 19 = 19 5 = 4 5 = 2 5 2 = 1. 5 (b) 35 67 = 5 67 7 67 = 67 5 · − 67 7 = − 2 5 4 7 = −(−1) · 12 = 1. (c) 46 97 = 2 97 23 97 = 97 23 = 5 23 , since 97 ≡1 (mod 4) and 97 ≡1 (mod 8). Now 5 23 = 23 5 = 3 5 = 5 3 = 2 3 = −1. (7) Suppose that q ≡3 (mod 4). If p ≡1 (mod 4), then −q p = −1 p q p = 1 · p q . On the other hand, if p ≡3 (mod 4), then −q p = −1 p q p = −1 · − p q . So −q p = p q in either case. (8) (a) We can transform x2 + x −3 ≡0 (mod 17) into (2x + 1)2 ≡13 (mod 17). Since 13 17 = 17 13 = 4 13 = 1, this congruence has two solutions. Trial-and-error shows that 13 ≡64 (mod 17), so we solve 2x + 1 ≡8 (mod 17) and 2x + 1 ≡−8 (mod 17) for x = 12 and x = 4 respectively. (b) 3x2+5x+8 ≡0 (mod 19) transforms into (6x+5)2 ≡−71 (mod 19), or (6x+5)2 ≡5 (mod 19). Here 5 19 = 19 5 = 4 5 = 1, so that two solutions exist, and we ﬁnd that 6x + 5 ≡±9 (mod 19). Solving these linear congruences, we ﬁnd that x = 4 and x = 7 are the two solutions of 3x2 + 5x + 8 ≡0 (mod 19). (c) 5x2−x+1 ≡0 (mod 23) transforms into (10x−1)2 ≡−19 (mod 23), or (10x−1)2 ≡4 (mod 23), with solutions given by solving 10x −1 ≡±2 (mod 23) for x = 16 and x = 21. Section 0.3. Quadratic Congruences Modulo Composite Integers. (1) Let f(x) = x2 + x −1 so that f′(x) = 2x + 1. Note that f(x) ≡0 (mod p) has the same solutions as (2x + 1)2 ≡5 (mod p) when p is an odd prime. Here 5 p = p 5 = 1 if and only if p is congruent to 1 or 4 modulo 5. To simplify calculations, we can note that if f(x) ≡0 (mod m) has a solution r, then −r −1 is also a solution. (a) f(x) ≡0 (mod 5) has one solution, x = 2. Here f′(2) = 5 and f(2) = 5, and since 5t ≡−1 (mod 5) has no solutions, it follows that f(x) ≡0 (mod 25) has no solutions. No solutions of f(x) ≡0 (mod 125) can exist either. (b) f(x) ≡0 (mod 11) has solutions 3 and −4 ≡7 (mod 11) by direct calculation. When r = 3, then f′(r) = 7 and f(r) = 11. The congruence 7t ≡−1 (mod 11) has one solution, t = 3, and so s = 3 + 11(3) = 36 is a solution of f(x) ≡0 (mod 121). A second solution, using r = −4 or r = 7, is s = −37 ≡84 (mod 121). Now f′(36) = 73 ≡7 (mod 11) and f(36) = 1331. The congruence 7t ≡−f(36) 112 ≡−11 (mod 11) has solution t = 0, and thus 36 + 0(121) = 36 is a solution of f(x) ≡0 (mod 1331). The second solution is −37 ≡1294 (mod 1331). (c) f(x) ≡0 (mod 19) has solutions 4 and −5 ≡14 (mod 19). When r = 4, then f′(r) = 9 and f(r) = 19. The congruence 9t ≡−1 (mod 19) has solution t = 2 and so s = 4 + 19(2) = 42 is a solution of f(x) ≡0 (mod 361). The second solution is −43 ≡318 (mod 361). Now when r = 42, we have f′(r) = 85 ≡9 (mod 19) and f(r) = 1805 = 5 · 361. The congruence 9t ≡−5 (mod 19) has solution t = 10, and thus 42 + 361(10) = 3652 satisﬁes f(x) ≡0 (mod 6859). A second solution is given by −3653 ≡3206 (mod 6859). (2) (a) Since f(x) ≡0 (mod 5) has one solution, q = 2, and f(x) ≡0 (mod 11) has two solutions, r = 3 and r = 7, then f(x) ≡0 (mod 55) has two solutions, obtained by solving x ≡2 (mod 5) and x ≡3 (mod 11) for x = −8 ≡47 (mod 55), and x ≡2 (mod 5) and x ≡7 (mod 11) for x ≡7 (mod 55). (Again, we can also use the fact that if x is a solution, then −x −1 is also a solution.) (b) f(x) ≡0 (mod 209) has four solutions, obtained by solving x ≡q (mod 11) and x ≡r (mod 19) where r = 3 or 7 and s = 4 or 14. We ﬁnd that these systems yield the solutions 14, 80, −81 ≡128 (mod 209) and −15 ≡194 (mod 209). 6 (c) f(x) ≡0 (mod 605) has two solutions, which we ﬁnd by solving x ≡2 (mod 5) and x ≡r (mod 121) with r = 36 or r = 84. We ﬁnd that x = 157 and x = −158 ≡447 (mod 605) are the solutions. (d) f(x) ≡0 (mod 2299) has four solutions, which we ﬁnd to be 641, 1125, −1126 ≡1173 (mod 2299), and −642 ≡1657 (mod 2299) by solving all systems of the form x ≡q (mod 121) and x ≡r (mod 19) with q = 36 or 84, and r = 4 or 14. (3) f(x) = x2 + 18x + 1 has discriminant ∆= 320 = 26 · 5. Here ℓ= 3 since 5 ≡1 (mod 4). If e > 2ℓ= 6, then n2e(f) = 0 since 5 2 = −1. On the other hand, if e ≤6 is written as e = 2k + r with r = 0 or 1, then n2e(f) = 2k. The following table lists all possibilities. e 0 1 2 3 4 5 6 7 8 · · · n2e(f) 1 1 2 2 4 4 8 0 0 · · · Since f′(x) = 2x + 18 is even for every x, the congruence f′(x) ≡−f(x) 2e (mod 2) always has either two solutions or no solutions, depending on whether the right-hand side is even or odd. Direct calculation shows that x = 1 is the only solution of f(x) ≡0 (mod 2). With f(1) = 20, we then ﬁnd two solutions of f(x) ≡0 (mod 4), that is, x = 1 + 2(0) = 1 and x = 1 + 2(1) = 3. Now −f(1)/4 = −5 is odd, but −f(3)/4 = −16 is even, and so there are two solutions of f(x) ≡0 (mod 8), x = 3 +4(0) = 3 and x = 3 +4(1) = 7. Then we ﬁnd that −f(3)/8 = −8 and −f(7)/8 = −22 are both even, and so 3, 7, 11, and 15 are all solutions of f(x) ≡0 (mod 16). Now −f(3)/16 = −4 and −f(11)/16 = −20 are even, while −f(7)/16 = −11 and −f(15)/16 = −31 are both odd, thus there are four solutions of f(x) ≡0 (mod 32), that is, 3, 11, 19, and 27. With −f(3)/32 = −2, −f(11)/32 = −10, −f(19)/32 = −22, and −f(27)/32 = −38 all even, we ﬁnd that 3, 11, 19, 27, 35, 43, 51, and 59 are all solutions of f(x) ≡0 (mod 64). But ﬁnally, −f(x)/64 is not even for any of these eight values of x, and so f(x) ≡0 (mod 128) has no solutions. Therefore, f(x) ≡0 (mod 2e) can have no solutions for e ≥7. (4) Let f(x) = x2 + x −1, which has discriminant ∆= 5. Here f(x) ≡0 (mod m) has no solutions if m is divisible by 25 or by any prime p ≡2, 3 (mod 5). If this is not the case, and m has t distinct prime divisors satisfying p ≡1, 4 (mod 5), then f(x) ≡0 (mod m) has 2t distinct solutions modulo m. (5) If q = q1 · q2 · · ·qk with each qi prime, and f(x) ≡0 (mod q) has a solution for some polynomial f(x) with discriminant ∆, then each f(x) ≡0 (mod qi) has a solution and so ∆ qi = 1 for each i. (Here we are assuming that q and ∆have no common divisors.) But then the product of all of those symbols, which is ∆ q , is also 1. On the other hand, this product can also be 1 if there are any even number of symbols equal to −1, in which case f(x) ≡0 (mod q) has no solutions. Section 1.1 Sums of Two Squares. (1) For each a, we test p a/2 < x < √a to ﬁnd a = x2 + y2. (a) a = 313 = 132 + 122. (b) a = 377 = 162 + 112 = 192 + 42. (c) a = 433 = 172 + 122. (2) (a) 481 = 13·37 = (32+22)(62+12) = (3·6+2·1)2+(3·1−2·6)2 = (3·6−2·1)2+(3·1+2·6)2. In standard form, 481 = 202 + 92 = 162 + 152. (b) 493 = 17 · 29 = (42 + 12)(52 + 22) = 222 + 32 = 182 + 132. (c) 697 = 17 · 41 = (42 + 12)(52 + 42) = 242 + 112 = 212 + 162. (d) 949 = 13 · 73 = (32 + 22)(82 + 32) = 302 + 72 = 252 + 182. (e) 1537 = 29 · 53 = (52 + 22)(72 + 22) = 392 + 42 = 312 + 242. 7 (f) 8633 = 89 · 97 = (82 + 52)(92 + 42) = 922 + 132 = 772 + 522. (3) (a) 1072 + 1 = 229 · 50. Here 107 ≡7 (mod 50) and 1 ≡1 (mod 50), and we ﬁnd that 72 + 12 = 50 = 50 · 1. So 229 · 50 · 50 · 1 = (1072 + 12)(72 + 12) = (107 · 7 + 1 · 1)2 + (107 · 1 −7 · 1)2 = 7502 + 1002. We can cancel 502 from both sides to conclude that 229 = 152 + 22. (b) 602+1 = 277·13. With 60 ≡−5 (mod 13) and 1 ≡1 (mod 13), and (−5)2+12 = 26 = 13·2, we then see that 277·13·13·2 = (602+12)((−5)2+12) = (−299)2+652. We cancel 132 to write 277 · 2 = (−23)2 + 52. Now with −23 ≡1 (mod 2) and 5 ≡1 (mod 2), and 12+12 = 2·1, we then have 277·2·2·1 = ((−23)2+52)(12+12) = (−18)2+(−28)2. Cancelling 22 yields 277 = (−9)2 + (−14)2, or 277 = 142 + 92 in standard form. (c) 1482+1 = 337·65. Here 148 ≡18 (mod 65), and 182+12 = 65·5. Thus 337·65·65·5 = (1482+12)(18212) = 26652+1302, and by cancelling 652 we ﬁnd that 337·5 = 412+22. But now 41 ≡1 (mod 5) and 2 ≡2 (mod 5), with 12 +22 = 5 · 1. Thus 337 · 5 · 5·1 = (412 + 22)(12 + 22) = 452 + 802, and we conclude that 337 = 92 + 162 = 162 + 92. (d) 522+1 = 541·5, with 52 ≡2 (mod 5). We ﬁnd that 541·5·5·1 = (522+12)(22+12) = 1052 + 502, and so 541 = 212 + 102. (4) For each prime p, we ﬁrst write p as a sum of two squares by testing p p/2 < x < √p. (a) p = 509 = 222 + 52. Here 1 = 22(−2) + 5(9) by the Euclidean algorithm, and then we ﬁnd that x = 22(9) −5(−2) = 208 is a solution of x2 ≡−1 (mod 509). (b) p = 757 = 262 + 92. Now 1 = 26(−1) + 9(3), and so x = 26(3) −9(−1) = 87 satisﬁes x2 ≡−1 (mod 757). (c) p = 953 = 282+132, and 1 = 28(−6)+13(13), and we ﬁnd that x = 28(13)−13(−6) = 442 satisﬁes x2 ≡−1 (mod 953). (d) p = 1009 = 282 + 152, and 1 = 28(7) + 15(−13), so that x = 28(−13) −15(7) = −469 satisﬁes x2 ≡−1 (mod 1009). (5) (a) 125 = 112 + 22 = 102 + 52. (b) 180 = 122 + 62. (c) 325 = 182 + 12 = 172 + 62 = 152 + 102. (d) 985 = 292 + 122 = 272 + 162. (e) 1000 = 302 + 102 = 262 + 182. (f) 27625 = 1642 + 272 = 1442 + 832 = 1412 + 882 = 1322 + 1012 = 1652 + 202 = 1602 + 452 = 1552 + 602 = 1202 + 1152. Section 1.2 Gaussian Integers. (1) If w = a + bi and z = c + di, then w + z = (a + c) −(b + d)i = (a −bi) + (c −di) = w + z. Likewise, wz = (ac −bd) −(ad + bc)i = (a −bi)(c −di) = w · z. (2) By deﬁnition, m divides q+ri in Z[i] if and only if q+ri = m(s+ti) for some s+ti ∈Z[i]. But this occurs if and only if q = ms and r = mt, that is, m divides q and m divides s in Z. (3) Let u, v, and w be Gaussian integers. (a) Suppose that v divides w, say with w = vz for some z in Z[i]. Then N(w) = N(vz) = N(v)·N(z). Since N(z) is a rational integers, it follows that N(v) divides N(w) in Z. (b) By deﬁnition, u is a unit in Z[i] if u divides 1. But then N(u) divides N(1) = N(1 + 0i) = 12 + 02 = 1 in Z, by part (a). The only divisors of 1 in Z are 1 and −1, and since the norm of a Gaussian integer cannot be negative, we conclude that N(u) = 1. Conversely, suppose that N(u) = u · u = 1 for some u ∈Z[i]. Here u is also a Gaussian integer, so it follows that u divides 1 in Z[i]. (c) By deﬁnition, v and w are associates in Z[i] if v divides w and w divides v. But then N(v) divides N(w) and N(w) divides N(v) in Z. For rational integers, this occurs if 8 and only if N(w) = ±N(v), but the norm of a Gaussian integer cannot be negative, so it follows that N(v) = N(w). (4) By the preceding exercise, u = a + bi is a unit in Z[i] if and only if N(u) = a2 + b2 = 1. We conclude that either a2 = 1 and b2 = 0, so a = ±1 and b = 0, or a2 = 0 and b2 = 1, so a = 0 and b = ±1. That is, u = 1, −1, i, or −i. (5) If w = uv for some unit u of Z[i], then v = u−1w, where u−1 = u is also a Gaussian integer. It follows that v divides w and w divides v, and so v and w are associates. Conversely, if v divides w and w divides v, say with w = vu and v = wz for some u, z ∈Z[i], then w = vu = w(zu). If w = 0, then v = wz = 0 and we have w = 1 · v. If w ̸= 0, we can cancel w from the equation w = w(zu) and conclude that zu = 1. But then u divides 1, so is a unit of Z[i], and we have that w = uv. Thus using Exercise 4, if v = a + bi, then its associates are 1 · v = a + bi, −1 · v = −a −bi, i · v = −b + ai, and −i · v = b −ai. (6) Deﬁne ∼on Z[i] by v ∼w if and only if v is an associate of w. By Exercise 5, we can say that v ∼w if and only if w = uv for some unit u of Z[i]. (a) Since v = 1 · v, then v ∼v and ∼is reﬂexive. If w = uv, then v = u−1w = uw, with u also a unit in Z[i]. Thus if v ∼w, then w ∼v and ∼is symmetric. If w = u1v and z = u2w for some units u1 and u2, then z = u2(u1v) = (u2u1)v. Here u2u1 is a unit since its norm is 1. Thus we can say that if v ∼w and w ∼z, then v ∼z, and so ∼ is transitive. (b) Suppose that v = u1u and z = u2w for some units u1 and u2. Then vz = (u1u2)uw, and u1u2 is a unit, as noted above. Therefore if u ∼v and w ∼z, then uw ∼vz. (c) Again suppose that v = u1u and z = u2w for some units u1 and u2. If u divides w, then w = uz1 for some z1 in Z[i]. But then z = u2w = u2uz1 = (u2u−1 1 z1)v, and it follows that v divides z. The converse is established similarly. (7) If N(w) = p is prime in Z, but w = uv in Z[i], then p = N(u) · N(v). Since each norm is a positive rational integer, one of them must equal 1. But if N(u) = 1, for instance, then u is a unit in Z[i]. So it is impossible to write w as a product of two Gaussian integers neither of which is a unit, and w is irreducible by deﬁnition. (8) If not every reducible Gaussian integer can be written as a product of irreducible elements of Z[i], we can assume that w is such an element with N(w) as small as possible. Since w is reducible, it is possible to write w = uv with neither u nor v a unit in Z[i]. But then N(w) = N(u) · N(v), with neither N(u) nor N(v) equal to 1, so that N(w) > N(u), N(v) > 1. But now by our assumption both u and v can be expressed as a product of irreducible elements, and thus w = uv can be written as such a product as well, contrary to assumption. (9) (a) We ﬁnd that v w = 37−10i 4+i = (37−10i)(4−i) 17 = 138 17 + −77 17 i. The closest integers to 138 17 and −77 17 are 8 and −4 respectively. If z = 8 −4i, we calculate that u = v −wz = 1 −2i. So 37 −10i = (4 + i)(8 −4i) + (1 −2i), with 5 = N(1 −2i) < N(4 + i) = 17. (b) By the same approach, 13 + 19i = (3 + 4i)(5) + (−2 −i), with 5 = N(−2 −i) < N(3 + 4i) = 25. (c) 41 + 9i = (9 + 7i)(3 −2i) + 6i, with 36 = N(6i) < N(9 + 7i) = 130. (10) If not every reducible Gaussian integer can be written uniquely as a product of irreducible elements of Z[i], we can assume that w is such an element with N(w) as small as possible. Let w = u1 · u2 · · ·uk and w = z1 · z2 · · ·zℓwith each ui and zi irreducible in Z[i]. We can assume that no ui is an associate of any zj, since otherwise we could cancel that common factor from both sides, and would have an element of smaller norm with distinct irreducible factorizations. Now u1 divides w, so must divide the product z1 · z2 · · ·zℓ. We can conclude, by properties established for irreducible elements in Z[i], that u1 divides 9 a term in this product, say zi. But now with zi irreducible and u1 not a unit, we must conclude that u1 and zi are associates, contrary to assumption. So it is impossible for w to have these distinct irreducible factorizations. (11) (a) 850 = 2 · 52 · 17 = (1 + i)(1 −i)(2 + i)2(2 −i)2(4 + i)(4 −i). (b) 4125 = 3·53·11 = 3·11·(2+i)3(2−i)3. (Here 3 and 11 are prime numbers congruent to 3 modulo 4, so are irreducible in Z[i].) (c) Here N(37 −12i) = 1513 = 17 · 89, and so vv = (4 + i)(4 −i)(8 + 5i)(8 −5i). These irreducible Gaussian integers are the only possible irreducible factors of v = 37 −12i. By trial-and-error, we ﬁnd that 37 −12i = (4 + i)(8 −5i). (d) Here N(−11 + 27i) = 850, so vv is made up of the same irreducible factors as in part (a). By trial-and-error, we ﬁnd that −11 + 27i = (1 + i)(2 + i)2(4 + i). Section 1.3 Ideal Form for Gaussian Integers. (1) Let v = g(q+ri) have ideal form g[a : k], so that gcd(q, r) = 1 with a = q2 +r2 and rk ≡q (mod a). In this case, v = g(q −ri), and since a = q2 +(−r)2 and (−r)(−k) ≡q (mod a), it follows that an ideal form for v is g[a : −k]. (2) (a) If w = 11 + 3i, then N(w) = 112 + 32 = 130 and 3x ≡11 (mod 130) has solution k = 47. So [130 : 47] is an ideal form for w. (b) w = 9 −7i has ideal form [130 : −57]. (To verify this, note that 92 +(−7)2 = 130 and −7(−57) ≡9 (mod 130).) (c) w = 13 + 5i has ideal form [194 : −75]. (d) w = −11 + 27i has ideal form [850 : 157]. (e) w = 14 + 16i = 2(7 + 8i) has ideal form 2[113 : 15] (f) w = 141 + 3i = 3(47 + i) has ideal form 3[2210 : 47]. (3) Let v = g(q + ri) have ideal form g[a : k], so that gcd(q, r) = 1 with a = q2 + r2 and rk ≡q (mod a). Then i · v = g(−r + qi). Here (−r)2 + q2 = a, and we have that qk ≡−r (mod a) by Proposition 1.3.1. So g[a : k] is also an ideal form for i · v. (4) An ideal form for v = 1 + i is [2 : 1]. So by Theorem 1.3.2, if w is a Gaussian integer with ideal form h[b : ℓ], then v divides w if and only if 2 divides bh and hℓ≡h (mod 2). Both requirements are met if h is even. If h is odd but b is even, then ℓmust be odd (so that b can divide ℓ2 + 1 as in Proposition 1.3.1), and so both requirements are met in that case also. But if b and h are both odd, then 2 does not divide bh. (5) Let g[a : k] be an ideal form for a Gaussian integer v, so that v = g(q+ri) with a = q2+r2 and rk ≡q (mod a). Likewise, let h[b : ℓ] be an ideal form for w = h(s + ti), so that b = s2 +t2 and tℓ≡s (mod b). If v divides w, then g divides h by Lemma 1.3.3, and then v g = q + ri and w g = h g (s + ti) are Gaussian integers having ideal form [a : k] and h g [b : ℓ] respectively. It is clear that w = vu if and only if w g = v gu. If g divides h, then bh = agq if and only if b · h g = aq. Similarly h(ℓ−k) = agr if and only if h g (ℓ−k) = ar. The claims of this exercise follow from these equations. (6) In each part, we write v and w in ideal form and apply the criterion for divisibility in Theorem 1.3.2. (a) [13 : −5] divides [130 : 47] since 13 divides 130 and 47 ≡−5 (mod 13). (b) [13 : 5] does not divide [130 : 47] since 47 ̸≡5 (mod 13). (c) [13 : −5] divides [130 : −57] since 13 divides 130 and −57 ≡−5 (mod 13). (d) [13 : 5] does not divide [130 : −57] since −57 ̸≡5 (mod 13). (e) [13 : −5] does not divide [194 : −75] since 13 does not divide 194. (f) [17 : 4] divides [850 : 157] since 17 divides 850 and 157 ≡4 (mod 17). (g) [17 : −4] does not divide [850 : 157] since 157 ̸≡−4 (mod 17). 10 (h) [25 : 7] divides [850 : 157] since 25 divides 850 and 157 ≡7 (mod 25). (i) [25 : −7] does not divide [850 : 157] since 157 ̸≡−7 (mod 25). (j) [2 : 1] divides 2[113 : 15] since 2 divides 2 · 113 and 2 · 15 ≡2 · 1 (mod 2). (k) 2[2 : 1] does not divide 2[113 : 15] since ag = 4 does not divide bh = 226. (l) [5 : −2] does not divide 3[2210 : 47] since 3 · 47 ̸≡3(−2) (mod 5). (m) [5 : 2] divides 3[2210 : 47] since 5 divides 3 · 2210 and 3 · 47 ≡3 · 2 (mod 5). (n) [13 : −5] divides 3[2210 : 47] since 13 divides 3 · 2210 and 3 · 47 ≡3(−5) (mod 13). (o) [13 : 5] does not divide 3[2210 : 47] since 3 · 47 ̸≡3 · 5 (mod 13). (p) [17 : 4] does not divide 3[2210 : 47] since 3 · 47 ̸≡3 · 4 (mod 17). (q) [17 : −4] divides 3[2210 : 47] since 17 divides 3 · 2210 and 3 · 47 ≡3(−4) (mod 17). Section 1.4. Factorization and Multiplication with Ideal Forms. (1) (a) [130 : 47] = [2 : 47][5 : 47][13 : 47] = [2 : 1][5 : 2][13 : −5]. These are ideal forms for (1 + i)(2 + i)(3 + 2i) = −3 + 11i = i(11 + 3i), an associate of 11 + 3i. (b) [130 : −57] = [2 : 1][5 : −2][13 : −5] = (1 + i)(2 −i)(3 + 2i) = 7 + 9i, an associate of 9 −7i. (c) [194 : −75] = [2 : 1][97 : 22] = (1 + i)(9 −4i) = 13 + 5i. (d) [850 : 157] = [2 : 1][5 : 2]2[17 : 4] = (1 + i)((2 + i)2(4 + i) = −11 + 27i. (e) 2[170 : 47] = [2 : 1]3[5 : 2][17 : −4] = (1 + i)3(2 + i)(4 −i) = −22 + 14i, an associate of 14 + 22i. (f) v = 141 +3i = 3[2210 : 47] = 3[2 : 1][5 : 2][13 : −5][17 : −4]. These are ideal forms for 3(1 +i)(2 +i)(3 +2i)(4 −i) = (−3 +141i), an associate of 141 +3i. (Here 3 = 3[1 : 0] is irreducible in Z[i].) (2) (a) [25 : 7][65 : −8] = [5 : 2]3[13 : 5] = [125 : 57][13 : 5] = [1625 : 57]. (Here 57 is the unique solution of x2 + 1 ≡0 (mod 125) congruent to 2 modulo 5, and 57 also satisﬁes x ≡5 (mod 13).) We calculate (3 + 4i)(1 + 8i) = −29 + 28i, and verify that (−29)2 + 282 = 1625, with 28 · 57 ≡−29 (mod 1625). (b) [25 : 7][65 : 8] = [5 : 2]2[5 : −2][13 : −5] = 5[5 : 2][13 : −5] = 5[65 : −18]. We verify that (3+4i)(8+i) = 20+35i = 5(4+7i) with 42 +72 = 65 and 7·−18 ≡4 (mod 65). (c) [13 : −5][65 : −8] = [13 : −5][13 : 5][5 : 2] = 13[5 : 2]. Here (3 + 2i)(1 + 8i) = −13 + 26i = 13(−1 + 2i), with [5 : 2] an ideal form for −1 + 2i. (d) [13 : −5][65 : 8] = [13 : −5]2[5 : −2] = [169 : −70][5 : −2] = [845 : 268]. We verify that (3 + 2i)(8 + i) = 22 + 19i, with 222 + 192 = 845 and 19 · 268 ≡22 (mod 845). (e) [29 : 12][41 : −9] = [1189 : 360], and (2 + 5i)(5 +4i) = −10 + 33i with (−10)2 + 332 = 1189 and 33 · 360 ≡−10 (mod 1189). (f) [29 : 12][58 : −17] = [29 : 12]2[2 : 1] = [1682 : 41], and (2 + 5i)(7 + 3i) = −1 + 41i with (−1)2 + 412 = 1682 and 41 · 41 ≡−1 (mod 1682). (g) [29 : 12][58 : 17] = [29 : 12][29 : −12][2 : 1] = 29[2 : 1], and (2 + 5i)(3 + 7i) = −29 + 29i = 29(−1 + i) with [2 : 1] an ideal form for −1 + i. (h) [65 : 18][65 : −8] = [5 : −2][5 : 2][13 : 5]2 = 5[169 : 70], while (7 + 4i)(1 + 8i) = −25 + 60i = 5(−5 + 12i) with [169 : 70] an ideal form for −5 + 12i. (i) [65 : 18][65 : 8] = [5 : −2]2[13 : 5][13 : −5] = 13[25 : −7], while (7 + 4i)(8 + i) = 52 + 39i = 13(4 + 3i) with [25 : −7] an ideal form for 4 + 3i. (j) [65 : 18][85 : 13] = [5 : −2]2[13 : 5][17 : −4] = [25 : −7][221 : −21] = [5525 : 1968]. We verify that (7 + 4i)(6 + 7i) = 14 + 73i with 142 + 732 = 5525 and 73 · 1968 ≡14 (mod 5525). Section 1.5. Reduction of Ideal Forms for Gaussian Integers. 11 (1) Note that 432 + 1 = 1850 = 370 · 5. Now [5 : −43] = [5 : 2] is an ideal form for w = 2 + i. Thus [370 : 43] is an ideal form for 1 5(43 + i)(2 + i) = 1 5(85 + 45i) = 17 + 9i. (2) In each part, we write [a : k] →[c : −k], with −k replaced by its minimal value modulo c, to summarize the data of the reduction algorithm. (a) [97 : 22] →[5 : −2] →[1 : 0] (that is, 222 + 1 = 97 · 5 and −22 ≡−2 (mod 5), and then (−2)2 + 1 = 5 · 1 with 2 ≡0 (mod 1)). So [97 : 22] is an ideal form for v = 1 5(22 + i) · 1 1(−2 + i) = −9 + 4i. (b) [145 : 17] →[2 : 1] →[1 : 0], so v = 1 2(17 + i) · 1 1(1 + i) = 8 + 9i. (c) [205 : 32] →[5 : −2] →[1 : 0], so v = 1 5(32 + i) · 1 1(−2 + i) = −13 + 6i. (d) [205 : 73] →[26 : 5] →[1 : 0], and v = 1 26(73 + i) · 1 1(5 + i) = 14 + 3i. (e) [377 : 70] →[13 : −5] →[2 : 1] →[1 : 0]; v = 1 13(70+i)· 1 2(−5+i)· 1 1(1+i) = −16−11i. (f) [377 : 99] →[26 : 5] →[1 : 0]; v = 1 26(99 + i) · 1 1(5 + i) = 19 + 4i. (g) [425 : 132] →[41 : −9] →[2 : 1] →[1 : 0]; v = 1 41(132 + i) · 1 2(−9 + i) · 1 1(1 + i) = −16 −13i. (h) [425 : 157] →[58 : 17] →[5 : −2] →[1 : 0]; v = 1 58(157 + i) · 1 5(17 + i) · 1 1((−2 + i) = −19 + 8i. (i) [493 : 157] →[50 : −7] →[1 : 0]; v = 1 50(157 + i) · 1 1(−7 + i) = −22 + 3i. (j) [493 : 191] →[74 : 31] →[13 : −5] →[2 : 1] →[1 : 0]; v = 1 74(191 + i) · 1 13(31 + i) · 1 2(−5 + i) · 1 1(1 + i) = −18 −13i. (3) The congruence x2 + 1 ≡0 (mod 290) has four solutions: x = ±17, x = ±133. Here [290 : 17] →[1 : 0], so that v = 17+i. On the other hand, [290 : 133] leads to v = −13−11i. The other solutions produce conjugates of these. (4) The ideal forms of Gaussian integers having norm 1625 are [1625 : ±57], [1625 : ±307], 5[65 : ±8], and 5[65 : 18]. These forms correspond to 28 + 29i, −37 −16i, 40 + 5i, 35 + 20i respectively, and their conjugates. Section 1.6. Sums of Two Squares Revisited. (1) Since 3532 + 1 = 733 · 170 with −353 ≡−13 (mod 170), and (−13)2 + 1 = 170, we ﬁnd that [733 : 353] is an ideal form for 1 170(353 + i)(−13 + i) = −27 + 2i, and conclude that 733 = 272 + 22. (2) From 1332 + 1 = 1769 · 10, we ﬁnd that 1 10(133 + i)(−3 + i) = −40 + 13i is a Gaussian integer with ideal form [1769 : 133]. Similar calculations show that [1769 : 621] is an ideal form for 1 218(621 + i) · 1 5(33 + i)(2 + i) = 37 + 20i. Thus 1769 = 402 + 132 = 372 + 202. (3) We ﬁnd eight proper representations of a = 10414625 by x2 + y2 from the following calculations. (a) (2 + i)3 · (3 + 2i)2 · (4 + i) · (5 + 2i) = −3223 −164i. (b) (2 + i)3 · (3 + 2i)2 · (4 + i) · (5 −2i) = −2447 + 2104i. (c) (2 + i)3 · (3 + 2i)2 · (4 −i) · (5 + 2i) = −2921 + 1372i. (d) (2 + i)3 · (3 + 2i)2 · (4 −i) · (5 −2i) = −1169 + 3008i. (e) (2 + i)3 · (3 −2i)2 · (4 + i) · (5 + 2i) = 2153 + 2404i. (f) (2 + i)3 · (3 −2i)2 · (4 + i) · (5 −2i) = 3217 + 256i. (g) (2 + i)3 · (3 −2i)2 · (4 −i) · (5 + 2i) = 3031 + 1108i. (h) (2 + i)3 · (3 −2i)2 · (4 −i) · (5 −2i) = 2959 −1288i. (4) (a) 305 = 5 · 61 = 172 + 42 = 162 + 72. (These arise from v = (2 + i)(6 −5i) and v = (2 + i)(6 + 5i) respectively, for which vv = 5 · 61.) (b) 493 = 17 · 29 = 222 + 32 = 182 + 132. (c) 754 = 2 · 13 · 29 = 272 + 52 = 232 + 152. (d) 1885 = 5 · 13 · 29 = 432 + 62 = 422 + 112 = 382 + 212 = 342 + 272. 12 (e) 1898 = 2 · 13 · 73 = 432 + 72 = 372 + 232. (f) 7565 = 5 · 17 · 89 = 862 + 132 = 832 + 262 = 822 + 292 = 622 + 612. (g) 15170 = 2 · 5 · 37 · 41 = 1212 + 232 = 1132 + 492 = 1072 + 612 = 912 + 832. (5) (a) a = 3250 = 2 · 53 · 13 = 572 + 12 = 552 + 152 = 532 + 212 = 452 + 352. (Here we use the following products of Gaussian integers respectively: (1 + i)(2 −i)3(3 + 2i), (1 + i)(2 + i)2(2 −i)(3 + 2i), (1 + i)(2 + i)3(3 + 2i), and (1 + i)(2 + i)(2 −i)2(3 + 2i).) With m = (3 + 1)(1 + 1) = 8, we expect four representations of a by x2 + y2 in total, with 22−1 = 2 of them proper, since a has two distinct prime divisors p ≡1 (mod 4). (b) a = 3825 = 32·52·17 = 602+152 = 572+242 = 482+392. Here m = (2+1)(1+1) = 6, so there are three representations of a by x2 + y2. None of them are proper, since a is divisible by a prime p ≡3 (mod 4). (c) a = 12025 = 52 · 13 · 37 = 1092 + 122 = 1082 + 192 = 1072 + 242 = 1002 + 452 = 962 + 532 = 802 + 752, with six total representations, four of them proper. (d) a = 357773 = 132 · 29 · 73 = 5982 + 132 = 5632 + 2022 = 5572 + 2182 = 5472 + 2422 = 5422 + 2532 = 4422 + 4032, with six total representations, four of them proper. (e) a = 359125 = 53 · 132 · 17. There are 23−1 = 4 proper representations of a by x2 + y2: a = 5992 + 182 = 5672 + 1942 = 5372 + 2662 = 4382 + 4092. There are twelve representations in total. The improper representations are a = 5982 + 392 = 5902+1052 = 5852+1302 = 5702+1852 = 5462+2472 = 5352+2702 = 4902+3452 = 4552 + 3902. (f) a = 585000 = 23 · 32 · 54 · 13 has ﬁve representations by x2 + y2, none of them proper: 7622 + 662 = 7502 + 1502 = 6902 + 3302 = 6782 + 3542 = 5702 + 5102. (g) a = 903125 = 55 · 172 has nine representations by x2 + y2, with two of them proper: 9502 +252 = 9352 +1702 = 9192 +2422 = 9052 +2902 = 8982 +3112 = 8502 +4252 = 7752 + 5502 = 7452 + 5902 = 6972 + 6462. (6) For a = 10414625 = 53·132·17·29, we have m = (3+1)(2+1)(1+1)(1+1) = 48, and so 24 representations by x2 + y2 in total. In addition to the eight proper representations listed in Exercise 3, we have the following improper representations: 32242 + 1432 = 32202 + 2152 = 32152 + 2802 = 31852 + 5202 = 31602 + 6552 = 31402 + 7452 = 30682 + 10012 = 30552 + 10402 = 29122 + 13912 = 28602 + 14952 = 27402 + 17052 = 27052 + 17602 = 26652 + 18202 = 24802 + 20652 = 24202 + 21352 = 23272 + 22362. Section 2.1. Quadratic Numbers and Quadratic Integers. (1) Let v = q + r √ d and w = s + t √ d be elements of a quadratic ﬁeld Q( √ d). (a) v + w = (q + s) + (r + t) √ d = (q + s) −(r + t) √ d = (q −r √ d) + (s −t √ d) = v + w. (b) vw = (qs + rtd) + (qt + rs) √ d = (qs+rtd)−(qt+rs) √ d = (q−r √ d)(s−t √ d) = v·w. (c) Using (b), N(vw) = vw · vw = (v · v)(w · w) = N(v) · N(w). (d) If d is negative, then N(v) = (q+r √ d)(q−r √ d) = q2−dr2 is a sum of two nonnegative real numbers, so can equal zero only when q2 = 0 and r2 = 0. In that case, v = q + r √ d = 0. If d > 1, write q = m k and r = n k with m, n, and k integers. Then N(v) = 0 implies that m2 = dn2. Since d > 1, then d has some prime divisor d. If n ̸= 0, then the exponent of p in dn2 is ep(dn2) = ep(d) +ep(n2) = 2ep(n) +1, since d is squarefree. This cannot equal ep(m2) = 2ep(m). So we must conclude that n and m equal 0. (2) Let Q( √ d) = {q + r √ d \| q, r ∈Q}, where d ̸= 1 is squarefree. Let v = q + r √ d and w = s + t √ d be elements of Q( √ d). 13 (a) v + w = (q + s) + (r + t) √ d and v −w = (q −s) + (r −t) √ d are elements of Q( √ d), since the sum and diﬀerence of rational numbers is rational. Also 0 = 0 + 0 √ d is an element of Q( √ d). (b) vw = (qs + rtd) + (qt + rs) √ d is an element of Q( √ d) since Q is closed under multiplication and addition. Also 1 = 1 + 0 √ d is in Q( √ d). (c) If v ̸= 0, then v N(v) = q q2−dr2 − r q2−dr2 √ d is an element of Q( √ d) since q2 −dr2 is a nonzero rational number by part (d) of Exercise 1. Here v· v N(v) = 1 since v·v = N(v). Thus v has an inverse under multiplication when v ̸= 0. (3) (a) If v = 1 3 −7 5 √ 3 = 5−21 √ 3 15 , then v = 5+21 √ 3 15 . We calculate that v + v = 2 5 and v · v = 25−3·441 225 = −1298 225 . Thus v is a root of x2 −2 5x −1298 225 , and has minimum polynomial f(x) = 225x2 −150x −1298. The discriminant of v is ∆(v) = (−150)2 − 4 · 225 · −1298 = 1190700 = 3 · 6302. (b) v = 3+5√−19 2 has minimum polynomial f(x) = x2 −3x+121 and discriminant ∆(v) = −475 = −19 · 52. (c) v = 2 7 + 1 3 √−29 = 6+7√−29 21 has minimum polynomial f(x) = 441x2 −252x + 1457 and discriminant ∆(v) = −2506644 = −29 · 2942. (d) v = 1 5 −2 3 √ 41 = 3−10 √ 41 15 has minimum polynomial f(x) = 225x2 −90x −4091 and discriminant ∆(v) = 3690000 = 41 · 3002. Section 2.2. Domains of Quadratic Integers. (1) (a) ∆(−7, 1) = −7 since −7 ≡1 (mod 4); z = 1+√−7 2 ; N(q + rz) = q2 + qr + 2r2; φ(x) = x2 + x + 2. (b) ∆(−7, 2) = −28; z = 1 + √−7; N(q + rz) = q2 + 2qr + 8r2; φ(x) = x2 + 2x + 8. (c) ∆(7, 1) = 28 since 7 ≡3 (mod 4); z = √ 7; N(q + rz) = q2 −7r2; φ(x) = x2 −7. (d) ∆(17, 1) = 17 since 17 ≡1 (mod 4); z = 1+ √ 17 2 ; N(q + rz) = q2 + qr −4r2; φ(x) = x2 + x −4. (2) (a) ∆= 45 = 32·5 = ∆(5, 3) since 5 ≡1 (mod 4); z = 3+ √ 45 2 ; N(q+rz) = q2+3qr−9r2; φ(x) = x2 + 3x −9. (b) ∆= −63 = 32 · −7 = ∆(−7, 3); z −3+√−63 2 ; N(q + rz) = q2 + 3qr + 18qr; φ(x) = x2 + 3x + 18. (c) ∆= −84 = 22 · −21 = ∆(−21, 1) since −21 ≡3 (mod 4); z = √−21; N(q + rz) = q2 + 21r2; φ(x) = x2 + 21. (d) ∆= 84 = 22 ·21 = ∆(21, 2) since 21 ≡1 (mod 4); z = 2+ √ 84 2 = 1+ √ 21; N(q +rz) = q2 + 2qr −20r2; φ(x) = x2 + 2x −20. (e) ∆= −88 = 22 · −22 = ∆(−22, 1); z = √−22; N(q + rz) = q2 + 22r2; φ(x) = x2 + 22. (f) ∆= 88 = 22 · 22 = ∆(22, 1); z = √ 22; N(q + rz) = q2 −22r2; φ(x) = x2 −22. (g) ∆= −99 = 32 · −11 = ∆(−11, 3); z = 3+√−99 2 ; N(q, r) = q2 + 3qr + 27r2; φ(x) = x2 + 3x + 27. (h) ∆= 297 = 32 · 33 = ∆(33, 3); z = 3+ √ 297 2 ; N(q, r) = q2 + 3qr −72r2; φ(x) = x2 + 3x −72. (i) ∆= 300 = 102 · 3 = 52 · 12 = ∆(3, 5) since 3 ≡3 (mod 4); z = √ 300 2 = √ 75; N(q, r) = q2 −75r2; φ(x) = x2 −75. (j) ∆= −300 = 102 · −3 = ∆(−3, 10) since −3 ≡1 (mod 4); z = 10+√−300 2 = 5 + √−75; N(q, r) = q2 + 10qr + 100r2; φ(x) = x2 + 10x + 100. (3) Let v = q + rz and w = s + tz be elements of a quadratic domain D∆. 14 (a) v + w = (q + s) + (r + t)z = (q + rz) + (s + tz) = v + w. (b) v · w = qs −ε2−∆ 4 · rt + (qt + rs + εrt)z = (q + rz) · (s + tz) = v · w. (Here we use the fact that z and z are both roots of x2 −εx + ε2−∆ 4 .) (c) N(v · w) = vw · vw = vv · ww = N(v) · N(w). (4) φ(−k−ε) = (−k−ε)2+ε(−k−ε)+ ε2−∆ 4 = k2+2εk+ε2 −εk−ε2 + ε2−∆ 4 = k2+εk+ ε2−∆ 4 , which equals φ(k). (5) We know that ∆(v) = (v −v)2 by Proposition 2.1.6, so then (v + v)2 −∆(v) = (v2 + 2vv + v2) −(v2 −2vv + v2) = 4vv = 4N(v). (6) If q + rz and s + tz are elements of a quadratic domain D = {a + bz \| a, b ∈Z}, then so is their sum, (q + s) + (r + t)z. Likewise 1 = 1 + 0z is in D, as is −(q + rz) = (−q) + (−r)z. (7) If φ(x) = x2 +εx+ ε2−∆ 4 is the principal polynomial of discriminant ∆= ∆(d, 1), then the principal polynomial of discriminant ∆γ = ∆(d, γ) = γ2∆is φγ(x) = x2 + εγx + ε2 γ−∆γ 4 = x2 + γεx + γ2 · ε2−∆ 4 , here using the fact that εγ = γε. So then φγ(γx) = γ2 x2 + εx + ε2 −∆ 4 = γ2φ(x). With φ′(x) = 2x + ε and φ′ γ(x) = 2x + εγ = 2x + γε, then φ′ γ(γx) = γ(2x + ε) = γφ′(x) is also true. (8) Let v = q + rzγ be an element of Dγ = Dγ2∆. Since zγ∆= γz∆, we can also write v = q + γrz in D = D∆. Now in Dγ, we have N(v) = q2 + εγqr + ε2 γ−γ2∆ 4 · r2, while in D we ﬁnd that N(v) = q2 + ε(γr) + ε2−∆ 4 (γr)2. But these are equal since εγ = γε. (9) (a) If w = vu for some u in D, then N(w) = N(vu) = N(v) · N(u). Since N(u) is a rational integer, then N(v) divides N(w) in Z. (b) If u divides 1, so that 1 = uv for some v in D, then N(1) = 1 = N(u) · N(v). So N(u) is a rational integer that divides 1 in Z, and so N(u) = ±1. Conversely, if N(u) = uu = ±1, then ±u is an inverse of u in D, and so u is a unit in D. (c) If u is a unit in D, then N(±un) = N(±1) · N(u)n = 1 · (±1)n = ±1, and so ±un is also a unit in D by part (b). (d) Let v and w be associates in D. First note that v = 0 if and only if w = 0, and in that case, N(v) = N(w). So assume that v ̸= 0. Now if v divides w and w divides v, then w = vu1 and v = wu2 for some u1 and u2 in D. Thus v = vu1u2, and so u1u2 = 1. But then u1 and u2 are units. Using part (b), then N(w) = N(v) · N(u1) = ±N(v). (10) By direct calculation, if z = 1+√−3 2 , then z2 = −1+√−3 2 = −1 + z, so that z3 = −1, and then z4 = −z, z5 = −(−1 + z) = 1 −z, and z6 = −(−1) = 1. Section 2.3. Ideal Form for Quadratic Integers. (1) Suppose that v = g(q + rz) has ideal form g[a : k], so that N(g + rz) = a and rk ≡q (mod a). Since N(w) = ww = N(w) for each element w in a quadratic domain D, then v = g(q +rz) has ideal form g[a : ℓ] for some ℓ. Since q +rz = q +r(ε−z) = (q +rε) −rz, we ﬁnd that rk ≡q (mod a) implies that −r(−k −ε) ≡q + rε (mod a), and so we can take ℓto equal −k −ε. (2) (a) In D−8, N(q + rz) = q2 + 2r2. If v = 3 + 5z, then N(v) = 59 and 5x ≡3 (mod 59) has k = −23 as a solution. So [59 : −23] is an ideal form for v. (b) N(3 + 5z) = 32 + 3 · 5 + 52 = 49 in D−3, and 5(−19) ≡3 (mod 49): v = [49 : −19]. (c) N(3 + 5z) = 32 −2 · 52 = −41 in D8, and 5(17) ≡3 (mod 41): v = [−41 : 17]. (d) N(3 + 5z) = 32 −3 · 52 = −66 in D12, and 5(27) ≡3 (mod 66): v = [−66 : 27]. 15 (e) N(5 + 2z) = 52 + 2 · 22 = 33 in D−8, and 2(−14) ≡5 (mod 33): v = [33 : −14]. (f) N(7 + 3z) = 72 −3 · 32 = 22 in D12, and 3(−5) ≡7 (mod 22): v = [22 : −5]. (g) N(5 −z) = 52 + 5(−1)2 = 30 in D−20, and −1(−5) ≡5 (mod 30): v = [30 : −5]. (h) v = 15 + 6z = 3(5 + 2z), with N(5 + 2z) = 52 + 5 · 2 + 6 · 22 = 59 in D−23. Since 2(−27) ≡5 (mod 59), then v = 3[59 : −27]. (i) v = 14 −6z = 2(7 −3z), with N(7 −3z) = 72 + 6(−3)2 = 103 in D−24. Here we have −3(32) ≡7 (mod 103), and so v = 2[103 : 32]. (j) v = 27 −12z = 3(9 −4z), with N(9 −4z) = 92 −6(−4)2 = −15 in D24. Here −4(−6) ≡9 (mod 15), so v = 3[−15 : −6]. (3) Let w = h(m + nz), with gcd(m, n) = 1, so that ms + nt = 1 for some integers s and t. If v = g(q+rz) divides w, then g divides both hm and hn, and so g divides (hm)s+(hn)t = h. (4) (a) v = 1 + z = [3 : 1] divides w = 5 + 2z = [33 : −14] in D−8, since 3 divides 33 and −14 ≡1 (mod 3). (b) v = [3 : 1] does not divide w = 5 −2z = [33 : 14] in D−8, since 14 ̸≡1 (mod 3). (c) v = 5 + 3z can be written as [−2 : 1], since N(5 + 3z) = 52 −3 · 32 = −2 with 3 ≡1 (mod 2), and w = 7 + 3z as [22 : −5] in D12, from part (f) of Exercise 2. Here v divides w since −2 divides 22 and −5 ≡1 (mod 2). (d) v = 1 + z = [6 : 1] divides w = 5 −z = [30 : −5] in D−20. (e) v = 1 + z = [6 : 1] does not divide w = 5 + z = [30 : 5] in D−20. (5) In D−8, N(q + rz) = q2 + 2r2. We ﬁnd that N(1 + 2z) = 9 with 2(−4) ≡1 (mod 9), and that N(1 + 3z) = 19 with 3(−6) ≡1 (mod 19). Since gcd(9, 19) = 1, we then ﬁnd that [9 : −4] · [19 : −6] = [171 : 32], since x = 32 satisﬁes x ≡−4 (mod 9) and x ≡−6 (mod 19). To verify this calculation directly, (1 + 2z)(1 + 3z) = 1 + 5z + 6z2 = −11 + 5z in D−8, where z2 = −2. We ﬁnd that N(−11 + 5z) = (−11)2 + 2 · 52 = 171, and that 5(32) ≡−11 (mod 171). (6) If ∆= −8, then φ(x) = x2 + 2. We ﬁnd that v = 1 + z has ideal form [3 : 1], and since φ(x) ≡0 (mod 3) has two solutions, we can use Proposition 2.3.7 to calculate powers of v in ideal form. Here v2 and v3 have ideal form [9 : 4] and [27 : −5] respectively. (For instance, φ(−5) = 27 so that x = −5 is the unique solution of x2 ≡0 (mod 27) that is congruent to 1 modulo 3.) To verify these claims, note that (1 + z)2 = 1 + 2z + z2 = −1 + 2z, and then (1+z)3 = (−1+2z)(1+z) = −1+z +2z2 = −5+z. Here N(−1+2z) = (−1)2+2·22 = 9, with 2(−4) ≡−1 (mod 9), and N(−5 + z) = (−5)2 + 2 · 12 = 27, with 1(−5) ≡−5 (mod 27). Section 2.4. Ideal Numbers. (1) We write [a : k] →[c : ℓ] if φ(ak) = ac and ℓ≡−k −ε (mod c). (a) With ∆= −8, then φ(x) = x2 + 2. We ﬁnd that [121 : 19] →[3 : −1] →[1 : 0] since φ(19) = 121 · 3 and φ(−1) = 3 · 1. Thus [121 : 19] is an ideal form for v = 1 3(19 + z)(−1 + z) = −7 + 6z. (b) For ∆= −11, φ(x) = x2+x+3. Here [111 : 23] →[5 : 1] →[1 : 0] since φ(23) = 111·5 with −k −ε = −24 ≡1 (mod 5), and φ(1) = 5 · 1. Thus [111 : 23] is an ideal form for v = 1 5(23 + z)(1 + z) = 4 + 5z. (Note that z2 = −3 + z in D−11.) (c) When ∆= −15, then φ(x) = x2+x+4. Here since φ(−10) = 94, we have immediately that v = −10 + z is an element of D−15 having ideal form [94 : −10]. (d) We ﬁnd that [94 : 37] →[15 : 7] →[4 : 0] →[1 : 0], and thus v = 1 15(37 + z) · 1 4(7 + z) · 1 1(0 + z) = 1 15(37 + z)(−1 + 2z) = −3 + 5z is an element of D−15 with ideal form [94 : 37]. (Note that z2 = −4 + z in D−15.) 16 (e) With φ−20(x) = x2 + 5, we ﬁnd that [123 : 47] →[18 : 7] →[3 : −1] →[2 : 1]. Here u−20 = 2, so we know that none of these expressions is an ideal form for an element of D−20. (f) With ∆= −20, we ﬁnd that [129 : 52] →[21 : −10] →[5 : 0][1 : 0]. Thus [129 : 52] is an ideal form for v = 1 21(52 + z) · 1 5(−10 + z) · 1 1(0 + z) = 1 21(52 + z)(−1 + 2z) = −2 −5z. (Here z2 = −5.) (g) For ∆= −23, we have φ(x) = x2 + x + 6. Since φ(15) = 246, we immediately have that [246 : 15] is an ideal form for v = 15 + z. (h) Here [246 : 56] →[13 : −5] →[2 : 0]. With u−23 = 2, it follows that [246 : 56] is not an ideal form for an element of D−23. (i) Here [246 : 66] →[18 : 5] →[2 : 0], so that [246 : 66] is not an ideal form for an element of D−23. (j) In this case, [246 : 107] →[47 : −14] →[4 : 1] →[2 : 0], and thus [246 : 107] is not an ideal form for an element of D−23. Section 2.5. Quadratic Domains with Unique Factorization. (1) Suppose that w is a prime element of a quadratic domain D, and that w = uv for some elements u and v of D. Then w divides uv, and so either w divides u or w divides v by the deﬁnition of prime elements. We can assume that w divides v without loss of generality, say that v = wx for some x in D. But now w = uv = w(ux) and so ux = 1 since w ̸= 0. Thus u is a unit of D, and it follows that w is irreducible in D by deﬁnition. (2) Suppose that D is a quadratic domain in which every irreducible element is prime. If D is not a unique factorization domain, then there is an element w of D that can be written both as w = u1 · · ·uk and as w = v1 · · ·vℓwith each ui and vi irreducible. We can assume that no ui is an associate of any vj by assuming that \|N(w)\| is as small as possible among all elements that can be written with distinct irreducible factorizations. But now u1 divides w, so that u1 divides the product v1 · · ·vℓ. Since u1 is irreducible, and so prime by assumption, it follows that u1 divides vj for some 1 ≤j ≤ℓ. With vj irreducible and u1 not a unit, we must conclude that u1 and vj are associates, contrary to assumption. Thus these distinct irreducible factorizations cannot exist. (3) Let D be a quadratic domain. Suppose that w is an irreducible element of D and that w divides uv for some u and v in D, say with uv = wx for some x in D. Assume that w divides neither u nor v. It follows that neither u nor v can be a unit, and so each can be written as a product of irreducible elements of D, say u = u1 · · ·uk and v = v1 · · ·vℓ. No ui nor vi can be an associate of w. But now if x is written as a product of irreducible elements, we have two irreducible factorizations of y = wx = uv, in which w appears as a factor in one expression (wx), but is not an associate of any irreducible factor in the other expression (uv = u1 · · ·uk · v1 · · ·vℓ). Thus D cannot be a unique factorization domain if it contains an irreducible element that is not prime. (4) It is noted in this section that D = D−8 is a principal ideal number domain, with −8 p = −1 if and only if p ≡5 or 7 (mod 8) and −8 p = 0 only for p = 2. Since φ(x, y) = x2 + 2y2 is the principal form of discriminant ∆= −8, the claim of this exercise is an immediate consequence of Theorem 2.5.3. (5) In an example in §13, it is established that D = D−3 is a principal ideal number domain. Here φ(x, y) = x2 + xy + y2 is the principal form of discriminant ∆= −3. In this case, 17 −3 p = p 3 = −1 if and only if p ≡2 (mod 3), and −3 p = 0 only when p = 3. So the claim of this exercise follows from Theorem 2.5.3. (6) If φ(q, r) = q2+qr+r2 = a, then φ(−r, q+r) = (−r)2−r(q+r)+(q+r)2 = q2+qr+r2 = a, and φ(−q −r, q) = (−q −r)2 −(q + r)q + q2 = q2 + qr + r2 = a. (7) Let φ(x, y) = x2 + xy + y2, and suppose that φ(q, r) = a with gcd(q, r) = 1. If r is odd and q is even, replace (q, r) by (−q −r, q). If r and q are both odd, replace (q, r) by (−r, q + r). In this way, we can assume that r is even. Now note that (q + r 2)2 + 3( r 2)2 = q2 + qr + r2 = a. Any prime common divisor of q + r 2 and r 2 would also divide both q and r, and so gcd(q + r 2, r 2) = 1. Thus every positive integer properly represented by φ(x, y) is also properly represented by x2 + 3y2. Section 2.6. Quadratic Domains without Unique Factorization. (1) (a) If ∆= −24, then N(q+rz) = q2 +6r2 and φ(x) = x2 +6. We ﬁnd that x = 2 satisﬁes the congruence φ(x) ≡0 (mod 5), and so [5 : 2] is an ideal number of discriminant ∆. This ideal number cannot be principal since q2 + 6r2 = 5 has no integer solutions. Then notice that φ(2) = 10 = 2 · 5, and that φ(2) = N(2 + z) = (2 + z)(2 −z). The equation 2 · 5 = (2 + z)(2 −z) is an example of distinct irreducible factorizations of the same element. (No two of these elements are associates, and each is irreducible because there are no elements v of D−24 with N(v) = 2 or N(v) = 5.) (b) If ∆= −40, then N(q + rz) = q2 + 10r2 and φ(x) = x2 + 10. Here [7 : 2] is an ideal number that cannot be principal, and since φ(2) = 14 = 2 · 7, but φ(2) = N(2 + z) = (2 + z)(2 −z), we ﬁnd that 14 = 2 · 7 = (2 + z)(2 −z) is an example of distinct irreducible factorizations. (c) If ∆= −52, then N(q + rz) = q2 + 13r2 and φ(x) = x2 + 13. We have that φ(1) = 14 = 2 · 7, and so [7 : 1] is an ideal number that is not principal. Since φ(1) = N(1 + z) = (1 + z)(1 −z), we obtain the following distinct irreducible factorizations: 14 = 2 · 7 = (1 + z)(1 −z). (2) For φ(x) = x2 + 5, we write [a : k] →[c : ℓ] if φ(k) = ac and ℓ≡−k (mod c). (a) Since [87 : −16] →[3 : 1] →[2 : 1], then [87 : −16] is not a principal ideal number. With 87 = 3 · 29, we can write [87 : −16] = [3 : −1] · [29 : 13]. (b) Here [161 : 31] →[6 : −1] →[1 : 0], so that [161 : 31] is an ideal form for the element v = 1 6(31 + z)(−1 + z) = −6 + 5z. Since 161 = 7 · 23, then [161 : 31] = [7 : 3] · [23 : 8]. (c) [161 : −38] →[9 : 2] →[1 : 0], so that [161 : −38] is an ideal form for v = 1 9(−38 + z)(2 + z) = −9 −4z. We can write [161 : −38] = [7 : −3] · [23 : 8]. (d) [203 : −74] →[27 : −7] →[2 : 1] is not an ideal form for an element of D−20. Since 203 = 7 · 29, we can write [203 : −74] = [7 : 3] · [29 : 13]. (e) [270 : 115] →[49 : −17] →[6 : −1] →[1 : 0], so that [270 : 115] is an ideal form for v = 1 49(115 + z) · 1 6(−17 + z)(−1 + z) = 5 −7z. Here 270 = 2 · 33 · 5, and thus [270 : 115] = [2 : 1] · [3 : 1]3 · [5 : 0]. (3) (a) Since z2 = −6, we ﬁnd that (3 + 2z)(2 + z) = −6 + 7z = (4 −z)(−3 + z) in D−24. Ideal form expressions for these factors are 3 + 2z = [33 : −15], 2 + z = [10 : 2], 4 −z = [22 : −4], and −3 + z = [15 : −3]. Each factor is irreducible in D−24 since there are no elements v = q + rz of that domain for which N(v) = q2 + 6r2 = 2 or 3. But we have the following ideal number factorizations: [33 : −15] · [10 : 2] = [3 : 0] · [11 : −4] · [2 : 0] · [5 : 2], and [22 : −4] · [15 : −3] = [2 : 0] · [11 : −4] · [3 : 0] · [5 : 2]. 18 (b) Here with z2 = −13, we have (5 + z)(8 + z) = 27 + 13z = (−3 + z)(4 −3z) in D−52. Ideal form expressions for the factors are 5 + z = [38 : 5] = [2 : 1] · [19 : 5], 8 + z = [77 : 8] = [7 : 1] · [11 : −3], −3 + z = [22 : −3] = [2 : 1] · [11 : −3], and 4 −3z = [133 : 43] = [7 : 1] · [19 : 5]. Each factor is irreducible in D−52 since q2 + 13r2 = 2 or 7 has no integer solutions, and we ﬁnd the same irreducible ideal number factors in both products. (c) Here (3 + 2z)(2 −z) = 34 + z = (4 + z)(5 −z) in D−56, where z2 = −14. We ﬁnd that 3 + 2z = [65 : −31] = [5 : −1] · [13 : −5], 2 −z = [18 : −2] = [2 : 0] · [3 : 1]2, 4 + z = [30 : 4] = [2 : 0] · [3 : 1] · [5 : −1], and 5 −z = [29 : −5] = [3 : 1] · [13 : −5]. The factors are irreducible in D−56 since q2 + 14r2 = 2, 3, or 5 has no integer solutions. (d) With z2 = −6 + z, we ﬁnd that (3 + 2z)(7 + z) = 9 + 19z = (−5 + z)(3 −4z) in D−23. Here 3 + 2z = [39 : −18] = [3 : 0] · [13 : −5], 7 + z = [62 : 7] = [2 : 1] · [31 : 7], −5 + z = [26 : −5] = [2 : 1] · [13 : −5], and 3 −4z = [93 : −24] = [3 : 0] · [31 : 7]. Here N(q + rz) = q2 + qr + 6r2 = (2q+r)2+23r2 4 = 2 or 3 has no integer solutions, so it follows that the factors are irreducible in D−23. Section 3.1. Ideals and Ideal Numbers. (1) Let v and w be elements of a quadratic domain D and let ⟨v, w⟩= {vx + wy \| x, y ∈D}. Using closure properties of D, we see that if vx + wy and vx′ + wy′ are in ⟨v, w⟩, then (vx + wy) −(vx′ + wy′) = v(x −x′) + w(y −y′) is in ⟨v, w⟩, and if u is an element of D, then (vx + wy)u = v(xu) + w(yu) is in ⟨v, w⟩. Thus ⟨v, w⟩is an ideal of D. (2) Let v be an element and A an ideal of D, and let vA = {vx \| x ∈A}. Using closure properties of A, we see that if vx and vx′ are in vA and y is in D, then vx−vx′ = v(x−x′) and (vx)y = v(xy) are in vA. If x is in A, then x is also in D, and so vx is in the principal ideal ⟨v⟩, and vA ⊆⟨v⟩. Suppose that v ̸= 0 and that vA = vB for some ideals A and B of D. If a is an element of A, then va is in vA = vB, and so has the form vb for some b in B. But if v ̸= 0, then va = vb implies that a = b, and thus A ⊆B. The reverse inclusion and the converse are similarly established. (3) Let A = {x ∈D \| vx ∈B}, where v is an element and B an ideal of a quadratic domain D. If x and y are in A, then vx and vy are in B, and so vx −vy = v(x −y) is in B. But then x −y is in A by deﬁnition. Likewise, if x is in A and u is in D, then vx is in B and so (vx)u = v(xu) is in B. But then xu is in A by deﬁnition. Thus A is an ideal of D. Now suppose that B is a subset of the principal ideal ⟨v⟩. If y is in B, then y = vx for some x in D. But then x is in A by deﬁnition, and so y = vx is in vA. Thus B ⊆vA. For the reverse inclusion, let x be an element of A. Then by deﬁnition vx is in B, and so vA ⊆B. (4) Let S = {v1, . . ., vt} be a ﬁnite subset of a quadratic domain D. Let m1v1 +· · ·+mtvt and n1v1 +· · ·+ntvt be Z-combinations of S. Then (m1v1 +· · ·+mtvt) −(n1v1 +· · ·+ntvt) = (m1 −n1)v1 + · + (mt −nt)vt is also a Z-combination of S. If S = {3, 1 + i} in Z[i], let A be the set of all Z-combinations of S. Note that 1 + i is in A and 1 −i is in Z[i], but that (1 + i)(1 −i) = 2 is not in A. (If 2 = 3m + (1 + i)n = (3m + n) + ni, then n = 0 and m is not a rational integer.) (5) Let w be an element in the Z-span of {a, k+z}, that is, let w = m(a)+n(k+z) for some m and n in Z. Then w = (−m)(−a)+n(k+z) is also in the Z-span of {−a, k+z}. The reverse inclusion is established in the same way. Suppose that ℓ≡k (mod a), say that ℓ−k = aq for some q in Z, and that w = m(a) + n(k + z) as above. Then w = (m −nq)a + n(ℓ+ z), so that w is also in the Z-span of {a, ℓ+ z}. Again the reverse inclusion is similar. (6) Let g be the divisor of an ideal B of a quadratic domain, so that B = gA for the ideal A = {x ∈D \| gx ∈B} of D. By deﬁnition, B contains an element m + gz, and the proof 19 of Proposition 3.1.2 shows that g divides m. But now m g + z is an element of A, and so the divisor of A must be 1, that is, A is primitive. (7) Let B be a nontrivial ideal of a quadratic domain D, with divisor g, subnorm a, and character k. By deﬁnition, then B contains the elements ga and g(k + z), and so contains every Z-combination of the set S = {ga, g(k + z)}. The proof of Theorem 3.1.5 shows that every element of B can be written as a Z-combination of S. Suppose that an element v of B can be written both as v = m(ga) + n(gk + gz) = (mga + ngk) + ngz and as s(ga) + t(gk + gz) = (sga + tgk) + tgz for some rational integers m, n, s, and t. Then ng = tg since otherwise we could solve for z as a rational number. With g positive, it follows that n = t. Now since mga + ngk = sga + tgk and a ̸= 0, we see that m = s. So expressions for elements of B as Z-combinations of S are unique, and S is a Z-basis for B. (8) Ideals of norm 75 = 3·52 can have divisor 1 or 5. Finding all solutions of φ(x) ≡0 (mod 3) and of φ(x) ≡0 (mod 75), with φ(x) = x2 −31 the principal polynomial of discriminant ∆= 124, we obtain the following six ideal of norm 75: [75 : 16], [75 : −16], [75 : 34], [75 : −34], 5[3 : 1], and 5[3 : −1]. (9) (a) The principal polynomial of discriminant ∆= −7 is φ(x) = x2+x+2. The congruence φ(x) ≡0 (mod 2) has two solutions, and so [2 : 0] and [2 : −1] are the distinct ideals of norm 2 in D−7. (b) Here φ(x) = x2 + x + 8, and with 70 = 2 · 5 · 7, we ﬁnd that there are eight distinct solutions of φ(x) ≡0 (mod 70). The ideals of norm 70 in D−31 can be written as [70 : 11], [70 : −12], [70 : 16], [70 : −17], [70 : 18], [70 : −19], [70 : 23], and [70 : −24]. (c) The divisor of an ideal of norm 100 = 102 can be any divisor of 10. The principal polynomial of discriminant ∆= 41 is φ(x) = x2+x−10. Finding solutions of φ(x) ≡0 (mod n) for n = 1, 4, 25, and 100, we obtain the following list of ideals of norm 100 in D41: [100 : 10], [100 : −11], [100 : 14], [100 : −15], 2[25 : 10], 2[25 : −11], 5[4 : 1], 5[4 : −2], and 10[1 : 0]. (10) If A is an ideal of a quadratic domain D, and v and w are elements of D, write v ≡w (mod A) to mean that v −w is an element of A. We show as follows that congruence modulo A is an equivalence relation on D. (a) If v is in D, then v ≡v (mod A) since v −v = 0 is an element of every ideal A. So congruence is reﬂexive. (b) If v ≡w (mod A) so that v−w is an element of A, then w−v = −(v−w) is an element of A by properties of ideals. Thus w ≡v (mod A) and congruence is symmetric. (c) If u ≡v (mod A) and v ≡w (mod A), so that u −v and v −w are elements of A, then u−w = (u−v)+(v −w) is also in A by closure properties of ideals. Thus u ≡w (mod A) and congruence is transitive. Now suppose that A = g[a : k] with g and a positive, so that S = {ga, gk +gz} is a Z-basis for A. Let v = s + tz be an element of D. Write t = gn + r with 0 ≤r < g, and then write s−gnk = (ga)m +q with 0 ≤q < ga. Then (s+tz) −(q +rz) = (s−q) +(t−r)z = m(ga) + n(gk + gz) is an element of A, so that s + tz ≡q + rz (mod A). Thus every element of D is congruent modulo A to an element q + rz with 0 ≤q < ga and 0 ≤r < g. Finally, suppose that q +rz ≡s+tz (mod A) with 0 ≤q, s < ga and 0 ≤r, t < g. We can assume without loss of generality that r ≥t. Then (q −s) + (r −t)z is an element of A. With 0 ≤r −t < g, we must conclude that r = t to avoid contradicting the deﬁnition of the divisor of A. So q −s is a rational integer in A, and with −ga < q −s < ga, we must likewise conclude that q = s. Thus every element of D is congruent modulo A to precisely one element q + rz with 0 ≤q < ga and 0 ≤r < g, and there are exactly g · ga = g2a distinct equivalence classes of elements of D under congruence modulo A. 20 Section 3.2. Writing Ideals as Ideal Numbers. (1) (a) In D13, N(q + rz) = q2 + qr −3r2. Since N(1 + 3z) = −23 and 3x ≡1 (mod 23) has solution k = 8, we can write A = ⟨1 + 3z⟩= [−23 : 8] = [23 : 8]. [To conﬁrm this claim directly, the typical element of A has the form (1+3z)(q+rz) = (q+9r)+(3q+4r)z, here using the fact that z2 = 3 + z. Such an element is in Z only when 3q + 4r = 0, so that q = 4u and r = −3u for some u ∈Z. Then q + 9r = −23u, and so a = 23 is the smallest positive integer in A. When q = −1 and r = 1, we have 8 + z in A, and so A = [23 : 8]. The following examples can be conﬁrmed in the same way.] (b) In D28, N(1 + 3z) = 12 −7 · 32 = −62 and 3x ≡1 (mod 62) has solution k = 21. So A = ⟨1 + 3z⟩= [−62 : 21] = [62 : 21]. (c) N(5 −3z) = 52 + 5(−3) −9(−3)2 = −71 in D = D37, and −3x ≡5 (mod 71) has solution k = 22. So A = ⟨5 −3z⟩= [−71 : 22] = [71 : 22]. (d) Here note ﬁrst that A = ⟨12 −9z⟩= 3 ⟨4 −3z⟩. In D = D−67, N(q + rz) = q2 + qr + 17r2, so that N(4−3z) = 157. We ﬁnd that −3x ≡4 (mod 157) has solution k = 51. Thus A = 3[157 : 51]. (e) N(13 + 5z) = 132 + 17 · 52 = 594 in D = D−68, and 5x ≡13 (mod 594) has solution k = −235. Thus A = ⟨13 + 5z⟩= [594 : −235]. (2) Let D be the quadratic domain and φ(x) the principal polynomial of discriminant ∆. Suppose that D is a principal ideal domain and that a and k are integers for which a divides φ(k). Then A = [a : k] is an ideal of D, so can be written as ⟨v⟩for some element v of D. But now either [a : k] or [−a : k] is an ideal form for v, using Theorem 3.2.2. Conversely, suppose that whenever a divides φ(k), then either [a : k] or [−a : k] is an ideal form of an element v of D. Then the ideal A = [a : k] equals the principal ideal ⟨v⟩. Since every ideal of D has the form g[a : k] where a divides φ(k), then every ideal of D is given by g ⟨v⟩= ⟨gv⟩for some v in D. That is, D is a principal ideal domain. (3) Let A and B be ideals of D, and let A + B = {a + b \| a ∈A and b ∈B}. If a1 + b1 and a2 +b2 are in A+B, then (a1 +b1)−(a2+b2) = (a1 −a2)+(b1−b2) is an element of A+B by the closure of both A and B under subtraction. Likewise, if a + b is in A + B and v is in D, then (a + b)v = av + bv is an element of A + B since av is in A and bv is in B. Thus A + B is an ideal of D. If a is an element of A, then a = a + 0 is in A + B, since 0 is an element of B. So A ⊆A + B, and by the same argument, B ⊆A + B. Finally, suppose that C is an ideal of D that contains both A and B as subsets. Then for any a in A and b in B, the sum a + b is an element of C by closure properties of ideals. It follows that A + B ⊆C. (4) Let v and w be elements of a quadratic domain D. The typical element of ⟨v, w⟩has the form vx + wy where x and y are in D. Since vx ∈⟨v⟩and wy ∈⟨w⟩, it follows that ⟨v, w⟩⊆⟨v⟩+ ⟨w⟩. The reverse inclusion is shown in the same way. (5) (a) In D13, we can write ⟨12⟩= 12[1 : 0] and ⟨5 + z⟩= 275 since N(5+z) = 52+5·1−3(1)2. Since gcd(12, 27, 12·5) = 3, with x ≡5 ≡−1 (mod 3), we ﬁnd that A = ⟨12, 5 + z⟩= [3 : −1] in D13. (b) In D−40, we have ⟨7⟩= 7[1 : 0] and ⟨3 −2z⟩= [49 : 23]. Here gcd(7, 49, 7 · 23) = 7 and we ﬁnd that A = ⟨7, 3 −2z⟩= [7 : 2] in D−40. (c) Since ⟨7⟩= 7[1 : 0] and ⟨3 + z⟩= [7 : 3], we ﬁnd that A = ⟨14, 6 + 2z⟩= 2 ⟨7, 3 + z⟩= 2[7 : 3] in D8. (Note that in this example, 7 is an element of ⟨3 + z⟩, so that ⟨7, 3 + z⟩= ⟨3 + z⟩.) (6) Suppose that A is an ideal of a quadratic domain D, and let A = {v \| v ∈A}. If v and w are elements of A, so that v and w are in A, then v −w = v −w is also in A, here using properties of conjugate elements and the fact that v −w is an element of A. Likewise, if 21 v is in A, so that v is in A, and w is an element of D, then v · w = v · w is an element of A. (Here w is also an element of D, so we use the multiplicative property of the ideal A.) Thus A is an ideal of D. (7) Let A = ⟨v⟩be a principal ideal of a quadratic domain D. If w is an element of A, then w = vx = v · x for some element x of D, since every element of A has the form vx. It follows that A ⊆⟨v⟩. Conversely, the typical element of ⟨v⟩has the form y = v ·x for some x in D. But then y = v · x is in A by deﬁnition, since x is in D so that v · x is in A = ⟨v⟩. Thus ⟨v⟩⊆A, and A = ⟨v⟩. Section 3.3. Prime Ideals of Quadratic Domains. (1) Let A = [5 : 2], A = [5 : −2], B = 5[1 : 0], and C = [65 : 8], ideals of the quadratic domain Z[i] = D−4. (a) B is a subset of A since g = 1 divides h = 5, ag = 5 divides bh = 5, and hℓ≡hk (mod ag), that is, 0 ≡10 (mod 5). (b) B is likewise a subset of A. (Here we also have 0 ≡−10 (mod 5).) (c) C is not a subset of A, and A is not a subset of C. Here 5 is an element of A not in C, and 8 + i is an element of C not in A. (d) C is a subset of A. Here with g = 1 = h, we note that a = 5 divides b = 65, and that ℓ≡k (mod a), that is, 8 ≡−2 (mod 5). (2) Let A = g[a : k] and B = h[b : ℓ] be ideals of a quadratic domain D. By Proposition 3.3.1, B ⊆A if and only if g divides h, ag divides bh, and hℓ≡hk (mod ag), and A ⊆B if and only if h divides g, bh divides ag, and gk ≡gℓ(mod bh). It follows that A = B if and only if g = h (since g and h are both positive), ag = ±bh so that a = ±b, and ℓ≡k (mod a), using the congruence cancellation property. (3) If B = h[b : ℓ] is a subset of A = g[a : k], then g divides h and ag divides bh. It follows immediately that N(A) = \|g2a\| divides N(B) = \|h2b\|. (4) If B = h[b : ℓ] is a subset of A = g[a : k], then g divides h, ag divides bh, and hℓ≡hk (mod ag). But if N(A) = \|g2a\| equals N(B) = \|h2b\|, then g = h and \|ag\| = \|bh\| so that b = ±a. Then hℓ≡hk (mod ag) implies that ℓ≡k (mod a), and by Exercise 2 we conclude that A = B. (5) (a) Since −7 23 = 23 7 = 1, there are two conjugate prime ideals of norm p = 23 in D−7, which we ﬁnd to be [23 : 9] and [23 : −10]. (b) With −23 ≡1 (mod 8), there are two prime ideals of norm p = 2 in D−23, namely [2 : 0] and [2 : −1]. (c) 29 5 = 4 5 = 1, so there are two prime ideals of norm p = 5 in D29: [5 : 1] and [5 : −2]. (d) −47 ≡1 (mod 8) so [2 : 0] and [2 : 1] are prime ideals of norm p = 2 in D−47. (e) −47 3 = 1 3 = 1, so there are two prime ideals of norm p = 3 in D−47: [3 : 0] and [3 : −1]. (f) −47 17 = 4 17 = 1, so there are two prime ideals of norm p = 17 in D−47: [17 : 7] and [17 : −8]. (g) −47 23 = −1 23 = −1, so there are no ideals of norm p = 23 in D−47. The only prime ideal containing 23 is the principal ideal ⟨23⟩. (h) −60 11 = −1, so there are no ideals of norm p = 11 in D−60. The only prime ideal containing 11 is ⟨11⟩. Section 3.4. Multiplication of Ideals. (1) Let A and B be ideals of a quadratic domain D. Let v1w1 + v2w2 + · · ·vnwn and x1y1 + x2y2 + · · ·+ xmym be elements of AB, where each vi and xi is in A, and each wi and yi is in B. Then (v1w1 + v2w2 + · · ·vnwn) −(x1y1 + x2y2 + · · ·+ xmym) = v1w1 + · · ·+ vnwn + 22 x1(−y1) +· · ·+xm(−ym) is a ﬁnite sum of products of element of A and elements of B, so is an element of AB. Likewise, if u is an element of D, then u(v1w1 + v2w2 + · · ·vnwn) = (uv1)w1+(uv2)w2+· · ·+(uvn)wn is a ﬁnite sum of products of elements of A with elements of B, so is in AB. Thus AB is an ideal of D. Note that each product viwi is an element of A, by closure properties of an ideal, and thus the sum of any ﬁnite number of such products is in A. Thus AB ⊆A. The proof that AB ⊆B is similar. (2) Let A, B, and C be ideals of a quadratic domain D. (a) The typical element of AB has the form v1w1 + v2w2 + · · ·vnwn with each vi in A and each wi in B. This expression equals w1v1 + w2v2 + · · ·wnvn, so is an element of BA. Thus AB ⊆BA. The proof of the reverse containment is similar. (b) The typical element of A(BC) is a sum of products u(v1w1+v2w2+· · ·vnwn) where u is in A, and each vi is in A and each wi is in B. If we distribute u through each sum, and write each product u(viwi) as (uvi)wi, we see that all such expressions are elements of (AB)C. Therefore A(BC) ⊆(AB)C. The proof of the reverse containment is similar. (c) If a is in A, then a = a · 1 is in AD, since 1 is in D. Thus A ⊆AD. For the reverse containment, we can use the observation from Exercise 1 that a product AD of A with any ideal of D is a subset of A. (d) If v1w1 +v2w2 +· · ·vnwn is in AB, then v1w1 + v2w2 + · · ·vnwn = v1 · w1 +· · ·vn · wn is an element of A · B. Thus AB ⊆A · B. The reverse containment uses the same equation. (e) The typical element of ⟨vw⟩is vwx where x is in D. But vwx = v(wx), with wx ∈⟨w⟩, and so ⟨vw⟩⊆⟨v⟩⟨w⟩. For the reverse containment, the typical element of ⟨v⟩⟨w⟩has the form (vx1)(wy1) + (vx2)(wy2) + · · · + (vxn)(wyn) = vw(x1y1 + x2y2 + · · ·xnyn), with each xi and yi in D. The ﬁnal equation shows that each such expression is an element of ⟨vw⟩, and so ⟨v⟩⟨w⟩⊆⟨vw⟩. (3) Let A be an ideal of a quadratic domain D∆, written as A = g[a : k] and as A = g[a : ℓ], where ℓ≡k (mod a), say ℓ= k + aq for some rational integer q. If φ(x) is the principal polynomial of discriminant ∆, then direct calculation shows that φ(ℓ) = φ(k + aq) = φ(k) + φ′(k) · aq + a2q2 and φ′(ell) = φ′(k) + 2aq. So if φ(k) = ac, φ(ℓ) = ac0, φ′(k) = b, and φ′(ℓ) = b0, we ﬁnd that c0 = c+bq+aq2 and b0 = b+2aq. Any common divisor of a, b, and c also is a common divisor of a, b0, and c0. But likewise, any common divisor of a, b0, and c0 also divides a, b = b0 −2aq, and c = c0 −bq −aq2. Thus gcd(a, b, c) = gcd(a, b0, c0), and the index of A is well-deﬁned. (4) The principal polynomial of discriminant ∆= 61 is φ(x) = x2 + x −15. Since φ(0) = −15 is divisible by 3 and 5, then A = [3 : 0] and B = [5 : 0] are ideals of D61. (a) AB = [3 : 0] · [5 : 0] = [15 : 0], since x = 0 is the unique solution modulo 15 of x ≡0 (mod 3) and x ≡0 (mod 5). (b) A2 = [3 : 0] · [3 : 0] = [9 : −3]. Here A ̸= A = [3 : −1], and we ﬁnd that −3 is the unique solution of φ(x) ≡0 (mod 9) that is congruent to 0 modulo 3. (c) B2 = [5 : 0] · [5 : 0] = [25 : −10]. Again B ̸= B = [5 : −1], and here −10 satisﬁes φ(x) ≡0 (mod 25) with −10 ≡0 (mod 5). (d) AB = [3 : 0] · [5 : −1] = [15 : −6], since x = −6 satisﬁes x ≡0 (mod 3) and x ≡−1 (mod 5). (e) A2B = [9 : −3] · [5 : 0] = [45 : 15], since x = 15 satisﬁes x ≡−3 (mod 9) and x ≡0 (mod 5). (f) A3 = [27 : 6]. Here φ(6) = 27 is divisible by 27, with 6 ≡0 (mod 3). (g) A3B2 = [27 : 6] · [25 : −10] = [675 : −210], since x = 210 satisﬁes x ≡6 (mod 27) and x ≡−10 (mod 25). 23 Section 3.5. Prime Ideal Factorization. (1) (a) The principal polynomial of discriminant ∆= −20 is φ(x) = x2 + 5. Since φ(35) = 1230 = 615 · 2, then [615 : 35] is an ideal of D−20. Since −20 = 4 · −5 is a primitive discriminant, and 615 = 3 · 5 · 41, we ﬁnd that [615 : 35] = [3 : 35] · [5 : 35] · [41 : 35] = [3 : −1] · [5 : 0] · [41 : −6]. (b) For ∆= 41, φ(x) = x2 + x −10. Here φ(39) = 1550 = 775 · 2, so [775 : 39] is an ideal of D41. With 775 = 5 · 5 · 31, we have that [775 : 39] = [5 : 39]2 · [31 : 39] = [5 : −1]2 · [31 : −8]. (c) For ∆= −39, φ(x) = x2 +x+10, and we ﬁnd that φ(29) = 880 = 220·4. So [220 : 29] is an ideal of D−39, which we can factor as [220 : 29] = [2 : 29]2 · [5 : 29] · [11 : 29] = [2 : 1]2 · [5 : −1] · [11 : −4]. (2) We found the following ideals of norm 75 = 3 · 52 in D124, and can factor them as shown since D124 is a complete quadratic domain. (a) [75 : 16] = [3 : 1] · [5 : 1] · [5 : 1]. (b) [75 : −16] = [3 : −1] · [5 : −1] · [5 : −1]. (c) [75 : 34] = [3 : 1] · [5 : −1] · [5 : −1]. (d) [75 : −34] = [3 : −1] · [5 : 1] · [5 : 1]. (e) 5 · [3 : 1] = [3 : 1] · [5 : 1] · [5 : −1]. (f) 5 · [3 : −1] = [3 : −1] · [5 : 1] · [5 : −1]. (3) (a) In D40, we ﬁnd that ⟨4 −6z⟩= 2 ⟨2 −3z⟩= 2 · [86 : 28], since N((2 −3z) = 22 − 10(−3)2 = −86, and k = 28 satisﬁes −3k ≡2 (mod 86). Since 2 divides 40, we ﬁnd that ⟨2⟩= [2 : 0]2, and so we calculate that ⟨4 −6z⟩= [2 : 0]3 · [43 : −15]. (b) In D−11, we can write ⟨4 + 3z⟩= [55 : −17], since N(4+3z) = 42+4·3+3·32 = 55 and 3·−17 ≡4 (mod 55). This ideal factors into prime ideals as [55 : 17] = [5 : 2]·[11 : −5]. (c) In D−68, we can write ⟨6 + 12z⟩= 6 ⟨1 + 2z⟩= 6 · [69 : −34], since N(1 + 2z) = 12 + 17 · 22 = 69, and 2 · −34 ≡1 (mod 69). Since 2 divides −68, we ﬁnd that ⟨2⟩= [2 : 1]2, and since −68 3 = 1, we ﬁnd that ⟨3⟩= [3 : 1] · [3 : −1]. Combining these results, we can write ⟨6 + 12z⟩= [2 : 1]2 · [3 : 1] · [3 : −1]2 · [23 : −11]. (d) In D−23, we can write ⟨−2 + 3z⟩as [52 : −18] = [2 : 0]2·[13 : −5]. (Here N(−2+3z) = (−2)2 + (−2)3 + 6(3)2 = 52, and 3 · −18 ≡−2 (mod 52).) (e) In D220, we can write ⟨23 + 3z⟩as [34 : −15] = [2 : 1] · [17 : 2]. (Here N(23 + 3z) = 232 −55 · 32 = 34, and 3 · −15 ≡23 (mod 34).) (4) Let D be the quadratic domain of discriminant ∆(−11, 5) = −275, for which φ(x) = x2 + 5x + 75. (a) The ideal A = [25 : 0] is not a complete ideal, and cannot be written as a product of prime ideals of D. Here P = [5 : 0] is the only prime ideal that contains A, but we ﬁnd that P 2 = 5P, so that P 2 ⊆A ⊆P with both containments proper. (b) A = [27 : 1] is a complete ideal since N(A) = 99 is relatively prime to γ∆= 5. We can factor A as [3 : 1]3. (c) A = [99 : 3] is also a complete ideal, and can be factored as [3 : 0]2 · [11 : 3]. (5) Applying Theorem 12.3, we ﬁnd that an element v having ideal form g[a : k] divides a rational integer m = m[1 : 0] in a quadratic domain if and only if ag divides m. In each part below, we ﬁnd divisors of m in D by testing for elements whose subnorm divides m. 24 (a) In D−40, with z = √−10, the norm of an element q + rz is q2 + 10r2. We ﬁnd that 49 = 7·7 = (3+2z)(3−2z), with each factor irreducible in D−40 since q2+10r2 = 7 has no integer solutions, so that no elements of norm 7 exist. In ideal number notation, we have that ⟨49⟩= [7 : 2]2 · [7 : −2]2. Here ⟨7⟩· ⟨7⟩= ([7 : 2] · [7 : −2])([7 : 2] · [7 : −2]), while ⟨3 + 2z⟩· ⟨3 −2z⟩= ([7 : −2] · [7 : −2])([7 : 2] · [7 : 2]). (Note that [7 : −2]2 = [49 : −23] = ⟨3 + 2z⟩since N(3 + 2z) = 49 and k = −23 satisﬁes 2k ≡3 (mod 49).) (b) In D−15, we can write 16 = 2 · 2 · 2 · 2 = z · z · (1 −z) · (1 −z) where z = 1+√−15 2 . (A third possibility, using the same factors, is 16 = 2 · 2 · z · (1 −z).) The norm of an element q + rz is q2 + qr + 4r2 = 1 4((2q + r)2 + 15r2), so it follows that there are no elements of norm 2 in D−15 (since there are no solutions of x2 + 15y2 = 8), and so each of these factors is irreducible. Here ⟨16⟩= [2 : 0]4 · [2 : −1]4, with ⟨2⟩· ⟨2⟩· ⟨2⟩· ⟨2⟩= ([2 : 0] · [2 : −1])2 · ([2 : 0] · [2 : −1])2 and ⟨z⟩· ⟨z⟩· ⟨1 −z⟩· ⟨1 −z⟩= ([2 : 0] · [2 : 0])2 · ([2 : −1] · [2 : −1])2. (For example, [2 : 0]2 = [4 : 0] = ⟨z⟩since N(z) = 4 and 1 · 0 ≡0 (mod 4).) (c) In D−84, we ﬁnd that there are three ways of writing 85 as a product of irreducible elements in D−84: 85 = 5 · 17 = (1 + 2z)(1 −2z) = (8 + z)(8 −z), where z = √−21. (These factors are irreducible since N(q + rz) = q2 + 21r2 cannot take on the value 5 or 17.) Here ⟨85⟩= [5 : 2] · [5 : −2] · [17 : 8] · [17 : −8], with ⟨5⟩· ⟨17⟩= ([5 : 2] · [5 : −2]) · ([17 : 8] · [17 : −8]), ⟨1 + 2z⟩· ⟨1 −2z⟩= ([5 : −2] · [17 : −8]) · ([5 : 2] · [17 : 8]), ⟨8 + z⟩· ⟨8 −z⟩= ([5 : −2] · [17 : 8]) · ([5 : 2] · [17 : −8]). For example, ⟨1 + 2z⟩= [85 : −42] = [5 : −2] · [17 : −8] since 2 · −42 ≡1 (mod 85), while ⟨8 + z⟩= [85 : 8] = [5 : −2] · [17 : 8]. (d) In D−23, we can write 12 = 2 · 2 · 3 = 2 · z · (1 −z) = (2 + z) · (3 −z), where z = 1+√−23 2 . Each factor is irreducible in D−23, since the norm of an element q + rz is q2 + qr + 6r2 = 1 4((2q + r)2 + 23r2). There are no elements of norm 2 or 3 in D−23 because x2 + 23y2 = 8 and x2 + 23y2 = 12 have no integer solutions. Here ⟨12⟩= [2 : 0]2 · [2 : −1]2 · [3 : 0] · [3 : −1], with ⟨2⟩· ⟨2⟩· ⟨3⟩= ([2 : 0] · [2 : −1]) · ([2 : 0] · [2 : −1]) · ([3 : 0] · [3 : −1]), ⟨2⟩· ⟨z⟩· ⟨1 −z⟩= ([2 : 0] · [2 : −1]) · ([2 : 0] · ⟨3⟩0) · ([2 : −1] · [3 : −1]), ⟨2 + z⟩· ⟨3 −z⟩= ([2 : 0] · [2 : 0] · [3 : −1]) · ([2 : −1] · [2 : −1] · [3 : 0]). For example, ⟨z⟩= [6 : 0] = [2 : 0]·[3 : 0] and ⟨2 + z⟩= [12 : 2] = [2 : 0]·[2 : 0]·[3 : −1]. (e) In D−120, 130 = 2 · 5 · 13 = (10 + z)(10 −z), where z = √−30. Since x2 −30y2 = n has no solutions for n = 2, n = 5, n = 10, and n = 13, we can see that each of these factors is irreducible. But ⟨130⟩= [2 : 0]2 · [5 : 0]2 · [13 : 3] · [13 : −3], and we ﬁnd that ⟨2⟩· ⟨5⟩· ⟨13⟩= ([2 : 0] · [2 : 0]) · ([5 : 0] · [5 : 0]) · ([13 : 3] · [13 : −3]), and ⟨10 + z⟩· ⟨10 −z⟩= ([2 : 0] · [5 : 0] · [13 : −3]) · ([2 : 0] · [5 : 0] · [13 : 3]). 25 (f) In D−132, 42 = 2 · 3 · 7 = (3 + z)(3 −z), where z = √−33. These factors are all irreducible in that domain. Here ⟨42⟩= [2 : 1]2 · [3 : 0]2 · [7 : 3] · [7 : −3], with ⟨2⟩· ⟨3⟩· ⟨7⟩= ([2 : 1] · [2 : 1]) · ([3 : 0] · [3 : 0]) · ([7 : 3] · [7 : −3]), and ⟨3 + z⟩· ⟨3 −z⟩= ([2 : 1] · [3 : 0] · [7 : 3]) · ([2 : 1] · [3 : 0] · [7 : −3]). (g) In D−35, 150 = 2 · 3 · 5 · 5 = 2 · 5 · (2 + z) · (3 −z), where z = 1+√−35 2 . With N(q + rz) = q2 + qr + 9r2 = 1 4((2q + r)2 + 35r2), we ﬁnd that each of these factors is irreducible. (For instance, to write 2 + z as a nontrivial product vw, we must have N(v) = 3 and N(w) = 5, or vice versa. Both of these are impossible.) Here the ideal ⟨150⟩factors as 2[3 : 0] · [3 : −1] · [5 : 2]4. (Here 2 does not factor in D−35 since −35 ≡5 (mod 8).) We ﬁnd that ⟨2⟩· ⟨3⟩· ⟨5⟩· ⟨5⟩= 2([3 : 0] · [3 : −1]) · ([5 : 2] · [5 : 2]) · ([5 : 2] · [5 : 2]), while ⟨2⟩· ⟨5⟩· ⟨2 + z⟩· ⟨3 −z⟩= 2([5 : 2] · [5 : 2]) · ([3 : −1] · [5 : 2]) · ([3 : 0] · [5 : 2]). Section 3.6. A Formula for Ideal Multiplication. (1) Let φ(x) = x2 + εx + ε2−∆ 4 . If t = k + ℓ+ φ′(0) = k + ℓ+ ε, then kt −φ(k) = k2 + kℓ+ εk −k2 −εk −ε2 −∆ 4 = kℓ−φ(0). Now if g = gcd(a, b, t), then g divides kt, and g divides φ(k) since a divides φ(k) when [a : k] is an ideal. Thus g divides kt −φ(k) = kℓ−φ(0). (2) Let P = [7 : 2], an ideal of D = D−31, so that P 2 = [49 : 9] as established in an example. Then P 4 = P 2·P 2 = [49 : 9]·[49 : 9]. Here γ = 1 since ∆= −31 is a primitive discriminant, and with φ(x) = x2 + x + 8, we ﬁnd that t = k + ℓ+ φ′(0) = 9 + 9 + 1 = 19. Now g = gcd(49, 49, 19) = 1 and c = 49·49 12 = 2401, so we know that P 4 = [2401 : m] for some integer m. Here m satisﬁes (among other congruences) tm ≡kℓ−φ(0) (mod c), that is, 19m ≡73 (mod 2401). Applying the Euclidean algorithm to write 1 = 19(1011) + 2401(−8), we calculate the unique solution of 19m ≡73 (mod 2401) as 73(1011) = 73803 ≡−628 (mod 2401). Thus P 4 = [2401 : −628]. (3) Each discriminant in this exercise is primitive, so that γ = 1 in each application of Theo-rem 3.6.1. (a) With ∆= −20, so that φ(x) = x2 + 5, we ﬁnd that [6 : 1] · [14 : 3] = 2[21 : 10]. Here t = 1 + 3 + 0 = 4 and so g = gcd(6, 14, 4) = 2. Then c = 6·14 2·2 = 21, and AB = 2[21 : m] for m satisfying t g · m ≡1 g(kℓ−φ(0)) (mod c), that is, 2m ≡1 2(3 −5) (mod 21), so that m = 10. (b) For ∆= −20, we ﬁnd [6 : 1][21 : 4] = [126 : 25]. In this case, t = 1 + 4 + 0 = 5, g = gcd(6, 21, 5) = 1, c = 6 · 21 = 126, and m satisﬁes 5m ≡−1 (mod 126). (c) If ∆= −20, then [6 : 1] · [21 : −4] = 3[14 : 3]. Here t = 1 −4 + 0 = −3, g = gcd(6, 21, −3) = 3, c = 6·21 3·3 = 14, and m satisﬁes −m ≡1 3(−4 −5) (mod 14). (d) If ∆= −23, with φ(x) = x2 + x + 6, then [36 : 14] · [142 : 20] = [5112 : 446]. In this case, t = 14 + 20 + 1 = 35 so that g = gcd(36, 142, 35) = 1. Then c = 36 · 142 = 5112 and m satisﬁes 35m ≡14 · 20 −6 (mod 5112). (e) If ∆= −23, then [36 : 14] · [142 : −21] = 2[1278 : 50]. Now t = 14 −21 + 1 = −6 and g = gcd(36, 142, −6) = 2. Then c = 36·142 2·2 = 1278, and m satisﬁes −3m ≡ 1 2(14 · −21 −6) (mod 1278). 26 (4) If ∆= ∆(−5, 3) = −180, then φ(x) = x2 + 45. In the following table, we list γ(A), γ(B), γ = gcd(γ(A), γ(B)), and t = k + ℓ+ φ′(0), for the calculation of AB as g[c : m] where g = gcd(a, b, t), c = abγ/g2, and m satisﬁes the congruences of Theorem 3.6.1. A B γ(A) γ(B) γ t AB [3 : 0] [3 : 0] 3 3 3 0 3[3 : 0] [9 : 0] [3 : 0] 1 3 1 0 3[3 : 0] [9 : 3] [3 : 0] 3 3 3 3 3[9 : 3] [9 : 0] [9 : 3] 1 3 1 3 3[9 : 3] [9 : 0] [9 : 0] 1 1 1 0 9[1 : 0] [9 : 3] [9 : 3] 3 3 3 6 3[27 : −6] [9 : −3] [9 : 3] 3 3 3 0 9[3 : 0] [27 : 3] [9 : 3] 1 3 1 6 3[27 : −6] (5) If ∆= ∆(−11, 5) = −275, then φ(x) = x2 + 5x + 75. In the following table, we list γ(A), γ(B), γ = gcd(γ(A), γ(B)), and t = k + ℓ+ φ′(0) = k + ℓ+ 5, for the calculation of AB as g[c : m] where g = gcd(a, b, t), c = abγ/g2, and m satisﬁes the congruences of Theorem 3.6.1. A B γ(A) γ(B) γ t AB [5 : 0] [25 : 0] 5 1 1 5 5[5 : 0] [5 : 0] [25 : 5] 5 5 5 10 5[25 : 5] [25 : 0] [25 : 0] 1 1 1 5 5[25 : 10] [25 : 0] [25 : 5] 1 5 1 5 5[25 : 5] [25 : 0] [25 : 10] 1 1 1 15 5[25 : −5] [25 : 5] [25 : 5] 5 5 5 15 5[125 : −45] [25 : 5] [25 : 10] 5 1 1 20 5[25 : 5] [25 : 10] [25 : 10] 1 1 1 25 25[1 : 0] (6) Let D1 be a subdomain of some quadratic domain D, let A be an ideal of D, and consider the set A1 = A ∩D1. If x and y are elements of A1, then x, y ∈A and x, y ∈D1. The subdomain D1 is closed under subtraction, as is the ideal A, so it follows that x −y is an element of both A and D1, so is in A1. Now let x be an element of A1 and let y be an element of D1, so also an element of D. Since x is in A and y is in D, then xy is an element of A by the deﬁnition of an ideal. But also x is in D1 and y is in D1, and so xy is in D1 by the closure of a subdomain under multiplication. Thus xy is in A1, and A1 is an ideal of D1. Section 4.1. Classiﬁcation of Quadratic Forms. (1) Let A = a b c d and B = e f g h , so that A = a −b −c d and B = e −f −g h . (a) The determinant of A is ad −(−b)(−c) = ad −bc, the same as the determinant of A. (b) A + B = a + e −b −f −c −g d + h = a + e −(b + f) −(c + g) d + h = A + B. (c) A · B = ae + bg −(af + bh) −(ce + dg) cf + dh = A · B. (2) Let f(x, y) = ax2 −bxy + cy2 and −f(x, y) = −ax2 −bxy −cy2 be the conjugate and negative of f(x, y) = ax2 + bxy + cy2 respectively. (a) If f(q, r) = aq2 + bqr + cr2 = m, then f(q, −r) = aq2 −bq(−r) + c(−r)2 = m also. 27 (b) Part (a) shows that (q, r) ↔(q, −r) describes a one-to-one correspondence between solutions of f(x, y) = m and solutions of f(x, y) = m for each integer m. So these two quadratic forms represent the same collection of integers. (c) If f(q, r) = aq2 + bqr + cr2 = m, then −f(q, r) = −aq2 −bqr −cr2 = −m, and vice versa. (3) (a) The discriminant of f(x, y) = 5x2 −3xy + 7y2 is ∆= (−3)2 −4 · 5 · 7 = −131 = ∆(−131, 1). With ε∆= 1, we ﬁnd that k = −3−1 2 = −2 and f = (5 : −2) in ideal notation. (b) The discriminant of f(x, y) = 3x2 −6xy + 2y2 is ∆= (−6)2 −4 · 3 · 2 = 12 = ∆(3, 1). Here ε = 0 and f = (3 : −3). (c) The discriminant of f(x, y) = 6x2 + 10xy + y2 is ∆= 102 −4 · 6 · 1 = 76 = ∆(19, 1). Here ε = 0 and f = (6 : 5). (4) If ∆= −31, then φ(x) = x2 + x + 8 is the principal polynomial of this discriminant. (a) φ(x) ≡0 (mod 5) has two solutions, x = 1 and x = −2. So (5 : 1 + 5q) and (5 : −2 + 5q) are quadratic forms in Q−31 for every integer q. (b) (7 : 2 + 7q) and (7 : −3 + 7q) are quadratic forms in Q−31 for all q. (c) (35 : 11 + 35q), (35 : 16 + 35q), (35 : −12 + 35q), and (35 : −17 + 35q) are elements of Q−31 for all q. (5) Let f(x, y) = ax2 + bxy + cy2 be a quadratic form of discriminant ∆, with f = (a : k) in ideal form. (So k = b−ε 2 .) (a) For f(x, y) = ax2 −bxy + cy2, we ﬁnd that −b−ε 2 = −b−ε 2 −ε = −k −ε, so that f = (a : −k −ε) in ideal notation. (b) For −f(x, y) = −ax2 −bxy −cy2, we have −b−ε 2 = −k −ε and −f = (−a : −k −ε). (c) For −f(x, y) = −ax2 + bxy −cy2, we have b−ε 2 = k and −f = (−a : k). Section 4.2. Equivalence of Quadratic Forms. (1) Assuming the standard fact that det(AB) = det(A) · det(B) for a pair of n × n matrices A and B, then Γ is closed under multiplication since det(UV ) = det(U) · det(V ) = 1 · 1 = 1. Also, Γ contains the inverse of each of its elements, using the fact that det(A−1) = 1/ det(A) when A is nonsingular. (2) If V = −U, then V T = (−U)T = −UT . So V T MfV = −UTMf · −U = UTMfU, and f ◦U = f ◦V . (3) Let f, g, and h be quadratic forms of some discriminant ∆. (a) Since IT = I, we ﬁnd that ITMfI = Mf, and so f ◦I = f. (b) If UT MfU = Mg, then Mf = (UT)−1MgU−1 = (U−1)TMgU−1, and so f = g ◦U−1. (Note that (UT)−1 = (U−1)T since UT · (U−1)T = (U−1 · U)T = IT = I, and likewise (U−1)T · UT = I.) (c) If UTMfU = Mg and V T MgV = Mh, then Mh = V T (UTMfU)V = (UV )T Mf(UV ), and so h = f ◦(UV ). (Here we use the fact that (AB)T = BT · AT for any two matrices A and B that can be multiplied.) (4) Let H be a subgroup of Γ, with f ∼H g if g = f ◦U for some U in H. (a) We know that f ◦I = f, and I is an element of any subgroup H. Thus f ∼H f and ∼H is reﬂexive. (b) If g = f ◦U, then f = g ◦U−1, and if U is in H, then U−1 is in H. Thus if f ∼H g, then g ∼H f and ∼H is symmetric. (c) If g = f ◦U and h = g ◦V , then h = f ◦(UV ), and if U and V are in H, then UV is also in H. Thus if f ∼H g and g ∼H h, then f ∼H h and ∼H is transitive. 28 Therefore ∼H is an equivalence relation. If f ∼H g so that g = f ◦U for some U in H, then U is also in Γ, and so f ∼g is also true. (5) The matrices of −f and −g are −Mf and −Mg respectively. If g = f ◦U, so that Mg = UT MfU, then −Mg = −(UTMfU) = UT (−Mf)U, and thus −g = −f ◦U. (6) The matrices of f and g are Mf and Mg respectively. If Mg = UT MfU, then Mg = UTMfU = U T · Mf · U. (Here we use Exercise 4.1.1, part (c).) Thus if g = f ◦U, then g = f ◦U. (7) Let ∆= −35, so that φ(x) = x2 + x + 9. Since φ(7) = 65 = 13 · 5 and φ′(7) = 15, then f = (13 : 7) is an element of Q−35, speciﬁcally f(x, y) = 13x2 + 15xy + 5y2. (a) If U = 2 3 3 5 , then f(2, 3) = 187 and 13(2)(3) + 15(3)(3) + 5(3)(5) + 7 = 295, so that f ◦U = (187 : 295). (b) If U = 7 3 −5 −2 , then f(7, −5) = 237 and 13(7)(3)+15(−5)(3)+5(−5)(−2)+7 = 105, and f ◦U = (237 : 105). (c) If U = 4 9 3 7 , then f(4, 3) = 433 and 13(4)(9) + 15(3)(9)+ 5(3)(7) + 7 = 985, and f ◦U = (433 : 985). (8) Since 1 u 0 1 · 1 v 0 1 = 1 u + v 0 1 and 1 u 0 1 −1 = 1 −u 0 1 , then H is closed under multiplication and contains the inverse of each of its elements, and thus is a subgroup of Γ. (9) Let ∆= −51, so that φ(x) = x2 + x + 13. Since φ(16) = 285 = 57 · 5 and φ′(16) = 33, we see that f = [57 : 16] is a quadratic form in Q−51, speciﬁcally f(x, y) = 57x2 +33xy +5y2. Here we ﬁnd that (57 : 16) ↔(5 : −17) →3 (5 : −2) ↔(3 : 1), so that f is equivalent to (3 : 1), that is, g(x, y) = 3x2 + 3xy + 5y2. Speciﬁcally, g = f ◦U where U = 0 −1 1 0 · 1 3 0 1 · 0 −1 1 0 = −1 0 3 −1 . Section 4.3. Representations of Integers by Quadratic Forms. (1) If f(x, y) = 3x2 −5xy + 4y2, then f(2, 1) = 6. With φ(x) = x2 + x + 6, we ﬁnd that f = (3 : −3) in ideal notation. (a) If U = 1 −3 0 1 , then g = f ◦U = (3 : −12), or g(x, y) = 3x2 −23xy + 46y2 in standard notation. Here U−1x = 1 3 0 1 · 2 1 = 5 1 , and we can verify that g(5, 1) = 6. (b) g = 2 −3 −3 5 ◦(3 : −3) = (78 : −126), that is, g(x, y) = 78x2 −251xy + 202y2. Here U−1x = 5 3 3 2 · 2 1 = 13 8 , and g(13, 8) = 6. 29 (c) g = 7 9 3 4 ◦(3 : −3) = (78 : 99), so g(x, y) = 78x2 + 199xy + 127y2. Here U−1x = 4 −9 −3 7 · 2 1 = −1 1 , and g(−1, 1) = 6. (2) If f is a quadratic form, then Aut(f) contains I, since f ◦I = f is always true. If U and V are in Aut(f), so that f ◦U = f and f ◦V = f, then we ﬁnd that f ◦U−1 = f and f ◦(UV ) = f. (See Exercise 4.2.3.) Thus U−1 and UV are elements of Aut(f), and Aut(f) is a subgroup of the group Γ of unimodular matrices. (3) Let f(x, y) = x2 +bxy +cy2, and suppose that ∆= b2 −4c < −4. By equation (4.1.2), we know that if f(q, r) = 1, then 4 = (2q + br)2 −∆r2 > (2q + br)2 + 4r2, and it follows that r = 0 and q = ±1. Then the characterization of automorphs in Proposition 4.3.4 shows that I and −I are the only automorphs of f. (4) For x and y in Z × Z (written as column matrices), we say that x ∼f y if there is an automorph U of f so that y = Ux. (a) We know that I is an element of Aut(f) and x = Ix. Thus ∼f is reﬂexive. (b) If y = Ux, then x = U−1y, and if U is in Aut(f), then U−1 is also in Aut(f). Thus ∼f is symmetric. (c) If y = Ux and w = V y, then w = (V U)x, and if U and V are in Aut(f), then V U is in Aut(f). Thus ∼f is transitive. (5) Let g = (87 : 39) in Q−23, with φ(x) = x2 + x + 6. Since φ(39) = 1566 = 87 · 18, and φ(−4) = 18, we ﬁnd that g = (87 : 39) ↔(18 : −40) →2 (18 : −4) ↔(1 : 3) →−3 (1 : 0) = h, with U = 0 −1 1 0 · 1 2 0 1 · 0 −1 1 0 · 1 −3 0 1 = −1 3 2 −7 a unimodular matrix for which h = g ◦U. Now since g(1, 0) = 87, we calculate that −1 3 2 −7 −1 · 1 0 = −7 −3 −2 −1 · 1 0 = −7 −2 , and verify that h(−7, −2) = 87. Section 4.4. Genera of Quadratic Forms. (1) In each part, we list all genus symbols deﬁned for a primitive quadratic form of discriminant ∆, the combinations of genus symbols that can appear in practice, and an example of a quadratic form having that collection of genus symbols (found by trial-and-error). (a) For ∆= 21 = 3 · 7, the genus symbols that exist are f 3 and f 7 . The product of these symbols must equal 1. We ﬁnd that f(x, y) = x2 +xy −5y2 is a quadratic form for which f 3 = 1 = f 7 , while for f(x, y) = −x2 −xy + 5y2 we have f 3 = −1 = f 7 . (b) For ∆= 28 = 22·7, the deﬁned genus symbols are f 7 and −1 f , since ∆ 4 ≡3 (mod 4). The product of these symbols equals 1. Here f(x, y) = x2 −7y2 has f 7 = 1 = −1 f , while f(x, y) = −x2 + 7y2 has f 7 = −1 = −1 f . (c) For ∆= 56 = 23·7, the deﬁned genus symbols are f 7 and −2 f , since ∆ 4 ≡6 (mod 8), and the product of these symbols is 1. Here f(x, y) = x2 −14y2 has f 7 = 1 = −2 f , while f(x, y) = −x2 + 14y2 has f 7 = −1 = −2 f . 30 (d) For ∆= 84 = 22 · 3 · 7, the deﬁned genus symbols are f 3 and f 7 . (Since ∆ 4 ≡1 (mod 4), the symbol −1 f is not well-deﬁned for forms f of discriminant ∆= 84.) Although ∆= ∆(7, 2), the product of these symbols equals 1, since both symbols are also deﬁned for a quadratic form of discriminant ∆(21, 1) = 21. We ﬁnd that f(x, y) = x2 −21y2 has f 3 = 1 = f 7 and f(x, y) = −x2 + 21y2 has f 3 = −1 = f 7 . (e) For ∆= 112 = 24 · 7, the deﬁned genus symbols are f 7 and −1 f , since ∆ 4 ≡0 (mod 4). Both symbols are deﬁned for a quadratic form of discriminant ∆(7, 1) = 28, so their product equals 1. Here f(x, y) = x2 −28y2 has f 7 = 1 = −1 f , while f(x, y) = −x2 + 28y2 has f 7 = −1 = −1 f . (f) For ∆= 224 = 25 ·7, the symbols f 7 , −1 f , 2 f , and −2 f are all deﬁned, since ∆ 4 ≡0 (mod 8). With ∆= ∆(14, 2), we have that the product of f 7 and −2 f must equal 1. (As noted in part (c), these are the deﬁned genus symbols for a quadratic form of discriminant ∆(14, 1) = 56.) The product of −1 f , 2 f , and −2 f must also equal 1. Thus there are four possible genera. Here we ﬁnd the following genus representatives, with symbols listed in order as f 7 , −2 f , −1 f , and 2 f . + + ++ : x2 −56y2 −−−+ : −x2 + 56y2 + + −−: −5x2 + 2xy + 11y2 −−+−: 5x2 −2xy −11y2. Section 5.1. Equivalence of Ideals. (1) Let B be an ideal of a quadratic domain D and let v be an element of D. Let x be an element of B, so that vx is an element of vB. Since v is an element of the principal ideal ⟨v⟩and x is in B, then vx is an element of ⟨v⟩B, and thus vB ⊆⟨v⟩B. Conversely, let w be an arbitrary element of ⟨v⟩B, so that w = (vy1)x1 + (vy2)x2 + · · · + (vyn)xn where each yi is an element of D and each xi is an element of B. Notice that we can write w = v(y1x1 + y2x2 + · · ·+ ynxn), where y1x1 + y2x2 + · · ·+ ynxn is an element of B by the closure properties of an ideal. So w is in vB and it follows that vB ⊆⟨v⟩B. Therefore, vB is the same as the product of ideals ⟨v⟩B. (2) Let A1, A2, B1, and B2 be nontrivial ideals of a quadratic domain D. Suppose that A1 ∼B1 and A2 ∼B2, say with m1A1 = v1B1 and m2A2 = v2B2 for some nonzero rational integers m1 and m2 and nonzero elements v1 and v2 of D. It follows that (m1m2)A1A2 = (v1v2)B1B2 and thus A1A2 ∼B1B2. (Here we use the fact that m1A1 = ⟨m1⟩A1, and so forth, by Exercise 1, and properties of ideal multiplication from Exercise 3.4.2.) (3) If A is a nontrivial ideal and v a nonzero element of a quadratic domain D, then the equation 1 · (vA) = v · A shows that vA ∼A by deﬁnition. (4) If mA = vB for some nonzero rational integer m and nonzero element v of D, then ⟨m⟩A = ⟨v⟩B by Exercise 1. It follows that ⟨m⟩A = ⟨v⟩B, so that ⟨m⟩· A = ⟨v⟩· B. Therefore mA = vB, and thus A ∼B, using properties of conjugate ideals and of ideal multiplication. (5) Let f(x, y) = ax2 + bxy + cy2, with index γ = gcd(a, b, c), and let φ(x) be the principal polynomial of discriminant ∆= b2 −4ac. Let k = b−ε 2 , where ε = φ′(0) is the basis index of ∆, so that f = (a : k) in ideal notation, and let Af = [a : k] be its corresponding ideal. Since φ(k) = ac and φ′(k) = b (see Proposition 4.1.1), then γ = gcd(a, b, c) is also the index of Af by deﬁnition. (6) Let D = D−79, with z = 1+√−79 2 and φ(x) = x2 + x + 20. 31 (a) We ﬁnd that A = [80 : 19] ∼[5 : 0] ∼[4 : −1] = B, using the fact that φ(19) = 400 = 80 · 5. It follows that 5 · 4 · A = (19 + z)(z)B, that is, A = (−1 + z)B, since z2 = −20 + z. (We can verify this equation by noting that ⟨−1 + z⟩= [20 : −1], and calculating [20 : −1] · [4 : −1] = [80 : 19].) (b) A = [80 : −36] ∼[16 : 3] ∼[2 : 0] = B, so that 16 · 2 · A = (−36 + z)(3 + z)B. This simpliﬁes to A = (−4 −z)B. (c) A = [178 : 59] ∼[20 : 0] ∼[1 : 0] = B, with 20 · A = (59 + z)(z)B. This simpliﬁes to A = (−1 + 3z)B. (d) A = [320 : −100] ∼[31 : 6] ∼[2 : −1] = B, so that 31 · 2 · A = (−100 + z)(6 + z)B, simplifying to 2A = (−20 −3z)B. (e) A = [325 : 80] ∼[20 : −1] ∼[1 : 0] = B, with 20 · A = (80 + z)(−1 + z)B. This simpliﬁes to A = (−5 + 4z)B. (f) A = [325 : −120] ∼[44 : −1] ∼[4 : 0] = B, with 44 · 4 · A = (−120 + z)(−1 + z)B, simplifying to 4A = (35 −3z)B. (g) A = [356 : 59] ∼[10 : 0] ∼[2 : −1] = B, with 10 · 2 · A = (59 + z)(z)B, simplifying to A = (−1 + 3z)B. (h) A = [712 : 59] ∼[5 : 0] ∼[4 : −1]=B, with 5 · 4 · A = (59 + z)(z)B, simplifying to A = (−1 + 3z)B. Section 5.2. Quadratic Forms Associated to an Ideal. (1) The principal polynomial of discriminant ∆= −111 is φ(x) = x2 + x + 28. Since φ(3) = 40 = 8 · 5, then A = [8 : 3] is an ideal of D = D−111, that is, S = {8, 3 + z} is an ordered basis for an ideal A of D. (Here z = 1+√−111 2 .) Let u = 8 and v = 3 + z. (a) For U = q s r t = 2 3 3 5 , we have qu + rv = 2(8) + 3(3 + z) = 25 + 3z and su + tv = 3(8) + 5(3 + z) = 39 + 5z. So S ◦U = {25 + 3z, 39 + 5z}. (b) If U = 4 −1 5 −1 , then S ◦U = {47 + 5z, −11 −z}. (c) For U = 3 5 1 2 , we ﬁnd S ◦U = {27 + z, 46 + 2z}. (2) If S = {u, v} = {q + rz, s + tz} is an ordered basis for the ideal A = [8 : 3] of D−111, then uu = q2 + qr + 28r2, uv + uv = 2qs + qt + rs + 56rt, vv = s2 + st + 28t2. In this case, N(A) = 8 divides each of these expressions, and fS(x, y) deﬁned by q2 + qr + 28r2 8 x2 + 2qs + qt + rs + 56rt 8 xy + s2 + st + 28t2 8 y2 is the quadratic form of this ordered basis. In particular, for S = {8, 3 + z}, with q = 8, r = 0, s = 3, and t = 1, we ﬁnd that fS(x, y) = 8x2 + 7xy + 5y2. In each part below, we calculate fS(x, y) for an ordered basis for A computed in Exercise 1. (a) If S = {25 + 3z, 39 + 5z}, we ﬁnd that fS(x, y) = 119x2 + 379xy + 302y2. (b) If S = {47 + 5z, −11 −z}, then fS(x, y) = 393x2 −177xy + 20y2. (c) If S = {27 + z, 46 + 2z}, then fS(x, y) = 98x2 + 337xy + 290y2. (3) Suppose that S = {u, v} is a Z-basis for an ideal A of a quadratic domain D so that every element of A can be written uniquely as mu + nv for some rational integers m and n. Let w be an element of D, and consider the ideal wA = {wx \| x ∈A}. Every element of wA can be written as w(mu + nv) = m(wu) + n(wv) for some rational integers m and n. If we write the same element of wA as m(wu) + n(wv) = q(wu) + r(wv), then 32 w(mu+nv) = w(qu+rv). If w ̸= 0, then mu+nv = qu+rv in A, from which we conclude that m = q and n = r. Thus {wu, wv} is a Z-basis for wA by deﬁnition. Section 5.3. Composition of Quadratic Forms. (1) Let f(x, y) = x2 + 5y2 and g(x, y) = 2x2 + 2xy + 3y2. (a) If f(q, r) = m and f(s, t) = n, then f(u, v) = mn for u = qs −5rt and v = qt + rs. Here f(2, 1) = 9 and f(3, 1) = 14, so with u = 2·3−5·1·1 = 1 and v = 2·1+1·3 = 5, it follows that f(1, 5) = 9 · 14 = 126. (b) If g(q, r) = m and g(s, t) = n, then f(u, v) = mn for u = 2qs + qt + rs −2rt and v = qt+rs+rt. With g(1, 1) = 7 and g(2, 1) = 15, we ﬁnd that f(5, 4) = 7·15 = 105. (c) If f(q, r) = m and g(s, t) = n, then g(u, v) = mn for u = qs −rs −3rt and v = qt + 2rs + rt. With f(2, 1) = 9 and g(1, 1) = 7, then g(−2, 5) = 9 · 7 = 63. (d) With f(3, 1) = 14 and g(1, 1) = 7, then g(−1, 6) = 14 · 7 = 98. (e) With f(2, 1) = 9 and g(2, 1) = 15, thn g(−1, 7) = 9 · 15 = 135. (f) With f(3, 1) = 14 and g(2, 1) = 15, then g(1, 8) = 14 · 15 = 210. (2) (a) If f1(x, y) = x2 + 2y2 = f2(x, y) in Q−8, then f1 = (1 : 0) = f2 in ideal notation, and f1 · f2 = f = (1 : 0) as well. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs −2rt and v = qt + rs. (b) If f1(x, y) = 2x2 + 3y2 = f2(x, y) in Q−24, then f1 = (2 : 0) = f2 and f1 · f2 = f = (1 : 0), that is, f(x, y) = x2 +6y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = 2qs −3rt and v = qt + rs. (c) For f1(x, y) = 2x2+xy+3y2 = f2(x, y) in Q−23, we have f = (2 : 0)·(2 : 0) = (4 : −2), that is, f(x, y) = 4x2 −3xy +2y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs + qt + rs −rt and v = 2qt + 2rs + rt. (d) For f1(x, y) = 2x2 + xy + 3y2 and f2(x, y) = 2x2 −xy + 3y2 in Q−23, we have f = (2 : 0) · (2 : −1) = (1 : 0), that is, f(x, y) = x2 + xy + 6y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = 2qs −qt −3rt and v = qt + rs. (e) For f1(x, y) = 2x2 +xy +6y2 = f2(x, y) in Q−47, we have f = (2 : 0)·(2 : 0) = (4 : 0), that is, f(x, y) = 4x2 + xy + 3y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs −3rt and v = 2qt + 2rs + rt. (f) For f1(x, y) = 2x2 + xy + 6y2 and f2(x, y) = 3x2 + xy + 4y2 in Q−47, we have f = (2 : 0) · (3 : 0) = (6 : 0), that is, f(x, y) = 6x2 + xy + 2y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs −2rt and v = 2qt + 3rs + rt. (g) For f1(x, y) = x2 −2y2 = f2(x, y) in Q8, we have f = (1 : 0) · (1 : 0) = (1 : 0), so that f(x, y) = x2 −2y2 also. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs + 2rt and v = qt + rs. (h) For f1(x, y) = x2 −10y2 = f2(x, y) in Q40, we ﬁnd that f(x, y) = x2 −10y2 also. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs + 10rt and v = qt + rs. (i) For f1(x, y) = 2x2 −5y2 = f2(x, y) in Q40, we have f = (2 : 0) · (2 : 0) = (1 : 0), so that f(x, y) = x2 −10y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = 2qs + 5rt and v = qt + rs. (j) For f1(x, y) = x2 −10y2 and f2(x, y) = 2x2 −5y2 in Q40, f = (1 : 0) · (2 : 0) = (2 : 0), so that f(x, y) = 2x2 −5y2. If f1(q, r) = m and f2(s, t) = n, then f(u, v) = mn for u = qs + 5rt and v = qt + 2rs. Section 5.4. Class Groups of Ideals and Quadratic Forms. (1) Let u be an irreducible element of a quadratic domain D, and suppose that ⟨u⟩⊆⟨v⟩for some element v of D. In particular, since u is an element of ⟨u⟩, then u is in ⟨v⟩, and so u = vw for some element w of D. But now by the deﬁnition of irreducible elements, then 33 either v or w is a unit in D. If w is a unit, then u and v are associates, and it follows that ⟨v⟩= ⟨u⟩. On the other hand, if v is a unit, then 1 = vv−1 is an element of ⟨v⟩, and thus ⟨v⟩= D. So ⟨u⟩is maximal among principal ideals when u is irreducible in D. (2) Suppose that D is a quadratic domain in which every ideal is a principal ideal, and let u be an irreducible element of D. If ⟨u⟩⊆B for some ideal B of D, then B = ⟨v⟩for some v in D, and Exercise 1 implies that either B = ⟨u⟩or B = D. But then ⟨u⟩is a prime ideal of D by deﬁnition. (3) Let u be an element of a quadratic domain D, and suppose that ⟨u⟩is a prime ideal of D. Suppose also that u divides some product vw of elements of D. It follows that vw is an element of ⟨u⟩. But then either v is in ⟨u⟩or w is in ⟨u⟩. (This uses the property of prime ideals of D established in Proposition 3.3.2.) If v is in ⟨u⟩, then u divides v, while if w is ⟨u⟩, then u divides w. It follows that u is prime as an element of D. (4) Suppose that D is a quadratic domain in which every ideal is a principal ideal. If u is an irreducible element of D, then Exercise 2 shows that ⟨u⟩is a prime ideal of D. But then Exercise 3 implies that u is a prime element of D. Thus D is a unique factorization domain, since every irreducible element of D is also prime. (5) Assuming that D = [1 : 0] and A = [2 : 1] represent all distinct classes of ideals of D = D−20, we ﬁnd the following operation table for C−20. · [1 : 0] [2 : 1] [1 : 0] [1 : 0] [2 : 1] [2 : 1] [2 : 1] [1 : 0] Here the class of D is an identity element for multiplication. Since A is its own conjugate, we ﬁnd that [2 : 1]2 = ⟨2⟩is principal, and so is in the class of the ideal D. On the other hand, there are four distinct classes of quadratic forms in F−20, represented by f = (1 : 0), g = (2 : 1), −f = (−1 : 0), and −g = (−2 : −1). An operation table for F−20 is · (1 : 0) (2 : 1) (−1 : 0) (−2 : −1) (1 : 0) (1 : 0) (2 : 1) (−1 : 0) (−2 : −1) (2 : 1) (2 : 1) (1 : 0) (−2 : −1) (−1 : 0) (−1 : 0) (−1 : 0) (−2 : −1) (1 : 0) (2 : 1) (−2 : −1) (−2 : −1) (−1 : 0) (2 : 1) (1 : 0) (6) Assuming that each ideal of D = D−56 is in the class of D = [1 : 0], A = [2 : 0], B = [3 : 1], or C = [3 : −1], we ﬁnd the following operation table for C−56. · [1 : 0] [2 : 0] [3 : 1] [3 : −1] [1 : 0] [1 : 0] [2 : 0] [3 : 1] [3 : −1] [2 : 0] [2 : 0] [1 : 0] [3 : −1] [3 : 1] [3 : 1] [3 : 1] [3 : −1] [2 : 0] [1 : 0] [3 : −1] [3 : −1] [3 : 1] [1 : 0] [2 : 0] For example, B2 = [9 : −2] ∼[2 : 0] using the fact that for the principal polynomial φ(x) = x2 + 14, we ﬁnd that φ(−2) = 18 = 9 · 2. Other calculations are similar. Section 6.1. Reduced Positive Deﬁnite Quadratic Forms. (1) Let f(x, y) = ax2 + bxy + cy2 be a reduced quadratic form of some negative discriminant ∆= b2 −4ac, so that −a < b ≤a < c or 0 ≤b ≤a = c. If g is a positive integer, then either −ga < gb ≤ga < gc or 0 ≤gb ≤ga = gc, and so the quadratic form f1(x, y) = gax2+gbxy+gcy2 is also reduced. The discriminant of f1 is (gb)2−4(ga)(gc) = g2(b2 −4ac) = g2∆. 34 (2) Let f1(x, y) = a1x2 + b1xy + c1y2 be a reduced quadratic form of negative discriminant ∆= b2 1−4a1c1. Let g = gcd(a1, b1, c1) with a1 = ga, b1 = gb, and c1 = gc. By deﬁnition, we have that −ga < gb ≤ga < gc or 0 ≤gb ≤ga = gc, and since g is positive, we can cancel g throughout without changing the order of the inequalities. Thus f(x, y) = ax2 +bxy +cy2 is also reduced, and has discriminant b2 −4ac = (b1/g)2 −4(a1/g)(c1/g) = ∆/g2. (3) Let f(x, y) = ax2 + bxy + cy2 be a positive deﬁnite quadratic form of discriminant ∆. Let φ(x) be the principal polynomial of discriminant ∆, and suppose that f = (a : k) in ideal notation, so that φ(k) = ac and φ′(k) = b. Then we have that −a < b ≤a < c if and only if −a < φ′(k) ≤a and a2 < ac = φ(k). Likewise 0 ≤b ≤a = c if and only if 0 ≤φ′(k) ≤a and a2 = ac = φ(k). (4) (a) If ∆= ∆(−29, 1) = −116, then φ(x) = x2 + 29 and u∆= jp 116/3 k = 6. We can ﬁnd all reduced forms of discriminant ∆by testing whether 1 ≤a ≤6 divides φ(k) for −a 2 < k ≤a 2. Using the following table k 0 ±1 ±2 ±3 φ(k) 29 30 33 38 we obtain the following six reduced forms in Q−116: (1 : 0), (2 : 1), (3 : 1), (3 : −1), (5 : 1), (5 : −1). Note that (6 : 1) and (6 : −1) are not reduced since 62 > φ(±1) = 30. (b) For ∆= ∆(−29, 2) = −464, we ﬁnd φ(x) = x2 + 116 and u∆= 12. From the table k 0 ±1 ±2 ±3 ±4 ±5 ±6 φ(k) 116 117 120 125 132 141 152 we obtain eighteen reduced forms in Q−464 (those in the third row are not primitive): (1 : 0), (3 : 1), (3 : −1), (4 : 0), (5 : 2), (5 : −2), (8 : 2), (8 : −2), (9 : 1), (9 : −1), (11 : 4), (11 : −1), (2 : 0), (4 : 2), (6 : 2), (6 : −2), (10 : 2), (10 : −2). (c) If ∆= ∆(−119, 1) = −119, then φ(x) = x2 + x + 30 and u∆= 6. The table k 0, −1 1, −2 2, −3 φ(k) 30 32 36 produces ten reduced forms in Q−119: (1 : 0), (2 : 0), (2 : −1), (3 : 0), (3 : −1), (4 : 1), (4 : −2), (5 : 0), (5 : −1), (6 : 2). Note that (6 : −3) is not reduced since 62 = φ(−3) but −3 < 0. (d) If ∆= ∆(−74, 1) = −296, then φ(x) = x2 + 74 and u∆= 9. From the table k 0 ±1 ±2 ±3 ±4 φ(k) 74 75 78 83 90 we obtain ten reduced forms in Q−296: (1 : 0), (2 : 0), (3 : 1), (3 : −1), (5 : 1), (5 : −1), (6 : 2)(6 : −2), (9 : 4), (9 : −4). (e) If ∆= ∆(−85, 1) = −340, then φ(x) = x2 + 85 and m = 10. The table k 0 ±1 ±2 ±3 ±4 ±5 φ(k) 85 86 89 94 101 110 produces four reduced forms in Q−340: (1 : 0), (2 : 1), (5 : 0), (10 : 5). 35 (f) If ∆= ∆(−86, 1) = −344, then φ(x) = x2 + 86 and u∆= 10. From the table k 0 ±1 ±2 ±3 ±4 ±5 φ(k) 86 87 90 95 102 111 we ﬁnd ten reduced forms in Q−344: (1 : 0), (2 : 0), (3 : 1), (3 : −1), (5 : 2), (5 : −2), (6 : 2), (6 : −2), (9 : 2), (9 : −2). (g) If ∆= ∆(−89, 1) = −356, then φ(x) = x2 + 89 and u∆= 10. From the table k 0 ±1 ±2 ±3 ±4 ±5 φ(k) 89 90 93 98 105 114 we ﬁnd twelve reduced forms in Q−356: (1 : 0), (2 : 1), (3 : 1), (3 : −1), (5 : 1), (5 : −1), (6 : 1), (6 : −1), (7 : 3), (7 : −3), (9 : 1), (9 : −1). (h) If ∆= ∆(−105, 1) = −420, then φ(x) = x2 + 105 and u∆= 11. From k 0 ±1 ±2 ±3 ±4 ±5 φ(k) 105 106 109 114 121 130 we ﬁnd eight reduced forms in Q−420: (1 : 0), (2 : 1), (3 : 0), (5 : 0), (6 : 3), (7 : 0), (10 : 5), (11 : 4). Note that (11 : −4) is not reduced since 112 = φ(−4) but −4 < 0. (5) (a) In Q−116, with φ(x) = x2 +x +30, we ﬁnd that (60 : 14) ↔(4 : −15) →4 (4 : 1) since φ(14) = 240 = 60 · 4. The form (4 : 1) is reduced, and the matrix U = 0 −1 1 0 · 1 4 0 1 = 0 −1 1 4 has the property that (60 : 14) ◦U = (4 : 1). (b) In Q−340, with φ(x) = x2 + 85, we see that (187 : 17) ↔(2 : −17) →9 (2 : 1), a reduced form. The matrix U = 0 −1 1 0 · 1 9 0 1 = 0 −1 1 9 is such that (187 : 17) ◦U = (2 : 1). (c) In Q−420, we have (179 : 49) ↔(14 : −49) →4 (14 : 7) ↔(11 : −7) →1 (11 : 4), a reduced form. The matrix U = 0 −1 1 0 · 1 4 0 1 · 0 −1 1 0 · 1 1 0 1 = −1 −1 4 3 has the property that (179 : 49) ◦U = (11 : 4). Section 6.2. Calculation of Ideal Class Groups. (1) (a) The ideal class group of discriminant ∆= −20 consists of the classes of D = [1 : 0] and A = [2 : 1]. The invariant factor type of C−20 is (2), as we can verify by noting that A2 = 2[1 : 0] ∼[1 : 0]. (b) C−40 = {[1 : 0], [2 : 0]}, with invariant factor type (2). (c) There are ﬁve reduced ideals of discriminant ∆= −47, so the class group C−47 must have invariant factor type (5). We can verify that if A = [2 : 0], then A2 ∼[3 : −1], A3 ∼[3 : 0], A4 ∼[2 : −1], and A5 ∼[1 : 0]. 36 (d) There are four reduced ideals of discriminant ∆= −56. The invariant factor type of C−56 might be either (4) or (2, 2), but by trial-and-error, we ﬁnd that if A = [3 : 1], then A2 ∼[2 : 0], A3 ∼[3 : −1], and A4 ∼[1 : 0], so that C−56 has invariant factor type (4). (e) There are four reduced ideals of discriminant ∆= −84: [1 : 0], [2 : 1], [3 : 1], and [5 : 2]. Each class is its own inverse in C−84, and so the ideal class group has invariant factor type (2, 2). (For example, if A = [5 : 2], then A2 = [25 : 2] ∼[1 : 0].) (f) The class group of ∆= −116 has six elements (listed in Exercise 6.1.4), and so must have invariant factor type (6). Direct calculation shows that if A = [3 : 1], then the powers of [A] in C−116 are, in order, [3 : 1], [5 : 1], [2 : 1], [5 : −1], [3 : −1], [1 : 0]. (g) There are ten reduced ideals of discriminant ∆= −119, found in Exercise 6.1.4, and so C−119 has invariant factor type (10). If A = [3 : 0], then the powers of [A] are: [3 : 0], [4 : −2], [5 : 0], [2 : −1], [6 : 2], [2 : 0], [5 : −1], [4 : 1], [3 : −1], [1 : 0]. (h) The class group of ∆= −296 has ten elements and invariant factor type (10). Direct calculation shows that if A = [5 : 1], then the powers of [A] are: [5 : 1], [3 : −1], [6 : −2], [9 : −4], [2 : 0], [9 : 4], [6 : 2], [3 : 1], [5 : −1], [1 : 0]. (i) The class group of ∆= −340 has four elements, [1 : 0], [2 : 1], [5 : 0], [10 : 5]. Each of these elements is its own inverse in C−340, so the class group has invariant factor type (2, 2). (j) The class group of ∆= −344 has invariant factor type (10). If A = [3 : 1], then the powers of [A] are: [3 : 1], [9 : −2], [5 : −2], [6 : 2], [2 : 0], [6 : −2], [5 : 2], [9 : 2], [3 : −1], [1 : 0]. (k) The class group of ∆= −356 has twelve elements, so C−356 might have invariant factor type (12) or (6, 2). By trial-and-error, we ﬁnd that every element of C−356 is a power of [A], where A = [3 : 1], and so has invariant factor type (12). The powers of [A] are: [3 : 1], [9 : 1], [7 : −3], [5 : 1], [6 : −1], [2 : 1], [6 : 1], [5 : −1], [7 : 3], [9 : −1], [3 : −1], [1 : 0]. (l) The class group of ∆= −420 has eight elements, [1 : 0], [2 : 1], [3 : 0], [5 : 0], [6 : 3], [7 : 0], [10 : 5], [11 : 4], so can have invariant factor type (8), (4, 2), or (2, 2, 2). In this case, each of these ideal classes is its own inverse, so C−420 has invariant factor type (2, 2, 2). (2) The upper bound u∆equals 1 for ∆= −3, −4, −7, −8, and −11, so there can be only one distinct ideal class in C∆in these cases, namely the class of D = [1 : 0]. (a) If ∆= −19, then φ(x) = x2 + x + 5 and u∆= 2. Since 2 does not divide φ(0) = 5 = φ(−1), we conclude that C−19 has only one element, the class of [1 : 0]. (b) If ∆= −43, then φ(x) = x2 + x + 11 and u∆= 3. Since neither 2 nor 3 divides φ(0) = 11 = φ(−1) or φ(1) = 13 = φ(−2), we see that C−43 = {[1 : 0]}. (c) If ∆= −67, then φ(x) = x2 + x + 17 and u∆= 4. Since 2, 3, and 4 do not divide φ(0) = 17 = φ(−1) or φ(1) = 19 = φ(−2), it follows that C−67 = {[1 : 0]}. (d) If ∆= −163, then φ(x) = x2 + x + 41 and u∆= 7. Since φ(0) = 41 = φ(−1), φ(1) = 43 = φ(−2), φ(2) = 47 = φ(−3), and φ(3) = 53 = φ(−4) are not divisible by any integer 2 through 7, we conclude that C−163 = {[1 : 0]}. 37 Section 6.3. Genera of Ideal Classes. (1) In parts (a)–(l), we use the computations of C∆compiled in Exercise 6.2.1. (a) For ideals A of ∆= −20, the deﬁned genus symbols are −1 A and A 5 . We ﬁnd that −1 A = 1 = A 5 for ideal classes in {[1 : 0]} (the principal genus) and −1 A = −1 = A 5 for {[2 : 1]}. If A = [2 : 1], then G = C−20 = {[A]0, [A]1}, and G2 = {[A]0} = {[1 : 0]}, the principal genus as noted above. (b) For ∆= −40, the deﬁned genus symbols are −2 A and A 5 . Here −2 A = 1 = A 5 for [A] in {[1 : 0]} and −2 A = −1 = A 5 for [A] in {[2 : 0]}. If A = [2 : 0], then G = C−20 = {[A]0, [A]1}, and G2 = {[A]0} = {[1 : 0]}. (c) For ∆= −47, the only deﬁned genus symbol is A 47 , and A 47 = 1 for all [A] in G = C−47 = {[1 : 0], [2 : 0], [2 : −1], [3 : 0], [3 : −1]}. If A = [2 : 0], then G = {[A]0, [A]1, [A]2, [A]3, [A]4} = G2. (For example, [A] = ([A]3)2.) (d) For ∆= −56, the deﬁned genus symbols are 2 A and A 7 , with 2 A = 1 = A 7 for [A] in {[1 : 0], [2 : 0]} and 2 A = −1 = A 7 for [A] in {[3 : 1], [3 : −1]}. If A = [3 : 1], then G = C−56 = {[A]0, [A]1, [A]2, [A]3} and G2 = {[A]0, [A]2}. (e) For ∆= −84, the deﬁned genus symbols are −1 A , A 3 , and A 7 (presented in that order below). The distinct genera are: + + + : {[1 : 0]}, −+ −: {[2 : 0]}, −−+ : {[3 : 1]}, + −−: {[5 : 2]}. If A = [2 : 0] and B = [3 : 1], then G = {[A]0[B]0, [A]1[B]0, [A]0[B]1, [A]1[B]1}, and G2 = {[A]0[B]0}. (f) For ∆= −116, we ﬁnd that −1 A = 1 = A 29 for [A] in {[1 : 0], [5 : 1], [5 : −1]} and −1 A = −1 = A 29 for [A] in {[2 : 1], [3 : 1], [3 : −1]}. If A = [3 : 1], then G = {[A]0, [A]1, [A]2, [A]3, [A]4, [A]5} and G2 = {[A]0, [A]2, [A]4}. (g) For ∆= −119, we have A 7 = 1 = A 17 in {[1 : 0], [2 : 0], [2 : −1], [4 : 1], [4 : −2]} and A 7 = −1 = A 17 in {[3 : 0], [3 : −1], [5 : 0], [5 : −1], [6 : 2]}. If A = [3 : 0], then the principal genus consists of the even powers of [A], and so equals G2. (h) For ∆= −296, we have −2 A = 1 = A 37 in {[1 : 0], [3 : 1], [3 : −1], [9 : 4], [9 : −4]} and −2 A = −1 = A 37 in {[2 : 0], [5 : 1], [5 : −1], [6 : 2], [6 : −2]}. If A = [3 : 1], then the principal genus consists of the even powers of [A]. (i) For ∆= −340, the deﬁned genus symbols are −1 A , A 5 , A 17 (presented in that order below). The distinct genera are: + + + : {[1 : 0]}, −−+ : {[2 : 1]}, + −−: {[5 : 0]}, −+ −: {[10 : 5]}. If A = [2 : 1] and B = [5 : 0], then G = {[A]0[B]0, [A]1[B]0, [A]0[B]1, [A]1[B]1}, and G2 = {[A]0[B]0}. (j) For ∆= −344, we have 2 A = 1 = A 43 in {[1 : 0], [6 : 2], [6 : −2], [9 : 2], [9 : −2]} and 2 A = −1 = A 43 in {[2 : 0], [3 : 1], [3 : −1], [5 : 2], [5 : −2]}. If A = [3 : 1], then the principal genus consists of the even powers of [A]. (k) If ∆= −356, then −1 A = 1 = A 89 for {[1 : 0], [2 : 1], [5 : 1], [5 : −1], [9 : 1], [9 : −1]} and −1 A = −1 = A 89 for {[3 : 1], [3 : −1], [6 : 1], [6 : −1], [7 : 3], [7 : −3]}. The principal genus consists of the even powers of [A], where A = [3 : 1]. (l) For ∆= −420, the deﬁned genus symbols are −1 A , A 3 , A 5 , and A 7 (presented in that order below). The distinct genera are: + + ++ : {[1 : 0]}, + −−+ : {[2 : 1]}, −−−−: {[3 : 0]}, + −+−: {[5 : 0]}, −+ +−: {[6 : 3]}, −+ −+ : {[7 : 0]}, + + −−: {[10 : 5]}, −−++ : {[11 : 4]}. 38 If A = [2 : 1], B = [3 : 0], and C = [5 : 0], then every element of G = C−420 has the form [A]i[B]j[C]k with i, j, and k either 0 or 1 (independently). The square of each such element is [A]0[B]0[C]0. (m) If ∆= −191, then φ(x) = x2 + x + 48 and u∆. From the table k 0, −1 1, −2 2, −3 3, −4 φ(k) 48 50 54 60 we ﬁnd thirteen reduced ideals of discriminant −191 (listed below), so that G = C−191 has invariant factor type (13). Since 191 is prime, there is only one genus symbol, A 191 , deﬁned for these ideals, and each ideal class is in the principal genus. Each ideal class is also the square of an element of G. For instance, if A = [2 : 0], then [A] = ([A]7)2. Direct calculation shows that the powers of [A] are, in order, [2 : 0], [4 : 0], [6 : −1], [3 : −1], [6 : 2], [5 : −2], [5 : 1], [6 : −3], [3 : 0], [6 : 0], [4 : −1], [2 : −1], [1 : 0]. (n) For ∆= −231 = −1 · 3 · 7 · 11, the deﬁned genus symbols of ideals are A 3 , A 7 , and A 11 (listed in that order below). Here φ(x) = x2 + x + 58 and u∆= 8, and from the table k 0, −1 1, −2 2, −3 3, −4 φ(k) 58 60 64 70 we ﬁnd twelve reduced ideals of discriminant −231 (listed in the genera below). There are four distinct genera, and G = C−231 has invariant factor type (6, 2). If A = [2 : 0] and B = [3 : 1], then every element of G can be written uniquely as [A]i[B]j with 0 ≤i < 6 and 0 ≤j < 2. Speciﬁcally, the distinct genera are + + + : {[A]0, [A]2, [A]4} = {[1 : 0], [4 : −2], [4 : 1]}, −+ −: {[A], [A]3, [A]5} = {[2 : 0], [8 : 2], [2 : −1]}, −−+ : {[B], [A]2[B], [A]4[B]} = {[3 : 1], [5 : 1], [5 : −2]}, + −−: {[A][B], [A]3[B], [A]5[B]} = {[6 : −2], [7 : 3], [6 : 1]}. Only the elements in the principal genus are squares of elements of G. (o) For ∆= −440, the deﬁned genus symbols are 2 A , A 5 , and A 11 . Calculating values of φ(x) = x2 + 110 for −6 < x ≤6, we ﬁnd twelve reduced ideals of discriminant −440, in four genera. The class group G = C−440 has invariant factor type (6, 2), and if A = [3 : 1] and B = [2 : 0], then every element of G can be written uniquely as [A]i[B]j with 0 ≤i < 6 and 0 ≤j < 2. The distinct genera are + + + : {[A]0, [A]2, [A]4} = {[1 : 0], [9 : 4], [9 : −4]}, −−+ : {[A], [A]3, [A]5} = {[3 : 1], [5 : 0], [3 : −1]}, + −−: {[B], [A]2[B], [A]4[B]} = {[2 : 0], [7 : 3], [7 : −3]}, −+ −: {[A][B], [A]3[B], [A]5[B]} = {[6 : −2], [10 : 0], [6 : 2]}. (p) For ∆= −724, the deﬁned genus symbols are −1 A and A 181 . From calculation of φ(x) = x2 + 181 for −7 ≤x ≤7, we obtain ten reduced forms of discriminant −724, so that G = C−724 has invariant factor type (10). If A = [7 : 1], the distinct genera are ++ : {[A]0, [A]2, [A]4, [A]6, [A]8} = {[1 : 0], [5 : 2], [13 : 1], [13 : −1], [5 : −2]}, −−: {[A], [A]3, [A]5, [A]7, [A]9} = {[7 : 1], [10 : 3], [2 : 1], [10 : −3], [7 : −1]}. 39 Section 7.1. Negative Discriminants with Trivial Class Groups. (1) The prime numbers p < 150 that satisfy p ≡1 or 3 (mod 8) are: 3, 11, 17, 19, 41, 43, 59, 67, 73, 83, 89, 97, 107, 113, 131, 137, and 139. In addition to 1 and 2, each of these primes appears exactly once in Table 4, as does each 2p < 150 (that is, 6, 22, 34, 38, 82, 86, 118, 134, and 146), pk with k > 1 (9, 27, 81, and 121), and 2pk with k > 1 (18 and 54). The numbers that appear twice in this table are products of two these primes, or prime powers, perhaps multiplied by 2. These are 33 = 3·11, 51 = 3·17, 57 = 3·19, 123 = 3·41, 129 = 3 · 43, 99 = 32 · 11, 66 = 2 · 3 · 11, 102 = 2 · 3 · 17, and 114 = 2 · 3 · 19. (2) Let f(x, y) = x2 + 2y2. If f(q, r) = m and f(s, t) = n, then f(u, v) = mn where u = \|qs −2rt\| and v = \|qt + rs\|, or u = \|qs + 2rt\| and v = \|qt −rs\|. (a) Since f(3, 1) = 11 and f(1, 3) = 19, we ﬁnd that f(3, 10) = 209 = f(9, 8). (b) From f(1, 3) = 19 and f(3, 4) = 41, we ﬁnd f(21, 13) = 779 = f(27, 5). (c) From f(3, 4) = 41 and f(5, 3) = 43, we ﬁnd f(9, 29) = 1763 = f(39, 11). (d) Using the preceding two solutions of f(x, y) = 1763 = 41·43, we obtain four solutions of f(x, y) = 5289 = 3 · 41 · 43. From f(1, 1) = 3 and f(9, 29) = 1763, we ﬁnd f(49, 38) = 5289 = f(67, 20). From f(1, 1) = 3 and f(39, 11) = 1763, we obtain f(17, 50) = 5289 = f(61, 28). (3) The prime numbers p < 150 with p ≡1 (mod 3) are: 7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, and 139. In addition to 1, 3, 4, and 12, these primes appear once in Table 5, as do 3p < 150 (that is, 21, 39, 57, 93, 111, and 129), and pk or 3pk with k > 1 (49 and 147). The numbers that appear twice in Table 5 are 4p (that is, 28, 52, 76, 124, and 148), or 3·4p (that is, 84), or products of two distinct primes congruent to 1 modulo p (91 = 7 · 13 and 133 = 7 · 19). (4) Let φ(x) = x2 +2x+4, the principal polynomial of discriminant ∆= −12. Since φ(x) ≡0 (mod 4) has two solutions, 0 and −2, we ﬁnd that if a > 3 has the form of equation (7.1.3), then φ(x) ≡0 (mod a) has 2n solutions if e = 0 and 2n+1 solutions if e = 2. Conjugate pairs of these values correspond to solutions (q, r) and (q, −r) of x2 + 3y2 = a, or their negatives. If we restrict our attention to primitive solutions in positive integers, then the number of resulting solutions is cut in half. (5) Let f(x, y) = x2 + 3y2 and suppose that f(q, r) = m and f(s, t) = n. If u = qs −3rt and v = qt + rs, then f(u, v) = (qs −3rt)2 + 3(qt + rs)2 = q2s2 −6qsrt + 9r2t2 + 3q2t2 + 6qtrs + 3r2s2 = q2s2 + 3q2t2 + 3r2s2 + 9r2t2 = (q2 + 3r2)(s2 + 3t2) = mn. (a) Using f(1, 2) = 13 = f(1, −2) and f(4, 1) = 19, we can then let u = 1·4−3·2·1 = −2 and v = 1 · 1 + 2 · 4 = 9, or u = 1 · 4 −3 · −2 · 1 = 10 and v = 1 · 1 −2 · 4 = −7, and conclude that f(2, 9) = 247 = f(10, 7). (b) From q = 4 and r = 1, so that f(q, r) = 19, and s = 2 and t = 3, with f(s, t) = 31, we ﬁnd that (u, v) = (qs−3rt, qt+rs) = (−1, 14) and (u, v) = (qs+3rt, qt−rs) = (17, 10) are solutions of f(x, y) = 19 · 31 = 589. (c) From q = 5 and r = 2, with f(q, r) = 37, and s = 4 and t = 3, with f(s, t) = 43, we have that (u, v) = (qs−3rt, qt+rs) = (2, 23) and (u, v) = (qs+3rt, qt−rs) = (38, 7) are solutions of f(x, y) = 37 · 43 = 1591. (6) The prime numbers p < 150 with p ≡1, 2, or 4 (mod 7) are: 11, 13, 19, 37, 43, 53, 67, 71, 79, 107, 109, 113, 127, 137, and 149. Along with 1 and 7, these primes appear once each in Table 6, as do p2 (121), 7p (77), 2e for e ≥3 (8, 16, 32, 64, and 128), and 2e · 7 for e ≥3 (56 and 112). The only number that appears twice in Table 6 is 88 = 23 · 11. 40 (7) Let φ(x) = x2 + 2x + 8, the principal polynomial of discriminant ∆= −28. For e ≥3, the congruence φ(x) ≡0 (mod 2e) has four solutions (by Theorem 3.4). But as noted in the proof of Proposition 7.1.5, one pair of these conjugate values corresponds to a solution of g(x) ≡0 (mod 2e), where g(x) = 2x2 + 2x + 4, and the other pair corresponds to a solution of φ(x) ≡0 (mod 2e). Thus if a is as given in Proposition 7.1.5, then there are 2n−1 solutions of x2 + 7y2 = a if e = 0, and 2n solutions if e ≥3. (8) Let f(x, y) = x2 + 7y2, with f(q, r) = m and f(s, t) = n. If u = qs −7rt and v = qt + rs, then f(u, v) = (qs −7rt)2 + 7(qt + rs)2 = q2s2 −14qsrt + 49r2t2 + 7q2t2 + 14qtrs + 7r2s2 = q2s2 + 7q2t2 + 7r2s2 + 49r2t2 = (q2 + 7r2)(s2 + 7t2) = mn. (a) From 11 = f(q, r) with q = 2 and r = 1, and 23 = f(s, t) with s = 4 and t = 1, we ﬁnd that (qs −7rt, qt + rs) = (1, 6) and (qs + 7rt, qt −rs) = (15, −2) are solutions of f(x, y) = 11 · 23 = 253. (b) From 16 = f(q, r) with q = 3 and r = 1, and 23 = f(s, t) with s = 4 and t = 1, then (qs −7rt, qt + rs) = (5, 7) and (qs + 7rt, qt −rs) = (19, −1) are solutions of f(x, y) = 16 · 23 = 368. (c) From 11 = f(q, r) with q = 2 and r = 1, and 37 = f(s, t) with s = 3 and t = 2, then (qs −7rt, qt + rs) = (−8, 7) and (qs + 7rt, qt −rs) = (20, 1) are solutions of f(x, y) = 11 · 37 = 407. (d) From 11 = f(q, r) with q = 2 and r = 1, and 71 = f(s, t) with s = 8 and t = 1, then (qs −7rt, qt + rs) = (9, 10) and (qs + 7rt, qt −rs) = (23, −6) are solutions of f(x, y) = 11 · 71 = 781. (e) We use the two solutions of f(x, y) = 11 · 23 from part (a), together with 29 = f(s, t) for s = 1 and t = 2. If q = 1 and r = 6, then (qs −7rt, qt + rs) = (−83, 8) and (qs + 7rt, qt−rs) = (85, −4) are solutions of f(x, y) = 11 · 23 · 29 = 7337. Likewise, if q = 15 and r = 2, we obtain (qs −7rt, qt + rs) = (−13, 32) and (qs + 7rt, qt −rs) = (43, 28) as a second pair of solutions. (9) Let ∆= −67, so that φ(x) = x2 + x + 17 and u∆= jp 67/3 k = 4. Since φ(0) = 17 = φ(−1) and φ(1) = 19 = φ(−2) are not divisible by 2, 3, or 4, then the principal form φ = (1 : 0), that is, φ(x, y) = x2 + xy + 17y2, is the only reduced form of discriminant ∆. A prime p is represented by φ(x, y) if and only if φ(x) ≡0 (mod p) has a solution. But φ(q, r) = q2 + qr + 17 ≥17 if r ̸= 0, so φ(x) ≡0 (mod p) can have no solutions for primes p < 17. Now φ(x) = x2+x+17 < 172 if 0 ≤x ≤15. If any of these values were composite, it would have a prime divisor p < 17. Our observation above shows that this is impossible. So φ(x) = x2 + x + 17 must be prime for 0 ≤x ≤15. Section 7.2. Principal Square Domains. (1) The integers properly represented by x2 + 6y2 are 1, 6 = 2 · 3, 7, 10 = 2 · 5, 15 = 3 · 5, 22 = 2 · 11, 25 = 52, 31, 33 = 3 · 11, 42 = 2 · 3 · 7, 49 = 72, 55 = 5 · 11, 58 = 2 · 29, 70 = 2 · 5 · 7, 73, 79, 87 = 3 · 29, 97, 103, 105 = 3 · 5 · 7, 106 = 2 · 53, 118 = 2 · 59, 121 = 112, 127, and 145 = 5 · 29. In each case, the number of primes p dividing the squarefree part of the integer for which p ≡5 or 11 (mod 24) (or p = 2 or p = 3) is even. On the other hand, the integers properly represented by 2x2 + 3y2 are 2, 3, 5, 11, 14 = 2 · 7, 21 = 3 · 7, 29, 30 = 2 · 3 · 5, 35 = 5 · 7, 50 = 2 · 52, 53, 59, 62 = 2 · 31, 66 = 2 · 3 · 11, 75 = 3 · 52, 77 = 7 · 11, 83, 93 = 3 · 31, 98 = 2 · 72, 101, 107, 110 = 2 · 5 · 11, 125 = 53, 131, 146 = 2 · 73, 147 = 3·72, and 149. In each case, the number of such primes dividing the squarefree part of the integer is odd. 41 (2) Since 1 = 3(3) + 8(−1), we ﬁnd that the solution of x ≡a (mod 8) and x ≡b (mod 3) is given by x ≡9a −8b (mod 24). (a) 2 p = 1 = p 3 if and only if p ≡1 or 7 (mod 8) and p ≡1 (mod 3). Substituting 1 or 7 for a and 1 for b in x ≡9a −8b (mod 24), we ﬁnd x ≡1 or 7 (mod 24). (b) 2 p = −1 = p 3 if and only if p ≡3 or 5 (mod 8) and p ≡2 (mod 3). With a = 3 or 5 and b = 2 in x ≡9a −8b (mod 24), we ﬁnd x ≡5 or 11 (mod 24). (c) 2 p = 1 and p 3 = −1 if and only if p ≡1 or 7 (mod 8) and p ≡2 (mod 3). Likewise, 2 p = −1 and p 3 = 1 if and only if p ≡3 or 5 (mod 8) and p ≡1 (mod 3). Solving all possible pairs of congruence using the formula established above, we ﬁnd that these cases occur if p ≡13, 17, 19, or 23 (mod 24). (3) (a) Since 223 ≡7 (mod 24), we have that 223 is represented by x2 + 6y2. By trial-and-error, 223 = 132 + 6 · 32. (b) Here 227 ≡11 (mod 24), so that 227 is represented by 2x2 + 3y2. In fact, we ﬁnd that 227 = 2 · 102 + 3 · 32. (c) 341 = 11 · 31 has one prime divisor congruent to 7 modulo 24 and one prime divisor congruent to 11 modulo 24. So 341 is represented by 2x2 +3 · y2. We ﬁnd in fact that 2 · 132 + 3 · 12 = 341 = 2 · 72 + 3 · 92. (d) Since 59 ≡11 (mod 24), the number of primes p dividing 354 = 2 · 3 · 59 for which p = 2, p = 3, or p ≡5 or 11 (mod 24) is odd. Thus 354 is represented by 2x2 + 3y2, and we ﬁnd that 354 = 2 · 92 + 3 · 82. (e) Since 409 ≡1 (mod 24), we ﬁnd that 409 is represented by x2 + 6y2, speciﬁcally 409 = 52 + 6 · 82. (f) 1015 = 5 · 7 · 11 has an even number of prime divisors p with p ≡5 or 11 (mod 24) (and no prime divisors with p ≡13, 17, 19, or 23 (mod 24)). So 1015 is represented by x2 + 6y2. We ﬁnd four such representations: 1015 = 312 + 6 · 32 = 232 + 6 · 92 = 172 + 6 · 112 = 12 + 6 · 132. (4) Let ∆= −20, so that φ(x) = x2 + 5 and u∆= 2. With φ(0) = 5 and φ(1) = 6, we ﬁnd that the reduced ideals of discriminant −20 are D = [1 : 0] and P = [2 : 1]. Thus C−20 has invariant factor type (2) and D−20 is a principal square domain of type one. An integer m is properly represented by f = (1 : 0) or g = (2 : 1) (that is, f(x, y) = x2 + 5y2 or g(x, y) = 2x2 + 2xy + 3y2) if and only if φ(x) ≡0 (mod m) has a solution. For a prime p ̸= 2, 5, this occurs if and only if −5 p = −1 p p 5 = 1. We can establish directly that −1 p = 1 = p 5 if p ≡1 or 9 (mod 20) and −1 p = −1 = p 5 if p ≡3 or 7 (mod 20), while −1 p ̸= p 5 if p ≡11, 13, 17, or 19 (mod 20). For p = 2 and p = 5, we ﬁnd that φ(x) ≡0 (mod p) has a solution but φ(x) ≡0 (mod p2) does not. The unique ideal of norm 5, Q = [5 : 0], is equivalent to D since φ(0) = 5 · 1, while P is the unique ideal of norm 2. We conclude that a positive integer m is properly represented by f or by g if and only if m is not divisible by 4, by 24, or by a prime p ≡11, 13, 17, or 19 (mod 20). If this is the case, let a be the squarefree part of m, and let t be the number of prime divisors of a such that p = 2 or p ≡3 or 7 (mod 20). Then m is properly represented by f if t is even, and is properly represented by g if t is odd. (5) For ∆= −40, we have φ(x) = x2 + 10 and u∆= 3. With φ(0) = 10 and φ(±1) = 11, we ﬁnd that D = [1 : 0] and P = [2 : 0] are the only reduced ideals of discriminant ∆, and so D−40 is a principal square domain of type one. Here P is the unique ideal of norm 2, and Q = [5 : 0], the unique ideal of norm 5, is equivalent to P. We ﬁnd that a positive integer m is properly represented by f(x, y) = x2 + 10y2 or by g(x, y) = 2x2 + 5y2 if and only if m is not divisible by 4, by 25, or by any prime p for which −2 p ̸= p 5 . (These primes are 42 congruent to 3, 17, 21, 27, 29, 31, 33, or 39 modulo 40.) If m has that property, and a is the squarefree part of m, let t be the number of prime divisors p of a for which p = 2, p = 5, or −2 p = −1 = p 5 (that is, p congruent to 7, 13, 23, or 37 modulo 40). If t is even, then m is properly represented by f. If t is odd, then m is properly represented by g. (Primes p with −2 p = 1 = p 5 , that is p congruent to 1, 9, 11, or 19 modulo 40, may appear any number of times in the factorization of m.) The integers m < 100 properly represented by f are 1, 10 = 2 · 5, 11, 14 = 2 · 7, 19, 26 = 2 · 13, 35 = 5 · 7, 41, 46 = 2 · 23, 49 = 72, 59, 65 = 5 · 13, 74 = 2 · 37, 89, 91 = 7 · 13, and 94 = 2 · 47. Those properly represented by g are 2, 5, 7, 13, 22 = 2 · 11, 23, 37, 38 = 2 · 19, 47, 53, 55 = 5 · 11, 70 = 2 · 5 · 7, 77 = 7 · 11, 82 = 2 · 41, 95 = 5 · 19, and 98 = 2 · 72. (6) Using the formula established in the proof of the Chinese Remainder Theorem in Appen-dix B, we ﬁnd that the unique solution of the congruences x ≡a (mod 8), x ≡b (mod 3), and x ≡c (mod 5) is given by x ≡−15a + 40b −24c (mod 120). (a) Substituting 1 or −1 for a, 1 for b, and 1 or −1 for c, we ﬁnd that 2 p = p 3 = p 5 = 1 if p is congruent to 1, 31, 49, or 79 modulo 120. (b) Substituting 1 or −1 for a, −1 for b, and 2 or −2 for c, we ﬁnd that 2 p = 1 and p 3 = −1 = p 5 if p is congruent to 17, 23, 47, or 113 modulo 120. (c) Substituting 3 or −3 for a, 1 for b, and 2 or −2 for c, we ﬁnd that p 3 = 1 and 2 p = −1 = p 5 if p is congruent to 13, 37, 43, or 67 modulo 120. (d) Substituting 3 or −3 for a, −1 for b, and 1 or −1 for c, we ﬁnd that p 5 = 1 and 2 p = −1 = p 3 if p is congruent to 11, 29, 59, or 101 modulo 120. (7) Let r, s, and t be deﬁned as in Proposition 7.2.4. (a) In the table for x2 + 30y2, we ﬁnd that r = s = t = 0 for 1, 31, 79, and 121 = 112; r = s = t = 1 for 30 = 2 · 3 · 5, 66 = 2 · 3 · 11, and 130 = 2 · 5 · 13; r = 2 and s = t = 0 for 34 = 2 · 17, 46 = 2 · 23, and 94 = 2 · 47; s = 2 and r = t = 0 for 39 = 3 · 13, 111 = 3 · 37, and 129 = 3 · 43; and t = 2 and r = s = 0 for 55 = 5 · 11 and 145 = 5 · 29. (b) In the table for 2x2 + 15y2, we have r = 1 and s = t = 0 for 2, 17, 23, 47, 62 = 2 · 31, 113, and 137; r = 0 and s = t = 1 for 15 = 3 · 5, 33 = 3 · 11, 65 = 5 · 13, 87 = 3 · 29, and 143 = 11 · 13; r = 1, s = 2, and t = 0 for 78 = 2 · 3 · 13; and r = 1, s = 0, and t = 2 for 110 = 2 · 5 · 11. (c) In the table for 3x2 + 10y2, we have s = 1 and r = t = 0 for 3, 13, 37, 43, 67, and 93 = 3 · 31; s = 0 and r = t = 1 for 10 = 2 · 5, 22 = 2 · 11, 58 = 2 · 29, 85 = 5 · 17, 115 = 5 · 23, and 118 = 2 · 59; and s = 1, r = 2, and t = 0 for 102 = 2 · 3 · 17 and 138 = 2 · 3 · 23. (d) In the table for 5x2 + 6y2, we have t = 1 and r = s = 0 for 5, 11, 29, 59, 101, 131, and 149; and t = 0 and r = s = 1 for 6 = 2 · 3, 26 = 2 · 13, 51 = 3 · 17, 69 = 3 · 23, 74 = 2 · 37, 86 = 2 · 43, 134 = 2 · 67, and 141 = 3 · 47. (8) If ∆= −84, then φ(x) = x2 + 21 and u∆= 5. Since φ(0) = 21, φ(±1) = 22, and φ(±2) = 25, we ﬁnd that the only reduced ideals of discriminant −84 are [1 : 0], [2 : 1], [3 : 0], and [5 : 2]. Each of these ideal classes is its own inverse in C−84 (for example, [5 : 2]2 = [25 : 2] ∼[1 : 0]), and so C−84 has invariant factor type (2, 2), and D−84 is a principal square domain of type two. A positive integer m is properly represented by a quadratic form of discriminant ∆= −84 if and only if m is not divisible by 4, by 9, by 49, or by any prime for which −21 p = −1. (These primes are congruent to 13, 29, 43, 47, 53, 59, 61, 65, 67, 73, 79, or 83 modulo 84.) If m satisﬁes this condition, let a be the squarefree part of m, let r be the number of prime divisors p of a with p = 2 or p ≡11, 23, or 71 (mod 8)4, let s be the number of those divisors with p = 3, p = 7, or p ≡19, 31, or 55 43 (mod 8)4, and let t be the number of those divisors with p ≡5, 17, or 41 (mod 8)4. Then m is properly represented by x2 + 21y2 if r, s, and t are all even or all odd; m is properly represented by 2x2 +2xy +11y2 if r has the opposite parity of both s and t; m is properly represented by 3x2 + 7y2 if s has the opposite parity of both r and t; and m is properly represented by 5x2 + 4xy + 5y2 if t has the opposite parity of both r and s. (9) In Exercise 6.2.1 and 6.3.1, parts (l), we determined that C−420 has eight elements, each in a distinct genus, and so has invariant factor type (2, 2, 2). Thus D−420 is a principal square domain of type three. The reduced quadratic forms of discriminant ∆= −420 are: (1 : 0) = x2 + 105y2, (2 : 1) = 2x2 + 2xy + 53y2, (3 : 0) = 3x2 + 35y2, (5 : 0) = 5x2 + 21y2, (6 : 3) = 6x2 + 6xy + 19y2, (7 : 0) = 7x2 + 15y2, (10 : 5) = 10x2 + 10xy + 13y2, (11 : 4) = 11x2 + 8xy + 11y2. If m is a positive integer not divisible by 4, by 9, by 25, by 49, or by a prime p with −105 p = −1, then m is properly represented by one of these forms. Let a be the squarefree part of m, and deﬁne integers q through w as follows. Let q be the number of prime divisors p of a so that p = 2 or −1 p = 1 = p 7 and p 3 = −1 = p 5 , let r be the number of these divisors with p = 3 or −1 p = p 3 = p 5 = p 7 = −1, s the number with p = 5 or −1 p = 1 = p 5 and p 3 = −1 = p 7 , t the number with −1 p = −1 = p 7 and p 3 = 1 = p 5 , u the number with −1 p = 1 = p 3 and p 5 = −1 = p 7 , v the number with p = 7 or −1 p = −1 = p 5 and p 3 = 1 = p 7 , and w the number with −1 p = −1 = p 3 and p 5 = 1 = p 7 . Then m is properly represented by 7x2 + 15y2 if and only if q + t + u + w is even, and r + t + v + w and s + u + v + w are both odd. [If Q = [2 : 1], R = [3 : 0], and S = [5 : 0], and A is an ideal of norm a, then we ﬁnd that in C−420 [A] = [Q]q+t+u+w · [R]r+t+v+w · [S]s+u+v+w. Here a and m are properly represented by 7x2 + 15y2 if and only if [A] = [R] · [S], from which our claim follows.] Section 7.3. Quadratic Domains that are not Principal Square Domains. (1) The unique solution of x ≡a (mod 8) and x ≡b (mod 7) satisﬁes x ≡−7a+8b (mod 56). (a) If 2 p = 1 = p 7 , then p ≡1 or 7 (mod 8) and p ≡1, 2, or 4 (mod 7). Substituting all possible pairs of these values into the formula above yields six possibilities for p: 1, 9, 15, 23, 25, or 39 modulo 56. (For instance, if p ≡7 (mod 8) and p ≡2 (mod 7), then p ≡−7(7) + 8(2) ≡−33 ≡23 (mod 56).) (b) If 2 p = −1 = p 7 , then p ≡3 or 5 (mod 8) and p ≡3, 5, or 6 (mod 7). Again by substitution of all possible pairs into our formula, we obtain six possibilities for p: 3, 5, 13, 19, 27, or 45 modulo 56. (2) As noted above, p = 3, 5, 13, and 19 all satisfy 2 p = −1 = p 7 . The possible products of two of these primes are 15, 39, 57, 65, 95, and 247. Table 9 veriﬁes that the ﬁrst ﬁve of these are properly represented by both x2 + 14y2 and 2x2 + 7y2. Direct calculation shows that 247 = 112 + 14 · 32 and 247 = 2 · 62 + 7 · 52. Table 9 likewise conﬁrms that 32 = 9 and 52 = 25 are properly represented by 2x2 + 7y2 but not by x2 + 14y2. We ﬁnd that 132 = 169 = 2 · 92 + 7 · 12 and 192 = 361 = 2 · 32 + 7 · 72, and can rule out proper representations of these squares by x2 + 14y2 using trial-and-error. 44 (3) (a) If ∆= −68, then φ(x) = x2 + 17 and u∆= 4. From φ(0) = 17, φ(±1) = 18, and φ(±2) = 21, we determine that [1 : 0], [2 : 1], [3 : 1], and [3 : −1] are the reduced ideals of discriminant −68. We ﬁnd that if A = [3 : 1], then A2 = [9 : 1] ∼[2 : 1], and then A3 ∼[6 : 1] ∼[3 : −1], and A4 ∼[1 : 0]. Thus C−68 has invariant factor type (4). The principal genus consists of [1 : 0] and [2 : 1]. (b) If p is an odd prime number for which −1 p = 1 = p 17 , then φ(x) ≡0 (mod p) has two solutions, say k and −k, and the prime ideals P = [p : k] and P = [p : −k] are in the principal genus of C−68, since −1 P = 1 = P 17 . So we must have P ∼[1 : 0] (and then P ∼[1 : 0]), or P ∼[2 : 1] (and P ∼[2 : 1]), but not both. If P ∼[1 : 0], then p is properly represented by x2 + 17y2. If P ∼[2 : 1], then P · [2 : 1] is an ideal of norm 2p that is equivalent to [1 : 0], then 2p is properly represented by x2 + 17y2. (c) If p > 3 is a prime number for which −1 p = −1 = p 17 , then φ(x) ≡0 (mod p) again has two solutions, and so there are two conjugate ideals of norm p in D−68, say P and P. Here −1 P = −1 = P 17 , and so P is not in the principal genus. We can assume that P ∼[3 : 1] and P ∼[3 : −1] by relabeling P if necessary. Now P ·[3 : −1] is an ideal of norm 3p that is equivalent to [1 : 0], so that 3p is properly represented by x2 + 17y2. (4) (a) If ∆= −152, then φ(x) = x2 + 38 and u∆= 7. There are six reduced ideals of discriminant −152: [1 : 0], [2 : 0], [3 : 1], [3 : −1], [6 : 2], [6 : −2], and so C−152 must have invariant factor type (6). If A = [3 : 1], then A2 = [9 : 4] ∼[6 : 2], A3 ∼[2 : 0], A4 ∼[6 : −2], A5 ∼[18 : 4] ∼[3 : −1], and A6 ∼[1 : 0]. The principal genus consists of [1 : 0], [6 : 2], and [6 : −2]. (b) If p > 3 is a prime number for which 2 p = 1 = p 19 , then φ(x) ≡0 (mod p) has two solutions, and so there are two conjugate ideals, P and P, of norm p in D−152. Here 2 P = 1 = P 19 , so that P is in the principal genus of C−152. If P ∼[1 : 0] (so that P ∼[1 : 0]), then p is properly represented by x2 + 38y2. Otherwise, we can assume, by relabeling P and P if necessary, that P ∼[6 : 2] and P ∼[6 : −2]. In that case, P · [6 : −2] is an ideal of norm 6p that is equivalent to [1 : 0], so that 6p is properly represented by x2 + 38y2. (c) If p > 3 is a prime number for which 2 p = −1 = p 19 , then φ(x) ≡0 (mod p) again has two solutions, and there are conjugate ideals P and P in D−152. Now 2 P = −1 = P 19 , so that P is not in the principal genus of C−152. If P ∼[2 : 0], then P · [2 : 0] is an ideal of norm 2p that is equivalent to [1 : 0]. In that case, x2 + 38y2 properly represents 2p. If not, we can assume that P ∼[3 : 1], and then P · [3 : −1] is an ideal of norm 3p equivalent to [1 : 0]. In that case, x2+38y2 properly represents 3p. Section 7.4. Construction of Representations. (1) (a) If ∆= −52, with φ(x) = x2 + 13, then g = (187 : 19) ↔10 (2 : 1) = f, with the latter form reduced. We conclude that if f(x, y) = 2x2 + 2xy + 7y2, then f(10, −1) = 187. (b) Again with ∆= −52, we see that g = (187 : 36) ↔5 (7 : −1) ↔0 [2 : 1] = f, so f(x, y) = 2x2 + 2xy + 7y2 is the reduced form equivalent to g. Here with 0 1 −1 0 · 5 1 −1 0 = −1 0 −5 −1 , it follows that f(−1, −5) = 187. (c) Let ∆= −56, so that φ(x) = x2 +14. Here g = (171 : 29) ↔6 (5 : 1) ↔0 (3 : −1) = f, with f(x, y) = 3x2 −2xy+5y2 reduced. Since 0 1 −1 0 · 6 1 −1 0 = −1 0 −6 −1 , it follows that f(−1, −6) = 171. 45 (d) Again with ∆= −56, we ﬁnd that g = (171 : 47) ↔4 (13 : 5) ↔2 (3 : 1), with f(x, y) = 3x2 + 2xy + 5y2 reduced. Here 2 1 −1 0 · 4 1 −1 0 = 7 2 −4 −1 , so that f(7, −4) = 171. (e) Let ∆= −84, so that φ(x) = x2 + 21. Then g = (185 : 33) ↔6 (6 : 3) ↔2 (5 : 2) = f, with f(x, y) = 5x2 +4xy+5y2 reduced. Since 1 1 −1 0 · 6 1 −1 0 = 5 1 −6 −1 , it follows that f(5, −6) = 185. (f) Again with ∆= −84, we have g = (185 : 78) ↔2 (33 : −12) ↔−2 (5 : 2) = f, with f(x, y) = 5x2 + 4xy + 5y2 reduced. Here −2 1 −1 0 · 2 1 −1 0 = −5 −2 −2 −1 , so that f(−5, −2) = 185. (2) (a) When ∆= −52, then M = [187 : 19] ↔[2 : 1] = A, a reduced ideal. Let f = (2 : 1) = 2x2+2xy+7y2. Here 2M = (19+z)A, where z = √−13, and with w = s+tz = 19+z, a = 2, and k = 1, we ﬁnd that q = s+tk a = 19+1(1) 2 = 10 and r = −t = −1 is a pair of integers for which f(q, r) = 187. (b) When ∆= −52, then M = [187 : 36] ↔[7 : −1] ↔[2 : 1] = A, again with f = (2 : 1) = 2x2 + 2xy + 7y2. In this case, 7 · 2M = (36 + z)(−1 + z) = (−36 + 35z + z2)A = (−49 + 35z)A, so that 2M = (−7 + 5z)A. With s + tz = −7 + 5z, a = 2, and k = 1, we ﬁnd that q = s+tk a = −7+5(1) 2 = −1 and r = −t = −5 are integers for which f(q, r) = 187. (c) Let ∆= −56, so that z = √−14. Here we ﬁnd that M = [171 : 29] ↔[5 : 1] ↔[3 : −1] = A, so that f = (3 : −1) = 3x2 −2xy + 5y2. Now 5 · 3M = (29 + z)(1 + z)A = (29+30z +z2)A = (15+30z)A, and so 3M = (3+6z)A. With s+tz = 3+6z, a = 3, and k = −1, we have that q = s+tk a = 3+6(−1) 3 = −1 and r = −t = −6 are integers for which f(q, r) = 171. (d) Again with ∆= −56, we have M = [171 : 47] ↔[13 : 5] ↔[3 : 1] = A, so that f = (3 : 1) = 3x2 + 2xy + 5y2. Now 13 · 3M = (47 + z)(5 + z)A, which simpliﬁes to 3M = (17 + 4z)A. With s + tz = 17 + 4z, a = 3, and k = 1, we ﬁnd that q = s+tk a = 17+4(1) 3 = 7 and r = −t = −4 are integers for which f(q, r) = 171. (e) Let ∆= −84, so that z = √−21. Here M = [185 : 33] ↔[6 : 3] ↔[5 : 2] = A, and so f = [5 : 2] = 5x2 + 4xy + 5y2. Now 6 · 5M = (33 + z)(3 + z)A, which simpliﬁes to 5M = (13 + 6z)A. With s + tz = 13 + 6z, a = 5, and k = 2, then q = s+tk a = 13+6(2) 5 = 5 and r = −t = −6 are integers for which f(q, r) = 185. (f) Again with ∆= −84, we have M = [185 : 78] ↔[6 : 3] ↔[5 : 2] = A, and f = [5 : 2] = 5x2 + 4xy + 5y2. Here 33 · 5M = (78 + z)(−12 + z)A, which simpliﬁes to 5M = (−29 + 2z)A. With s + tz = −29 + 2z, a = 5, and k = 2, we ﬁnd that q + s+tk a = −29+2(2) 5 = −5 and r = −t = −2 are integers for which f(q, r) = 185. Section 8.1. Constructing Class Groups of Subdomains. (1) (a) For ∆= −20, we ﬁnd that G = F∆consists of (1 : 0) and (2 : 1), which is equivalent to (3 : −1). For p = 2, we can use (1 : 0) and (3 : −2) as representatives for G in G2 = Fp2∆; for p = 3, we can use (1 : 0) and (2 : 3) as these representatives; for p = 5, we can use (1 : 0) and (2 : 5). (b) When ∆= −40, then (1 : 0) and (2 : 0) ∼(5 : 0) are the elements of G = F∆. We can use (1 : 0) and (5 : 0) as representatives for G in Fp2∆when p = 2; we can use 46 (1 : 0) and (2 : 0) as these representatives when p = 3; we can use (1 : 0) and (2 : 0) when p = 5. (c) For ∆= −56, we ﬁnd that G = F∆consists of (1 : 0), (2 : 0), (3 : 1), and (3 : −1). When p = 2, then (1 : 0), (7 : 0), (3 : 1), and (3 : −1) are representatives for G in Fp2∆(since (2 : 0) ∼(7 : 0)); when p = 3, we can use (1 : 0), (2 : 0), (5 : −3), and (5 : 3) as these representatives (here using the fact that (3 : 1) ∼(5 : −1) and (3 : −1) ∼(5 : 1)); when p = 5, then (1 : 0), (2 : 0), (3 : 5), and (3 : −5) are these representatives. (d) For ∆= −84, we have that G = F∆consists of (1 : 0), (2 : 1), (3 : 0), and (5 : 2). Since (2 : 1) ∼(11 : −1), we then ﬁnd that (1 : 0), (11 : −2), (3 : 0), and (5 : 4) are representatives for G in Fp2∆when p = 2. With (3 : 0) ∼(7 : 0), we likewise ﬁnd that (1 : 0), (2 : 3), (7 : 0), and (5 : 6) are representatives for G in Fp2∆when p = 3. Finally, since (5 : 2) ∼(5 : −3) ∼(6 : 3), we have that (1 : 0), (2 : 5), (3 : 0), and (6 : 15) are representatives for G in Fp2∆when p = 5. (e) For ∆= −116, class representatives for G = F∆are (1 : 0), (2 : 1), (3 : 1), (3 : −1), (5 : 1), and (5 : −1). For p = 2, with (2 : 1) ∼(15 : 1), we have that (1 : 0), (15 : 2), (3 : 2), (3 : −2), (5 : 2), and (5 : −2) are representatives for G in Fp2∆. For p = 3, with (3 : ±1) ∼(10 : ∓1), we can use (1 : 0), (2 : 3), (10 : −3), (10 : 3), (5 : 3), and (5 : −3) as these representatives. For p = 5, with (5 : ±1) ∼(6 : ∓1), we ﬁnd these representatives as (1 : 0), (2 : 5), (3 : 5), (3 : −5), (6 : −5), and (6 : 5). (f) For ∆= −119, we ﬁnd that G = F∆has ten elements, with class representatives (1 : 0), (2 : 0), (2 : −1), (3 : 0), (3 : −1), (4 : 1), (4 : −2), (5 : 0), (5 : −1), (6 : 2). Using various equivalences, we ﬁnd the following representatives for G in Fp2∆. When p = 2, (1 : 0), (15 : −2), (15 : 0), (3 : 0), (3 : −2), (9 : 4), (9 : −6), (5 : 0), (5 : −2), (7 : 6); when p = 3, (1 : 0), (2 : 0), (2 : −3), (10 : −3), (10 : 0), (4 : 3), (4 : −6), (5 : 0), (5 : −3), (7 : 9); and when p = 5, (1 : 0), (2 : 0), (2 : −5), (3 : 0), (3 : −5), (4 : 5), (4 : −10), (6 : −5), (6 : 0), (6 : 10). (2) Let ∆= −3. Then p − ∆ p = 2 −(−1) = 3 when p = 2; p − ∆ p = 3 −0 = 3 when p = 3; p − ∆ p = p −1 when p > 3 is congruent to 1 modulo 3; and p − ∆ p = p −(−1) = p + 1 when p > 3 is congruent to 2 modulo 3. In each case, p −∆ p is divisible by 3, and so \|Kp\| = 1 3 p − ∆ p is an integer when ∆= −3. Now let ∆= −4. Then p− ∆ p = 2−0 = 2 when p = 2; p − ∆ p = p −1 if p ≡1 (mod 4); and p − ∆ p = p −(−1) = p + 1 if p ≡3 (mod 4). In each case, p − ∆ p is even, and so \|Kp\| = 1 2 p − ∆ p is an integer when ∆= −4. (3) For each discriminant ∆, we list the elements of Kp for p = 2, p = 3, and p = 5, and the reduced form of discriminant p2∆to which each is equivalent, showing that the classes of these forms are distinct in Cp2∆. (a) For ∆= −20, then K2 contains (1 : 0) and (4 : 0), both reduced; K3 contains (1 : 0) and (9 : 0) ∼(5 : 0); and K5 contains (1 : 0), (25 : 5) ∼(6 : 1), (25 : −5) ∼(6 : −1), (25 : 10) ∼(9 : −1), and (25 : −10) ∼(9 : 1). 47 (b) For ∆= −40, K2 contains (1 : 0) and (4 : 2); K3 contains (1 : 0), (9 : 0), (9 : 3), and (9 : −3); and K5 contains (1 : 0), (25 : 5) ∼(11 : −5), (25 : −5) ∼(11 : 5), (25 : 10) ∼(14 : 4), and (25 : −10) ∼(14 : −4). (c) For ∆= −56, K2 contains (1 : 0) and (4 : 2); K3 contains (1 : 0) and (9 : 0); and K5 contains (1 : 0), (25 : 0) ∼(14 : 0), (25 : 10) ∼(18 : 8), and (25 : −10) ∼(18 : −8). (d) For ∆= −84, K2 contains (1 : 0) and (4 : 0); K3 contains (1 : 0), (9 : 3), and (9 : −3); and K5 contains (1 : 0), (25 : 0) ∼(21 : 0), (25 : 5) ∼(22 : −5), and (25 : −5) ∼(22 : 5). (e) For ∆= −116, K2 contains (1 : 0) and (4 : 0); K3 contains (1 : 0) and (9 : 0); and K5 contains (1 : 0), (25 : 0), (25 : 10), and (25 : −10). All of these forms are reduced. (f) For ∆= −119, K2 contains only (1 : 0) ∼(1 : −1); K3 contains (1 : 0) ∼(1 : −1) and (9 : 3); and K5 contains (1 : 0) ∼(1 : −2), (25 : 5), (25 : 10), and (25 : −10). (4) If ∆= −3, so that φ(x) = x2 + x + 1, then φ(x) ≡0 (mod 11) has no solutions, and thus K11 could have as many as twelve distinct elements, of the form (1 : 0) or (121 : 11k) with −5 ≤k ≤5. But with φ11(x) = x2 + 11x + 121, we ﬁnd that (121 : 0) ∼(121 : −11) ∼(1 : 0) ∼(1 : −5), a reduced form of discriminant ∆11 = −363. Likewise (121 : 11) ∼(121 : −22) ∼(121 : −55) ∼(3 : −4), (121 : 22) ∼(121 : −44) ∼(121 : 44) ∼(7 : −5), and (121 : −33) ∼(121 : 33) ∼(121 : −55) ∼(7 : −6). So in fact there are only four distinct elements in K11. (5) The principal polynomial of discriminant ∆p = −47 · 52 = −1175 is φ(x)x2 + 5x + 300. We use this polynomial below to ﬁnd the reduced form equivalent to each representative of F−1175 listed. For example, we have that (75 : 30) ∼(18 : −35) = (18 : 1) since φ(30) = 1350 = 75 · 18. But then with φ(1) = 306 = 18 · 17, then (18 : 1) ∼(17 : −6), a reduced form. We list the reduced form representatives (a : k) of F−1175 in increasing order of a, to help verify that all such forms are distinct. (1 : 0) ∼(1 : −2), (2 : 0) ∼(2 : −2), (2 : −5) ∼(2 : −3), (3 : 0) ∼(3 : −3), (3 : −5) ∼(3 : −2), (75 : 0) ∼(4 : −1), (75 : −5) ∼(4 : −4), (50 : 0) ∼(6 : −5), (50 : −5) ∼(6 : 0), (75 : −15) ∼(6 : −2), (75 : 10) ∼(6 : −3), (50 : 5) ∼(7 : −3), (50 : −10) ∼(7 : −2), (75 : −20) ∼(8 : −1), (75 : 15) ∼(8 : −4), (50 : 10) ∼(9 : −6), (50 : −15) ∼(9 : 1), (25 : −5) ∼(12 : 0), (25 : 0) ∼(12 : −5), (50 : −20) ∼(12 : 3), (50 : 15) ∼(12 : −8), (75 : 25) ∼(14 : −2), (75 : −30) ∼(14 : −3), (25 : 5) ∼(14 : 4), (25 : −10) ∼(14 : −9), (50 : −25) ∼(16 : 4), (50 : 20) ∼(16 : −9), (75 : 30) ∼(17 : −6), (75 : −35) ∼(17 : 1), (25 : 10) ∼(18 : 3). (6) In parts (a)–(i), we use calculations of Sp and Kp from Exercises 1 and 3 in this section. In each array, the ﬁrst row lists the elements of Sp and the ﬁrst column lists the elements of Kp, with products of those elements as the remaining entries of the table. We then note the reduced form of discriminant p2∆to which each one is equivalent. In parts (j)–(o), the group C∆is trivial, so that Sp contains only (1 : 0) and the class group of discriminant p2∆is identical to Kp. 48 (a) ∆= −20 and p = 2. (1 : 0) (3 : −2) (4 : 0) (12 : 4) Here (1 : 0) and (4 : 0) are reduced, while (3 : −2) ∼(3 : 1) and (12 : 4) ∼(3 : −1). One can verify that these are the only (primitive) reduced forms of discriminant −80. (b) ∆= −20 and p = 3. (1 : 0) (2 : 3) (9 : 0) (18 : 9) with (2 : 3) ∼(2 : 1), (9 : 0) ∼(5 : 0), and (18 : 9) ∼(7 : 2). (c) ∆= −20 and p = 5. (1 : 0) (2 : 5) (25 : 5) (50 : 5) (25 : −5) (50 : −5) (25 : 10) (50 : −15) (25 : −10) (50 : 15) with (2 : 5) ∼(2 : 1), (25 : 5) ∼(6 : 1), (25 : −5) ∼(6 : −1), (25 : 10) ∼(9 : −1), (25 : −10) ∼(9 : 1), (50 : 5) ∼(3 : 1), (50 : −5) ∼(3 : −1), (50 : −15) ∼(7 : 1), and (50 : 15) ∼(7 : −1) as reduced form representatives. (d) ∆= −40 and p = 2. (1 : 0) (5 : 0) (4 : 2) (20 : 10) with (20 : 10) ∼(7 : 3) and all other forms reduced. (e) ∆= −40 and p = 3. (1 : 0) (2 : 0) (9 : 0) (18 : 0) (9 : 3) (18 : −6) (9 : −3) (18 : 6) with (18 : 0) ∼(5 : 0), (18 : −6) ∼(7 : −1), (18 : 6) ∼(7 : 1), and all other forms reduced. (f) ∆= −56 and p = 2. (1 : 0) (7 : 0) (3 : 2) (3 : −2) (4 : 2) (28 : 14) (12 : 2) (12 : −2) with (3 : 2) ∼(3 : −1), (3 : −2) ∼(3 : 1), (28 : 14) ∼(8 : 4), (12 : 2) ∼(5 : −2), and (12 : −2) ∼(5 : 2). (g) ∆= −56 and p = 3. (1 : 0) (2 : 0) (5 : −3) (5 : 3) (9 : 0) (18 : 0) (45 : −18) (45 : 18) with (5 : −3) ∼(5 : 2), (5 : 3) ∼(5 : −2), (18 : 0) ∼(7 : 0), (45 : −18) ∼(10 : −2), and (45 : 18) ∼(10 : 2). (h) ∆= −84 and p = 2. (1 : 0) (11 : −2) (3 : 0) (5 : 4) (4 : 0) (44 : 20) (12 : 0) (20 : 4) with (11 : −2) ∼(8 : 2), (5 : 4) ∼(5 : −1), (44 : 20) ∼(8 : −2), (12 : 0) ∼(7 : 0), and (20 : 4) ∼(5 : 1). 49 (i) ∆= −116 and p = 2. (1 : 0) (15 : 2) (3 : 2) (3 : −2) (5 : 2) (5 : −2) (4 : 0) (60 : −28) (12 : −4) (12 : 4) (20 : −8) (20 : 8) with (15 : 2) ∼(8 : −2), (3 : 2) ∼(3 : −1), (3 : −2) ∼(3 : 1), (60 : −28) ∼(8 : 2), (12 : −4) ∼(11 : 4), (12 : 4) ∼(11 : −4), (20 : −8) ∼(9 : −1), and (20 : 8) ∼(9 : 1). (j) For ∆= −3 and p = 7, we ﬁnd that (49 : 0) ∼(49 : −7) ∼(1 : 0) ∼(1 : −3) and (49 : 7) ∼(49 : −14) ∼(49 : 21) ∼(3 : −2), and so F−147 = {(1 : −3), (3 : −2)}. (k) F−363 = {(1 : −5), (3 : −4), (7 : −5), (7 : −6)}. (l) F−507 = {(1 : −6), (3 : −5), (7 : −4), (7 : −9)}. (m) F−196 = {(1 : 0), (2 : 1), (5 : 1), (5 : −1)}. (n) F−484 = {(1 : 0), (2 : 1), (5 : 2), (5 : −2), (10 : 3), (10 : −3)}. (o) F−676 = {(1 : 0), (2 : 1), (5 : 1), (5 : −1), (10 : 1), (10 : −1)}. (7) Let ∆= ∆(−1, 12) = −576, so that φ(x) = x2 + 144 and u∆= 13. From the table k 0 ±1 ±2 ±3 ±4 ±5 ±6 φ(k) 144 145 148 153 160 169 180 we compile the following list of reduced forms of discriminant ∆. (We group these forms by index, which is a divisor of 12.) 1 (1 : 0) (4 : 2) (5 : 1) (5 : −1) (9 : 0) (9 : 3) (9 : −3) (13 : 5) 2 (2 : 0) (8 : 0) (10 : 4) (10 : −4) 3 (3 : 0) (12 : 6) 4 (4 : 0) 6 (6 : 0) 12 (12 : 0) The form class group F−576 has eight elements. Section 8.2. Projection Homomorphisms. (1) For each ∆and p, we list the reduced quadratic forms of discriminant ∆and the (primitive) reduced forms of discriminant p2∆, and describe the projection homomorphism ψ from Fp2∆to F∆. (a) The only reduced form of discriminant ∆= −4 is (1 : 0); if p = 7, the primitive reduced forms of discriminant p2∆= −196 are (1 : 0), (2 : 1), (5 : 1), and (5 : −1). To apply the projection homomorphism, we note that (2 : 1) ∼(2 : 7), (5 : 1) ∼ (5 : −14), and (5 : −1) ∼(5 : 14). Then ψ((1 : 0)) = (1 : 0), ψ((2 : 7)) = (2 : 1), ψ((5 : −14)) = (5 : −2), and ψ((5 : 14)) = (5 : 2). Each of the resulting forms is equivalent to (1 : 0) in F∆. (b) The primitive reduced forms of discriminant 132 · −4 = −676 are (1 : 0), (2 : 1) ∼ (2 : 13), (5 : 1) ∼(5 : 26), (5 : −1) ∼(5 : −26), (10 : 1) ∼(10 : −39), and (10 : −1) ∼(10 : 39). These are sent by the projection homomorphism to (1 : 0), (2 : 1), (5 : 2), (5 : −2), (10 : −3), and (10 : 3) respectively. Each of those forms is equivalent to (1 : 0) in F−4. (c) The only reduced form of discriminant ∆= −7 is (1 : 0); when p = 3, the primitive reduced forms of discriminant p2∆= −63 are (1 : −1) ∼(1 : 0), (2 : −1) ∼(2 : −3), (2 : −2) ∼(2 : 0), and (4 : −1) ∼(4 : 3), sent by ψ to (1 : 0), (2 : −1), (2 : 0), and (4 : 1) respectively. Each of these forms is equivalent to (1 : 0) in F−7. (d) The primitive reduced forms of discriminant 52 · −7 = −175 are (1 : −2) ∼(1 : 0), (2 : −2) ∼(2 : 0), (2 : −3) ∼(2 : −5), (4 : −2) ∼(4 : −10), (4 : −3) ∼(4 : 5), and 50 (7 : 1) ∼(7 : 15), sent by ψ to (1 : 0), (2 : 0), (2 : −1), (4 : −2), (4 : 1), and (7 : 3) respectively, each equivalent to (1 : 0) in F−7. (e) The primitive reduced forms of discriminant 112 · −7 = −847 are (1 : −5) ∼(1 : 0), (2 : −5) ∼(2 : −11), (2 : −6) ∼(2 : 0), (4 : −5) ∼(4 : 11), (4 : −6) ∼(4 : −22), (7 : −2) ∼(7 : 33), (8 : −2) ∼(8 : 22), (8 : −9) ∼(8 : −33), (14 : −2) ∼(14 : −44), and (14 : −9) ∼(14 : 33). These elements are sent by ψ to (1 : 0), (2 : −1), (2 : 0), (4 : 1), (4 : −2), (7 : 3), (8 : 2), (8 : −3), (14 : −4), and (14 : 3) respectively, each equivalent to (1 : 0) in F−7. (f) There are two reduced forms of discriminant ∆= −15, (1 : 0) and (2 : 0). When p = 3, we ﬁnd the following reduced forms of discriminant p2∆= −135: (1 : −1) ∼(1 : 0), (2 : −1) ∼(2 : −3), (2 : −2) ∼(2 : 0), (4 : 0), (4 : −3), and (5 : 1) ∼(5 : 6). These are sent by ψ to (1 : 0), (2 : −1), (2 : 0), (4 : 0), (4 : −1), and (5 : 6). We then ﬁnd that (4 : 0) ∼(4 : −1) ∼(1 : 0), while (5 : 2) ∼(2 : −1) ∼(2 : 0) in F−15. (g) There are ten primitive reduced forms of discriminant 52 · −15 = −375. We list these forms below, along with equivalent forms to which ψ can be applied, and then the resulting forms in F−15. Those in the ﬁrst column are equivalent to (1 : 0), and those in the second column are equivalent to (2 : 0). (1 : −2) ∼(1 : 0) →(1 : 0) (2 : −2) ∼(2 : 0) →(2 : 0) (4 : −1) ∼(4 : −5) →(4 : −1) (2 : −3) ∼(2 : −5) →(2 : −1) (4 : −4) ∼(4 : 0) →(4 : 0) (3 : −1) ∼(3 : 5) →(3 : 1) (6 : −1) ∼(6 : 5) →(6 : 1) (8 : −1) ∼(8 : 15) →(8 : 3) (6 : −4) ∼(6 : −10) →(6 : −2) (8 : −4) ∼(8 : −20) →(8 : −4) (h) There are sixteen primitive reduced forms of discriminant 72 · −15 = −735. We note the eﬀect of the projection homomorphism below, with those forms sent to (1 : 0) in the ﬁrst column and those sent to (2 : 0) in the second column. (1 : −3) ∼(1 : 0) →(1 : 0) (2 : −3) ∼(2 : −7) →(2 : −1) (4 : −3) ∼(4 : −7) →(4 : −1) (2 : −4) ∼(2 : 0) →(2 : 0) (4 : −4) ∼(4 : 0) →(4 : 0) (3 : −2) ∼(3 : 7) →(3 : 1) (6 : −2) ∼(6 : −14) →(6 : −2) (5 : −1) ∼(5 : 14) →(5 : 2) (6 : −5) ∼(6 : 7) →(6 : 1) (8 : −3) ∼(8 : 21) →(8 : 3) (10 : −1) ∼(10 : −21) →(10 : −3) (8 : −4) ∼(8 : −28) →(8 : −4) (10 : −6) ∼(10 : 14) →(10 : 4) (12 : 1) ∼(12 : −35) →(12 : −5) (15 : 4) ∼(15 : 49) →(15 : 7) (12 : −8) ∼(12 : 28) →(12 : 4) (i) There are two reduced forms of discriminant ∆= −20, (1 : 0) and (2 : 1). When p = 3, we ﬁnd the following primitive reduced forms of discriminant p2∆= −180: (1 : 0), (2 : 1) ∼(2 : 3), (5 : 0), and (7 : 2) ∼(7 : 9). The projection homomorphism sends these forms to (1 : 0), (2 : 1), (5 : 0), and (7 : 3) respectively. Here (5 : 0) ∼(1 : 0) and (7 : 3) ∼(2 : 1) in F−20. (j) There are twelve primitive reduced forms of discriminant 72 · −20 = −980. These forms are listed below, along with an equivalent form to which ψ can be applied. Those in the ﬁrst column are sent to a form equivalent to (1 : 0) in F−20; those in 51 the second column are sent to (2 : 1). (1 : 0) (2 : 1) ∼(2 : 7) (5 : 0) (3 : 1) ∼(3 : 7) (6 : 1) ∼(6 : 7) (3 : −1) ∼(3 : −7) (6 : −1) ∼(6 : −7) (10 : 5) ∼(10 : 35) (9 : 4) ∼(9 : −14) (15 : 5) ∼(15 : 35) (9 : −4) ∼(9 : 14) (15 : −5) ∼(15 : −35) (k) There are three reduced forms of discriminant ∆= −23: (1 : 0), (2 : 0), and (2 : −1). When p = 3, we ﬁnd six primitive reduced forms of discriminant p2∆= −207: (1 : −1) ∼(1 : 0), (2 : −1) ∼(2 : −3), (2 : −2) ∼(2 : 0), (4 : −1) ∼(4 : 3), (4 : −2) ∼(4 : −6), and (8 : 2) ∼(8 : −6). The projection homomorphism sends these forms to (1 : 0), (2 : −1), (2 : 0), (4 : 1) ∼(2 : 0), (4 : −2) ∼(2 : −1), and (8 : −2) ∼(1 : 0) respectively. (l) There are eighteen primitive reduced forms of discriminant 52 · −23 = −575. We list these forms below, along with an equivalent form to which ψ can be applied. Those in the ﬁrst column are sent to a form equivalent to (1 : 0) in F−23; those in the second column are sent to (2 : 0); those in the third column are sent to (2 : −1). (1 : −2) ∼(1 : 0) (2 : −2) ∼(2 : 0) (2 : −3) ∼(2 : −5) (6 : 0) (3 : −2) ∼(3 : −5) (3 : −3) ∼(3 : 0) (6 : −5) (4 : −3) ∼(4 : 5) (4 : −2) ∼(4 : −10) (8 : −2) ∼(8 : −10) (6 : −3) ∼(6 : −15) (6 : −2) ∼(6 : 10) (8 : −3) ∼(8 : 5) (9 : −3) ∼(9 : 15) (9 : −2) ∼(9 : −20) (12 : −2) ∼(12 : 10) (12 : −6) ∼(12 : −30) (12 : 1) ∼(12 : 25) (m) There are two reduced forms of discriminant ∆= −24, (1 : 0) and (2 : 0). When p = 3, we ﬁnd six primitive reduced forms of discriminant p2∆= −216: (1 : 0), (2 : 0), (5 : 1) ∼(5 : 6), (5 : −1) ∼(5 : −6), (7 : 3), and (7 : −3). The projection homomorphism sends these forms to (1 : 0), (2 : 0), (5 : 2) ∼(2 : 0), (5 : −2) ∼(2 : 0), (7 : 1) ∼(1 : 0), and (7 : −1) ∼(1 : 0) respectively. (n) There are eight primitive reduced forms of discriminant 52·−24 = −600. These forms are listed below, along with an equivalent form to which ψ can be applied. Those in the ﬁrst column are sent to a form equivalent to (1 : 0) in F−24; those in the second column are sent to (2 : 0). (1 : 0) (2 : 0) (6 : 0) (3 : 0) (7 : 2) ∼(7 : −5) (11 : 2) ∼(11 : −20) (7 : −2) ∼(7 : 5) (11 : −2) ∼(11 : 20) (2) Let Γp be the set of all unimodular 2×2 matrices having the lower left-hand entry divisible by p. The identity matrix is an element of Γp; the inverse of a matrix q s r t in Γp is t −s −r q , which is also in Γp; the product of two elements of Γp is also in Γp, since the lower left-hand entry of q s r t · u w v x is ru+tv, which is divisible by p if both r and v are divisible by p. The remaining claims of this exercise were established in Exercise 4 of §4.2. 52 Section 8.3. The Kernel of a Projection Homomorphism. (1) Let φ = (1 : 0) = x2 + bxy + cy2, where b = ε and c = ε2−∆ 4 for some discriminant ∆, and let Vk = −k −b −1 1 0 . Applying Proposition 4.2.1, we have that φ ◦Vk = (m : ℓ), where m = φ(−k −b, 1) = (−k −b)2 + b(−k −b) + c = k2 + bk + c = φ(k) and ℓ= (−k −b)(−1) + b(−1)(1) + c(1)(0) + 0 = k. So φ ◦Vk = (φ(k) : k), which is denoted as φk. (2) Let φk = (φ(k) : k) = φ(k)x2 + φ′(k)xy + y2 and Vℓ,k = 1 0 ℓ−k 1 for some k and ℓ. Again applying Proposition 4.2.1, we ﬁnd that φk ◦Vℓ,k = (φ(ℓ) : ℓ) since φk(1, ℓ−k) = φ(k) + φ′(k)(ℓ−k) + (ℓ−k)2 = (k2 + bk + c) + (2k + b)(ℓ−k) + (ℓ2 −2kℓ+ k2) = ℓ2 + bℓ+ c and φ(k)(1)(0) + φ′(k)(0)(1) + 1(ℓ−k)(1) + k = ℓ. That is, φℓ= φ(k) ◦Vℓ,k. Section 9.1. Introduction to Continued Fractions. (1) Let F0 = 0, F1 = 1 and Fn = Fn−1 +Fn−2 for n ≥2. Let r0 = 1 and ri+1 = ri+1 ri for i ≥0. Note that r0 = F2/F1 since F2 = F1 + F0 = 1 + 0 = 1. Suppose that rk = Fk+2/Fk+1 for some k ≥0. Then notice that rk + 1 = Fk+2 Fk+1 + 1 = Fk+2 + Fk+1 Fk+1 = Fk+3 Fk+1 . So now rk+1 = rk + 1 rk = Fk+3 Fk+1 · Fk+1 Fk+2 = F(k+1)+2 F(k+1)+1 . It follows that rn = Fn+2/Fn+1 for all n ≥0 by induction. (2) (a) If v = v0 = 249 89 = 2 + 71 89, then v1 = 89 71 = 1 + 18 71, v2 = 71 18 = 3 + 17 18, v3 = 18 17 = 1 + 1 17, and v4 = 17 1 = 17 + 0 1, with v5 undeﬁned. The continued fraction for v is 2 + 1 1 + 1 3+ 1 1+ 1 17 , with the sequence of convergents r0 = 2, r1 = 3, r2 = 11 4 , r3 = 14 5 , r4 = 249 89 . (b) For v = v0 = √ 2 = 1+(−1+ √ 2), we ﬁnd that v1 = 1 −1+ √ 2 = 1+ √ 2 = 2+(−1+ √ 2), and then vi = 1 + √ 2 for all i ≥1. Now r0 = 1, then r1 = 1 + 1 2 = 3 2, so that r2 = 1 + 1 2+ 1 2 = 7 5, and so forth. The sequence of convergents ri with i ≤5 is r0 = 1, r1 = 3 2, r2 = 7 5, r3 = 17 12, r4 = 41 29, r5 = 99 70. (c) For v = √ 3, the sequence of convergents begins r0 = 1, r1 = 2, r2 = 5 3, r3 = 7 4, r4 = 19 11, r5 = 26 15. 53 (d) The sequence of convergents for v = √ 7 begins r0 = 2, r1 = 3, r2 = 5 2, r3 = 8 3, r4 = 37 14, r5 = 45 17. (e) The sequence of convergents for v = 4+ √ 3 7 begins r0 = 0, r1 = 1, r2 = 4 5, r3 = 5 6, r4 = 9 11, r5 = 104 127. (3) (a) If w = 1, 2 , we can assume that w = 1+ 1 2+ 1 w = 1+ w 2w+1 = 3w+1 2w+1, so that w satisﬁes (2w2 + w) −(3w + 1) = 2w2 −2w −1 = 0. Since w is larger than 1, we conclude that w = 2+ √ 12 4 = 1+ √ 3 2 . (b) If w = 1, 2, 3 , then w = 1 + 1 2+ 1 3+ 1 w = 1 + 1 2+ w 3w+1 = 1 + 3w+1 7w+2 = 10w+3 7w+2 . We ﬁnd that w is the larger solution of (7w2 + 2w) −(10w + 3) = 7w2 −8w −3 = 0, that is, w = 4+ √ 37 7 . (c) If w = 1, 2 = 1+ √ 3 2 as in part (a), then v = 3, 1, 2 = 3 + 1 w = 5+3 √ 3 1+ √ 3 = 2 + √ 3. (d) If w = 1 = 1 + 1 w, we ﬁnd that w2 −w −1 = 0 and that w = 1+ √ 5 2 . So then v = 3, 2, 1 = 3 + 1 2+ 1 w = 3 + w 2w+1 = 7w+3 2w+1, which simpliﬁes to v = 9− √ 5 2 . (e) If w = 2, 5 = 2 + 1 5+ 1 w , then w is the larger solution of (5w2 + w) −(11w + 2) = 5w2 −10w −2 = 0, that is, w = 10+ √ 140 10 . Then v = 1, 2, 5 = 1 + 1 w, which simpliﬁes to v = −3+ √ 35 2 . (f) If w = 1, 1, 1, 4, we ﬁnd that w = 14w+3 9w+2 . Then w satisﬁes 3w2 −4w −1 = 0, and so w = 2+ √ 7 3 . Now v = 1, 4, 1, 1, 1, 4 = 1 + 1 4+ 1 w , which simpliﬁes to v = 1+ √ 7 3 . Section 9.2. Pell’s Equation. (1) For each d, we present the data from the algorithm of Theorem 9.2.2 as a table. (a) For d = 11: i 0 1 2 a 1 2 1 k 0 −3 −3 q 3 3 m 3 10 n 1 3 Here beginning with a0 = 1 and k0 = 0, we select k1 to be the smallest integer greater than − √ 11 ≈−3.3 congruent to −k0 = 0 modulo a0 = 1, that is, k1 = −3. So then a1 = 11−k2 1 a0 = 2. Now we select k2 > − √ 11 congruent to −k1 = 3 modulo a1 = 2, that is, k2 = −3. Thus a2 = 11−k2 2 a1 = 1. We terminate the algorithm when we obtain an ℓ> 0 so that aℓ= 1. For 0 ≤i < ℓ, we calculate qi = −ki+ki+1 ai , and then mi = qimi−1 + mi−2 and ni = qini−1 + ni−2 where m−2 = 0, m−1 = 1, n−2 = 1, and n−1 = 0. If ℓis even, then (x, y) = (mℓ−1, nℓ−1) is the fundamental solution of x2 −dy2 = 1. In this example, (x, y) = (10, 3) and we verify that 102 −11 · 32 = 1. 54 (b) For d = 13: i 0 1 2 3 4 5 a 1 4 3 3 4 1 k 0 −3 −1 −2 −1 −3 q 3 1 1 1 1 m 3 4 7 11 18 n 1 1 2 3 5 Here ℓ= 5 is odd. If we let v = mℓ−1 + nℓ−1 √ d, then v2 = m2ℓ−1 + n2ℓ−1 √ d and (x, y) = (m2ℓ−1, n2ℓ−1) is the fundamental solution of x2 −dy2 = 1. In this example, v = 18 + 5 √ 13 and v2 = 649 + 180 √ 13, so that (x, y) = (649, 180). (c) For d = 14: i 0 1 2 3 4 a 1 5 2 5 1 k 0 −3 −2 −2 −3 q 3 1 2 1 m 3 4 11 15 n 1 1 3 4 So (x, y) = (15, 4) is the fundamental solution of x2 −14y2 = 1. (d) For d = 20: i 0 1 2 a 1 4 1 k 0 −4 −4 q 4 2 m 4 9 n 1 2 Thus (x, y) = (9, 2) is the fundamental solution of x2 −20y2 = 1. (e) For d = 23: i 0 1 2 3 4 a 1 7 2 7 1 k 0 −4 −3 −3 −4 q 4 1 3 1 m 4 5 19 24 n 1 1 4 5 The fundamental solution of x2 −23y2 = 1 is (x, y) = (24, 5). (f) For d = 31: i 0 1 2 3 4 5 6 7 8 a 1 6 5 3 2 3 5 6 1 k 0 −5 −1 −4 −5 −5 −4 −1 −5 q 5 1 1 3 5 3 1 1 m 5 6 11 39 206 657 863 1520 n 1 1 2 7 37 118 155 273 The fundamental solution of x2 −31y2 = 1 is (x, y) = (1520, 273). 55 (g) For d = 33: i 0 1 2 3 4 a 1 8 3 8 1 k 0 −5 −3 −3 −5 q 5 1 2 1 m 5 6 17 23 n 1 1 3 4 The fundamental solution of x2 −33y2 = 1 is (x, y) = (23, 4). (h) For d = 41: i 0 1 2 3 a 1 5 5 1 k 0 −6 −4 −6 q 6 2 2 m 6 13 32 n 1 2 5 Here ℓis odd. Since (32+5 √ 41)2 = 2049+320 √ 41, we have that (x, y) = (2049, 320) is the fundamental solution of x2 −41y2 = 1. (i) For d = 46: i 0 1 2 3 4 5 6 7 8 9 10 11 12 a 1 10 3 7 6 5 2 5 6 7 3 10 1 k 0 −6 −4 −5 −2 −4 −6 −6 −4 −2 −5 −4 −6 q 6 1 3 1 1 2 6 2 1 1 3 1 m 6 7 27 34 61 156 997 2150 3147 5297 19038 24335 n 1 1 4 5 9 23 147 317 464 781 2807 3588 The fundamental solution of x2 −46y2 = 1 is (x, y) = (24335, 3588). (j) For d = 53: i 0 1 2 3 4 5 a 1 4 7 7 4 1 k 0 −7 −5 −2 −5 −7 q 7 3 1 1 3 m 7 22 29 51 182 n 1 3 4 7 25 Since (182 + 25 √ 53)2 = 66249 + 9100 √ 53, the fundamental solution of x2 −53y2 = 1 is (x, y) = (66249, 9100). (2) (a) Let d = t2 + 1 for some positive integer t. The smallest integer greater than − √ d is −t, and we ﬁnd in applying Theorem 9.2.2 that k1 = −t and a1 = 1, so that ℓ= 1 in the notation of that algorithm. Then q0 = t = m0 and n0 = 1, and the fundamental solution of x2−dy2 = 1 is given by (t+ √ d)2 = (t2+d)+2t √ d = (2t2+1)+2t √ d. That is, (x, y) = (2t2 + 1, 2t) is the fundamental solution of x2 −dy2 = 1 when d = t2 + 1. (b) Let d = t2 −1 for some t > 1. Here the smallest integer greater than − √ d is −t + 1. Since d −(−t + 1)2 = 2t −2, we can compile the following data from Pell’s equation 56 algorithm. i 0 1 2 a 1 2t −2 1 k 0 −t + 1 −t + 1 q t −1 1 m t −1 t n 1 1 (For example, t −1 ≡−t + 1 (mod 2t −2), which produces the calculation of k2 and a2.) We conclude that (x, y) = (t, 1) is the fundamental solution of x2 −dy2 = 1 when d = t2 −1 is positive. (c) Let d = t2 + 2 for some positive integer t. Then −t is the smallest integer greater than − √ d, and we compile the following data from Pell’s equation algorithm. (Note that d −(−t)2 = 2 and that t ≡−t (mod 2t).) i 0 1 2 a 1 2 1 k 0 −t −t q t t m t t2 + 1 n 1 t We conclude that (x, y) = (t2 + 1, t) is the fundamental solutions of x2 −dy2 = 1 when d = t2 + 2. (d) Let d = t2 +t for some positive integer t. We ﬁnd again that −t is the smallest integer greater than − √ d, since t2 < d < (t + 1)2 = t2 + 2t + 1. With d −(−t)2 = t and t ≡−t (mod t), we compile the following data from Pell’s equation algorithm. i 0 1 2 a 1 t 1 k 0 −t −t q t 2 m t 2t + 1 n 1 2 Thus (x, y) = (2t + 1, 2) is the fundamental solution of x2 −dy2 = 1 when d = t2 + t. (3) Let v = q + r √ d where (q, r) is the fundamental solution of x2 −dy2 = 1. For all integers n, write vn as qn + rn √ d. Note that vk+1 = vk · v = (qk + rk √ d)(q + r √ d) = (qqk + drrk) + (qrk + rqk) √ d for all integers k. Now let U = q dr r q and suppose that Uk = qk drk rk qk for some positive integer k. Then Uk+1 = U · Uk = q dr r q · qk drk rk qk = qqk + drrk qdrk + drqk rqk + qrk rdrk + qqk = qk+1 drk+1 rk+1 qk+1 , using the preceding observation. It follows that Un = qn drn rn qn for all positive integers n by induction. The claim also holds for n = 0 and for negative integers n, using the 57 fact that if m is positive, then v−m = qm −rm √ d (since (qm + rm √ d)(qm −rm √ d) = q2 m −dr2 m = 1) while U−m = qm drm rm qm −1 = qm −drm −rm qm by properties of matrix inverses. Section 9.3. Convergence of Continued Fractions. (1) (a) ⟨−3, 1, 5, 3⟩= −3, 1, 5 + 1 3 = −3, 1 + 3 16 = −3 + 16 19 = −41 19. (b) ⟨0, 2, 2, 2, 4⟩= 0, 2, 2, 2 + 1 4 = 0, 2, 2 + 4 9 = 0, 2 + 9 22 = 0 + 22 53 = 22 53. (c) ⟨2, 1, 3, 1, 4, 3⟩= 2, 1, 3, 1, 4 + 1 3 = 2, 1, 3, 1 + 3 13 = 2, 1, 3 + 13 16 = 2, 1 + 16 61 = 2 + 61 77 = 215 77 . (2) We calculate ri = mi ni for 0 ≤i ≤5 for each continued fraction using the following tables. (a) For ⟨2, 1, 2, 1, 2, 1, . . .⟩: i 0 1 2 3 4 5 q 2 1 2 1 2 1 m 2 3 8 11 30 41 n 1 1 3 4 11 15 (b) For ⟨3, 3, 3, 3, 3, 3, . . .⟩: i 0 1 2 3 4 5 q 3 3 3 3 3 3 m 3 10 33 109 360 1189 n 1 3 10 33 109 360 (c) For ⟨1, 2, 3, 1, 2, 3, . . .⟩: i 0 1 2 3 4 5 q 1 2 3 1 2 3 m 1 3 10 13 36 121 n 1 2 7 9 25 84 (3) (a) ⟨1, 2, w⟩= 1, 2 + 1 w = 1 + w 2w+1 = 3w+1 2w+1. (b) ⟨2, 3, 5, w⟩= 2, 3, 5 + 1 w = D 2, 3 + w 5w+1 E = 2 + 5w+1 16w+3 = 37w+7 16w+3. (c) ⟨2, 1, 3, 1, w⟩= 2, 1, 3, 1 + 1 w = D 2, 1, 3 + w w+1 E = D 2, 1 + w+1 4w+3 E = 2 + 4w+3 5w+4 = 14w+11 5w+4 . (4) (a) ⟨1, 2⟩= 1 + 1 2 = 3 2 while ⟨1, 3⟩= 1 + 1 3 = 4 3. So ⟨1, 2⟩is larger than ⟨1, 3⟩. (b) ⟨1, 2, 3⟩= 1, 2 + 1 3 = 1 + 3 7 = 10 7 while ⟨1, 2, 4⟩= 1, 2 + 1 4 = 1 + 4 9 = 13 9 . Here we ﬁnd that ⟨1, 2, 4⟩is larger that ⟨1, 2, 3⟩. (c) ⟨1, 2⟩= 3 2 is larger than ⟨1, 2, 3⟩= 10 7 . (d) ⟨1, 1, 1, 1, 1, 2⟩= 1, 1, 1, 1, 3 2 = 1, 1, 1, 5 3 = 1, 1, 8 5 = 1, 13 8 = 21 13 ≈1.615 while ⟨1, 1, 1, 1, 1, 1, 2⟩= 1, 1, 1, 1, 1, 3 2 = 1, 1, 1, 1, 5 3 = 1, 1, 1, 8 5 = 1, 1, 13 8 = 1, 21 13 = 34 21 ≈1.619. So ⟨1, 1, 1, 1, 1, 1, 2⟩is larger than ⟨1, 1, 1, 1, 1, 2⟩. (5) Let x and y be real numbers with x > y ≥1, let q0, q1, . . ., qi be integers with qj > 0 if j > 0, and let v, w, and z be as follows: v = ⟨q0, q1, . . ., qi⟩, w = ⟨q0, q1, . . ., qi, x⟩, and z = ⟨q0, q1, . . ., qi, y⟩. If i = 0, then v = q0, w = q0 + 1 x, and z = q0 + 1 y. Since 0 < 1 x < 1 y ≤1 if x > y ≥1, then v < w < z. Now if i = 1, we have that v = ⟨q0, q1⟩, w = q0, q1 + 1 x , and z = D q0, q1 + 1 y E . 58 Now 1 ≤q1 < q1 + 1 x < q1 + 1 y, so from the case of i = 0, it follows that v > w > z. Continuing in this way, we see inductively that v < w < z if i is even and v > w > z if i is odd. Section 9.4. Continued Fraction Expansions of Real Numbers. (1) The statement ⟨q0, q1, q2, . . .⟩= ⟨q0, ⟨q1, q2, . . .⟩⟩is true by Lemma 9.4.3. Suppose that we have established that ⟨q0, q1, q2, . . .⟩= ⟨q0, q1, . . ., qk, ⟨qk+1, qk+2, . . .⟩⟩for some k ≥0. Let w = ⟨qk+2, qk+3, . . .⟩. Then again applying Lemma 9.4.3 and the deﬁnition of continued fractions, we have ⟨q0, q1, q2, . . .⟩= ⟨q0, q1, . . ., qk, ⟨qk+1, qk+2, . . .⟩⟩ = ⟨q0, q1, . . ., qk, ⟨qk+1, w⟩⟩ = q0, q1, . . ., qk, qk+1 + 1 w = ⟨q0, q1, . . ., qk, qk+1, w⟩ = ⟨q0, q1, . . ., qk, qk+1, ⟨qk+2, qk+3, . . .⟩⟩ The result follows for all k ≥0 by induction. (2) In each part, we apply calculations from Exercise 9.3.3. (a) If v = ⟨1, 2, v⟩, then v = 3v+1 2v+1. So then v is a solution of (2v2 + v) −(3v + 1) = 2v2 −2v −1 = 0. Since v > 1, we conclude that v = 2+ √ 12 4 = 1+ √ 3 2 . (b) If v = ⟨2, 3, 5, v⟩, then v = 37v+7 16v+3. So v > 2 satisﬁes (16v2 + 3v) −(37v + 7) = 16v2 −34v −7 = 0. We conclude that v = 17+ √ 401 16 . (c) If v = ⟨2, 1, 3, 1, v⟩, then v = 14v+11 5v+4 . So v > 2 satisﬁes (5v2 + 4v) −(14v + 11) = 5v2 −10v −11 = 0, and we conclude that v = 5+4 √ 5 5 . (3) Suppose that ⟨q0, q1, . . ., qk⟩= ⟨r0, r1, . . ., rk, rk+1, . . .⟩with each qi and ri a positive in-teger for i ≥1. By the same argument as in the proof of Theorem 9.4.4, we obtain the equation qk = ⟨rk, rk+1, . . .⟩= rk + 1 w, where w = ⟨rk+1, rk+2, . . .⟩. But since w is a real number with w > rk+1 ≥1, we know that 0 < 1 w < 1. Thus the equation 1 w = qk −rk is impossible. (4) If n is an integer and v = ⟨q0, q1, q2, . . .⟩, then v + n = ⟨n + q0, q1, q2, . . .⟩. Here if w = ⟨q1, q2, . . .⟩, then v = ⟨q0, w⟩= q0 + 1 w. But then v + n = (n + q0) + 1 w = ⟨n + q0, w⟩= ⟨n + q0, q1, q2, . . .⟩, using the deﬁnition of continued fractions and Lemma 9.4.3. (5) If v = ⟨q0, q1, q2, . . .⟩is larger than 1, then 1 < v < q0 + 1, so that q0 is positive. Thus ⟨0, q0, q1, q2, . . .⟩is a continued fraction for a real number w, and w = ⟨0, ⟨q0, q1, q2, . . .⟩⟩= ⟨0, v⟩= 0 + 1 v = 1 v, using Lemma 9.4.3. (6) Let v = ⟨q0, q1, q2, . . .⟩be larger than 1, so that 1 v = ⟨0, q0, q1, q2, . . .⟩. If ri is the i-th convergent of v, then ri = ⟨q0, q1, . . ., qi⟩by Corollary 9.3.3. Similarly, if si is the i-th convergent of 1 v, then si+1 = ⟨0, q0, q1, . . ., qi⟩= ⟨0, ⟨q0, q1, . . ., qi⟩⟩= ⟨0, ri⟩= 1 ri by Lemma 9.4.3 and the deﬁnition of continued fractions. Section 9.5. Purely Periodic Continued Fractions. (1) (a) Let v = 2, 5 = ⟨2, 5, v⟩= 2, 5 + 1 v = 2 + v 5v+1 = 11v+2 5v+1 . Then v > 2 satisﬁes the equation (5v2+v)−(11v+2) = 5v2−10v−2 = 0, so that v = 10+ √ 140 10 = 5+ √ 35 5 ≈2.18. Here v = 5− √ 35 5 ≈−0.18, conﬁrming that v is reduced. 59 (b) For v = 2, 1, 2 = 2, 1, 2 + 1 v = D 2, 1 + v 2v+1 E = 2 + 2v+1 3v+1 = 8v+3 3v+1, we ﬁnd that v satisﬁes (3v2 + v) −(8v + 3) = 3v2 −7v −3 = 0. So v = 7+ √ 85 6 ≈2.70 and v = 7− √ 85 6 ≈−0.37. (c) For v = 4, 3, 1, 3 = 4, 3, 1, 3 + 1 v = D 4, 3, 1 + v 3v+1 E = D 4, 3 + 3v+1 4v+1 E = 4+ 4v+1 15v+4 = 64v+17 15v+4 , we see that v satisﬁes 15x2 −60v −17 = 0. So v = 30+ √ 1155 15 ≈4.27 and v = 30− √ 1155 15 ≈−0.27. (2) (a) To ﬁnd reduced quadratic numbers v = −b+ √ b2−4ac 2a with ∆= b2 −4ac = 5, we must have b an odd integer with − √ 5 < b < 0, that is, b = −1. Now −1+ √ 5 2 < a < 1+ √ 5 2 forces a = 1, and b2 −4ac = 5 implies that c = −1. The only reduced quadratic number of discriminant ∆= 5 is the larger root of ax2 + bx + c = x2 −x −1, that is, v = 1+ √ 5 2 . (To verify that v is reduced, note that v ≈1.62 and v = 1− √ 5 2 ≈−0.62.) (b) For ∆= 12, we follow the same approach as in part (a), compiling the results in the following table. a b c v v 1 −2 −2 1 + √ 3 ≈2.73 1 − √ 3 ≈−0.73 2 −2 −1 1+ √ 3 2 ≈1.37 1− √ 3 2 ≈−0.37 (c) For ∆= 21: a b c v v 1 −3 −3 3+ √ 21 2 ≈3.79 3− √ 21 2 ≈−0.79 3 −3 −1 3+ √ 21 6 ≈1.26 3− √ 21 6 ≈−0.26 (d) For ∆= 76: a b c v v 1 −8 −3 4 + √ 19 ≈8.36 4 − √ 19 ≈−0.36 2 −6 −5 3+ √ 19 2 ≈3.68 3− √ 19 2 ≈−0.68 3 −4 −5 2+ √ 19 3 ≈2.12 2− √ 19 3 ≈−0.79 3 −8 −1 4+ √ 19 3 ≈2.79 4− √ 19 3 ≈−0.12 5 −4 −3 2+ √ 19 5 ≈1.27 2− √ 19 5 ≈−0.47 5 −6 −2 3+ √ 19 5 ≈1.47 3− √ 19 5 ≈−0.27 60 (e) For ∆= 145: a b c v v 1 −11 −6 11+ √ 145 2 ≈11.52 11− √ 145 2 ≈−0.52 2 −11 −3 11+ √ 145 4 ≈5.76 11− √ 145 4 ≈−0.26 2 −9 −8 9+ √ 145 4 ≈5.26 9− √ 145 4 ≈−0.76 3 −11 −2 11+ √ 145 6 ≈3.84 11− √ 145 6 ≈−0.17 3 −7 −8 7+ √ 145 6 ≈3.17 7− √ 145 6 ≈−0.84 4 −9 −4 9+ √ 145 8 ≈2.63 9− √ 145 8 ≈−0.38 4 −7 −6 7+ √ 145 8 ≈2.38 7− √ 145 8 ≈−0.63 5 −5 −6 5+ √ 145 10 ≈1.70 5− √ 145 10 ≈−0.70 6 −11 −1 11+ √ 145 12 ≈1.92 11− √ 145 12 ≈−0.09 6 −7 −4 7+ √ 145 12 ≈1.59 7− √ 145 12 ≈−0.42 6 −5 −5 5+ √ 145 12 ≈1.42 5− √ 145 12 ≈−0.59 6 −1 −6 1+ √ 145 12 ≈1.09 1− √ 145 12 ≈−0.92 8 −9 −2 9+ √ 145 16 ≈1.32 9− √ 145 16 ≈−0.19 8 −7 −3 7+ √ 145 16 ≈1.19 7− √ 145 16 ≈−0.32 (3) For each v, we write v →q w if q = ⌊v⌋and w = (v −q)−1, and follow this process until we return to the original v. These cycles produce the continued fraction expansion of each value that is obtained in the cycle. (a) For ∆= 5, 1+ √ 5 2 →1 1+ √ 5 2 (that is, 1+ √ 5 2 −1 −1 = 1+ √ 5 2 ) and thus 1+ √ 5 2 = 1. (b) For ∆= 12, 1 + √ 3 →2 1+ √ 3 2 →1 1 + √ 3. We ﬁnd that 1 + √ 3 = 2, 1 and 1+ √ 3 2 = 1, 2 . (c) For ∆= 21, 3+ √ 21 2 →3 3+ √ 21 6 →1 3+ √ 21 2 , with 3+ √ 21 2 = 3, 1 and 3+ √ 21 2 = 1, 3 . (d) For ∆= 76, 4 + √ 19 →8 4+ √ 19 3 →2 2+ √ 19 5 →1 3+ √ 19 2 →3 3+ √ 19 5 →1 2+ √ 19 3 →2 4 + √ 19. Starting this cycle at diﬀerent numbers, we obtain the following continued fraction expansions: 4 + √ 19 = 8, 2, 1, 3, 1, 2 , 4+ √ 19 3 = 2, 1, 3, 1, 2, 8 , 2+ √ 19 5 = 1, 3, 1, 2, 8, 2 , 3+ √ 19 2 = 3, 1, 2, 8, 2, 1 , 3+ √ 19 5 = 1, 2, 8, 2, 1, 3 , 2+ √ 19 3 = 2, 8, 2, 1, 3, 1 . (e) For ∆= 145, we ﬁnd four distinct cycles, which we combine with the continued fraction expansions of each number as follows. 11+ √ 145 2 = 11, 1, 1 →11 11+ √ 145 12 = 1, 1, 11 →1 1+ √ 145 12 = 1, 11, 1 . (This cycle returns to the original number at the next step, as do the following cycles.) 11+ √ 145 4 = 5, 1, 3 →5 9+ √ 145 16 = 1, 3, 5 →1 7+ √ 145 6 = 3, 5, 1 . 11+ √ 145 6 = 3, 1, 5 →3 7+ √ 145 16 = 1, 5, 3 →1 9+ √ 145 4 = 5, 3, 1 . 9+ √ 145 8 = 2, 1, 1, 1, 2 →2 7+ √ 145 12 = 1, 1, 1, 2, 2 →1 5+ √ 145 10 = 1, 1, 2, 2, 1 →1 5+ √ 145 12 = 1, 2, 2, 1, 1 →1 7+ √ 145 8 = 2, 2, 1, 1, 1 . 61 Section 9.6. Continued Fractions of Irrational Quadratic Numbers. (1) (a) Let v = 1, 1, 2 = ⟨1, 1, 2, v⟩= 1, 1, 2 + 1 v = D 1, 1 + v 2v+1 E = 1 + 2v+1 3v+1 = 5v+2 3v+1. Then v > 1 is a solution of (3v2 + v) −(5v + 2) = 3v2 −4v −2 = 0, so that v = 4+ √ 40 6 = 2+ √ 10 3 . (b) If v = 4, 3, 1, 1, 2 , then v = ⟨4, 3, w⟩= 4, 3 + 1 w = 4 + w 3w+1 = 13w + 43w + 1, where w = 2+ √ 10 3 from part (a). We ﬁnd that v = 38+13 √ 10 3 · 1 3+ √ 10 = 38+13 √ 10 3 · (−3 + √ 10) = 16− √ 10 3 . (c) If w = 2, 5 = ⟨2, 5, w⟩= 2, 5 + 1 w = 2 + w 5w+1 = 11w+2 5w+1 , we ﬁnd that w is the larger root of 5x2 −10x −2, that is, w = 10+ √ 140 10 = 5+ √ 35 5 . Then v = 3, 7, 6, 2, 5 = ⟨3, 7, 6, w⟩= 135w+22 43w+7 , from which we calculate that v = 1385+ √ 35 443 . (d) If w = 3, 1, 5 = ⟨3, 1, 5, w⟩= 23w+4 6w+1 , then w is the larger root of 6x2 −22x −4, that is, w = 11+ √ 145 6 . Then v = 2, 1, 3, 1, 5 = ⟨2, 1, w⟩= 3w+2 w+1 , from which we calculate that v = 55+ √ 145 24 . (e) If w = 1, 2, 3 , we ﬁnd that w is the larger root of 7x2 −8x −3, that is, w = 4+ √ 37 7 . Then v = 1, 2, 1, 1, 2, 3 = ⟨1, 2, 1, w⟩= 4w+3 3w+2, from which we calculate that v = 74− √ 37 49 . (2) (a) If v = 1+ √ 5 2 , so that v = 1− √ 5 2 , then v +v = 1 and v · v = −1. Thus f(x) = x2 −x −1 is the minimum polynomial of v, and ∆= 5 is the discriminant of v. We ﬁnd that (v −⌊v⌋)−1 = 2 −1+ √ 5 = 1+ √ 5 2 = v, and conclude that v = 1 . (b) If v = 11+5 √ 7 3 and v = 11−5 √ 7 3 , then v + v = 22 3 and v · v = −6. Thus v has minimum polynomial f(x) = 3x2 −22x −18 and discriminant ∆= (−22)2 −4 · 3 · −18 = 700. Using the notation introduced in Exercise 3 of §9.5, we have 11+5 √ 7 3 →8 13+5 √ 7 2 →13 13+5 √ 7 3 →8 11+5 √ 7 18 →1 7+5 √ 7 7 →2 7+5 √ 7 18 →1 11+5 √ 7 3 , which implies that v = 8, 13, 8, 1, 2, 1 . (c) If v = 7− √ 11 4 and v = 7+ √ 11 4 , then v + v = 7 2 and v · v = 19 8 . Thus the minimum polynomial of v is f(x) = 8x2−28x+19 and v has discriminant ∆= (−28)2−4·8·19 = 176. Here we ﬁnd 7− √ 11 4 →0 14+2 √ 11 19 →1 5 + 2 √ 11 →11 3+ √ 11 4 →1 2+2 √ 11 5 →1 3+2 √ 11 7 →1 2+ √ 11 2 →2 4+2 √ 11 7 →1 3+2 √ 11 5 →1 1+ √ 11 4 →1 6 + 2 √ 11 →12 3+ √ 11 4 , and conclude that v = 0, 1, 11, 1, 1, 1, 2, 1, 1, 1, 12 . (3) (a) If d = t2 +1, then j√ d k = t and 1 √ d−t = √ d+t. Then j√ d + t k = 2t and 1 ( √ d+t)−2t = 1 √ d−t = √ d + t. The continued fraction algorithm now repeats this value indeﬁnitely, and so √ d = t, 2t. (Using the notation of the preceding exercise, the sequence √ d →t ( √ d + t) →2t ( √ d + t) produces this periodic continued fraction.) (b) If d = t2 −1 for some t > 1, then j√ d k = t −1 and 1 √ d−(t−1) = √ d+(t−1) 2t−2 , here using the fact that d −(t −1)2 = (t2 −1) −(t2 −2t + 1) = 2t −2. Now since t −1 < √ d < t, we ﬁnd that j√ d+(t−1) 2t−2 k = 1 and √ d+(t−1) 2t−2 −1 = √ d−(t−1) 2t−2 . Then 62 2t−2 √ d−(t−1) = (2t−2)( √ d+(t−1)) 2t−2 = √ d + (t −1). But now j√ d + (t −1) k = 2t −2, with ( √ d + (t −1)) −(2t −2) = √ d −(t −1) and we are back to a previous calculation. In other words, we obtain the sequence √ d →(t−1) √ d+(t−1) 2t−2 →1 √ d + (t −1) →(2t−2) √ d+(t−1) 2t−2 , which produces the continued fraction √ d = t −1, 1, 2t −2 . (c) If d = t2 + 2, then, following the same method as in parts (a) and (b), we obtain the sequence √ d →t √ d+t 2 →t √ d + t →2t √ d+t 2 , which produces the continued fraction √ d = t, t, 2t . (d) If d = t2 + t, then we obtain the sequence √ d →t √ d+t t →2 √ d + t →2t √ d+t t , which produces the continued fraction √ d = t, 2, 2t . (4) Let ni be the denominator of the i-th convergent of the continued fraction ⟨q0, q1, q2, . . .⟩. Since n−2 = 1 and n−1 = 0, we have that n0 = q0(0) + 1 = 1 and n1 = q1(1) + 0 = q1. Thus ⟨q1⟩= q1 = n1 n0 . Suppose now that ⟨qk, . . ., q1⟩= nk nk−1 for some k ≥1. Then ⟨qk+1, qk, . . ., q1⟩= ⟨qk+1, ⟨qk, . . ., q1⟩⟩= qk+1 + nk−1 nk = qk+1nk+nk−1 nk = nk+1 nk , which proves that ⟨qi, . . ., q1⟩= ni ni−1 for all i ≥1 by induction. (5) v = ⟨1, 1, 3, 1, v⟩= 1, 1, 3, v+1 v = D 1, 1, 4v+3 v+1 E = D 1, 5v+4 4v+3 E = 9v+7 5v+4. Thus v is the larger root of f(x) = 5x2 −5x −7, that is, v = 5+ √ 165 10 . We know that v is semi-reduced by Proposition 9.6.4, and then v is palindromic since v + v = 5+ √ 165 10 + 5− √ 165 10 = 1. (6) v = ⟨1, 2, 1, v⟩= 1, 2, v+1 v = D 1, 3v+2 v+1 E = 4v+3 3v+2, and so v is the larger root of f(x) = 3x2 −2x −3, that is, v = 1+ √ 10 3 . Since v + v = 1+ √ 10 3 + 1− √ 10 3 = 2 3 is not an integer, then v is not palindromic. (7) Since v is reduced, then v is semi-reduced, and we have that v = ⟨q0, q1, . . ., qℓ−1⟩= ⟨q0, q1, . . ., qℓ−1, q0⟩. As noted following Proposition 9.6.7, v is then palindromic if and only if the sequence q1, . . ., qℓ−1 is a palindrome. (8) Suppose that v is a semi-reduced number with minimum polynomial f(x) = ax2 + bx + c. Then v = −b+ √ b2−4ac 2a and v = −b− √ b2−4ac 2a , using the fact that a semi-reduced number is the larger root of its minimum polynomial. Then v + v = −b a is an integer if and only if a divides b. We know that g = ⌊−v⌋is the unique integer so that g + v is reduced. But g = ⌊−v⌋= j b+ √ b2−4ac 2a k = j −b+ √ b2−4ac 2a + b a k = v + b a = ⌊v⌋+ b a, if b a is an integer. (9) Let v = 3+ √ 30 3 , the larger root of f(x) = 3x2 −6x −7. From the sequence 3+ √ 30 3 →2 3+ √ 30 7 →1 4+ √ 30 2 →4 4+ √ 30 7 →1 3+ √ 30 3 , we ﬁnd that v = 2, 1, 4, 1 . As noted in Exercise 7, this number is semi-reduced and palindromic. (We can verify this directly by noting that v + ⌊−v⌋> 1 and that v + v is an integer.) 63 (10) If v = ⟨−2, w⟩where w = ⟨1, 2, 3, 2, 1, 4, w⟩, we can calculate directly that v = −44+ √ 880 11 . Here ⌊−v⌋= j 44+ √ 880 11 k = 6, so that v + ⌊−v⌋= 22+ √ 880 11 > 1, and v + v = −8 is an integer. Thus v is both semi-reduced and palindromic. Section 10.1. Class Groups of Indeﬁnite Quadratic Forms. (1) For convenience, we combine the calculation of candidate forms and of equivalence classes of forms and ideals from Exercises 1 and 2 in each part below. (2) (a) If ∆= 17, then φ(x) = x2 + x −4 and u∆= 2. From the following table, x −2, 1 −1, 0 φ(x) −2 −4 we determine that (±1 : −2), (±2 : −2), and (±2 : −1) are candidate forms of discriminant 17. The equivalence algorithm of Theorem 10.1.2 applied to (1 : −2) produces the following data. i 0 1 2 3 a 1 2 2 1 k −2 −2 −1 −2 So (1 : −2) ∼(−2 : −2) ∼(2 : −1) ∼(−1 : −2) ∼(2 : −2) ∼(−2 : −1). All candidate forms are equivalent, and there is precisely one class of quadratic forms of discriminant 17. The same is true for ideals of discriminant 17. (b) If ∆= 28, then φ(x) = x2 −7 and u∆= 2. From the table, x ±2 ±1 0 φ(x) −3 −6 −7 we ﬁnd the candidate forms (±1 : −2) and (±2 : −1). Applying the equivalence algorithm to (1 : −2), i 0 1 2 3 4 a 1 3 2 3 1 k −2 −2 −1 −1 −2 we determine that (1 : −2) ∼(2 : −1) and (−1 : −2) ∼(−2 : −1), but that these two classes of quadratic forms are distinct. However, since [1 : −2] = [−1 : −2], there is only one class of ideals of discriminant 28. (c) If ∆= 33, then φ(x) = x2 + x −8 and u∆= 2. x −3, 2 −2, 1 −1, 0 φ(x) −2 −6 −8 The candidate forms are (±1 : −3), (±2 : −3), and (±2 : −2). Applying the equiva-lence algorithm to (1 : −3), i 0 1 2 3 4 a 1 2 3 2 1 k −3 −3 −2 −2 −3 we determine that (1 : −3) ∼(−2 : −3) ∼(−2 : −2) and (−1 : −3) ∼(2 : −3) ∼ (2 : −2), with these two classes of quadratic forms distinct. There is only one class of ideals of discriminant 33. 64 (d) If ∆= 37, then φ(x) = x2 + x −9 and u∆= 3. x −3, 2 −2, 1 −1, 0 φ(x) −3 −7 −9 The candidate forms are (±1 : −3), (±3 : −3), and (±3 : −1). Applying the equiva-lence algorithm to (1 : −3), i 0 1 2 3 a 1 3 3 1 k −3 −3 −1 −3 we determine that all of these candidate forms are equivalent. There is only one class of quadratic forms, and one class of ideals of discriminant 37. (e) If ∆= 57, then φ(x) = x2 + x −14 and u∆= 3. x −4, 3 −3, 2 −2, 1 −1, 0 φ(x) −2 −8 −12 −14 The candidate forms are (±1 : −4), (±2 : −4), (±2 : −3), and (±3 : −2). Applying the equivalence algorithm to (1 : −4), i 0 1 2 3 4 5 6 a 1 2 4 3 4 2 1 k −4 −4 −3 −2 −2 −3 −4 we determine that (1 : −4) ∼(−2 : −4) ∼(−3 : −2) ∼(−2 : −3) and that (−1 : −4) ∼(2 : −4) ∼(3 : −2) ∼(2 : −3). There are two classes of quadratic forms of discriminant 57, but only one class of ideals. (f) If ∆= 65, then φ(x) = x2 + x −16 and u∆= 4. x −4, 3 −3, 2 −2, 1 −1, 0 φ(x) −4 −10 −14 −16 The candidate forms are (±1 : −4), (±2 : −4), (±2 : −3), (±4 : −4), and (±4 : −1). Applying the equivalence algorithm to (1 : −4), i 0 1 2 3 a 1 4 4 1 k −4 −4 −1 −4 we ﬁnd that (1 : −4) ∼(−4 : −4) ∼(4 : −1) ∼(−1 : −4) ∼(4 : −4) ∼(−4 : −1). Applying the same algorithm to (2 : −4), i 0 1 2 3 a 2 5 2 2 k −4 −3 −3 −4 we (2 : −4) ∼(2 : −3) ∼(−2 : −4) ∼(−2 : −3). Thus there are two distinct classes of quadratic forms of discriminant 65, and likewise two classes of ideals. (g) If ∆= 88, then φ(x) = x2 −22 and u∆= 4. x ±4 ±3 ±2 ±1 0 φ(x) −6 −13 −18 −21 −22 65 The candidate forms are (±1 : −4), (±2 : −4), (±3 : −4), and (±3 : −2). Applying the equivalence algorithm to (1 : −4), i 0 1 2 3 4 5 6 a 1 6 3 2 3 6 1 k −4 −4 −2 −4 −4 −2 −4 we determine that (1 : −4) ∼(3 : −2) ∼(−2 : −4) ∼(3 : −4) and that (−1 : −4) ∼ (−3 : −2) ∼(2 : −4) ∼(−3 : −4). There are two classes of quadratic forms of discriminant 88 and one class of ideals. (h) If ∆= 93, then φ(x) = x2 + x −23 and u∆= 4. x −5, 4 −4, 3 −3, 2 −2, 1 −1, 0 φ(x) −3 −11 −17 −21 −23 The only candidate forms are (±1 : −5) and (±3 : −5). From the equivalence algorithm applied to (1 : −5), i 0 1 2 a 1 3 1 k −5 −5 −5 we see that (1 : −5) ∼(−3 : −5) and (−1 : −5) ∼(3 : −5). There are two classes of quadratic forms of discriminant 93 and one class of ideals. (i) If ∆= 104, then φ(x) = x2 −26 and u∆= 5. x ±5 ±4 ±3 ±2 ±1 0 φ(x) −1 −10 −17 −22 −25 −26 The candidate forms are (±1 : −5), (±2 : −4), (±5 : −4), and (±5 : −1). Applying the equivalence algorithm to (1 : −5) and to (2 : −4), i 0 1 a 1 1 k −5 −5 i 0 1 2 3 a 2 5 5 2 k −4 −4 −1 −4 we ﬁnd that (1 : −5) ∼(−1 : −5) and that (2 : −4) ∼(−5 : −4) ∼(5 : −1) ∼ (−2 : −4) ∼(5 : −4) ∼(−5 : −1). There are two classes of quadratic forms of discriminant 104 and two classes of ideals. (j) If ∆= 152, then φ(x) = x2 −38 and u∆= 6. x ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −2 −13 −22 −29 −34 −37 −38 The only candidate forms are (±1 : −6) and (±2 : −6). Applying the equivalence algorithm to (1 : −6), i 0 1 2 a 1 2 1 k −6 −6 −6 we see that (1 : −6) ∼(−2 : −6) and that (−1 : −6) ∼(2 : −6). There are two classes of quadratic forms of discriminant 152 and one class of ideals. 66 Section 10.2. Genera of Quadratic Forms and Ideals. (1) (a) If ∆= 37, then φ(x) = x2 + x −9 and u∆= 3. The only genus symbol deﬁned for quadratic forms of discriminant 37 is f 37 , which must equal 1. From the table x −3, 2 −2, 1 −1, 0 φ(x) −3 −7 −9 we ﬁnd the candidate forms (±1 : −3), (±3 : −3), and (±3 : −1). Applying the equivalence algorithm to (1 : −3), i 0 1 2 3 a 1 3 3 1 k −3 −3 −1 −3 we ﬁnd that all of these forms are equivalent. So F37 has invariant factor type (1), containing only one class and only one genus. The same is true for the ideal class group C37. (b) If ∆= 40, then φ(x) = x2 −10 and u∆= 3. The deﬁned genus symbols are 2 f and f 5 , whose product equals 1. x ±3 ±2 ±1 0 φ(x) −1 −6 −9 −10 Candidate forms are (±1 : −3), (±2 : −2), (±3 : −2), and (±3 : −1). Applying the equivalence algorithm to (1 : −3) and to (2 : −2) i 0 1 a 1 1 k −3 −3 i 0 1 2 3 a 2 3 3 2 k −2 −2 −1 −2 we ﬁnd that (1 : −3) ∼(−1 : −3), with both forms in the genus for which 2 f = 1 = f 5 , and that (2 : −2) ∼(−3 : −2) ∼(3 : −1) ∼(−2 : −2) ∼(3 : −2) ∼(−3 : −1), with each of these forms in the genus for which 2 f = −1 = f 5 . So F40 has invariant factor type (2), with two classes in two genera. The same is true for the ideal class group C40. (c) If ∆= 61, then φ(x) = x2 + x −15 and u∆= 3. The only genus symbol deﬁned for quadratic forms of discriminant 61 is f 61 . From the table x −4, 3 −3, 2 −2, 1 −1, 0 φ(x) −3 −9 −13 −15 we ﬁnd the candidate forms (±1 : −4), (±3 : −4), and (±3 : −3). Applying the equivalence algorithm to (1 : −4), i 0 1 2 3 a 1 3 3 1 k −4 −4 −3 −4 we ﬁnd that all of these forms are equivalent. So F61 has invariant factor type (1), containing only one class and only one genus. The same is true for the ideal class group C61. 67 (d) If ∆= 92, then φ(x) = x2 −23 and u∆= 4. The deﬁned genus symbols are −1 f and f 23 . x ±4 ±3 ±2 ±1 0 φ(x) −7 −14 −19 −22 −23 Candidate forms are (±1 : −4) and (±2 : −3). Applying the equivalence algorithm to (1 : −4) i 0 1 2 3 4 a 1 7 2 7 1 k −4 −4 −3 −3 −4 we ﬁnd that (1 : −4) ∼(2 : −3) and (−1 : −4) ∼(−2 : −3). The class of (1 : −4) is in the genus for which −1 f = 1 = f 23 , and the class of (−1 : −4) is in the genus for which −1 f = −1 = f 23 . So F92 has invariant factor type (2), with two classes in two genera. On the other hand, since [1 : −4] = (−1 : −4), the ideal class group has only one element. (e) If ∆= 93, then the deﬁned genus symbols are f 3 and f 31 . In part (h) of Exercise 10.1.2, we found that F93 has two elements, represented by (1 : −5) and (−1 : −5). Both genus symbols equal 1 for the ﬁrst form, and both equal −1 for the second. So F93 has invariant factor type (2) with two forms in two genera. But since [1 : −5] = [−1 : −5], the ideal class group C93 has only one element. (f) If ∆= 105, then φ(x) = x2 + x −26 and u∆= 5. The deﬁned genus symbols are f 3 , f 5 , and f 7 , whose product equals 1. From the table x −5, 4 −4, 3 −3, 2 −2, 1 −1, 0 φ(x) −6 −14 −20 −24 −26 and the equivalence algorithm applied to (1 : −5) and (2 : −5) i 0 1 2 3 4 5 6 a 1 6 4 5 4 6 1 k −5 −5 −2 −3 −3 −2 −5 i 0 1 2 3 4 a 2 7 2 3 2 k −5 −4 −4 −5 −5 we determine that there are precisely four classes of quadratic forms, each in a separate genus, with symbols f 3 , f 5 , and f 7 as follows. + + + : (1 : −5), −+ −: (−1 : −5), −−+ : (2 : −5), + −−: (−2 : −5). (For example, (1 : −5) ∼(4 : −2) ∼(−5 : −3) ∼(4 : −3), listing only candidate forms here.) The form class group F105 has invariant factor type (2, 2). On the other hand, since [1 : −5] = [−1 : −5] and [2 : −5] = [−2 : −5], the ideal class group C105 has only two elements and has invariant factor type (2). (g) If ∆= 156, then φ(x) = x2 −39 and u∆= 6. The deﬁned genus symbols are −1 f , f 3 , and f 13 , written in that order below. From the table x ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −3 −14 −23 −30 −35 −38 −39 and the equivalence algorithm applied to (1 : −6) and (2 : −5) i 0 1 2 a 1 3 1 k −6 −6 −6 i 0 1 2 3 4 5 6 a 2 7 5 6 5 7 2 k −5 −5 −2 −3 −3 −2 −5 68 we ﬁnd four classes of quadratic forms, each in a separate genus as follows. + + + : (1 : −6), −−+ : (−1 : −6), + −−: (2 : −5), −+ −: (−2 : −5). The form class group F156 has invariant factor type (2, 2), while the ideal class group C156 has invariant factor type (2). (h) If ∆= 328, then φ(x) = x2 −82 and u∆= 9, with genus symbols 2 f and f 41 . x ±9 ±8 ±7 ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −1 −18 −33 −46 −57 −66 −73 −78 −81 −82 From the following tables for the equivalence algorithm, i 0 1 a 1 1 k −9 −9 i 0 1 2 3 a 2 9 9 2 k −8 −8 −1 −8 i 0 1 2 3 a 3 11 6 3 k −8 −7 −4 −8 i 0 1 2 3 a 3 6 11 3 k −7 −8 −4 −7 we ﬁnd four classes of quadratic forms in two distinct genera. ++ : (1 : −9), (2 : −8) −−: (3 : −8), (3 : −7). The form class group F328 and the ideal class group C328 both have invariant factor type (4). (For example, if A = [3 : −8], then A2 = [9 : −8] ∼[2 : −8] as we can see from the equivalence algorithm applied to (2 : −8). Then A3 ∼[6 : −8] ∼[3 : −7] and A4 ∼[1 : −9].) (i) If ∆= 440, then φ(x) = x2 −110 and u∆= 10, with genus symbols −2 f , f 5 , and f 11 (in that order below). From the following table, x ±10 ±9 ±8 ±7 ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −10 −29 −46 −61 −74 −85 −94 −101 −106 −109 −110 we ﬁnd candidate forms (±1 : −10), (±2 : −10), (±5 : −10), and (±10 : −10). The equivalence algorithm applied to (1 : −10) and (2 : −10), i 0 1 2 a 1 10 1 k −10 −10 −10 i 0 1 2 a 2 5 2 k −10 −10 −10 shows that there are four classes of quadratic forms, each in a separate genus as follows. + + + : (1 : −10), −+ −: (−1 : −10), + −−: (2 : −10), −−+ : (−2 : −10). The form class group F440 has invariant factor type (2, 2), while the ideal class group C440 has invariant factor type (2). (j) If ∆= 568, then φ(x) = x2 −142 and u∆= 11, with genus symbols −2 f and f 71 . x ±11 ±10 ±9 ±8 ±7 ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −21 −42 −61 −78 −93 −106 −117 −126 −133 −138 −141 −142 From the following tables for the equivalence algorithm, i 0 1 2 3 4 a 1 21 2 21 1 k −11 −11 −10 −10 −11 69 i 0 1 2 3 4 5 6 a 3 14 9 13 6 7 3 k −11 −10 −4 −5 −8 −10 −11 i 0 1 2 3 4 5 6 a 3 7 6 13 9 14 3 k −10 −11 −10 −8 −5 −4 −10 we ﬁnd six classes of quadratic forms in two distinct genera. ++ : (1 : −11), (3 : −11), (3 : −10) −−: (−1 : −11), (−3 : −11), (−3 : −10). The form class group F568 has invariant factor type (6), while the ideal class group C568 has invariant factor type (3). (Note that (−3 : −11) · (−3 : −11) ∼(9 : −5) ∼ (3 : −10), as shown in the third application of the equivalence algorithm above. Then (−3 : −11) · (3 : −10) ∼(−1 : −11) and so forth, showing that f = (−3 : −11) generates F568.) Section 10.3. Continued Fractions of Irrational Quadratic Numbers. (1) (a) v = √ 41 is the larger root of x2 −41, with discriminant ∆= 164. Since 41 ≡1 (mod 4),the principal polynomial of discriminant 164 is φ(x) = x2 + 2x −40, with smaller root w = −1 − √ 41 ≈−7.4. We can calculate the following values of φ(x). x −7, 5 −6, 4 −5, 3 −4, 2 −3, 1 −2, 0 −1 φ(x) −5 −16 −25 −32 −37 −40 −41 To ﬁnd the continued fraction of √ 41, we apply the quadratic continued fraction algorithm with a0 = 1 and k0 = b−ε 2 = −1. The smallest k1 > w that satisﬁes k1 ≡ −k0 −ε (mod a0) is k1 = −7, and then a1 = −φ(k1) a0 = 5. Now the smallest k2 > w with k2 ≡−k1−ε (mod a1) is k2 = −5, and a2 = −φ(k2) a1 = 5. Continuing in this way, we ﬁnd eventually that a4 = a1 and k4 = k1. Now q0 = −k0+k1+ε a0 = −−1−7+2 1 = 6, then q1 = −k1+k2+ε a1 = −−7−5+2 5 = 2 and so forth. The following table summarizes these calculations and shows that √ 41 = 6, 2, 2, 12 . i 0 1 2 3 4 a 1 5 5 1 5 k −1 −7 −5 −7 −7 q 6 2 2 12 (b) v = √ 43 has minimum polynomial x2 −43, which is also the principal polynomial of discriminant ∆= 4 · 43 = 172. The smaller root of φ(x) is w = − √ 43 ≈−6.6, and we calculate the following values of φ(x). x ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −7 −18 −27 −34 −39 −42 −43 Applying the quadratic continued fraction algorithm with a0 = 1 and k0 = 0, we obtain the following data. i 0 1 2 3 4 5 6 7 8 9 10 11 a 1 7 6 3 9 2 9 3 6 7 1 7 k 0 −6 −1 −5 −4 −5 −5 −4 −5 −1 −6 −6 q 6 1 1 3 1 5 1 3 1 1 12 Thus √ 43 = 6, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12. 70 (c) Since 53 ≡1 (mod 4), we apply the quadratic continued fraction algorithm to a0 = 1 and k0 = −1, with the following values of φ(x) = x2 + 2x −52. x −8, 6 −7, 5 −6, 4 −5, 3 −4, 2 −3, 1 −2, 0 −1 φ(x) −4 −17 −28 −37 −44 −49 −52 −53 From the following data, i 0 1 2 3 4 5 6 a 1 4 7 7 4 1 4 k −1 −8 −6 −3 −6 −8 −8 q 7 3 1 1 3 14 we conclude that √ 53 = 7, 3, 1, 1, 3, 14 . (d) For v = √ 89, with φ(x) = x2 + 2x −88 having smaller root w = −1 − √ 89 ≈−10.4, we obtain the following data from the quadratic continued fraction algorithm i 0 1 2 3 4 5 6 a 1 8 5 5 8 1 8 k −1 −10 −8 −9 −8 −10 −10 q 9 2 3 3 2 18 and conclude that √ 89 = 9, 2, 3, 3, 2, 18. (e) For v = √ 97, with φ(x) = x2 + 2x −96 having smaller root w = −1 − √ 97 ≈−10.8, the quadratic continued fraction algorithm i 0 1 2 3 4 5 6 7 8 9 10 11 12 a 1 16 3 11 8 9 9 8 11 3 16 1 16 k −1 −10 −8 −9 −4 −6 −5 −6 −4 −9 −8 −10 −10 q 9 1 5 1 1 1 1 1 1 5 1 18 shows that √ 97 = 9, 1, 5, 1, 1, 1, 1, 1, 1, 5, 1, 18 . (f) For v = √ 111, with φ(x) = x2 −111, we obtain the following data from the quadratic continued fraction algorithm i 0 1 2 3 4 5 6 7 a 1 11 10 3 10 11 1 11 k 0 −10 −1 −9 −9 −1 −10 −10 q 10 1 1 6 1 1 20 which shows that √ 111 = 10, 1, 1, 1, 6, 1, 1, 20 . (2) (a) v = 3+ √ 23 and v = 3− √ 23 are roots of the minimum polynomial f(x) = x2−6x−14, with discriminant ∆= 92 = ∆(23, 1). The principal polynomial of discriminant 92 is φ(x) = x2 −23. To ﬁnd the continued fraction of v we apply the quadratic continued fraction algorithm with a0 = 1 and k0 = b−ε 2 = −3. i 0 1 2 3 4 5 a 1 7 2 7 1 7 k −3 −4 −3 −3 −4 −4 q 7 1 3 1 8 71 Since a5 = a1 and k5 = k1, we conclude that v = 7, 1, 3, 1, 8 . For v, we start the same algorithm with a0 = −1 and k0 = 3. i 0 1 2 3 4 5 6 a −1 2 7 1 7 2 7 k 3 −5 −3 −4 −4 −3 −3 q −2 4 1 8 1 3 Here since a0 is negative, we select k1 to be the largest integer less than w = − √ 23 with k1 ≡−3 (mod 1), that is, k1 = −5. Now a1 = −φ(−5) −1 = 2 is positive, so we select k2 to be the smallest integer greater than w with k2 ≡5 (mod 2), that is, k2 = −3. Continuing the algorithm in this way, we ﬁnd that a6 = a2 and k6 = k2, and conclude that v = −2, 4, 1, 8, 1, 3 . (b) The minimum polynomial of v = 3+ √ 17 5 and v = 3− √ 17 5 is f(x) = 25x2 −30x−8. The discriminant of f(x) is ∆= 1700 = ∆(17, 10), so that φ(x) = x2 + 10x −400. For v, we apply the quadratic continued fraction algorithm with a0 = 25 and k0 = −30−10 2 = −20. For v, we apply the same algorithm with a0 = −2 and k0 = 30−10 2 = 10. i 0 1 2 3 4 5 6 7 a 25 13 13 25 8 17 8 25 k −20 −15 −21 −15 −20 −22 −22 −20 q 1 2 2 1 4 2 4 Here k1 = −15 is the smallest integer greater than w = −10− √ 1700 2 ≈−25.6 that is congruent to −k0−ε = 10 modulo a0 = 25. Note that q0 = −k0+k1+ε a0 = −−20−15+10 25 = 1, and so forth. Since a7 = a0 and k7 = k0, then v = 1, 2, 2, 1, 4, 2, 4 . For v, we apply the same algorithm with a0 = −2 and k0 = 30−10 2 = 10. i 0 1 2 3 4 5 6 7 8 9 10 a −25 47 8 17 8 25 13 13 25 8 17 k 10 −45 −12 −22 −22 −20 −15 −21 −15 −20 −22 q −1 2 3 2 4 1 2 2 1 4 In this case, k1 = −45 is the largest integer less than w congruent to −k0 −ε = −20 modulo 25. So a1 = −φ(−45) −25 = 47 and now k2 = −12 is the smallest integer greater than w congruent to −k1 −ε = 35 modulo 47. Continuing in this way, we obtain a10 = a3 and k10 = k3, and conclude that v = −1, 1, 3, 2, 4, 1, 2, 2, 1, 4. (c) v = −5+2 √ 13 3 and v = −5−2 √ 13 3 have minimum polynomial f(x) = 3x2 + 10x −9. The discriminant of f(x) is ∆= 208 = ∆(13, 4), so that φ(x) = x2 + 4x −48. For v, we apply the quadratic continued fraction algorithm with a0 = 3 and k0 = 10−4 2 = 3. i 0 1 2 3 4 5 6 7 a 3 9 4 9 3 1 3 9 k 3 −7 −6 −6 −7 −9 −9 −7 q 0 1 2 1 4 14 4 72 We conclude that v = 0, 1, 2, 1, 4, 14, 4 . For v, we begin the algorithm with a0 = −3 and k0 = −10−4 2 = −7. i 0 1 2 3 4 5 6 7 8 9 a −3 16 1 3 9 4 9 3 1 3 k −7 −12 −8 −9 −7 −6 −6 −7 −9 −9 q −5 1 13 4 1 2 1 4 14 We ﬁnd that v = −5, 1, 13, 4, 1, 2, 1, 4, 14 . (d) v = 13 + 4 √ 113 and v = 13 −4 √ 113 have minimum polynomial f(x) = x2 −26x −7. The discriminant of f(x) is ∆= 704 = ∆(11, 4), and φ(x) = x2 −176. For v, we apply the quadratic continued fraction algorithm with a0 = 1 and k0 = −26−0 2 = −13, and conclude from the following data that v = 26, 3, 1, 3. i 0 1 2 3 4 a 1 7 16 7 1 k −13 −13 −8 −8 −13 q 26 3 1 3 For v, we begin the algorithm with a0 = −1 and k0 = 13, and ﬁnd that v = −1, 1, 2, 1, 3, 26, 3. i 0 1 2 3 4 5 6 7 a −1 20 7 16 7 1 7 16 k 13 −14 −6 −8 −8 −13 −13 −8 q −1 1 2 1 3 26 3 (3) (a) v = 1 + √ 94 is the larger root of f(x) = x2 −2x −93, with discriminant ∆= 376 = ∆(94, 1). We compile the following values of φ(x) = x2 −94, the principal polynomial of discriminant 376. x ±9 ±8 ±7 ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −13 −30 −45 −58 −69 −78 −85 −90 −93 −94 Applying the quadratic continued fraction algorithm with a0 = 1 and k0 = −2−0 2 = −1 yields the following data. i 0 1 2 3 4 5 6 7 8 9 a 1 13 6 5 9 10 3 15 2 15 k −1 −9 −4 −8 −7 −2 −8 −7 −8 −8 q 10 1 2 3 1 1 5 1 8 1 i 10 11 12 13 14 15 16 17 a 3 10 9 5 6 13 1 13 k −7 −8 −2 −7 −8 −4 −9 −9 q 5 1 1 3 2 1 18 We obtain the ﬁrst repetition with a17 = a1 and k17 = k1, and thus conclude that v = 10, 1, 2, 3, 1, 1, 5, 1, 8, 1, 5, 1, 1, 3, 2, 1, 18 . (b) v = 5− √ 114 3 is the smaller root of f(x) = 9x2−30x−89, with discriminant ∆= 4104 = ∆(114, 3). We apply the quadratic continued fraction algorithm with φ(x) = x2−1026, 73 a0 = −9, and k0 = 30−0 2 = 15. i 0 1 2 3 4 5 6 7 8 a −9 7 18 25 14 9 50 7 18 k 15 −33 −30 −24 −26 −30 −24 −26 −30 q −2 9 3 2 4 6 1 8 Here k1 = −33 is the largest integer smaller than w = − √ 1026 ≈−32.03 congruent to −k0 −ε = −15 modulo 9. Then a1 = −φ(−33) −9 = 7 and k2 = −30 is the smallest integer greater than w congruent to −k1 −ε = 33 modulo 7. Continuing in this way, we ﬁnd that a8 = a2 and k8 = k2, and conclude that v = −2, 9, 3, 2, 4, 6, 1, 8 . (c) v = −4+ √ 65 2 is the larger root of f(x) = 4x2+16x−49, with discriminant ∆= 1040 = ∆(65, 4). So φ(x) = x2 + 4x −256, and we apply the quadratic continued fraction algorithm with a0 = 4 and k0 = 16−4 2 = 6. The following data i 0 1 2 3 a 4 1 4 1 k 6 −18 −18 −18 q 2 32 8 shows that v = 2, 32, 8 . (d) v = 3+ √ 67 2 is the larger root of f(x) = 2x2 −6x −29, with discriminant ∆= 268 = ∆(67, 1). We compile the following values of φ(x) = x2 −67, the principal polynomial of discriminant 268. x ±8 ±7 ±6 ±5 ±4 ±3 ±2 ±1 0 φ(x) −3 −18 −31 −42 −51 −58 −63 −66 −67 Applying the quadratic continued fraction algorithm with a0 = 2 and k0 = −3 yields the following data. i 0 1 2 3 4 5 6 7 8 9 10 11 a 2 9 7 6 3 1 3 6 7 9 2 9 k −3 −7 −2 −5 −7 −8 −8 −7 −5 −2 −7 −7 q 5 1 1 2 5 16 5 2 1 1 7 Since a11 = a1 and k11 = k1, we have v = 5, 1, 1, 2, 5, 16, 5, 2, 1, 1, 7 . (e) v = 4− √ 83 7 is the smaller root of f(x) = 49x2 −56x −67, with discriminant ∆= 16268 = ∆(83, 7). Applying the quadratic continued fraction algorithm with φ(x) = x2 −4067, and with a0 = −49 and k0 = 28, we obtain the following data. i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 a −49 38 71 41 43 61 58 49 2 49 58 61 43 41 k 28 −77 −37 −34 −48 −38 −23 −35 −63 −63 −35 −23 −38 −48 q −1 3 1 2 2 1 1 2 63 2 1 1 2 2 i 14 15 16 17 18 19 20 21 22 23 24 25 26 a 71 38 67 49 74 17 98 17 74 49 67 38 71 k −34 −37 −39 −28 −21 −53 −49 −49 −53 −21 −28 −39 −37 q 1 2 1 1 1 6 1 6 1 1 1 2 The ﬁrst repetition occurs with a26 = a2 and k26 = k2, so that v = −1, 3, 1, 2, 2, 1, 1, 2, 63, 2, 1, 1, 2, 2, 1, 2, 1, 1, 1, 6, 1, 6, 1, 1, 1, 2 . 74 (f) v = −7+ √ 113 4 is the larger root of f(x) = 2x2 + 7x −8, with discriminant ∆= 113. Using the following values of φ(x) = x2 + x −28, x −5, 4 −4, 3 −3, 2 −2, 1 −1, 0 φ(x) −8 −16 −22 −26 −28 we apply the quadratic continued fraction algorithm with a0 = 2 and k0 = 7−1 2 = 3. i 0 1 2 3 4 5 6 7 8 a 2 8 1 8 2 4 4 2 8 k 3 −4 −5 −5 −4 −5 −4 −5 −4 q 0 1 9 1 4 2 2 4 We conclude that v = 0, 1, 9, 1, 4, 2, 2, 4 . Section 10.4. Equivalence of Indeﬁnite Quadratic Forms. (1) For each i, write φ(ki) = −ai · −ci and φ′(ki) = bi. If Ui = qi 1 −1 0 and we let (−ai : ki) ◦Ui = (m : ℓ), then m = −ai(qi)2 + biqi(−1) −ci(−1)2 = −aiq2 i −biqi −ci and ℓ= −ai(qi)(1) + bi(1)(−1) −ci(−1)(0) + ki = −aiqi −ki −ε = ki+1, using the fact that bi = 2ki + ε along with equation (10.3.2). From equation (10.3.3), we have that φ(ki+1) = ai(aiq2 i + biqi + ci) = −ai(−aiq2 i −biqi −ci) = −aim. But then by equation (10.3.1), ai+1 = −φ(ki+1) ai = m, and so (−ai : ki) ◦Ui = (ai+1 : ki+1). (2) If i is odd, then WiUi = mi−1 −mi−2 ni−1 −ni−2 · qi 1 −1 0 = mi−1qi + mi−2 mi−1 ni−1qi + ni−2 ni−1 = mi mi−1 ni ni−1 = Wi+1 since i + 1 is even. (3) By deﬁnition of the numerator and denominator sequences, we have that W0 = m−1 m−2 n−1 n−2 = 1 0 0 1 and W1 = m0 −m−1 n0 −n−1 = q0 −1 1 0 = U0. Now W2 = W1 ·U1 = U0 ·U1 by Exercise 2, then W3 = W2 ·U2 = U0 ·U1 ·U2 as in the proof of Theorem 10.4.2, and so forth. We obtain the general expression for Wt as claimed. Section 11.1. The Continued Fraction of a Quadratic Form. (1) For 1 ≤r ≤20, the following are the values of q r that are closest to √ 2 in absolute value. 1 1, 3 2, 4 3, 6 4, 7 5, 8 6, 10 7 , 11 8 , 13 9 , 14 10, 16 11, 17 12, 18 13, 20 14, 21 15, 23 16, 24 17, 25 18, 27 19, 28 20. 75 Direct calculation shows that \| q r − √ 2\| < 1 2r2 only for the following fractions in this list: 1 1, 2 3, 7 5, and 17 12. The continued fraction of √ 2 is 1, 2 , and the following table shows that each value listed above is a convergent of this continued fraction. i 0 1 2 3 4 q 1 2 2 2 2 m 1 3 7 17 41 n 1 2 5 12 29 (2) For 1 ≤r ≤20, the following are the values of q r that are closest to √ 3 in absolute value. 2 1, 3 2, 5 3, 7 4, 9 5, 10 6 , 12 7 , 14 8 , 16 9 , 17 10, 19 11, 21 12, 23 13, 24 14, 26 15, 28 16, 29 17, 31 18, 33 19, 35 20. Here \| q r − √ 3\| < 1 2r2 only for the following fractions in this list: 2 1, 7 4, and 26 15. The continued fraction of √ 3 is 1, 1, 2 , and we verify by the following table that each value listed above is a convergent of this continued fraction. i 0 1 2 3 4 5 q 1 1 2 1 2 1 m 1 2 5 7 19 26 n 1 1 3 4 11 15 (Note that not all convergents satisfy this condition however. For instance, we ﬁnd that \| 19 11 − √ 3\| ≈0.0048 while 1 2·112 ≈0.0041.) Section 11.2. Units and Automorphs. (1) Suppose that u = q+rz ̸= ±1 is a unit in D = D∆for some positive discriminant ∆. With −u, u−1, and −u−1 also units, we can assume that u > 1. Since N(u) = uu = ±1, either u−1 or −u−1 equals u. In any case, we can assume that q + rz > q + rz, which implies that r √ ∆is positive. ((q + rz) −(q + rz) = r(z −z) = r √ ∆in every case.) Likewise, q + rz > −q −rz, implying that 2q + εr > 0. Since √ ∆and ε are positive for every discriminant ∆> 0, it follows that q and r must both be positive. There are ﬁnitely many elements s+tz < u with s and t both positive, and so there must be a smallest unit u > 1 in D (assuming again that there is at least one such unit). (2) Suppose that D = D∆has a unit u > 1 and that u is the smallest such unit. Let v be some unit of D, and assume that v > 1. (We can replace v by −v, v−1, or −v−1 if not.) If u > 1 then the positive integer powers of u get arbitrarily large, so there must be some positive integer n so that un ≤v < un+1. But now 1 ≤u−nv < u, and u−nv is a unit of D since its norm is ±1. We must conclude that u−nv = 1, that is, v = un, to avoid contradicting the deﬁnition of u. Including negatives and inverses of these units, together with ±1, we then see that every unit of D must be equal to ±un for some integer n. (3) In each part, we calculate the fundamental unit of D∆by applying the equivalence algo-rithm to (1 : 0), the principal form of discriminant ∆. (a) For ∆= 21, with principal polynomial φ(x) = x2+x−5 and basis element z = 1+ √ 21 2 , we obtain the following data. i 0 1 2 3 a 1 3 1 3 k 0 −2 −2 −2 q 1 1 3 m 1 2 n 1 1 76 We conclude that the fundamental unit of D21 is 2 + z = 5+ √ 21 2 . (b) For ∆= 37, with φ(x) = x2 + x −8 and z = 1+ √ 37 2 , we ﬁnd from the table i 0 1 2 3 4 a 1 3 3 1 3 k 0 −3 −1 −3 −3 q 2 1 1 5 m 2 3 5 n 1 1 2 that the fundamental unit of D37 is 5 + 2z = 6 + √ 37. (c) For ∆= 40, with φ(x) = x2 −10 and z = √ 10: i 0 1 2 a 1 1 1 k 0 −3 −3 q 3 6 m 3 n 1 The fundamental unit of D40 is 3 + z = 3 + √ 10. (d) For ∆= 41, with φ(x) = x2 + x −10 and z = 1+ √ 41 2 : i 0 1 2 3 4 5 6 a 1 4 2 2 4 1 4 k 0 −3 −2 −3 −2 −3 −3 q 2 1 2 2 1 5 m 2 3 8 19 27 n 1 1 3 7 10 The fundamental unit of D41 is 27 + 10z = 32 + 5 √ 41. (e) For ∆= 53, with φ(x) = x2 + x −13 and z = 1+ √ 53 2 : i 0 1 2 a 1 1 1 k 0 −4 −4 q 3 7 m 3 n 1 The fundamental unit of D53 is 3 + z = 7+ √ 53 2 . (f) For ∆= 76, with φ(x) = x2 −19 and z = √ 19: i 0 1 2 3 4 5 6 7 a 1 3 5 2 5 3 1 3 k 0 −4 −2 −3 −3 −2 −4 −4 q 4 2 1 3 1 2 8 m 4 9 13 48 61 170 n 1 2 3 11 14 39 The fundamental unit of D76 is 170 + 39z = 170 + 39 √ 19. 77 (g) For ∆= 92, with φ(x) = x2 −23 and z = √ 23: i 0 1 2 3 4 5 a 1 7 2 7 1 7 k 0 −4 −3 −3 −4 −4 q 4 1 3 1 8 m 4 5 10 24 n 1 1 4 5 The fundamental unit of D92 is 24 + 5z = 24 + 5 √ 23. (h) For ∆= 104, with φ(x) = x2 −26 and z = √ 26: i 0 1 2 a 1 1 1 k 0 −5 −5 q 5 10 m 5 n 1 The fundamental unit of D104 is 5 + z = 5 + √ 26. (i) For ∆= 232, with φ(x) = x2 −58 and z = √ 58: i 0 1 2 3 4 5 6 7 8 a 1 9 6 7 7 6 9 1 9 k 0 −7 −2 −4 −3 −4 −2 −7 −7 q 7 1 1 1 1 1 1 14 m 7 8 15 23 38 61 99 n 1 1 2 3 5 8 13 The fundamental unit of D232 is 99 + 13z = 19 + 13 √ 58. (j) For ∆= 276, with φ(x) = x2 −69 and z = √ 69: i 0 1 2 3 4 5 6 7 8 9 a 1 5 4 11 3 11 4 5 1 5 k 0 −8 −7 −5 −6 −6 −5 −7 −8 −8 q 8 3 3 1 4 1 3 3 16 m 8 25 83 108 515 623 2384 7775 n 1 3 10 13 62 75 287 936 The fundamental unit of D276 is 7775 + 936z = 7775 + 936 √ 69. (4) We determined a collection of class representatives for quadratic forms of each discriminant ∆in Exercise 10.2.1. The tables compiled in that exercise are extended below, with Theorem 11.2.2 applied to construct a generator U for the group of automorphs for each representative (that is, so that Aut(f) = {±Un \| n ∈Z}). (a) Each element of F37 is equivalent to f = (1 : −3). Using φ(x) = x2 +x−9, we obtain the following data from the equivalence/continued fraction algorithm applied to this form. i 0 1 2 3 4 5 6 a 1 3 3 1 3 3 1 k −3 −3 −1 −3 −3 −1 −3 q 5 1 1 5 1 1 m 5 6 11 61 72 133 n 1 1 2 11 13 24 78 We conclude that Aut(f) = {±Un \| n ∈Z} for U = 133 72 24 13 . (b) Each form in F40 is equivalent to f = (1 : −3) or to g = (2 : −2). With φ(x) = x2−10, we compile the following data. i 0 1 1 a 1 1 1 k −3 −3 −3 q 6 6 6 m 6 37 n 1 6 i 0 1 2 3 4 5 6 a 2 3 3 2 3 3 2 k −2 −2 −1 −2 −2 −1 −2 q 2 1 1 2 1 1 m 2 3 5 13 18 31 n 1 1 2 5 7 12 So U = 37 6 6 1 and V = 31 18 12 7 are generating automorphs of f and of g respectively. (c) Each form in F61 is equivalent to f = (1 : −4). With φ(x) = x2+x−15, the following data i 0 1 2 3 4 5 6 a 1 3 3 1 3 3 1 k −4 −4 −3 −4 −4 −3 −4 q 7 2 2 7 2 2 m 7 15 37 274 585 1444 n 1 2 5 37 79 195 shows that Aut(f) = {±Un \| n ∈Z} for U = 1444 585 195 79 . (d) Each form in F92 is equivalent to f = (1 : −4) or to −f = (−1 : −4). Using φ(x) = x2 −23, we compile the following data. i 0 1 2 3 4 a 1 7 2 7 1 k −4 −4 −3 −3 −4 q 8 1 3 1 m 8 9 35 44 n 1 1 4 5 By Theorem 11.2.2, U = 44 35 5 4 and U = 44 −35 −5 4 are generating auto-morphs for f and for −f respectively. (e) Each form in F93 is equivalent to f = (1 : −5) or to −f = (−1 : −4). With φ(x) = x2 + x −23, we compile the following data i 0 1 2 a 1 3 1 k −5 −5 −5 q 9 3 m 9 28 n 1 3 which shows that U = 28 9 3 1 and U = 28 −9 −3 1 are generating automorphs for f and for −f respectively. 79 (f) Each form in F105 is equivalent to f = (1 : −5), to −f = (−1 : −5), to g = (2 : −5), or to −g = (−2 : −5). With φ(x) = x2 + x −26, we compile the following tables. i 0 1 2 3 4 5 6 a 1 6 4 5 4 6 1 k −5 −5 −2 −3 −3 −2 −5 q 9 1 1 1 1 1 m 9 10 19 29 48 77 n 1 1 2 3 5 8 i 0 1 2 3 4 a 2 7 2 3 2 k −5 −4 −4 −5 −5 q 4 1 4 3 m 4 5 24 77 n 1 1 5 16 We conclude that U = 77 48 8 5 , U = 77 −48 −8 5 , V = 77 24 16 5 , and V = 77 −24 −16 5 are generating automorphs for f, for −f, for g, and for −g respectively. Section 11.3. Existence of Representations by Indeﬁnite Forms. (1) Using the Chinese Remainder Theorem (in the form of Theorem ??), we ﬁnd that for arbitrary a, b, and c, the system of congruences x ≡a (mod 4), x ≡b (mod 5), x ≡c (mod 7) has a simultaneous solution x ≡−35a + 56b −20c (mod 140). (a) −1 p = p 5 = p 7 = 1 if and only if p ≡1 (mod 4), p ≡1 or 4 (mod 5), and p ≡ 1, 2, or 4 (mod 7). Substituting each possible triple for a, b, and c in the system of congruences, the formula above produces six possibilities for p modulo 140. a b c x = −35a + 56b −20c x mod 140 1 1 1 1 1 1 1 2 −19 121 1 1 4 −59 81 1 4 1 169 29 1 4 2 149 9 1 4 4 109 109 (b) −1 p = 1 and p 5 = −1 = p 7 if and only if p ≡1 (mod 4), p ≡2 or 3 (mod 5), and p ≡3, 5, or 6 (mod 7). a b c x = −35a + 56b −20c x mod 140 1 2 3 17 17 1 2 5 −23 117 1 2 6 −43 97 1 3 3 73 73 1 3 5 33 33 1 3 6 13 13 80 (c) p 5 = 1 and −1 p = −1 = p 7 if and only if p ≡3 (mod 4), p ≡1 or 4 (mod 5), and p ≡3, 5, or 6 (mod 7). a b c x = −35a + 56b −20c x mod 140 3 1 3 −109 31 3 1 5 −149 131 3 1 6 −169 111 3 4 3 59 59 3 4 5 19 19 3 4 6 −1 139 (d) p 7 = 1 and −1 p = −1 = p 5 if and only if p ≡3 (mod 4), p ≡2 or 3 (mod 5), and p ≡1, 2, or 4 (mod 7). a b c x = −35a + 56b −20c x mod 140 3 2 1 −13 127 3 2 2 −33 107 3 2 4 −73 67 3 3 1 43 43 3 3 2 23 23 3 3 4 −17 123 (2) (a) It was established in Exercise 10.1.2 part (b) that every quadratic form of discriminant ∆= 28 is equivalent either to (1 : −2) or to (−1 : −2), with these two not equivalent to each other. Note that (1 : −2) ∼(1 : 0) and that (−1 : −2) ∼(−1 : 0), so we can use the forms f(x, y) = x2 −7y2 and g(x, y) = −x2 + 7y2 as representatives of F28. (b) If p ̸= 2, 7 is prime, then p is represented by a form of discriminant 28 if and only if 28 p = 7 p = −1 p −7 p = −1 p p 7 = 1. Note that −1 f = 1 = f 7 and that −1 g = −1 = g 7 . Thus if −1 p = 1 = p 7 , then p is represented by f, while if −1 p = −1 = p 7 , then p is represented by g. Since g = −f, it is clear that f represents p if and only if g represents −p. (3) (a) Exercise 10.2.1 part (b) shows that F40 consists of the classes of (1 : −3) and (2 : −2), with (1 : −3) ∼(−1 : −3) and (2 : −2) ∼(−2 : −2). Since (1 : −3) ∼(1 : 0) and (2 : −2) ∼(2 : 0), we can use f(x, y) = x2 −10y2 and g(x, y) = 2x2 −5y2 as representatives of these classes. (b) A prime p ̸= 2, 5 is represented by one of these forms if and only if 40 p = 2 p p 5 = 1. If 2 p = 1 = p 5 , then p is represented by f (since 2 f = 1 = f 5 ). With f ∼−f, then −p is also represented by f. Likewise, if 2 p = −1 = p 5 , then p and −p are represented by g. (4) (a) By Exercise 10.1.2 part (e), F57 consists of the classes of (1 : −4) and (−1 : −4), with (−1 : −4) ∼(2 : −4). We can use f = (1 : 0) = x2 + xy −14y2 and g = (2 : 0) = 2x2 + xy −7y2 as representatives of these classes. Note that f ∼−g and g ∼−f. (b) A prime p ̸= 3, 19 is represented by one of these forms if and only if 57 p = −3 p −19 p = p 3 p 19 = 1. If p 3 = 1 = p 19 , then p is represented by f and −g, so that −p is represented by g. Similarly, if p 3 = −1 = p 19 , then p is represented by g and −f, and so −p is represented by f. (5) (a) By Exercise 10.1.2 part (f), F65 consists of the classes of (1 : −4) and (2 : −4), with (1 : −4) ∼(−1 : −4) and (2 : −4) ∼(−2 : −4). We can use f = (1 : 0) = 81 x2 + xy −16y2 and g = (2 : 0) = 2x2 + xy −8y2 as representatives of these classes, with f ∼−f and g ∼−g. (b) A prime p ̸= 5, 13 is represented by one of these forms if and only if 65 p = 5 p 13 p = p 5 p 13 = 1. If p 5 = 1 = p 13 , then p is represented by f and −f, so that −p is represented by f. If p 5 = −1 = p 13 , then p is represented by g and −g, and so −p is represented by g. (6) (a) By Exercise 10.2.1 part (d), F92 consists of the classes of (1 : −4) and (−1 : −4). We can use f(x, y) = x2 −23y2 and g(x, y) = −x2 + 23y2 as representatives of these classes. Since g = −f, then f represents an integer n if and only if g represents −n. (b) A prime p ̸= 2, 23 is represented by one of these forms if and only if 92 p = −1 p −23 p = −1 p p 23 = 1. If −1 p = 1 = p 23 , then p is represented by f and so −p is represented by g. If −1 p = −1 = p 23 , then p is represented by g and −p is represented by f. Section 11.4. Constructing Representations by Indeﬁnite Forms. (1) In each part, we apply the equivalence algorithm to (a : k), where a = p and φ(x) = x2−6, until we obtain an i for which ai = 1. If φ(x, y) = x2 −6y2, we use Theorem 11.4.1 to construct a solution of φ(x, y) = p or φ(x, y) = −p, as illustrated. (Other answers are possible.) (a) With p = 23 and k = 11, we obtain the following data from the equivalence/continued fraction algorithm. (As noted in Theorem 11.4.1, we need only the denominator terms in the convergents of the continued fraction.) i 0 1 2 3 a 23 −6 5 1 k 11 12 −6 1 q −1 1 1 n 1 1 2 Here k1 = 12 is the smallest integer for which k1 ≡−11 (mod 23) and k1 > − √ 6, and then a1 = −φ(12) 23 = −6. Now with a2 negative, we need k2 < − √ 6 as large as possible with k2 ≡−12 (mod 6), and ﬁnd that k2 = −6 and then a2 = −φ(−6) −6 = 5. Finally, k3 = 1 is the smallest k3 > − √ 6 for which k3 ≡6 (mod 5), so that a3 = −φ(1) 5 = 1. Now with i = 3 odd, we ﬁnd that f3(n1, n2) = f3(1, 2) = 23, where f3 = (−1 : 1), that is, f3(x, y) = −x2+2xy+5y2. Note that −f3(x, y) = x2−2xy−5y2 = (x−y)2−6y2 = φ(x −y, y). Since −f3(1, 2) = −23, then φ(1 −2, 2) = φ(−1, 2) = −23. (b) With p = 53 and k = 18: i 0 1 2 3 a 53 −23 6 1 k 18 35 −12 0 q −1 1 2 n 1 1 3 Here i = 3 is odd. If f3 = (−1 : 0) = −x2 + 6y2, then f3(n1, n2) = f3(1, 3) = 53. Since f3 = −φ, we immediately conclude that φ(1, 3) = −53. 82 (c) With p = 67 and k = 26: i 0 1 2 3 4 a 67 −25 10 −3 1 k 26 41 −16 6 −3 q −1 1 1 1 n 1 1 2 3 Since i = 4 is even, we let f4 = (1 : −3) = x2−6xy+3y2, and have that f4(n2, −n3) = f4(2, −3) = 67. Note that f4(x, y) = x2 −6xy + 3y2 = (x −3y)2 −6y2 = φ(x −3y, y). We conclude that φ(2 −3(−3), −3) = φ(11, −3) = 67. (d) With p = 73 and k = 15: i 0 1 2 3 4 a 73 −46 3 2 1 k 15 58 −12 0 −2 q −1 1 4 1 n 1 1 5 6 Here f4 = (1 : −2) = x2 −4xy −2y2, and f4(n2, −n3) = f4(5, −6) = 73. Since f4(x, y) = x2 −4xy −2y2 = (x −2y)2 −6y2 = φ(x −2y, y), we conclude that φ(5 −2(−6), −6) = φ(17, −6) = 73. (2) In each part, we apply Theorem 11.4.1 using a solution of φ(x) ≡0 (mod p) found by trial-and-error. Other approaches and other ﬁnal answers are possible. (a) If p = 53, then −1 p = 1 = p 7 . Thus we know by Exercise 11.3.2 that p is represented by φ(x, y) = x2 −7y2. The congruence x2 −7 ≡0 (mod 53) has x = 22 as a solution. We apply Theorem 11.4.1 with a = 53 and k = 22. i 0 1 2 3 a 53 −18 9 −1 k 22 31 −13 4 q −1 1 1 n 1 1 2 Here f3 = (−(−1) : 4), that is, f3(x, y) = x2 + 8xy + 9y2 = (x + 4y)2 −7y2 = φ(x+4y, y), and Theorem 11.4.1 implies that f3(n1, n2) = f3(1, 2) = 53. We conclude that φ(1 + 4(2), 2) = φ(9, 2) = 53. (b) If p = 47, then −1 p = −1 = p 7 . Exercise 11.3.2 shows that −p is represented by φ(x, y) = x2 −7y2, and we ﬁnd that x = 17 is a solution of x2 −7 ≡0 (mod 47). Let a = 47 and k = 17. i 0 1 2 3 a 47 −19 6 1 k 17 30 −11 −1 q −1 1 2 n 1 1 3 If f3 = (−1 : −1), that is, f3(x, y) = −x2 −2xy +6y2, then f3(n1, n2) = f3(1, 3) = 47. Then we see that −f3(x, y) = x2 +2xy−6y2 = (x+y)2 −7y2 = φ(x+y, y). Therefore −f3(1, 3) = φ(4, 3) = −47. As another option, we can apply the equivalence algorithm 83 with a = −47 and k = 17, obtaining the following data. i 0 1 2 a −47 6 1 k 17 −17 −1 q 0 3 n 1 3 Here with f2 = (1 : −1), that is, f2(x, y) = x2−2xy−6y2 = (x−y)2−7y2 = φ(x−y, y), then f2(n0, −n1) = f2(1, −3) = −47, and so φ(1 −(−3), −3) = φ(4, −3) = −47. (c) If p = 31, then 2 p = 1 = p 5 . By Exercise 11.3.3, then p and −p are both represented by φ(x, y) = x2 −10y2. We ﬁnd that x = 14 is a solution of x2 −10 ≡0 (mod 31), and apply Theorem 11.4.1 with a = 31 and k = 14. i 0 1 2 3 4 a 31 −9 6 1 1 k 14 17 −8 2 −3 q −1 1 1 1 n 1 1 2 3 Here f4 = (1 : −3), that is, f4(x, y) = x2 −6xy −y2 = (x−3y)2 −10y2 = φ(x−3y, y), and Theorem 11.4.1 implies that f4(n2, −n3) = f4(2, −3) = 31. We conclude that φ(2 −3(−3), −3) = φ(11, −3) = 31. (d) Using the data of Exercise 3, we also see that f3 = (−1 : 2) and f3(n1, n2) = f3(1, 2) = 31. Here −f3(x, y) = x2 −4xy −6y2 = (x −2y)2 −10y2 = φ(x −2y, y), so we see that −31 = −f3(1, 2) = φ(−3, 2). (e) If p = 43, then 2 p = −1 = p 5 , so that p and −p are both represented by g(x, y) = 2x2 −5y2. We ﬁnd that x = 15 is a solution of x2 −10 ≡0 (mod 43). i 0 1 2 3 4 5 6 a 43 −18 5 2 3 3 2 k 15 28 −10 0 −2 −1 −2 q −1 1 2 1 1 1 n 1 1 3 4 7 11 Here f6 = (2 : −2), that is, f6(x, y) = 2x2 −4xy −3y2 = 2(x2 −2xy + y2) −5y2 = g(x−y, y), and f6(n4, −n5) = f6(7, −11) = 43. We conclude that g(7−(−11), −11) = g(18, −11) = 43. (f) From the data of Exercise 3, we have that f3(n1, n2) = f3(1, 3) = 43, where f3 = (−2 : 0) = −g. Thus g(1, 3) = −43. (g) If p = 41, then p 3 = −1 = p 19 . By Exercise 11.3.4, p is represented by g(x, y) = 2x2 +xy−7y2 and −p is represented by f(x, y) = x2+xy−14y2. We ﬁnd that x = 18 is a solution of x2 + x −14 ≡0 (mod 41). Following the alternative approach of part (b) of this exercise, we apply the equivalence algorithm to (a : k) = (−41 : 18). i 0 1 2 a −41 8 1 k 18 −19 2 q 0 2 n 1 2 84 We ﬁnd that f2 = (1 : 2), that is, f2(x, y) = x2 + 5xy −8y2 = (x + 2y)2 + (x + 2y)y −14y2 = f(x + 2y, y). Theorem 11.4.1 shows that f2(n0, −n1) = f2(1, −2) = −41, and so f(1+2(−2), −2) = f(−3, −2) = −41. (h) With ∆= 57, we apply the equivalence algorithm to (a : k) = (41 : 18). i 0 1 2 3 4 a 41 −12 8 1 2 k 18 22 −11 2 −4 q −1 1 1 1 n 1 1 2 3 If f4 = (2 : −4), so that f4(x, y) = 2x2 −7xy −y2 = 2(x −2y)2 + (x −2y)y −7y2 = g(x −2y, y) we have that f4(n2, −n3) = f4(2, −3) = 41. Thus g(2 −2(−3), −3) = g(8, −3) = 41. (i) If p = 29, then p 5 = 1 = p 13 . By Exercise 11.3.5, φ(x, y) = x2 +xy−16y2 represents both p and −p. We ﬁnd that x = 11 is a solution of x2 + x −16 ≡0 (mod 29). We can solve both φ(x, y) = 29 and φ(x, y) = −29 by applying the equivalence algorithm to (a : k) = (29 : 11). i 0 1 2 3 4 5 6 7 a 29 −10 4 4 1 4 4 1 k 11 17 −8 −1 −4 −4 −1 −4 q −1 1 2 1 7 1 1 n 1 1 3 4 31 35 66 Here f4 = (1 : −4), that is, f4(x, y) = x2 −7xy −4y2 = (x −4y)2 + (x −4y)y −16y2 = φ(x −4y, y). Now f4(n2, −n3) = f4(3, −4) = 29, and so φ(3 −4(−4), −4) = φ(19, −4) = 29. (j) From part (i), we also have that if f7 = (−1 : −4), then f7(n5, n6) = f7(35, 66) = 29. Now −f7(x, y) = x2 + 7xy −4y2 = (x + 3y)2 + (x + 3y)y −16y2 = φ(x + 3y, y), so we also ﬁnd that φ(35 + 3(66), 66) = φ(233, 66) = −29. (k) If p = 83, then p 5 = −1 = p 13 , so that g(x, y) = 2x2 + xy −8y2 represents both p and −p. We can solve both g(x, y) = 83 and g(x, y) = −83, beginning with the calculation that x = 13 satisﬁes x2 + x −16 ≡0 (mod 83). i 0 1 2 3 4 5 a 83 −58 2 5 2 2 k 13 69 −12 −3 −3 −4 q −1 1 7 1 3 n 1 1 8 9 35 For f2 = (2 : −12), that is, f2(x, y) = 2x2 −23xy + 58y2 = 2(x −6y)2 + (x −6y)y −8y2 = g(x −6y, y), we have that f2(n0, −n1) = f2(1, −1) = 83. Thus g(1 −6(−1), −1) = g(7, −1) = 83. 85 (l) From part (k), we also have that if f5 = (−2 : −4), then f5(n3, n3) = f5(9, 35) = 83. Now −f5(x, y) = 2x2 + 7xy −2y2 = 2(x + 2y)2 + (x + 2y)(−y) −8(−y)2 = g(x + 2y, −y), so we also ﬁnd that g(9 + 2(35), −35) = g(79, −35) = −83. (m) If p = 29, then −1 p = 1 = p 23 . So x2 −23 ≡0 (mod 29) has solutions, which we ﬁnd to be x = ±9, and Exercise 11.3.6 shows that φ(x, y) = x2 −23y2 represents p. We apply the equivalence algorithm to (a : k) = (29 : −9) below. (This turns out to require fewer steps than if we use (29 : 9).) i 0 1 2 a 29 −2 1 k −9 9 −5 q 0 2 n 1 2 Here f2 = (1 : −5), so f2(x, y) = x2 −10xy + 2y2 = (x −5y)2 −23y2 = φ(x −5y, y). We have that f2(n0, −n1) = f2(1, −2) = φ(1 −5(−2), −2) = φ(11, −2) = 29. (n) If p = 79, then −1 p = −1 = p 23 . Here x2 −23 ≡0 (mod 79) has solutions, which we ﬁnd to be x = ±24, and φ(x, y) = x2 −23y2 represents −p. We apply the equivalence algorithm to (a : k) = (−79 : 24) below. i 0 1 2 a −79 7 1 k 24 −24 −4 q 0 4 n 1 4 Here f2 = (1 : −4), so f2(x, y) = x2 −8xy −7y2 = (x −4y)2 −23y2 = φ(x −4y, y). We have that f2(n0, −n1) = f2(1, −4) = φ(1 −4(−4), −4) = φ(17, −4) = −79. Section 12.1. Divisibility Properties of Quadratic Recursive Sequences. (1) In each part, we use the fact that Fn is divisible by Fn/p if p is a prime that divides n. Thus Fn is divisible by the least common multiple of these expressions. (a) F15 is divisible by lcm(F3, F5) = lcm(2, 5) = 2 · 5. We have that F15 = 610 = 2 · 5 · 61. (b) F18 is divisible by lcm(F6, F9) = lcm(8, 34) = 23 · 17. We ﬁnd that F18 = 2584 = 23 · 17 · 19. (c) F20 is divisible by lcm(F4, F10) = lcm(3, 55) = 3·5·11. Here F20 = 6765 = 3·5·11·41. (d) F24 is divisible by lcm(F8, F12) = lcm(21, 144) = 24·32·7. We ﬁnd that F24 = 46368 = 25 · 32 · 7 · 23. (e) F30 is divisible by lcm(F6, F10, F15) = lcm(8, 55, 610) = 23 · 5 · 11 · 61. We ﬁnd that F30 = 832040 = 23 · 5 · 11 · 31 · 61. (f) F36 is divisible by lcm(F12, F18) = lcm(144, 2584) = 24 · 32 · 17 · 19. We ﬁnd that F36 = 14930352 = 24 · 33 · 17 · 19 · 107. (2) In each part, let r0 = 0 and r1 = 1. We list the values of rn for 0 ≤n ≤10 and obtain a formula for rn in general. Veriﬁcation of Theorem 12.1.8 is omitted, aside from part (a) and special cases in other parts. (a) If rn = 2rn−2, then the sequence begins 0, 1, 0, 2, 0, 4, 0, 8, 0, 16, 0, . . .. The charac-teristic polynomial of the sequence is x2 −2, with roots √ 2 and − √ 2. By Theorem 86 12.1.2, rn = ( √ 2)n −(− √ 2)n √ 2 −(− √ 2) = ( 0, if n is even, 2q, if n = 2q + 1 is odd. Here −t = −2. Note that if m and n are both even or both odd, then rmrn+1 −rm+1rn = 0 = (−2)nrm−n. If m = 2k + 1 and n = 2ℓwith k ≥ℓ, then rmrn+1 −rm+1rn = 2k · 2ℓ= (−2)n · 2k−ℓ= (−2)nrm−n. If m = 2k and n = 2ℓ+ 1 with k > ℓso that m −n = 2(k −ℓ−1) + 1, then rmrn+1 −rm+1rn = −2k+1 · 2ℓ= (−2)n · 2k−ℓ−1 = (−2)nrm−n. (b) The characteristic polynomial of rn = 2rn−1 + rn−2 is x2 −2x −1 with roots 1 + √ 2 and 1− √ 2. This sequence begins 0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, . . ., with n-th term rn = (1 + √ 2)n −(1 − √ 2)n 2 √ 2 . Here −t = −1. To illustrate Theorem 12.1.8, note that if m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = 408 · 169 −985 · 70 = 2 = (−1)6 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 985 · 2 −2378 · 1 = −408 = (−1)1 · r8. (c) The characteristic polynomial of rn = 2rn−1 −3rn−2 is x2 −2x+3 with roots 1+√−2 and 1 −√−2. This sequence begins 0, 1, 2, 1, −4, −11, −10, 13, 56, 73, −22, . . ., with n-th term rn = (1 + √−2)n −(1 −√−2)n 2√−2 . Here −t = 3. If m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = 56 · 13 −73 · (−10) = 1458 = 36 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 73 · 2 −(−22) · 1 = 168 = 31 · r8. (d) The characteristic polynomial of rn = 2rn−1 + 3rn−2 is x2 −2x −3 = (x −3)(x + 1). This sequence begins 0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, . . ., with n-th term rn = 3n −(−1)n 3 −(−1) = 3n −(−1)n 4 . Here −t = −3. If m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = 1640 · 547 −4921 · 182 = 1458 = (−3)6 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 4921 · 2 −14762 · 1 = −4972 = (−3)1 · r8. 87 (e) The characteristic polynomial of rn = 2rn−1 −2rn−2 is x2 −2x + 2 with roots 1 + i and 1 −i. This sequence begins 0, 1, 2, 2, 0, −4, −8, −8, 0, 16, 32, . . ., with n-th term rn = (1 + i)n −(1 −i)n 2i . Here −t = 2. If m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = 0(−8) −16(−8) = 128 = 26 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 16 · 2 −32 · 1 = 0 = 21 · r8. (f) The characteristic polynomial of rn = 5rn−1−6rn−2 is x2−5x+6 = (x−3)(x−2). This sequence begins 0, 1, 5, 19, 65, 211, 665, 2059, 6305, 19171, 58025, . . ., with n-th term rn = 3n −2n 3 −2 = 3n −2n. Here −t = 6. If m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = 6305 · 2059 −19171 · 665 = 233280 = 66 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 19171 · 5 −58025 · 1 = 37830 = 61 · r8. (g) The characteristic polynomial of rn = 3rn−1 −4rn−2 is x2 −3x + 4 with roots 3+√−7 2 and 3−√−7 2 . This sequence begins 0, 1, 3, 5, 3, −11, −45, −91, −93, 85, 627, . . ., with n-th term rn = 3+√−7 2 n − 3−√−7 2 n √−7 . Here −t = 4. If m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = −93(−91) −85(−45) = 12288 = 46 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 85 · 3 −627 · 1 = −372 = 41 · r8. (h) The characteristic polynomial of rn = 3rn−1 −5rn−2 is x2 −3x +5 with roots 3+√−11 2 and 3−√−11 2 . This sequence begins 0, 1, 3, 4, −3, −29, −72, −71, 147, 796, 1653, . . ., with n-th term rn = 3+√−11 2 n − 3−√−11 2 n √−11 . Here −t = 5. If m = 8 and n = 6, then rmrn+1 −rm+1rn = r8 · r7 −r9 · r6 = 147(−71) −796(−72) = 46875 = 56 · r2, while if m = 9 and n = 1, then rmrn+1 −rm+1rn = r9 · r2 −r10 · r1 = 796 · 3 −1653 · 1 = 735 = 51 · r8. 88 Section 12.2. Periodicity of Quadratic Recursive Sequences. (1) (a) If f(x) = x2 + 1 is the characteristic polynomial of the quadratic recursive sequence rn, that is, if rn = −rn−2 with r0 = 0 and r1 = 1, then the sequence begins 0, 1, 0, −1, 0, 1, . . .. The discriminant of f(x) is ∆= −4. If p divides ∆, that is, p = 2, then subp(rn) = 2 divides p and ordp(rn) = 2 divides p(p −1) = 2. (Note that −1 ≡1 (mod 2).) If ∆ p = 1 so that p ≡1 (mod 4), then subp(rn) = 2 and ordp(rn) = 4 both divide p−1. If ∆ p = −1 so that p ≡3 (mod 4), then subp(rn) = 2 divides p + 1 and ordp(rn) = 4 divides (p + 1)(p −1). Thus the claims of Corollary 12.2.7 are satisﬁed by this sequence. (b) The discriminant of f(x) = x2 + 2 is ∆= −8, and the quadratic recursive sequence rn with characteristic polynomial f(x) begins 0, 1, 0, −2, 0, 4, 0, −8, 0, 16, 0, −32, 0, 64, . . .. Here subp(rn) = 2 for every prime p. Since −2 ≡1 (mod 3), 16 ≡1 (mod 5), and 64 ≡1 (mod 7), we ﬁnd that ord3(rn) = 2 (dividing p−1 = 2), ord5(rn) = 8 (dividing (p + 1)(p −1) = 24), and ord7(rn) = 12 (dividing (p + 1)(p −1) = 48). These results are consistent with Corollary 12.2.7 since −8 3 = 1 while −8 5 = −1 = −8 7 . (c) Let f(x) = x2 + x + 1 with discriminant ∆= −3. The quadratic recursive sequence with characteristic polynomial f(x) begins 0, 1, −1, 0, 1, −1, . . ., so that subp(rn) = 3 and ordp(rn) = 3 for every prime p. These values are consistent with Corollary 12.2.7 since 3 divides p if p divides ∆(that is, p = 3), 3 divides p −1 if ∆ p = 1 (this occurs when p ≡1 (mod 3)), and 3 divides p + 1 if ∆ p = −1 (that is, p ≡2 (mod 3)). (d) Let f(x) = x2 + x + 2 with discriminant ∆= −7. The quadratic recursive sequence with characteristic polynomial f(x) begins as follows, when reduced modulo various primes. Modulo 3: 0, 1, 2, 2, 0, 2, 1, 1, 0, 1, . . ., so that sub3(rn) = 4 and ord3(rn) = 8. Modulo 5: 0, 1, 4, 4, 3, 4, 0, 2, . . ., with sub5(rn) = 6 and (using Proposition 12.2.2 and the fact that ord5(2) = 4) ord5(rn) = 24. Modulo 7: 0, 1, 6, 6, 3, 6, 2, 0, 3, . . ., with sub7(rn) = 7 and (since ord7(3) = 6) ord7(rn) = 42. These results are consistent with Corollary 12.2.7 since −7 3 = −1 = −7 5 and 7 divides −7. (e) Let f(x) = x2 + 2x + 1 with discriminant ∆= 0. If rn is the quadratic recursive sequence with characteristic polynomial f(x), then we ﬁnd, using Theorem 12.1.2, that rn = (−1)n−1n for all n ≥0. If follows that subp(rn) = p for every prime p, while ordp(rn) = 2p for every odd prime p. (This is true because rp+1 = (−1)p(p + 1) ≡−1 (mod p) and ordp(−1) = 2.) These claims are consistent with Corollary 12.2.7 since every prime p divides ∆. (f) Let f(x) = x2 + 2x + 2 with discriminant ∆= −4. We ﬁnd that the corresponding quadratic recursive sequence begins as follows, when it is reduced modulo various primes. Modulo 3: 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, . . ., so that sub3(rn) = 4 and ord3(rn) = 8. Modulo 5: 0, 1, 3, 2, 0, 1, 3, 2, 0, . . ., 89 with sub5(rn) = 4 and ord5(rn) = 4. Modulo 7: 0, 1, 5, 2, 0, 3, 1, 6, 0, . . ., with sub7(rn) = 4 and (since ord7(3) = 6) ord7(rn) = 24. These results are consistent with Corollary 12.2.7 since −4 3 = −1 = −4 7 , while −4 5 = 1. (g) If f(x) = x2 −x + 2, with discriminant ∆= −7, then the corresponding quadratic recursive sequence begins as follows. Modulo 3: 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, . . ., so that sub3(rn) = 4 and ord3(rn) = 8. Modulo 5: 0, 1, 1, 4, 2, 4, 0, 2, . . ., with sub5(rn) = 6 and ord5(rn) = 24 (since ord5(2) = 4). Modulo 7: 0, 1, 1, 6, 4, 6, 5, 0, 4, . . ., with sub7(rn) = 7 and ord7(rn) = 21 (since ord7(4) = 3). Here −7 3 = −1 = −7 5 and 7 divides −7. (h) If f(x) = x2 + x −2, with discriminant ∆= 9, then: Modulo 3: 0, 1, 2, 0, 1, 2, 0, 1, . . ., with sub3(rn) = 3 = ord3(rn). Modulo 5: 0, 1, 4, 3, 0, 1, 4, 3, . . ., with sub5(rn) = 4 = ord5(rn). Modulo 7: 0, 1, 6, 3, 2, 4, 0, 1, . . ., with sub7(rn) = 6 = ord7(rn). Here 3 divides ∆but ∆ p = 1 for every other odd prime p. (i) If f(x) = x2 −2x −1, with discriminant ∆= 8, then: Modulo 3: 0, 1, 2, 2, 0, 2, 1, 1, 0, 1, . . ., with sub3(rn) = 4 and ord3(rn) = 8. Modulo 5: 0, 1, 2, 0, 2, 4, 0, 4, 3, 0, 3, 1, 0, 1, . . ., with sub5(rn) = 3 and ord5(rn) = 12. Modulo 7: 0, 1, 2, 5, 5, 1, 0, 1, . . ., with sub7(rn) = 6 = ord7(rn). Here 8 3 = −1 = 8 5 , while 8 7 = 1. (j) If f(x) = x2 −2x −2, with discriminant ∆= 12, then: Modulo 3: 0, 1, 2, 0, 1, 2, 0, 1, . . ., with sub3(rn) = 3 = ord3(rn). Modulo 5: 0, 1, 2, 1, 1, 4, 0, 3, . . ., with sub5(rn) = 6 and ord5(rn) = 24 (since ord5(3) = 4). Modulo 7: 0, 1, 2, 6, 2, 2, 1, 6, 0, 5, . . ., with sub7(rn) = 8 and ord7(rn) = 48 (since ord7(5) = 6. Here 3 divides 12 while 12 5 = −1 = 12 7 . 90 Section 12.3. Suborder Functions. (1) Using Propositions 12.3.2 and 12.3.4, it suﬃces to ﬁnd subp(c) for 3 ≤c ≤p−1 2 . (a) If p = 7 and c = 3, then c+2 p = 5 7 = −1 and c−2 p = 1 7 = 1. By Theorem 12.3.5, m = sub7(3) divides p+1 = 8 with e2(m) = e2(8). The only possibility is sub7(3) = 8 and then sub7(−3) = 8 also. We compile the values of the suborder function on F7 as follows. c 0 1 2 3 sub7(c) 4 6 1 8 sub7(−c) 3 2 8 (b) Let p = 13. For each of c = 3, c = 5, and c = 6, we ﬁnd that c+2 p = −1 and c−2 p = 1. In those cases, m = sub13(c) divides p + 1 = 14 with e2(m) = e2(14). We must have sub13(c) = 14 and then sub13(−c) = 7, since sub13(c) = 2 only for c = −2. If c = 4, then c+2 p = −1 and c−2 p = −1, so that m = sub13(4) divides p −1 = 12 with e2(m) = e2(12). We conclude that sub13(4) = 12 = sub13(−4) since sub13(c) = 4 only for c = 0. We summarize these results as follows. c 0 1 2 3 4 5 6 sub13(c) 4 6 1 14 12 14 14 sub13(−c) 3 2 7 12 7 7 (c) Let p = 17. We ﬁnd that c+2 p = −1 and c−2 p = 1 for c = 3 and c = 4. Here m = sub17(c) divides p + 1 = 18 with e2(m) = e2(18), and so sub17(c) = 18 and sub17(−c) = 9. (Both sub17(c) = 6 and sub17(c) = 2 occur only in one instance.) For c = 5 and c = 8, we have c+2 p = −1 and c−2 p = −1. Here m = sub17(c) divides p −1 = 16 with e2(m) = e2(16), and we conclude that sub17(c) = 16 = sub17(−c). If c = 6, then c+2 p = 1 and c−2 p = 1, so that m = sub17(c) divides p −1 = 16 with e2(m) < e2(16). Here we conclude that sub17(c) = 8 = sub17(−c). Finally, if c = 7, then c+2 p = 1 and c−2 p = −1. Here m = sub17(c) divides p + 1 = 18 with e2(m) < e2(18), and we conclude that sub17(c) = 9 and sub17(−c) = 18. To summarize: c 0 1 2 3 4 5 6 7 8 sub17(c) 4 6 1 18 18 16 8 9 16 sub17(−c) 3 2 9 9 16 8 18 16 (d) Let p = 19. Here if c = 4 or c = 5, we ﬁnd that c+2 p = 1 and c−2 p = −1. So then m = sub19(c) divides p + 1 = 20 with e2(m) < e2(20). This occurs if m = 10 or m = 5, and we cannot immediately rule out either possibility. Instead we can calculate sub19(c) as the order of the quadratic sequence deﬁned by the characteristic polyno-mial x2 −cx + 1. Modulo 19, the sequence with c = 4 begins 0, 1, 4, −4, −1, 0, 1, . . ., while the sequence with c = 5 begins 0, 1, 5, 5, 1, 0, −1, −5, −5, −1, 0, 1, . . .. We con-clude that sub19(4) = 5 and sub19(5) = 10. For all other values of c, we can determine sub19(c) with the same approach as in the previous parts. We omit the details and compile the results as follows. c 0 1 2 3 4 5 6 7 8 9 sub19(c) 4 6 1 9 5 10 20 9 20 9 sub19(−c) 3 2 18 10 5 20 18 20 18 91 (2) In parts (a)–(e), the characteristic polynomial of rn is x2 −2x −1, with suborder number cp(f) = s2(−t)−1 −2 = 4(−1)−1 −2 = −6 for each prime p. In parts (f)–(j), the charac-teristic polynomial of rn is x2 −x + 3, and we ﬁnd that cp(f) = s2(−t)−1 −2 = 3−1 −2 is the solution of the congruence 3c ≡−5 (mod p). (a) If rn = 2rn−1 + rn−2 with r0 = 0 and r1 = 1, then this sequence modulo 7 begins 0, 1, 2, −2, −2, 1, 0, . . .. So sub7(rn) = 6. Since c = −6 = 1 in F7, we ﬁnd, using the table in Exercise 1 part (a), that sub7(c) = 6 as well. (b) Modulo 11, the same sequence begins 0, 1, 2, 5, 1, −4, 4, 4, 1, −5, 2, −1, 0, . . ., so that sub11(rn) = 12. Since −6 = 5 in F11, an example in §12.3 shows that sub13(c) = 12 also. (c) Modulo 13, the sequence begins 0, 1, 2, 5, −1, 3, 5, 0, . . ., with sub13(rn) = 7. The table in Exercise 1 part (b) shows that sub13(−6) = 7 also. (d) Modulo 17: 0, 1, 2, 5, −5, −5, 2, −1, 0 . . ., with sub17(rn) = 8. We ﬁnd that sub17(−6) = 8 as well, from the table in part (c) of Exercise 1. (e) Modulo 19: 0, 1, 2, 5, −7, −9, −6, −2, 9, −3, 3, 3, 9, 2, −6, 9, −7, −5, 2, −1, 0, . . ., with sub19(rn) = 20. By Exercise 1 part (d), we ﬁnd that sub19(−6) = 20 as well. (f) In the remaining parts, let rn = rn−1 −3rn−2 with r0 = 0 and r1 = 1. Modulo 7, this sequence begins 0, 1, 1, −2, 2, 1, 2, −1, 0 . . ., so that sub7(rn) = 8. The solution of 3c ≡−5 (mod 7) is c = 3, and sub7(3) = 8 as well, using part (a) of Exercise 1. (g) Modulo 11: 0, 1, 1, −2, −5, 1, 5, 2, −2, 3, −2, 0, . . ., with sub11(rn) = 11. The solution of 3c ≡−5 (mod 11) is c = 2. We have that sub11(2) = 1, and that the suborder of rn modulo p = 11 equals p, as in Theorem 12.3.3. (h) Modulo 13: 0, 1, 1, −2, −5, 1, 3, 0 . . ., with sub13(rn) = 7. The solution of 3c ≡−5 (mod 13) is c = −6, and sub13(−6) = 7 also. (i) Modulo 17: 0, 1, 1, −2, −5, 1, −1, −4, −1, −6, −3, −2, 7, −4, −8, 4, −6, −1, 0 . . ., with sub11(rn) = 18. The solution of 3c ≡−5 (mod 17) is c = 4, and sub17(4) = 18 as well from part (c) of Exercise 1. 92 (j) Modulo 19: 0, 1, 1, −2, −5, 1, −3, −6, 3, 2, −7, 6, 8, 9, 4, −4, 3, −4, 6, −1, 0 . . ., with sub13(rn) = 20. The solution of 3c ≡−5 (mod 19) is c = −8, and sub19(−8) = 20 also, from part (d) of Exercise 1. Section 12.4. Suborder Sequences. (1) The suborder sequence of c = 3 (that is, cn+1 = 3cn −cn−1 with c0 = 2 and c1 = 3) modulo p = 17 begins as follows: 2, 3, 7, 1, −4, 4, −1, −7, −3, −2, −3, −7, −1, 4, −4, 1, 7, 3, 2, . . .. We ﬁnd that m = 18 is the smallest positive integer for which cm = 2, and conclude that sub17(3) = 18. Furthermore, sub17(ck) = 18 gcd(18,k) for all k ≥0. For instance, c5 = 4 and c7 = −7 also have suborder 18; c2 = 7, c4 = −4, and c8 = −3 have suborder 9; c3 = 1 has suborder 6; and so forth. Similarly, the suborder sequence of c = 5 modulo p = 17 begins: 2, 5, 6, 8, 0, −8, −6, −5, −2, −5, −6, −8, 0, 8, 6, 5, 2, . . .. Here c16 = 2, and we conclude that c1 = 5, c3 = 8, c5 = −8, and c7 = −5 have suborder 16; c2 = 6 and c6 = −6 have suborder 8; and so forth. From these two sequences, we determine all elements whose suborder is a divisor of 18 or 16, that is, all elements of F17. (2) The suborder sequence of c = 3 modulo p = 37 begins as follows: 2, 3, 7, 18, 10, 12, −11, −8, −13, 6, −6, 13, 8, 11, −12, −10, −18, −7, −3, −2, −3, −7, −18, −10, −12, 11, 8, 13, −6, 6, −13, −8, −11, 12, 10, 18, 7, 3, 2, . . .. Here c38 = 2. Values of cn with gcd(n, 38) = 1 have suborder 38 (for example c1 = 3, c3 = 18, c5 = 12, and so forth); those with gcd(n, 38) = 2 have suborder 19 (such as c2 = 7, c4 = 10, c6 = −11, and so forth). The suborder sequence of c = 4 modulo p = 37 begins 2, 4, 14, 15, 9, −16, 1, −17, 5, 0, −5, 17, −1, 16, −9, −15, −14, −4, −2, −4, −14, −15, −9, 16, −1, 17, −5, 0, 5, −17, 1, −16, 9, 15, 14, 4, 2, . . .. This time we ﬁnd that c36 = 2. Values of cn with gcd(n, 36) = 1 (such as c1 = 4, c5 = −16, c7 = −17, and so forth) have suborder 36; those with gcd(n, 36) = 2 (c2 = 14, c10 = −5, c14 = −9) have suborder 18; those with gcd(n, 36) = 3 (c3 = 15, c15 = −15) have suborder 12; those with gcd(n, 36) = 4 (c4 = 9, c8 = 5, c16 = −14) have suborder 9, and so forth. All elements of F37 appear in one of these sequences, so we determine the entire suborder function on F37. (3) The suborder sequence of c = 3 modulo p = 43 begins as follows: 2, 3, 7, 18, 4, −6, 21, −17, 14, 16, −9, 0, 9, −16, −14, 17, −21, 6, −4, −18, −7, −3, −2, −3, −7, −18, −4, 6, −21, 17, −14, −16, 9, 0, −9, 16, 14, −17, 21, −6, 4, 18, 7, 3, 2, . . ., with c44 = 2. Values of cn with gcd(n, 44) = 1 (for example c1 = 3, c3 = 18, c5 = −6, and so forth) have suborder 44; those with gcd(n, 44) = 2 (c2 = 7, c6 = 21, c10 = −9, c14 = −14, c18 = −4) have suborder 22; those with gcd(n, 44) = 4 (c4 = 4, c8 = 14, c12 = 9, c16 = −21, c20 = −7) have suborder 11, and so forth. The suborder sequence of c = 5 modulo p = 43 begins 2, 5, −20, −19, 11, −12, 15, 1, −10, −8, 13, −13, 8, 10, −1, −15, 12, −11, 19, 20, −5, −2, −5, 20, 19, −11, 12, −15, −1, 10, 8, −13, 13, −8, −10, 1, 15, −12, 11, −19, −20, 5, 2, . . ., 93 with c42 = 2. Values of cn with gcd(n, 42) = 1 (c1 = 5, c5 = −12, c11 = −13, c13 = 10, c17 = −11, c19 = 20) have suborder 42; those with gcd(n, 42) = 2 (c2 = −20, c4 = 11, c8 = −10, c10 = 13, c16 = 12, c20 = −5) have suborder 21; those with gcd(n, 36) = 3 (c3 = −19, c9 = −8, c15 = −15) have suborder 14; those with gcd(n, 42) = 6 (c6 = 15, c12 = 8, c18 = 19) have suborder 7, and so forth. All elements of F43 appear in one of these sequences. (4) Let ck be the k-th term in the suborder sequence of c modulo an odd prime p, and let m be the suborder of c modulo p. (a) Suppose that m = 2k + 1 for some integer k. Then k + 1 ≡−k (mod m), and so ck = ck+1 by part (2) of Theorem 12.4.1. On the other hand, suppose that ck = ck+1 for some integer k. Then then either k + 1 ≡k (mod m) or k + 1 ≡−k (mod m), again by part (2) of Theorem 12.4.1. The ﬁrst case is possible only when m = 1 and c = c1 = 2. The second possibility implies that m divides 2k+1. But then m = 2ℓ+1 is odd, and we must conclude that cℓ+1 = cℓas above. Thus m = 2k + 1 for some integer k if and only if k is the smallest integer for which ck = ck+1. (b) Suppose that m = 4k for some integer k. Using part (5) of Theorem 12.4.1, we have that 2 = c2(2k) = c2 2k −2, so that c2 2k = 4. Since 2k < m, we cannot have c2k = 2, and thus c2k = −2. But now using part (5) of Theorem 12.4.1 again, we see that −2 = c2k = c2 k −2 and conclude that ck = 0. Conversely, if ck = 0 for some k, then c2k = 02 −2 = −2 and c4k = (−2)2 −2 = 2. Thus m = 4k if and only if k is the smallest integer for which ck = 0. (c) Suppose that m = 6k for some integer k. Then 4k ≡−2k (mod m) so that c4k = c2k by part (2) of Theorem 12.4.1. But now we have c2k = c2(2k) = c2 2k −2 by part (5) of the same theorem. Thus c2k satisﬁes the equation x2 −x −2 = (x −2)(x + 1) = 0, so that either c2k = 2 or c2k = −1 in the ﬁeld Fp. The ﬁrst case is impossible if m = 6k, and thus c2k = −1. But now −1 = c2k = c2 k −2 and so ck = 1 or ck = −1. It is established in part (4) of Corollary 12.4.3 that the second case implies that c3k = 2, contrary to our assumption. Thus ck = 1. Conversely, if ck = 1, then c2k = c2 k −2 = −1, so that c3k = c2k · ck −ck = −2 by part (4) of Theorem 12.4.1, and then c6k = c2 3k −2 = 2. So m = 6k if and only if k is the smallest positive integer for which ck = 1. (5) (a) The suborder sequence of c = 4 modulo p = 61 (that is, cn+1 = 4cn−cn−1 with c0 = 2 and c1 = 4) begins as follows: 2, 4, 14, −9, 11, −8, 18, 19, −3, 30, 1, −26, 17, −28, −7, 0, . . .. From the fact that c10 = 1, or using c15 = 0, we determine that sub61(4) = 60 (by parts (5) and (3) of Corollary 12.4.3). Now we know that c30−k = −ck for 0 ≤k ≤15, and c60−k = ck for 0 ≤k ≤30 (by parts (3) and (2) of Theorem 12.4.1). By Corollary 12.4.2, we then ﬁnd that the suborders of c1 = 4, c7 = 19, c11 = −26, c13 = −28, c17 = 28, c19 = 26, c23 = −19, and c29 = −4 are all equal to 60; the suborders of c2 = 14, c14 = −7, c22 = −c8 = 3, and c26 = −c4 = −11 all equal 30; the suborders of c4 = 11, c8 = −3, c16 = −c14 = 7, and c28 = −c2 all equal 15; the suborders of c3 = −9, c9 = 30, c21 = −30, and c27 −9 all equal 20; the suborders of c5 = −8 and c25 = 8 equal 12; the suborders of c6 = 18 and c18 = −c12 = −17 equal 10; and the suborders of c12 = 17 and c24 = −c6 = −18 equal 5 (along with the usual elements of suborder 6, 4, 3, 2, and 1). (b) The suborder sequence of c = 5 modulo p = 61 begins: 2, 5, 23, −12, −22, 24, 20, 25, −6, 16, 25, −13, −29, −10, −21, 27, −27, . . ., 94 with c15 = −c16. It follows that sub61(5) = 4(15) + 2 = 62 by part (2) of Corollary 12.4.3, and that c31−k = −ck for 0 ≤k ≤15 by part (3) of Theorem 12.4.1. Thus sub61(ck) = 62 if 1 ≤k < 31 is odd, that is, for the following values: 5, −12, 24, 25, 16, −13, −10, 27, 21, 29, −25, 6, −20, 22, and −23, while sub61(ck) = 31 if 1 ≤k < 31 is even, that is, for 23, −22, 20, −6, 25, −29, −21, −27, 10, 13, −16, −25, −24, 12, and −5. (c) The suborder sequence of c = 3 modulo p = 67 begins: 2, 3, 7, 18, −20, −11, −13, −28, −4, 16, −15, 6, 33, 26, −22, −25, 14, 0, . . .. Since c17 = 0, then sub67(3) = 68, and c34−k = −ck for 0 ≤k ≤17. It follows that sub67(ck) = 68 for odd values of k with 1 ≤k ≤33 and k ̸= 17, that is, for 3, 18, −11, −28, 16, 6, 26, −25, 25, −26, −6, −16, 28, 11, −18, and −3; while c2 = 7, c6 = −13, c10 = −15, c14 = −22, c18 = −14, c22 = −33, c26 = 4, and c30 = 20 have suborder 34; and c4 = −20, c8 = −4, c12 = 33, c16 = 14, c20 = 22, c24 = 15, c28 = 13, and c32 = −7 have suborder 17. (d) The suborder sequence of c = 5 modulo p = 67 begins: 2, 5, 23, −24, −9, −21, −29, 10, 12, −17, −30, 1, −32, −27, 31, −9, 8, −8, . . .. Here c11 = 1 so that sub67(5) = 6(11) = 66. (Also, since c16 = −c17, we know that sub67(5) = 4(16) + 2 = 66.) Thus c33−k = −ck for 1 ≤k ≤16, and we can compute the following values of the suborder function: c1 = 5, c5 = −21, c7 = 10, c13 = −27, c17 = −8, c19 = −31, c23 = 30, c25 = −12, c29 = 9, and c31 = −23 have suborder 66; c2 = 23, c4 = −9, c8 = 12, c10 = −30, c14 = 31, c16 = 8, c20 = 27, c26 = −10, c28 = 21, and c32 = −5 have suborder 33; c3 = −24, c9 = −17, c15 = −19, c21 = 32, and c27 = 29 have suborder 22; and c6 = −29, c12 = −32, c18 = 19, c24 = 17, and c30 = 24 have suborder 11. (6) The characteristic polynomial of the quadratic sequence rn = 2rn−1 + rn−2 is f(x) = x2 −2x −1, with suborder number cp(f) = 22(−1)−1 −2 = −6 for every prime p. (a) Modulo p = 29, the sequence begins 0, 1, 2, 5, 12, 0, . . ., so that sub29(rn) = 5. Table 12.1 shows that sub29(6) = 10, and so sub29(−6) = 5 by Proposition 12.3.4. (b) Modulo p = 31, the sequence begins 0, 1, 2, 5, 12, −2, 8, 14, 5, −7, −9, 6, 3, 12, −4, 4, 4, 12, −3, 6, 9, −7, −5, 14, −8, −2, −12, 5, −2, 1, 0, . . ., with sub31(rn) = 30. This is consistent with the calculation of sub31(6) = 15 in Table 12.1. (c) Modulo p = 37, the sequence begins 0, 1, 2, 5, 12, −8, −4, −16, 1, −14, 10, 6, −15, 13, 11, −2, 7, 12, −6, 0, . . . so that sub37(rn) = 19, consistent with the calculation of sub37(6) = 38 in Table 12.1. (d) Modulo p = 41, we ﬁnd 0, 1, 2, 5, 12, −12, −12, 5, −2, 1, 0, . . ., with sub41(rn) = 10. From Table 12.1, we see that sub41(6) = 5. (e) Modulo p = 43: 0, 1, 2, 5, 12, −14, −16, −3, 21, −4, 13, −21, 14, 7, −15, 20, −18, −16, −7, 13, 19, 8, −8, −8, 19, −13, −7, 16, −18, −20, −15, −7, 14, 21, 13, 4, 21, 3, −16, 14, 12, −5, 2, −1, 0, . . ., so that sub43(rn) = 44. This is consistent with the Table 12.1 value of sub43(6) = 44, using Proposition 12.3.4, since this number is divisible by 4. 95 (f) Modulo p = 47: 0, 1, 2, 5, 12, −18, 23, −19, −15, −2, −19, 7, −5, −3, −11, 22, −14, −6, 21, −11, −1, −13, 20, −20, −20, −13, 1, −11, −21, −6, 14, 22, 11, −3, 5, 7, 19, −2, 15, −19, −23, −18, −12, 5, −2, 1, 0, . . ., so that sub47(rn) = 46, consistent with the Table 12.1 value of sub47(6) = 23. (7) In each part, we use Theorem 12.3.5 to determine preliminary possibilities for subp(c), then calculate the suborder subsequence (dn) of c modulo p and describe the implications of that sequence via Theorem 12.4.5. Finally we calculate the suborder sequence (cn) of c modulo p to conﬁrm these claims. (a) If c = 3 and p = 53, then c+2 p = 5 53 = −1 and c−2 p = 1 53 = 1, so that m = sub53(3) divides p+1 = 54 with e2(m) = e2(54) = 1. We can rule out m = 2 and m = 6 by Proposition 12.3.2, and so m = 18 or m = 54. The suborder subsequence with d0 = 3 and dn = d2 n−1 −2 for n > 0 begins 3, 7, −6, −19, −12, −17, 22, 5, 23, −3, 7, . . .. We see that d10 = d1, conﬁrming that e2(m) = 1. Furthermore t = 10 −1 = 9 is the smallest positive integer with 2t ≡±1 (mod m/2). Since 23 ≡−1 (mod 9), we eliminate the possibility that m = 18, and conclude that m = 54. (One can verify that 29 ≡−1 (mod 27).) The suborder sequence of c = 3 modulo p = 53 (that is, c0 = 2, c1 = 3, and cn+1 = 3 · cn −cn−1 for n > 1) begins 2, 3, 7, 18, −6, 17, 4, −5, −19, 1, . . .. Since c9 = 1, we know that m = sub53(3) = 6(9) = 54 by part (5) of Corollary 12.4.3. (b) Let c = 8 and p = 53. Then c+2 p = 10 53 = 1 and c−2 p = 6 53 = 1, so that m = sub53(8) divides p −1 = 52 with e2(m) < e2(52). We can eliminate m = 1 and 2, and conclude that m = 13 or 26. The suborder subsequence of c begins 8, 9, 26, −15, 11, 13, 8, . . ., with d6 = d0. Thus e2(m) = 0 and we must have m = 13. We can conﬁrm that t = 6 is the smallest positive integer with 2t ≡−1 (mod 13). The suborder sequence of c = 8 modulo p = 53 begins 2, 8, 9, 11, 26, −15, 13, 13, . . .. With c6 = c7, we conﬁrm that m = sub53(8) = 2(6) + 1 = 13 by part (1) of Corol-lary 12.4.3. (c) Let c = 6 and p = 59. Since c+2 p = 8 59 = −1 and c−2 p = 4 59 = 1, we know that m = sub59(6) divides p + 1 = 60 with e2(m) = e2(60) = 2. So m = 12, 20, or 60. (subp(c) = 4 only when c = 0 in Fp.) The suborder subsequence with d0 = 6 and dn = d2 n−1 −2 for n > 0 begins 6, −25, −26, 25, −26, . . ., with d4 = d0. (This conﬁrms that e2(m) = 2.) Here t = 4 −2 = 2 is the smallest positive integer for which 2t ≡±1 (mod m/4). Since 21 = 2 ≡−1 (mod 3), we can eliminate the possibility that m = 12. Then since 22 = 4 ≡−1 (mod 5) but 4 ̸≡±1 (mod 15), we conclude that m = 20. The suborder sequence of c = 6 modulo p = 59 begins 2, 6, −25, 21, −26, 0, . . ., which conﬁrms that m = sub59(6) = 4(5) = 20 by part (3) of Corollary 12.4.3. 96 (d) Let c = 7 and p = 59. Here c+2 p = 9 59 = 1 and c−2 p = 5 59 = 1, so we know that m = sub59(7) divides p −1 = 58 with e2(m) < e2(58). The only possibility is m = 29 (since m = 1 only for c = 2 in Fp). The suborder subsequence with d0 = 8 and dn = d2 n−1 −2 for n > 0 begins 7, −12, 24, −16, 18, 27, 19, 5, 23, −4, 14, 17, −8, 3, 7, . . ., with d14 = d0. We can conﬁrm that t = 14 is the smallest positive integer for which 2t ≡±1 (mod 29). From the suborder sequence for c = 7, 2, 7, −12, 27, 24, 23, 19, −8 −16, 14, −4, 17, 5, 18, 3, 3, . . ., with c14 = c15, we also see that m = sub59(7) = 2(14) + 1 = 29. (e) Let c = 3 and p = 61. Then c+2 p = 5 61 = 1 and c−2 p = 1 61 = 1, so that m = sub61(3) divides p −1 = 60 with e2(m) < e2(60). We can eliminate m = 1, 2, 3, and 6 by Proposition 12.3.2, but remaining possibilities are m = 5, 10, 15, or 30. The suborder subsequence with d0 = 3 and dn = d2 n−1 −2 for n > 0 begins 3, 7, −14, 11, −3, 7, . . ., with d5 = d1. It follows that e2(m) = 1 (so that m = 10 or 30) and that t = 5−1 = 4 is the smallest positive integer for which 2t ≡±1 (mod m/2). Since 22 = 4 ≡−1 (mod 5), we can eliminate the possibility that m = 10, and conclude that m = 30. The suborder sequence with c = 3, 2, 3, 7, 18, −14, 1, . . ., conﬁrms that sub61(3) = 6(5) = 30. (f) If c = 4 and p = 61, then c+2 p = 6 61 = −1 and c−2 p = 2 61 = −1. Thus m = sub61(4) divides p −1 = 60 with e2(m) = e2(60). Possibilities are m = 12, 20, or 60. The suborder subsequence with d0 = 4 and dn = d2 n−1 −2 for n > 0 begins 4, 14, 11, −3, 7, −14, 11, . . ., with d6 = d2. This conﬁrms that e2(m) = 2, and we know further that t = 6 −2 = 4 is the smallest positive integer for which 2t ≡±1 (mod m/4). Since 21 = 2 ≡−1 (mod 3) and 22 = 4 ≡−1 (mod 5), the only possibility is m = 60 (with 24 = 16 ≡1 (mod 15)). The suborder sequence with c = 4, 2, 4, 14, −9, 11, −8, 18, 19, −3, 30, 1, . . ., conﬁrms that sub61(4) = 6(10) = 60. (8) In each part, we consider divisors m of p −1 or p + 1 other than 1, 2, 3, 4, and 6, and conﬁrm, using Table 13.1, that the number of elements c of Fp with subp(c) = m is φ(m) 2 . (There is precisely one element with m = 1, 2, 3, 4, 6 for p > 3 by Proposition 12.3.2.) (a) Let p = 17. For divisors of p −1 = 16, we ﬁnd that sub17(c) = 8 for c = ±6 and sub17(c) = 16 for c = ±5, ±8. We verify that φ(8) 2 = 2 and φ(16) 2 = 4. For divisors of p + 1 = 18, we have sub17(c) = 9 for c = −3, −4, 7 and sub17(c) = 18 for c = 3, 4, −7. Here φ(9) 2 = 3 = φ(18) 2 . (b) Let p = 19. For divisors of p −1 = 18, we ﬁnd sub19(c) = 9 for c = 3, 7, 9 and sub19(c) = 18 for c = −3, −7, −9. For divisors of p +1 = 20, we have sub19(c) = 5 for c = 4, −5, sub19(c) = 10 for c = −4, 5, and sub19(c) = 20 for c = ±6, ±8. These are the predicted number of values in each case. 97 (c) Let p = 29. For divisors of p −1 = 28, we have sub29(c) = 7 for c = 3, 7, −11, sub29(c) = 14 for c = −3, −7, 11, and sub29(c) = 28 for c = ±10, ±12, ±13. For divisors of p+1 = 30, we have sub29(c) = 5 for c = 5, −6, sub29(c) = 10 for c = −5, 6, sub29(c) = 15 for c = 4, −8, −9, 14, and sub29(c) = 30 for c = −4, 8, 9, −14. (d) Let p = 31. For divisors of p −1 = 30, we have sub31(c) = 5 for c = 12, −13, sub31(c) = 10 for c = −12, 13, sub31(c) = 15 for c = 3, 6, 7, −15, and sub31(c) = 30 for c = −3, −6, −7, 15. For divisors of p + 1 = 32, we have sub31(c) = 8 for c = ±8, sub31(c) = 16 for c = ±5, ±14, and sub31(c) = 32 for c = ±4, ±9, ±10, ±11. (e) Let p = 37. For divisors of p −1 = 36, we have sub37(c) = 9 for c = 5, 9, −14, sub37(c) = 12 for c = ±15, sub37(c) = 18 for c = −5, −9, 14, and sub37(c) = 36 for c = ±4, ±16, ±17. For divisors of p + 1 = 38, we have sub37(c) = 19 for c = −3, −6, 7, 8, 10, −11, −12, −13, −18 and sub37(c) = 38 for c = 3, 6, −7, −8, −10, 11, 12, 13, 18. (Note that φ(19) 2 = 9 = φ(38) 2 .) (f) Let p = 43. For divisors of p −1 = 42, we have sub43(c) = 7 for c = 8, 15, 19, sub43(c) = 14 for c = −8, −15, −19, sub43(c) = 21 for c = −5, −10, 11, 12, 13, −20, and sub43(c) = 42 for c = 5, 10, −11, −12, −13, 20. For divisors of p +1 = 44, we have sub43(c) = 11 for c = 4, −7, 9, 14, −21, sub43(c) = 22 for c = −4, 7, −9, −14, 21, and sub43(c) = 44 for c = ±3, ±6, ±16, ±17, ±18. (g) Let p = 47. For divisors of p −1 = 46, we have sub47(c) = 23 for c = 4, 5, 6, 10, −13, 14, 16, −17, 19, −21, 23 and sub47(c) = 46 for c = −4, −5, −6, −10, 13, −14, −16, 17, −19, 21, −23. For divisors of p+1 = 48, we have sub47(c) = 8 for c = ±7, sub47(c) = 12 for c = ±12, sub47(c) = 16 for c = ±3, ±18, sub47(c) = 24 for c = ±15, ±22, and sub47(c) = 48 for c = ±8, ±9, ±11, ±20. Section 13.1. Recursive Sequences and Automorphs. (1) In each part, applying the equivalence algorithm of Theorem 10.1.2 to the quadratic form φ = (1 : 0) produces the smallest solution (x, y) = (q, r) in positive integers of φ(x, y) = 1, as in Theorem 11.1.1. We omit those calculations. (a) For ∆= 8, the smallest positive solution of φ(x, y) = x2 −2y2 = 1 is (q, r) = (3, 2). Here b = 0 (for φ(x, y) = x2 + bxy + cy2), and so the sequence of Theorem 13.1.2 is deﬁned by an = 6an−1 −an−2 with a0 = 0 and a1 = 1. The following table lists the terms of this sequence for 1 ≤n ≤4, along with the four smallest positive solutions (x, y) = (qn, rn) of x2 −2y2 = 1, produced by the equations qn = anq −an−1 = 3an −an−1 and rn = anr = 2an of x2 −2y2 = 1. n 1 2 3 4 an 1 6 35 204 qn 3 17 99 577 rn 2 12 70 408 (b) For ∆= 12, (q, r) = (2, 1) is the smallest positive solution of φ(x, y) = x2 −3y2 = 1, and we let an = 4an−1 −an−2 with a0 = 0 and a1 = 1. We obtain all other positive 98 solutions of φ(x, y) = 1 as (qn, rn) with qn = 2an −an−1 and rn = an. The ﬁrst four solutions are as in the following table. n 1 2 3 4 an 1 4 15 56 qn 2 7 26 97 rn 1 4 15 56 (c) For ∆= 13, φ(x, y) = x2 + xy −3y2 = 1 has smallest positive solution (q, r) = (4, 3). Here we let an = (2q +br)an−1 −an−2 = 11an−1 −an−2 (where b = 1 is the coeﬃcient of xy in φ(x, y)) and with a0 = 0 and a1 = 1 as always. All positive integer solutions (qn, rn) of φ(x, y) = 1 are given by qn = 4an −an−1 and rn = 3an. n 1 2 3 4 an 1 11 120 1309 qn 4 43 469 5116 rn 3 33 360 3927 (d) For ∆= 17, φ(x, y) = x2+xy−4y2 = 1 has smallest positive solution (q, r) = (25, 16). So an = 66an−1 −an−2, with qn = 25an −an−1 and rn = 16an producing all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 66 4355 287364 qn 25 1649 108809 7179745 rn 16 1056 69680 4597824 (e) For ∆= 21, (q, r) = (2, 1) is the smallest positive solution of φ(x, y) = x2+xy−5y2 = 1, and an = 5an−1 −an−2. Here qn = 2an −an−1 and rn = an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 5 24 115 qn 2 9 43 206 rn 1 5 24 115 (f) For ∆= 24, (q, r) = (5, 2) is the smallest positive solution of φ(x, y) = x2 −6y2 = 1, and an = 10an−1 −an−2. So qn = 5an −an−1 and rn = 2an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 10 99 980 qn 5 49 485 4801 rn 2 20 198 1960 (g) For ∆= 28, (q, r) = (8, 3) is the smallest positive solution of φ(x, y) = x2 −7y2 = 1, and an = 16an−1 −an−2. Here qn = 8an −an−1 and rn = 3an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 16 255 4064 qn 8 127 2024 32257 rn 3 48 765 12192 99 (h) For ∆= 33, φ(x, y) = x2 +xy −8y2 = 1 has smallest positive solution (q, r) = (19, 8). So an = 46an−1 −an−2, and qn = 19an −an−1 and rn = 8an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 46 2115 97244 qn 19 873 40139 1845521 rn 8 368 16920 777952 (i) For ∆= 37, φ(x, y) = x2+xy−9y2 = 1 has smallest positive solution (q, r) = (61, 24). So an = 146an−1 −an−2, and qn = 61an −an−1 and rn = 24an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 146 21315 3111844 qn 61 8905 1300069 189801169 rn 24 3504 511560 74684256 (j) For ∆= 40, (q, r) = (19, 6) is the smallest positive solution of φ(x, y) = x2−10y2 = 1, and an = 38an−1 −an−2. Here qn = 19an −an−1 and rn = 6an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 38 1443 54796 qn 19 721 27379 1039681 rn 6 228 8658 328776 (k) For ∆= 41, φ(x, y) = x2 + xy −10y2 = 1 has smallest positive solution (q, r) = (1729, 640). So an = 4098an−1 −an−2, and qn = 1729an −an−1 and rn = 640an produce all positive solutions of φ(x, y) = 1. n 1 2 3 4 an 1 4098 16793603 68820180996 qn 1729 7085441 29036135489 118990076148481 rn 640 2622720 10747905920 44044915837440 Section 13.2. An Application to Pell’s Equation. (1) We apply Pell’s equation algorithm (Theorem 9.2.2) to d = 847 compiling the following data. i 0 1 2 3 4 5 6 7 8 9 10 a 1 6 37 19 9 7 9 19 37 6 1 k 0 −29 −25 −12 −26 −28 −28 −26 −12 −25 −29 q 29 9 1 2 6 8 5 2 1 9 58 m 29 262 291 844 5355 43684 267459 578602 846061 8193151 n 1 9 10 29 184 1501 9190 19881 29071 281520 Here k1 = −29 is the smallest integer with d −k2 1 is positive, and so a1 = 847−(−29)2 1 = 6. Now k2 = −25 is the smallest integer with k2 ≥−29 for which k2 ≡−k1 (mod a1), and then a2 = 847−(−25)2 6 = 37. Continuing this process, we ﬁnd that a10 = 1, and the terms of the qi, mi, and ni sequences are as deﬁned in Theorem 9.2.2. The fundamental solution of x2 −847y2 = 1 is given by (m9, n9) = (8193151, 281520). (2) In parts (a)–(e), the fundamental solution of x2 −2y2 = 1 is (q, r) = (3, 2), and so we let an = 6an−1 −an−2 with a0 = 0 and a1 = 1. 100 (a) Let d = 18 = 2 · 32, so that p = 3. Since p divides q, we know that p divides r2. Thus (3 + 2 √ 2)2 = 17 + 12 √ 2 = 17 + 4 √ 18 is the fundamental solution of x2 −18y2 = 1. We can verify this with the Pell’s equation algorithm of Theorem 9.2.2 applied to d = 18. i 0 1 2 a 1 2 1 k 0 −4 −4 q 4 4 8 m 4 17 n 1 4 (b) Let d = 50 = 2 · 52, with p = 5. Here 2 5 = −1, so the smallest m for which p divides rm is a divisor of p+1 2 = 3. Since p does not divide r1 = 2, we conclude that m = 3, and so (3 + 2 √ 2)3 = 99 + 70 √ 2 = 99 + 14 √ 50 is the fundamental solution of x2 −50y2 = 1. The following table veriﬁes this claim. i 0 1 2 a 1 1 1 k 0 −7 −7 q 7 14 14 m 7 99 n 1 14 (c) Let d = 98 = 2 · 72, with p = 7. Since 2 7 = 1, the smallest m for which p divides rm is a divisor of p−1 2 = 3. Again, p does not divide r1 = 2, so we conclude that m = 3, and (3 + 2 √ 2)3 = 99 + 70 √ 2 = 99 + 10 √ 98 is the fundamental solution of x2 −98y2 = 1. To verify: i 0 1 2 3 4 a 1 17 2 17 1 k 0 −9 −8 −8 −9 q 9 1 8 1 18 m 9 10 89 99 n 1 1 9 10 (d) Let d = 242 = 2 · 112, with p = 11. Here 2 11 = −1, so the smallest m for which p divides rm is a divisor of p+1 2 = 6. From previous calculations, we see that p does not divide r1 = 2, r212, or r3 = 70, and so we conclude that m = 6. (3 + 2 √ 2)6 = (99 + 70 √ 2)2 = 19601 + 13860 √ 2 = 19601 + 1260 √ 242 is the fundamental solution of x2 −242y2 = 1. To verify: i 0 1 2 3 4 5 6 7 8 9 10 a 1 17 14 7 23 2 23 7 14 17 1 k 0 −15 −2 −12 −9 −14 −14 −9 −12 −2 −15 q 15 1 1 3 1 14 1 3 1 1 30 m 15 16 31 109 140 2069 2209 8696 10905 19601 n 1 1 2 7 9 133 142 559 701 1260 101 (e) Let d = 450 = 2·152, with p = 11. In part (a), we saw that 3 divides r2 and so divides r2k for every k. Likewise from part (b), we have that 5 divides r3 and so divides r3k for every k. Thus 15 divides r6. (3 + 2 √ 2)6 = 19601 + 13860 √ 2 = 19601 + 924 √ 450 is the fundamental solution of x2 −450y2 = 1. i 0 1 2 3 4 5 6 7 8 a 1 9 25 14 9 14 25 9 1 k 0 −21 −15 −10 −18 −18 −10 −15 −21 q 21 4 1 2 4 2 1 4 42 m 21 85 106 297 1294 2885 4179 19601 n 1 4 5 14 61 136 197 924 (f) The fundamental solution of x2 −5y2 = 1 is (q, r) = (9, 4), or v = 9 + 4 √ 5. Since 2 divides r, then 9 + 2 √ 20 is the fundamental solution of x2 −20y2 = 1. We verify this with the Pell’s equation algorithm as follows. i 0 1 2 a 1 4 1 k 0 −4 −4 q 4 2 8 m 4 9 n 1 2 (g) Since 3 divides q in the fundamental solution of x2 −5y2 = 1, then (9 + 4 √ 5)2 = 161 + 72 √ 5 = 161 + 24 √ 45 is the fundamental solution of x2 −45y2 = 1. To verify: i 0 1 2 3 4 5 6 a 1 9 4 5 4 9 1 k 0 −6 −3 15 −5 −3 −6 q 6 1 2 2 2 1 12 m 6 7 20 47 114 161 n 1 1 3 7 17 24 (h) Since 2 divides r and 3 divides q in the fundamental solution of x2 −5y2 = 1, we also ﬁnd that (9 + 4 √ 5)2 = 161 + 72 √ 5 = 161 + 12 √ 180 is the fundamental solution of x2 −180y2 = 1. To verify: i 0 1 2 3 4 a 1 11 9 11 1 k 0 −13 −9 −9 −13 q 13 2 2 2 26 m 13 27 67 161 n 1 2 5 12 102 (i) The fundamental solution of x2 −7y2 = 1 is q + r √ 7 = 8 + 3 √ 7. Since 3 divides r, we also have that 8 + √ 63 is the fundamental solution of x2 −63y2 = 1. To verify: i 0 1 2 a 1 14 1 k 0 −7 −7 q 7 1 14 m 7 8 n 1 1 (j) The fundamental solution of x2 −10y2 = 1 is q + r √ 7 = 19 + 3 √ 10. Again 3 divides r, so we also have that 19 + √ 90 is the fundamental solution of x2 −90y2 = 1. i 0 1 2 a 1 9 1 k 0 −9 −9 q 9 2 18 m 9 19 n 1 1 Section 13.3. Quadratic Subdomains of Positive Discriminant. (1) For each ∆, we let (q, r) be the smallest positive solution of φ(x, y) = 1, where φ(x, y) = x2 + εxy + ε2−∆ 4 y2. If an = (2q + εr)an−1 −an−2 with a0 = 0 and a1 = 1, then sp is the smallest positive integer so that p divides rasp, and the order of the kernel of the projection homomorphism from Fp2∆to F∆is given by \|K(∆, p)\| = p−(∆ p) sp . If p divides r, then sp = 1; otherwise sp is the suborder of the sequence an modulo p. Notice that for each ∆, the suborder number of the characteristic polynomial of an (as in Theorem 12.3.3) is c = (2q + εr)2 −2 for every prime p. (a) If ∆= 8, then (q, r) = (3, 2) and an = 6an−1 −an−2, with suborder number c = 34. We list c modulo odd primes p, along with sp, and \|K(∆, p)\| for each prime p < 20 in the following table. p 2 3 5 7 11 13 17 19 c − −2 −1 −1 1 −5 0 −4 sp 1 2 3 3 6 7 4 10 \|K\| 2 2 2 2 2 2 4 2 Here s2 = 1 since r = 2, and calculation of sp for all other primes comes from Proposition 12.3.2 or Table 12.1. (b) If ∆= 12, then (q, r) = (2, 1) and an = 4an−1 −an−2, with suborder number c = 14. p 2 3 5 7 11 13 17 19 c − 2 −1 0 3 1 −3 −5 sp 2 3 3 4 5 6 9 5 \|K\| 1 1 2 2 2 2 2 4 When 2 does not divide r, as in this case, then s2 = 2 or 3 depending on whether 2q + εr is even or odd. Note that when c ≡2 (mod p), then the suborder of the sequence an modulo p equals p. (See Theorem 12.3.3.) 103 (c) If ∆= 13, then (q, r) = (4, 3) and an = 11an−1−an−2, with suborder number c = 119. p 2 3 5 7 11 13 17 19 c − 2 −1 0 −2 2 0 5 sp 3 1 3 4 2 13 4 10 \|K\| 1 2 2 2 6 1 4 2 Here s3 = 1 since 3 divides r. (d) If ∆= 17, then (q, r) = (25, 16) and an = 66an−1 −an−2, with c = 4354. p 2 3 5 7 11 13 17 19 c − −2 −1 0 −2 −1 2 3 sp 1 2 3 4 2 3 17 9 \|K\| 1 2 2 2 6 4 1 2 (e) If ∆= 21, then (q, r) = (2, 1) and an = 5an−1 −an−2, with c = 23. p 2 3 5 7 11 13 17 19 c − 2 −2 2 1 −3 6 4 sp 3 3 2 7 6 7 8 5 \|K\| 1 1 2 1 2 2 2 4 (f) If ∆= 24, then (q, r) = (5, 2) and an = 10an−1 −an−2, with c = 98. p 2 3 5 7 11 13 17 19 c − 2 −2 0 −1 −6 −4 3 sp 1 3 2 4 3 7 9 9 \|K\| 2 1 2 2 4 2 2 2 (g) If ∆= 28, then (q, r) = (8, 3) and an = 16an−1 −an−2, with c = 254. p 2 3 5 7 11 13 17 19 c − 2 −1 2 1 −6 −1 7 sp 2 1 3 7 6 7 3 9 \|K\| 1 2 2 1 2 2 6 2 (h) If ∆= 33, then (q, r) = (19, 8) and an = 46an−1 −an−2, with c = 2114. p 2 3 5 7 11 13 17 19 c − 2 −1 0 2 −5 6 5 sp 1 3 3 4 11 7 8 10 \|K\| 1 1 2 2 1 2 2 2 (i) If ∆= 37, then (q, r) = (61, 24) and an = 146an−1 −an−2, with c = 21314. p 2 3 5 7 11 13 17 19 c − 2 −1 −1 −4 −6 −4 −4 sp 1 1 3 3 5 7 9 10 \|K\| 3 2 2 2 2 2 2 2 (j) If ∆= 40, then (q, r) = (19, 6) and an = 38an−1 −an−2, with c = 1442. p 2 3 5 7 11 13 17 19 c − 2 2 0 1 −1 −3 −2 sp 1 1 5 4 6 3 9 2 \|K\| 2 3 1 2 2 4 2 10 104 (k) If ∆= 41, then (q, r) = (1729, 640) and an = 4098an−1 −an−2, with c = 16793602. p 2 3 5 7 11 13 17 19 c − −2 2 0 1 −6 −1 −4 sp 1 2 1 4 6 7 3 10 \|K\| 3 2 5 2 2 2 6 2
1234	http://web.mit.edu/ngh-group/pubs%20final%20website/2-28_DFD05paper.pdf	The limits of Navier-Stokes theory and kinetic extensions for describing small-scale gaseous hydrodynamics Nicolas G. Hadjiconstantinoua Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, Massachusetts 02139 Received 28 June 2006; accepted 1 September 2006; published online 27 November 2006 This paper reviews basic results and recent developments in the ﬁeld of small-scale gaseous hydrodynamics which has received signiﬁcant attention in connection with small-scale science and technology. We focus on the modeling challenges arising from the breakdown of the Navier-Stokes description, observed when characteristic lengthscales become of the order of, or smaller than, the molecular mean free path. We discuss both theoretical results and numerical methods development. Examples of the former include the limit of applicability of the Navier-Stokes constitutive laws, the concept of second-order slip and the appropriate form of such a model, and how to reconcile experimental measurements of slipping ﬂows with theory. We also review a number of recently developed theoretical descriptions of canonical nanoscale ﬂows of engineering interest. On the simulation front, we review recent progress in characterizing the accuracy of the prevalent Boltzmann simulation method known as direct simulation Monte Carlo. We also present recent variance reduction ideas which address the prohibitive cost associated with the statistical sampling of macroscopic properties in low-speed ﬂows. © 2006 American Institute of Physics. DOI: 10.1063/1.2393436 I. INTRODUCTION A. Overview Small-scale, atmospheric pressure, internal gaseous ﬂows have received signiﬁcant attention in recent years in connection with microscale and nanoscale science and technology.1,2 In addition to applications of practical interest, small-scale gaseous hydrodynamics continues to attract sig-niﬁcant attention due to the number of scientiﬁc challenges it poses. In this paper we discuss some of the recent progress in modeling small-scale internal gaseous ﬂows of engineering interest in which extensions or alternatives to the Navier-Stokes description are required. In small-scale ﬂows, the Navier-Stokes description is expected to fail2,3 when the characteristic hydrodynamic lengthscale approaches the ﬂuid “internal lengthscale;” in dilute gases this scale is associated with the molecular mean free path, the average distance trav-eled between intermolecular collisions. Until recently, and with the exception of the classical shear, pressure-driven, and thermal-creep-driven duct ﬂows, very few ﬂows of engineer-ing interest have been theoretically characterized, primarily because previous efforts had focused on external, high-speed ﬂows associated with ﬂight in the upper atmosphere. Gaining a fundamental understanding in this new low-speed regime is important for facilitating the design of small-scale devices but also for educational purposes. For this reason, particular emphasis will be given here to basic theoretical results for monoatomic gases, which can facilitate a fundamental under-standing of the ﬂow physics. B. Background In the small-scale, low-speed ﬂows of interest here, the failure of the Navier-Stokes description can be quantiﬁed by the Knudsen number, Kn=/H, where is the molecular mean free path and H is the characteristic hydrodynamic lengthscale. The Navier-Stokes description in the interest of simplicity, limiting cases of this description e.g., Stokes ﬂow will not be denoted separately but will be understood to apply under the appropriate conditions corresponds to the collision-dominated transport limit of small Knudsen number Kn1. More speciﬁcally, in the presence of hydrodynamic gradients in a homogeneous gas over lengthscales character-ized by Kn1, the Chapman-Enskog4,5 or Grad-Hilbert;6 see Ref. 7 for a critical comparison expansion procedure can be applied to the governing kinetic Boltzmann equation with Kn as a small parameter to show that the gas response can be described by linear-gradient constitutive relations which lead to the Navier-Stokes description.4,5 Close to the system boundaries, however, the inhomogeneity introduced by the former results in kinetic effects that cannot be cap-tured by the Navier-Stokes description. Although kinetic effects are always present adjacent to system boundaries in the presence of nonequilibrium, in the Kn1 limit the former remain localized to thin layers thickness of order known as Knudsen layers. Analysis of the governing kinetic Boltzmann equation shows8 that the inhomogeneity induced by the wall presence may be ac-counted for by a boundary-layer-type theory in which the Navier-Stokes description, albeit subject to slip boundary aThis paper is based on the invited talk presented by the author at the 58th Annual American Physical Society Division of Fluid Dynamics Meeting, which was held 20–22 November 2005 in Chicago, IL. PHYSICS OF FLUIDS 18, 111301 2006 1070-6631/2006/1811/111301/19/$23.00 © 2006 American Institute of Physics 18, 111301-1 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see conditions, is formally the outer, collision-dominated, solu-tion. The inner description, which amounts to a Knudsen layer correction to the ﬂowﬁeld, will be discussed further in Secs. II A and II C 1. In other words, for small Knudsen numbers, kinetic effects at the walls manifest themselves, at the Navier-Stokes description level, in the form of “appar-ent” hydrodynamic property slip/jump at the boundaries which can be captured by slip-ﬂow boundary conditions see Sec. II A. For Kn0.1 a ﬁrst-order asymptotic theory is sufﬁciently accurate, especially for practical purposes. For this reason, the regime Kn0.1 is known as slip ﬂow. It is noteworthy that for Kn1 Knudsen layers are present ir-respective of the characteristic system lengthscale, H; how-ever, as Kn decreases, their effect becomes less pronounced, as one would expect, to the extent that in the limit Kn≪1 their effect is, for all practical purposes, negligible and the classical no-slip boundary condition becomes an excellent approximation. When the Knudsen number becomes appreciable Kn 0.1, one expects kinetic effects to be important in a large part of the domain and the Navier-Stokes description to fail. When the Knudsen number is large Kn→, the rate of intermolecular collisions is very small compared to the rate of molecule-wall collisions. As a result, transport at high Knudsen numbers is ballistic. Ballistic transport is typically assumed to take place for Kn10. The regime 0.1Kn 10 is known as the transition regime, and is typically the most challenging to model. In this regime, nonlocal transport is important while collisions between molecules are not negligible. C. Kinetic description Gaseous hydrodynamics beyond Navier-Stokes can be captured using a kinetic description. Under the assumption of a dilute gas air at atmospheric pressure satisﬁes the dilute-gas criteria9, a kinetic description characterizes the state of the gas in terms of the single-particle distribution function f = fx,c,t, which is proportional to the probability of ﬁnd-ing a particle at a location x with velocity c at time t Ref. 4. Within this description, connection to hydrodynamics is made through the moments of f. For example, the gas den-sity is given by x,t = allc mfdc, 1 while the gas ﬂow velocity is given by ux,t = 1 x,t allc mcfdc. 2 Here m is the molecular mass; n=/m is the gas number density. The gas pressure and temperature will be denoted P=Px,t and T=Tx,t, respectively, and are given by higher moments of f Ref. 4. The evolution of the distribution function is governed by the Boltzmann equation7,9 f t + c · f x + a · f c = df dt coll , 3 df dt coll = ff1 −ff1gd2d3c1. Here, f1 fx,c1,t, f fx,c,t, f1 fx,c1,t, where a prime indicates postcollision velocities. Postcollision ve-locities depend on the precollision velocities and the scatter-ing solid angle . Additionally, g= c−c1 is the relative precollision speed, is the collision cross section, and a is the acceleration due to body forces acting on a molecule. The Boltzmann equation describes a balance between phase-space advection left-hand side LHS and collisions right-hand side RHS. The RHS of the Boltzmann equa-tion is typically referred to as the collision integral and con-tributes signiﬁcantly to the intransigence of this equation. This led Bhatnagar et al.10,4 to propose a model equation, known now as the BGK model, which signiﬁcantly simpli-ﬁes the mathematics by replacing the collision integral with the simple relaxation model, namely −f −f0/ ; here, f0 is a Maxwellian distribution at the local velocity and temperature and is an inverse collision frequency that does not depend on the molecular velocity. The value of the collision fre-quency is usually chosen to match the viscosity of the mod-eled gas. The BGK model has proven itself as a useful quali-tative model, because it retains most qualitative features of the Boltzmann equation. One of its major disadvantages is that it predicts a Prandtl number Pr of 1 Refs. 4 and 7, rather than a value close to 2/3 appropriate for a mono-atomic gas; this means that the BGK model cannot match the viscosity and the thermal conductivity of a real gas at the same time, and thus ﬂow calculations involving heat transfer need to be interpreted with great care. The most prevalent solution method for Eq. 3 is a sto-chastic particle simulation method known as direct simula-tion Monte Carlo DSMC. Comprehensive descriptions of this method can be found in Ref. 11 or the monograph by the inventor of this method, Bird;9 a brief description is given in Sec. III A. The majority of theoretical developments pre-sented here use DSMC for veriﬁcation purposes. In some cases, however, DSMC provides the only solution available to the problem of interest. In the interest of simplicity, unless otherwise stated, the theoretical results and DSMC simulations will be using the hard-sphere HS model =d2/4, where d is the hard-sphere diameter.9 The hard-sphere model provides reasonably ac-curate models of rareﬁed gas ﬂows,7 and for the purposes of this discussion it provides a good compromise between sim-plicity and realistic modeling. The mean free path of a hard-sphere gas is given by = 1 2 nd2 , 4 while the ﬁrst-order approximations to the viscosity and ther-mal conductivity of the hard-sphere gas within the Chapman-Enskog theory are given by 111301-2 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see = 5 16d2 mkbT 5 and = 75kb 64d2 kbT m , 6 respectively.5 Here, T is the gas temperature, kb is Boltz-mann’s constant. The above rational approximations to the transport coefﬁcients are typically preferred over the more accurate “inﬁnite-order” expressions from which they only differ by approximately 2% Ref. 5. One of the disadvan-tages of the hard-sphere model is that it predicts transport coefﬁcients which are proportional to T0.5, whereas real gases exhibit a slightly higher exponent of approximately T0.7. To remedy this, collision models with more complex collision cross sections have been proposed;9 one example is the variable hard-sphere VHS model in which the collision cross section is a function of the relative velocity of the colliding molecules. The discussion in this paper can easily be extended to these modiﬁed collision models. D. Further remarks Gaseous ﬂows beyond the Navier-Stokes regime Kn 0.1 are frequently referred to as rareﬁed, most likely due to historical reasons, viz the rareﬁed gas dynamics literature,9,12 where deviation from Navier-Stokes was exten-sively studied in the context of high altitude aerodynamics. Flows for which the Navier-Stokes description breaks down are also frequently referred to as “noncontinuum;” this ter-minology is very common within the rareﬁed gas dynamics9 and now the microelectromechanical systems MEMS lit-erature, and may, unfortunately, lead to confusion in a me-chanics setting where the expression “noncontinuum” will most likely be associated with a breakdown of the continuum assumption. One may surmise that use of the term “noncon-tinuum” derives from the view that because the continuum approach culminates in the Navier-Stokes equations, when the latter fails, the continuum approach fails, without this necessarily implying failure of the continuum assumption. We would thus like to emphasize that for a large class of the problems of practical interest, including the ones discussed here, hydrodynamic ﬁelds and the associated conservation laws remain well deﬁned. The view taken in this paper is the one typically adopted within the rareﬁed gas dynamics com-munity and described in Ref. 4: conservation laws for mass, momentum, and energy follow naturally from moments of the Boltzmann equation. The implied averaging procedure can be performed using an ensemble of realizations with no restrictions on space or time averaging intervals, thus lead-ing to a meaningful description in terms of said conservation laws i.e., both in the presence and absence of Navier-Stokes closures, for quite a wide range of conditions, including very small lengthscales and time scales. II. FLOW PHYSICS In this section we will review theoretical results focusing on the physics of problems of engineering interest; both re-cent and basic fundamental results will be discussed. One of the basic geometries that we will visit frequently is a two-dimensional channel such as the one shown in Fig. 1. The two-dimensional channel geometry has been widely studied in the context of small-scale ﬂows due to its direct relevance to typical small-scale applications, but also due to its sim-plicity, which enables investigations aimed at the physics of transport at small scales. A. First-order slip-ﬂow theory When the Knudsen number is small, solution of the lin-earized Boltzmann equation can be obtained using asymptotic approaches.6,8 By taking advantage of the fact that for Kn0.1 kinetic effects are limited to the vicinity of the walls, one can obtain solutions of the linearized Boltz-mann equation by superposing a Navier-Stokes solution with a kinetic boundary Knudsen layer correction, where the latter is signiﬁcant only up to distances of the order of one mean free path from the wall see Fig. 2. The matching procedure between inner and outer solu-tions leads to “effective” boundary conditions for the Navier-Stokes description, known as slip-ﬂow relations. Using these boundary conditions, the Navier-Stokes description is able to capture the bulk ﬂowﬁeld away from the walls. These rela-tions, to ﬁrst order in Kn, can be written as follows: FIG. 1. Two-dimensional channel geometry and nomenclature. FIG. 2. Schematic of the Knudsen layer in the vicinity of the wall =0. 111301-3 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see us NS wall −Us = vdus NS d wall + v 2kb mTdTNS ds wall . 7 Here, s is the coordinate tangent to the wall, is the coor-dinate normal to the wall and pointing into the gas, and U denotes wall velocity. The superscript NS is used to differ-entiate between the linearized Boltzmann equation solution no superscript, which is the “true” ﬂowﬁeld, and the Navier-Stokes slip-ﬂow approximation to this solution; as stated above, the two are related through the superposition of a Knudsen layer correction, e.g., us=us NS+us Kn, T=TNS +TKn,..., with these corrections decaying fast away from the walls, i.e., us Kn, TKn→0 as /v→. Also, v is a viscosity-based mean free path deﬁned by v = P RT 2 . 8 This deﬁnition is very useful for molecular models without a well-deﬁned range of interaction where a mean free path is difﬁcult to deﬁne.7 Note that this deﬁnition is such that v== 2 nd2−1 for the hard-sphere gas. The hard-sphere results presented below have been developed using v 2 nd2−1; the slight numerical difference v 2 nd2−1 observed when one uses 5 into 8 is due to the use of a rational form for 8 in the interests of simplicity. The temperature jump at the wall is given by the follow-ing analogous expression: T ˆ wall −Tw = 2 + 1 v Pr dT ˆ d wall , 9 where is the ratio of speciﬁc heats. The coefﬁcients , , and are determined by matching the outer and inner descriptions within the asymptotic solu-tion. Maxwell was able to estimate these coefﬁcients without solving the Boltzmann equation using insightful physical ar-guments; he found = 2 − , = 0.75, 10 where is the accommodation coefﬁcient which param-etrizes the gas-wall-surface interaction and denotes the frac-tion of diffusely reﬂected molecules 1− being the fraction of molecules reﬂected specularly at the surface. In a similar fashion, was estimated by von Smoluchowski13 to equal = 2 − . 11 The amount of slip at the surface is clearly a strong function of the accommodation coefﬁcient. Our discussion will be limited to the interpretation given above—known as Maxwell’s gas-surface interaction model14—primarily be-cause the majority of theoretical results have been obtained using this model. A number of more complex models exist, typically relying on more than one accommodation coefﬁ-cient; for example, the use of separate momentum and en-ergy accommodation coefﬁcients is common.2,13 Scattering kernels based on more than one accommodation coefﬁcient have been developed,7,15 but theoretical results on the behav-ior of slip coefﬁcients are very few and scattered.7 A discus-sion of accommodation coefﬁcients can be found in Ref. 14. Solutions of the Boltzmann equation for the slip coefﬁ-cients were originally obtained for the signiﬁcantly simpler BGK model. Early work by Cercignani,7,16 and recent results for the hard-sphere gas, show that the ﬁrst-order coefﬁcients are fairly insensitive to the gas model e.g., hard sphere, BGK. Thus, although hard-sphere results are more realistic, in our discussion below we will also include BGK results, especially since in some cases they are the only ones available. Solutions of the Boltzmann equation have mostly fo-cused on the fully accommodating case =1, primarily be-cause for most “engineering surfaces” of practical interest 1 Refs. 9 and 17. In this limit, it is known that BGK,=1=1.1466, BGK,=1=1.149,BGK, =11.168. When reporting BGK results for and , a number of authors introduce a correction for the fact that for the BGK model Pr=1 rather than the more realistic for a monoatomic gas value of 2/3. This correction amounts to multiplying these coefﬁcients by 3/2 as a result of enhanc-ing the collision frequency by the same amount in order to match the value of thermal conductivity.7 This correction has been applied throughout this paper. On the other hand, HS,=1=1.11, HS,=1=1.015, HS,=1=1.13 Ref. 18, verifying that ﬁrst-order slip coefﬁcients are fairly insensitive to the gas model and that Maxwell’s estimate is approximately 10%–15% in error in the diffuse reﬂection =1 limit. This discrepancy has implications in the inter-pretation of experimental results and will be discussed be-low. In the limit →0, BGK→2/ Ref. 19, in agree-ment with Maxwell’s approximate solution. Very few results exist for intermediate values of accom-modation coefﬁcients. Loyalka20 used a variational approach to show that for the BGK model a linear interpolation be-tween the =1 and →0 limits provides reasonable accu-racy, i.e., BGK, 2 − 1 + BGK, = 1 −1 = 2 − 1 + 0.1466. 12 In the case of the temperature jump coefﬁcient, Loyalka20 ﬁnds that similar interpolation between the fully accommo-dating and specular reﬂection limits may be applied for in-termediate values of . For the thermal creep slip coefﬁcient, Loyalka and Cipolla ﬁnd14 BGK, = 0.75 + 0.399. 13 Discussion: 1 Slip-ﬂow theory naturally reduces to no-slip boundary conditions in the limit Kn≪1. This can be easily seen by nondimensionalizing in Eqs. 7 and 9 using the characteristic lengthscale H. 111301-4 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see 2 The above slip-ﬂow relations are formally valid for steady ﬂows. Comparison of transient slip-ﬂow solutions to solutions of the Boltzmann equation suggest that the above slip-ﬂow relations remain accurate for time-dependent ﬂows, provided the latter are evolving at suf-ﬁciently long time scales. One may expect that this ef-fectively “quasistatic” behavior would be observed when the hydrodynamic evolution time scale is long compared to the molecular collision time that character-izes the evolution of kinetic effects at the wall. This is veriﬁed by theoretical treatments of the Boltzmann equation, at least in the BGK approximation,21 where slip-ﬂow relations equivalent at least formally to the above are obtained by assuming that the evolution time scale is long compared to the molecular collision time c=/c ¯; here c ¯= 8kbT/ m is the mean thermal speed. As shown and discussed further in Sec. II C 1, our results obtained using the second-order slip model of that section conﬁrm that. 3 Corrections to the above slip relations due to wall cur-vature are formally of second order in the Knudsen number7 and are given in Ref. 6. 4 First-order Knudsen layer corrections for the most com-mon state variables and their ﬂuxes are tabulated in Ref. 6 for the BGK and HS models; the lack of analytical descriptions is the primary reason for their neglect in engineering analysis. In Sec. II C 1 we will discuss the Knudsen layer correction for the tangential ﬂow veloc-ity; we will illustrate that its contribution to the average ﬂow velocity is of order Kn2 Ref. 22, and show how it can be incorporated into a second-order slip model. 5 Conveniently, the ﬁrst-order Knudsen layer correction for the stress tensor is zero in isothermal ﬂow; this means that within slip-ﬂow theory the stress ﬁeld is given by the Navier-Stokes solution no modiﬁcation to the Navier-Stokes constitutive relation is required or al-lowed; in fact, this is one of the basic premises and undoubtedly one of the great advantages of slip-ﬂow theory. In a similar fashion, the ﬁrst-order Knudsen layer correction to the heat ﬂux in the direction normal to the wall in the presence of a temperature gradient in the same direction is zero. 6 In addition to the experimental difﬁculties associated with accurate measurement of accommodation coefﬁ-cients, theoretical difﬁculties make this an even harder task. One such difﬁculty is the uncertainty associated with the gas-surface interaction model; although Max-well’s accommodation model is frequently used in the-oretical studies, experimental results have not reached a consensus on whether it accurately describes gas-surface interaction, or whether a more complex model is re-quired. This uncertainty is compounded by the fact that experimental results are typically compared to Max-well’s estimates for the slip coefﬁcients Eqs. 10 and 11, which can lead to errors in accommodation coef-ﬁcients of the order of 5%. From a number of experi-mental studies conducted to date see, for example, Refs. 2, 9, 17, 23, and 25, it appears that for “engineering surfaces” the accommodation coefﬁcient is close to 1. Recent experiments using air on silicon ﬁnd accommo-dation coefﬁcients in the range 0.85–0.95 Ref. 2; for these measurements, the actual accommodation coefﬁ-cients are probably even closer to 1 for two reasons: a Experimental results are typically interpreted using Eqs. 10 and 11, which underestimate the value of the slip coefﬁcient and would thus lead to lower estimates of . b A number of experimental studies extend to Kn 0.1 while still using ﬁrst-order slip-ﬂow rela-tions. In pressure-driven ﬂows, second-order slip effects increase the amount of slip present see Sec. II C 1. Neglecting second-order slip effects would again lead to lower estimates of . 7 Thermal creep phenomena extend beyond the slip-ﬂow regime; thermal creep ﬂow for all Knudsen numbers for the hard-sphere model has been characterized in Ref. 18. Unless otherwise stated, in what follows we will assume that =1 and thus =1.11, =1.015, and =1.13. B. Isothermal pressure-driven ﬂows in two-dimensional channels Isothermal pressure-driven ﬂow in two-dimensional ducts for Kn0.1 was originally studied by Knudsen.26 He studied the ﬂow through capillaries and showed the existence of a minimum in the ﬂow rate when the latter is normalized by the pressure difference driving the ﬂow and plotted against the average pressure in the channel inverse Knudsen number;2 this minimum cannot be predicted by the Navier-Stokes description. Following Knudsen’s discovery, a theo-retical description of this phenomenon remained a signiﬁcant challenge for a number of years. Following the development of semianalytical solutions of simple models of the Boltzmann equation e.g., BGK,27 numerical solutions of the linearized Boltzmann equation for the more realistic hard-sphere gas for various two-dimensional geometries were ﬁnally developed.18 For two-dimensional channels as in Fig. 1 the gas response for ar-bitrary Knudsen numbers is typically expressed in kinetic terms through the following expression: Q ˙ = ubH = −1 P dP dx H2 RT 2 Q ¯ , 14 where Q ˙ is the volumetric ﬂow rate per unit depth, ub is the bulk average over the channel width velocity, R=kb/m is the gas constant, and Q ¯ =Q ¯ Kn is a proportionality coefﬁ-cient whose dependence on Kn is shown in Fig. 3. Similarly deﬁned Q ¯ parameters have now been tabulated for a variety of two-dimensional duct geometries.2 Note that the linear-ized treatment assumes a constant pressure gradient; in other words these solutions do not include the effects of stream-wise acceleration due to the gas compressibility. As shown in Fig. 3, Q ¯ Kn for a two-dimensional chan-nel in the transition regime varies slowly about its minimum value occurring at Kn1. Numerical solutions, such as lin-111301-5 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see earized solutions of the Boltzmann equation for hard spheres7,18,28 and molecular simulations,28 have been shown to be in good agreement with experiments.7,28 C. Second-order velocity slip Slip-ﬂow solutions to a variety of problems of practical interest suggest that slip-ﬂow theory is remarkably robust, in the sense that it continues to be reasonably accurate, at least in a qualitative sense, well beyond its expected limits of applicability Kn0.1; examples in this paper include the pressure-driven ﬂow of Sec. II B and the convective heat-transfer problem of Sec. II G. Robust slip-ﬂow models will always be preferable to alternatives such as molecular simu-lations or solutions of the Boltzmann equation, since the dif-ﬁculty associated with solving the Navier-Stokes equations is negligible compared with the cost of these alternative meth-ods. For this reason, a variety of researchers22,24,25,28–30 have attempted to develop or evaluate slip models that can be used beyond Kn0.1. A review of a number of these approaches can be found in Ref. 2. In this paper we will discuss ap-proaches which fall within the rigorous framework of asymptotic analysis of the Boltzmann equation. Although the general structure of slip-ﬂow relations up to second order in Kn has been known for some time,8,31 and the second-order coefﬁcient for one-dimensional, isothermal ﬂow for the BGK model was calculated by Cercignani22 as early as 1964, the calculation of other second-order slip co-efﬁcients has proven to be a signiﬁcantly harder task. Second-order slip coefﬁcients for the BGK model have even-tually been calculated;6,7,22 these, however, unlike the ﬁrst-order slip coefﬁcients, are not necessarily approximately equal to the corresponding coefﬁcients of the more realistic hard-sphere model. This has been one of the main reasons for the lack of interest in second-order slip-ﬂow theory, since agreement with experiments and solutions of the Boltzmann equation e.g., DSMC simulations could not be achieved using the BGK theory. Another reason for the lack of interest in the second-order slip-ﬂow theory is the special care needed in interpreting slip-ﬂow solutions and comparing those to numerical solutions of the Boltzmann equation or experimental data due to the presence of Knudsen layers in the ﬂow. This is illustrated in Fig. 4, which shows the extent of Knudsen layers in a one-dimensional ﬂow at Kn=0.2 by comparing the Boltzmann solution u=uNS+uKN to the Navier-Stokes solution uNS obtained using the second-order slip model to be introduced in the next section; this ﬁgure clearly illustrates that, at this Knudsen number, Knudsen lay-ers from both walls penetrate approximately 60% of the physical domain. This means that the effective width of the Knudsen layer is approximately 1.5v this will also be dis-cussed in more detail below, and thus for Kn0.3 direct comparison between the Navier-Stokes and the true ﬂowﬁeld is impossible. 1. A second-order slip model for the hard-sphere gas To address this deﬁciency, the author has developed32 a second-order slip model for the hard-sphere gas. According to this model, for one-dimensional variations only in di-rection, fully accommodating =1 ﬂows, the second-order slip is given by us NS wall −Us = vus NS wall −v 22us NS 2 wall , 15 where =0.61 is the second-order slip coefﬁcient =1.11. Validation of this model for a variety of ﬂows shows excel-lent agreement up to Kn0.4 provided the effect of the Knudsen layers is properly accounted for. The ﬁrst-order Knudsen layer correction in the tangential ﬂow velocity us KN can be written as FIG. 3. Color online Nondimensional ﬂow rate as a function of the Knud-sen number for fully developed pressure-driven ﬂow. The solid line denotes Q ¯ as determined by solution of the linearized Boltzmann equation for a hard-sphere gas Ref. 18, and the dash-dotted line denotes the second-order slip model discussed in Sec. II C 1. The stars denote DSMC simulation results, and the dashed line a ﬁrst-order slip model. FIG. 4. Snapshot of ﬂow velocity in an oscillatory Couette ﬂow at Kn =0.2. The stars represent the DSMC solution u and the solid line the Navier-Stokes approximation to this solution uNS. The vertical dotted lines illustrate the approximate thickness of the Knudsen layers. Uo is the wall-velocity amplitude. 111301-6 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see us KN /v = −vdus NS d wall I1 /v, 16 where I1 /v is a positive function that decays to zero for /v1; it has been numerically calculated for both the BGK Ref. 7 and HS gas.33 The implication of this relation is that the Knudsen layer correction is parametrized by the Navier-Stokes solution. In other words, despite the fact that the underlying distribution in the Knudsen layer differs from the Chapman-Enskog solution, this correction becomes a functional of, and is thus effectively determined by, the local Navier-Stokes solution after a short transient of the order of the collision time; the length of time over which this relax-ation takes place has implications in the validity of slip-ﬂow theory in transient ﬂows, and is discussed in Sec. II A and below. Tabulated values show that us KN /v decays to ap-proximately 3% of its maximum value us KN /v=0 at =1.5v; we will refer to this distance as the effective width of the Knudsen layer. Due to the lack of an analytical description for us KN /v, the contribution of the Knudsen layer can be most conveniently accounted for in an average sense, i.e., when calculating averages over the domain. In a one-dimensional geometry such as that of Fig. 1, the average bulk ﬂow velocity is given by ub = 1 H −H/2 H/2 uxdy = 1 H −H/2 H/2 ux NS + ux KNdy. 17 Recalling that v/H1 we deﬁne = 0 I1 /vd /v to obtain ub = 1 H −H/2 H/2 ux NSdy −v 2 H dux NS dy −H/2 + v 2 H dux NS dy H/2 = 1 H −H/2 H/2ux NS + v 22ux NS y2 dy, 18 where =0.296 Ref. 32. In other words, the contribution of the Knudsen layer to the average ﬂow velocity is OKn2. The above value for was obtained by using the fact that the Knudsen layer function I1 for the hard-sphere gas is very similar to the BGK Knudsen layer function whose inte-gral is known exactly. A slightly more precise value for could have been obtained if one integrated the hard-sphere Knudsen layer function; unfortunately, this value can only be obtained approximately by numerically integrating the tabu-lated solution for this function. While the contribution of the Knudsen layer can always be found by a Boltzmann equation analysis, the value of Eq. 18 lies in the fact that it relates this contribution to the Navier-Stokes solution and thus it requires no solution of the Boltzmann equation. This correction makes comparison to Boltzmann equation solutions possible when Knudsen layers occupy a large fraction of the domain; additionally, it makes comparison possible with experiments which report average ﬂow rates. In fact, it has already been used to explain recent experimental data: a direct consequence of the above relation is that in Poiseuille-type ﬂows, where 2ux NS/y2 is a con-stant, experimental measurement of the ﬂow rate mean ﬂow velocity yields an “effective” second-order slip coefﬁcient − see also Ref. 32. In other words, while the average value of a Navier-Stokes Poiseuille proﬁle subject to second-order slip of the form 15 is given by ub NS = 1 H −H/2 H/2 ux NSdy = −H2 2 dP dx 1 6 + Kn + 2Kn2, 19 the true bulk ﬂow speed as inferred by an experiment mea-suring the ﬂow rate is given by Eq. 18, which leads to ub = 1 H −H/2 H/2ux NS + v 22ux NS y2 dy = −H2 2 dP dx 1 6 + Kn + 2Kn2 20 or Q ¯ = 4 15 1 + 6Kn + 12Kn2 Kn 12 1 + 6Kn + 12Kn2 Kn , 21 with =−=0.31. The above two expressions for Q ¯ differ by less than 2%; the difference between them is due to the use of slightly different approximations for the hard-sphere gas viscosity.5,22 As shown in Fig. 3, the above equation captures the ﬂow rate in isothermal pressure-driven ﬂow very accurately up to Kn0.4. This is also demonstrated in Sec. II E, where the pressure-driven ﬂow rate is used to determine the wave propagation constant in two-dimensional channels under the long-wavelength approximation. Most importantly, the above model explains the ﬁndings of recent experiments25 on helium and nitrogen ﬂow in small-scale channels; these experiments ﬁnd the second-order slip coefﬁcient to be approximately 0.25±0.1. Of course, since the slip coefﬁcient was determined by measur-ing the ﬂow rate, these experiments were in fact determining the effective second-order slip coefﬁcient , which is in good agreement with the value 0.31 given above. Discussion: 1 The second-order Knudsen layer correction for the hard-sphere model has not been calculated. Its contribution to the mean ﬂow velocity was not required in 18 since it is of higher order. 2 The second-order Knudsen layer correction to the stress tensor in isothermal ﬂow is zero.6 Thus, no modiﬁcation to the Navier-Stokes constitutive relation is required al-lowed within isothermal slip-ﬂow theory. 3 In higher spatial dimensions, additional terms appear in Eqs. 15 and 18. In the case of Eq. 15, these terms also include contributions from the curvature of the boundary. As a result, the “effective” second-order slip coefﬁcient for ﬂow in tubes is different see experimen-111301-7 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see tal results of Ref. 30, but note that this fact was over-looked in the paper than the one for two-dimensional channels . 4 The assumption of variations only in the normal-to-the-wall direction, inherent in the model described above, is not very restrictive. Approaches based on assumptions of slow variation in the axial direction x in Fig. 1, such as the widely used locally fully developed assumption or long-wavelength approximation, are expected to yield excellent approximations when appropriately used for two-dimensional problems. This is veriﬁed by compari-son of solutions of such problems to DSMC simulations see Sec. II E for example or experiments e.g., Ref. 25. 5 The linearized conditions Ma1 under which the second-order slip framework is derived,6,8,31 imply Re 1 since MaReKn and Kn0.1. Here Ma is the Mach number and Re is the Reynolds number, based on the same characteristic lengthscale as Kn. 6 It appears that the steady-ﬂow assumption on which this model is based does not signiﬁcantly limit its applicabil-ity. As discussed in Sec. II A, slip-ﬂow theory appears to extend to ﬂows that evolve at time scales that are long compared to the molecular collision time; in fact, our results, some of which will be shown below, suggest that the quasistatic behavior of slip at the wall seems to hold for time scales 5 c, which correspond to viscous evo-lution time scales for lengthscales characterized by Kn 0.4; this restriction on evolution timescale is easily satisﬁed by the vast majority of practical ﬂows of interest. 7 Comparisons with solutions of the Boltzmann equation for a variety of problems34–36 show that the above slip model provides reasonably accurate approximations to solutions of the Boltzmann equation up to Kn0.4 maximum error at Kn0.4 is of the order of a few percent for both ﬂow speed and stress. Moreover, it remains qualitatively robust well beyond this Knudsen number. This is rather remarkable for a number of rea-sons: ﬁrst, for Kn0.3 the Knudsen layers, treated within the theory as thin boundary layers at the wall see, for example, the development of Eq. 18, merge, leading to kinetic corrections that overlap but apparently may still be superposed. Moreover, although a kinetic correction is required throughout the physical domain for Kn0.3, it appears that the underlying Navier-Stokes constitutive relation remains robust the stress ﬁeld is accurately captured for arbitrary ﬂows with no adjustable parameters up to Kn0.4 at Kn=0.4 the domain width is 2.5. 2. Example To illustrate some of the discussion items above, we present a solution of a one-dimensional model problem. Con-sider the two-dimensional channel of Fig. 1 in which both channel walls impulsively start to move at time t=0 in the x direction with velocity Ux; this velocity is small compared to the most probable molecular velocity Ma1. Figure 5 shows a comparison between second-order-slip-corrected Navier-Stokes solution and DSMC simulations for Kn =0.21. The top plot shows the ﬂowﬁeld at three different times; the existence of the Knudsen layer, manifested by the difference between the DSMC solution and the Navier-Stokes solution, can be seen clearly. The three snapshots also clearly illustrate how the magnitude of the Knudsen layer correction scales with uNS/ y wall. The middle plot shows the stress ﬁeld at the same three times, and demonstrates that no Knudsen layer correction exists close to the wall. The bottom plot shows the bulk ﬂow velocity ub as a function of time: as this plot shows, this quantity can be calculated to excellent accuracy using Eq. 18. This last comparison as well as a comparison of the stress ﬁeld becomes particularly useful as Kn increases to Kn0.3 and beyond, where the Knudsen layer essentially covers the whole physical domain. This is illustrated in Fig. 6, where a comparison at Kn =0.35 is shown. The ﬁgure veriﬁes that although the Knud-sen layers cover the physical domain, the second-order slip remains reasonably accurate in predicting the stress ﬁeld and bulk ﬂow velocity, especially considering that the compari-son takes place for times as low as ﬁve collision times. Com-parisons at higher Knudsen numbers34,35 show that, even though the Knudsen layers have penetrated to the middle of the domain, the slip model remains quantitatively accurate up to Kn0.4, and qualitatively accurate well beyond that. In some cases see next section quantitative accuracy ex-tends up to Kn1. D. Oscillatory shear ﬂows Oscillatory shear ﬂows one wall in Fig. 1 oscillates in the x direction with velocity amplitude Uo and frequency are very common in MEMS and are considered to be of “tremendous importance in MEMS devices.”37 A comprehen-sive simulation study of rarefaction effects on oscillatory shear Couette ﬂows was recently conducted by Park et al.38 Park et al. used an extended ﬁrst-order slip-ﬂow relation with a number of adjustable coefﬁcients to describe the amount of slip at the wall for all Knudsen numbers, provided the ﬂow was quasistatic i.e., the Stokes number S = H2/ is much smaller than 1. Note that the quasistatic assumption is not at all restrictive in practical applications where the gap size, H, is very small. Also note that in the quasistatic regime, shear ﬂow results in a linear velocity pro-ﬁle at the Navier-Stokes level of approximation and thus describing the amount of slip completely characterizes the ﬂowﬁeld. The oscillatory shear ﬂow problem was used in Refs. 34 and 35 as a validation test problem for the second-order slip model of Sec. II C 1. Excellent agreement with DSMC solu-tions was found for all S for Kn0.4 for both the ﬂowﬁeld and the stress ﬁeld. In fact, for sufﬁciently low frequencies S1 the agreement is very good up to Kn1 note that for this ﬂow, in the quasistatic regime S1, the second-order slip model reduces to the ﬁrst-order slip model, which is known to describe the stress very well up to Kn1. Figure 7 shows a comparison between DSMC solutions and the second-order slip model of Sec. II C 1 for the magnitude of 111301-8 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see the shear stress at the wall. This comparison shows that for S1 the second-order slip model in conjunction with a col-lisionless model shown in dashed line may be used to bridge the transition regime. The collisionless result of Fig. 7 was given by Park et al., who showed that the velocity and shear stress at the wall are independent of and equal to the values of the steady problem. Analytical results for the ﬂowﬁeld and the bounded ballistic shear layers appearing at high frequency in the Kn1 limit—analogous to Stokes layers in the Kn1 limit—were given by the present author in Ref. 35. E. Axial wave propagation in small-scale channels In this section we discuss a theory of axial plane-wave propagation in two-dimensional channels Fig. 1 for arbi-trary Knudsen numbers. The theory is based on the long-wavelength approximation and on the observation that, in the Navier-Stokes limit, the propagation of disturbances in small-scale channels for most frequencies of practical inter-est is viscous dominated. The importance of viscosity can be quantiﬁed by a narrow channel criterion, S= H2/1. When S1 whereby the channel is termed narrow the vis-cous diffusion length based on the oscillation frequency is much larger than the channel height; viscosity is expected to be dominant and inertial effects will be negligible. This ob-servation has two corollaries; ﬁrst, since the inertial effects are negligible the ﬂow is governed by the steady equation of motion, that is, the ﬂow is effectively quasisteady.39 Second, since for gases the Prandtl number is of order one, the ﬂow is also isothermal for a discussion see Ref. 40. This was ﬁrst realized by Lamb,41 who used this approach to describe wave propagation in small-scale channels using the Navier-Stokes description. Lamb’s prediction for the propagation constant using this theory is identical to the more general theory of FIG. 5. Comparison between DSMC results symbols and second-order-slip-corrected Navier-Stokes solution lines for Kn=0.21, for the impulsive start problem described in Sec. II C 1. Due to the problem symmetry about y=0, only half domain is shown; the wall is at y/H=−0.5. Flowﬁeld and stress plots shown at three times: 11.1 c , 16.2 c , and 22.3 c . Top plot: comparison of ﬂow velocities; the Knudsen layer con-tribution u−uNS is visible close to the wall the verti-cal dotted line delimits the approximate extent of the Knudsen layer −0.5y/H−0.185. Middle plot: comparison of the shear stress. Bottom plot: compari-son of the bulk velocity as a function of time. 111301-9 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see Kirchhoff42 when the narrow channel limit is taken in the latter. The author has recently39 used the fact that wave propa-gation in the narrow channel limit the narrow channel crite-rion needs to be suitably redeﬁned in the transition regime where viscosity loses its meaning; however, the work in Refs. 39 and 40 shows that S remains a conservative criterion for the neglect of inertia and thermal effects for Kn0.1 is governed by the steady equation of motion to provide a pre-diction for the propagation constant for arbitrary Knudsen numbers without explicitly solving the Boltzmann equation. This is achieved by rewriting Eq. 14 in the form u ˜b = −1 R dP ˜ dx , 22 where tilde denotes the amplitude of a sinusoidally time-varying quantity and R=RKn, the “ﬂow resistance,” is given by R = P0 HQ ¯ RT0/2 . 23 Here, P0 and T0 denote the mean pressure and temperature, respectively. Equation 22 locally describes wave propaga-tion since, as we argued above, in the narrow channel limit the ﬂow is isothermal and quasistatic and governed by the steady-ﬂow equation of motion. Using the long-wavelength approximation allows us to integrate mass conservation, written here as a kinematic condition,39 P x = − P T 0 2 x2 24 across the channel height. Here P/T indicates that this derivative is evaluated under isothermal conditions appropri-ate to a narrow channel. Additionally, 0 is the average den-sity and is the ﬂuid-particle displacement deﬁned by FIG. 6. Comparison between DSMC results symbols and second-order-slip-corrected Navier-Stokes solution lines for Kn=0.35, for the impulsive start problem described in Sec. II C 1. Due to the problem symmetry about y=0, only half domain is shown; the wall is at y/H=−0.5. Flowﬁeld and stress plots shown at three times: 4.9 c , 7.9 c , and 12.8 c . Top plot: comparison of ﬂow velocities. Middle plot: comparison of the shear stress. Bottom plot: comparison of the bulk velocity as a function of time. 111301-10 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see uxx,y,t = x,y,t t . 25 Combining Eqs. 22 and 24 we obtain39 ib = 0P/T R 2b x2 , 26 where b is the bulk average over the channel width ﬂuid-particle displacement. From the above we can obtain the propagation constant mm + ik2 = iR P0 , 27 where mm is the attenuation coefﬁcient and k is the wave-number. Using Eq. 23 we obtain mm + ik22 = 8i Kn Q ¯ c T , 28 where T=2 / is the oscillation period. This result is expected to be of general use because the narrow channel requirement is easily satisﬁed in the transi-tion regime.39 A more convenient expression for use in the early transition regime, which does not require a lookup table for Q ¯ , can be obtained using the second-order slip model discussed in Sec. II C 1. Using this model we obtain mm + ik22 = 96iKn2 1 + 6Kn + 12Kn2 c T , 29 which, as can be seen in Fig. 8, remains reasonably accurate up to Kn1 aided by the square-root dependence of the propagation constant on R. This expression for Kn→0 re-duces to the well-known narrow-channel result obtained us-ing the no-slip Navier-Stokes description.43 Figure 8 shows a comparison between Eq. 29 Eq. 28, DSMC simulations, and the Navier-Stokes result. DSMC simulations of wave propagation are discussed in Ref. 39. It can be seen that the theory is in excellent agree-ment with DSMC results. As noted above, the second-order slip model provides an excellent approximation for Kn 0.5 and a reasonable approximation up to Kn1. The no-slip Navier-Stokes result clearly fails as the Knudsen number increases. The theory presented here can be easily general-ized to ducts of arbitrary cross-sectional shape, and has been extended40 to include the effects of inertia and heat transfer in the slip-ﬂow regime where closures for the shear stress tensor and heat ﬂux vector exist. F. Reynolds equation for thin ﬁlms The approach of Sec. II E is reminiscent of lubrication theory analyses used in describing the ﬂow in thin ﬁlms.44 This type of approach, typically using the Reynolds equa-tion, is fairly common in small-scale devices whose geom-etry lends itself naturally to this type of analysis. An exten-sive discussion of the Reynolds equation and its applications to small-scale ﬂows can be found in Ref. 37. Our objective FIG. 7. Normalized wall shear stress magnitude as a function of the Knud-sen number. Symbols denote DSMC results. Solid lines denote the second-order slip model result. Dashed line denotes the collisionless result. Here, the left wall oscillates with velocity amplitude Uo. FIG. 8. Comparison between the theoretical predictions of Eq. 28, shown as a solid line, and the simulation results denoted by stars as a function of the Knudsen number at a constant frequency given by T/ c6400. The dash-dotted line denotes the prediction of Eq. 29. The no-slip Navier-Stokes solution dashed lines is also included for comparison. 111301-11 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see here is to brieﬂy discuss the opportunities provided by the lubrication approximation for obtaining solutions for arbi-trary Knudsen numbers to various small-scale problems, without explicitly solving the Boltzmann equation. In a one-dimensional gap see Fig. 1 with one plate moving in the x direction with velocity Ux, the Reynolds equation reads x−H3 12 dP dx + HUx 2 = −H t . 30 This formulation is convenient because it requires only knowledge of the ﬂow rate average ﬂow speed through the gap; this has the fortunate consequence that it can be easily generalized to arbitrary Knudsen numbers in a fashion that is exactly analogous to the procedure used in Sec. II E. This was realized by Fukui and Kaneko,45 who formulated such a generalized Reynolds equation by: • Realizing that the “Couette” ﬂow contribution to the mass ﬂow rate HUx/2 does not change with the Knudsen number. • Replacing the Navier-Stokes Poiseuille mass ﬂow rate −H3 12 dP dx by the form −H2 P RT 2 Q ¯ ,KndP dx valid for all Knudsen numbers and accommodation coefﬁ-cients. • Including the ﬂow rate due to thermal creep H2 T RT 2 Q ¯ T,KndTw dx Ref. 45 into the total ﬂow rate, and thus accounting for the effects of an axial temperature gradient. Comparison between the formulation of Fukui and Kaneko and DSMC simulations can be found in Ref. 46. A number of approaches using ﬁts of Q ¯ Kn to deﬁne an “effective viscosity” for integrating the resulting “general-ized” Reynolds equation have appeared. It is hoped that the discussion of Sec. II C 1 illustrates that the concept of an “effective viscosity” is not very robust for a number of rea-sons this discussion is not limited to the context of Reynolds equation applications. For Kn0.1 the physical mechanism of transport changes and there is no reason to expect the concept of linear-gradient transport to hold. Even in the early transition regime, the concept of an “effective viscosity” is contradicted by a variety of ﬁndings see Sec. II C 1. To be more speciﬁc, an “effective viscosity” can be viewed as the particular choice of absorbing the kinetic corrections to the Poiseuille ﬂow rate in Eq. 20, namely 1+6Kn+12Kn2, into one of the proportionality constants, namely the viscos-ity. However, Sec. II C 1 has shown that the origin of these corrections ﬁrst- and second-order slip and Knudsen layer contribution to the ﬂow rate is not consistent with a variable viscosity. In fact, the correct way of interpreting Eq. 20 is that the constitutive relation, including the value of the vis-cosity, remain unchanged up to at least Kn0.4. Moreover, when the “effective viscosity” approach is adopted, the fol-lowing problems arise: • Kinetic corrections to the ﬂow rate are problem dependent ﬂow, geometry; as a result, an effective viscosity ap-proach cannot be predictive. For example, the “effective viscosity” ﬁtted from the Poiseuille ﬂow rate in a tube is different from the “effective viscosity” ﬁtted from the Poiseuille ﬂow rate in a channel due to the curvature cor-rections to the second-order slip coefﬁcient discussed in Sec. II C 1. • The ﬁtted “effective viscosity” does not give the correct stress through the linear constitutive law. The “effective viscosity” approach has another disadvan-tage when used in the Reynolds equation: the complex ex-pressions used to ﬁt Q ¯ typically cannot be directly integrated, unless the assumption KnKnP is made. Use of Eq. 21 for Kn0.5, on the other hand, should not suffer from this disadvantage. G. Flows involving heat transfer In this section we review ﬂows in which heat transfer is important. We give particular emphasis to convective heat transfer in internal ﬂows, which has only recently been in-vestigated within the context of Navier-Stokes failure in small-scale gaseous ﬂows. We also summarize the investiga-tion of Gallis and co-workers on thermophoretic forces on small particles in gas ﬂows. 1. The Graetz problem for arbitrary Knudsen numbers Since its original solution in 1885 Ref. 47, the Graetz problem has served as an archetypal convective heat-transfer problem both from a process modeling and an educational viewpoint. In the Graetz problem a ﬂuid is ﬂowing in a long channel whose wall temperature changes in a step fashion. The channel is assumed to be sufﬁciently long so that the ﬂuid is in an isothermal and hydrodynamically fully devel-oped state before the wall temperature changes. One is typi-cally interested in the energy exchange between the walls and the gas, which can be quantiﬁed by the Nusselt number NuT = q2H Tw −Tb , 31 which is a nondimensional heat-transfer coefﬁcient.48 Here q is the wall heat ﬂux and Tb is the bulk temperature deﬁned by Tb = −H/2 H/2 uxTdy ubH . 32 Although the work in Ref. 50 focused on the fully developed Nusselt number, the developing tube entrance part of the ﬂow can also be treated using the same approach. The gas-phase Graetz problem subject to slip-ﬂow boundary conditions was studied originally by Sparrow;49 this study, however, did not include the effects of axial heat 111301-12 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see conduction which cannot be neglected in small-scale ﬂows. Here we review the solution by the author,50 in which the extended Graetz problem including the effect of axial heat conduction parametrized by the Peclet number Pe=RePr =ub2H/Pr is solved in the slip-ﬂow regime. The solu-tion is compared to DSMC simulations in a wide range of Knudsen numbers; the DSMC solutions serve to verify the slip-ﬂow solution but also extend the Graetz solution to the transition regime. The DSMC simulations were performed at sufﬁciently low speeds for the effects of viscous heat dissi-pation to be small; this is very important since high speeds typically used in DSMC simulations to alleviate signal-to-noise limitations a new method for solving the Boltzmann equation, which does not suffer from this limitation, is dis-cussed in Sec. III E may introduce sufﬁcient viscous heat dissipation effects to render the simulation results invalid. The effect of viscous dissipation on convective heat transfer is brieﬂy discussed at the end of this section. In Ref. 50 a semianalytical solution of the Graetz prob-lem in the slip-ﬂow regime for all Peclet numbers was pre-sented. This solution showed that in the presence of axial heat conduction Pe10 the Nusselt number is larger than the corresponding Nusselt number in the absence of axial heat conduction Pe→. In particular, it was found that for typical small-scale applications where Pe1 the Nusselt number is fairly insensitive to the Peclet number, but higher by about 10% than NuPe→. The semianalytical slip-ﬂow solution of Ref. 50 was complemented by low-speed DSMC simulations in both the slip-ﬂow and transition regimes see Fig. 9. Comparison of the two solutions in the slip-ﬂow regime shows that the ef-fects of thermal creep are negligible for typical conditions, and also that the velocity slip and temperature jump coefﬁ-cients of Sec. II A provide good accuracy in this regime. The DSMC solutions in the transition regime showed that for fully accommodating walls the Nusselt number decreases monotonically with increasing Knudsen number. Unpub-lished DSMC solutions for accommodation coefﬁcients smaller than one exhibit the same qualitative behavior as partially accommodating slip-ﬂow results; in other words, decreasing the thermal accommodation coefﬁcient increases the thermal resistance and decreases the Nusselt number, while decreasing the momentum accommodation coefﬁcient increases the ﬂow velocity close to the wall leading to a small increase in the Nusselt number.50 The similarity be-tween the Nusselt number dependence on the Knudsen num-ber and the dependence of the skin-friction coefﬁcient on the Knudsen number50 suggests that some form of Reynolds analogy between the two nondimensional numbers may exist for Kn0.1. The effect of viscous dissipation on convective heat transfer is typically quantiﬁed by the Brinkman number Br =ub 2/ T, where T is the characteristic temperature dif-ference in the ﬂow. It is well known that viscous dissipation in the ﬂuid affects convective heat transfer both in terms of bulk temperature ﬁelds and resulting Nusselt numbers. Dis-sipation in small-scale ﬂows is interesting both from practi-cal and theoretical points of view. It becomes especially rel-evant in view of the limitations associated with DSMC, which require artiﬁcially high ﬂow velocities in order to ob-tain a discernible hydrodynamic signal. From a theoretical point of view, small-scale slipping ﬂows differ from their large-scale nonslipping counterparts because the slip present at the system boundaries leads to an additional mode of dissipation, namely, shear work at the boundary. In Ref. 51 the author presented an analysis of a model convective heat-transfer problem, namely convective heat transfer under constant wall heat ﬂux48—a dual to the Graetz problem de-scribed above—in the presence of dissipation: it was shown that, in the slip-ﬂow regime, shear work on the boundary scales with the Brinkman number Br, similarly to viscous heat dissipation in the bulk; also, although shear work at the boundary must be included in the total heat exchange with the system walls, it has no direct inﬂuence on the bulk tem-perature ﬁeld, because it occurs at the system boundaries. Consequently, it was demonstrated how shear work at the boundary can be accounted for in convective heat-transfer calculations under the assumption of locally fully devel-oped conditions. It was also shown that, as the Knudsen number increases, this mode of dissipation can be as impor-tant as dissipation in the bulk of the ﬂow. 2. Thermophoretic force on small particles Small particles in a gas through which heat ﬂows expe-rience a thermophoretic force in the direction of the heat ﬂux; this force is a result of the net momentum transferred to the particle due to the asymmetric velocity distribution of the surrounding gas52 in the presence of a heat ﬂux. This phe-nomenon was ﬁrst described by Tyndall53 and has become of signiﬁcant interest in connection with contamination of mi-crofabrication processes by small solid particles. This prob-lem appears to be particularly important in plasma-based processes which generate small particles.52 FIG. 9. Variation of Nusselt number NuT with Knudsen number Kn from Ref. 50. The stars denote DSMC simulation data for increasing wall tem-perature and the circles denote DSMC simulation data for decreasing wall temperature. The overlying solid lines denote hard-sphere slip-ﬂow results for Pe=0.01, 0.1, and 1.0. 111301-13 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see Considerable progress has been made in describing this phenomenon by assuming a spherical radius R and inﬁ-nitely conducting particle in a quiescent monoatomic gas. Provided that the particle is sufﬁciently small, such that it has no effect on the molecular distribution function of the surrounding gas, the thermophoretic force can be calculated by integrating the momentum ﬂux imparted by the molecules striking the particle. The particle can be considered sufﬁ-ciently small when the Knudsen number based on the par-ticle radius, KnR=/R, implies a free-molecular ﬂow around the particle, i.e., KnR1. Based on these assumptions, Gallis and his collaborators54 have also developed a general method for calculating forces on particles in DSMC simulations of arbitrary gaseous ﬂows, provided the particle concentration is dilute. This method is brieﬂy discussed in Sec. III C. In the cases where the molecular velocity distribution function is known, such as free-molecular ﬂow or the Navier-Stokes limit, the thermophoretic force can be ob-tained analytically. Performing the calculations in these two extremes and under the assumption that the particle surface is fully accommodating, reveals that the thermophoretic force can be expressed in the following form: Fth = R2q/c ¯, 33 where is a thermophoresis proportionality parameter which obtains the values FM=0.75 for free-molecular ﬂow and CE=32/15 =0.679 for a Chapman-Enskog distribution for Maxwellian molecules. Maxwellian molecules are de-ﬁned such that the interaction force between them scales with the ﬁfth power of the distance between their centers.4 Although not very realistic, this model was introduced by Maxwell because it makes the product gd2 independent of g, allowing analytical evaluation of certain functionals involving the collision integral. Here, q is the local heat ﬂux. Writing the thermophoretic force in the above form is, in fact, very instructive.52 It shows that the force is only very mildly dependent on the velocity distribution function with only a change of order 10% observed between Kn1 and Kn1. These conclusions extend to other collision models; for example, for a hard-sphere gas, CE=0.698 Ref. 52. The two limiting values can be used to provide bounds for the value of the thermophoretic force on fully accommo-dating particles close to system walls. Using the weak depen-dence of on the distribution function, Gallis et al.52 pro-vided an estimate of this quantity in the Knudsen layer, KN, by assuming that the distribution function can be written as a superposition of a Chapman-Enskog incoming and outgoing molecules and Maxwellian distribution outgoing mol-ecules, with the relative proportions determined by the ac-commodation coefﬁcient at the wall surface. More speciﬁ-cally, they consider a wall at temperature Tw with accommodation coefﬁcient . For Maxwell molecules, they ﬁnd KN = 1 2CE + 2 −FM 2 1 + Tw/T, 34 which simpliﬁes to KN = 1 2CE + 2 −FM 35 in the limit T→Tw, where, here T is the temperature of the gas outside the Knudsen layer. In other words, the presence of a Knudsen layer has a very small effect on the thermo-phoresis parameter, with KN=0.5CE+FM for a fully ac-commodating wall and KN=FM in the specular reﬂection limit. DSMC simulations show52 that the deviation from CE increases with proximity to the wall, as expected; they also show see Fig. 10 that Eq. 35 serves as an upper bound to the actual thermophoresis parameter within the Knudsen layer; this is presumably because the assumed distribution function overestimates the deviation from the actual distribution. III. NUMERICAL METHODS In this section we brieﬂy discuss recent developments in the numerical solution of the Boltzmann equation. The ma-jority of these developments is associated with the direct simulation Monte Carlo, brieﬂy discussed below, since this is by far the most popular approach for dilute gases. In the interests of brevity we will not discuss hybrid Boltzmann-Navier-Stokes methods, which increase computational efﬁ-ciency by limiting the use of the Boltzmann treatment to the regions where it is needed; discussions of hybrid methods can be found in Refs. 55 and 56. We will close this section with a description of a new variance reduction technique developed to address the computational intransigence of DSMC in low-speed ﬂows resulting from the slow conver-gence rate associated with statistical sampling of macro-scopic properties. FIG. 10. Comparison between the approximate theory of Gallis and co-workers for the thermophoretic force in the Knudsen layer and DSMC re-sults. The theoretical value for the thermophoretic force ratio for Maxwell-ian molecules, KN/CE, is shown as a solid line. The DSMC results represent the average value over ﬁve cells of size x=0.042 adjacent to the wall there are two wall locations in a Kn=0.0475 calculation. 111301-14 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see A. The direct simulation Monte Carlo The direct simulation Monte Carlo is a stochastic par-ticle simulation technique9 for solving the nonlinear Boltz-mann equation.57 DSMC solves the Boltzmann equation by applying a splitting approach to the motion of the particle simulators each particle simulates a large number of real molecules comprising the system; the time evolution of the system is approximated by a sequence of discrete time steps of duration t, in which particles successively undergo col-lisionless advection and collisions. Collisions take place be-tween collision partners selected randomly within cells of size x. The introduction of the above approach with the associated discretization ingredients t,x eliminates the computational cost associated with calculating exact particle trajectories and leads to a simulation method that is signiﬁ-cantly more efﬁcient than “brute force” molecular dynamics approaches. An augmented DSMC formulation which extends the applicability of DSMC to gases of moderate densities, where molecular size effects are not negligible, has also been developed58 and is known as the “consistent Boltzmann algorithm.” B. The effect of ﬁnite discretization in DSMC DSMC has been used to capture and predict nonequilib-rium hydrodynamic phenomena in all Knudsen regimes9 for more than three decades. However, it is only recently that signiﬁcant progress has been made in its characterization as a numerical method and in understanding the numerical errors associated with it. Recently, Wagner57 has shown that DSMC simulations approach solutions of the nonlinear Boltzmann equation in the limit of zero cell size and time step and inﬁnite number of molecules. This result essentially proves “consistency.” Convergence results for the transport coefﬁcients have been recently obtained by Alexander et al.59 for the cell size, and Hadjiconstantinou60 for the time step. Alexander et al.59 used the Green-Kubo theory to evalu-ate the transport coefﬁcients in DSMC when the cell size is ﬁnite but the time step is negligible. They found that because DSMC allows collisions between molecules at a distance as long as they are within the same cell the transport coefﬁcients—when measured using the heat ﬂux at the wall61—deviate from the dilute-gas Enskog values quadrati-cally with the cell size. For example, for the viscosity Alex-ander et al.59 ﬁnd = 5 162 mkT 1 + 16 45 x2 2 . 36 In Ref. 60, the author considered the convergence with respect to a ﬁnite time step when the cell size is negligible. Because DSMC is discrete in time, in order to apply the Green-Kubo formulation the author developed a continuous-time analog of DSMC. Using this continuous-time analog, and the fact that the DSMC dynamics appears symmetric at the long times associated with diffusive transport coefﬁ-cients, the author was able to show that the transport coefﬁ-cients deviate from the dilute-gas Enskog values proportion-ally to the square of the time step. For example, for the viscosity he found = 5 162 mkT 1 + 16 75 cot2 2 , 37 where co= 2kbT/m is the most probable molecular speed. This prediction for the viscosity, and similar predictions for the thermal conductivity and diffusion coefﬁcient derived in Ref. 60, were veriﬁed by DSMC simulations.62 Good agree-ment was found between theory and simulation as illustrated in the example of Fig. 11. The theoretical result for the ther-mal conductivity was also recently veriﬁed by Rader et al.;61 by studying a number of variants of the DSMC algorithm, these authors also verify that the above results are observed when sampling is performed in a fashion which is consistent with the symmetry in the dynamics. The study by Rader et al.61 also reports a discretization error due to a ﬁnite number of particles in a cell. The best ﬁt to their numerical result for the deviation from the theoretical value of the thermal conductivity, in the limit x,t→0, gives the leading-order term to be −0.083/N. C. Forces on small spherical particles in DSMC One of the most important challenges associated with semiconductor manufacturing is the presence of contami-nants, sometimes produced during the manufacturing pro-cess, in the form of small particles. Understanding the trans-port of these particles is very important for their removal or their handling in ways which ensure that they do not inter-fere with the manufacturing process. Recently, Gallis and his co-workers54 developed a method for calculating the force on small particles in rareﬁed ﬂows simulated by DSMC. This method is based on the assumption that the particle concen-tration is very small and the observation that particles with FIG. 11. Error in coefﬁcient of viscosity as a function of normalized time step cot/ from Ref. 62. Circles denote the normalized error in momen-tum ﬂux in the simulations of Garcia and Wagner Ref. 62, and the solid line is the prediction of 37. Note that as the time step increases, the trans-port rate asymptotes to the collisionless limit value. 111301-15 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see sufﬁciently small radius such that KnR=/R1 will have a very small effect on the ﬂowﬁeld; in this case, the effect of the ﬂowﬁeld on the particles can be calculated from DSMC simulations which do not include the particles themselves. Gallis and his co-workers deﬁne appropriate “Green’s functions” which quantify the momentum Fgc ˜ and energy Qgc ˜ transfer rates of individual molecules to the particle surface as a function of the molecule mass, momentum and energy, and degree of accommodation on the particle surface. These can then be integrated over the molecular velocity distribution function, fc ˜, to yield the average force F = Fgc ˜fc ˜dc ˜ 38 or heat ﬂux q = Qgc ˜fc ˜dc ˜ 39 to the particle, where c ˜=c−up and up is the particle speed. For the case of Maxwell’s gas-surface interaction, Gallis et al.54 ﬁnd Fgc ˜ = R2c ˜ c ˜ + 1/2/3co p, 40 Qgc ˜ = R2 c ˜ 1/2 c ˜ 2 −co p2, 41 where co p= 2kbTp/m and Tp is the particle temperature. More complex accommodation models can also be treated; in Ref. 54 an extended Maxwell accommodation model is presented. In the DSMC implementation, integration of Eqs. 38 and 39 is achieved by summing the contributions of mol-ecules within a cell. This yields the force and heat ﬂux to a particle as a function of position. Because the force and heat ﬂux are a function of up, the former are calculated as a function of a number of values of the latter; the values of the force and heat ﬂux at intermediate values of up can be sub-sequently obtained by interpolation.54 D. Statistical noise in low-speed ﬂows In a recent paper, Hadjiconstantinou et al.63 used equi-librium statistical mechanics to obtain theoretical results for the relative statistical sampling error in hydrodynamic quantities in molecular simulations of ﬂows close to equilib-rium. These results characterize the dependence of the rela-tive statistical error EQ=Q/Q of hydrodynamic quantity Q on gas properties and the number of samples taken; here Q is the mean value of Q and Q is the standard deviation in the error in estimating Q. Such results are useful because DSMC, the prevalent solution method for the Boltzmann equation, relies on statistical sampling for extracting hydro-dynamic ﬁelds from particle data. In fact, perhaps the biggest disadvantage associated with DSMC stems from the large relative statistical error present in low-speed ﬂows where the deviation from equilibrium is small. A variance reduction method developed to reduce statistical uncertainty is de-scribed in the next section. In Ref. 63 a variety of expressions for the relative statis-tical error for the most common hydrodynamic state vari-ables and their ﬂuxes shear stress, heat ﬂux was derived. For example, it was shown that the statistical uncertainty in the ﬂow velocity is given by Eu = u u = 1 Ma NM , 42 where N is the number of particles in the sampling volume and M is the number of independent samples per particle. For the hydrodynamic ﬂuxes, expressions were derived when those are measured as volume averages and when measured as surface ﬂux averages. The main ﬁndings of this work can be summarized as follows: 1 The two averaging methods for hydrodynamic ﬂuxes volume, surface yield comparable relative statistical errors, provided that xcot. Here t is the averaging time used in the ﬂux method; x is the linear dimension, in the direction normal to the ﬂux, of the cell in which volume averaging is performed. 2 For Kn1, the relative error in a particular hydrody-namic ﬂux e.g., shear stress is signiﬁcantly larger than the relative error in the “conjugate” state variable e.g., velocity. 3 A simple theory for incorporating the effects of correla-tions in volume averaging was presented. This theory is based on the theory of persistent random walks. 4 It was shown that not only the number of molecules per unit volume in an ideal gas is Poisson distributed, but also arbitrary number ﬂuctuations of an inﬁnite ideal gas in equilibrium are Poisson distributed. Good agreement was found with DSMC simulations of low-speed, low Knudsen number ﬂows where statistical noise presents the biggest challenges. This is expected since the deviation from equilibrium is small under these condi-tions. The results for state variables were also veriﬁed for dense ﬂuids using molecular dynamics simulations. E. Variance reduction Equation 42 shows that for a given statistical uncer-tainty, the computational cost in DSMC scales as Ma−2. As a result, the simulation of low-speed ﬂows at negligible statis-tical uncertainty is prohibitively expensive unless massively parallel computer resources are available: for example, to obtain 1% statistical uncertainty in a 1 m/s ﬂow at typical temperatures, one would need on the order of 5108 inde-pendent samples per cell.63 To address the inefﬁciency of DSMC for low-speed ﬂows, Baker and Hadjiconstantinou developed a variance re-duction technique for calculating the collision integral of the Boltzmann equation which exploits the fact that, in these ﬂows, the deviation from equilibrium is small. More speciﬁ-cally, they showed that one can construct methods that are signiﬁcantly more efﬁcient by focusing the computational effort on calculating the value of the collision integral due to the deviation from equilibrium, since the value of the colli-111301-16 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see sion integral at equilibrium is zero. This can be achieved by evaluating the following “variance-reduced” form of the col-lision integral: df dt coll = 1 2 1 + 2 −1 −2 2f1 MB + f1 df2 dgd2d3v1d3v2, 43 where fd= f −fMB is the deviation from any chosen equilib-rium Maxwell-Boltzmann distribution fMB. The above expression is obtained64 by expanding f about this equilibrium state in the following form7 of the collision integral: df dt coll = 1 2 1 + 2 −1 −2 f1f2gd2d3v1d3v2 44 and noticing that the value of the collision integral for any equilibrium distribution is identically zero, i.e., 1 2 1 + 2 −1 −2f1 MBf2 MBgd2d3v1d3v2 = 0. 45 Here, f1 fx,v1,t, f2 fx,v2,t, 1v−v1, 2v −v2, 1 v−v1 , 2 v−v2 , where a prime indicates postcollision velocities and is the Dirac delta function. When the deviation from equilibrium is small, the vari-ance reduction achieved by evaluating 43 by a Monte Carlo method instead of 44 is large. Moreover, the degree of vari-ance reduction is larger for distributions that are closer to the equilibrium distribution, leading to a method that can practi-cally capture very small deviations from equilibrium. This is in sharp contrast to methods that do not use fd in evaluating the collision integral such as DSMC. In DSMC in particu-lar, as fd→0, fMB dominates the integrand landscape and thus leads to a constant statistical noise,63 which in turn means that the signal-to-noise ratio decreases linearly63 with decreasing Mach number see Eq. 42. On the other hand, in the method presented here, the integrand landscape and consequently the statistical error65 scale with fd in the limit fd→0; as a result, in this limit, the statistical error decreases linearly with the signal leading to a constant signal-to-noise ratio.64 This is shown in Fig. 12, which compares the relative statistical uncertainty of the present method with that of DSMC as a function of the characteristic ﬂow velocity for a shear ﬂow. It was shown in Ref. 64 that the Boltzmann equation can be numerically solved using this approach for evaluating the collision integral and by discretizing the advection operator using a ﬁnite volume or ﬁnite difference method. In Ref. 66 the same authors show how variance reduction ideas can be used to derive a DSMC-like particle scheme for solving the Boltzmann equation, namely by simulating particles which represent deviation from equilibrium. IV. DISCUSSION Theoretical solutions of various phenomena involving isothermal and nonisothermal ﬂows suggest that slip ﬂow is remarkably robust. In channel ﬂows, slip ﬂow seems to cor-rectly predict average quantities of interest ﬂow rates, wave propagation constants, heat-transfer coefﬁcients even be-yond its typically acknowledged limit of applicability of Kn0.1 with acceptable error; moreover, in some cases it can qualitatively describe the behavior of such average quantities well into the transition regime. Methods that extend the range of applicability of the Navier-Stokes description even beyond ﬁrst-order slip ﬂow are highly desirable. The simplicity and signiﬁcant computa-tional efﬁciency advantage enjoyed by the Navier-Stokes de-scription compared to molecular approaches, coupled with the effort already invested in continuum methods, make the former the approach of choice. Despite the lack of general closure models for transport in the transition regime, analyti-cal solutions are sometimes possible through the use of the lubrication approximation and judicious use of already exist-ing analytical results for simple ﬂows. Rigorous high-order slip models such as the one presented in Sec. II C 1 are prov-ing to be valuable in this respect. Direct simulation Monte Carlo has played and will con-tinue to play a central role in the analysis of small-scale, internal gaseous ﬂows. The statistical sampling employed by this method and the slow convergence associated with it, is, perhaps, the most serious limitation of DSMC in the context of small-scale, low-speed ﬂows. Variance reduction tech-niques such as the one described in Sec. III E have the po-tential to completely eliminate this limitation. Signiﬁcant effort should be invested in carefully evalu-ating commonly used gas-surface interaction models such as Maxwell’s model; although more sophisticated gas-surface interaction models have been developed,15 and have been shown to be sufﬁciently tractable for numerical FIG. 12. Relative statistical uncertainty in ﬂow velocity averaged over the ﬂow domain as a function of the wall velocity, U, in Couette ﬂow. Note that the number of samples required to make the statistical uncertainty of the two methods the same, scales with the square of the ratio of statistical uncertainties. 111301-17 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. 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Wachman Academic, New York, 1969, Vol. 1, pp. 667–680. 31Y. Sone, “Asymptotic theory of ﬂow of rareﬁed gas over a smooth bound-ary I,” in Proceedings of the Sixth International Symposium on Rareﬁed Gas Dynamics Academic, New York, 1969, Vol. 1, pp. 243–253. 32N. G. Hadjiconstantinou, “Comment on Cercignani’s second order slip coefﬁcient,” Phys. Fluids 15, 2352 2003. 33T. Ohwada, Y. Sone, and K. Aoki, “Numerical analysis of the shear and thermal creep ﬂows of a rareﬁed gas over a plane wall on the basis of the linearized Boltzmann equation for hard-sphere molecules,” Phys. Fluids A 1, 1588 1989; Y. Sone, T. Ohwada, and K. Aoki, “Temperature jump and Knudsen layer in a rareﬁed gas over a plane wall: Numerical analysis of the linearized Boltzmann equation for hard-sphere molecules,” ibid. 1, 363 1989. 34N. G. Hadjiconstantinou, “Validation of a second-order slip model for dilute gas ﬂows,” Microscale Thermophys. Eng. 9, 137 2005. 35N. G. Hadjiconstantinou, “Oscillatory shear-driven gas ﬂows in the tran-sition and free-molecular-ﬂow regimes,” Phys. Fluids 17, 100611 2005. 36N. G. Hadjiconstantinou and H. A. Al-Mohssen, “A linearized kinetic formulation including a second-order slip model for an impulsive start problem at arbitrary Knudsen numbers,” J. Fluid Mech. 533, 47 2005. 37K. S. Breuer, “Lubrication in MEMS,” in Handbook of MEMS, edited by M. Gad-el-Hak CRC, Boca Raton, FL, 2002. 38J. H. Park, P. Bahukudumbi, and A. Beskok, “Rarefaction effects on shear driven oscillatory gas ﬂows: A direct simulation Monte Carlo study in the entire Knudsen regime,” Phys. Fluids 16, 317 2004. 39N. G. Hadjiconstantinou, “Sound wave propagation in transition-regime micro- and nanochannels,” Phys. Fluids 14, 802 2002. 40N. G. Hadjiconstantinou and O. Simek, “Sound propagation at small scales under continuum and non-continuum transport,” J. Fluid Mech. 488, 399 2003. 41I. B. Crandall, Theory of Vibrating Systems and Sound Van Nostrand, New York, 1926. 42G. Kirchhoff, “Ueber den Einﬂufs der Wärmeleitung in einem Gase auf die Schallbewegung,” Ann. Phys. Chem. 134, 177 1868. 43J. W. S. Rayleigh, The Theory of Sound Macmillan, New York, 1896, Vol. 2. 44B. J. Hamrock, Fundamentals of Fluid Film Lubrication McGraw-Hill, New York, 1994. 45S. Fukui and R. Kaneko, “Analysis of ultra thin gas ﬁlm lubrication based on linearized Boltzmann equation: First report-derivation of a generalized lubrication equation including thermal creep ﬂow,” J. Tribol. 110, 253 1988. 46F. J. Alexander, A. L. Garcia, and B. J. Alder, “Direct simulation Monte Carlo for thin-ﬁlm bearings,” Phys. Fluids 6, 3854 1994. 47L. Graetz, “On the thermal conduction of liquids,” Ann. Phys. Chem. 25, 337 1885. 48A. F. Mills, Heat Transfer Irwin, Homewood, IL, 1992. 49E. M. Sparrow and S. H. Lin, “Laminar heat transfer in tubes under slip ﬂow conditions,” J. Heat Transfer 84, 363 1962. 50N. G. Hadjiconstantinou and O. Simek, “Constant-wall-temperature Nus-selt number in micro and nano channels,” J. Heat Transfer 124, 356 2002. 51N. G. Hadjiconstantinou, “Dissipation in small scale gaseous ﬂows,” J. Heat Transfer 125, 944 2003. 52M. A. Gallis, D. J. Rader, and J. R. Torczynski, “Calculations of the 111301-18 Nicolas G. Hadjiconstantinou Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see near-wall thermophoretic force in rareﬁed gas ﬂow,” Phys. Fluids 14, 4290 2002. 53J. Tyndall, “On dust and disease,” Proc. R. Inst. G. B. 6, 1 1870. 54M. A. Gallis, J. R. Torczynski, and D. J. Rader, “An approach for simu-lating the transport of spherical particles in a rareﬁed gas ﬂow via the direct simulation Monte Carlo,” Phys. Fluids 13, 3482 2001. 55H. S. Wijesinghe and N. G. Hadjiconstantinou, “A discussion of hybrid atomistic-continuum methods for multiscale hydrodynamics,” Int. J. Mul-tiscale Comp. Eng. 2, 189 2004. 56N. G. Hadjiconstantinou, “Discussion of recent developments in hybrid atomistic-continuum methods for multiscale hydrodynamics,” Bull. Pol. Acad. Sci.: Tech. Sci. 53, 335 2005. 57W. Wagner, “A convergence proof for Bird’s direct simulation Monte Carlo method for the Boltzmann equation,” J. Stat. Phys. 66, 1011 1992. 58F. J. Alexander, A. L. Garcia, and B. J. Alder, “A consistent Boltzmann algorithm,” Phys. Rev. Lett. 74, 5212 1995. 59F. J. Alexander, A. L. Garcia, and B. J. Alder, “Cell size dependence of transport coefﬁcients in stochastic particle algorithms,” Phys. Fluids 10, 1540 1998; erratum, ibid. 12, 731 2000. 60N. G. Hadjiconstantinou, “Analysis of discretization in the direct simula-tion Monte Carlo,” Phys. Fluids 12, 2634 2000. 61D. J. Rader, M. A. Gallis, J. R. Torczynski, and W. Wagner, “DSMC convergence behavior of the hard-sphere-gas thermal conductivity for Fourier heat ﬂow,” Phys. Fluids 18, 077102 2006. 62A. L. Garcia and W. Wagner, “Time step truncation error in direct simu-lation Monte Carlo,” Phys. Fluids 12, 2621 2000. 63N. G. Hadjiconstantinou, A. L. Garcia, M. Z. Bazant, and G. He, “Statis-tical error in particle simulations of hydrodynamic phenomena,” J. Com-put. Phys. 187, 274 2003. 64L. L. Baker and N. G. Hadjiconstantinou, “Variance reduction for Monte Carlo solutions of the Boltzmann equation,” Phys. Fluids 17, 051703 2005. 65W. H. Press, S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery, Nu-merical Recipes in C Cambridge University Press, New York, 1994. 66L. L. Baker and N. G. Hadjiconstantinou, “Variance reduction in particle methods for solving the Boltzmann equation,” in Proceedings of the Fourth International Conference on Nanochannels, Microchannels and Minichannels, Limerick, Ireland, 2006 ASME, New York, 2006, paper ICNMM2006–96089. 67R. G. Lord, “Some further extensions of the Cercignani-Lampis gas-surface interaction model,” Phys. Fluids 7, 1159 1995. 68F. Sharipov, “Application of the Cercignani-Lampis scattering kernel to calculations of rareﬁed gas ﬂows. I. Plane ﬂow between two parallel plates,” Eur. J. Mech. B/Fluids 21, 113 2002. 111301-19 The limits of Navier-Stokes theory Phys. Fluids 18, 111301 2006 Downloaded 05 Dec 2006 to 18.80.0.144. Redistribution subject to AIP license or copyright, see
1235	https://www.nature.com/scitable/definition/codon-155/	This page has been archived and is no longer updated codon A codon is a sequence of three DNA or RNA nucleotides that corresponds with a specific amino acid or stop signal during protein synthesis. DNA and RNA molecules are written in a language of four nucleotides; meanwhile, the language of proteins includes 20 amino acids. Codons provide the key that allows these two languages to be translated into each other. Each codon corresponds to a single amino acid (or stop signal), and the full set of codons is called the genetic code. The genetic code includes 64 possible permutations, or combinations, of three-letter nucleotide sequences that can be made from the four nucleotides. Of the 64 codons, 61 represent amino acids, and three are stop signals. For example, the codon CAG represents the amino acid glutamine, and TAA is a stop codon. The genetic code is described as degenerate, or redundant, because a single amino acid may be coded for by more than one codon. When codons are read from the nucleotide sequence, they are read in succession and do not overlap with one another. Further Exploration Concept Links for further exploration mRNA \| frameshift mutation \| nonsense mutation \| DNA \| RNA \| translation \| genetic code \| mutation \| tRNA \| hairpin loop (mRNA) \| peptide \| amino acid \| frameshift mutation Related Concepts (13) Genetics Gene Inheritance and Transmission Gene Expression and Regulation Nucleic Acid Structure and Function Chromosomes and Cytogenetics Evolutionary Genetics Population and Quantitative Genetics Genomics Genes and Disease Genetics and Society Cell Biology Cell Origins and Metabolism Proteins and Gene Expression Subcellular Compartments Cell Communication Cell Cycle and Cell Division Scientific Communication Career Planning Loading ... LearnCast You have authorized LearnCasting of your reading list in Scitable. Do you want to LearnCast this session? Yes No This article has been posted to your Facebook page via Scitable LearnCast. Close Change LearnCast Settings © 2014 Nature Education About\| Contact\| Press Room\| Sponsors\| Terms of Use\| Privacy Notice\| Glossary\| Catalog\| Home\| Library\| Blogs Scitable Chat
1236	https://statisticsbyjim.com/probability/geometric-distribution/	Geometric Distribution: Uses, Calculator & Formula - Statistics By Jim Skip to secondary menu Skip to main content Skip to primary sidebar Menu My Store Glossary Home About Me Contact Me Statistics By Jim Making statistics intuitive Graphs Basics Hypothesis Testing Regression ANOVA Probability Time Series Fun Calculators Geometric Distribution: Uses, Calculator & Formula By Jim FrostLeave a Comment What is a Geometric Distribution? The geometric distribution is a discrete probability distribution that calculates the probability of the first success occurring during a specific trial. In other words, during a series of attempts, what is the probability of success first occurring during each attempt? Use this distribution when you need to understand how many attempts are necessary to produce the first successful outcome. For example, the geometric distribution can answer the following questions. What is the probability of the first: Six in a series of die rolls? Year of catching the flu over the years? Person to support a law during a repeated sampling for an interview? Product to have a defect in a random sample from an assembly line? Successful attempt for a project or task? This distribution is an example of a Probability Mass Function (PMF) because it calculates likelihoods for discrete random variables. In this post, learn when to use the geometric distribution and its cumulative form, about its formula, and how to calculate probabilities by hand. I also include a geometric distribution calculator that you can use with what you learn. For information about other distributions for binary data, read my posts, Maximize the Value of Your Binary Data, the Bernoulli, Binomial, Negative Binomial, and the Hypergeometric Distribution. The Geometric Distribution is Memoryless The geometric distribution is “memoryless.” Memoryless is a distribution attribute indicating that the occurrence of the next success does not depend on when the last success occurred or when you start looking for successes. This type of process has independent events that occur with a constant probability. For example, when you start observing die rolls looking for the next six, the probability of when the next six appears does not depend on when the last six appeared or when you start watching for sixes. Both the geometric distribution and exponential distribution are memoryless. Learn more about the Exponential Distribution. Using the Geometric Distribution The geometric distribution models the probabilities for the first event occurring during various trials when the likelihood of an event is known. Let’s bring it to life with an example! I’ll start by using statistical software to calculate the geometric distribution probabilities and create distribution plots. This progression will help you know what you can learn from this distribution. Imagine you’re playing a game where rolling sixes on a die is beneficial. You’d like to know the probability of when you will throw the first six. Suppose you want to learn the chance of getting the first six on precisely the third roll. In this example, the probability (p) is 1/6 = 0.1667, and we’re interested in the third roll. My software displays the probability as the following: The geometric distribution calculates that the probability of the first six occurring on the third die roll is 0.115755. Please note the full implications of this scenario. The final probability incorporates the fact that the first two rolls are not sixes, and then the third roll is a six—that’s the only sequence of events allowing the first success to occur on the third attempt. Consequently, the calculations factor in the probability of the event not occurring on particular trials (failures) and then appearing on a specific attempt (success). Later, you’ll see that the geometric distribution formula incorporates both the probabilities of failure and success. That single probability is interesting but perhaps not so useful by itself. We’re also interested in the chances of rolling the first six on other rolls. Graphing the Full Distribution of Outcomes The geometric distribution is valuable because it can describe the probability of the first event occurring on all possible numbers of trials. Graphing the distribution excels at displaying the larger context for various attempts. Theoretically, the distribution goes up to an infinite number of trials, but we’ll stop before then! Returning to the die example, we’ll use the geometric distribution to find the probability of rolling the first six on various rolls. The geometric distribution graph below displays the probability of rolling the first six in precisely 1, 2, 3, etc. rolls, up to 30. Each bar in the geometric distribution graph indicates the probability of rolling the first six on a specific trial. For instance, the likelihood of rolling the first six on the third roll is 0.115755, corresponding to the preceding statistical output. The distribution continues beyond 30 rolls, but the probabilities become infinitesimal. Related post: Understanding Probability Distributions Geometric Cumulative Distribution Function The geometric distribution is superb for understanding when an event might first occur. However, many times you’re not interested in a specific trial. For example, you might be interested in the probability of getting your first six within the first six rolls. In that case, you’re interested in the total probability across multiple trials. Let me introduce you to the geometric cumulative distribution function. The cumulative distribution simply sums the probabilities for a range of trials. Again, a geometric distribution graphs brings it to life. Technically, the geometric cumulative probability calculates the likelihood of obtaining the first event in less than or equal toN trials. If you need a ≥ probability, use the inverse geometric cumulative distribution. These days, most statistical software lets you specify the direction. For our example, we want to learn the cumulative probability that the first 6 appears within the first six rolls (i.e., ≤ 6). In the geometric distribution graph below, the shaded region displays the cumulative probability. Interestingly, you might think you’re virtually guaranteed to get a 6 when you roll the die six times. However, the red shaded region in the geometric distribution graph indicates you have a two-thirds (66%) cumulative chance of getting the first six within the first six rolls. This cumulative probability sums the individual likelihoods of the first six die rolls. For a real-world example, learn how I use the geometric distribution to model the number of years to the first flu infection for the vaccinated vs. unvaccinated. Learn more about Cumulative Distribution Functions: Uses, Graphs & vs PDF. Geometric Distribution Assumptions and Notation The geometric distribution models the probabilities for the first success occurring on the X th trial. However, your data must meet the following requirements for the geometric distribution to be appropriate. Your data must be binary: For example, infected or uninfected, 6 or not 6, pass or fail, etc. Independent trials: One trial’s result does not affect the next trial. For example, a coin toss doesn’t affect the following coin toss. The probability remains constant over time. In some contexts, this supposition is true due to the physical attributes of the process, such as coin tosses and die rolls. However, the likelihood won’t necessarily remain constant in other contexts. For example, the probability of a product defect at a manufacturing plant can change over time. Use a P chart (a control chart) to verify this assumption when the chance can change. Geometric Experiments and Bernoulli Trials Typically, you’ll use the geometric distribution when you have Bernoulli Trials. These trials satisfy the binomial distribution assumptions above. In these trials, analysts label one outcome a success and the other a failure. A Bernoulli trial contains attempts where the probability of success is constant. Geometric experiments perform a series of attempts until the first success. Two Forms of the Geometric Distribution Note that there are two forms of the geometric distribution. The two forms model: The first success occurring on the X th trial. The number of failures before the first success. Fortunately, these two forms are equivalent. The difference comes down to how you count the trials. The former definition counts all trials, including the final, successful one. The latter definition counts only the failures before the success. Success on the X th trial is equivalent to having X – 1 failures. Statisticians sometimes refer to the first form as the shifted geometric distribution because it moves the final trial over by counting the success. For the die example, I calculated the probability of first rolling a six on the third trial. Using the number of failures form, we don’t count the success trial, just the preceding failures. Hence, if we have success on the third roll, that means we have two failures, and those two forms of the geometric distribution calculate the same probability. I use the success on the X th trial version throughout this post. Parameter and Notation The geometric distribution has one parameter, p = the probability of success for each trial. You denote the distribution as G(p), which indicates a geometric distribution with a success probability of p. Geometric Distribution Calculator Use my geometric distribution calculator below to calculate probabilities and cumulative probabilities. Click the link for its standalone page that you might want to bookmark. Geometric Distribution Calculator Geometric Distribution Calculator▼ Probability of Success (p): Number of Trials (x): [x] Include success trial Calculate Clear Print For the die example, I calculated the probability of the first success of rolling a six on the third trial as 0.115755. Using the distribution form that includes the successful trial, we add the preceding failure trials and the success trial. In short, if we have success on the third roll, that means we have three trials (x = 3), which we enter in this calculator. In the calculator, enter Probability of Success (p) = 0.1667 and Number of Trials (x) = 3. Be sure that the Include success trail box is checked. The calculator displays a probability of 0.1158, matching our results above within rounding error. Now, try one yourself. Suppose you have a team working on a project, and you give them four attempts to complete the project successfully due to time constraints. It’s a very complex task, and they have only a 30% chance of correctly completing the project on each attempt. If they fail, they’ll need to try again. What is the cumulative probability of succeeding on attempts 1 – 4? Answer at the end! Geometric Distribution Formula Generally, you’ll use statistical software or online calculators to calculate probabilities for the geometric distribution. However, I’ll teach you how to use the geometric distribution formula so you can calculate them manually. Additionally, I’ll walk you through the formula with a worked example. The geometric distribution formula for the probability of the first success occurring on the X th trial is the following: where: x is the number of trials. p is the probability of a success for each trial. The geometric distribution formula takes the probability of failure (1 – p) and raises it by the number of failures (x – 1). That produces the likelihood of having failures for all trials before the trial of interest (x). Then the equation multiplies the probability of failure by the probability of success (p) occurring on the trial of interest. That gives you the probability of the first success happening on the X th trial. The mean of a geometric distribution is 1 / P. This mean is the expected value for a geometric distribution. Learn more about Expected Values: Definition, Formula & Finding. Let’s work through an example calculation to bring the formula to life! Worked Example of Finding a Geometric Distribution Probability We’ll use the geometric distribution formula to calculate the probability of rolling the first six on the third roll. That’s the example we used before, and now we’ll calculate it by hand. Here are the values to enter into the formula: x = 3 p = 0.1667 This calculation by hand confirms the previous statistical software results within rounding error. If you need to calculate the cumulative geometric probability for a range of trials, calculate the probability for each attempt and then sum the probabilities. For example, if you need to calculate the probability of the first six occurring in the first six rolls, calculate the likelihood of it happening on rolls 1, 2, 3, 4, 5, and 6. Then sum those probabilities. My calculator provides the cumulative probabilities automatically, which leads to solving the practice problem. Finally, the solution for the problem in the geometric distribution calculator section is that the cumulative probability of completing the project successfully during attempts 1 – 4 is 0.7599. To solve this problem: Enter 0.3 for the Probability of Success. In Number of Trials, enter 1, 2, 3, and 4 individually and record each probability. Be sure that the Include success trial box is checked. In the answer table, the row for P(X ≤ 4) indicates that the probability for success in less than or equal to 4 trials is 0.7599. Share this: Tweet Like this: Like Loading... ### Related Filed Under: ProbabilityTagged With: distributions, graphs Reader Interactions Comments and QuestionsCancel reply Primary Sidebar Meet Jim I’ll help you intuitively understand statistics by focusing on concepts and using plain English so you can concentrate on understanding your results. Read More... Search this website Buy My Introduction to Statistics Book! Buy My Hypothesis Testing Book! 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1237	https://artofproblemsolving.com/community/c2701h1062795_challenge_aops_problem_3__trigonometric_inequality?srsltid=AfmBOor2oZCJHe53qIMwIcH20gyDeH7sJQ0e0WhRG1HHfLCIcqX7p5yt	Bits of Math : Challenge AoPS, Problem #3 - Trigonometric Inequality Community » Blogs » Bits of Math » Challenge AoPS, Problem #3 - Trigonometric Inequality Sign In • Join AoPS • Blog Info Bits of Math ============ Challenge AoPS, Problem #3 - Trigonometric Inequality by MSTang, Mar 13, 2015, 10:46 PM Welcome back to Challenge AoPS, the ~~regular, weekly~~ series of posts I write ~~so I can claim that my entries have some amount of variety~~ to give nice problems for AoPSers to try out and write solutions for! This problem is from 103 Trig, and it's a very nice inequality problem that ties together algebraic and trigonometric concepts! Problem #3: Let and be real numbers. Prove that Please feel free to post solutions, thoughts on the problem, different approaches, anything you can think of regarding this problem in comments. PS: Ever since the new site came about and this blog's "table of contents" were removed, I've been trying to find a good way to replace it so we can still navigate efficiently throughout. For now, I have standardized all of the tags on the posts - each post has a subject (alg, geo, etc.), a contest or book that it came from, or other labels related to things I've done here (Challenge AoPS, Explanations, ...) You can find a listing of all tags in the sidebar on the left. Hope that helps! This post has been edited 4 times. Last edited by MSTang, Apr 17, 2015, 11:19 PM trigonometryinequalitiesChallenge AoPS103 Trigonometry Problems Read 5 bits Post your bit! Comment 5 Comments The post below has been deleted. Click to close. This post has been deleted. Click here to see post. thoughts Well, C-S fails as we get . Because then it remains to show that or which is not true (no matter how you try to AM-GM it it keeps on failing ) I also tried applying the well-known and similarly for . Now it gets kind of tricky though realsol Take the derivative of the LHS w.r.t. x and you get that it is . So max/min occurs when Now sub it back in and bash. I'll edit in a legit non-calc solution if I find it. by tastymath75025, Mar 14, 2015, 1:32 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Yeah, C-S alone has no chance of working sadly. Why? Hint Equality cases. Though yeah, calculus sort of kills it ;P Thanks for commenting! by MSTang, Mar 14, 2015, 5:30 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Err, if I'm not mistaken, this is simply equal to (sorry for the bad latex), which is easy to optimize using Cauchy? This post has been edited 1 time. Last edited by MSTang, Mar 14, 2015, 10:12 PM Reason: LaTeX by AnonymousBunny, Mar 14, 2015, 8:55 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Yeah, that's it The double-angle is the key part there. by MSTang, Mar 14, 2015, 10:12 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Here's my proof (yes I know this post is 10 years old lol). I think it's what AnonymousBunny outlined. Let and . Then: . Since the RHS of the above inequality is clearly non-negative, it suffices to show that:By C-S, we just need to prove , which holds. by busy-beaver, Jan 19, 2025, 5:47 AM Report Where I do math! I try to show the motivation behind my solutions and teach HOW to think about hard problems. MSTang Archives January 2016 New blog! December 2015 [Blitz 8] Cayley's Formula! 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January 2015 Problem #40: Quadratic Inequality II - Attack of the Roots Problem #39: Divisibility Card Game (Putnam, 1993) Problem #38: Sides and Sines Cubed (ARML Individual, 1984) Problem #37: Inscribed Square Inequality (Math. Ref., 2013.6) December 2014 Year Recap and Review (TED #10) Problem #36: Isosceles Triangle Trigonometry (TED #9) One of my number theory problems (TED #8) Six Nice AIME #1s! (TED #7) Problem #35: Equilateral, Cartesian Hexagon (TED #6) Back from vacation! 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May 2014 Problem #17(a): A Circle and a Triangle Problem #16: Primes in Number Theory Problem #15: 103 Trigonometry Problems, #7 April 2014 Good luck on the USA(J)MO! Problem #14: Fractional Power System (Interm Alg, 20.72) Problem #13: Canada 2011.1 March 2014 Geometry Revisited, 1.6.4 Geometry Revisited, 1.4.5, 1.4.6 Two Geometric Lemmas Euler's Phi Function: An Introduction February 2014 Problem #12: Set Satisfies GCD Property (104 NT, #28) Problem #11: What's the Missing Digit!? (104 NT, #29) January 2014 Problem #10: Diophantine on a Quadratic (South Africa, 1995) Problem #9: Cycling Sequence of Integers (AMC 10, 2007) Problem #8: Late-Night Number Theory Square Ray Extension (Prasalov, 1.19) Challenge AoPS: Problem 2 - Polygon Bowl December 2013 Problem #7: Expressions Both Perfect Cubes (USAJMO, 2013) Problem #6: Canada 2001.1 The entire history of MSTang. Problem #5: (AoPS Forums) Algebra Geometry Revisited: 1.1.2, 1.2.4 November 2013 Problem #4: Proving Bernoulli's Inequality October 2013 Geometry Revisited: 1.1.1 (again) and 1.1.4 Challenge AoPS! Problem #1 Geometry Revisited: 1.1.3 Geometry Revisited: 1.1.1 Problem #2: Three-Digit Integer and Digit Sum (ITAMO, 2002) Problem #1: Unique Base Gives Power of Prime (ITAMO, 2013) Shouts Submit This is amazing by sayemsub15, Aug 19, 2022, 11:03 AM hello mstang! by Williamgolly, Aug 14, 2020, 2:10 AM shout!!! by bulbasaur4life, Jan 13, 2019, 6:01 AM by haha0201, Aug 20, 2016, 4:46 PM Unfortunately didn't get to meet you at ARML by champion999, Jun 9, 2016, 1:26 AM This blog is dead. Thank BQ by q14, Mar 25, 2016, 8:25 PM In a convex heptagon, the degree measures of the interior angles are and degrees. What is the degree measure of the largest interior angle? by zainabgirl, Mar 24, 2016, 1:04 AM by MSTang, Dec 9, 2015, 9:37 PM no this blog is beyond op cuz it's MSTang by phi_ftw1618, Dec 9, 2015, 5:07 AM oh dang this blog is OP by Cube1, Dec 4, 2015, 3:16 AM npnpnpnp by Eugenis, Oct 22, 2015, 12:13 AM tytytyty by MSTang, Oct 12, 2015, 10:27 PM shoutbox rev by Generic_Username, Oct 11, 2015, 1:26 AM Wow this is amazing, you kept this blog up for 2 years and the problems you post are very high quality, thank you sir! by measurement, Sep 7, 2015, 1:29 AM Wow! This really is a great blog! Keep blogging forever! by hexagram, Aug 25, 2015, 1:46 AM thank you so much for this blog! I learn something new every day! by CornSaltButter, Jan 15, 2014, 3:04 PM Guys, I made a special post about my "story" of learning math - It's on the blog, and it's called "the story of MSTang" or something. by MSTang, Jan 4, 2014, 8:30 PM I have the same question for you MSTang like joboboish97 had. by walmartswagdaddy, Jan 2, 2014, 7:51 PM i don't know you but my science teachers naame is ms.tang, just saying. by isaaclo, Dec 18, 2013, 2:27 AM Thanks for the compliment by MSTang, Dec 10, 2013, 4:57 AM Please PM me too. You're too good. Not jk by HYP135peppers, Dec 8, 2013, 8:52 PM I'll PM you. by MSTang, Dec 8, 2013, 7:30 PM Hey Mstang, I'm also on Brilliant but I'm not as good as you are. I'd like to know what do you do in order to learn maths, read books?, lot of time in front of problems or what? by joboboish97, Dec 8, 2013, 2:52 PM Not not not not not not contrib pls? Not not not not not not not not not not jk by nowunkie, Dec 1, 2013, 6:32 PM Contrib pls? jk by El_Ectric, Nov 19, 2013, 1:18 AM Not Contrib pls? Not jk by crastybow, Oct 7, 2013, 1:58 AM Contrib pls? jk by sunny2000, Oct 6, 2013, 7:13 PM Contrib pls? jk by nsun48, Oct 5, 2013, 4:04 AM Hi everyone - I unfortunately don't accept contributors on this blog. Please don't ask me! by MSTang, Oct 3, 2013, 6:25 PM 137 shouts Tags algebrageometrynumber theoryMiscellanoustrigonometryTED ProjectAIME problemscombinatoricsBlitzinequalitiesClevermathGeometry Revisited103 Trigonometry Problems104 Number Theory Problemsanalytic geometryCanadian MOHMMT Problemsown problemsUSAMO Problems105 Algebra ProblemsAMC ProblemsBridging the GapChallenge AoPSPolynomialsPrasalovARML ProblemsarticleBAMO Problemsbrilliant.orgExplanationsITAMO ProblemsMathematical ReflectionsPUMaC ProblemsPutnam Problemssolid geometryUKMT Book Problems101 Algebra Problems102 Combinatorics ProblemsAoPS Intermediate Algebracomplex numbersgeometric lemmasgraph theoryMath Olympiad TreasuresMWMT Year 3OaNI ProblemsPaTiCPEN ProblemsProblems from the BookPurple Comet ProblemsSlovenian MOSouth African MO ProblemsSpainTurkish NMOWOOT About Owner Posts: 6012 Joined: Oct 1, 2012 Blog Stats Blog created: Sep 29, 2013 Total entries: 113 Total visits: 196306 Total comments: 385 Search Blog Something appears to not have loaded correctly. 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1238	https://fiveable.me/key-terms/united-states-history-since-1945/rapprochement	printables 🧸us history – 1945 to present review key term - Rapprochement Citation: MLA Definition Rapprochement refers to the process of improving or establishing friendly relations between countries, particularly after a period of conflict or hostility. This term is significant in the context of international diplomacy, as it often signifies a shift towards cooperation and dialogue, moving away from adversarial postures. In the case of the Cold War, rapprochement was a crucial strategy used by leaders to ease tensions and foster better relations, especially between the United States and major communist nations. 5 Must Know Facts For Your Next Test Rapprochement played a key role in President Nixon's foreign policy, particularly in easing tensions with China and the Soviet Union during the 1970s. Nixon's visit to China in 1972 marked a pivotal moment in rapprochement, signaling a shift from isolation to engagement with communist nations. The policy aimed at reducing the risk of nuclear confrontation and fostering stability in international relations during the Cold War. Rapprochement led to increased trade and cultural exchanges, which helped build mutual understanding between adversarial nations. The success of rapprochement with China influenced other nations and encouraged a broader trend towards dialogue and negotiation over military confrontation. Review Questions How did rapprochement impact U.S. relations with China during Nixon's presidency? Rapprochement fundamentally transformed U.S.-China relations during Nixon's presidency by facilitating a shift from decades of hostility to engagement. Nixon's historic visit in 1972 opened diplomatic channels, allowing for dialogue on various issues, including trade and security. This change not only altered China's perception of the U.S. but also shifted global power dynamics as both countries began to collaborate on mutual interests. Analyze how rapprochement with the Soviet Union differed from that with China in terms of objectives and outcomes. While both efforts aimed at easing tensions during the Cold War, rapprochement with the Soviet Union was primarily focused on arms control and managing nuclear threats, leading to treaties like SALT I. In contrast, rapprochement with China was about opening diplomatic ties and economic cooperation. The outcomes reflected these differences: while U.S.-Soviet rapprochement led to structured negotiations, U.S.-China rapprochement resulted in a broader integration into global economic systems. Evaluate the long-term implications of rapprochement on U.S. foreign policy beyond Nixon's era. The long-term implications of rapprochement significantly reshaped U.S. foreign policy by establishing a precedent for engaging with adversaries through diplomacy rather than military confrontation. This approach laid the groundwork for future administrations to pursue similar strategies with other nations like Vietnam and Iran. As a result, the idea of using diplomatic engagement as a tool for conflict resolution became integral to U.S. foreign policy philosophy, influencing international relations well into the 21st century. Related terms Détente: A period of relaxed tensions and improved relations between the United States and the Soviet Union during the Cold War, characterized by arms control agreements and diplomatic exchanges. Ping Pong Diplomacy: A significant diplomatic event in the early 1970s where a series of table tennis matches between the United States and China facilitated improved relations between the two countries. Normalization: The process of establishing formal diplomatic relations between countries, often following a period of estrangement or conflict.
1239	https://www.utmel.com/tools/time-constant-calculator?id=18	Contact us Hong Kong: +852-53488308 Canada: +1-4388377556 Email: info@utmel.com Privacy PolicyTerm & ConditionCookies Policy Products Blog Components Sign In Hello , Welcome to UTMEL My Profile Order Status & History RFQ History My Favorite Address Management Sign InOrSign Up All Products ManufacturersBlogComponentsToolsAlternative PartsCertificationAbout UsContact Us Products Products Blog Components Sign In Hello , Welcome to UTMEL My Profile Order Status & History RFQ History My Favorite Address Management Sign InOrSign Up Home /Tools /Time Constant Calculator / Time Constant Calculator This Capacitor Energy and Time Constant calculator help electrician to calculate the energy stored in a capacitor and compute the time constant of a capacitor for the given voltage across it. With this online calculator for the capacitor energy and time constant calculation, you are able to get the energy (E) and Time Constant (T) just by inputting 3 values: Voltage across capacitor, Capacitance and Load Resistance. Input Voltage (V) Capacitance (C) Load Resistance (R) Output Time Constant () s Energy (E) J Formula Introduction RC Time Constant Circuit - Calculations Introduction of Capacitor Energy and Time Constant Calculator This online calculator tool calculates the RC time constant, which is the product of resistance and capacitance values. This number, which appears in the equation describing the charging or discharging of a capacitor via a resistor, describes the time it takes for the voltage across the capacitor to reach approximately 63.2 percent of its final value after a voltage shift is applied to such a circuit. It's also possible to measure the total energy contained in a capacitor charged to a specific voltage. Definition of Capacitor Energy Energy is equals to product of capacitance and voltage is reciprocal of two. Capacitor Energy Formula Calculation Where,C is Capacitance in faradsV is Voltage across capacitor in volts Example for Calculating Capacitor Energy Let's consider capacitance C as 1000 micro-farad and voltage V as 10 volts. Let's apply formula.E= 1000102 /2E= 0.0500 joules Definition of Time Constant Time constant is a degree of put off in an electrical circuit on account of both an inductor and resistor or capacitor and resistor. I will talk the maximum not unusual place case that is resistor and capacitor, but the inductor resistor aggregate behaves in a comparable manner. The time constant is identical to the cost of the resistance in ohms elevated with the aid of using the cost of capacitance in Farads. The time constant is measured in seconds . It represents the time for the voltage to decay to 1/2.72. A capacitor shops electrons (price), the greater voltage is implemented, the greater price is stored. If you observed of plumbing, voltage is like strain and a capacitor is sort of a balloon complete of water. A resistor restricts the the glide of electrons (cutting-edge ). Current is like water glide price and a resistor is sort of a slender piece of pipe. If a voltage is implemented to a capacitor it shops price. If a resistor is switched in among the 2 ends of the capacitor the price flows via the resistor, discharging the capacitor. As this takes place the voltage at the capacitor decreases, inflicting much less cutting-edge to glide via the resistor. As time is going at the capacitor discharges greater slowly. After one time constant the voltage decays to 1/2.72 of the authentic amount. Time Constant Formula Calculation Time constant is equals to product of resistance and capacitance. τ=RC where R= Resistance in ohmsC= Capacitance in faradsτ= Time constant in seconds The above formula is for calculating the time constant of RC element. And the product of R C is the time constant of RC circuit. Meanwhile the measurements unit of the capacitors is seconds, with the symbol of the Greek letter Τ (tau). Example for Time Constant Calculation Let's consider capacitance C as 2000 micro-farad and reactance R as 10000 ohms. Let's apply formula.Τ τ = 200010000Τ τ = 20 seconds More Info About Capacitor Enegry and Time Constant Calculation Using the supply voltage, load resistance, and capacitor capacitance, this calculator calculates the capacitor charge time and capacity. Since it is very useful when measuring capacitor value for RC filters or calculating the energy stored in a capacitor, the Time Constant Calculator is also known as an RC filter calculator or a Capacitor Charge Time Calculator. The RC time constant is a measurement that can be used to determine how long a capacitor will take to charge to a certain amount. The size of the capacitor and the resistance of the circuit determine how long it takes for a capacitor to reach a certain stage. The voltage over a capacitor of Value C increases slowly when a voltage is applied through a resistance of Value R. The time constant is the amount of time it takes to charge to 63.21% of the final voltage value. How to Work Out the Time Constant and Stored Energy: The capacitance (C) and load resistance (R) values can be used to calculate the time constant (τ). (R). Voltage (V) and capacitance can be used to calculate the energy stored on a capacitor (E): = R.C. CV2/2 = E Energy Stored = (Voltage (V)2 x Capacitance (µF)) ÷ 2 Time Constant = Capacitance (µF) x Load Resistance (Ω) When and where: Voltage (V) is the volts of the capacitor's input voltage. Capacitance (C) is the measure of capacitance in microfarads. Load Resistance (RL) is defined as the resistance in ohms. Constant of Time () = Constant of Time in seconds Frequently Asked Questions How is time constant calculated? The time constant, τ is found using the formula T = R x C in seconds. What is time constant of a capacitor? The time constant of a resistor-capacitor series combination is defined as the time it takes for the capacitor to deplete 36.8% (for a discharging circuit) of its charge or the time it takes to reach 63.2% (for a charging circuit) of its maximum charge capacity given that it has no initial charge. What is the time constant equal to? The RC time constant, also called tau, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads), i.e. What is meant by time constant of a circuit? So time constant is the duration in seconds during which the current through a capacities circuit becomes 36.7 percent of its initial value. This is numerically equal to the product of resistance and capacitance value of the circuit. The time constant is normally denoted by τ (tau). What is time constant in control system? Time Constant is the “how fast” variable. It describes the speed with which the measured Process Variable (PV) responds to changes in the Controller Output (CO). More specifically it represents the time needed for the PV to reach 63.2% of its total and final change. How is RC calculated? Calculating the RC is straight forward -- multiply the capacitance C, in Farads, by the resistance R, in Ohms. Remember to take care of your powers of 10 -- a micro-Farad is 10-6F, while a pico-Farad is 10-9F. What is the unit of RC? seconds The units of RC are seconds, units of time. This quantity is known as the time constant: τ=RC. At time t=τ=RC, the charge equal to 1−e−1=1−0.368=0.632 of the maximum charge Q=Cϵ. Why we use RC circuit? The RC circuit is used in camera flashes, pacemaker, timing circuit etc. The RC signal filters the signals by blocking some frequencies and allowing others to pass through it. It is also called first-order RC circuit and is used to filter the signals bypassing some frequencies and blocking others. 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1240	https://cs.stackexchange.com/questions/145972/find-two-non-overlapping-subarrays-with-total-sum-equal-to-k	Skip to main content Find two non-overlapping subarrays, with total sum equal to k Ask Question Asked Modified 2 years, 10 months ago Viewed 2k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Given an array of N non-negative integers, and a number K, we need to find two non-overlapping contiguous subarrays that have a total sum of K. Our algorithm is supposed to find the minimum total length of the two subarrays that have total sum K. Also, after I checked the biggest test cases that we got, N doesn't go above 104 and K doesn't go above 106 (but specific subarrays can have a sum way larger than that) The algorithm implementation (using C++), must have a time complexity of at most O(logN N^2). The only idea I've had that is better than O(n^3) is to find all possible contiguous subarrays and their sum (O(N^2) using a prefix sum array), then sort them by sum (O(logN N^2)), and for each subarray I do a binary search to find the subarray that has the remaining sum (which is K - (sum of first subarray)). While this idea has better time complexity than O(n^3), its space complexity is pretty bad, because I have to save three arrays of size N(N+1)/2, and the C++ program ends up using 1-2 GB of RAM for large N (>1000). So my question is: Is there any way to solve the problem efficiently, without the need of the longer arrays?? Or is there any other way I can implement the idea above with C++ so that I don't use up so much memory? Thanks algorithms time-complexity space-complexity Share CC BY-SA 4.0 Improve this question Follow this question to receive notifications edited Dec 3, 2021 at 20:41 D.W.♦ 168k2222 gold badges233233 silver badges509509 bronze badges asked Nov 21, 2021 at 11:38 NiXtNiXt 7377 bronze badges 4 I don't "see" the 3rd array to keep. Rather than starting to argue about a constant factor of, say, 6 here, I suggest asking for a review of your code on Code Review@SE - they even have a guide How to get the best value out of Code Review - Asking Questions. greybeard – greybeard 2021-11-25 08:18:18 +00:00 Commented Nov 25, 2021 at 8:18 It really makes no difference, since one single array of 108 integers will be more than 64MB. The problem here isn't the implementation, but the asymptotic complexity, and trying not to store an array of size O(n2). I still haven't found an acceptable solution, but as I said, I will post the answer tonight probably, or tomorrow morning. NiXt – NiXt 2021-11-25 08:30:19 +00:00 Commented Nov 25, 2021 at 8:30 2 Somebody commented an entirely similar problem on SO is similar to this one. greybeard – greybeard 2021-11-29 17:55:12 +00:00 Commented Nov 29, 2021 at 17:55 Yep, I'm pretty sure we are from the same class :) Our prof said he will upload a O(n2) solution this week. I'll let you know when I have it. NiXt – NiXt 2021-11-29 21:29:26 +00:00 Commented Nov 29, 2021 at 21:29 Add a comment \| 5 Answers 5 Reset to default This answer is useful 1 Save this answer. Show activity on this post. The following algorithm works but does not achieve the required time complexity of O(n2logn), since I misread the time constraint as just being subcubic. Let ai be the i-th element in the array and ε>0 be an arbitrary small constant. Construct a prefix sum array (in time O(n)) and then, for each pair i,j with j−i≥n1−ε generate the tuple (∑jh=iah,i,j). This requires time O(n1+ε). Next, construct two sorted lists L+,L− containing the above tuples. Both lists sort the tuples w.r.t. the first entry. Additionally, the first (resp. second) list breaks ties w.r.t. the second entry while the second list breaks ties w.r.t. the third entry. This requires time O(n1+εlogn). Next use your approach with a slight variation: guess the start position i and the end position j of the first subarray (there are O(n2) guesses). For each guess (i,j): Compute the sum S=∑jh=iah (in constant time). Binary search L+ and L− to decide whether there is a disjoint array of length at least n1−ε with sum K−S. If this is the case you are done. Otherwise exhaustively search all disjoint subarrays of length smaller than n1−ε. Overall this requires time O(logn+n1−ε)=O(n1−ε) per guess and time O(n3−ε) in total, which falls within the time requirements. The space requirement is only O(n1+ϵ). Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Nov 21, 2021 at 13:54 answered Nov 21, 2021 at 13:20 StevenSteven 29.7k22 gold badges2929 silver badges4949 bronze badges 2 Thanks for the answer!! Do you happen to have any Wikipedia link or something similar where I can read about how/why this technique works?? I have never seen any algorithm that uses the same technique, so any extra info would definitely help. NiXt – NiXt 2021-11-21 15:49:38 +00:00 Commented Nov 21, 2021 at 15:49 1 I'm sorry I don't know if this technique even has a name. In general splitting the work into a "small parts" and "large parts" and using different approaches to handle each part is a quite common strategy. The actual details vary from problem to problem. Here we are getting a trade-off between time and space that depends on the threshold where we switch from exhaustive search to binary search. Steven – Steven 2021-11-21 16:27:26 +00:00 Commented Nov 21, 2021 at 16:27 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Compute the sums of all subarrays (O(n2)), sort all distinct sums (O(n2logn)), then find all pairs of sums that sum to K (O(n2)). Note each sum may correspond to multiple subarrays. Given two sums that sum to K, you have to check for all pairs of subarrays corresponding to the two sums whether they overlap. To solve this, for each sum, pre-find the subarray with the leftmost end position and the subarray with the rightmost start position. You only need to store and check these two subarrays (also note to store a subarray, you only need to store its start position and end position). Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Nov 24, 2021 at 2:18 answered Nov 24, 2021 at 2:09 xskxzrxskxzr 7,62355 gold badges2424 silver badges4848 bronze badges 5 start position and end position for the checks, it should do nicely to just know the end of the leftmost and the start of the rightmost subarray. The other end can be re-established once a pair is found. greybeard – greybeard 2021-11-24 02:26:49 +00:00 Commented Nov 24, 2021 at 2:26 This is exactly the algorithm described in the third paragraph of the question. As the question states, this has running time O(n2logn) and requires space O(n2). The question asks whether it is possible to achieve the same running time with lower space complexity. As such, I don't see how this answers the question that was asked. D.W. – D.W. ♦ 2021-11-24 17:39:24 +00:00 Commented Nov 24, 2021 at 17:39 exactly the algorithm described in the 3rd paragraph of the question with notable details such as keeping no more than two positions for each sum. There have been two characterisations of the storage requirement: 1) three arrays of size N(N+1)/2 2) 1-2 GB. While details may not improve asymptotic growth, they may or may not alleviate a problem in one/any use case. greybeard – greybeard 2021-11-24 18:36:40 +00:00 Commented Nov 24, 2021 at 18:36 1 I tried storing only the distinct sums, but it still doesn't save much space (we can only use 64 MB as temporary storage while the program is running, so I think that any solution that stores information about all the subarrays is out of the question). Anyhow, in a couple of days we will learn the solution, and I will let you guys know. NiXt – NiXt 2021-11-24 19:02:10 +00:00 Commented Nov 24, 2021 at 19:02 leftmost end position [and] rightmost start position would have done if the original question asked for any pair. Alas, it asked for minimum total length. greybeard – greybeard 2021-11-25 05:47:55 +00:00 Commented Nov 25, 2021 at 5:47 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Our professor gave us this solution, which has O(N2+K) time complexity, and O(N+K) space complexity (since for the biggest test cases K is about N1.5, it is as good as O(N2) for these upper bounds). We start by initializing an array Lengths of size K, every value set to N+1, and calculating a prefix-sum for the initial array. We also keep a variable min for the final answer (initially set to N+1). Then we iterate the initial array with an index y, starting from N−1 down to 0. For each iteration we do the following: We iterate the array with an index z starting from y+1 up to N. For each subarray, we calculate its sum, and if its length is less than Lengths[sum] (where sum is the one we just computed), we set Lengths[sum] to z−y, the length of this subarray. We iterate the array with an index x starting from 0 up to y. For each subarray, we calculate its sum, and if this subarray's length PLUS Lengths[K−sum] is less than min, then we update min with the new value. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 11, 2022 at 21:36 喜欢算法和数学 39.2k44 gold badges3535 silver badges9595 bronze badges answered Dec 3, 2021 at 8:49 NiXtNiXt 7377 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. The restriction to non-negative values simplifies solving the problem drastically: Once the sum has reached K, adding elements increases length, if not sum. This reduces the number of subarrays to know from O(n2) (start and end) to O(K2) (length and subtotal). If K is not greater than zero, you dont even need to inspect the array. Check if K is the value of an element of the array upfront: With 0<K, the combined length won't go below 1. With a largish K and hell bent going lowest/o(n2) space, how about the following? let array be the input array of n integral numbers no less than 0; best, left and combined for the best pair known the subtotal of the left subarray, its length and the combined length; length and end arrays of n candidate subarray lengths and ends, respectively for target from 0 to K−1 1.1. from index 0 up, for each subarray totaling target of record brevity, keep length and end 1.2. next target if none found 1.3. for each subarray right with sum K−target from the end of array if sum of its length and the shortest one kept "to the left of right" ∗ bests best, update best re-find the rightmost subarray with sum K−best and length combined−left O(n) storage, O(nK) time: O(n2logn) where K∈O(nlogn) Exchange the roles of subarray length and and target subtotal for O(K) storage, O(n2) time ∗ linear search from the end should do nicely, if the last index used is kept Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Nov 25, 2021 at 22:22 answered Nov 25, 2021 at 6:02 greybeardgreybeard 1,18622 gold badges1010 silver badges2424 bronze badges 5 I'd try to start from both ends: before meeting in the middle, there can be no overlap. greybeard – greybeard 2021-11-25 06:04:11 +00:00 Commented Nov 25, 2021 at 6:04 Generally, my experience with algorithms that depended on K (e.g. O(nK)) is that they are really slow, so I guess that K can get a few orders bigger than n. Unless I manage to put it in as a logK (some kind of binary search I suppose, but I haven't worked the details) it won't help. NiXt – NiXt 2021-11-25 07:26:40 +00:00 Commented Nov 25, 2021 at 7:26 You disclosed a lower limit for n/N - can you add an upper limit and limits for K, in the question? greybeard – greybeard 2021-11-25 07:30:05 +00:00 Commented Nov 25, 2021 at 7:30 1 I have derived that from a bunch of test cases that we got, but I'll check the biggest ones, and update the question. NiXt – NiXt 2021-11-25 07:32:17 +00:00 Commented Nov 25, 2021 at 7:32 @greybeard Could you elaborate a bit on the exchanging of roles of "subbaray length" & "target subtotal" ? If you mean that we are initially fixing the left subarray length to L, for 1≤L≤n−1, I don't see how we could "check" the candidate right subbarays in O(n), for each L, (i.e. O(n2) for the total algorithm). Could you explain that part? entechnic – entechnic 2021-11-29 15:44:21 +00:00 Commented Nov 29, 2021 at 15:44 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Slight modification to other answers that may help: Since one subarray must have a sum from 0 to K/2, and the other must have a sum from K/2 to K, only look at subarrays with a sum <= K. This may help a lot, especially if K is much smaller than the total size of all array elements. If K is large, but heuristically you expect only few solutions, pick a prime p and find solutions modulo p, picking p large enough so there are still few solutions, but small enough to reduce your memory requirements. Then check whether your solution modulo p is a solution of the original problem. Once you found a solution with length M, only consider subarrays with fewer than M elements. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 13, 2022 at 23:04 answered Nov 24, 2021 at 13:07 gnasher729gnasher729 32.5k3636 silver badges5656 bronze badges 0 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithms time-complexity space-complexity See similar questions with these tags. Featured on Meta stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 9 How to develop an O(N) algorithm solve the 2-sum problem? 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1241	https://math.stackexchange.com/questions/2927923/partitions-created-by-intersections-of-circles	combinatorics - Partitions created by intersections of Circles - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Partitions created by intersections of Circles [duplicate] Ask Question Asked 7 years ago Modified7 years ago Viewed 127 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. This question already has answers here: N circles in the plane (2 answers) Closed 2 years ago. N N circles in a plane all intersect each other such that every circle intersects every other circle at exactly 2 2 points. Find in terms of N N the minimum and maximum number of disjoint closed regions that can be formed. Hint: You may wish to check your answers visually for greater N N, e.g N=4 N=4. combinatorics geometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 23, 2018 at 18:10 sc_ 365 1 1 silver badge 15 15 bronze badges asked Sep 23, 2018 at 17:16 InquirerInquirer 875 2 2 gold badges 9 9 silver badges 22 22 bronze badges 0 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The answer will be: N(n)=n 2−n+2 N(n)=n 2−n+2 This can be proved as: When a new circle is introduced in a collection of n n circles, 2 n 2 n new regions will be created (To see this, check the number of parts in which the new circle will be cut by the existing circles). So, S(n+1)=S(n)+2 n S(n+1)=S(n)+2 n Solving this recurrence with the initial condition S(1)=2 S(1)=2 gives the required answer Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 23, 2018 at 17:35 sc_sc_ 365 1 1 silver badge 15 15 bronze badges 1 mathworld.wolfram.com/PlaneDivisionbyCircles.htmlsc_ –sc_ 2018-09-23 17:35:58 +00:00 Commented Sep 23, 2018 at 17:35 Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 5N circles in the plane Related 10Minimum number of circles with 3 neighbors 0Locus of centres of circles that touch two intersecting circles 9Venn diagram with rectangles: How many among (n k)(n k) regions created by intersections of exactly k k rectangles can be represented? 12Hard Combinatorical Geometric Problem on Intersecting Circles 1n 2−n+2 n 2−n+2 plane divisions by n n circles 1Construct n n black and red circles 3Number of disjoint regions = number of intersections + 1 2n n lines meet at a point in the area of intersection of m m circles. Do the number of nonoverlapping plane regions formed only depend on m m and n n? Hot Network Questions Fundamentally Speaking, is Western Mindfulness a Zazen or Insight Meditation Based Practice? 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1242	https://www.geogebra.org/m/BndFHyNy	Introduction to Slope – GeoGebra Google Classroom GeoGebra Classroom Sign in Search Google Classroom GeoGebra Classroom Outline Introduction to Slope Introduction to Slope Introduction to the Point-Slope Equation slope Slope Intercept Slider Constructing a slope triangle Slope of a Line Slope Between 2 Points Activity Writing an Equation of Line from Graph in Point-Slope Form Using Slope to Find New Points on a Line CME 4.05 Marcuccio Graphing point-slope form Practice using the Slope Formula. Introduction to Slope Author:Daniel Morgan ### Introduction to Slope ### Introduction to the Point-Slope Equation ### slope ### Slope Intercept Slider ### Constructing a slope triangle ### Slope of a Line ### Slope Between 2 Points Activity ### Writing an Equation of Line from Graph in Point-Slope Form ### Using Slope to Find New Points on a Line CME 4.05 Marcuccio ### Graphing point-slope form ### Practice using the Slope Formula. Next Introduction to Slope New Resources [z`]]]( 判斷錐體 Untitled רישום חופשי bewijs stelling van Pythagoras Discover Resources asdfasdf Turning Wheels Hole #2 Bearings Example Copia de suma de angulos internos poligono Discover Topics Integral Calculus Confidence Interval Definite Integral LCM and GCD Pythagoras or Pythagorean Theorem AboutPartnersHelp Center Terms of ServicePrivacyLicense Graphing CalculatorCalculator SuiteMath Resources Download our apps here: English / English (United States) © 2025 GeoGebra®
1243	https://www.sciencedirect.com/science/article/pii/S036054422301527X	Inverse methods: How feasible are spatially low-resolved capacity expansion modelling results when disaggregated at high spatial resolution? - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Highlights Abstract Keywords 1. Introduction 2. Data and methods 3. Results 4. Conclusions 5. Limitations of this study 6. Glossary CRediT authorship contribution statement Declaration of Competing Interest Acknowledgements Appendix. Data availability References Show full outline Cited by (11) Figures (8) Show 2 more figures Tables (5) Table 1 Table 2 Table 3 Table 4 Table 5 Energy Volume 281, 15 October 2023, 128133 Inverse methods: How feasible are spatially low-resolved capacity expansion modelling results when disaggregated at high spatial resolution?☆ Author links open overlay panel Martha Maria Frysztacki a, Veit Hagenmeyer a, Tom Brown b Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Highlights •Methodology to disaggregate spatially low-resolved capacity expansion model results. •Feasibility analysis of the 100% renewable, disaggregated, highly-resolved model. •One-node-per-country models of Europe are not fit for accurate investment decisions. •Results from 100–200 node models of Europe lead to more than 3%–7% of load-shedding. Abstract Spatially highly-resolved capacity expansion models are often simplified to a lower spatial resolution because they are computationally intensive. The simplification mixes sites with different renewable features while ignoring transmission lines that can cause congestion. As a consequence, the results may represent an infeasible system when the capacities are fed back at higher spatial detail. Thus far there has been no detailed investigation of how to disaggregate results and whether the spatially highly-resolved disaggregated model is feasible. This is challenging since there is no unique way to invert the clustering. This article is split into two parts to tackle these challenges. First, methods to disaggregate spatially low-resolved results are presented: (a) an uniform distribution of regional results across its original highly-resolved regions, (b) a re-optimisation for each region separately, (c) an approach that minimises the “excess electricity”. Second, the resulting highly-resolved models’ feasibility is investigated by running an operational dispatch. While re-optimising yields the best results, the third inverse method provides comparable results for less computational effort. Feasibility-wise, the study design strengthens that modelling countries by single regions is insufficient. State-of-the-art reduced models with 100–200 regions for Europe still yield 3%–7% of load-shedding, depending on model resolution and inverse method. Previous article in issue Next article in issue Keywords Electricity system optimisation Renewable energy Investment planning Spatial clustering Inverse methods Disaggregation methods 1. Introduction Capacity expansion models are computer programs for the dimensioning of energy generators, storage or the expansion of transmission lines, typically utilised to simulate ambitious climate change mitigation or carbon-dioxide reduction targets. Such models have gained recognition over the past years as their results are often taken as a reference to formulate energy transition road maps. They are used by non-governmental organisations, political institutions, transmission system operators and large companies. Latest trends in model utilisation are analysed by Lopion et al. , who outline that today’s models must accurately portray renewable potentials and thus capture the weather-driven variability of wind and solar photovoltaic generation in order to provide reliable investment recommendations for renewable generator, storage and transmission installations. Recent literature has identified that a high spatial resolution is required for the modelling to produce an accurate representation of renewable generation. Schlachtberger et al. found that modelling a large geographical area at the scale of Europe allows the model to exploit very good continental renewable potentials and strongly impacts the composition of the generation and storage fleet of individual regions. Frew and Jacobson demonstrated similar findings for the United States. For models at continental scope, Aryanpur et al. stress the incompleteness of single region optimisation models that do not capture sufficient spatial detail. They demonstrate that a high spatial granularity is particularly relevant for models that include heterogeneous regions (such as regions with a high share of renewable generation), or regions with higher variability in energy demands. In such modelling scenarios, results significantly change when the spatial resolution of the model is varied. Martínez-Gordón et al. explain the different modelling results by the ability of a spatially detailed model resolution to detect bottlenecks in the transmission grid and, therefore, to adequately assess renewable potentials based on local weather conditions and to identify regional variations in electricity demand. The granularity of this data-driven information significantly improves the design and composition of the electricity mix and the routing of new grid infrastructure predicted by the model. Similar results have been found by Miranda et al. , who show that a detailed transmission infrastructure significantly affects capacity deployments and electricity prices. On the resource side, Frew and Jacobson reveal significant differences in wind and solar development when modelling renewable generation sites independently for optimal site diversity, compared to assuming an aggregated buildout across all sites uniformly as is done in many grid integration studies. Frysztacki et al. disentangle the effects of sourcing renewable generation versus routing the electricity to locations of high demand using large transmission networks, revealing that routing dominates the system effects and forces the model to build renewable assets closer to demand centres at potentially worse capacity factors when the model has a high spatial resolution. These findings are not unique to the electricity sector; they can also be observed when modelling the heat decarbonisation . The downside of higher resolution is that processing large data volumes inevitably results in a computational burden that arises from solving the associated mathematical formulation in the model. In order for the spatially aggregated models to better represent the highly-resolved system, more effective clustering methods have been developed. Methods developed by Scaramuzzino et al. and Siala and Mahfouz show that aggregating based on political zones and borders is not suited to accurately portray the electricity system because there is per se no correlation between the distribution of solar radiation, wind speed, electrical load on the one hand and the administrative divisions on the other hand. Instead, they suggest to aggregate regions based on their similarities in renewable feed-in or demand-patterns, such as load density distribution and solar and wind potentials. Another approach by Biener and Garcia Rosas suggests to cluster regions with high electrical connectivity to minimise load flow deviations after the aggregation. A recent survey by Frysztacki et al. improves previous methods and suggests which of them is most suitable for which modelling scenario. The overall consensus of most studies is to model the European electricity system that contains more than 5000 electrical substations at and above 220 kV at a spatial model resolution of at least 100–200 regions, depending on the model configuration. Previous literature did not provide answers if the spatial resolution of the models impacts the optimality of the solution, or whether these spatially simplified modelling results are feasible with respect to the original, spatially highly-resolved model. Moreover, as of now, there exist no or only limited approaches in previous research to disaggregate the spatially simplified modelling results back at its original high resolution. There exist only few publications to the authors knowledge, such as from who disaggregate simplified modelling results at higher spatial detail. In their approach only the power dispatch of spatially low-resolved model was disaggregated to a spatially higher resolved model, investment variables were not considered. Beyond that, the study did not analyse the overall feasibility of the resulting disaggregated model at a higher spatial resolution and focuses only on a German model. Neither the assumptions on the disaggregation, or the resulting highly-resolved systems were analysed in detail with respect to their plausibility. Another disaggregation method is presented by Reinert et al. , who propose an iterative disaggregation process and allow transmission grid expansion in the final iteration, so that solving the highly-resolved model becomes feasible. It remains unclear how feasible the disaggregated model is when omitting transmission grid expansion in the final iteration. Finally, Grochowicz et al. propose three disaggregation methods. At highest resolution, they generate results for a model where every country is represented as a single region, which is already below the sufficient resolution suggested by prior research. Moreover, the proposed disaggregation follows a pre-defined iterative approach that guarantees that the resulting highly-resolved one-region-per-country model is feasible. Therefore, they do not provide an analysis on whether spatially low-resolved capacity planning results can represent a feasible solution to a spatially more complex system. This contribution focuses on the unresolved inverse problem of mapping spatially low-resolved optimisation variables at a higher resolution. The first comparison of different disaggregation methods in a European continent-scale model is presented and new algorithms for disaggregation options are provided in Section 2. Spatial decomposition methods such as optimality cut or feasibility cut were specifically not applied, because this approach would aim to solve the fully resolved model, and provide no insight if the spatially aggregated model results are feasible with respect to the higher dimensioned model formulation. First, the capacity expansion model used for this study is presented, followed by a detailed description of three new methods to disaggregate the resulting coarse model variables at high resolution in Section 2. In Section 3 a feasibility analysis is carried out where it is tested if the spatially low-resolved capacity planning results are capable to meet electricity demand at all places at all times when disaggregated at a spatially highly-resolved model. The feasibility of the resulting system is not intuitive, because the spatially low-resolved models omit transmission constraints that can result in congestion and, at the same time, smooth variable renewable resources by aggregating potentially heterogeneous resource sites. The novelty of this analysis is that none of the very limited disaggregation methods from the literature was previously tested with respect to model feasibility. Therefore, the error in investment decisions made by the low-resolved models compared to a spatially highly-resolved model is quantified for the first time. Conclusions are drawn in Section 4. Finally, the limitations of this study are discussed in Section 5. 2. Data and methods In this section the underlying data is presented. The overall modelling process in described in 2.1, where the novelty of this study is highlighted. A selection of the most important data and methods of the model employed for this study are presented in Section 2.2. The three proposed inverse methods on how spatially low-resolved capacity expansion model results can be disaggregated back at higher spatial detail are presented in Sections 2.3–2.5 and are summarised in Table 1. The treatment of inter-cluster powerflows is discussed in Section 2.6. The study design and evaluation of results are presented in Section 2.7. 2.1. Modelling overview Fig. 1 displays the overall approach of electricity system modelling. It typically executes in the following order: (i) Creating the model. This includes collecting data of the system to be analysed, for example the network topology of the transmission system, capacities of generators that are to be included in the model, land-use constraints, time-series of electricity demand, wind speeds, solar radiation, etc. and assigning the data to the correct locations. (ii) Clustering the spatially highly-resolved network down to a smaller approximation to gain computational advantages. (iii) Formulating a set of mathematical equations associated with the problem and solving it. 1. Download: Download high-res image (349KB) 2. Download: Download full-size image Fig. 1. Illustration of the typical model process: (i) Data collection and model creation, (ii) spatially clustering the network down to a smaller number of nodes to obtain a computationally tractable model, (iii) optimise the model, i.e. find the cost-optimal configuration of generators, storage units and power flows. Normally, the modelling task is completed at this stage. In this work, two additional steps are proposed: (iv) disaggregating the spatially reduced model results into higher spatial resolution and (v) an operational optimisation is conducted, testing the resulting network model for feasibility. Here, the additional fourth and fifth steps are introduced to this queue: (iv) Disaggregating the low-resolved results (i.e. the resulting renewable capacities G v,s) back at high spatial resolution. As the clustering f:R m→R n reduces the (spatial) dimension of the data (n<m, in many cases even n≪m), the mapping is surjective but not injective, hence not bijective. Therefore, finding an inverse that maps the results back at high dimension f−1:R n→R m is a challenging task and the inverse is not unique. Therefore three different approaches to tackle the disaggregation of the generation and storage capacities for each technology from the low-resolved capacity expansion model to the highly-resolved operational model are presented and adequate inverse methods are suggested. The proposed methods are summarised in Table 1 and are explained in detail in the following three sections. (v) Running an optimisation with fixed capacity that is derived from step (iii), applying the disaggregation from (iv). (v) is also referred to as operational optimisation. It allows to gain insights into the dynamics of the electricity system, particularly to analyse its feasibility. 2.2. Model and input data For this study, the openly available European Electricity System Model at transmission substation level, PyPSA-Eur is employed. It is used to model a future fully renewable electricity system that consists of today‘s transmission grid, and can build solar pv, onshore wind, offshore wind, hydrogen storage and batteries to cover European electricity demands. The model is described in detail in its original publication by Hörsch et al. , therefore only its major functionality are summarised here. The mathematical model formulation and solving is based on the python package PyPSA, originally developed by Brown et al. . This section focuses on features that are relevant for the specific use-case for this study. The full model covers 33 European countries with a full spatial resolution of 5323 nodes (3609 substations), 6640 transmission HVAC and HVDC lines at and above 220 kV. The transmission grid is assumed as it was installed by 2020, but includes planned over-sea HVDC lines that strengthen the connection between continental Europe and the British Isles. Electricity demand is embedded from the Open Power System Data (OPSD) project. Historical weather data to account for the variability of renewable resources is openly provided by the European Centre for Medium-Range Weather Forecasts (ECMWF) in the “ERA5 Reanalysis” dataset, and by Pfeifroth et al. in the second edition of the “Surface Radiation Data Set (SARAH-2)”. This raw data contains for example solar radiation, wind-speeds or temperature, and is processed using the open source software “atlite”, developed by Hofmann et al. , to translate it into capacity factors for the modelling. The main objective of the capacity expansion model is to minimise the annual system costs that consist of the sum of all investments in new capacity G v,s at each node v of every technology s, as well as the variable costs related to the dispatch of the generators g v,s,t at each time t. The weight w t relates to the duration of dispatch and is fixed to 2 in our application, i.e. w t≡2 to balance computational model fidelity against accuracy of the modelling results [see ]. Mathematically, this can be formulated as (1)min G v,s,g v,s,t,f(v,w),t[∑v∈V∑s∈S(c v,s G v,s+∑t∈T w t o v,s g v,s,t)], with additional constraints to guarantee network security, to cover demand at all times and places, to account for the system to be physically plausible including Kirchhoff’s circuit laws and upper and lower bounds for generator dispatch. Such model is called “capacity expansion model”, because the variables that are subject to the optimisation represent the dimensioning of renewable generators or storage units, G v,s. Furthermore, generator dispatch and energy storage behaviour as well as electricity power flows are also subject to the optimisation, constrained by the size of the respective unit. The optimisation is solved using an interior point method to find the minimum using the Gurobi Optimisation python interface. All variables appearing in this article are explained in the Glossary, Section 6, Table 4. Cost assumptions are based on suggestions by the Danish Energy Agency for wind technologies, Schröder et al. in case of open cycle gas turbines, pumped hydro storage, hydro, run-of-river, Budischak et al. for storage technologies and Vartiainen et al. for solar. For details, see A.1. To enable computational feasibility, the whole model with more than 5000 nodes must be reduced to a computationally tractable size by spatially clustering the nodes. Then, the reduced optimisation problem (1), now with a smaller set of nodes V, can be solved. There exist many approaches to spatially reduce the size of such model. Latest research from , , motivated by insights from , have demonstrated that the best suited clustering method for capacity expansion models with highly renewable scenarios are of hierarchical nature. Therefore, for this study, a hierarchical agglomerative clustering is chosen for the spatial scale. It is a bottom-up approach, where each node is treated as a singleton cluster. Then, in every iteration, two clusters that have the highest similarity and that are connected by a transmission line are aggregated. The same similarity measure as previously analysed is used. It is defined such that the aggregated nodes have the most similar renewable time-series ḡv,s,t throughout the whole year, i.e. (2)ḡv∈V,s∈{solar,wind},t∈T∈[0,1]2\|T\|. Note that such spatially aggregated models are not capable to account for all transmission lines that can be constraining in terms of the load-flow. For example, transmission lines within an aggregated region are ignored in the simulation. As a result, the optimal solution of a spatially clustered model can still be inaccurate compared to a highly-resolved solution. Eventually, this inaccuracy can lead to an infeasibility when coarse modelling results are implemented in a spatially more complex system. This inaccuracy is quantified in Section 3. For the remaining part of this section, the focus lies on the three proposed disaggregation methods including the treatment of inter-cluster transmission flows. In terms of temporal scale, every two consecutive hours are aggregated. Temporal aggregation, such as spatial aggregation, can lead to a different type of modelling error. Therefore a high temporal resolution is maintained. The approach to aggregate only two consecutive hours has shown to yield good results compared to hourly-resolution while reducing the model size by a factor of 2. A general overview of temporal aggregation and accompanying errors are discussed by Kotzur et al. or Jacobson et al. . Table 1. Summary of the proposed disaggregation methods. \| Short name \| Method description \| --- \| \| Optimal capacities retrieved from an optimised low-resolution model are distributed ... \| \| uniform \| ... uniformly across all nodes within a cluster of the highly-resoled network while accounting for land-use restrictions by imposing an upper bound. \| \| re-optimise \| ... anew by re-optimising capacities within each cluster with full formulation, while enforcing the same build-out capacity totals per technology as in the clustered model. \| \| min excess \| ... according to a local optimisation which seeks to concentrate generation at nodes with higher demand and grid capacity and thus to minimise load-shedding. \| 2.3. Uniform distribution The first approach to disaggregate spatially low-resolved modelling results is inspired by suggestions made by Müller et al. , where a similar disaggregation was applied on the power dispatch g c,s,t. Here, the focus lies on the investment variables G c,s and the disaggregated dispatch is determined only after running an operational problem (see Section 2.1). The method is simple and computationally inexpensive. For each generation and storage technology, the capacities retrieved from the spatially low-resolved model is distributed within every cluster uniformly across all highly-resolved nodes: (3)G c,s↦1\|V c\|G c,s...G c,s∈R\|V c\|.An additional constraint is formulated to account for land-use constraints. This upper limit is formally defined as (4)G v,s≤G v,s max∀v∈V c and is enforced by selecting the generators where (4) is not satisfied and uniformly distributing the residual capacity over the remaining nodes within the cluster, i.e. across the following set of generators: (5){G v,s:v∈V c∧G v,s<G v,s max}.The last step is repeated until all nodes satisfy constraint (4). 2.4. Regional re-optimisation The second considered method is inspired by suggestions made by Reinert et al. , who have proposed a similar, iterative disaggregation method with adaptations to the transmission grid in the final iteration. Further adaptations to the optimal solution or the original transmission grid are not allowed, and the problem is not iterative. However, it may be computationally challenging. For each cluster V c⊂V the re-optimisation is conducted at high resolution using the original objective function (1) with all associated mathematical constraints. Additionally, a set of constraints to incorporate the low-resolved modelling results is imposed: (6)∑v∈V c G v,s=G c,s∀s∈S.This set of constraints ensures that the amount of installed capacity is the same for every technology in every region. Depending on the size of the cluster, the disaggregation may blow up the problem beyond the computational capacity of the machine and can result in an even larger problem than the clustered one. On the positive side, the re-optimisations for each cluster can be run in parallel. 2.5. Minimal excess electricity Our third Ansatz for disaggregation is motivated by finding a compromise in terms of computational resources. It is not evident that solving the full optimisation problem is necessary to distribute renewable capacity. Instead, a simpler new objective function that minimises (renewable) excess electricity is defined. It is designed to spatially align renewable generation with demand and possible flexibility options inside the cluster: (7)min G v,s∑v∈V c,s∈S re,t∈T[ḡv,s,t G v,s−d v,t−0.7∑l(v,w)∈L:v=c∨w=c F(v,w)]+. The bracket [x]+≔max{0,x} yields the positive part of the sum. Compared to the regional re-optimisation introduced in Section 2.4, the set of additional constraints to the optimisation problem is much smaller. Only Eqs. (4), (6) are imposed. The choice of this objective function is motivated by prior research carried out by Lund et al. or Frysztacki and Brown where Eq. (7) was invoked as a measure to balance the spatial resolution of the model and accurate modelling results. Similarly to the previous methods, this disaggregation method can be run in parallel for each cluster. 2.6. Modelling power flows between clusters For two of the three proposed disaggregation methods (‘re-optimize’ and ‘min excess’), additional boundary constraints on the inter-cluster transmission lines are added to simulate electricity im- and exports. This can be done by extracting the optimal power flows of the low-resolved network f(c,d),t and distributing them proportional to the capacities of the inter-cluster highly-resolved transmission lines F(v,w) following (8)f(v,w),t=F(v,w)F(c,d)f(c,d),t∈[−F(v,w),F(v,w)]∀(v,w)∈E:v∈V c,w∈V d∧c≠d. The resulting fine power flow f(v,w),t is modelled as additional demand imposed on all nodes v∈V c and w∈V d that are connected by a transmission line (v,w)∈E, with c≠d, i.e. d v,t↦d v,t+f(v,w),t∀v∈V c,d w,t↦d w,t−f(v,w),t∀w∈V d(c≠d), where positive power flows represent electricity imports and negative ones electricity exports. These results do not deviate strongly from those where each region is treated as an island, meaning that no powerflows retrieved from the coarse model are considered in the disaggregation. However it is plausible to include them when inverting modelling results, therefore this article focuses on this approach. Islanded results can be found in the Appendix, see A.2. 2.7. Study design To investigate the quality of the proposed disaggregation methods, the model is solved as a pure operational problem, where no further capacity can be built. This is equivalent to solving Eq. (1) with its associated constraints, however the technology capacities G v,s are removed from the set of optimisation variables and replaced by a fixed number that is the result of the disaggregation. Load-shedding generators with high but non-extendable capacity are added to the network to guarantee physical feasibility. The operational problem is computationally less extensive to solve because the inter-temporal capacity expansion has been removed from the problem. Therefore, solving a spatially highly-resolved model becomes computationally feasible. The amount of load-shedding in the highly-resolved operational network model with the disaggregated investment variables taken from the low-resolved capacity expansion model, and the amount of renewable curtailment are considered as main quality measures. Curtailment describes how much abundant electricity the low-resolved model chooses to generate which, when highly-resolved, cannot be transported to locations with high electricity demand. This is mostly due to an inaccurate choice of siting capacity due to missing information about possible transmission bottlenecks in the low-resolved model. Load-shedding is chosen because it indicates how much capacity is underestimated by the low-resolved model due to averaging capacity factors and removing grid bottlenecks from the network. Load-shedding could stem from different reasons: (i) the disaggregation of solar and wind capacities to multiple sites with different capacity factors could result in a lower overall yield compared to the aggregated site, or (ii) the grid bottlenecks inside the clusters could cause congestion, such that power generated at locations with surplus of electricity cannot be transported to locations with high net load. As load-shedding is a greater risk in terms of energy security, a test is designed to better understand its origin. To rule out reason (ii), a second operational scenario is run (which is referred to as “copper-plate”), where the capacity of all transmission lines that have vanished in the low-resolved network due to aggregation are set to ∞, i.e. (9)F(v,w)→∞∀(v,w):v,w∈V c,∀c.This modification is only applied to solve the operational problem, not for the disaggregation of results. Note that all inter-cluster transmission capacity is still finite, meaning (10)F(v,w)≪∞∀(v,w):v∈V c,w∈V d,c≠d. A summary of the two considered scenarios (“regular” and “copperplate”) is provided in Table 2. Table 2. Investigated intra-cluster scenarios for each of the disaggregation methods. This means that additional constraints on the transmission lines within each cluster are formulated. \| Short name \| Scenario description \| --- \| \| The highly-resolved network with disaggregated capacities is solved as an operational problem where ... \| \| regular \| ... no further adaptations are made. \| \| copperplate \| ... the intra-cluster transmission capacity is infinitely high. Note, that the inter-cluster transmission capacity is still bound. \| 3. Results First, the feasibility of low-resolved modelling results is presented when disaggregated at high spatial resolution in Section 3.1 using the three proposed disaggregation methods. It is distinguished between the “regular” set-up where the intra-cluster transmission capacity is not changed (Section 3.1.1). An additional focus lies on where the curtailment and load-shedding measures are spatially located (Section 3.1.2). The “copperplate” set-up sets the transmission capacity within clusters of the spatially highly-resolved model to infinity to approximate the clustered copperplate optimisation model (Section 3.1.3). Then, computational Trade-Offs of the presented disaggregation methods are discussed in Section 3.2. Results on how the generation mix and build-out rates are affected as a function of the modelling resolution are discussed in , . 3.1. Feasibility considerations In this section the feasibility of the spatially highly-resolved operational model is discussed that consists of the disaggregated spatially low-resolved modelling investment variables. Here, “feasibility” means the notion that electricity demand can be covered by 100% renewable electricity and no additional generation of conventional plants is necessary for a stable operation of the grid. Finally, the model is modified to study possible reasons of necessary intervention measures to secure electricity supply at all times and places. 3.1.1. Regular intra-cluster transmission capacity The results presented here are derived from the “regular” set-up, meaning with no adjustment to the intra-cluster grid capacities in the high resolved optimisation model (see Table 2). The resulting amounts of load-shedding and curtailment are presented in Fig. 2. 1. Download: Download high-res image (210KB) 2. Download: Download full-size image Fig. 2. Amounts of annual load-shedding and curtailment after solving the highly-resolved operational problem with the three proposed disaggregation methods. The number of clusters on the x-axis refers to the spatial resolution of the model of which the investment variables are disaggregated at high resolution. Transmission capacity within every cluster is not adjusted. (“regular” setup in Table 2). For any of the three proposed disaggregation methods it can be seen that the amounts of both curtailment and load-shedding decrease as the resolution of the underlying low-resolved capacity expansion model increases. This can be explained by a better approximation of the spatially highly-resolved model. The higher the spatial resolution of the capacity expansion model, the better is its approximation of the original highly-resolved model. Therefore, at higher spatial resolution, less costly load-shedding measures are necessary when disaggregating investment variables. It can also be seen that load-shedding is caused by high curtailment rates, that are likely to be provoked by transmission congestion. However, there are substantial differences in the performance of the disaggregation methods. Curtailment rates of the different disaggregation methods deviate by 1–3% from one another on average, depending on the underlying capacity expansion model resolution. Distributing coarse investment variables across the spatially highly-resolved operational model using the ‘min excess’ approach yields the highest curtailment rates of 11%–22% of annual electricity demand, depending on the low-resolved capacity expansion model resolution and the disaggregation method. The lowest resolution has the highest curtailment. Uniformly distributing results performs similar to the ‘min excess’ method at a very low resolution of 37 nodes (one node per country), resulting in 22% of curtailed electricity. The ‘re-optimise’ method performs better in this regard, resulting only in 19% of curtailed electricity. But the curtailment rates decrease to approximately 14.5% (‘min excess’), 13% (‘uniform’) and 11% (‘re-optimised’) of annual electricity demand, as the capacity expansion model resolution increases. Re-optimising the local problem yields the lowest curtailment for every low-resolved model resolution, which is approximately 2–3% lower compared to the results of the ‘uniform’ approach. Regarding load-shedding, for a low-resolved capacity expansion network where every country is represented by a single node (37 clusters in Fig. 2), ‘re-optimise’ performs best as it results in the lowest load-shedding rates. Re-optimising yields approximately 265 TWh or 8.2% of the annual electricity demand that cannot be covered by renewable generation. If this gap were filled with gas to satisfy electricity demand, annual carbon emissions would rise from 0% to 3.2% of 1990s levels. Compensating the unmet demand when disaggregating results with the ‘min excess’ method yields approximately 400 TWh of load-shedding, resulting in 4.8% of carbon emissions (1.6% more compared to ‘re-optimised’) if gas is dispatched for the load-shedding measure. When uniformly disaggregating renewable capacity within the clusters, the operational problem returns 500 TWh of load-shedding measures. Compensating with gas would result in 6% of carbon emissions of 1990, 1.2% more compared to ‘min excess’. When increasing the capacity expansion model resolution, the amount of load-shedding decreases for all the disaggregation methods. It can be seen that at a model resolution of 67 or more nodes, the amount of load-shedding is in the same range for the methods ‘re-optimised’ and ‘min excess’ deviating by only 0.5% on average. When uniformly distributing the retrieved low-resolved, optimal capacities, load-shedding measures are higher than those of the competing disaggregation methods by initially 2.8% at a capacity expansion resolution of 37 nodes and linearly decreases as the resolution of the capacity expansion model increases. At around 187 nodes, the difference for all three methods is below 0.5% in terms of necessary load-shedding measures. At an underlying capacity expansion model resolution of 217 nodes, the amounts of load-shedding are all within the range 94–110 TWh, corresponding to 3–3.5% of annual electricity demand in Europe. 3.1.2. Localisation of load-shedding and curtailment In this section it is analysed where curtailment and load-shedding is spatially localised. Recall that the highly-resolved model yields load-shedding measures because of (i) disaggregating capacity factors results in a different overall yield of renewable electricity or (ii) grid bottlenecks that did not occur in the low-resolved model, as described in detail in Section 2.7. 1. Download: Download high-res image (869KB) 2. Download: Download full-size image Fig. 3. Spatial distribution of curtailment and load-shedding measures across Europe in the highly-resolved operational 1250 node model for the three proposed disaggregation methods after running the operational dispatch. Optimal capacity installations are taken from a model resolution of 97 nodes, and are disaggregated at 1250 regions. Fig. 3 displays the regions of curtailment and load-shedding spatially distributed after running the operational highly-resolved 1250 node model for all three disaggregation methods for a reference capacity expansion model resolution of 97 nodes. In all three cases it can be seen that the load-shedding is scattered in central European regions such as southern Poland, central and southern Germany, Switzerland and Austria and thus far from coastal areas and southern locations, such as e.g. northern Germany and France, Italy and Spain. At the same time, coastal and southern locations have high amounts of curtailment. Transmission lines connecting regions with high amounts of curtailment and regions with high load-shedding show high congestion rates. Thus, the results suggest that the low-resolved capacity expansion model favours investments in wind turbines at locations with good wind conditions at coastal areas, and in solar panels in the southern regions with good solar radiation, while it is blind to transmission bottlenecks that prohibit transporting the electricity to demand centres. 3.1.3. Infinite intra-cluster transmission capacity To verify why load-shedding measures are necessary as well as to better understand the high curtailment rates, a setting is considered where within each cluster the transmission capacity is set to infinity, following the description provided in the beginning of Section 2.7, see Eqs. (9)–(10). This means that in the highly-resolved model, only the capacity between clusters is limited. Results on load-shedding and curtailment are presented in Fig. 4. 1. Download: Download high-res image (194KB) 2. Download: Download full-size image Fig. 4. Amounts of annual load-shedding and curtailment after solving the highly-resolved operational problem with the three proposed disaggregation methods. The number of clusters on the x-axis refers to the spatial resolution of the model of which the investment variables are disaggregated at high resolution. Transmission capacity within every cluster is set to infinite to approximate the copperplate clustered network (“copperplate” set-up in Table 2). It can be seen that the amounts of renewable curtailment in the disaggregated operational models deviate by less than 5% from the curtailment rates of the capacity expansion model. They can mainly be explained by varying capacity factors. In the highly-resolved operational models, larger deviations of capacity factors within each clustered region become available compared to the spatially low-resolved capacity expansion model. As the resolution of the capacity expansion model increases, the amount of curtailment also tends to increase slightly. This can be explained by the fact that more total capacity is installed for a high capacity expansion model resolution. In terms of congestion, the necessary amount of load-shedding when uniformly disaggregating renewable capacity drops to 0 for a capacity-expansion model resolution of above 100 nodes. For a one-node-per-country model (37 nodes), there remains a relatively low amount of load-shedding of approximately 50 TWh, resembling about 1.5% of annual electricity demand. In case the ‘re-optimise’ disaggregation Ansatz is invoked, load-shedding decreases to 0% of annual electricity demand for every low-resolved capacity expansion model. ‘min excess’ yields less than 5 TWh (<0.5%) of load-shedding measures for any low-resolved capacity expansion model. At peak (97 nodes) this amount of gas would emit around 800 kg of CO 2 (0.05% of 1990s emissions). One can conclude that these results are consistent with the main cause of load-shedding being the transmission restrictions within the clusters. 3.2. Trade-offs of the disaggregation approaches There are four main qualities that can be considered when evaluating trade-offs of the different disaggregation methods. First, the quality of results: How well do the proposed methods solve the problem at hand? Second and third, the computational efforts can be considered. These mainly focus on the question: Are the proposed methods computationally legitimate for the considered problem? This consideration includes not only the memory requirements needed to solve the problem, but also the time it takes to solve. Fourth, depending on the results of the methods or the problem formulation, it might also be worth considering the efforts to implement a solver. For the proposed methods in this paper, a summary of these four qualities is provided in Table 3. The performance with respect to the quality of results of the proposed disaggregation methods was already discussed in Section 3.1. Now, the performance of the proposed methods is analysed from a computational point of view. Rating the efforts of implementation is a subjective task, therefore it is solely related to the fact that uniformly distributing a number across a set of nodes does not involve mathematical optimisation. Therefore applying an uniform distribution is rated “easier” than formulating a mathematical constraint to an existing optimisation problem as proposed in ‘re-optimise’, or a whole optimisation problem including both objective function and associated constraints, as proposed in ‘min-excess’. Table 3. Trade-Offs of the three proposed approaches to disaggregate results. Marked with a indicate a reasonable trade-off, entries with a indicate an inadequate compromise. Computational resources and solving times for disaggregating spatially low-resolved model results at high spatial resolution are presented in Fig. 5 for every proposed method. Resource-wise, re-optimising the local model consumes up to 13 times (1.7 times in average) the amount of resources compared to minimising a simpler objective in “min excess”, and up to 26 times (2.7 times in average) compared to uniformly distributing the capacity obtained from the low-resolved model (“uniform”). In absolute numbers, the method “re-optimise” consumes up to 22.2 GB RAM at peak, compared to 2.2 GB RAM for “min excess” and only 1.5 GB RAM in case of “uniform”. Today‘s average state-of-the-art personal computers are able to solve both the “uniform” and “min excess” problem formulations for any model resolutions, while solving the “re-optimise” approach needs more computational power and, therefore, requires a more advanced machine or even a high-computational cluster access. All local disaggregation runs were carried out in parallel. 1. Download: Download high-res image (187KB) 2. Download: Download full-size image Fig. 5. Memory resource requirements (left) and solving times (right) for executing the proposed disaggregation methods for individual regions. The number of clusters on the x-axis refers to the spatial resolution of the model of which the investment variables are disaggregated at high resolution. Possible memory consumption and solving times are marked by the corresponding colour shades, depending on the size of the cluster that is to be disaggregated. The memory resource requirement and solving time for the largest and smallest local problem can be taken from the edges of the marked area. Median requirements are displayed with a dotted line. The black dotted line denotes the requirements of solving the spatially low-resolved network that has the spatial resolution as given on the x-axis. Note that for ‘min excess’ and ‘uniform’, the median values coincide with the lower bounds, and are thus hidden in the plot. Moreover, for ‘uniform’, there upper and lower bounds collapse, thus no distribution is visible. Considering the computational times, these trade-offs are similar. The method “uniform” is up to 4000 times faster at peak than “min excess” and 70 times faster in average. In turn, “min excess” is up to 20 times faster than “re-optimise” and 11 times faster in average. Note that computational times might change when allowing a lower accuracy of the results. Here, a barrier convergence tolerance of 10−9 is chosen and a feasibility tolerance of 10−6, which is not necessarily required. A tolerance of 10−3 might suffice in most applications. However, lowering the tolerance of the solver reduces solving times, but the memory consumption persists. All experiments presented in this article were carried out on a high-computational cluster with 5 nodes, each having an allocatable capacity of 48 cpu’s and 256 GB memory. 4. Conclusions From these results, several conclusions on the methodology of the disaggregation methods can be drawn as well as on the insights of disaggregating coarse modelling results at a higher spatial detail. Table 4. Glossary. Variables and their description. \| Abbrev. \| Description \| --- \| \| V \| Set of all nodes contained in the model. \| \| E \| Set of all edges representing transmission lines contained in the model. \| \| v,w \| Representative names for highly-resolved nodes. \| \| c,d \| Representative names for clustered nodes. \| \| V c \| Set of highly-resolved nodes that are aggregated to node c. \| \| (v,w) \| a HVAC or HVDC line connecting nodes v and w. \| \| S \| Set of available technologies in the model, for example wind generator or battery storage. \| \| s \| Generator or storage technology. \| \| T \| Set of snapshots in the model. \| \| t \| snapshot, typically covering a duration of 2 hours. \| \| G v,s \| Capacity in node v of generators of type s. \| \| g v,s,t \| Dispatch in node v of technology type s at time t. \| \| ḡv,s,t \| Capacity factor at node v for technology s at time t. \| \| w t \| Weighting for time, here w t≡2∀t∈T, representing a 2-hourly model run. \| \| L \| Set of transmission lines in the model. \| \| F(v,w) \| Capacity of the transmission line connecting nodes v and w. \| \| f(v,w),t \| Power flow from node v to node w at time t. \| \| d v,t \| Electricity demand in node v at time t. \| \| c v,s \| Capital costs at node v of technology s. \| \| o v,s,t \| Operational costs at node v for technology s at time t. \| \| Υ V(t), Υ C(t) \| Curtailment of the highly-resolved model (V) or the clustered model (C) at time t. \| \| Δ V(t), Δ C(t) \| Load-shedding measure of the highly-resolved model (V) or the clustered model (C) at time t. \| The presented methods to disaggregate optimal infrastructure investment of renewable generation technologies and flexibility options have significant differences in their quality of results, simplicity of implementation and computational resource consumption. It has been shown that it is not necessary to locally solve the full optimisation problem to disaggregate coarse results at higher spatial detail, as it was conducted in previous research. Instead, it can be sufficient to formulate a suitable alternative objective which reduces computational cost and is able to preserve the quality of the disaggregation. In this paper, a novel function “min excess” has been suggested that performs just as well, for lower computational burden. Further inverse functions could be considered in future research. Regarding the disaggregated highly-resolved modelling results, the result presented in this article once again stress that modelling a fully renewable European electricity system at a resolution of one node per country is insufficient to retrieve reliable capacity expansion suggestions. Moreover, results retrieved from models that simulate a fully renewable electricity system that are clustered to a spatial resolution of around 100–200 nodes using state-of-the-art evaluated aggregation methods fail to cover approximately 100 TWh of Europe’s electricity demand, approximating 3–5% of its annual consumption. Instead of consuming the excess electricity, curtailment rates rise by approximately the shed amount, additional to what would have been expected for an economic optimum. Our analysis reveals that the electricity shortage is due to local transmission constraints. Spatially low-resolved models assume that power can be transferred without limit to all locations that are represented within a single region. Therefore, intra-nodal transmission constraints are ignored in the aggregated model. Thus, disaggregated results at higher spatial detail are confronted with power flow restrictions, resulting in transmission congestion and imply necessary load-shedding measures, eventually making the investment decisions retrieved from a coarse model sub-optimal and technically infeasible, if no additional investments can be assumed. These findings imply that accurately representing transmission and power-flows in the model is of high relevance to find a cost-optimal or low-cost solution that is technically feasible. Our results do not show that a fully renewable system is not possible. Conversely, the insight from our results together with lessons learned from spatial clustering studies provide valuable insights that emphasise the relevance of high resolution modelling, such that a fully renewable system can be achieved at low cost. 5. Limitations of this study Removing the set of optimisation variables that accounts for the capacity expansion allows solving an operational dispatch model at a higher model resolution. Nevertheless, due to a persisting computational burden, the presented operational model results are based on model runs retrieved from a model resolution of 1250 nodes, i.e. approximately 25% of the original network size. Therefore, the resulting amounts of load-shedding and curtailment resulting in the disaggregated operational model runs are likely to increase if the operational model was spatially higher resolved (for example at 5000 nodes), and, thus, strengthen our main argument. 1. Download: Download high-res image (205KB) 2. Download: Download full-size image Fig. 6. Results as displayed in Fig. 2 of an island model, meaning that no inter-cluster electricity imports or exports are considered for the disaggregation. 1. Download: Download high-res image (190KB) 2. Download: Download full-size image Fig. 7. Results as displayed in Fig. 7 of an island model, meaning that no inter-cluster electricity imports or exports are considered for the disaggregation. This study analysed methods to disaggregate spatially low-resolved optimal generation variables. However, the study did not investigate methods to disaggregate transmission capacity expansion modelling results, or how additional transfer capacity obtained from a transmission expansion problem could improve the overall results. Such an analysis could build on our presented methods and extend them on an additional optimisation variable. Moreover, all results presented in this paper were carried out for a fully self-sufficient and fully renewable Europe. Lowering the carbon emission target could relax the findings and would not make as strong implications. Therefore, in a future study, different carbon emission targets could be analysed more carefully. Results of this study are all based on the MIT-licensed models PyPSA v0.18.0 and PyPSA-EUR v0.3.0. Therefore, nearly all of the limitations that apply for this version of the model also apply for this study. These include for example retrieving optimal capacities that rely on weather data from a single weather year, applying only a linearised power flow model or neglecting dynamic line rating. Some of these simplifications might improve in future model releases. 6. Glossary Abbreviations and variables are documented in Table 4. CRediT authorship contribution statement Martha Maria Frysztacki: Conceptualization, Methodology, Software, Formal analysis, Data curation, Writing – original draft, Writing – review & editing, Visualization. Veit Hagenmeyer: Writing – review & editing, Funding acquisition. Tom Brown: Conceptualization, Writing – review & editing, Project administration, Funding acquisition. 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 MF and VH acknowledge funding from the Helmholtz Association under the program “Energy System Design”, MF and TB acknowledge funding from the Helmholtz Association, Germany under Grant No. VH-NG-1352. Appendix. A.1. Cost assumptions Cost assumptions used for the capacity expansion models to make cost-optimal investment decisions can be taken from Table 5. A.2. Results of island-ed disaggregation method In this setting each cluster is treated as an island, meaning that no electricity trade between other clusters is considered for the disaggregation. Results on load-shedding and curtailment for this scenario are displayed in Fig. 6. The overall trend of the results is similar to the simulations where inter-cluster power flows were considered in the simulations. However, there are minor differences mainly affecting the “re-optimise” results. These result in an overall higher curtailment of 2–3%, and lower load-shedding of 1–2%. A.3. Analysing the source for load-shedding Fig. 8 additionally displays if load-shedding measures occur at times where the curtailment of the highly-resolved operational model is higher compared to the lower resolved reference results. If true, this indicates that the load-shedding measures are due to underestimated within-cluster transmission bottlenecks. To precisely evaluate this statement, the Figure displays the following hypothesis: (11)δ{t:(Υ V−Υ C)(t)>0}⋅(Υ V−Υ C)(t)≥0.5⋅Δ V(t)(12)δ{t:(Υ V−Υ C)(t)<0}⋅(Υ C−Υ V)(t)≤0.5⋅Δ V(t), where Δ V(t) represents the amount of load-shedding measures in the highly-resolved disaggregated model at snapshot t, and Υ V(t)≔∑s∈S v∈V ḡv,s,t G v,s−g v,s,t the amount of curtailment in the highly- resolved disaggregated model at snapshot t. Accordingly, Υ C(t) represents the amount of curtailment in the lower resolved reference model. 1. Download: Download high-res image (466KB) 2. Download: Download full-size image Fig. 8. Evaluation if load-shedding measures of the disaggregated highly-resolved operational models occur at times with higher curtailment compared to the low-resolved capacity expansion model, see Eqs. (11)–(12). Table 5. Technology investment costs with 1$=0.7532€. \| Asset \| Cost \| Unit \| --- \| Onshore wind \| 1110 \| €/kW \| \| Offshore wind \| 1640 \| €/kW \| \| (AC/DC grid connection separate) \| \| \| \| Solar PV utility \| 425 \| €/kW \| \| Solar PV rooftop \| 725 \| €/kW \| \| Open cycle gas turbine \| 400 \| €/kW \| \| Run of river \| 3000 \| €/kW \| \| Pumped hydro storage \| 2000 \| €/kW \| \| Hydro storage \| 2000 \| €/kW \| \| Battery storage \| 192 \| $/kWh \| \| Battery power conversion \| 411 \| $/kW el \| \| Hydrogen storage \| 11.3 \| $/kWh \| \| Hydrogen power conversion \| 689 \| €/kW el \| \| HVAC overhead transmission \| 400 \| €/(MW km) \| \| HVAC underground transmission \| 1342 \| €/(MW km) \| \| HVAC subsea transmission \| 2685 \| €/(MW km) \| \| HVDC underground transmission \| 1000 \| €/(MW km) \| \| HVDC subsea transmission \| 2000 \| €/(MW km) \| It can be seen that, as the reference model resolution increases, there are more and more times t where the hypothesis is wrong. The amount of curtailed electricity is higher than load-shedding in 85% of the times on average for all of the three disaggregation methods for a very low-resolved reference model of 37 nodes. As the reference model resolution increases to 217 nodes, the statement is only true in average for 65% of the times for all three disaggregation methods. This indicates that transmission resolution is starting to saturate, however is still the major bottleneck preventing to feed-in the extra green electricity that is being curtailed. Recommended articles Data availability All data is open-source. References Lopion P., Markewitz P., Robinius M., Stolten D. A review of current challenges and trends in energy systems modeling Renew Sustain Energy Rev, 96 (2018), pp. 156-166, 10.1016/j.rser.2018.07.045 URL: View PDFView articleView in ScopusGoogle Scholar Schlachtberger D., Brown T., Schramm S., Greiner M. 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In: Proceedings of 17th international conference on the European energy market (EEM 2020). 2020, Google Scholar Cited by (11) The importance of spatial resolution in large-scale, long-term planning models 2025, Applied Energy Show abstract Capacity expansion models are important tools for examining the evolution of the electric power sector. Embedded in these tools are many modeling choices with consequential impacts on computational burden and associated analysis. In this study, the spatial resolution of the national-scale Regional Energy Deployment System (ReEDS) model is adjusted to understand the implications of higher-fidelity modeling on energy system projections and model solve times. The default ReEDS regions capture the contiguous United States in 134 balancing areas, while the regions in the higher resolution version are defined by more than 3000 U.S. counties. Using both resolutions, a case study is conducted in the United States for the Texas Interconnection (ERCOT) and the Western Interconnection (WI) to explore how differences in spatial resolution impact model projections and to inform appropriate applications of high spatial resolution in a large-scale capacity expansion model. In both interconnections, the higher spatial resolution model achieves a lower-cost solution, attributed to the more detailed representation of variable renewable resources and transmission. Shifts in land-based wind capacity between the balancing-area-level model and the county-level model are more prominent than the changes in solar, in part because of the heterogeneity of wind resource across the United States and the stronger dependency of wind on transmission. Furthermore, at higher spatial resolution there is a locational shift in the installed capacity toward regions characterized by resource profiles that are better aligned to contribute to resource adequacy. Beyond the nuances in the results, running the high-fidelity ReEDS model introduces a significant computational burden, with an order of magnitude increase in the number of model regions leading to at least an order of magnitude increase in the runtime. Spatial flexibility can offer users and developers the opportunity to perform high-fidelity analysis, however the benefits of high-resolution modeling must be weighed against the availability of the necessary data and the scope of the research question. ### A solar and wind clustering framework with downscaling and bias correction of reanalysis data using singular value decomposition 2025, Energy Show abstract Solar and wind energy have emerged as pivotal elements driving the transition towards sustainable power generation sources. To sustain this momentum, reanalysis data provide a critical foundation for spatial planning of renewable sources. However, effectively structuring large-scale, multi-decade reanalysis data into well-delineated clusters for resource planning remains a challenge. Additionally, the relatively coarse resolution of reanalysis data can limit the spatial precision needed for detailed resource planning. These challenges can be addressed by employing dimensionality reduction and clustering techniques alongside high-resolution datasets, thereby enhancing the characterization of renewable resources. This paper introduces a novel framework that applies singular value decomposition to reduce the dimensionality of multi-decade hourly datasets from ERA5, preserving 99% of the original variance. By identifying spatial clusters from these reduced-dimension datasets and subsequently applying statistical downscaling and bias-correction methods, high-resolution time series can be derived to perform simulations using a mesh of subgrids suitable for renewable generation. A case study using a three-decade dataset from Brazil indicates that 15 solar and 16 wind clusters capture fundamental hourly generation patterns, enabling the estimation of energy outputs for each cluster and the crafting of renewable portfolios that leverage hourly complementarities. This research fills a critical gap by linking long-term hourly patterns to spatial clusters with global applicability. It advances the methodological toolkit for handling reanalysis datasets and enables the use of more detailed spatial information in energy models, including capacity expansion and portfolio selection models. ### Applicability of a flow-based grid representation in spatially resolved generation expansion planning models 2025, Applied Energy Show abstract Generation expansion planning models (GEPMs) support decision-makers in developing the future energy system. Research indicates that the accuracy of a GEPM improves with increased spatial granularity, as these models are able to leverage geographic variations in weather conditions and electricity demand. The outcome of such spatially resolved models, however, can be significantly distorted when employing a simplified representation of the transmission grid. Then again, considering detailed power flows and individual transmission limits is often computationally intractable. We propose a novel approach to incorporate transmission constraints in spatially resolved GEPMs, inspired by the flow-based market coupling algorithm in the European Core Region. This approach is benchmarked against two commonly used grid representations in GEPMs, i.e. a nodal and net transfer capacity representation. The flow-based approach significantly enhances computational efficiency, while closely approximating the accuracy of a nodal GEPM. ### This is SpArta: Rigorous Optimization of Regionally Resolved Energy Systems by Spatial Aggregation and Decomposition 2024, Applied Energy Citation Excerpt : However, while aggregation is an effective tool to reduce problem size and computational effort, the solution resulting from the aggregated problem is usually infeasible for the full-scale problem. Frysztacki et al. systematically evaluate the feasibility of spatially aggregated models and underline that the results of aggregated models can be infeasible at higher spatial resolution. Reinert et al. , Syranidou et al. thus discuss the necessity to disaggregate the resulting aggregated energy system and propose heuristic methods for disaggregation. Show abstract Energy systems with high shares of renewable energy are characterized by local variability and grid limitations. The synthesis of such energy systems, therefore, requires models with high spatial resolution. However, high spatial resolution increases the computational effort. Here, we present the SpArta method for rigorous optimization of regionally resolved energy systems by Sp atial A gg r ega t ion a nd decomposition. SpArta significantly reduces computational effort while providing a near-optimal solution at the full spatial resolution of sector-coupled energy systems. SpArta first reduces problem size by spatially aggregating the energy system using clustering. The aggregated problem is then relaxed and restricted to obtain a lower and an upper bound. The spatial resolution is iteratively increased until the difference between upper and lower bound satisfies a predefined optimality gap. Finally, each cluster of the aggregated problem is redesigned at full resolution. For this purpose, SpArta decomposes the original synthesis problem into subproblems for each cluster. Combining the redesigned cluster solutions yields an optimal feasible solution of the full-scale problem within a predefined optimality gap. SpArta thus optimizes large-scale energy systems rigorously with significant reductions in computational effort. We apply SpArta to a case study of the sector-coupled German energy system, reducing the computational time by a factor of 7.5, compared to the optimization of the same problem at full spatial resolution. As SpArta shows a linear increase in computational time with problem size, SpArta enables computing larger problems allowing to resolve energy system designs with improved accuracy. ### An Integrated and Iterative Multiscale Modeling Framework for Robust Capacity Expansion Planning 2024, Current Sustainable Renewable Energy Reports ### Integration of Land Use Potential in Energy System Optimization Models at Regional Scale: The Pantelleria Island Case Study 2024, Sustainability Switzerland View all citing articles on Scopus ☆ This document is the result of the research project funded by the Helmholtz Association, Germany under the program “Energy System Design” and under grant no. VH-NG-1352. © 2023 The Authors. Published by Elsevier Ltd. 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1244	https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/pages/unit-ii-chemical-bonding-structure/lecture-8/	Browse Course Material Course Info Instructor Prof. Catherine Drennan Departments Chemistry As Taught In Fall 2014 Level Undergraduate Topics Science Chemistry Inorganic Chemistry Organic Chemistry Physical Chemistry Learning Resource Types theaters Lecture Videos notes Lecture Notes assignment_turned_in Problem Sets with Solutions grading Exams with Solutions theaters Other Video co_present Instructor Insights Download Course search GIVE NOW about ocw help & faqs contact us 5.111SC \| Fall 2014 \| Undergraduate Principles of Chemical Science Unit II: Chemical Bonding & Structure Lecture 8: The Periodic Table and Periodic Trends « Previous \| Next » Topics \| \| \| --- \| \| \| 1. Ionization Energy and Photoelectron Spectroscopy 2. Electron Affinity \| Lecture Video The periodic table is to chemistry like the laws of motion are to physics. In this lecture, we discover the secrets of the periodic table and meet the elements. View video page Download video Download transcript Lecture Notes Notes for Lecture 8 (PDF) Clicker Questions Lecture 8 Clicker Questions (PDF) Textbook Reading \| TOPICS \| 5th EDITION \| 4th EDITION \| --- \| Electronic Structure and the Periodic Table \| Section 1.14 \| Section 1.14 \| \| Atomic Radius \| Section 1.15 \| Section 1.15 \| \| Ionic Radius \| Section 1.16 \| Section 1.16 \| \| Ionization Energy \| Section 1.17 \| Section 1.17 \| \| Electron Affinity \| Section 1.18 \| Section 1.18 \| \| Diagonal Relationships \| Section 1.20 \| Section 1.20 \| Problems and Solutions Problems for Lecture 8 (PDF) Solutions for Lecture 8 (PDF) Acknowledgement Excerpt of “Meet the Elements” by They Might be Giants, courtesy of Idlewild / Disney Sound, used with permission. From Here Comes Science, 2009. « Previous \| Next » Course Info Instructor Prof. Catherine Drennan Departments Chemistry As Taught In Fall 2014 Level Undergraduate Topics Science Chemistry Inorganic Chemistry Organic Chemistry Physical Chemistry Learning Resource Types theaters Lecture Videos notes Lecture Notes assignment_turned_in Problem Sets with Solutions grading Exams with Solutions theaters Other Video co_present Instructor Insights Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
1245	https://www.sciencedirect.com/topics/engineering/dispersion-force	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Adhesives and Sealants 1.3.2.3 Dispersion Forces The weakest van der Waals forces occur between non-polar entities, and are termed dispersion or London forces. They arise because such species have temporary fluctuating dipoles such that at any instant the centre of gravity of the electrons does not coincide with the nucleus. Dispersion forces are responsible for many phenomena including the liquefaction of argon (boiling point, − 186°C; enthalpy of vaporisation, 6.43 kJ mol− 1). The approximate potential energy of a pair of such molecules, separated by distance r, is given by Eq. (8), where α1 and α2 are their polarisibilities and I1, and I2 are their ionisation potentials. (8) The results of some calculations using Eqs (5), (7) and (8) are shown in Table 4. Table 4. Potential energies for molecules in contact \| Molecules \| Type of interaction \| r0 (nm) \| E (J mol− 1) \| --- --- \| \| H2O–H2O \| polar–polar \| 0.38 \| − 1120 \| \| H2O–CH4 \| polar–nonpolar \| 0.43 \| −85 \| \| CH4–CH4 \| nonpolar–nonpolar \| 0.30 \| − 7800 \| The calculation for the attraction between two methane molecules used I = 1133 kJ mol− 1 and α = 2.593 × 10− 30 m3, and the result is quite close to the enthalpy of evaporation for methane which is 8190 J mol− 1. Data were taken from Ref. . The potential energies for the molecules in contact (i.e. r = r0) are shown. Because these are all proportional to r−6, potential energies will fall to 1/64th of these values if the molecules are separated by a further intermolecular diameter (i.e. r = 2r0). This demonstrates that these forces of adhesion are very short range and are experienced only by one or two layers of molecules in the interfacial layers. View chapterExplore book Read full chapter URL: Chapter Working mechanisms of water reducers and superplasticizers 11.2 Dispersion forces Dispersion forces, also called van der Waals forces, ‘arise because local fluctuations in the polarization within one particle induce, via the propagation of electromagnetic waves, a correlated response in the other’ (Russel et al., 1992). For two particles made of the same isotropic matter, this interaction is always attractive. The range of interparticle dispersion forces is much larger than that of the individual dipoles. This is because in condensed media, the fluctuations of dipoles are correlated. The first description of the dispersion forces was given by the microscopic theory, which is based on pairwise summation of interacting dipoles. The attractive potential due to dispersion forces between two spherical particles of radius a1 and a2 separated by a distance h was approximated by Hamaker (1937) by: (11.1) where A is the Hamaker constant and H a geometric factor. The Hamaker constant depends on the optical properties of particles and the intervening medium, whereas the geometric factor depends on the particle size and the separation distance, and can be written as: (11.2) The dispersion force is equal to the derivative with respect to separation distance of the interaction potential: (11.3) If the radii of the spheres are substantially larger than the separation distance (a1, a2 > h), the expression for the van der Waals force can be simplified to: (11.4) where is the harmonic average radius: (11.5) In the microscopic theory, the Hamaker constant A is considered independent of the separation distance. However, beyond a certain distance (>5–10 nm) the dipoles of the particles may no longer be correlated, so the value of the Hamaker constant decreases with increasing separation distance because of this effect, known as retardation. Retardation of the Hamaker constant is taken into account in the continuum theory formulated by Lifshitz (1956). In cement suspensions the maximum attractive force, which is relevant for yield stress calculations, occurs at a very small separation. Even though this separation can be increased by the presence of adsorbed polymers, the values remain small. Hence for the size range of interest to cement suspension, the microscopic theory can be used to estimate the magnitude of the dispersion forces. The Hamaker constant depends on the electrolyte concentration. The intervening electrolyte can reduce the attractive force by interacting with the fluctuating electromagnetic field. This screening effect can be taken into account by calculating a screened (retarded or non-retarded) Hamaker constant. More details on the dispersion forces in cement suspension can be found in Flatt (2004). View chapterExplore book Read full chapter URL: Book2016, Science and Technology of Concrete AdmixturesG. Gelardi, R.J. Flatt Review article Two-phase bubbly flow simulation using CFD method: A review of models for interfacial forces 2020, Progress in Nuclear EnergyIrfan Khan, ... Suizheng Qiu 2.3 Turbulent dispersion force The turbulent dispersion force is also a hydrodynamic force that consists of the combined effect of continuous phase turbulent eddies on the dispersed phase and has an important role in the radial distribution of bubbles volume fraction profile (Moraga et al., 2003). This force is responsible for the dispersion of bubbles from the high concentration region to the low concentration region and smooths the local void fraction peak (Laviéville et al., 2017). In the condition of having a negative lift force coefficient, turbulent dispersion force is responsible for the transfer of bubbles from the neighborhood of the wall to the center of the flow channel. Generally, the turbulent dispersion force is given by (Lubchenko et al., 2018) (8) The (Bertodano and Martin, 1991) model was one of the first empirical models for turbulent dispersion force (TDF). They proposed the TDF as a gradient of dispersed phase (bubbles) volume fraction. Different researchers used different values of form constant to the variable value as shown in Table 3. Up to a few millimeters bubble diameter , could have the value of to . Table 3. Popular interfacial turbulent dispersion force models. \| Model \| Equation \| Parametric Dependence \| --- \| Bertodano and Martin( 1991) \| Where: for bubbly flow for few millimeters \| \| \| Lahey et al. (1993) \| \| \| \| Drew (2001) \| Where: \| \| \| Lucas et al. (2001) \| \| \| \| Burns et al. (2004) \| = turbulent viscosity, = turbulent Prandtl number. \| \| \| Hosoi and Yoshida (2010) \| , , , \| \| \| Tabib and Schwarz (2011) \| = sub-grid-scale turbulent kinetic energy per unit volume. = 0.1 \| \| \| Laviéville et al. (2017) \| \| , \| (Lahey et al., 1993) performed experimental and CFD analysis for the lateral phase distribution of bubbly flow. For CFD studies they used the same expression for turbulent dispersion force as (Bertodano and Martin, 1991) but with a constant value of 0.1. (Drew, 2001) suggested a model for turbulent dispersion force based on two-fluid models by studying the dispersed two-phase flow. His model incorporated the liquid turbulence time scale and dispersed unit relaxation time scale. (Lucas et al., 2001) studied the radial gas volume fraction profile in the pipe and emphasized the addition of an extra turbulence dispersion force based on the turbulence due to bubble deformation. According to the authors, deformed bubbles have more turbulence, therefore would be responsible for more smoothing of radial gas profiles. The proposed model is based on the Eötvös number . (Burns et al., 2004) performed mathematically study and proposed the expression for turbulent dispersion force by considering the double-time averaging of drag term in the momentum transport equation of Eulerian multiphase flow. This is called Favre-Averaged Drag (FAD) turbulent dispersion model. The general applicability of FAD turbulent dispersion model is within all Reynolds-Averaged turbulence models. Furthermore, when the suitable value of is not readily available, the FAD turbulent dispersion model is preferable. (Hosoi and Yoshida, 2010) developed the turbulent dispersion force model applicable to bubbly flow regime based on the kinetic theory of molecules. The time effect of bubbles Brownian motion was introduced into the formulation. (Tabib and Schwarz, 2011) established the sub-grid-scale turbulent dispersion force model by incorporating the sub-grid-scale turbulent kinetic energy in (Bertodano and Martin, 1991) model. (Laviéville et al., 2017) derived mathematically the generalized turbulent dispersion (GTD) force mode which is more applicable to pressurized water reactor (PWR) flow conditions as compared to standard turbulent dispersion models. They also incorporated the statistical mean of drag coefficient, , and virtual mass coefficient, in their GTD force model. The selection of turbulence dispersion force model requires the prescription of suitable turbulence model and near-wall treatment because it is the function of turbulent viscosity and turbulent intensity. Moreover, turbulence in the continuum also influences the redistribution of small bubbles from regions with low turbulence to regions with high turbulence. The model purposed by (Bertodano and Martin, 1991) incorporated the effect of continuum turbulence in the model with the local turbulent intensity in the flow. Similarly, (Burns et al., 2004) model included the effect of liquid turbulence on the bubble through the turbulent viscosity and turbulent Prandtl number for the void fraction. Although, (Laviéville et al., 2017) derived the generalized turbulent dispersion (GTD) model on the major assumption of the weak influence of liquid turbulence but suggested further investigation on this issue. View article Read full article URL: Journal2020, Progress in Nuclear EnergyIrfan Khan, ... Suizheng Qiu Review article Understanding the relationships between rheology and chemistry of asphalt binders: A review 2022, Construction and Building MaterialsYajian Wang, ... Linbing Wang 5.1.3 The London dispersion force The LD force is the attractive force caused by a temporary polarity. When the electron position of one atom fluctuates, that atom will form a temporary dipole and produce a temporary attraction on the electron layers of the surrounding atoms, inducing more electron fluctuations and thus creating an overall attraction. Although the LD is the weakest van der Waals force, it usually dominates the interactions between molecules due to its presence between all molecules, both polar and non-polar . The magnitude of the LD depends on the size and shape of the molecule. On the one hand, large molecules imply that more electrons are sloshing around and are more likely to be polarized, so the dispersion grows with increasing molecular weight, and on the other hand, molecules with long, thin shapes possess more contact surfaces, thus create stronger dispersion compared to spherical molecules . View article Read full article URL: Journal2022, Construction and Building MaterialsYajian Wang, ... Linbing Wang Chapter Interface Science and Composites 2011, Interface Science and TechnologySoo-Jin Park, Min-Kang Seo 4.3.1 London Dispersion Force The London dispersion force (also known as dispersion forces, London forces, induced dipole–induced dipole forces) is a type of force, acting between atoms and molecules . They are part of the van der Waals forces. The London dispersion force is named after the German-American physicist Fritz London. The London dispersion force is a weak intermolecular force, arising from the quantum induced instantaneous polarization multipoles in molecules. It is can therefore, act between molecules without permanent multipole moments. London forces are confirmed by non-polar molecules, due to the correlated movements of the electrons in interacting molecules. Electron density in a molecule becomes redistributed in the proximity of another molecule; because the electrons from different molecules start feeling and avoiding each other (see quantum mechanical theory of dispersion forces). This is frequently mentioned as the formation of instantaneous dipoles that attract each other. London forces are present between all chemical functional groups, and usually represent the main part of the total interaction force in condensed matter, even though they are largely weaker, than ionic bonds and covalent bonds. This is the only attractive intermolecular force exhibited between neutral atoms (e.g., a noble gas). Without London forces, there could be no attractive force between noble gas atoms, and they could not exist in liquid form. London forces become stronger as the atom or molecule in question becomes larger . This is due to the increased deformation polarizability of molecules with larger more dispersed electron clouds. This trend is exemplified by the halogens (from smallest to largest: F2, Cl2, Br2, I2). Fluorine and chlorine are gases at 25 °C, bromine is a liquid, and iodine is a solid. London forces also become stronger with larger amounts of surface contact. Greater surface area means closer interaction between different molecules. (See Chapter 1.2.1.1.) 4.3.1.1 Quantum mechanical theory of dispersion force The first explanation of the attraction between noble gas atoms was given by Fritz London in 1930 [15, 16]. He used a quantum mechanical theory based on a second-order perturbation theory. The perturbation is the Coulomb interaction (V) between the electrons and nuclei of the two monomers (atoms or molecules) that constitute the dimer. The second-order perturbation expression of the interaction energy contains a sum over states. The states appearing in this sum are simple products of the excited electronic states of the monomers. Thus, no intermolecular antisymmetrization of the electronic states is included and the Pauli exclusion principle is only partially satisfied. London developed the perturbation V in a Taylor series in 1/R, where R is the distance between the nuclear centers of a mass of monomers. This Taylor expansion is known as the multipole expansion of V, because the terms in this series can be regarded as energies of two interacting multipoles, one on each monomer. Substitution of the multipole-expanded form of V into the second-order energy yields an expression that resembles, somewhat, an expression describing the interaction between instantaneous multipoles (see the qualitative description above). Additionally an approximation, named after Albrecht Unsöld, must be introduced in order to obtain a description of London dispersion in terms of deformation polarizability. In this manner, the following approximation is obtained for the dispersion interaction between two atoms A and B. Here αA and αB are the deformation polarizabilities, of the respective atoms. The quantities IA and IB are the first ionization potentials of the atoms, and R is the intermolecular distance. (4-1) Note that this final London equation does not contain instantaneous dipoles (see molecular dipoles). The explanation of the dispersion force as the interaction between two such dipoles was invented after London gave the proper quantum mechanical theory. See the authoritative work , for a criticism of the instantaneous dipole model, and for a modern and thorough exposition of the theory of intermolecular forces. The London theory has much similarity to the quantum mechanical theory of light dispersion, which is why London coined the phrase ‘dispersion effect’. View chapterExplore book Read full chapter URL: Book series2011, Interface Science and TechnologySoo-Jin Park, Min-Kang Seo Chapter Van der Waals Forces 2011, Intermolecular and Surface Forces (Third Edition)Jacob N. Israelachvili 6.1 Origin of the Van der Waals-dispersion Force between Neutral Molecules: the London Equation The various types of physical forces described so far are fairly easy to understand, since they arise from straightforward electrostatic interactions involving charged or dipolar molecules. But there is a another type of force that like the gravitational force—acts between all atoms and molecules, even totally neutral ones such as helium, carbon dioxide, and hydrocarbons. These forces have been variously known as dispersion forces, London forces, charge-fluctuation forces, electrodynamic forces, and induced-dipole-induced-dipole forces. We shall refer to them as dispersion forces, since it is by this name that they are most widely known. The origin of this name has to do with their relation to the dispersion of light in the visible and UV regions of the spectrum, as we shall see. The literature on this subject is quite voluminous, and the reader is referred to books and reviews by1 London (1937), Hirschfelder et al., (1954), Moelwyn-Hughes (1961), Margenau and Kestner (1971), Israelachvili (1974), Mahanty and Ninham (1976), and Parsegian (2006). Dispersion forces make up the third and perhaps most important contribution to the total van der Waals force between atoms and molecules, and because they are always present (in contrast to the other types of forces that may or may not be present, depending on the properties of the molecules), they play a role in a host of important phenomena such as adhesion; surface tension; physical adsorption; wetting; the properties of gases, liquids, and thin films; the strengths of solids; the flocculation of particles in liquids; and the structures of condensed macromolecules such as proteins and polymers. Their main features may be summarized as follows: 1. : They are long-range forces and, depending on the situation, can be effective from large distances (greater than 10 nm) down to interatomic spacings (about 0.2 nm). 2. : These forces may be repulsive or attractive, and in general the dispersion force between two molecules or large particles does not follow a simple power law. 3. : Dispersion forces not only bring molecules together but also tend to mutually align or orient them, though this orienting effect is usually weaker than with dipolar interactions. 4. : Dispersion forces are not additive; that is the force between two bodies is affected by the presence of other bodies nearby. This is called the nonadditivity of an interaction. Dispersion forces are quantum mechanical in origin and amenable to a host of theoretical treatments of varying complexity, the most rigorous of which would take us into the world of quantum electrodynamics. Their origin may be understood intuitively as follows: For a nonpolar atom such as helium, the time average of its dipole moment is zero, but at any instant there exists a finite dipole moment given by the instantaneous positions of the electrons about the nuclear protons. This instantaneous dipole generates an electric field that polarizes any nearby neutral atom, inducing a dipole moment in it. The resulting interaction between the two dipoles gives rise to an instantaneous attractive force between the two atoms, and the time average of this force is finite. For a simple semiquantitative understanding of how these forces arise, we may consider the following model based on the interaction between two Bohr atoms. In the Bohr atom an electron is pictured as orbiting around a proton. The smallest distance between the electron and proton is known as the first Bohr radius a0 and is the radius at which the Coulomb energy e2/4πɛ0a0 is equal to 2hν—that is, (6.1) where h is the Planck constant and ν the orbiting frequency of the electron.2 For a Bohr atom, ν = 3.3 × 1015 s−1, so that hν = 2.2 × 10−18 J. This is the energy of an electron in the first Bohr radius and is equal to the energy needed to ionize the atom—the first ionization potential, I. The Bohr atom has no permanent dipole moment. However, at any instant there exists an instantaneous dipole of moment, u = a0e, whose field will polarize a nearby neutral atom, giving rise to an attractive interaction that is entirely analogous to the dipole-induced dipole (Debye) interaction discussed in Chapter 5. The energy of this interaction in a vacuum will therefore be given by Eq. (5.23) as where α0 is the electronic polarizability of the second Bohr atom, which from Eq. (5.5) is approximately 4πɛ0a03. Using this expression for α0 and Eq. (6.1) for a0, we immediately find that the preceding interaction energy can be written approximately as (6.2) Except for a numerical factor, Eq. (6.2) is the same as that derived by London in 1930 using quantum mechanical perturbation theory. London’s famous expression for the dispersion interaction energy between two identical atoms or molecules is (London, 1937) (6.3) and for two dissimilar atoms,3 (6.4) London’s equation has since been superseded by more exact, though more complicated, expressions (see Section 6.6), but it can be relied upon to give fairly accurate values for interactions in a vacuum although these are usually lower than more rigorously determined ones. From the preceding simple model we see that while dispersion forces are quantum mechanical (in determining the instantaneous, but fluctuating, dipole moments of neutral atoms), the ensuing interaction is still essentially electrostatic—a sort of quantum mechanical polarization force. And we may further note that the 1/r6 distance dependence is the same as that of the two other polarization interactions, the Keesom and Debye forces discussed in Section 5.6, that together contribute to the net van der Waals force. But before we consider these three interactions collectively, let us first investigate the nature of dispersion forces. View chapterExplore book Read full chapter URL: Book2011, Intermolecular and Surface Forces (Third Edition)Jacob N. Israelachvili Chapter FUNDAMENTAL PRINCIPLES GOVERNING SOLVENTS USE 2014, Handbook of Solvents (Second Edition), Volume 1ESTANISLAO SILLA, ... GEORGE WYPYCH 2.1.2.3 Dispersion Even when solvent and solute are constituted of nonpolar molecules, there is interaction between them. It was F. London who was first to face up to this problem, for which reason these forces are known as London's forces, but also as dispersion forces, charge-fluctuation forces, or electrodynamic forces. Their origin is as follows: when a substance is nonpolar it indicates that the distribution of charges of its molecules is symmetrical throughout a wide average time span. But, without doubt, in an interval of time sufficiently restricted, the molecular movements generate displacements of their charges which break that symmetry giving birth to instantaneous dipoles. Since the orientation of the dipolar moment vector is varying constantly due to the molecular movement, the average dipolar moment is zero, which does not prevent the existence of these interactions between momentary dipoles. Starting with two instantaneous dipoles, they will be oriented to reach a disposition which will favor them energetically. The energy of this dispersion interaction can be given, to the first approximation, by: [2.1.6] where \| \| \| --- \| \| Ii, Ij \| ionization potentials \| \| αi, αj \| polarizabilities \| \| r \| intermolecular distance \| From equation [2.1.6] it becomes evident that dispersion is an interaction, which is more noticeable the greater the volume of molecules involved. The dispersion forces are often more intense than the electrostatic forces and, in any case, are universal for all the atoms and molecules, given that they are not seen to be subjected to the requirement that permanent dipoles should exist beforehand. These forces are responsible for the aggregation of substances which possess neither free charges nor permanent dipoles, and are also the protagonists of phenomena such as surface tension, adhesion, flocculation, physical adsorption, etc. Although the origin of the dispersion forces may be understood intuitively, it has a quantum mechanical nature. View chapterExplore book Read full chapter URL: Book2014, Handbook of Solvents (Second Edition), Volume 1ESTANISLAO SILLA, ... GEORGE WYPYCH Chapter Interface Science and Composites 2011, Interface Science and TechnologySoo-Jin Park, Min-Kang Seo 4.3 London Dispersion and Acid–Base Interaction 4.3.1 London Dispersion Force The London dispersion force (also known as dispersion forces, London forces, induced dipole–induced dipole forces) is a type of force, acting between atoms and molecules . They are part of the van der Waals forces. The London dispersion force is named after the German-American physicist Fritz London. The London dispersion force is a weak intermolecular force, arising from the quantum induced instantaneous polarization multipoles in molecules. It is can therefore, act between molecules without permanent multipole moments. London forces are confirmed by non-polar molecules, due to the correlated movements of the electrons in interacting molecules. Electron density in a molecule becomes redistributed in the proximity of another molecule; because the electrons from different molecules start feeling and avoiding each other (see quantum mechanical theory of dispersion forces). This is frequently mentioned as the formation of instantaneous dipoles that attract each other. London forces are present between all chemical functional groups, and usually represent the main part of the total interaction force in condensed matter, even though they are largely weaker, than ionic bonds and covalent bonds. This is the only attractive intermolecular force exhibited between neutral atoms (e.g., a noble gas). Without London forces, there could be no attractive force between noble gas atoms, and they could not exist in liquid form. London forces become stronger as the atom or molecule in question becomes larger . This is due to the increased deformation polarizability of molecules with larger more dispersed electron clouds. This trend is exemplified by the halogens (from smallest to largest: F2, Cl2, Br2, I2). Fluorine and chlorine are gases at 25 °C, bromine is a liquid, and iodine is a solid. London forces also become stronger with larger amounts of surface contact. Greater surface area means closer interaction between different molecules. (See Chapter 1.2.1.1.) 4.3.1.1 Quantum mechanical theory of dispersion force The first explanation of the attraction between noble gas atoms was given by Fritz London in 1930 [15, 16]. He used a quantum mechanical theory based on a second-order perturbation theory. The perturbation is the Coulomb interaction (V) between the electrons and nuclei of the two monomers (atoms or molecules) that constitute the dimer. The second-order perturbation expression of the interaction energy contains a sum over states. The states appearing in this sum are simple products of the excited electronic states of the monomers. Thus, no intermolecular antisymmetrization of the electronic states is included and the Pauli exclusion principle is only partially satisfied. London developed the perturbation V in a Taylor series in 1/R, where R is the distance between the nuclear centers of a mass of monomers. This Taylor expansion is known as the multipole expansion of V, because the terms in this series can be regarded as energies of two interacting multipoles, one on each monomer. Substitution of the multipole-expanded form of V into the second-order energy yields an expression that resembles, somewhat, an expression describing the interaction between instantaneous multipoles (see the qualitative description above). Additionally an approximation, named after Albrecht Unsöld, must be introduced in order to obtain a description of London dispersion in terms of deformation polarizability. In this manner, the following approximation is obtained for the dispersion interaction between two atoms A and B. Here αA and αB are the deformation polarizabilities, of the respective atoms. The quantities IA and IB are the first ionization potentials of the atoms, and R is the intermolecular distance. (4-1) Note that this final London equation does not contain instantaneous dipoles (see molecular dipoles). The explanation of the dispersion force as the interaction between two such dipoles was invented after London gave the proper quantum mechanical theory. See the authoritative work , for a criticism of the instantaneous dipole model, and for a modern and thorough exposition of the theory of intermolecular forces. The London theory has much similarity to the quantum mechanical theory of light dispersion, which is why London coined the phrase ‘dispersion effect’. 4.3.2 Acid–Base Interactions 4.3.2.1 Introduction When other interactions are derived from the strength of an interface, one should precisely indicate their fundamental properties. Forming of interactions can take place and lead to the formation of acid–base bonds or to covalent bonding. We will consider here the importance of acid–base interactions in the improvement of adhesion. From the idiom of molecular orbital theory, acid–base definitions are as follows: A base is a species that employs an empty orbital in initiating a reaction. The donor orbital is commonly the highest occupied molecular orbital (HOMO), and the acceptor orbital is commonly the lowest unoccupied molecular orbital (LUMO). To sum up how acid–base reactions work on the basis of molecular orbitals perturbation theory, Fig. 4-1 presents the relative energies of the frontier orbitals HOMO and LUMO of a hypothetical species A, and the frontier orbitals of several hypothetical reaction partners B, C, D, E, and F. Fig. 4-1 demonstrates that with respect to B, electron transfer from B to A will be favorable, and A will act as an oxidizing agent. With respect to C, the A (LUMO)-C (HOMO) perturbation will be favorable, and A will perform as an acid. With respect to D, the A (HOMO)-D (LUMO) perturbation will be favorable and A will perform as a base. With respect to E, complete electron transfer from A to E will be favorable and A will perform as a reducing agent. For F species, the frontier orbitals are degenerated with A. Neither of these species are obviously the donor or acceptor, and species may display, both the behaviors simultaneously. Moreover, from the general discussion of Lewis acid–base interactions, hydrogen bonding represents a particular case. It is apparent that electrostatic and charge transfer interactions are predominant, for hydrogen bonds. Generally, the principal charge transfer contribution originates from the proton acceptor–proton donor charge transfer complex, through the σ-type interactions. 4.3.2.2 Hydrogen Bonding A hydrogen bond is the interaction of a hydrogen atom with an electronegative atom, such as nitrogen, oxygen, or fluorine from another molecule. The hydrogen must form covalent bond with another electronegative atom to create the bond. These bonds can occur between molecules or within different parts of a single molecule . The hydrogen bond that was generally from 5 to 30 kJ /mol is stronger than a van der Waals interaction, but weaker than covalent or ionic bonds. A hydrogen attached to carbon can also participate in hydrogen bonding when the carbon atom is bound to electronegative atoms, as is the case in chloroform, CHCl3. The electronegative atom attracts the electron cloud, from around the hydrogen nucleus and, by decentralizing the cloud, leaves the atom with a positive partial charge. Because of the small size of hydrogen relative to other atoms and molecules, the resulting charge, though only partial, nevertheless, represents a large charge density. A hydrogen bond results, when this strong positive charge density attracts a lone pair of electrons on another heteroatom, which becomes the hydrogen-bond acceptor. Usually, the partially covalent nature of a hydrogen bond is simple to determine on the basis of inter-atomic distances in the X-H-Y system. X-H distance is typically about 110 pm, whereas H-Y distance is from 160 to 200 pm. Hydrogen bonds can vary in strength from very weak (1–2 kJ/mol) to extremely strong (over 155 kJ/mol) . The length of hydrogen bonds depends on bond strength, temperature, and pressure. The bond strength itself is dependent on temperature, pressure, bond angle, and environment. Many materials are strengthened by hydrogen bonds in their main chains. The best known example is nylon, where hydrogen bonds occur in the repeat unit, and plays a major role in the crystallization of the material. The bonds occur between carbonyl and amine groups in the amide repeat unit. They effectively link the adjacent chains to create crystals, which help reinforce the materials. The effect is greatest in the structure of aramid fiber, where hydrogen bonds stabilize the linear chains laterally, as shown in Fig. 4-2. The chain axes are aligned along the fiber axis, making the fibers extremely stiff and strong. The hydrogen bond networks make both the natural and sensitive materials to humidity levels in the atmosphere because water molecules can diffuse into the surface and disrupt the network. Some materials are more sensitive than others. The acid–base bonds including hydrogen bonding are exothermic and their energy ranges do not exceed 55 kJ/mol [21, 22]. This compares with the energy of London dispersion forces but exceeds that of Debye interactions, and Keesom interactions. With a large and negative value, the heat of an acid–base interaction can overcome the positive or the negligibly small negative entropic term, −TΔS, so that adhesion and mixing can be substantially improved. The high energy associated with the acid–base interactions is due to intermolecular forces, acting at intermolecular distances of about 0.3 nm. The acid–base interactions are thus of the long-range type rather than the short-range type. For a series of single-fiber composites, Nardin and Schultz demonstrated that the maximal work of adhesion (W) was obtained for fiber-matrix systems interacting via both dispersive and acid–base interactions. Hydrogen bonds are involved in the species, which are combined bond distances of the A–H···B compared to the sum of the van der Waals radius, as listed in Table 4-1. The types of various hydrogen bonding between atoms A and B are located closer than the estimated distance of the sum of the van der Waals radii. One of several features is that a hydrogen atom can approach fairly close over a distance to gain access to the atom B, compared with the estimated distance of the sum of the van der Waals radii, which suggests that a hydrogen atom is penetrated in the electron cloud of atom B. (See Chapter 1.2.2.) Table 4-1. van der Waals Bonding Ddistances vs. the Observed Distances for the Typical Hydrogen Honding \| Bonding state \| A·Ba (Calculated value) \| A·B (Observed value) \| H·B (Calculated value) \| H·B (Observed value) \| --- --- \| F–H–F \| 270 \| 240 \| 260 \| 120 \| \| O–H–O \| 280 \| 270 \| 260 \| 170 \| \| O–H–F \| 280 \| 270 \| 260 \| 170 \| \| O–H–N \| 290 \| 280 \| 270 \| 190 \| \| O–H–Cl \| 320 \| 310 \| 300 \| 220 \| \| N–H–O \| 290 \| 290 \| 260 \| 200 \| \| N–H–F \| 290 \| 280 \| 260 \| 190 \| \| N–H–Cl \| 330 \| 330 \| 300 \| 240 \| \| N–H–N \| 300 \| 310 \| 270 \| 220 \| \| N–H–S \| 340 \| 340 \| 310 \| 240 \| \| C–H–O \| 300 \| 320 \| 260 \| 230 \| a : Not be calculated by Bondi's method , but using the calculated by method of Hamilton et al. . 4.3.2.3 Work of Adhesion The work that is needed to separate reversibly the interface between the two bulk phases 1 and 2 from their equilibrium interacting distance to infinity is called the ‘work of adhesion’. (4-2) where γ1 and γ2 are the surface free energies of phases 1 and 2, respectively, and γ12, the interfacial free energy, between phases 1 and 2. When an interface comprises from two surfaces, the work of adhesion, is the decrease of Gibbs free energy per unit area. Therefore, greater the interfacial attraction greater is the work of adhesion, and the smaller the interfacial free energy, between phases 1 and 2. Thereby, γ12 can be related, to the character of the two phases. In the study of low surface energy materials such as polymers, many researchers have evaluated the thermodynamic surface free energy from wetting contact angle measurements between the interface, phases 1 and 2, which are commonly presented as the solid phase (S) and liquid phase (L) so that the interfacial free energy between the liquid and the solid may be written as: (4-3) The wetting equilibrium is defined from the profile of a liquid drop (L) in the presence of the vapor phase (V) on a planar, smooth, and homogeneous solid surface (S). The surface energies of materials and the equilibrium contact angle θ at the three-phase contact point are related by Young's equation : (4-4) For a first approximation and ignoring the spreading pressure of the vapor phase of the liquid onto the solid, the solid-liquid work of adhesion is (4-5) (4-6) To consider the contribution of the fundamental property of the forces, across the interface it has been suggested that the surface free energies, and the work of adhesion may be expressed by the sum of two terms: The London's dispersion (L) interactions and the non-dispersion forces (SP), that include Debye, Keesom, and acid–base interactions [25, 26]. Note that three different types of forces contribute to the total long-range interaction, between the polar molecules. Generally known as ‘van der Waals force’ these are the dispersion force (London interaction) the induction force (Debye interactions), and the orientation force (Keesom interactions). (4-7) (4-8) Fowkes proposes that the London dispersive part of the work of adhesion, is measured as twice the geometric mean of the dispersive component of the surface energy, of solids 1 and 2 . The total work of adhesion is reduced to (4-9) In the particular case such as London's forces across the interface, the work of adhesion is determined because the non-dispersive part becomes zero. Clearly, and can be determined, by wettability experiments. The only existence of London interactions at the interface between two adherents leads to (4-10) where G is the adhesive strength of the assembly. For knowledge of the strength of an acid–base interaction both the specific Gibbs free energy of interaction and the polar interaction (Debye, Keesom, and Hydrogen bonding) across the interface allow the determination of the reversible acid–base work of adhesion. In the case of adhesion between different solids, the contribution of the polar interactions to the thermodynamic work of adhesion occurs rapidly. Therefore, the acid–base of the adhesion energy is expressed as the following equation : (4-11) where D is the Debye interaction, K the Keesom interaction, and H the Hydrogen bonding. The quantitative determination of these quantities could become either a theoretical or experimental value. Theoretical determinations are mostly based on the hypothesis that the initial perturbation of the orbital of interaction species, determines the process of a reaction or an interaction. In the equation of this state, the development of perturbation theory was studied by Hudson and Klopman . From the energy gap between HOMO and LUMO, acids and bases were defined by their hardness and softness . The Hard Soft Acid Base principle explains some basic rules about kinetics, and equilibrium of the acid–base interactions and is explained on the basis of the density-functional theory, and the frontier orbital method . The results of γL and γSP are investigated by water and diiodomethane, on each carbon black and thermoplastic polymer, as shown in Table 4-2. From the matrix results, the γSP on water was significantly changed in the high density polyethylene specimen. It was clearly expected that the positive temperature coefficient properties of composites made with high density polyethylene polymer would be greatly influenced. Table 4-2. Results of and on the Carbon Black and Thermoplastic Polymers Studied \| Positive temperature coefficient materials \| (mJ/m2) \| (mJ/m2) \| γS (mJ/m2) \| --- --- \| \| CB (Carbon black) \| 36.8 ± 1.8 \| 34.4 ± 1.7 \| 71.2 ± 3.5 \| \| EVA (Ethylene vinyl acetate) \| 27.6 ± 0.5 \| 13.1 ± 0.7 \| 40.7 ± 1.2 \| \| LDPE (Low density polyethylene) \| 28.0 ± 0.3 \| 14.3 ± 0.8 \| 42.3 ± 1.1 \| \| LLDPE (Linear low density polyethylene) \| 30.2 ± 0.7 \| 14.6 ± 0.5 \| 44.8 ± 1.2 \| \| HDPE (High density polyethylene) \| 27.5 ± 0.5 \| 10.5 ± 0.5 \| 38.0 ± 1.0 \| \| PP (Polypropylene) \| 30.3 ± 0.4 \| 15.0 ± 0.3 \| 45.3 ± 0.7 \| The results of the work of adhesion of carbon black/thermoplastic polymer composites derived from Eq. (4-3) are shown in Fig. 4-3. It is interesting to note that carbon black/high density polyethylene composites produce a minimum value of work of adhesion, but carbon black/linear low density polyethylene composites produce a relatively higher work of adhesion, due to the abundant terminal groups of the linear low density polyethylene molecular chains that can be bonded with carbon black. A good correlation between the work of adhesion and the maximum resistivity, is shown in Fig. 4-4. It can be clearly observed that the resistivity is increased as the work of adhesion between carbon blacks and thermoplastic polymers. It is then noted that the decrease in the work of adhesion induced by interactions between carbon black surfaces and polymer chains, plays an important role in improving the positive temperature coefficient behaviors of a composite in this system. As a result, the largest positive temperature coefficient intensity and the maximum work of adhesion occur on the carbon black/high density polyethylene composites since the resistivity of carbon black/high density polyethylene composites is higher than that of carbon black/low density polyethylene, carbon black/linear low density polyethylene, and carbon black/polypropylene composites at room temperature. However, the carbon black/linear low density polyethylene composites showed a relatively higher work of adhesion, due to the abundant terminal groups, of the linear low density polyethylene molecular chains that can be bonded with carbon black. Consequently, a good relationship between the work of adhesion and the maximum resistivity is shown in these positive temperature coefficient composites . 4.3.2.4 Drago's Approach Drago and Wayland suggested an equation similar to the perturbation theory, for predicting reaction enthalpies between the acid and base. Both the acid and base are defined by two independent parameters: The E value measures their ability to participate in electrostatic bonding and the C value measures their capability to participate in covalent bonding leading to (4-12) where ΔHAB is the enthalpy of the acid–base adduct formation, EA the capability of the acid to participate in electrostatic bonding, EB the ability of the base to participate in electrostatic bonding, CA the ability of the acid to participate in covalent bonding, and CB the capability of the base to participate in covalent bonding. A self-consistent set of E and C values is now available from 33 acids and 48 bases, allowing ΔH prediction for over 1584 adducts . The cogency of Eq. (4-12) was clearly demonstrated for polymer adsorption on various substrates . (See Chapter 1.3.7.) 4.3.2.5 Gutmann's Numbers In addition to the previous discussion of the dual-polarity properties of Lewis acids and bases, it should be noted that many of them are amphoteric, by definition. The donor number (DN) was developed to correlate the behavior of a solute in various donor solvents with a given basicity or the degree of the donor electron. A relative measurement of the basicity of a solvent D is given by the enthalpy of its reaction with the reference to the acid which was chosen as SbCl5 by Gutmann's scale. Subsequently, Mayer et al. introduced an acceptor number (AN) as the relative P NMR shift induced by triethylphosphine and relative to the acidic strength between hexane and SbCl5. In 1989, Riddle and Fowkes modified this acceptor number in order to express from these, to acceptor number (AN) number in correct enthalpic unit (kcal/mol) . The most important assumption of Gutmann's approach is that the order of the determined base strengths remains constant for the other acids the Gibbs free energy of formation of a given adduct, being linearly related to the donor number (DN) of the base through Eq. (4-13): (4-13) where ΔG is the Gibbs free energy of the acid–base adducts formation, DNB the donor number of the base, and aA and bA the constants characteristic of the acid. This means that a plot of the DN for a series of donors vs. –ΔG of their adduct formation with a given acid brings up a straight line, allowing the determination of aA and bA. By experimentally measuring the formation free energy of only two adducts from a given acid, one can predict through the resulting aA and bA values, the adduct formation free energy of this acid with any other donor for which DN is known. Gutmann also suggested that the free energy of an acid–base interaction could be approximated by a two-parameter equation of the form (4-14) where ΔG12AB is the Gibbs free energy of the acid–base adduct formation, DNB the donor number of the base, and ANA the acceptor number of the acid. The factor of 100 converts the acceptor number value from a percentage of the SbCl5 value to a decimal fraction. However, it is important that of the 171 donor number values reported in the literature only 50 were determined precisely by calorimetry . In a recent research , Gutmann's semi-empirical scale was used in terms of acceptor number and donor number values, which described the ability of the polar component or the Lewis acid–based or electron acceptor-donor interaction, and these are given in Table 4-3 (See Chapter 1.3.8). Table 4-3. Characteristics of Polar Probes Used \| Probe \| DN \| AN \| --- \| chloroform \| 0 \| 23.1 \| \| acetonitrile \| 14.1 \| 18.9 \| \| acetone \| 17.0 \| 12.5 \| \| ethylacetate \| 17.1 \| 9.3 \| \| diethylether \| 19.2 \| 3.9 \| \| tetrahydrofuran \| 20.0 \| 8.0 \| \| nitromethane \| 2.7 \| 20.5 \| 4.3.2.6 Approaches of van Oss, Good, and Chaudhury van Oss et al. introduced the notion of acidic and basic components of the surface energy (γ+ and γ−, respectively) to characterize the acid–base properties of materials and predict WAB: (4-15) The γ+ and γ− for a solid, can be determined by the contact angle measurements using three reference liquids of known , , and . The and for test liquids were established with model surfaces and test liquids, on the arbitrary assumption, that for water and (4-16) Application of Eq. (4-16), to water (w) yields (4-17) (4-18) where γSP is the non-dispersive or the so-called polar component of the surface energy of water. Once γ+ and γ− are determined for a given material the overall acid–base (‘polar’) contribution to its γ can be evaluated for water in Eq. (4-16). Table 4-4 shows γ+ and γ− values for liquids, polymers, and other materials. Table 4-4. Surface Energy Components for Commonly Used Test Liquids and for Polymers and Other Materials (mJ/m2) \| \| \| \| \| \| \| --- --- --- \| \| Empty Cell \| γ \| γL \| γSP \| γ+ \| γ− \| \| TestingLiquid \| \| \| \| \| \| \| Water \| 72.8 \| 21.8 \| 51.0 \| 25.5 \| 25.5 \| \| Diiodomethane \| 50.8 \| 50.42 \| 0.38 \| 0 \| 0 \| \| Ethylene glycol \| 47.7 \| 31.0 \| 16.7 \| 1.92 \| 47.0 \| \| Polymer \| \| \| \| \| \| \| Poly(ethylene oxide) 6000 \| 43 \| 43 \| 0 \| 0 \| 64 \| \| Dextran 10000 \| 61.2 \| 47.4 \| 13.8 \| 1.0 \| 47.4 \| \| Poly(methyl methacrylate) \| 39–43 \| 39–43 \| 0 \| 0 \| 9.5–22.4 \| \| Poly(vinyl chloride) \| 43.7 \| 43 \| 0.7 \| 0.04 \| 3.5 \| \| Polystyrene \| 42.0 \| 42 \| 0 \| 0 \| 1.1 \| \| Polyethylenea \| 33 \| 33 \| 0 \| 0 \| 0.1 \| \| Polyethyleneb \| 57.9–62.5 \| 42 \| 15.9–20.5 \| 2.1 \| 30–50 \| \| Iodinated polyacetylene \| 52.6 \| 42.8 \| 9.8 \| 1.04 \| 23.15 \| \| Cellulose acetate \| 40.2 \| 35 \| 5.2 \| 0.3 \| 22.7 \| \| Cellulose nitrate \| 45 \| 45 \| 0 \| 0 \| 16 \| \| Agarose \| 44.1 \| 41 \| 3.1 \| 0.1 \| 24 \| \| Gelatin \| 38 \| 38 \| 0 \| 0 \| 19 \| \| Human serum albumin \| 44.3 \| 41 \| 3.3 \| 0.15 \| 18 \| \| Inorganic filler \| \| \| \| \| \| \| Red mud \| 55.4 \| 39.1 \| 16.2 \| 2.27 \| 28.98 \| \| Acid-treated red mud \| 55.9 \| 40.6 \| 15.3 \| 2.37 \| 24.77 \| \| Base-treated red mud \| 54.8 \| 38.6 \| 16.2 \| 2.19 \| 29.94 \| Source: Data in Ref. [1, 48, 49]. a : Based on advancing angles. b : Based on receding angles. van Oss and Good have shown that their approach can predict the solubility of poly (ethylene oxide) and dextran in water but the ‘γPolar approach’ cannot, and this leads to a positive interfacial free energy for the polymer-water systems (5.8 and 6 mJ/m2 for poly(ethylene oxide), and dextran, respectively) and thus to the insolubility of these polymers in water. Although the novel approaches to interfacial inter-facial interactions just described were shown to be more valid than the ‘polar interactions’ approach in predicting adhesion, solubility, and miscibility phenomena [42–45], they must be viewed with caution, as the controversial publications have shown their shortcomings [45, 46]. Nevertheless the qualitative and/or quantitative determination of the acid–base properties for polymers and other materials is an important step toward the understanding and the prediction of their interfacial interactions. This is a very delicate task, because it requires the determination of the polar component of heat or the free energy changes of acid–base interactions of reference acidic and basic chemical species (of known acid–base properties) with the materials . The choice of reference test acids and bases is also crucial and usually depends on the nature of the material under investigation and the experimental conditions associated with the technique used for the assessment of acid–base properties. 4.3.2.7 IR spectroscopic tools to access acid–base strength Fowkes proposed that the enthalpy of acid–base adduct formation is related to the shift of infrared frequency (Δv) of its absorption band for specific functional groups involved in the acid–base interaction, according to the following equation: (4-19) where ΔG12AB is the Gibbs free energy of the acid–base adduct formation, Δk12AB the correlation constant between the IR wave number shift and enthalpy, and Δv12AB the infrared frequency shift. kAB is a characteristic constant of the functional group determined on the basis of infrared and micro-calorimetrical from the results of adduct formation. For example, the latter is equal to –0.99 kJ/mol cm in the carbonyl group. The stretching frequency of the C=O vibration band is reduced by an amount of proportional ΔvAB to the Gibbs free energy of the acid–base bonding (ΔGAB), according to kAB. This methodology has recently been proved not only for polymer-solvent adduction but for polymer-polymer and polymer-metal adduction. Furthermore, the infrared shifts can be used to determine Drago's E and C constants of unknown polymers or substrates . If a test of acid of known CA and EA gives ΔvAB shifts, allowing the determination of ΔGAB, then Eq. (4-12), is reduced to (4-20) in which E′B and C′B are ‘the trial’ values. The actual values of EB and CB exist at one point of the straight line acquired from Eq. (4-20). The operation of various tests of acids shows that straight lines intersect at EB and CB. 4.3.2.8 Density of interacting sites On the foundation of reflectance spectroscopy , the section of functional groups that develop acid–base adducts at an interface can be estimated according to (4-21) where A represents the integrated intensity of the considered stretching mode for free, F, or acid–base, AB, species, and ɛ the molar absorption coefficient of Beer's law. As an explaining example, we present the case of the adsorption of ethylene-vinyl-acetate copolymers. From various amounts of acetate comonomer, we expect to affect the growth of acid–base interactions of the acetate CO groups and aluminols of the substrate. Fig. 4-5 showed the evolution of the fraction of interfacial acid–base interactions vs. the 12 nm thick vinyl-acetate copolymer, which was adsorbed on aluminum coated silicon wafers. This result shows two major regimes. The first one, with a low content of vinylacetate, consists of a sharp increase of connector density, with co-monomer density, until a maximum is obtained. Second, a constant decrease of interfacial connecting sites is studied, until the density of the co-monomer increases. The first important point, is, that the co-monomer density allows us to control the level of interfacial acid–base bonds. The result of the foundation of interfacial polymer-metal bonds that is the failure strength of the polymer-metal assemblies can be estimated by evaluating peel forces in different media, such as air, or ethanol, in 180 peel test geometry. Peel analyzes were performed for a large range of separation rates Rs from 8.4 × 10−6 to 8.4 × 10−3 m/s. The rheological model of adhesion states that the peel energy G can be expressed as (4-22) where Go is the intrinsic fracture energy of the interface Φ a dissipative function corresponding to the energy irreversibly expanded in visco-elastic and plastic deformations in the bulk adhesive. The function of Φ depends on both the separation rate (Rs) and temperature (T), and follows a time-temperature equivalence law . The superscripts A and L refer to air or liquid. Under specific conditions , the dissipative function Φ is identical in air and in liquid media. Therefore, supposing as a first assumption that the intrinsic fracture energy of the interface (Go) is proportional, to the reversible energy of adhesion (Wo) Eq. (4-22) is obtained: (4-23) (4-24) Given that the dissipation factors cancel, changes in the interfacial characteristic can be immediately predicted, from peel measurements. The ratios of Eqs. (4-23) and (4-24), can be determined promptly, either experimentally (GA–GL/GA) or by calculation (WA0–WLo/WA0) according to thermodynamic considerations. Presenting the definition of the thermodynamic work of adhesion Eq. (4-9), we can indicate the reduction of adhesion strength due to the presence of a liquid medium as (4-25) where superscripts P, S, and L is the polymer, the substrate, and the liquid, respectively. Physical interactions occur only during the solid-solid and solid-liquid contact. The values of the dispersive component of the surface free energy of polymers and substrates as well as and are determined by the wettability measurements. A comparison between the experimental values and the calculated values allows us to demonstrate the properties of interfacial polymer-substrate bonds and to evaluate the magnitude of the interfacial acid–base work of adhesion according to (4-26) Understanding the enthalpy of interaction at the interface, and the reversible acid–base work of adhesion, allows us to evaluate the number of acid–base interacting sites, at the interface as in Eq. (4-11). Fig. 4-6 showed the evolution of the fraction of vinyl-acetate involved in acid–base adducts as well as the density of interacting sites vs. the acid–base work of adhesion for EVA copolymer-aluminum assemblies. The acid–base work of adhesion is directly proportional to the fraction of acid–base adducts determined from IR analysis. Two regimes are observed. The first one demonstrates, that a distinct increase of interfacial connecting sites contributes to low acid–base adhesion energy until a threshold value is obtained. Also, the acid–base adhesion energy rapidly increases for a low variation the density of the connecting sites. We suppose that these phenomena are largely related to the interfacial mobility reduction and thus molecular dissipation . View chapterExplore book Read full chapter URL: Book series2011, Interface Science and TechnologySoo-Jin Park, Min-Kang Seo Review article Interfacial forces used in two-phase flow numerical simulation 2017, International Journal of Heat and Mass TransferTien-Juei Chuang, Takashi Hibiki 2.5 Turbulent dispersion force The turbulent dispersion force is a force due to the fluctuating component of the forces acting on bubbles. 2.5.1 Lopez de Bertodano’s model Eq. (34) represents the simple turbulent dispersion force modeled by assuming the bubble motion approximated by the thermal diffusion of air molecules in the atmosphere . (34) In Eq. (34), denotes the turbulent kinetic energy of liquid phase. The value of recommended coefficient, , is between 0.1 and 0.5. Eq. (35) represents extended Lopez de Berodano’s model, which can be applicable to two-phase turbulence jet . (35) Eq. (36) represents the definition of turbulent Stoke number, St. (36) In Eq. (36), and denote the relaxation time of bubbles and relaxation time of eddies, respectively. Eq. (37) represents the extended Lopez de Bertodano’s model, which can be applicable to cap bubbly flow . (37) In Eq. (37), and , respectively, denote the virtual mass coefficient and bubble induce kinetic turbulence viscosity computed by Sato’s model . 2.5.2 Burns’ model (or Favre average drag model) Eq. (38) represents the turbulence dispersion force modeled by applying Favre average to the drag force . (38) Eq. (39) represents the definition of turbulent Prandtl number for volume fraction dispersion, , and kinematic eddy viscosity of liquid phase, . (39) In Eq. (39), denotes the turbulence dispassion rate. View article Read full article URL: Journal2017, International Journal of Heat and Mass TransferTien-Juei Chuang, Takashi Hibiki Chapter Airlift loop reactors 2014, Numerical Simulation of Multiphase Reactors with Continuous Liquid PhaseChao Yang, Zai-Sha Mao 4.3.2.3 Turbulent dispersion force The turbulent dispersion force is used to describe the spread of bubbles due to the effect of flow turbulence, though arguments exist on its physical soundness. This force is partly physical in the sense that turbulent vortices exert an impact onto the bubbles and push them to move in the fluid. There are many expressions for this but they vary greatly, and two models that have been widely used in recent years will be addressed here. The simplest and most common approach consists of modeling the effect of turbulence by analogy with the Einstein theory of Brownian motion. This approach was derived by Lahey et al. (1993) for the force per unit volume, and it reads (4.72) where k is the turbulent kinetic energy. The key problem is to correlate the coefficient CTD with flow conditions. As pointed out by Mao and Yang (2009), there is no general agreement so far on the physical basis and the estimation of the coefficient for the turbulent dispersion force. Therefore, many values for this coefficient have been adopted in the literature even for the bubbly flow in an air–water system to fit predictions to the experimental data. For the bubbly flow in air–water systems, the coefficient is in the range of 0.1–1.0. It is noteworthy that this model is an analogy with the thermal diffusion of air molecules in the atmosphere, and has no direct connection to the nonlinearity that it is meant to model. Another typical expression for the turbulent dispersion force was developed in the frame of homogeneous turbulence. Talvy et al. (2007a) proposed that the turbulent dispersion force can be modeled as (4.73) (4.74) (4.75) (4.76) where CA = 0.5 denotes the added mass coefficient. Three time scales, i.e., the characteristic time of the particle entrainment by the continuous fluid motion , the characteristic time of energetic turbulent eddies and the characteristic time of the fluid turbulence viewed by bubbles , are given similar to Podila (2005) and Chen, X. H. (2004) by (4.77) (4.78) (4.79) (4.80) (4.81) where θ is the angle between the gas velocity and the slip velocity. The added mass coefficient is taken as 0.5, and the parameter Cβ is set as 0.45 in the direction parallel to the mean relative velocity and as 1.80 in the normal direction. In general, although great efforts have been made in developing a mathematical model to improve the prediction of bubbly flow in recent years, there is still much room for more accurate accounts of its mechanism. The turbulent dispersion force results from the turbulence of the continuous phase and the radial profiles of the gas void fraction, which leads to a more uniform distribution of gas in the radial direction. View chapterExplore book Read full chapter URL: Book2014, Numerical Simulation of Multiphase Reactors with Continuous Liquid PhaseChao Yang, Zai-Sha Mao Related terms: Functional Group Energy Engineering Adsorption Electrostatics Surfactant Adsorbate Dipole Interaction Mols Hydrophobic Induced Dipole View all Topics
1246	https://www.cantorsparadise.com/deriving-the-area-of-a-triangle-using-a-quadrilateral-07a29697928e	Deriving the Area of a Triangle using a Quadrilateral \| by Archie Smith \| Cantor’s Paradise Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Cantor’s Paradise ----------------- · Follow publication Medium’s #1 Math Publication Follow publication Deriving the Area of a Triangle using a Quadrilateral Obtaining Heron’s formula from quadrilaterals Archie Smith Follow 4 min read · Apr 3, 2024 150 3 Listen Share Most people are taught that the area of a triangle is 1/2 × base × perpendicular height. Many also will be taught at some point 1/2×ab×sinC. However, I think a less-known formula for the area of a triangle is Heron’s formula for a triangle ABC with side lengths a, b and c Press enter or click to view image in full size Notice s is half the perimeter of the triangle ABC Today I want to go through a lovely proof of Heron’s formula, which surprisingly starts with a quadrilateral. The maths used to prove the result is not complex, requiring only GCSE/high school knowledge to complete. Hopefully we can gain an appreciation for the beauty of the proof. The Proof Start with a cyclic quadrilateral ABCD where the lengths AB, BC, CD and DA are a, b, c and d respectively, that is a quadrilateral who’s vertices all lie on the circumference of a circle. Press enter or click to view image in full size Now let ∠BAD=θ. We can use our circle theorems to deduce that ∠BCD=π-θ as opposite angles in a cyclic quadrilateral are supplementary, sum to π. We will also add the line BD to our diagram. Press enter or click to view image in full size I want us to consider the length BD. How could we write this in terms of a, b, c, d and θ? I think by drawing the line BD, we can more clearly see we could use the cosine rule on either triangle ABD and BCD. So let’s try to find two different expressions for the length of BD Press enter or click to view image in full size Note that cos(π-θ)=-cosθ ∀θ In doing this, we can set the two expressions for the square of the magnitude of BD and rearrange for cosθ in terms of a, b, c and d. Press enter or click to view image in full size Perhaps the start of this proof seems ambiguous as to where we are going or maybe you can’t see the motivation behind why we are finding cosθ in terms of the side lengths. Hopefully in the next part we can start to put together an idea of where to go. I challenge you to think of what we are going to consider next, remember that we are trying to derive the area of a triangle. Get Archie Smith’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe You might have been able to see that we will consider the area of the quadrilateral next. Notice thje area of the quadrilateral is equal to the sum of the areas of the triangles ABD and BCD. So call the area of our quadrilateral ABCD, Q. We can find an expression for Q in terms of a, b, c, d and θ by performing the 1/2×ab×sinC rule for the area of a triangle. From this we will rearrange for sinθ Press enter or click to view image in full size Now we have gotten an expression for sinθ and cosθ, I want to make use of the infamous identity sin²θ+cos²θ ≡ 1. Let’s substitute our obtained expressions and rearrange for the area of ABCD, Q. Press enter or click to view image in full size This looks somewhat promising. Let’s quickly perform some simplification by making use of the difference of two squares Press enter or click to view image in full size Finally introduce the variable s, where s=1/2(a+b+c+d). Some keen eyed mathematicians will also notice that s-a=1/2(b+c+d-a) and similar results for s-b, s-c and s-d. Press enter or click to view image in full size This looks very promising. All that is left to do is let d approach 0. As we do this, the quadrilateral ABCD ‘becomes’ a triangle ABC as the points C and D become coincident. Letting d tend to 0 in our formula for the area of the shape, and in s such that s=1/2(a+b+c), we obtain Heron’s formula Press enter or click to view image in full size You may think this formula is restricted by the fact we used a cyclic quadrilateral to start with and as we know not all quadrilaterals are cyclic. However, remember that every triangle is cyclic, therefore the formula holds for all triangles. Thank you for reading. Feel free to comment your thoughts on the proof Math Education Science Mathematics Programming 150 150 3 Follow Published in Cantor’s Paradise ------------------------------ 39K followers ·Last published Sep 15, 2025 Medium’s #1 Math Publication Follow Follow Written by Archie Smith ----------------------- 911 followers ·24 following Follow Responses (3) Write a response What are your thoughts? Cancel Respond Nicky Mac Apr 4, 2024 always loved Heron's formula as an amazing Greek achievement. this was an interesting proof -- Reply Nomeranduz Jun 28, 2024 Wow! Beautiful. -- Reply Astoria Bob Apr 3, 2024 Great stuff! Interesting that you used another formula for the area of a triangle to prove this formula. Is that ok in math proof law? And I wonder whether there is another proof for Heron’s formula? I personally love this formula! -- 1 reply Reply More from Archie Smith and Cantor’s Paradise In Cantor’s Paradise by Archie Smith How I got an Offer from Cambridge University -------------------------------------------- ### Commenting on my journey through the application process to study mathematics at Cambridge University Apr 28, 2024 6 In Cantor’s Paradise by Kasper Müller Numbers the Way God Intended: The Surreal Numbers ------------------------------------------------- ### Are we using the wrong number system? Aug 4 16 In Cantor’s Paradise by Kasper Müller A Beautiful and Unexpected Connection Between Generating Functions and Dirichlet Series --------------------------------------------------------------------------------------- ### And a surprising limit formula! 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1247	https://artofproblemsolving.com/wiki/index.php/Special_right_triangles?srsltid=AfmBOorSRsZRIbcP2nYcP5bSC0dPU4cxn_63Eew9OfYwo5MRrtbbD6ns	Art of Problem Solving Special right triangles - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Special right triangles Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Special right triangles A special right triangle is a right triangle whose angles and ratios of sides make the triangle useful. 45-45-90 Triangles Main article: 45-45-90 triangle A 45-45-90 triangle is a right triangle that has two angles. All 45-45-90 triangles are isosceles, so let's call both legs of the triangle . If that is the case, then the hypotenuse will always be . 30-60-90 Triangles Main article: 30-60-90 triangle A 30-60-90 triangle is a right triangle that has a angle and a angle. Let's call the side opposite of the angle . Then, the side opposite of the angle would have a length of . Finally, the hypotenuse of a 30-60-90 Triangle would have a length of . There is also the ratio of . With 2 as the hypotenuse and 1 opposite of the 30 degrees. That leaves as the only length left. See Also Pythagorean triple This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1248	https://www.principiosdeeconomia.com/slides/3.%20Principios%20de%20Econom%C3%ADa%20-%20Teor%C3%ADa%20del%20productor.pdf	Principios de Econom´ ıa Dentro de la firma Introduzca su nombre aqu´ ı Principios de Econom´ ıa 1 / 56 Principios de Econom´ ıa Cap´ ıtulo 11: El objetivo de la firma Introduzca su nombre aqu´ ı Principios de Econom´ ıa 2 / 56 El comportamiento de la firma ¿Cu´ al es el objetivo de las empresas? ¡¡¡GANAR DINERO!!! Seguramente los empresarios tienen otros objetivos, pero quiz´ as los satisfacen usando parte de sus ganancias. Las empresas que no buscan maximizar el beneficio econ´ omico van a tener problemas para sobrevivir (¿C´ omo hacer una gran fortuna en la industria del vino? Empezando con una grande!) Las decisiones que toman las empresas dependen de... ▶las caracter´ ısticas del mercado (la demanda) ▶y de los costos de producci´ on, ▶... las pol´ ıticas gubernamentales (por ejemplo impuestos) van a afectar las decisiones. Principios de Econom´ ıa 3 / 56 Restricciones y decisiones La firma se enfrenta a dos grandes restricciones: ▶Las caracter´ ısticas del mercado. - ¿Qu´ e es lo que los consumidores est´ an dispuestos a comprar? - ¿C´ omo se comportan las otras empresas? ▶Sus capacidades tecnol´ ogicas. - Las caracter´ ısticas de su funci´ on de producci´ on dada por sus costos de producci´ on. Tomando en cuenta estas restricciones, la empresa toma decisiones... ▶sobre el precio al que ofrece sus productos... ▶... y la cantidad de este producto que va a producir Principios de Econom´ ıa 4 / 56 ¿Que informaci´ on necesita la firma para decidir qu´ e precio cobrar? Para decidir qu´ e precio cobrar, una empresa necesita informaci´ on sobre la demanda. ▶¿Qu´ e es la demanda? La demanda es cu´ anto est´ an dispuestos a pagar los consumidores potenciales por su producto. Recuerden como estimamos la demanda de pizza de Eco I... ¡con una encuesta sobre la disposici´ on a pagar! Principios de Econom´ ıa 5 / 56 El problema de la firma Una vez que conocemos la demanda... ¿c´ omo se elige cu´ anto producir y el precio que cobrar? El problema principal de la empresa es el de la maximizaci´ on del beneficio. ▶¿Qu´ e es el beneficio? Beneficio = Ingresos totales – Costos totales ▶¿Qu´ e es el ingreso total? El valor de la producci´ on al precio ofrecido (p q) ▶¿Qu´ e es el costo total? Los costos por unidad, por la cantidad de unidades producidas (c q) Principios de Econom´ ıa 6 / 56 Vamos a ser maestros pizzeros Principios de Econom´ ıa 7 / 56 Pensando en los costos Costos expl´ ıcitos ▶Harina, levadura, sal, agua, etc. ▶Alquiler del negocio ▶Salario de los empleados. ▶Servicios que hay que pagar (agua corriente, luz, gas, etc.). Costos impl´ ıcitos ▶El salario que perdemos por la actividad alternativa (costo de oportunidad). ▶El costo del capital invertido (que podr´ ıa ser utilizado en otra actividad econ´ omica). Ingresos anuales por ventas de porciones de pizza $300.000 Alquilar 5 hornos pizzeros $45.000 Gasto en harina, tomate y mozzarella $75.000 Salario de dos empleados adicionales $40.000 Gasto en alquiler y servicios p´ ublicos $55.000 Ingresos y costos de la pizzer´ ıa Principios de Econom´ ıa 8 / 56 Elecci´ on de Tommy Tommy debe elegir entre invertir en un plazo fijo o abrir la pizzer´ ıa Al decidir entre dos alternativas se desea elegir la que da mayor ganancia. ▶El costo de oportunidad asociado con la elecci´ on de una decisi´ on se mide por los beneficios que se obtendr´ ıan en la siguiente mejor alternativa, ya que el costo de oportunidad de una alternativa es a lo que se renuncia cuando se opta por esa alternativa. Para calcular los costos tenemos que tener en cuenta: ▶inversi´ on de capital (costo de oportunidad del dinero) ▶trabajo gerencial (costo de oportunidad del tiempo) Principios de Econom´ ıa 9 / 56 Alta ganancia contable, baja ganancia econ´ omica Principios de Econom´ ıa 10 / 56 Elecci´ on Ingreso marginal: ingreso total adicional que se obtiene de vender una unidad m´ as Costo marginal: costo total adicional en que se incurre cuando se produce esa unidad adicional. Las empresas producen hasta el punto en el que el costo marginal se iguala al ingreso marginal ▶¡Es el punto en el que se maximiza el beneficio! Las decisiones que toma una empresa dependen de tres aspectos: ▶Condiciones de demanda ▶Caracter´ ısticas del mercado ▶Costos de producci´ on de la empresa Principios de Econom´ ıa 11 / 56 Principios de Econom´ ıa Cap´ ıtulo 12: Producci´ on Introduzca su nombre aqu´ ı Principios de Econom´ ıa 12 / 56 La funci´ on de producci´ on Las empresas, en general, tienen un amplio men´ u de factores productivos entre los que pueden elegir para producir el producto que desean vender. En general, estos factores incluyen: ▶Capital ▶Fuerza laboral/ Trabajo ▶Factor tierra ▶Capacidad empresarial o de organizaci´ on Principios de Econom´ ıa 13 / 56 La funci´ on de producci´ on La funci´ on de producci´ on es una expresi´ on matem´ atica que indica la cantidad m´ axima de bienes que una empresa puede producir, dada su tecnolog´ ıa, a partir de la combinaci´ on una cantidad espec´ ıfica de insumos. La funci´ on de producci´ on puede expresarse de la siguiente manera: Q = f(L, K) Estamos asumiendo, para simplificar, que los factores de producci´ on son trabajo (L) y capital (K). A medida que aumente la cantidad de trabajadores y/o la cantidad de capital, va a aumentar la producci´ on final de la empresa. ▶la funci´ on de producci´ on de esta empresa es mon´ otona creciente en L y en K∗. Principios de Econom´ ıa 14 / 56 Corto vs Largo Plazo El problema de producci´ on se puede pensar en dos momentos diferentes: ▶Corto plazo ▶Largo plazo En el corto plazo se considera que al menos un factor de producci´ on est´ a fijo, es decir, no puede cambiar. En cambio, en el largo plazo, todos los factores son variables y la empresa puede contratar cualquier cantidad de ellos. ▶Vamos a considerar que en el corto plazo el factor variable va a ser la cantidad de trabajadores, mientras que el factor fijo ser´ a el capital. Principios de Econom´ ıa 15 / 56 Funci´ on de producci´ on de Tommy Q Empleados Funci´ on de producci´ on A B C D E F 1 2 3 4 5 6 7 8 9 10 25 50 90 180 Conjunto de producci´ on Funci´ on de producci´ on Principios de Econom´ ıa 16 / 56 Mejora tecnol´ ogica Manteniendo la cantidad de trabajadores constante, un cambio tecnol´ ogico provoca un desplazamiento de la funci´ on de producci´ on. Producci´ on Empleados A G 1 2 3 4 5 6 7 8 9 25 50 Mejora Tecnol´ ogica Desplazamiento de la funci´ on de producci´ on Principios de Econom´ ıa 17 / 56 La funci´ on de producci´ on en el corto plazo Producci´ on Empleados A B C D E F G H I 1 2 3 4 5 6 7 8 9 10 25 50 90 140 180 210 220 Funci´ on de producci´ on Principios de Econom´ ıa 18 / 56 La funci´ on de producto medio El producto medio representa la cantidad de producto que cada trabajador genera en promedio durante un periodo de tiempo. ▶Pme = Q L PMe L PMe A B C D E F G H I 30 1 2 3 4 5 6 7 8 9 Derivaci´ on gr´ afica del producto medio Principios de Econom´ ıa 19 / 56 La funci´ on de producto marginal El producto marginal es la cantidad de producci´ on adicional que se genera al aumentar la cantidad de trabajadores en una unidad. ▶Pmg = ∆Q ∆L = Q1−Q0 L1−L0 PMe L PMg y1/x1 y4/x4 q7/x7 PMg 1 2 3 4 5 6 7 8 9 A B C D E F G H I Principios de Econom´ ıa 20 / 56 Rendimientos Marginales Decrecientes La ley de rendimientos marginales decrecientes especifica que, cuando se tiene un factor productivo fijo, se llega un momento en el que el producto marginal de cada trabajador ser´ a cada vez menor. ▶Esta ley indica que en el corto plazo, cuando tengo un factor fijo, va a llegar un momento donde el producto marginal que aporte cada trabajador va a ser cada vez menor. ▶Esta ley se da en todas las empresas o sectores. ⋆Pensemos que si no se cumpliera la ley de rendimientos marginales decrecientes estar´ ıamos se˜ nalando que podr´ ıa aumentar siempre la producci´ on sin importar que la cantidad de capital, maquinarias y tierras sea fija. ⋆Que suceda eso es improbable. Principios de Econom´ ıa 21 / 56 Equilibrio en el corto plazo El productor elige la cantidad de empleados a contratar (recuerden que estamos en el corto plazo y el capital est´ a fijo) tal que maximice sus beneficios Teniendo en cuenta que el precio (P) y el salario (W) est´ an fijos, los beneficios van a depender de la cantidad de trabajadores contratados en este caso. La cantidad ´ optima de empleados que se debe contratar para maximizar los beneficios debe cumplir la siguiente condici´ on necesaria: ▶P × Pmg = W ▶Pmg = W P Para alcanzar el ´ optimo, las empresas van a contratar empleados hasta el momento en el que el producto marginal de cada uno sea igual al salario real. Principios de Econom´ ıa 22 / 56 ¿Qu´ e sucede en el largo plazo? En el largo plazo, todos los factores de producci´ on van a ser variables. ▶Esto quiere decir que en el largo plazo la cantidad total de producci´ on puede aumentar no solo modificando la cantidad de trabajadores, sino tambi´ en la cantidad de capital. ▶La combinaci´ on eficiente de factores a largo plazo se logra cuando la relaci´ on entre el producto marginal de un factor y el precio de un factor es id´ entica para todos los factores: ⋆ P mgL W = P mgK r ▶La interpretaci´ on es la misma que antes: en el punto donde se maximizan los beneficios de la empresa, la relaci´ on entre el valor de lo que generan los factores productivos en relaci´ on a su costo debe igualarse para todos los factores. Principios de Econom´ ıa 23 / 56 Principios de Econom´ ıa Cap´ ıtulo 13: Costos Introduzca su nombre aqu´ ı Principios de Econom´ ıa 24 / 56 Funciones de costos Para producir se requieren insumos y, para obtener insumos, hay que realizar desembolsos de dinero, ya sea contratando personal, alquilando o comprando m´ aquinas e inmuebles, comprando bienes intermedios, etc. Vimos en el cap´ ıtulo anterior que la funci´ on de producci´ on nos ayuda a definir cuanto se produce contratando factores productivos, pero ¿cu´ anto hay que contratar? ▶Son las funciones de costos, junto con la informaci´ on de la demanda del mercado, las que brindan a la empresa toda la informaci´ on que requiere para tomar decisiones de producci´ on, de forma tal de maximizar sus beneficios. Principios de Econom´ ıa 25 / 56 Costos en el corto vs largo plazo El corto plazo es un periodo de tiempo tan limitado que la empresa es incapaz de variar el uso de algunos de sus insumos. ▶En el corto plazo, por ejemplo, consideramos que el capital es un insumo fijo (es decir, no puede variar) y que el trabajo es el ´ unico insumo variable. ▶Esto implica que la producci´ on adicional se logra utilizando horas adicionales de trabajo en combinaci´ on con un stock fijo de capital en la planta de la empresa. En el largo plazo, todos los insumos ser´ an variables, lo que implica que la empresa podr´ a ajustar las cantidades que adquiere de todos ellos. ▶Esto significa que en el largo plazo, la producci´ on puede aumentar ya sea por un aumento del trabajo, por uno del capital o por ambos. Principios de Econom´ ıa 26 / 56 ¿Qu´ e sucede en el corto plazo? La funci´ on de costos de una empresa especifica el costo total de producir una cantidad determinada de bienes para cada una de las cantidades posibles ▶Es una funci´ on de las cantidades: ▶C = C(Q) En el corto plazo, el trabajo es el ´ unico insumo variable (el capital est´ a fijo). ▶Supongamos que Tommy alquila el local y 2 hornos y, adem´ as, contrata empleados. En el corto plazo Tommy puede aumentar la cantidad de pizzas producidas s´ olo contratando m´ as personal. ⋆Los costos relacionados con el alquiler del local y los hornos son costos fijos. Principios de Econom´ ıa 27 / 56 Producci´ on y costos La funci´ on de producci´ on tiene una representaci´ on como funci´ on de costos. Los costos son de dos tipos: ▶Los costos fijos (no cambian con la producci´ on) ▶Los costos variables (cambian con la producci´ on) Principios de Econom´ ıa 28 / 56 Veamos los costos fijos... Costos Pizza CF $400 Principios de Econom´ ıa 29 / 56 y ahora los costos variables Costos Pizza CF CV Principios de Econom´ ıa 30 / 56 Los costos totales Costos Pizza CF CV CT Principios de Econom´ ıa 31 / 56 Los costos de hacer pizzas Se observan dos tipos de costos: ▶Los costos fijos ▶Los costos variables Para evaluar la cantidad de pizzas que queremos producir nos vamos a hacer dos preguntas: ▶¿Cu´ anto cuesta en promedio una pizza? (Costo medio) ▶¿Cu´ anto cuesta hacer una pizza adicional? (Costo marginal) Principios de Econom´ ıa 32 / 56 Costos: ¡para completar! Principios de Econom´ ıa 33 / 56 Los costos medios y marginales El costo medio es el costo promedio por unidad producida. ▶Gr´ aficamente es la pendiente del rayo que sale desde el origen a un punto dado de la funci´ on de costo - En el ejemplo, disminuye al principio pero luego aumenta. El costo marginal mide el efecto sobre el costo total al producir una unidad adicional. ▶Gr´ aficamente, es la pendiente de la funci´ on de costo en un punto dado - En el ejemplo, los costos marginales aumentan a medida que aumenta la producci´ on. Principios de Econom´ ıa 34 / 56 ¿C´ omo construir los costos medios fijos? Principios de Econom´ ıa 35 / 56 ¿C´ omo construir los costos medios variables? Principios de Econom´ ıa 36 / 56 ¿C´ omo construir el costo marginal? Principios de Econom´ ıa 37 / 56 Graficando costo medio y marginal en el mismo gr´ afico Principios de Econom´ ıa 38 / 56 Las propiedades de las curvas El costo marginal eventualmente aumenta debido a los rendimientos marginales decrecientes. La curva de costo total promedio suele tener forma de U. Bajan los costos fijos medios y aumentan los costos fijos medios variables. La curva de costo marginal corta la curva de costo total promedio en su nivel m´ ınimo. Si el costo marginal es menor al costo medio, el costo medio disminuye al aumentar la producci´ on. Principios de Econom´ ıa 39 / 56 El error de confundir Cme con Cmg Nos enteramos que la universidad cuenta con un centro de llamadas que ya no usa. Al consultar a las autoridades, responden: ≪Bueno, una llamada cuesta $200, mientras que mandar una carta cuesta $100 ≫. ¿Como calculo el costo? Para estimar el costo la universidad sum´ o el costo del sistema inform´ atico de telefon´ ıa que hab´ ıa comprado a˜ nos antes y el coste de pagar a estudiantes para realizar las llamadas, y dividieron esa cifra entre el n´ umero total de llamadas. ¿Cual es el error? Principios de Econom´ ıa 40 / 56 Entonces... ¿C´ omo decidimos cuanto producir? Por un momento, supongamos que el precio de la pizza es p. ¿Me conviene producir una pizza adicional? ▶Si uno aumenta una unidad de producto, recibe p. ▶Pero al mismo tiempo sus costos se incrementan en Cmg. ▶Entonces, el beneficio extra que se recibe es p – CMg. La diferencia entre el precio p y el costo marginal CMg se denomina ganancia marginal. Principios de Econom´ ıa 41 / 56 Conclusiones con un P fijo ¿Me conviene producir una pizza adicional? ▶Si el beneficio marginal es positivo entonces S´ I conviene producir una pizza adicional. ▶Pero si el beneficio marginal es negativo entonces NO conviene producir una pizza adicional. ¿Qu´ e significa que el beneficio marginal sea igual 0? Principios de Econom´ ıa 42 / 56 Vamos a ver que pasa con diferentes precios.... Principios de Econom´ ıa 43 / 56 La curva de oferta de la empresa Costos q CMg Oferta CMeV CMeT qmin CMeV Punto de Cierre Medios Punto de Cierre Principios de Econom´ ıa 44 / 56 En el largo plazo tenemos m´ as flexibilidad Los costos de una empresa dependen de su escala y el tipo de tecnolog´ ıa de producci´ on. Empresas grandes pueden ser m´ as rentables que las peque˜ nas debido diversas ventajas: ▶Ventajas tecnol´ ogicas - Producci´ on a gran escala permite mejorar la especializaci´ on y bajar los costos. ▶Ventajas de costos - Por ejemplo, empresas grandes, con mayor poder de negociaci´ on, pueden comprar recursos en t´ erminos m´ as favorables. ▶Ventajas de demanda - Por ejemplo, efectos de red (valor de la producci´ on aumenta con el n´ umero de usuarios). Principios de Econom´ ıa 45 / 56 Rendimientos a escala ¿Qu´ e sucede con la producci´ on cuando aumentamos la cantidad de insumos productivos en la misma proporci´ on? ▶La producci´ on aumenta pero... ¿cu´ anto aumenta? Si la producci´ on aumenta m´ as que proporcionalmente, entonces la funci´ on de producci´ on exhibe rendimientos crecientes a escala (econom´ ıas de escala o costos decrecientes a escala). Principios de Econom´ ıa 46 / 56 Rendimientos a escala Si la producci´ on aumenta proporcionalmente, entonces la funci´ on de producci´ on exhibe rendimientos constantes a escala (Costos constantes a escala). Si la producci´ on aumenta menos que proporcionalmente, entonces la funci´ on de producci´ on exhibe rendimientos decrecientes a escala (Deseconom´ ıas de escala o costos crecientes a escala). Principios de Econom´ ıa 47 / 56 Costos en el largo plazo Principios de Econom´ ıa 48 / 56 Costos en el largo plazo Principios de Econom´ ıa 49 / 56 Costos en el largo plazo Principios de Econom´ ıa 50 / 56 Costos en el largo plazo CMe LP q CMeLP CMeCP1 CMeCP3 CMeCP2 Rendimientos constantes a escala Econom´ ıas de escala Deseconom´ ıas de escala Principios de Econom´ ıa 51 / 56 Principios de Econom´ ıa Cap´ ıtulo 14: Construyendo la oferta Introduzca su nombre aqu´ ı Principios de Econom´ ıa 52 / 56 La curva de oferta del mercado P Q 20 10 30 40 50 60 20 30 40 50 60 4 8 1216 116 94 72 28 Oferta Emp 1 Oferta Emp 2 Oferta Emp 3 Oferta del mercado La curva de oferta del mercado Principios de Econom´ ıa 53 / 56 La curva de oferta del mercado P Q 30 40 40 72 Oferta del mercado A B La curva de oferta del mercado Principios de Econom´ ıa 54 / 56 ¿Qu´ e factores afectan la funci´ on de oferta? Precio de los insumos Tecnolog´ ıa Precios de bienes relacionado N´ umero de vendedores Pol´ ıtica Gubernamental Influencias especiales Principios de Econom´ ıa 55 / 56 Desplazamientos de la curva de oferta volver Principios de Econom´ ıa 56 / 56
1249	https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/water.php	Water, Acids, and Bases Water, Acids, and Bases Nahkleh Group Robinson Group Weaver Group Bodner Group Acid-Base Equilibria Water, Acids, and Bases pH, pOH, and Ka Equilibrium Problems Involving Strong Acids Weak Acids and Equilibrium Equilibrium Problems Involving Bases Mixtures and Buffers Diprotic and Triprotic Acids and Bases Compounds that could be either Acids or Bases Water, Acids, and Bases The Acid-Base Chemistry of WaterStrong Acids and the H 3 O+ and OH- Ion Concentrations The Acid-Base Chemistry of Water The chemistry of aqueous solutions is dominated by the equilibrium between neutral water molecules and the ions they form. 2 H 2 O(l) H 3 O+(aq) + OH-(aq) Strict application of the rules for writing equilibrium constant expressions to this reaction produces the following result. This is a legitimate equilibrium constant expression, but it fails to take into account the enormous difference between the concentrations of neutral H 2 O molecules and H 3 O+ and OH- ions at equilibrium. Measurements of the ability of water to conduct an electric current suggest that pure water at 25 o C contains 1.0 x 10-7 moles per liter of each of these ions. [H 3 O+] = [OH-] = 1.0 x 10-7 M At the same temperature, the concentration of neutral H 2 O molecules is 55.35 molar. The ratio of the concentration of the H+ (or OH-) ion to the concentration of the neutral H 2 O molecules is therefore 1.8 x 10-9. In other words, only about 2 parts per billion (ppb) of the water molecules dissociate into ions at room temperature. The equilibrium concentration of H 2 O molecules is so much larger than the concentrations of the H 3 O+ and OH- ions that it is effectively constant. We therefore build the [H 2 O] term into the equilibrium constant for the reaction and thereby greatly simplify equilibrium calculations. We start by rearranging the equilibrium constant expression for the dissociation of water to give the following equation. [H 3 O+][OH-] = K c x [H 2 O]2 We then replace the term on the right side of this equation with a constant known as the waterdissociation equilibrium constant, K w. [H 3 O+][OH-] = K w In pure water, at 25C, the [H 3 O+] and [OH-] ion concentrations are 1.0 x 10-7 M. The value of K w at 25C is therefore 1.0 x 10-14. [1.0 x 10-7][1.0 x 10-7]=1.0 x 10-14(at 25C) Although K w is defined in terms of the dissociation of water, this equilibrium constant expression is equally valid for solutions of acids and bases dissolved in water. Regardless of the source of the H 3 O+ and OH- ions in water, the product of the concentrations of these ions at equilibrium at 25C is always 1.0 x 10-14. Strong Acids and the H3 O + and OH - Ion Concentrations Suppose we add enough strong acid to a beaker of water to raise the H 3 O+ ion concentration to 0.010 M. According to LeChatelier's principle, this should drive the equilibrium between water and its ions to the left, reducing the number of H 3 O+ and OH- ions in the solution. 2 H 2 O(l) H 3 O+(aq) + OH-(aq) Because there are so many H 3 O+ ions in this solution, the change in the concentration of this ion is too small to notice. When the system returns to equilibrium, the H 3 O+ ion concentration is still about 0.010 M. Furthermore, when the reaction returns to equilibrium, the product of the H 3 O+ and OH- ion concentrations is once again equal to K w. [H 3 O+][OH-] = 1 x 10-14 The solution therefore comes back to equilibrium when the dissociation of water is so small that the OH- ion concentration is only 1.0 x 10-12 M. Adding an acid to water therefore has an effect on the concentration of both the H 3 O+ and OH- ions. Because it is a source of this ion, adding an acid to water increases the concentration of the H 3 O+ ion. Adding an acid to water, however, decreases the extent to which water dissociates. It therefore leads to a significant decrease in the concentration of the OH- ion. As might be expected, the opposite effect is observed when a base is added to water. Because we are adding a base, the OH- ion concentration increases. Once the system returns to equilibrium, the product of the H 3 O+ and OH- ion concentrations is once again equal to K w. The only way this can be achieved, of course, is by decreasing the concentration of the H 3 O+ ion.
1250	https://www.maneuveringthemiddle.com/how-to-teach-integer-operations/	Skip to primary sidebar CLICK HERE TO LEARN ABOUT MTM ALL ACCESS MEMBERSHIP FOR GRADES 6-ALGEBRA 1 Maneuvering the Middle Student-Centered Math Lessons Posted by Tyne Brack How to Teach Integer Operations 8 Sep 1.5K shares Share Some of the links in this post are affiliate links that support the content on this site. Read our disclosure statement for more information. Are you eager for students in your classroom to conceptually understand integers and master the skills required for integer operations? Keep reading and get individual sized and classroom sized vertical number line. STUDENT Background WITH INTEGERS If you are in a state that teaches Common Core, 6th grade is when the conceptual understanding of integers begins. Students are expected to be able to place and order integers on a number line, as well as determine opposites (6.NS.C.6.C). In Texas, students will be expected to do that and solve addition, subtraction, multiplication, and division problems with integers (6.3D). Mastering rational number addition, subtraction, multiplication, and division is required by the end of 7th grade in most states (7.NS.A.1). Integer Operations Activity Bundle 6th Grade TEKS Add to cart Integer Operations Unit 6th Grade TEKS Add to cart Students’ previous knowledge includes operations for positive whole numbers and positive rational numbers. If you are a little unfamiliar with this vocabulary, then I included a diagram to help. And you can real more about how to teach the real number system here. Students will not be asked to just solve these integers problems in isolation. They will be required to solve integer problems that include multiple steps, in order of operations problems, in solving for variables, and graphing. Why are integers such a struggle? Integers can be so deceiving. Students often think they are rocking it (it’s simple math, after all) only to be making the same mistakes over and over again. Unfortunately, as teachers, we can often say or do things that are actually detrimental to our students’ understanding of integers. For example, math students have heard from a young age that, “You cannot take a larger number from a smaller number” when, in fact, you can! My weakness was my scope and sequence. I had one day to teach models, so I would cram the models for all of the operations into one or two days, and then make students practice using the algorithm over and over again. Each year, I spent more time investing students into the models, and each year, I saw my students’ confidence in working with integers improve. No shame if you have taught tricks over conceptual understanding; I have! They are well intended. The tic-tac-toe to help with multiplication and division rules was taught to me! The problem with these tricks and shortcuts is that 100% of students will forget the trick or mix up the tricks with something else. If you have ever heard a student say, “Keep, change, flip” in response to how to solve an integer subtraction problem, then you know what I mean! Integer Operations Activity Bundle 6th Grade TEKS Add to cart Numbers and Operations Unit 7th Grade TEKS Add to cart Integer Operations Unit 6th Grade TEKS Add to cart Ideas for Supporting Conceptual Understanding Provide Context While the term “integers” and the concept of using operations to solve integer problems are new to students, the idea of an integer is not new. Before jumping into integer operations, provide real-world examples of integers. Students are likely familiar with the following real-life scenarios: Owing and depositing money Losing and gaining yards in football A temperature below zero Above and below sea level VERTICAL NUMBER LINES If you don’t use vertical number lines, I would highly recommend you start using them! They give more context to above and below zero. Most real-world examples are more vertically inclined: a thermometer and above and below sea level. (I use a border to clean up my not so beautiful edges.) Side note: Before jumping into showing students integer addition and subtraction using number lines, show students examples of addition and subtraction problems with positive whole numbers using number lines. I learned that some students had not been exposed to number lines for any operations. By modeling a few familiar problems, you avoid trying to teach two new skills at the same time. Geogebra is a great tool to show addition and subtraction with vertical number lines. It’s interactive, so you could have students play with the website before introducing models or the algorithm. You could ask students what patterns they see and to derive some of the integer rules. Subtracting integers has always been my Achilles’ heel. I had used “keep, change, change” to little success, but it wasn’t until I used the hot air balloon example that I saw students completely grasp why subtracting a negative would cause the answer to increase. Dropping sandbags off a hot air balloon would result in the hot air balloon moving up! (I made a very short video to demonstrate it) Download, print, and cut out vertical number lines for your students here. Counters to Show Multiple Representations Another way for students to gain more of a conceptual understanding of integers is using counters to show multiple representations of a number. For example, using counters, ask students to show -4. They might grab four red counters. Ask them to show -4 another way, and they might not know what to do, or they might say there isn’t another way. However, they could grab five red counters and one yellow counter, which is still a -4. Once students get the hang of it, you could ask them to work in pairs to produce as many representations as possible within a short time frame. By doing this, you are essentially introducing zero pairs without introducing zero pairs. Possible prompts: How can we make positive 4 with eight counters? How many different ways can you show -6? This helps build integer fluency. It also allows students to be more flexible in their thinking. Moving Toward the Algorithm Once students are feeling more confident with models, here is another hands-on way for students to dive deeper into integers. Give students a value they are trying to reach. Provide sticky notes or cards marked with a variety of integers. Students match integers to equal the given value. Similar to using counters, this allows for students to practice their fluency but also to be flexible problem solvers. What best practices do you have for teaching integers? Don’t forget to grab your printable vertical number lines here. 1.5K shares Share Vertical Number Line Grab free vertical number lines for both student and classroom use Check Out These Related Products From My Shop View All Products Previous Post Classroom Ideas from an Experienced Teacher Next Post Making the Most of Your Schoology Account Reader Interactions 3 Comments Brianna says Great ideas! Thanks for sharing the hot air balloon example, I’ve never heard that but think it will help to make sense of integer subtraction. Tyne Brack says Hi, Brianna – I’m so happy to help! 2. Brittany Clements says I love how you incorporated why students are struggling with the concept of integers and how you listed multiple ways we can support students’ conceptual understanding of integers. You mentioned that before teachers jump right into integer operations, they should provide real-world examples of integers. I absolutely agree! It is important to connect classrooms to the real world. Incorporating real world connections into a lesson helps students understand the purpose for what they are learning and how the concepts they learn are useful beyond school.
1251	https://people.uncw.edu/hermanr/mat463/odebook/book/systems.pdf	Chapter 2 Linear Systems of Differential Equations “Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater.” - Albert Einstein (1879-1955) 2.1 Coupled Systems In this chapter we will begin our study of systems of differential equa-tions. After defining first order systems, we will look at constant coefficient systems and the behavior of solutions for these systems. Also, most of the discussion will focus on planar, or two dimensional, systems. For such sys-tems we will be able to look at a variety of graphical representations of the family of solutions and discuss the qualitative features of systems we can solve in preparation for the study of systems whose solutions cannot be found in an algebraic form. However, we first turn to some simple physical problems. There are many problems in physics that can result in systems of equa-tions. This is because the most basic law of physics is given by Newton’s Second Law, which states that if a body experiences a net force, it will ac-celerate. Thus, ∑F = ma. Since a = ¨ x we have a system of second order differential equations in general for three dimensional problems, or one second order differential equation for one dimensional problems for a single mass. x k m Figure 2.1: Spring-Mass system. We have already seen reminded in the last chapter of the simple problem of a mass on a spring. This is shown in Figure 2.1. The mass slides on a frictionless surface and reacts to the restoring force of the spring attached to a wall. The restoring force of the spring given by Hooke’s Law, Fs = −kx, where k > 0 is the spring constant and x is the elongation of the spring. When the spring elongation is positive, the spring force is negative and when the spring elongation is negative the spring force is positive. The 38 differential equations equation for simple harmonic motion for the mass-spring system is found from Newton’s Second Law as m ¨ x + kx = 0. This second order equation, constant coefficient equation is easily solved using the methods in the previous chapter. However, it can also be written as a system of two first order equations in terms of the unknown position and velocity. We first set y = ˙ x. Noting that ¨ x = ˙ y, we rewrite the second order equation in terms of x and ˙ y. Thus, we have ˙ x = y ˙ y = −k m x. (2.1) One can look at more complicated spring-mass systems. Consider two blocks attached with two springs as in Figure 2.2. In this case we apply Newton’s second law for each block. We will designate the elongations of each spring from equilibrium as x1 and x2. These are shown in Figure 2.2. For mass m1, the forces acting on it are due to each spring. The first spring with spring constant k1 provides a force on m1 of −k1x1. The second spring is stretched, or compressed, based upon the relative locations of the two masses. So, the second spring will exert a force on m1 of k2(x2 −x1). Figure 2.2: System of two masses and two springs. x k m x m k 1 1 1 2 2 2 Similarly, the only force acting directly on mass m2 is provided by the restoring force from spring 2. So, that force is given by −k2(x2 −x1). The reader should think about the signs in each case. Putting this all together, we apply Newton’s Second Law to both masses. We obtain the two equations m1 ¨ x1 = −k1x1 + k2(x2 −x1) m2 ¨ x2 = −k2(x2 −x1). (2.2) Thus, we see that we have a coupled system of two second order differential equations. Each equation depends on the unknowns x1 and x2. One can rewrite this system of two second order equations as a system of four first order equations by letting x3 = ˙ x1 and x4 = ˙ x2. This leads to linear systems of differential equations 39 the system ˙ x1 = x3 ˙ x2 = x4 ˙ x3 = −k1 m1 x1 + k2 m1 (x2 −x1) ˙ x4 = −k2 m2 (x2 −x1). (2.3) As we will see in the next chapter, this system can be written more com-pactly in matrix form: d dt      x1 x2 x3 x4     =      0 0 1 0 0 0 0 1 −k1+k2 m1 k2 m1 0 0 k2 m2 −k2 m2 0 0           x1 x2 x3 x4      (2.4) We can solve this system of first order equations using matrix methods. However, we will first need to recall a few things from linear algebra. This will be done in the next chapter. For now, we will return to simpler systems and explore the behavior of typical solutions in planar systems. 2.2 Planar Systems 2.2.1 Introduction We now consider examples of solving a coupled system of first order differential equations in the plane. We will focus on the theory of linear sys-tems with constant coefficients. Understanding these simple systems will help in the study of nonlinear systems, which contain much more interest-ing behaviors, such as the onset of chaos. In the next chapter we will return to these systems and describe a matrix approach to obtaining the solutions. A general form for first order systems in the plane is given by a system of two equations for unknowns x(t) and y(t) : x′(t) = P(x, y, t) y′(t) = Q(x, y, t). (2.5) An autonomous system is one in which there is no explicit time dependence: Autonomous systems. x′(t) = P(x, y) y′(t) = Q(x, y). (2.6) Otherwise the system is called nonautonomous. A linear system takes the form x′ = a(t)x + b(t)y + e(t) y′ = c(t)x + d(t)y + f (t). (2.7) 40 differential equations A homogeneous linear system results when e(t) = 0 and f (t) = 0. A linear, constant coefficient system of first order differential equations is given by x′ = ax + by + e y′ = cx + dy + f. (2.8) We will focus on linear, homogeneous systems of constant coefficient first order differential equations: A linear, homogeneous system of con-stant coefficient first order differential equations in the plane. x′ = ax + by y′ = cx + dy. (2.9) As we will see later, such systems can result by a simple translation of the unknown functions. These equations are said to be coupled if either b ̸= 0 or c ̸= 0. We begin by noting that the system (2.9) can be rewritten as a second or-der constant coefficient linear differential equation, which we already know how to solve. We differentiate the first equation in system (2.9) and system-atically replace occurrences of y and y′, since we also know from the first equation that y = 1 b(x′ −ax). Thus, we have x′′ = ax′ + by′ = ax′ + b(cx + dy) = ax′ + bcx + d(x′ −ax). (2.10) Rewriting the last line, we have x′′ −(a + d)x′ + (ad −bc)x = 0. (2.11) This is a linear, homogeneous, constant coefficient ordinary differential equation. We know that we can solve this by first looking at the roots of the characteristic equation r2 −(a + d)r + ad −bc = 0 (2.12) and writing down the appropriate general solution for x(t). Then we can find y(t) using Equation (2.9): y = 1 b(x′ −ax). We now demonstrate this for a specific example. Example 2.1. Consider the system of differential equations x′ = −x + 6y y′ = x −2y. (2.13) linear systems of differential equations 41 Carrying out the above outlined steps, we have that x′′ + 3x′ −4x = 0. This can be shown as follows: x′′ = −x′ + 6y′ = −x′ + 6(x −2y) = −x′ + 6x −12 x′ + x 6 = −3x′ + 4x (2.14) The resulting differential equation has a characteristic equation of r2 + 3r −4 = 0. The roots of this equation are r = 1, −4. Therefore, x(t) = c1et + c2e−4t. But, we still need y(t). From the first equation of the system we have y(t) = 1 6(x′ + x) = 1 6(2c1et −3c2e−4t). Thus, the solution to the system is x(t) = c1et + c2e−4t, y(t) = 1 3c1et −1 2c2e−4t. (2.15) Sometimes one needs initial conditions. For these systems we would specify conditions like x(0) = x0 and y(0) = y0. These would allow the determination of the arbitrary constants as before. Solving systems with initial conditions. Example 2.2. Solve x′ = −x + 6y y′ = x −2y. (2.16) given x(0) = 2, y(0) = 0. We already have the general solution of this system in (2.15). In-serting the initial conditions, we have 2 = c1 + c2, 0 = 1 3c1 −1 2c2. (2.17) Solving for c1 and c2 gives c1 = 6/5 and c2 = 4/5. Therefore, the solution of the initial value problem is x(t) = 2 5 3et + 2e−4t , y(t) = 2 5 et −e−4t . (2.18) 2.2.2 Equilibrium Solutions and Nearby Behaviors In studying systems of differential equations, it is often useful to study the behavior of solutions without obtaining an algebraic form for the solution. This is done by exploring equilibrium solutions and solutions 42 differential equations nearby equilibrium solutions. Such techniques will be seen to be useful later in studying nonlinear systems. We begin this section by studying equilibrium solutions of system (2.8). For equilibrium solutions the system does not change in time. Therefore, equilibrium solutions satisfy the equations x′ = 0 and y′ = 0. Of course, this can only happen for constant solutions. Let x0 and y0 be the (constant) equilibrium solutions. Then, x0 and y0 must satisfy the system Equilibrium solutions. 0 = ax0 + by0 + e, 0 = cx0 + dy0 + f. (2.19) This is a linear system of nonhomogeneous algebraic equations. One only has a unique solution when the determinant of the system is not zero, i.e., ad −bc ̸= 0. Using Cramer’s (determinant) Rule for solving such systems, we have x0 = − e b f d a b c d , y0 = − a e c f a b c d . (2.20) If the system is homogeneous, e = f = 0, then we have that the origin is the equilibrium solution; i.e., (x0, y0) = (0, 0). Often we will have this case since one can always make a change of coordinates from (x, y) to (u, v) by u = x −x0 and v = y −y0. Then, u0 = v0 = 0. Next we are interested in the behavior of solutions near the equilibrium solutions. Later this behavior will be useful in analyzing more complicated nonlinear systems. We will look at some simple systems that are readily solved. Example 2.3. Stable Node (sink) Consider the system x′ = −2x y′ = −y. (2.21) This is a simple uncoupled system. Each equation is simply solved to give x(t) = c1e−2t and y(t) = c2e−t. In this case we see that all solutions tend towards the equilibrium point, (0, 0). This will be called a stable node, or a sink. Before looking at other types of solutions, we will explore the stable node in the above example. There are several methods of looking at the behavior of solutions. We can look at solution plots of the dependent versus the independent variables, or we can look in the xy-plane at the parametric curves (x(t), y(t)). Solution Plots: One can plot each solution as a function of t given a set of initial conditions. Examples are shown in Figure 2.3 for several initial linear systems of differential equations 43 conditions. Note that the solutions decay for large t. Special cases result for various initial conditions. Note that for t = 0, x(0) = c1 and y(0) = c2. (Of course, one can provide initial conditions at any t = t0. It is generally easier to pick t = 0 in our general explanations.) If we pick an initial condition with c1=0, then x(t) = 0 for all t. One obtains similar results when setting y(0) = 0. Figure 2.3: Plots of solutions of Example 2.3 for several initial conditions. Phase Portrait: There are other types of plots which can provide addi-tional information about the solutions even if we cannot find the exact so-lutions as we can for these simple examples. In particular, one can consider the solutions x(t) and y(t) as the coordinates along a parameterized path, or curve, in the plane: r = (x(t), y(t)) Such curves are called trajectories or orbits. The xy-plane is called the phase plane and a collection of such orbits gives a phase portrait for the family of solutions of the given system. One method for determining the equations of the orbits in the phase plane is to eliminate the parameter t between the known solutions to get a relationship between x and y. Since the solutions are known for the last example, we can do this, since the solutions are known. In particular, we have x = c1e−2t = c1 y c2 2 ≡Ay2. Another way to obtain information about the orbits comes from noting that the slopes of the orbits in the xy-plane are given by dy/dx. For au-tonomous systems, we can write this slope just in terms of x and y. This leads to a first order differential equation, which possibly could be solved analytically or numerically. First we will obtain the orbits for Example 2.3 by solving the correspond-ing slope equation. Recall that for trajectories defined parametrically by x = x(t) and y = y(t), we have from the Chain Rule for y = y(x(t)) that dy dt = dy dx dx dt . Therefore, The Slope of a parametric curve. dy dx = dy dt dx dt . (2.22) Figure 2.4: Orbits for Example 2.3. For the system in (2.21) we use Equation (2.22) to obtain the equation for the slope at a point on the orbit: dy dx = y 2x. The general solution of this first order differential equation is found using separation of variables as x = Ay2 for A an arbitrary constant. Plots of these solutions in the phase plane are given in Figure 2.4. [Note that this is the same form for the orbits that we had obtained above by eliminating t from the solution of the system.] Once one has solutions to differential equations, we often are interested in the long time behavior of the solutions. Given a particular initial condition 44 differential equations (x0, y0), how does the solution behave as time increases? For orbits near an equilibrium solution, do the solutions tend towards, or away from, the equilibrium point? The answer is obvious when one has the exact solutions x(t) and y(t). However, this is not always the case. Let’s consider the above example for initial conditions in the first quad-rant of the phase plane. For a point in the first quadrant we have that dx/dt = −2x < 0, meaning that as t →∞, x(t) get more negative. Similarly, dy/dt = −y < 0, indicating that y(t) is also getting smaller for this problem. Thus, these orbits tend towards the origin as t →∞. This qualitative information was obtained without relying on the known solutions to the problem. x y (1, 2) (−1, 2) (−1, −2) (1, −2) (1, 1) (−1, 1) (−1, −1) (1, −1) Figure 2.5: Sketch of tangent vectors us-ing Example 2.3. Direction Fields: Another way to determine the behavior of the solutions of the system of differential equations is to draw the direction field. A direction field is a vector field in which one plots arrows in the direction of tangents to the orbits at selected points in the plane. This is done because the slopes of the tangent lines are given by dy/dx. For the general system (2.9), the slope is dy dx = cx + dy ax + by. This is a first order differential equation which can be solved as we show in the following examples. Example 2.4. Draw the direction field for Example 2.3. Figure 2.6: Direction field for Example 2.3. We can use software to draw direction fields. However, one can sketch these fields by hand. We have that the slope of the tangent at this point is given by dy dx = −y −2x = y 2x. For each point in the plane one draws a piece of tangent line with this slope. In Figure 2.5 we show a few of these. For (x, y) = (1, 1) the slope is dy/dx = 1/2. So, we draw an arrow with slope 1/2 at this point. From system (2.21), we have that x′ and y′ are both negative at this point. Therefore, the vector points down and to the left. We can do this for several points, as shown in Figure 2.5. Sometimes one can quickly sketch vectors with the same slope. For this example, when y = 0, the slope is zero and when x = 0 the slope is infinite. So, several vectors can be provided. Such vectors are tangent to curves known as isoclines in which dy dx =constant. Figure 2.7: Phase portrait for Example 2.3. This is a stable node, or sink It is often difficult to provide an accurate sketch of a direction field. Com-puter software can be used to provide a better rendition. For Example 2.3 the direction field is shown in Figure 2.6. Looking at this direction field, one can begin to “see” the orbits by following the tangent vectors. linear systems of differential equations 45 Of course, one can superimpose the orbits on the direction field. This is shown in Figure 2.7. Are these the patterns you saw in Figure 2.6? In this example we see all orbits “flow” towards the origin, or equilibrium point. Again, this is an example of what is called a stable node or a sink. (Imagine what happens to the water in a sink when the drain is unplugged.) This is another uncoupled system. The solutions are again simply gotten by integration. We have that x(t) = c1e−t and y(t) = c2et. Here we have that x decays as t gets large and y increases as t gets large. In particular, if one picks initial conditions with c2 = 0, then orbits follow the x-axis towards the origin. For initial points with c1 = 0, orbits originating on the y-axis will flow away from the origin. Of course, in these cases the origin is an equilibrium point and once at equilibrium, one remains there. In fact, there is only one line on which to pick initial conditions such that the orbit leads towards the equilibrium point. No matter how small c2 is, sooner, or later, the exponential growth term will dominate the solution. One can see this behavior in Figure 2.8. Figure 2.8: Plots of solutions of Example 2.5 for several initial conditions. Example 2.5. Saddle Consider the system x′ = −x y′ = y. (2.23) Similar to the first example, we can look at plots of solutions orbits in the phase plane. These are given by Figures 2.8-2.9. The orbits can be obtained from the system as dy dx = dy/dt dx/dt = −y x. The solution is y = A x . For different values of A ̸= 0 we obtain a family of hyperbolae. These are the same curves one might obtain for the level curves of a surface known as a saddle surface, z = xy. Thus, this type of equilibrium point is classified as a saddle point. From the phase portrait we can verify that there are many orbits that lead away from the origin (equilibrium point), but there is one line of initial conditions that leads to the origin and that is the x-axis. In this case, the line of initial conditions is given by the x-axis. Figure 2.9: Phase portrait for Example 2.5. This is a saddle. Example 2.6. Unstable Node (source) x′ = 2x y′ = y. (2.24) This example is similar to Example 2.3. The solutions are obtained by replacing t with −t. The solutions, orbits, and direction fields can be seen in Figures 2.10-2.11. This is once again a node, but all orbits lead away from the equilibrium point. It is called an unstable node or a source. Figure 2.10: Plots of solutions of Exam-ple 2.6 for several initial conditions. 46 differential equations Example 2.7. Center x′ = y y′ = −x. (2.25) Figure 2.11: Phase portrait for Example 2.6, an unstable node or source. Figure 2.12: Plots of solutions of Exam-ple 2.7 for several initial conditions. This system is a simple, coupled system. Neither equation can be solved without some information about the other unknown function. However, we can differentiate the first equation and use the second equation to obtain x′′ + x = 0. We recognize this equation as one that appears in the study of simple harmonic motion. The solutions are pure sinusoidal oscillations: x(t) = c1 cos t + c2 sin t, y(t) = −c1 sin t + c2 cos t. In the phase plane the trajectories can be determined either by look-ing at the direction field, or solving the first order equation dy dx = −x y. Performing a separation of variables and integrating, we find that x2 + y2 = C. Thus, we have a family of circles for C > 0. (Can you prove this using the general solution?) Looking at the results graphically in Figures 2.12-2.13 confirms this result. This type of point is called a center. Figure 2.13: Phase portrait for Example 2.7, a center. Example 2.8. Focus (spiral) x′ = αx + y y′ = −x. (2.26) In this example, we will see an additional set of behaviors of equi-librium points in planar systems. We have added one term, αx, to the system in Example 2.7. We will consider the effects for two spe-cific values of the parameter: α = 0.1, −0.2. The resulting behaviors are shown in the Figures 2.15-2.18. We see orbits that look like spi-rals. These orbits are stable and unstable spirals (or foci, the plural of focus.) We can understand these behaviors by once again relating the sys-tem of first order differential equations to a second order differential equation. Using the usual method for obtaining a second order equa-tion form a system, we find that x(t) satisfies the differential equation x′′ −αx′ + x = 0. linear systems of differential equations 47 We recall from our first course that this is a form of damped simple harmonic motion. The characteristic equation is r2 −αr + 1 = 0. The solution of this quadratic equation is r = α ± √ α2 −4 2 . Figure 2.14: Plots of solutions of Ex-ample 2.8 for several initial conditions, α = −0.2. Figure 2.15: Plots of solutions of Ex-ample 2.8 for several initial conditions, α = 0.1. There are five special cases to consider as shown in the below clas-sification. Classification of Solutions of x′′ −αx′ + x = 0 1. α = −2. There is one real solution. This case is called critical damping since the solution r = −1 leads to exponential decay. The solution is x(t) = (c1 + c2t)e−t. 2. α < −2. There are two real, negative solutions, r = −µ, −ν, µ, ν > 0. The solution is x(t) = c1e−µt + c2e−νt. In this case we have what is called overdamped motion. There are no oscillations 3. −2 < α < 0. There are two complex conjugate solutions r = α/2 ± iβ with real part less than zero and β = √ 4−α2 2 . The solution is x(t) = (c1 cos βt + c2 sin βt)eαt/2. Since α < 0, this consists of a decaying expo-nential times oscillations. This is often called an underdamped oscillation. 4. α = 0. This leads to simple harmonic motion. 5. 0 < α < 2. This is similar to the underdamped case, except α > 0. The solutions are growing oscillations. 6. α = 2. There is one real solution. The solution is x(t) = (c1 + c2t)et. It leads to unbounded growth in time. 7. For α > 2. There are two real, positive solutions r = µ, ν > 0. The solution is x(t) = c1eµt + c2eνt, which grows in time. Figure 2.16: Phase portrait for 2.9. This is a degenerate node. For α < 0 the solutions are losing energy, so the solutions can oscil-late with a diminishing amplitude. (See Figure 2.14.) For α > 0, there is a growth in the amplitude, which is not typical. (See Figure 2.15.) Of course, there can be overdamped motion if the magnitude of α is too large. Example 2.9. Degenerate Node For this example, we will write out the solutions. It is a coupled system for which only the second equa-tion is coupled. x′ = −x y′ = −2x −y. (2.27) There are two possible approaches: 48 differential equations a. We could solve the first equation to find x(t) = c1e−t. Inserting this solution into the second equation, we have y′ + y = −2c1e−t. This is a relatively simple linear first order equation for y = y(t). The integrating factor is µ = et. The solution is found as y(t) = (c2 − 2c1t)e−t. Figure 2.17: Phase portrait for Example 2.8 with α = −0.2. This is a stable focus, or spiral. b. Another method would be to proceed to rewrite this as a second order equation. Computing x′′ does not get us very far. So, we look at y′′ = −2x′ −y′ = 2x −y′ = −2y′ −y. (2.28) Therefore, y satisfies y′′ + 2y′ + y = 0. The characteristic equation has one real root, r = −1. So, we write y(t) = (k1 + k2t)e−t. This is a stable degenerate node. Combining this with the solution x(t) = c1e−t, we can show that y(t) = (c2 −2c1t)e−t as before. Figure 2.18: Phase portrait for Example 2.9. This is a degenerate node. In Figure 2.16 we see several orbits in this system. It differs from the stable node show in Figure 2.4 in that there is only one direction along which the orbits approach the origin instead of two. If one picks c1 = 0, then x(t) = 0 and y(t) = c2e−t. This leads to orbits running along the y-axis as seen in the figure. x K 3 K 2 K 1 0 1 2 3 y K 3 K 2 K 1 1 2 3 Figure 2.19: Plots of direction field of Ex-ample 2.10. Example 2.10. A Line of Equilibria, Zero Root x′ = 2x −y y′ = −2x + y. (2.29) In this last example, we have a coupled set of equations. We rewrite it as a second order differential equation: x′′ = 2x′ −y′ = 2x′ −(−2x + y) = 2x′ + 2x + (x′ −2x) = 3x′. (2.30) So, the second order equation is x′′ −3x′ = 0 and the characteristic equation is 0 = r(r −3). This gives the general solution as x(t) = c1 + c2e3t linear systems of differential equations 49 and thus y = 2x −x′ = 2(c1 + c2e3t) −(3c2e3t) = 2c1 −c2e3t. In Figure 2.19 we show the direction field. The constant slope field seen in this example is confirmed by a simple computation: dy dx = −2x + y 2x −y = −1. Furthermore, looking at initial conditions with y = 2x, we have at t = 0, 2c1 −c2 = 2(c1 + c2) ⇒ c2 = 0. Therefore, points on this line remain on this line forever, (x, y) = (c1, 2c1). This line of fixed points is called a line of equilibria. 2.2.3 Polar Representation of Spirals In the examples with a center or a spiral, one might be able to write the solutions in polar coordinates. Recall that a point in the plane can be described by either Cartesian (x, y) or polar (r, θ) coordinates. Given the polar form, one can find the Cartesian components using x = r cos θ and y = r sin θ. Given the Cartesian coordinates, one can find the polar coordinates using r2 = x2 + y2 and tan θ = y x. (2.31) Since x and y are functions of t, then naturally we can think of r and θ as functions of t. Converting a system of equations in the plane for x′ and y′ to polar form requires knowing r′ and θ′. So, we first find expressions for r′ and θ′ in terms of x′ and y′. Differentiating the first equation in (2.31) gives rr′ = xx′ + yy′. Inserting the expressions for x′ and y′ from system 2.9, we have rr′ = x(ax + by) + y(cx + dy). In some cases this may be written entirely in terms of r’s. Similarly, we have that θ′ = xy′ −yx′ r2 , which the reader can prove for homework. In summary, when converting first order equations from rectangular to polar form, one needs the relations below. 50 differential equations Derivatives of Polar Variables r′ = xx′ + yy′ r , θ′ = xy′ −yx′ r2 . (2.32) Example 2.11. Rewrite the following system in polar form and solve the resulting system. x′ = ax + by y′ = −bx + ay. (2.33) We first compute r′ and θ′: rr′ = xx′ + yy′ = x(ax + by) + y(−bx + ay) = ar2. r2θ′ = xy′ −yx′ = x(−bx + ay) −y(ax + by) = −br2. This leads to simpler system r′ = ar θ′ = −b. (2.34) This system is uncoupled. The second equation in this system in-dicates that we traverse the orbit at a constant rate in the clockwise direction. Solving these equations, we have that r(t) = r0eat, θ(t) = θ0 −bt. Eliminating t between these solutions, we finally find the polar equation of the orbits: r = r0e−a(θ−θ0)t/b. If you graph this for a ̸= 0, you will get stable or unstable spirals. Example 2.12. Consider the specific system x′ = −y + x y′ = x + y. (2.35) In order to convert this system into polar form, we compute rr′ = xx′ + yy′ = x(−y + x) + y(x + y) = r2. r2θ′ = −xy′ −yx′ = x(x + y) −y(−y + x) = r2. This leads to simpler system r′ = r θ′ = 1. (2.36) Solving these equations yields r(t) = r0et, θ(t) = t + θ0. Eliminating t from this solution gives the orbits in the phase plane, r(θ) = r0eθ−θ0. linear systems of differential equations 51 A more complicated example arises for a nonlinear system of differential equations. Consider the following example. Example 2.13. x′ = −y + x(1 −x2 −y2) y′ = x + y(1 −x2 −y2). (2.37) Transforming to polar coordinates, one can show that in order to convert this system into polar form, we compute r′ = r(1 −r2), θ′ = 1. This uncoupled system can be solved and this is left to the reader. 2.3 Applications In this section we will describe some simple applications leading to systems of differential equations which can be solved using the methods in this chapter. These systems are left for homework problems and the as the start of further explorations for student projects. 2.3.1 Mass-Spring Systems The first examples that we had seen involved masses on springs. Re-call that for a simple mass on a spring we studied simple harmonic motion, which is governed by the equation m ¨ x + kx = 0. This second order equation can be written as two first order equations ˙ x = y ˙ y = −k m x, (2.38) or ˙ x = y ˙ y = −ω2x, (2.39) where ω2 = k m. The coefficient matrix for this system is A = 0 1 −ω2 0 ! . We also looked at the system of two masses and two springs as shown in Figure 2.20. The equations governing the motion of the masses is m1 ¨ x1 = −k1x1 + k2(x2 −x1) m2 ¨ x2 = −k2(x2 −x1). (2.40) 52 differential equations Figure 2.20: System of two masses and two springs. x k m x m k 1 1 1 2 2 2 We can rewrite this system as four first order equations ˙ x1 = x3 ˙ x2 = x4 ˙ x3 = −k1 m1 x1 + k2 m1 (x2 −x1) ˙ x4 = −k2 m2 (x2 −x1). (2.41) The coefficient matrix for this system is A =      0 0 1 0 0 0 0 1 −k1+k2 m1 k2 m1 0 0 k2 m2 −k2 m2 0 0     . We can study this system for specific values of the constants using the meth-ods covered in the last sections. Writing the spring-block system as a sec-ond order vector system. Actually, one can also put the system (2.40) in the matrix form m1 0 0 m2 ! ¨ x1 ¨ x2 ! = −(k1 + k2) k2 k2 −k2 ! x1 x2 ! . (2.42) This system can then be written compactly as M¨ x = −Kx, (2.43) where M = m1 0 0 m2 ! , K = k1 + k2 −k2 −k2 k2 ! . This system can be solved by guessing a form for the solution. We could guess x = aeiωt or x = a1 cos(ωt −δ1) a2 cos(ωt −δ2) ! , where δi are phase shifts determined from initial conditions. linear systems of differential equations 53 Inserting x = aeiωt into the system gives (K −ω2M)a = 0. This is a homogeneous system. It is a generalized eigenvalue problem for eigenvalues ω2 and eigenvectors a. We solve this in a similar way to the standard matrix eigenvalue problems. The eigenvalue equation is found as det (K −ω2M) = 0. Once the eigenvalues are found, then one determines the eigenvectors and constructs the solution. Example 2.14. Let m1 = m2 = m and k1 = k2 = k. Then, we have to solve the system ω2 m 0 0 m ! a1 a2 ! = 2k −k −k k ! a1 a2 ! . The eigenvalue equation is given by 0 = 2k −mω2 −k −k k −mω2 = (2k −mω2)(k −mω2) −k2 = m2ω4 −3kmω2 + k2. (2.44) Solving this quadratic equation for ω2, we have ω2 = 3 ± 1 2 k m. For positive values of ω, one can show that ω = 1 2 ±1 + √ 5 r k m. The eigenvectors can be found for each eigenvalue by solving the homogeneous system 2k −mω2 −k −k k −mω2 ! a1 a2 ! = 0. The eigenvectors are given by a1 = − √ 5+1 2 1 ! , a2 = √ 5−1 2 1 ! . We are now ready to construct the real solutions to the problem. Similar to solving two first order systems with complex roots, we take the real and imaginary parts and take a linear combination of the so-lutions. In this problem there are four terms, giving the solution in the form x(t) = c1a1cosω1t + c2a1sinω1t + c3a2cosω2t + c4a2sinω2t, where the ω’s are the eigenvalues and the a’s are the corresponding eigenvectors. The constants are determined from the initial conditions, x(0) = x0 and ˙ x(0) = v0. 54 differential equations 2.3.2 Circuits In the last chapter we investigated simple series LRC circuits. More complicated circuits are possible by looking at parallel connections, or other combinations, of resistors, capacitors and inductors. This results in several equations for each loop in the circuit, leading to larger systems of differential equations. An example of another circuit setup is shown in Figure 2.21. This is not a problem that can be covered in the first year physics course. There are two loops, indicated in Figure 2.22 as traversed clockwise. For each loop we need to apply Kirchoff’s Loop Rule. There are three oriented currents, labeled Ii, i = 1, 2, 3. Corresponding to each current is a changing charge, qi such that Ii = dqi dt , i = 1, 2, 3. We have for loop one I1R1 + q2 C = V(t) (2.45) and for loop two I3R2 + LdI3 dt = q2 C . (2.46) − + V(t) R1 R2 L C Figure 2.21: A circuit with two loops containing several different circuit ele-ments. There are three unknown functions for the charge. Once we know the charge functions, differentiation will yield the three currents. However, we only have two equations. We need a third equation. This equation is found from Kirchoff’s Point (Junction) Rule. − + V(t) R1 R2 L A C B I1 I3 I2 1 2 Figure 2.22: The previous parallel circuit with the directions indicated for travers-ing the loops in Kirchoff’s Laws. Consider the points A and B in Figure 2.22. Any charge (current) entering these junctions must be the same as the total charge (current) leaving the junctions. For point A we have I1 = I2 + I3, (2.47) or ˙ q1 = ˙ q2 + ˙ q3. (2.48) Equations (2.45), (2.46), and (2.48) form a coupled system of differential equations for this problem. There are both first and second order derivatives involved. We can write the whole system in terms of charges as R1 ˙ q1 + q2 C = V(t) R2 ˙ q3 + L ¨ q3 = q2 C ˙ q1 = ˙ q2 + ˙ q3. (2.49) The question is whether, or not, we can write this as a system of first order differential equations. Since there is only one second order derivative, we can introduce the new variable q4 = ˙ q3. The first equation can be solved for ˙ q1. The third equation can be solved for ˙ q2 with appropriate substitutions linear systems of differential equations 55 for the other terms. ˙ q3 is gotten from the definition of q4 and the second equation can be solved for ¨ q3 and substitutions made to obtain the system ˙ q1 = V R1 − q2 R1C ˙ q2 = V R1 − q2 R1C −q4 ˙ q3 = q4 ˙ q4 = q2 LC −R2 L q4. So, we have a nonhomogeneous first order system of differential equa-tions. 2.3.3 Mixture Problems There are many types of mixture problems. Such problems are standard in a first course on differential equations as examples of first order differential equations. Typically these examples consist of a tank of brine, water con-taining a specific amount of salt with pure water entering and the mixture leaving, or the flow of a pollutant into, or out of, a lake. We first saw such problems in Chapter 1. In general one has a rate of flow of some concentration of mixture enter-ing a region and a mixture leaving the region. The goal is to determine how much stuff is in the region at a given time. This is governed by the equation Rate of change of substance = Rate In −Rate Out. This can be generalized to the case of two interconnected tanks. We will pro-vide an example, but first we review the single tank problem from Chapter 1. Example 2.15. Single Tank Problem A 50 gallon tank of pure water has a brine mixture with concentra-tion of 2 pounds per gallon entering at the rate of 5 gallons per minute. [See Figure 2.23.] At the same time the well-mixed contents drain out at the rate of 5 gallons per minute. Find the amount of salt in the tank at time t. In all such problems one assumes that the solution is well mixed at each instant of time. Figure 2.23: A typical mixing problem. Let x(t) be the amount of salt at time t. Then the rate at which the salt in the tank increases is due to the amount of salt entering the tank less that leaving the tank. To figure out these rates, one notes that dx/dt has units of pounds per minute. The amount of salt entering per minute is given by the product of the entering concentration times the rate at which the brine enters. This gives the correct units: 2pounds gal 5 gal min = 10pounds min . Similarly, one can determine the rate out as x pounds 50 gal 5 gal min = x 10 pounds min . 56 differential equations Thus, we have dx dt = 10 −x 10. This equation is easily solved using the methods for first order equations. Example 2.16. Double Tank Problem One has two tanks connected together, labeled tank X and tank Y, as shown in Figure 2.24. Figure 2.24: The two tank problem. X Y Let tank X initially have 100 gallons of brine made with 100 pounds of salt. Tank Y initially has 100 gallons of pure water. Pure water is pumped into tank X at a rate of 2.0 gallons per minute. Some of the mixture of brine and pure water flows into tank Y at 3 gallons per minute. To keep the tank levels the same, one gallon of the Y mixture flows back into tank X at a rate of one gallon per minute and 2.0 gallons per minute drains out. Find the amount of salt at any given time in the tanks. What happens over a long period of time? In this problem we set up two equations. Let x(t) be the amount of salt in tank X and y(t) the amount of salt in tank Y. Again, we carefully look at the rates into and out of each tank in order to set up the system of differential equations. We obtain the system dx dt = y 100 −3x 100 dy dt = 3x 100 −3y 100. (2.50) This is a linear, homogenous constant coefficient system of two first order equations, which we know how to solve. The matrix form of the system is given by ˙ x = −3 100 1 100 3 100 −3 100 ! x, x(0) = 100 0 ! . The eigenvalues for the problem are given by λ = −3 ± √ 3 and the eigenvectors are 1 ± √ 3 ! . linear systems of differential equations 57 Since the eigenvalues are real and distinct, the general solution is easily written down: x(t) = c1 1 √ 3 ! e(−3+ √ 3)t + c2 1 − √ 3 ! e(−3− √ 3)t. Finally, we need to satisfy the initial conditions. So, x(0) = c1 1 √ 3 ! + c2 1 − √ 3 ! = 100 0 ! , or c1 + c2 = 100, (c1 −c2) √ 3 = 0. So, c2 = c1 = 50. The final solution is x(t) = 50 1 √ 3 ! e(−3+ √ 3)t + 1 − √ 3 ! e(−3− √ 3)t ! , or x(t) = 50 e(−3+ √ 3)t + e(−3− √ 3)t y(t) = 50 √ 3 e(−3+ √ 3)t −e(−3− √ 3)t . (2.51) 2.3.4 Chemical Kinetics There are many problems in the chemistry of chemical reactions which lead to systems of differential equations. The simplest reaction is when a chemical A turns into chemical B. This happens at a certain rate, k > 0. This reaction can be represented by the chemical formula A k / B. In this case we have that the rates of change of the concentrations of A, [A], and B, [B], are given by The chemical reactions used in these ex-amples are first order reactions. Second order reactions have rates proportional to the square of the concentration. d[A] dt = −k[A] d[B] dt = k[A] (2.52) Think about this as it is a key to understanding the next reactions. A more complicated reaction is given by A k1 / B k2 / C. Here there are three concentrations and two rates of change. The system of equations governing the reaction is d[A] dt = −k1[A], d[B] dt = k1[A] −k2[B], d[C] dt = k2[B]. (2.53) 58 differential equations The more complication rate of change is when [B] increases from [A] chang-ing to [B] and decrease when [B] changes to [C]. Thus, there are two terms in the rate of change equation for concentration [B]. One can further consider reactions in which a reverse reaction is possible. Thus, a further generalization occurs for the reaction A k1 / B k3 o k2 / C. The reverse reaction rates contribute to the reaction equations for [A] and [B]. The resulting system of equations is d[A] dt = −k1[A] + k3[B], d[B] dt = k1[A] −k2[B] −k3[B], d[C] dt = k2[B]. (2.54) Nonlinear chemical reactions will be discussed in the next chapter. 2.3.5 Predator Prey Models Another common population model is that describing the coexistence of species. For example, we could consider a population of rabbits and foxes. Left to themselves, rabbits would tend to multiply, thus dR dt = aR, with a > 0. In such a model the rabbit population would grow exponentially. Similarly, a population of foxes would decay without the rabbits to feed on. So, we have that dF dt = −bF for b > 0. Now, if we put these populations together on a deserted island, they would interact. The more foxes, the rabbit population would decrease. However, the more rabbits, the foxes would have plenty to eat and the pop-ulation would thrive. Thus, we could model the competing populations as dR dt = aR −cF, dF dt = −bF + dR, (2.55) where all of the constants are positive numbers. Studying this coupled system would lead to a study of the dynamics of these populations. The nonlinear version of this system, the Lotka-Volterra model, will be discussed in the next chapter. linear systems of differential equations 59 2.3.6 Love Affairs The next application is one that was introduced in 1988 by Stro-gatz as a cute system involving relationships.1 One considers what happens 1 Steven H. Strogatz introduced this problem as an interesting example of systems of differential equations in Mathematics Magazine, Vol. 61, No. 1 (Feb. 1988) p 35. He also describes it in his book Nonlinear Dynamics and Chaos (1994). to the affections that two people have for each other over time. Let R de-note the affection that Romeo has for Juliet and J be the affection that Juliet has for Romeo. Positive values indicate love and negative values indicate dislike. One possible model is given by dR dt = bJ dJ dt = cR (2.56) with b > 0 and c < 0. In this case Romeo loves Juliet the more she likes him. But Juliet backs away when she finds his love for her increasing. A typical system relating the combined changes in affection can be mod-eled as dR dt = aR + bJ dJ dt = cR + dJ. (2.57) Several scenarios are possible for various choices of the constants. For ex-ample, if a > 0 and b > 0, Romeo gets more and more excited by Juliet’s love for him. If c > 0 and d < 0, Juliet is being cautious about her relationship with Romeo. For specific values of the parameters and initial conditions, one can explore this match of an overly zealous lover with a cautious lover. 2.3.7 Epidemics Another interesting area of application of differential equation is in predicting the spread of disease. Typically, one has a population of sus-ceptible people or animals. Several infected individuals are introduced into the population and one is interested in how the infection spreads and if the number of infected people drastically increases or dies off. Such models are typically nonlinear and we will look at what is called the SIR model in the next chapter. In this section we will model a simple linear model. Let us break the population into three classes. First, we let S(t) represent the healthy people, who are susceptible to infection. Let I(t) be the number of infected people. Of these infected people, some will die from the infection and others could recover. We will consider the case that initially there is one infected person and the rest, say N, are healthy. Can we predict how many deaths have occurred by time t? We model this problem using the compartmental analysis we had seen for mixing problems. The total rate of change of any population would be 60 differential equations due to those entering the group less those leaving the group. For example, the number of healthy people decreases due infection and can increase when some of the infected group recovers. Let’s assume that a) the rate of infection is proportional to the number of healthy people, aS, and b) the number who recover is proportional to the number of infected people, rI. Thus, the rate of change of healthy people is found as dS dt = −aS + rI. Let the number of deaths be D(t). Then, the death rate could be taken to be proportional to the number of infected people. So, dD dt = dI Finally, the rate of change of infected people is due to healthy people getting infected and the infected people who either recover or die. Using the corresponding terms in the other equations, we can write the rate of change of infected people as dI dt = aS −rI −dI. This linear system of differential equations can be written in matrix form. d dt    S I D   =    −a r 0 a −d −r 0 0 d 0       S I D   . (2.58) The reader can find the solutions of this system and determine if this is a realistic model. 2.4 First Order Matrix Differential Equations 2.4.1 Matrix Formulation We have investigated several linear systems in the plane and in the next chapter we will use some of these ideas to investigate nonlinear systems. We need a deeper insight into the solutions of planar systems. So, in this section we will recast the first order linear systems into matrix form. This will lead to a better understanding of first order systems and allow for extensions to higher dimensions and the solution of nonhomogeneous equations later in this chapter. We start with the usual homogeneous system in Equation (2.9). Let the unknowns be represented by the vector x(t) = x(t) y(t) ! . Then we have that x′ = x′ y′ ! = ax + by cx + dy ! = a b c d ! x y ! ≡Ax. linear systems of differential equations 61 Here we have introduced the coefficient matrix A. This is a first order vector differential equation, x′ = Ax. Formerly, we can write the solution as x = x0eAt. You can verify that this is a solution by simply differentiating, dx dt = x0 d dt eAt = Ax0eAt = Ax. 2.4.2 Exponentiating a Matrix However, there remains the question, “What does it mean to ex-ponentiate a matrix?” The exponential of a matrix is defined using the Maclaurin series expansion ex = ∞ ∑ k=0 = 1 + x + x2 2! + x3 3! + · · · . We define The exponential of a matrix is defined us-ing the Maclaurin series expansion ex = ∞ ∑ k=0 xn n! = 1 + x + x2 2! + x3 3! + · · · . So, we define eA = I + A + A2 2! + A3 3! + · · · . (2.59) In general, it is difficult computing eA unless A is diagonal. eA = ∞ ∑ k=0 1 n! An = I + A + A2 2! + A3 3! + · · · . (2.60) In general it is difficult to sum this series, but it is doable for some simple examples. Example 2.17. Evaluate etA for A = 1 0 0 2 ! . etA = I + tA + t2 2! A2 + t3 3! A3 + · · · . = 1 0 0 1 ! + t 1 0 0 2 ! + t2 2! 1 0 0 2 !2 + t3 3! 1 0 0 2 !3 + · · · = 1 0 0 1 ! + t 1 0 0 2 ! + t2 2! 1 0 0 4 ! + t3 3! 1 0 0 8 ! + · · · = 1 + t + t2 2! + t3 3! · · · 0 0 1 + 2t + 2t2 2! + 8t3 3! · · · ! = et 0 0 e2t ! (2.61) Example 2.18. Evaluate etA for A = 0 1 1 0 ! . We first note that A2 = 0 1 1 0 ! 0 1 1 0 ! = 1 0 0 1 ! = I. 62 differential equations Therefore, An = ( A, n odd, I, n even. Then, we have etA = I + tA + t2 2! A2 + t3 3! A3 + · · · . = I + tA + t2 2! I + t3 3! A + · · · . = 1 + t2 2! + t4 4! · · · t + t3 3! + t5 5! · · · t + t3 3! + t5 5! · · · 1 + t2 2! + t4 4! · · · ! = cosh sinh t sinh t cosh t ! . (2.62) Since summing these infinite series might be difficult, we will now inves-tigate the solutions of planar systems to see if we can find other approaches for solving linear systems using matrix methods. We begin by recalling the solution to the problem in Example (2.16). We obtained the solution to this system as x(t) = c1et + c2e−4t, y(t) = 1 3c1et −1 2c2e−4t. (2.63) This can be rewritten using matrix operations. Namely, we first write the solution in vector form. x = x(t) y(t) ! = c1et + c2e−4t 1 3c1et −1 2c2e−4t ! = c1et 1 3c1et ! + c2e−4t −1 2c2e−4t ! = c1 1 1 3 ! et + c2 1 −1 2 ! e−4t. (2.64) We see that our solution is in the form of a linear combination of vectors of the form x = veλt with v a constant vector and λ a constant number. This is similar to how we began to find solutions to second order constant coefficient equations. So, for the general problem (2.4.1) we insert this guess. Thus, x′ = Ax ⇒ λveλt = Aveλt. (2.65) For this to be true for all t, we have that Av = λv. (2.66) linear systems of differential equations 63 This is an eigenvalue problem. A is a 2 × 2 matrix for our problem, but could easily be generalized to a system of n first order differential equa-tions. We will confine our remarks for now to planar systems. However, we need to recall how to solve eigenvalue problems and then see how solutions of eigenvalue problems can be used to obtain solutions to our systems of differential equations.. 2.4.3 Eigenvalue Problems We seek nontrivial solutions to the eigenvalue problem Av = λv. (2.67) We note that v = 0 is an obvious solution. Furthermore, it does not lead to anything useful. So, it is called a trivial solution. Typically, we are given the matrix A and have to determine the eigenvalues, λ, and the associated eigenvectors, v, satisfying the above eigenvalue problem. Later in the course we will explore other types of eigenvalue problems. For now we begin to solve the eigenvalue problem for v = v1 v2 ! . Inserting this into Equation (2.67), we obtain the homogeneous algebraic system (a −λ)v1 + bv2 = 0, cv1 + (d −λ)v2 = 0. (2.68) The solution of such a system would be unique if the determinant of the system is not zero. However, this would give the trivial solution v1 = 0, v2 = 0. To get a nontrivial solution, we need to force the determinant to be zero. This yields the eigenvalue equation 0 = a −λ b c d −λ = (a −λ)(d −λ) −bc. This is a quadratic equation for the eigenvalues that would lead to nontrivial solutions. If we expand the right side of the equation, we find that λ2 −(a + d)λ + ad −bc = 0. This is the same equation as the characteristic equation (2.12) for the gen-eral constant coefficient differential equation considered in the first chapter. Thus, the eigenvalues correspond to the solutions of the characteristic poly-nomial for the system. Once we find the eigenvalues, then there are possibly an infinite number solutions to the algebraic system. We will see this in the examples. So, the process is to a) Write the coefficient matrix; b) Find the eigenvalues from the equation det(A −λI) = 0; and, c) Find the eigenvectors by solving the linear system (A −λI)v = 0 for each λ. 64 differential equations 2.5 Solving Constant Coefficient Systems in 2D Before proceeding to examples, we first indicate the types of solutions that could result from the solution of a homogeneous, constant coefficient system of first order differential equations. We begin with the linear system of differential equations in matrix form. dx dt = a b c d ! x = Ax. (2.69) The type of behavior depends upon the eigenvalues of matrix A. The pro-cedure is to determine the eigenvalues and eigenvectors and use them to construct the general solution. If we have an initial condition, x(t0) = x0, we can determine the two arbitrary constants in the general solution in order to obtain the particular solution. Thus, if x1(t) and x2(t) are two linearly independent solutions2, 2 Recall that linear independence means c1x1(t) + c2x2(t) = 0 if and only if c1, c2 = 0. The reader should derive the condition on the xi for linear indepen-dence. then the general solution is given as x(t) = c1x1(t) + c2x2(t). Then, setting t = 0, we get two linear equations for c1 and c2: c1x1(0) + c2x2(0) = x0. The major work is in finding the linearly independent solutions. This de-pends upon the different types of eigenvalues that one obtains from solving the eigenvalue equation, det(A −λI) = 0. The nature of these roots indicate the form of the general solution. In Table 2.1 we summarize the classifica-tion of solutions in terms of the eigenvalues of the coefficient matrix. We first make some general remarks about the plausibility of these solutions and then provide examples in the following section to clarify the matrix methods for our two dimensional systems. The construction of the general solution in Case I is straight forward. However, the other two cases need a little explanation. Let’s consider Case III. Note that since the original system of equations does not have any i’s, then we would expect real solutions. So, we look at the real and imaginary parts of the complex solution. We have that the complex solution satisfies the equation d dt [Re(y(t)) + iIm(y(t))] = A[Re(y(t)) + iIm(y(t))]. Differentiating the sum and splitting the real and imaginary parts of the equation, gives d dt Re(y(t)) + i d dt Im(y(t)) = A[Re(y(t))] + iA[Im(y(t))]. Setting the real and imaginary parts equal, we have d dt Re(y(t)) = A[Re(y(t))], linear systems of differential equations 65 and d dt Im(y(t)) = A[Im(y(t))]. Therefore, the real and imaginary parts each are linearly independent so-lutions of the system and the general solution can be written as a linear combination of these expressions. Classification of the Solutions for Two Linear First Order Differential Equations 1. Case I: Two real, distinct roots. Solve the eigenvalue problem Av = λv for each eigenvalue obtain-ing two eigenvectors v1, v2. Then write the general solution as a linear combination x(t) = c1eλ1tv1 + c2eλ2tv2 2. Case II: One Repeated Root Solve the eigenvalue problem Av = λv for one eigenvalue λ, obtaining the first eigenvector v1. One then needs a second linearly independent solution. This is obtained by solving the nonhomogeneous problem Av2 −λv2 = v1 for v2. The general solution is then given by x(t) = c1eλtv1 + c2eλt(v2 + tv1). 3. Case III: Two complex conjugate roots. Solve the eigenvalue problem Ax = λx for one eigenvalue, λ = α + iβ, obtaining one eigenvector v. Note that this eigenvector may have com-plex entries. Thus, one can write the vector y(t) = eλtv = eαt(cos βt + i sin βt)v. Now, construct two linearly independent solutions to the problem using the real and imaginary parts of y(t) : y1(t) = Re(y(t)) and y2(t) = Im(y(t)). Then the general solution can be written as x(t) = c1y1(t) + c2y2(t). Table 2.1: Solutions Types for Planar Sys-tems with Constant Coefficients We now turn to Case II. Writing the system of first order equations as a second order equation for x(t) with the sole solution of the characteristic equation, λ = 1 2(a + d), we have that the general solution takes the form x(t) = (c1 + c2t)eλt. This suggests that the second linearly independent solution involves a term of the form vteλt. It turns out that the guess that works is x = teλtv1 + eλtv2. Inserting this guess into the system x′ = Ax yields (teλtv1 + eλtv2)′ = A h teλtv1 + eλtv2 i . eλtv1 + λteλtv1 + λeλtv2 = λteλtv1 + eλtAv2. eλt (v1 + λv2) = eλtAv2. (2.70) 66 differential equations Noting this is true for all t, we find that v1 + λv2 = Av2. (2.71) Therefore, (A −λI)v2 = v1. We know everything except for v2. So, we just solve for it and obtain the second linearly independent solution. 2.6 Examples of the Matrix Method Here we will give some examples for typical systems for the three cases mentioned in the last section. Example 2.19. A = 4 2 3 3 ! . Eigenvalues: We first determine the eigenvalues. 0 = 4 −λ 2 3 3 −λ (2.72) Therefore, 0 = (4 −λ)(3 −λ) −6 0 = λ2 −7λ + 6 0 = (λ −1)(λ −6) (2.73) The eigenvalues are then λ = 1, 6. This is an example of Case I. Eigenvectors: Next we determine the eigenvectors associated with each of these eigenvalues. We have to solve the system Av = λv in each case. Case λ = 1. 4 2 3 3 ! v1 v2 ! = v1 v2 ! (2.74) 3 2 3 2 ! v1 v2 ! = 0 0 ! (2.75) This gives 3v1 + 2v2 = 0. One possible solution yields an eigenvector of v1 v2 ! = 2 −3 ! . Case λ = 6. 4 2 3 3 ! v1 v2 ! = 6 v1 v2 ! (2.76) −2 2 3 −3 ! v1 v2 ! = 0 0 ! (2.77) linear systems of differential equations 67 For this case we need to solve −2v1 + 2v2 = 0. This yields v1 v2 ! = 1 1 ! . General Solution: We can now construct the general solution. x(t) = c1eλ1tv1 + c2eλ2tv2 = c1et 2 −3 ! + c2e6t 1 1 ! = 2c1et + c2e6t −3c1et + c2e6t ! . (2.78) Example 2.20. A = 3 −5 1 −1 ! . Eigenvalues: Again, one solves the eigenvalue equation. 0 = 3 −λ −5 1 −1 −λ (2.79) Therefore, 0 = (3 −λ)(−1 −λ) + 5 0 = λ2 −2λ + 2 λ = −(−2) ± p 4 −4(1)(2) 2 = 1 ± i. (2.80) The eigenvalues are then λ = 1 + i, 1 −i. This is an example of Case III. Eigenvectors: In order to find the general solution, we need only find the eigenvector associated with 1 + i. 3 −5 1 −1 ! v1 v2 ! = (1 + i) v1 v2 ! 2 −i −5 1 −2 −i ! v1 v2 ! = 0 0 ! . (2.81) We need to solve (2 −i)v1 −5v2 = 0. Thus, v1 v2 ! = 2 + i 1 ! . (2.82) Complex Solution: In order to get the two real linearly indepen-dent solutions, we need to compute the real and imaginary parts of veλt. eλt 2 + i 1 ! = e(1+i)t 2 + i 1 ! 68 differential equations = et(cos t + i sin t) 2 + i 1 ! = et (2 + i)(cos t + i sin t) cos t + i sin t ! = et (2 cos t −sin t) + i(cos t + 2 sin t) cos t + i sin t ! = et 2 cos t −sin t cos t ! + iet cos t + 2 sin t sin t ! . General Solution: Now we can construct the general solution. x(t) = c1et 2 cos t −sin t cos t ! + c2et cos t + 2 sin t sin t ! = et c1(2 cos t −sin t) + c2(cos t + 2 sin t) c1 cos t + c2 sin t ! . (2.83) Note: This can be rewritten as x(t) = et cos t 2c1 + c2 c1 ! + et sin t 2c2 −c1 c2 ! . Example 2.21. A = 7 −1 9 1 ! . Eigenvalues: 0 = 7 −λ −1 9 1 −λ (2.84) Therefore, 0 = (7 −λ)(1 −λ) + 9 0 = λ2 −8λ + 16 0 = (λ −4)2. (2.85) There is only one real eigenvalue, λ = 4. This is an example of Case II. Eigenvectors: In this case we first solve for v1 and then get the second linearly independent vector. 7 −1 9 1 ! v1 v2 ! = 4 v1 v2 ! 3 −1 9 −3 ! v1 v2 ! = 0 0 ! . (2.86) Therefore, we have 3v1 −v2 = 0, ⇒ v1 v2 ! = 1 3 ! . linear systems of differential equations 69 Second Linearly Independent Solution: Now we need to solve Av2 −λv2 = v1. 7 −1 9 1 ! u1 u2 ! −4 u1 u2 ! = 1 3 ! 3 −1 9 −3 ! u1 u2 ! = 1 3 ! . (2.87) Expanding the matrix product, we obtain the system of equations 3u1 −u2 = 1 9u1 −3u2 = 3. (2.88) The solution of this system is u1 u2 ! = 1 2 ! . General Solution: We construct the general solution as y(t) = c1eλtv1 + c2eλt(v2 + tv1). = c1e4t 1 3 ! + c2e4t " 1 2 ! + t 1 3 !# = e4t c1 + c2(1 + t) 3c1 + c2(2 + 3t) ! . (2.89) 2.7 Planar Systems - Summary The reader should have noted by now that there is a connection between the behavior of the solutions obtained in Section 2.2.2 and the eigenvalues found from the coefficient matrices in the previous examples. In Table 2.2 we summarize some of these cases. Type Eigenvalues Stability Node Real λ, same signs λ < 0, stable λ > 0, unstable Saddle Real λ opposite signs Mostly Unstable Center λ pure imaginary — Focus/Spiral Complex λ, Re(λ) ̸= 0 Re(λ) < 0, stable Re(λ) > 0, unstable Degenerate Node Repeated roots, λ > 0, stable Lines of Equilibria One zero eigenvalue λ < 0, stable Table 2.2: List of typical behaviors in pla-nar systems. The connection, as we have seen, is that the characteristic equation for the associated second order differential equation is the same as the eigen-value equation of the coefficient matrix for the linear system. However, one should be a little careful in cases in which the coefficient matrix in not diag-onalizable. In Table 2.3 are three examples of systems with repeated roots. The reader should look at these systems and look at the commonalities and 70 differential equations differences in these systems and their solutions. In these cases one has un-stable nodes, though they are degenerate in that there is only one accessible eigenvector. Table 2.3: Three examples of systems with a repeated root of λ = 2. System 1 System 2 System 3 x K 3 K 2 K 1 0 1 2 3 y K 3 K 2 K 1 1 2 3 a = 2, b = 0, c = 0, d = 2 x K 3 K 2 K 1 0 1 2 3 y K 3 K 2 K 1 1 2 3 a = 0, b = 1, c = -4, d = 4 x K 3 K 2 K 1 0 1 2 3 y K 3 K 2 K 1 1 2 3 a = 2, b = 1, c = 0, d = 2 x′ = 2 0 0 2 ! x x′ = 0 1 −4 4 ! x x′ = 2 1 0 2 ! x Another way to look at the classification of these solution is to use the determinant and trace of the coefficient matrix. Recall that the determinant and trace of A = a b c d ! are given by detA = ad −bc and trA = a + d. We note that the general eigenvalue equation, λ2 −(a + d)λ + ad −bc = 0, can be written as λ2 −(trA)λ + detA = 0. (2.90) Therefore, the eigenvalues are found from the quadratic formula as λ1,2 = trA ± p (trA)2 −4detA 2 . (2.91) The solution behavior then depends on the sign of discriminant, (trA)2 −4detA. If we consider a plot of where the discriminant vanishes, then we could plot (trA)2 = 4detA in the detAtrA)-plane. This is a parabolic cure as shown by the dashed line in Figure 2.25. The region inside the parabola have a negative discriminant, leading to complex roots. In these cases we have oscillatory solutions. If trA = 0, then one has centers. If trA < 0, the solutions are stable spirals; otherwise, they are unstable spirals. If the discriminant is positive, then the roots are real, leading to nodes or saddles in the regions indicated. 2.8 Theory of Homogeneous Constant Coefficient Systems There is a general theory for solving homogeneous, constant coefficient sys-tems of first order differential equations. We begin by once again recalling linear systems of differential equations 71 detA trA tr2A=4detA Saddles Centers Unstable Spirals Stable Spirals Unstable Nodes Stable Nodes Figure 2.25: Solution Classification for Planar Systems. the specific problem (2.16). We obtained the solution to this system as x(t) = c1et + c2e−4t, y(t) = 1 3c1et −1 2c2e−4t. (2.92) This time we rewrite the solution as x = c1et + c2e−4t 1 3c1et −1 2c2e−4t ! = et e−4t 1 3et −1 2e−4t ! c1 c2 ! ≡ Φ(t)C. (2.93) Thus, we can write the general solution as a 2 × 2 matrix Φ times an arbi-trary constant vector. The matrix Φ consists of two columns that are linearly independent solutions of the original system. This matrix is an example of what we will define as the Fundamental Matrix of solutions of the system. So, determining the Fundamental Matrix will allow us to find the general solution of the system upon multiplication by a constant matrix. In fact, we will see that it will also lead to a simple representation of the solution of the initial value problem for our system. We will outline the general theory. Consider the homogeneous, constant coefficient system of first order dif-ferential equations dx1 dt = a11x1 + a12x2 + . . . + a1nxn, dx2 dt = a21x1 + a22x2 + . . . + a2nxn, 72 differential equations . . . dxn dt = an1x1 + an2x2 + . . . + annxn. (2.94) As we have seen, this can be written in the matrix form x′ = Ax, where x =       x1 x2 . . . xn       and A =       a11 a12 · · · a1n a21 a22 · · · a2n . . . . . . ... . . . an1 an2 · · · ann       . Now, consider m vector solutions of this system: ϕ1(t), ϕ2(t), . . . ϕm(t). These solutions are said to be linearly independent on some domain if c1ϕ1(t) + c2ϕ2(t) + . . . + cmϕm(t) = 0 for all t in the domain implies that c1 = c2 = . . . = cm = 0. Let ϕ1(t), ϕ2(t), . . . ϕn(t) be a set of n linearly independent set of solutions of our system, called a fundamental set of solutions. We construct a matrix from these solutions using these solutions as the column of that matrix. We define this matrix to be the fundamental matrix solution. This matrix takes the form Φ = ϕ1 . . . ϕn =       ϕ11 ϕ12 · · · ϕ1n ϕ21 ϕ22 · · · ϕ2n . . . . . . ... . . . ϕn1 ϕn2 · · · ϕnn       . What do we mean by a “matrix” solution? We have assumed that each ϕk is a solution of our system. Therefore, we have that ϕ′ k = Aϕk, for k = 1, . . . , n. We say that Φ is a matrix solution because we can show that Φ also satisfies the matrix formulation of the system of differential equations. We can show this using the properties of matrices. d dtΦ = ϕ′ 1 . . . ϕ′ n = Aϕ1 . . . Aϕn = A ϕ1 . . . ϕn = AΦ. (2.95) Given a set of vector solutions of the system, when are they linearly independent? We consider a matrix solution Ω(t) of the system in which we have n vector solutions. Then, we define the Wronskian of Ω(t) to be W = det Ω(t). linear systems of differential equations 73 If W(t) ̸= 0, then Ω(t) is a fundamental matrix solution. Before continuing, we list the fundamental matrix solutions for the set of examples in the last section. (Refer to the solutions from those examples.) Furthermore, note that the fundamental matrix solutions are not unique as one can multiply any column by a nonzero constant and still have a fundamental matrix solution. Example 2.19 A = 4 2 3 3 ! . Φ(t) = 2et e6t −3et e6t ! . We should note in this case that the Wronskian is found as W = det Φ(t) = 2et e6t −3et e6t = 5e7t ̸= 0. (2.96) Example 2.20 A = 3 −5 1 −1 ! . Φ(t) = et(2 cos t −sin t) et(cos t + 2 sin t) et cos t et sin t ! . Example 2.21 A = 7 −1 9 1 ! . Φ(t) = e4t e4t(1 + t) 3e4t e4t(2 + 3t) ! . So far we have only determined the general solution. This is done by the following steps: Procedure for Determining the General Solution 1. Solve the eigenvalue problem (A −λI)v = 0. 2. Construct vector solutions from veλt. The method depends if one has real or complex conjugate eigenvalues. 3. Form the fundamental solution matrix Φ(t) from the vector solution. 4. The general solution is given by x(t) = Φ(t)C for C an arbitrary con-stant vector. We are now ready to solve the initial value problem: x′ = Ax, x(t0) = x0. 74 differential equations Starting with the general solution, we have that x0 = x(t0) = Φ(t0)C. As usual, we need to solve for the ck’s. Using matrix methods, this is now easy. Since the Wronskian is not zero, then we can invert Φ at any value of t. So, we have C = Φ−1(t0)x0. Putting C back into the general solution, we obtain the solution to the initial value problem: x(t) = Φ(t)Φ−1(t0)x0. You can easily verify that this is a solution of the system and satisfies the initial condition at t = t0. The matrix combination Φ(t)Φ−1(t0) is useful. So, we will define the resulting product to be the principal matrix solution, denoting it by Ψ(t) = Φ(t)Φ−1(t0). Thus, the solution of the initial value problem is x(t) = Ψ(t)x0. Furthermore, we note that Ψ(t) is a solution to the matrix initial value problem x′ = Ax, x(t0) = I, where I is the n × n identity matrix. Matrix Solution of the Homogeneous Problem In summary, the matrix solution of dx dt = Ax, x(t0) = x0 is given by x(t) = Ψ(t)x0 = Φ(t)Φ−1(t0)x0, where Φ(t) is the fundamental matrix solution and Ψ(t) is the principal matrix solution. Example 2.22. Let’s consider the matrix initial value problem x′ = 5x + 3y y′ = −6x −4y, (2.97) satisfying x(0) = 1, y(0) = 2. Find the solution of this problem. We first note that the coefficient matrix is A = 5 3 −6 −4 ! . The eigenvalue equation is easily found from 0 = −(5 −λ)(4 + λ) + 18 = λ2 −λ −2 = (λ −2)(λ + 1). (2.98) linear systems of differential equations 75 So, the eigenvalues are λ = −1, 2. The corresponding eigenvectors are found to be v1 = 1 −2 ! , v2 = 1 −1 ! . Now we construct the fundamental matrix solution. The columns are obtained using the eigenvectors and the exponentials, eλt : ϕ1(t) = 1 −2 ! e−t, ϕ1(t) = 1 −1 ! e2t. So, the fundamental matrix solution is Φ(t) = e−t e2t −2e−t −e2t ! . The general solution to our problem is then x(t) = e−t e2t −2e−t −e2t ! C for C is an arbitrary constant vector. In order to find the particular solution of the initial value problem, we need the principal matrix solution. We first evaluate Φ(0), then we invert it: Φ(0) = 1 1 −2 −1 ! ⇒ Φ−1(0) = −1 −1 2 1 ! . The particular solution is then x(t) = e−t e2t −2e−t −e2t ! −1 −1 2 1 ! 1 2 ! = e−t e2t −2e−t −e2t ! −3 4 ! = −3e−t + 4e2t 6e−t −4e2t ! (2.99) Thus, x(t) = −3e−t + 4e2t and y(t) = 6e−t −4e2t. 2.9 Nonhomogeneous Systems Before leaving the theory of systems of linear, constant coefficient systems, we will discuss nonhomogeneous systems. We would like to solve systems of the form x′ = A(t)x + f(t). (2.100) We will assume that we have found the fundamental matrix solution of the homogeneous equation. Furthermore, we will assume that A(t) and f(t) are continuous on some common domain. 76 differential equations As with second order equations, we can look for solutions that are a sum of the general solution to the homogeneous problem plus a particular so-lution of the nonhomogeneous problem. Namely, we can write the general solution as x(t) = Φ(t)C + xp(t), where C is an arbitrary constant vector, Φ(t) is the fundamental matrix solution of x′ = A(t)x, and x′ p = A(t)xp + f(t). Such a representation is easily verified. We need to find the particular solution, xp(t). We can do this by applying The Method of Variation of Parameters for Systems. We consider a solution in the form of the solution of the homogeneous problem, but replace the constant vector by unknown parameter functions. Namely, we assume that xp(t) = Φ(t)c(t). Differentiating, we have that x′ p = Φ′c + Φc′ = AΦc + Φc′, or x′ p −Axp = Φc′. But the left side is f. So, we have that, Φc′ = f, or, since Φ is invertible (why?), c′ = Φ−1f. In principle, this can be integrated to give c. Therefore, the particular solu-tion can be written as xp(t) = Φ(t) Z t Φ−1(s)f(s) ds. (2.101) This is the variation of parameters formula. The general solution of Equation (2.100) has been found as x(t) = Φ(t)C + Φ(t) Z t Φ−1(s)f(s) ds. (2.102) We can use the general solution to find the particular solution of an ini-tial value problem consisting of Equation (2.100) and the initial condition x(t0) = x0. This condition is satisfied for a solution of the form x(t) = Φ(t)C + Φ(t) Z t t0 Φ−1(s)f(s) ds (2.103) provided x0 = x(t0) = Φ(t0)C. linear systems of differential equations 77 This can be solved for C as in the last section. Inserting the solution back into the general solution (2.103), we have x(t) = Φ(t)Φ−1(t0)x0 + Φ(t) Z t t0 Φ−1(s)f(s) ds (2.104) This solution can be written a little neater in terms of the principal matrix solution, Ψ(t) = Φ(t)Φ−1(t0) : x(t) = Ψ(t)x0 + Ψ(t) Z t t0 Ψ−1(s)f(s) ds (2.105) Finally, one further simplification occurs when A is a constant matrix, which are the only types of problems we have solved in this chapter. In this case, we have that Ψ−1(t) = Ψ(−t). So, computing Ψ−1(t) is relatively easy. Example 2.23. x′′ + x = 2 cos t, x(0) = 4, x′(0) = 0. This example can be solved using the Method of Undetermined Coefficients. However, we will use the matrix method described in this section. First, we write the problem in matrix form. The system can be written as x′ = y y′ = −x + 2 cos t. (2.106) Thus, we have a nonhomogeneous system of the form x′ = Ax + f = 0 1 −1 0 ! x y ! + 0 2 cos t ! . Next we need the fundamental matrix of solutions of the homoge-neous problem. We have that A = 0 1 −1 0 ! . The eigenvalues of this matrix are λ = ±i. An eigenvector associated with λ = i is easily found as 1 i ! . This leads to a complex solution 1 i ! eit = cos t + i sin t i cos t −sin t ! . From this solution we can construct the fundamental solution matrix Φ(t) = cos t sin t −sin t cos t ! . So, the general solution to the homogeneous problem is xh = Φ(t)C = c1 cos t + c2 sin t −c1 sin t + c2 cos t ! . 78 differential equations Next we seek a particular solution to the nonhomogeneous prob-lem. From Equation (2.103) we see that we need Φ−1(s)f(s). Thus, we have Φ−1(s)f(s) = cos s −sin s sin s cos s ! 0 2 cos s ! = −2 sin s cos s 2 cos2 s ! . (2.107) We now compute Φ(t) Z t t0 Φ−1(s)f(s) ds = cos t sin t −sin t cos t ! Z t t0 −2 sin s cos s 2 cos2 s ! ds = cos t sin t −sin t cos t ! −sin2 t t + 1 2 sin(2t) ! = t sin t sin t + t cos t ! . (2.108) therefore, the general solution is x = c1 cos t + c2 sin t −c1 sin t + c2 cos t ! + t sin t sin t + t cos t ! . The solution to the initial value problem is x = cos t sin t −sin t cos t ! 4 0 ! + t sin t sin t + t cos t ! , or x = 4 cos t + t sin t −3 sin t + t cos t ! . 2.10 Appendix: More on Exponentiation of Matrices We have seen that in general that exponentiating a matrix using the Maclau-rin series (2.60) can be difficult. If the matrix is diagonal, then the exponen-tiation is simple since powers of a diagonal matrix involves the powers of the diagonal elements. In other words, if A =          λ1 0 · · · · · · 0 0 λ2 0 · · · 0 . . . 0 ... ... . . . . . . ... ... 0 0 · · · · · · 0 λn          , linear systems of differential equations 79 then eA =          eλ1 0 · · · · · · 0 0 eλ2 0 · · · 0 . . . 0 ... ... . . . . . . ... ... 0 0 · · · · · · 0 eλn          . If A is not diagonal, then if it is diagonalizable, then there is an invertible matrix P such that P−1AP = D is a diagonal matrix. Then, we have eA = ePDP−1 = ∞ ∑ k=0 1 n!(PDP−1)n = I + PDP−1 + PDP−1PDP−1 2! + PDP−1PDP−1PDP−1 3! + · · · = I + PDP−1 + PD2P−1 2! + PD3P−1 3! + · · · = P I + D + D2 2! + D3 3! + · · · P−1 = PeDP−1. (2.109) Therefore, we have a prescription for exponentiating a diagonalizable ma-trix. But when are matrices diagonalizable? We learn from a linear algebra course that real symmetric matrices are diagonalizable.3 A real symmetric 3 This is a simple example of a more gen-eral theory where real symmetric matri-ces are replaced by complex Hermitian matrices which are self-adjoint and or-thogonal matrices are replaced with uni-tary matrices. As matrices are represen-tations of operators, the more general theory of self-adjoint operators are dis-cussed later. matrix A is one which is a matrix of real numbers such that A equals its transpose, AT = A. Such matrices satisfy several properties, which we will not prove here. • The eigenvalues of a real symmetric matrix are real. • The eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are orthogonal. • Every real symmetric n × n matrix posses a complete set of orthonormal vectors. • For every real symmetric n × n matrix A there is an n × n real orthogonal matrix P such that P−1AP = D, where D is a diagonal matrix. A real matrix P is orthogonal if P−1 = PT. Therefore PTP = I. Such orthogonal matrices have column (and row) vectors which are orthogonal. Thus, one can use the orthonormal eigenvectors of A to form the matrix P. Namely, we let Avk = λkvk and P = (v1, . . . , vn). Then, P−1AP = (v1, . . . , vn)−1A(v1, . . . , vn), = (v1, . . . , vn)T(λ1v1, . . . , λnvn), =     v1 . . . vn     λ1v1 . . . λnvn 80 differential equations =          λ1 0 · · · · · · 0 0 λ2 0 · · · 0 . . . 0 ... ... . . . . . . ... ... 0 0 · · · · · · 0 λn          . (2.110) Thus, we have shown that P−1AP is a diagonal matrix and this is called a similarity transformation. Similarity transformation. Example 2.24. Consider the matrix A = 1 3 3 1 ! . Find etA using the diagonalization of A. First, we solve the eigenvalue problem. The eigenvalue equation is given by 0 = \|A −λI\| = (1 −λ)2 −9. So, λ = 4, −2. Next, we seek the eigenvectors. Letting v = (v1, v2)T, we have (1 −λ)v1 + 3v2 = 0. For λ = 4, we need to solve (−3)v1 + 3v2 = 0, or v1 = v2. For λ = −2, we need to solve (3)v1 + 3v2 = 0, or v1 = −v2. Therefore, the eigenvalues and normalized eigenvectors are given by λ = 4, v1 = 1 √ 2 1 1 ! , λ = −2, v2 = 1 √ 2 −1 1 ! . This gives4 4 Since P is orthogonal, it is a rotation matrix. In two dimensions a rotation of coordinate axes about the origin by an-gle θ is given by R(θ) = cos θ sin θ −sin θ cos θ . Therefore, in the example θ = −π 4 . P = 1 √ 2 1 −1 1 1 ! . Then, we have P−1AP = 1 √ 2 1 1 −1 1 ! 1 3 3 1 ! 1 √ 2 1 −1 1 1 ! = 1 2 1 1 −1 1 ! 4 2 4 −2 ! = 1 2 8 0 0 −4 ! = 4 0 0 −2 ! . (2.111) Finally, we can exponentiate tA as etA = etPDP−1 = PetDP−1 = 1 √ 2 1 −1 1 1 ! e4t 0 0 e−2t ! 1 √ 2 1 1 −1 1 ! = 1 2 1 −1 1 1 ! e4t e4t −e−2t e−2t ! = 1 2 e4t + e−2t e4t −e−2t e4t −e−2t e4t + e−2t ! . (2.112) linear systems of differential equations 81 As a word of caution, we note that real symmetric matrices can be di-agonalized. what about real nonsymmetric matrices? A real matrix can be transformed into a diagonal matrix through a similarity transformation as long as it has a full set of linearly independent eigenvectors. We need only look at Example 2.22. In that case A = 5 3 −6 −4 ! . The eigenvalues were found as λ = −1, 2 and the corresponding eigenvec-tors were v1 = 1 −2 ! , v2 = 1 −1 ! . This gives P = 1 1 −2 −1 ! . Then, we have P−1AP = −1 −1 2 1 ! 5 3 −6 −4 ! 1 1 −2 −1 ! = −1 −1 2 1 ! −1 2 2 −2 ! = −1 0 0 2 ! . (2.113) So, we have a similarity transformation, but note that P is not an orthogonal matrix in this case. We can use this diagonalization to compute etA as we show in the next example. Example 2.25. Find etA for A = 5 3 −6 −4 ! . etA = etPDP−1 = PetDP−1 = 1 1 −2 −1 ! e−t 0 0 e2t ! −1 −1 2 1 ! = 1 1 −2 −1 ! −e−t −e−t 2e2t e2t ! = −e−t + 2e2t −e−t + 2e2t 2e−t −2e2t 2e−t −e2t ! . (2.114) From Example 2.22, the fundamental matrix solution is given as Φ(t) = e−t e2t −2e−t −e2t ! and Φ−1(0) = −1 −1 2 1 ! . 82 differential equations So, the principal matrix solution is found as Φ(t) = Φ(t)Φ−1(0) = e−t e2t −2e−t −e2t ! −1 −1 2 1 ! = −e−t + 2e2t −e−t + 2e2t 2e−t −2e2t 2e−t −e2t ! , (2.115) which is the same as our evaluation of etA in the previous example. There are other methods for exponentiating a matrix. These involve more sophisticated matrix methods and rely on the Cayley-Hamilton Theorem. The Cayley-Hamilton Theorem Matrix A satisfies its own characteristic (eigenvalue) equation. This means that any power of A, and therefore eA, can be expressed as a polynomial of A of degree n −1. We can also use this to evaluate A−1. We demonstrate this by example. Example 2.26. First, we note that A = 5 3 −6 −4 ! , A2 = 7 3 −6 −2 ! . Inserting these into the characteristic equation [see Example 2.22], we have A2 −A −2I = 7 3 −6 −2 ! − 5 3 −6 −4 ! −2 1 0 0 1 ! = 0. Now, we rearrange the equation as A(A −I) = 2I and find that A−1 = 1 2(A −I). Thus, A−1 = 1 2 4 3 −6 −5 ! = 2 3 2 −3 −5 2 ! . This agrees with the usual method for computing A−1. We can use the Cayley Hamilton Theorem to evaluate functions of A. Let f (A) be a matrix polynomial of A of degree n Also, let d(λ) = det(A −λI). If the characteristic equation for A is d(λ) = 0, then we know by the Cayley-Hamilton Theorem that d(A) = 0. Given polynomials f (x) and d(x), we know that we can write f (x) = d(x)q(x) + r(x), where r(x) is a polynomial of degree n −1. So, we also have f (A) = d(A)q(A) + r(A) = r(A). Therefore, f (A) is a polynomial in A of degree n −1. So, we can write f (A) = αn−1An−1 + αn−1An−1 + . . . α0I. Example 2.27. Let A = 4 1 2 3 ! . Find A10. The eigenvalues are λ = 5, 2. So, for f (A) = A10, then f (λ) = λ10 and r(A) is a polynomial of degree 1. So, we have A10 = r(A) = α1A + α0I. linear systems of differential equations 83 Evaluating f (λi) = r(λi), we obtain two equations for the two un-knowns, 510 = 5α1 + α0 210 = 2α1 + α0. (2.116) Solving, we find α1 = 3254867, α0 = −6508710. Then, A10 = 6510758 3254867 6509734 3255891 ! . These methods can be generalized to evaluate arbitrary functions. As an example, we can exponentiate a matrix Example 2.28. Find etA for A = 5 3 −6 −4 ! . This is Example 2.25. However, in this case we let etA = α1A + α0I. The eigenvalues are λ = −1, 2. So, for f (A) = etA, then f (λ) = etλ. We have r(A) = α1A + α0I and r(λ) = α1λ + α0. Substituting the eigenvalues, we obtain e−t = −α1 + α0 e2t = 2α1 + α0. (2.117) Solving, we have α0 = 1 3 2e−t + e2t , α1 = 1 3 e2t −e−t . Then, etA = α1A + α0I = −e−t + 2e2t −e−t + 2e2t 2e−t −2e2t 2e−t −e2t ! . Again, we arrive at the same result. The Putzer Algorithm. Suppose that A is an n × n matrix with eigenval-ues λk, k = 1, . . . , n, not necessarily distinct. Then, etA = n−1 ∑ j=0 rj+1(t)Pj where Pj = j ∏ k=1 (A −λkI) 84 differential equations and the rj(t)’s are the components of the solutions of p′(t) =          λ1 0 · · · 0 0 1 λ2 · · · 0 0 0 1 ... 0 0 0 0 ... λn−1 0 0 0 · · · 1 λn          p(t) satisfying p(0) = (1, 0, . . . , 0)T. Example 2.29. Let’s try a different example. Consider A = 5 3 −6 −4 ! . Find etA using the Putzer Algorithm. Here n = 2 and the eigenvalues are λ = −1, −3. First we find the P′ j s. P0 = I = 1 0 0 1 ! , P1 = A + I = 2 −2 4 −4 ! . Next, we solve the tridiagonal system r′ 1(t) r′ 2(t) ! = −1 0 1 −3 r1(t) r2(t) !! with r1(0) = 1, r2(0) = 0. Expanding into a coupled system of dif-ferential equations, r′ 1 = −r1, r1(0) = 1, r′ 2 = r1 −3r2, r2(0) = 0. (2.118) The first equation is readily solved to find r1(t) = e−t. The second equation becomes r′ 2 + 3r2 = e−t, r2(0) = 0. This is solved using and integrating factor. The result is r2(t) = 1 2 e−te−3t . Now that we have the pieces for the algorithm, we can construct the answer. etA = n−1 ∑ j=0 rj+1(t)Pj = r1(t)P0 + r2(t)P1 = e−t 1 0 0 1 ! + 1 2 e−te−3t 2 −2 4 −4 ! = 2e−t −e−3t −e−t2e−3t 2e−t −2e−3t −e−t + 2e−3t ! . (2.119) linear systems of differential equations 85 Problems 1. Consider the system x′ = −4x −y y′ = x −2y. a. Determine the second order differential equation satisfied by x(t). b. Solve the differential equation for x(t). c. Using this solution, find y(t). d. Verify your solutions for x(t) and y(t). e. Find a particular solution to the system given the initial conditions x(0) = 1 and y(0) = 0. 2. Consider the following systems. Determine the families of orbits for each system and sketch several orbits in the phase plane and classify them by their type (stable node, etc.) a. x′ = 3x y′ = −2y. b. x′ = −y y′ = −5x. c. x′ = 2y y′ = −3x. d. x′ = x −y y′ = y. e. x′ = 2x + 3y y′ = −3x + 2y. 3. Use the transformations relating polar and Cartesian coordinates to prove that dθ dt = 1 r2 x dy dt −ydx dt . 86 differential equations 4. Consider the system of equations in Example 2.13. a. Derive the polar form of the system. b. Solve the radial equation, r′ = r(1 −r2), for the initial values r(0) = 0, 0.5, 1.0, 2.0. c. Based upon these solutions, plot and describe the behavior of all solutions to the original system in Cartesian coordinates. 5. Consider the following systems. For each system determine the coeffi-cient matrix. When possible, solve the eigenvalue problem for each matrix and use the eigenvalues and eigenfunctions to provide solutions to the given systems. Finally, in the common cases which you investigated in Problem 2, make comparisons with your previous answers, such as what type of eigenvalues correspond to stable nodes. a. x′ = 3x −y y′ = 2x −2y. b. x′ = −y y′ = −5x. c. x′ = x −y y′ = y. d. x′ = 2x + 3y y′ = −3x + 2y. e. x′ = −4x −y y′ = x −2y. f. x′ = x −y y′ = x + y. linear systems of differential equations 87 6. For the given matrix, evaluate etA, using the definition etA = ∞ ∑ n=0 tn n! An = I + tA + t2 2 A2 + t3 3! A3 + . . . , and simplifying. a. A = 1 0 0 2 ! . b. A = 1 0 −2 2 ! . c. A = 0 −1 0 1 ! . d. A = 0 1 1 0 ! . e. A = 0 −i i 0 ! . f. A =    0 1 0 0 0 1 0 0 0   . 7. Find the fundamental matrix solution for the system x′ = Ax where matrix A is given. If an initial condition is provided, find the solution of the initial value problem using the principal matrix. a. A = 1 0 −2 2 ! . b. A = 12 −15 4 −4 ! , x(0) = 1 0 ! c. A = 2 −1 5 −2 ! . d. A = 4 −13 2 −6 ! , x(0) = 2 0 ! e. A = 4 2 3 3 ! . f. A = 3 5 −1 1 ! . g. A = 8 −5 16 8 ! , x(0) = 1 −1 ! . h. A = 1 −2 2 −3 ! . 88 differential equations i. A =    5 4 2 4 5 2 2 2 2   . 8. Solve the following initial value problems using Equation (2.105), the solution of a nonhomogeneous system using the principal matrix solution. a. x′ = 2 −1 3 −2 ! x + et t ! , x(0) = 1 2 ! b. x′ = 5 3 −6 −4 ! x + 1 et ! , x(0) = 1 0 ! c. x′ = 2 −1 5 −2 ! x + cos t sin t ! , x(0) = 0 1 ! 9. Add a third spring connected to mass two in the coupled system shown in Figure 2.2 to a wall on the far right. Assume that the masses are the same and the springs are the same. a. Model this system with a set of first order differential equations. b. If the masses are all 2.0 kg and the spring constants are all 10.0 N/m, then find the general solution for the system. c. Move mass one to the left (of equilibrium) 10.0 cm and mass two to the right 5.0 cm. Let them go. find the solution and plot it as a function of time. Where is each mass at 5.0 seconds? d. Model this initial value problem with a set of two second order differential equations. Set up the system in the form M¨ x = −Kx and solve using the values in part b. 10. In Example 2.14 we investigated a couple mass-spring system as a pair of second order differential equations. a. In that problem we used q 3± √ 5 2 = √ 5±1 2 . Prove this result. b. Rewrite the system as a system of four first order equations. c. Find the eigenvalues and eigenfunctions for the system of equa-tions in part b to arrive at the solution found in Example 2.14. d. Let k = 5.00 N/m and m = 0.250 kg. Assume that the masses are initially at rest and plot the positions as a function of time if initially i) x1(0) = x2(0) = 10.0 cm and i) x1(0) = −x2(0) = 10.0 cm. Describe the resulting motion. 11. Consider the series circuit in Figure 1.6 with L = 1.00 H, R = 1.00 × 102 Ω, C = 1.00 × 10−4 F, and V0 = 1.00 × 103 V. a. Set up the problem as a system of two first order differential equa-tions for the charge and the current. b. Suppose that no charge is present and no current is flowing at time t = 0 when V0 is applied. Find the current and the charge on the capacitor as functions of time. linear systems of differential equations 89 c. Plot your solutions and describe how the system behaves over time. 12. Consider the series circuit in Figure 2.21 with L = 1.00 H, R1 = R2 = 1.00 × 102 Ω, C = 1.00 × 10−4 F, and V0 = 1.00 × 103 V. a. Set up the problem as a system of first order differential equations for the charges and the currents in each loop. b. Suppose that no charge is present and no current is flowing at time t = 0 when V0 is applied. Find the current and the charge on the capacitor as functions of time. c. Plot your solutions and describe how the system behaves over time. 13. Initially a 100 gallon tank is filled with pure water. At time t = 0 water with a half a pound of salt per two gallons is added to the container at the rate of 3 gallons per minute, and the well-stirred mixture is drained from the container at the same rate. a. Find the number of pounds of salt in the container as a function of time. b. How many minutes does it take for the concentration to reach 2 pounds per gallon? c. What does the concentration in the container approach for large values of time? Does this agree with your intuition? 14. You make two quarts of salsa for a party. The recipe calls for five teaspoons of lime juice per quart, but you had accidentally put in five table-spoons per quart. You decide to feed your guests the salsa anyway. Assume that the guests take a quarter cup of salsa per minute and that you replace what was taken with chopped tomatoes and onions without any lime juice. [1 quart = 4 cups and 1 Tb = 3 tsp.] a. Write down the differential equation and initial condition for the amount of lime juice as a function of time in this mixture-type problem. b. Solve this initial value problem. c. How long will it take to get the salsa back to the recipe’s suggested concentration? 15. Consider the chemical reaction leading to the system in (2.54). Let the rate constants be k1 = 0.20 ms−1, k2 = 0.05 ms−1, and k3 = 0.10 ms−1. What do the eigenvalues of the coefficient matrix say about the behavior of the system? Find the solution of the system assuming A = A0 = 1.0 µmol, B = 0, and C = 0. Plot the solutions for t = 0.0 to 50.0 ms and describe what is happening over this time. 16. Find and classify any equilibrium points in the Romeo and Juliet prob-lem for the following cases. Solve the systems and describe their affections as a function of time. 90 differential equations a. a = 0, b = 2, c = −1, d = 0, R(0) = 1, J(0) = 1. b. a = 0, b = 2, c = 1, d = 0, R(0) = 1, J(0) = 1. c. a = −1, b = 2, c = −1, d = 0, R(0) = 1, J(0) = 1. Figure 2.26: Figure for Problem 17. A 500L B 1000L 10L/min 10L/min 5L/min 15L/min 17. Two tanks contain a mixture of water and alcohol with tank A contain-ing 500 L and tank B 1000L. Initially, the concentration of alcohol in Tank A is 0% and that of tank B is 80%. Solution leaves tank A into B at a rate of 15 liter/min and the solution in tank B returns to A at a rate of 5 L/min while well mixed solution also leaves the system at 10 liter/min through an outlet. A mixture of water and alcohol enters tank A at the rate of 10 liter/min with the concentration of 10% through an inlet. What will be the concentration of the alcohol of the solution in each tank after 10 mins? 18. Consider the tank system in Problem 17. Add a third tank (C) to tank B with a volume of 300 L. Connect C with 8 L/min from tank B and 2 L/min flow back. Let 10 L/min flow out of the system. If the initial concentration is 10% in each tank and a mixture of water and alcohol enters tank A at the rate of 10 liter/min with the concentration of 20% through an inlet, what will be the concentration of the alcohol in each of the tanks after an hour? 19. Consider the epidemic model leading to the system in (2.58). Choose the constants as a = 2.0 days−1, d = 3.0 days−1, and r = 1.0 days−1. What are the eigenvalues of the coefficient matrix? Find the solution of the system assuming an initial population of 1, 000 and one infected individual. Plot the solutions for t = 0.0 to 5.0 days and describe what is happening over this time. Is this model realistic?
1252	https://www.datacamp.com/tutorial/bernoulli-distribution	What Is a Bernoulli Distribution? A Deep Dive \| DataCamp Skip to main content EN EnglishEspañolPortuguêsDeutsch BetaFrançais Beta More InformationFound an Error? Log inGet Started Tutorials Blogs Tutorials docs Podcasts Cheat Sheets code-alongs Newsletter Category Category Technologies Discover content by tools and technology AI AgentsArtificial IntelligenceAWSAzureBusiness IntelligenceChatGPTDatabricksdbtDockerExcelGenerative AIGitGoogle Cloud PlatformHugging FaceJavaJuliaKafkaKubernetesLarge Language ModelsMongoDBMySQLNoSQLOpenAIPostgreSQLPower BIPySparkPythonRScalaSnowflakeSpreadsheetsSQLSQLiteTableau Category Topics Discover content by data science topics AI for BusinessBig DataCareer ServicesCloudData AnalysisData EngineeringData LiteracyData ScienceData VisualizationDataLabDeep LearningMachine LearningMLOpsNatural Language Processing Browse Courses category Home Tutorials Data Science Bernoulli Distribution: A Complete Guide with Examples Discover how the Bernoulli distribution captures binary outcomes and is applied in everything from coin flips to customer predictions. Contents Aug 22, 2024· 11 min read Contents What is a Bernoulli Distribution? Bernoulli trials Bernoulli distributions Properties of Bernoulli Distributions Binary outcomes Mean and variance Symmetry and asymmetry Practical Applications of Bernoulli Distributions Binary classification in machine learning Hypothesis testing Simulation and modeling Performance Considerations Handling imbalanced data Computational efficiency Common Misconceptions Misinterpreting the probability of success Bernoulli vs. binomial distribution Alternatives to Bernoulli Distributions Binomial distribution Geometric distribution Conclusion Frequently Asked Questions Training more people? Get your team access to the full DataCamp for business platform.For BusinessFor a bespoke solution book a demo. The Bernoulli distribution is a fundamental concept in statistics and data science. Named after the Swiss mathematician Jacob Bernoulli, this distribution is important in probability theory and it serves as a building block for more complex statistical models- everything from predicting customer behavior to developing machine learning algorithms. As we get started, consider our Introduction to Statistics course for a refresher on probability distributions. Also, consider our Foundations of Probability in Python course to study problems with two possible outcomes and practice with the scipy library. Our Statistical Thinking in Python (Part 1) course is another great option because it will teach you how to write your own Python functions to perform Bernoulli trials. What is a Bernoulli Distribution? A Bernoulli distribution is a discrete probability distribution that models a random variable with only two possible outcomes. These outcomes are typically labeled as "success" and "failure," or else they are represented numerically as 1 and 0. Bernoulli trials Let’s start with Bernoulli trials. A Bernoulli trial is a random experiment with exactly two possible outcomes. Classic examples include: Flipping a coin (heads or tails) Answering a true/false question Determining if a customer will make a purchase (buy or not buy) Each Bernoulli trial is independent, meaning the outcome of one trial does not affect the probability of success in subsequent trials. Bernoulli distributions A Bernoulli distribution describes the probability of success in a single Bernoulli trial. It is characterized by a single parameter, p, which represents the probability of success. The probability of failure is consequently 1 - p. Mathematically, we can express a Bernoulli distribution as follows, w here X is the random variable representing the outcome of the Bernoulli trial. To illustrate how Bernoulli distributions change with different probabilities of success, let’s compare three examples. Comparison of Bernoulli distributions for different values of p. Image by Author. Here, each graph shows a different Bernoulli distribution. Left: p = 0.3 (30% chance of success) Center: p = 0.5 (50% chance of success) Right: p = 0.7 (70% chance of success) As we can see, the probability mass function of a Bernoulli distribution always has two bars: one for the probability of failure (X = 0) and one for the probability of success (X = 1). The height of these bars changes based on the value of p, but they always sum to 1. Properties of Bernoulli Distributions Let's explore the key characteristics that define the Bernoulli distribution. Binary outcomes The most distinctive feature of a Bernoulli distribution is its binary nature. Each trial can result in only one of two possible outcomes: Success (usually denoted as 1) Failure (usually denoted as 0) This binary property makes Bernoulli distributions particularly useful when we're interested in yes/no, true/false, or success/failure outcomes. Here are some key advantages of using Bernoulli distributions in data science and machine learning: Simplicity in Modeling: Binary outcomes allow for straightforward modeling of probabilities and odds. Ease of Interpretation: Results are often easily interpretable as probabilities or proportions. Versatility: Many complex scenarios can be broken down into a series of binary decisions or outcomes. Foundation for More Complex Distributions: Bernoulli trials form the basis for other important distributions, like the binomial distribution, which counts successes across multiple Bernoulli trials, and the geometric distribution, which tracks the number of Bernoulli trials until the first success. Mean and variance The mean (μ) and variance (σ ²) of a Bernoulli distribution are directly related to the probability of success, p. Mean T he mean is equal to the probability of success. This implies that if you were to repeat a Bernoulli trial many times, the average outcome would converge to p. Variance The variance is given by: σ ² = p(1 - p). The variance reaches its maximum when p = 0.5 and decreases as p approaches 0 or 1. This relationship provides insights into the spread of the data and the uncertainty associated with the outcomes. In terms of data spread, the variance in a Bernoulli distribution signifies: Predictability: A low variance (when p is close to 0 or 1) indicates more predictable outcomes, while a high variance (when p is close to 0.5) suggests more uncertainty. Informational Content: The closer p is to 0.5, the more information each trial provides, as outcomes are less predictable. Sampling Considerations: The variance informs sample size decisions in experimental design. Higher variance typically requires larger sample sizes to achieve the same level of precision in estimates. Risk Assessment: In applications like finance or insurance, the variance can be interpreted as a measure of risk or volatility. Symmetry and asymmetry The shape of a Bernoulli distribution depends on the value of p: When p = 0.5, the distribution is symmetric. One example is a fair coin toss, where the probabilities of success and failure are equal. When p ≠ 0.5, the distribution becomes asymmetric. As p approaches 0 or 1, the asymmetry increases. The symmetry or asymmetry of the distribution affects interpretation in several ways: Expectation Setting: In a symmetric distribution, neither outcome is more likely, which can inform decision-making processes where no bias toward one outcome exists, such as in a fair game. Bias Detection: Asymmetry in observed outcomes when symmetry is expected can indicate bias in the process or measurement. Threshold Determination: In classification tasks, the symmetry or asymmetry of the distribution can inform the choice of classification thresholds. Model Selection: The degree of asymmetry can influence the choice of statistical models or machine learning algorithms. Sampling Strategies: In highly asymmetric cases (p very close to 0 or 1), special sampling techniques might be needed to ensure rare events are adequately represented in the data. Practical Applications of Bernoulli Distributions Bernoulli distributions find widespread use in various fields, particularly in data science and statistics. Let's explore some of the most common applications: Binary classification in machine learning In machine learning, Bernoulli distributions play a central role in binary classification problems. These are scenarios where we need to categorize data into one of two classes. Examples include: Spam detection in emails (spam or not spam) Fraud detection in financial transactions (fraudulent or legitimate) Disease diagnosis based on symptoms (present or absent) Algorithms that leverage Bernoulli distributions for binary classification include: Logistic Regression: Logistic regression assumes that the binary outcome follows a Bernoulli distribution. It models the probability of the outcome using the logit function, which is the inverse of the logistic function. The resulting output can be interpreted as a Bernoulli probability.Read our Understanding Logistic Regression in Python and Logistic Regression in R tutorials to learn the specifics. Bernoulli Naive Bayes: This classifier is based on the Naive Bayes algorithm and is particularly suited for document classification tasks where features are binary (word presence or absence). Become an ML Scientist Upskill in Python to become a machine learning scientist. Start Learning for Free Hypothesis testing Bernoulli distributions are fundamental in hypothesis testing, particularly when dealing with proportions or success rates. Common applications include: A/B testing in Marketing: Comparing the success rates of two different marketing strategies. Learn more about A/B testing with our Customer Analytics and A/B Testing in Python course and also our A/B Testing in R code-along. Quality Control: Testing whether the defect rate in a manufacturing process exceeds a certain threshold. Medical trials: Assessing the efficacy of a new treatment compared to a placebo. In these scenarios, the null hypothesis often assumes a specific value for the probability of success p, and the alternative hypothesis challenges this assumption based on observed data. Simulation and modeling Bernoulli distributions are handy in simulations and probabilistic modeling, especially for scenarios with binary outcomes. They are used in: Monte Carlo Simulations: Modeling complex systems with many binary decision points. Learn more with our Introduction to Monte Carlo Methods tutorial. Risk Analysis: Assessing the probability of success or failure in business scenarios. Population Genetics: Modeling the inheritance of traits in genetic studies. By generating random samples from a Bernoulli distribution, researchers and data scientists can create realistic simulations of complex systems and processes. Performance Considerations When working with Bernoulli distributions in data analysis and machine learning, there are several performance considerations to keep in mind: Handling imbalanced data In many real-world scenarios, the probability of success p in a Bernoulli distribution is not 0.5, leading to imbalanced datasets. This imbalance can pose challenges for machine learning algorithms, potentially biasing them towards the majority class. Strategies to address this include: Oversampling the Minority Class: Techniques like SMOTE (Synthetic Minority Oversampling Technique) can be used to generate synthetic samples of the minority class. Undersampling the Majority Class: Randomly removing instances from the majority class to balance the dataset. Adjusting Class Weights: Giving more importance to the minority class during model training. Using Ensemble Methods: Techniques like Random Forests or Gradient Boosting can often handle imbalanced data better than single models. Read our What is Boosting tutorial to learn all about boosting and ensemble methods more generally. Choosing Appropriate Evaluation Metrics: Accuracy alone can be misleading for imbalanced datasets. Consider metrics like F1-score, precision, recall, or area under the ROC curve. Computational efficiency Bernoulli distributions are computationally efficient, especially for large datasets in binary classification tasks. However, there are still considerations to keep in mind: Vectorization: When implementing Bernoulli-based algorithms, use vectorized operations (e.g., NumPy in Python) for improved performance. Sparsity: In text classification tasks using the Bernoulli Naive Bayes classifier, leveraging sparse matrix representations can significantly reduce memory usage and computation time. Trade-offs in Model Complexity: While more complex models might offer slightly better performance, the computational cost may not always justify the marginal improvement. Consider the balance between model accuracy and processing time. Common Misconceptions To fully grasp Bernoulli distributions, it's important to address some common misconceptions: Misinterpreting the probability of success One frequent misconception is assuming that the probability of success p in a Bernoulli distribution is always 0.5. In reality, p can take any value between 0 and 1, inclusive. The value of p depends on the specific scenario or process being modeled. For example: In a fair coin toss, p = 0.5. In a loaded die where rolling a 6 is considered success, p might be 1/4. In a quality control process where the defect rate is 1%, p would be 0.01 for a "defect" outcome. Accurately estimating p from data is crucial for proper statistical inference and modeling. This often involves collecting a representative sample and calculating the proportion of successes. Bernoulli vs. binomial distribution Bernoulli distributions are sometimes confused with Binomial distributions. While related, they are distinct: A Bernoulli distribution models a single trial with two possible outcomes. A binomial distribution models the number of successes in a fixed number of independent Bernoulli trials. In other words, a binomial distribution is the sum of multiple independent Bernoulli distributions. For instance, if you flip a coin 10 times and count the number of heads, that's a binomial distribution. Each individual flip, however, follows a Bernoulli distribution. L et's compare them visually: Bernoulli distribution versus binomial distribution. Image by Author. This comparison clearly shows the fundamental difference between these two distributions: The Bernoulli distribution (left) represents a single trial with only two possible outcomes: 0 (failure) or 1 (success). In this example, with p = 0.3, we see a 70% chance of failure and a 30% chance of success. The binomial distribution (right) represents the number of successes in multiple trials (in this case, 10 trials). It shows the probability of getting each possible number of successes, from 0 to 10. The shape of this distribution is determined by both n (the number of trials) and p (the probability of success in each trial). While a Bernoulli distribution always has exactly two possible outcomes, a binomial distribution can have n+1 possible outcomes, where n is the number of trials. Alternatives to Bernoulli Distributions While Bernoulli distributions are widely used, there are scenarios where alternative distributions might be more appropriate. Let's explore two common alternatives: Binomial distribution The binomial distribution is useful when you're dealing with multiple trials as opposed to a single event. This alternative is often considered when you're interested in the number of successes across several attempts rather than just one outcome. Geometric distribution The geometric distribution comes into play when you're interested in how many attempts it will take to achieve the first success. It's often applied in situations where the focus is on the waiting time or number of attempts before success happens. Conclusion Bernoulli distributions form the foundation of many statistical concepts and are essential components in data science. Their simple binary nature, combined with wide-ranging applications in machine learning, hypothesis testing, and simulation, makes them a key framework for analyzing and modeling binary outcomes. Here are the takeaways: Bernoulli distributions model single trials with two possible outcomes. The probability of success p is the key parameter that defines the distribution. They have many applications in binary classification, hypothesis testing, and simulations. Understanding their properties and common misconceptions is useful for effective application. While powerful, alternatives like binomial and geometric distributions may be more appropriate in certain scenarios. If you are interested in applying these concepts in specific programming environments, our Introduction to Statistics in Python course and Introduction to Statistics in R course offer practical, hands-on learning experiences. And if you are ready to look into more advanced topics, the Mixture Models in R course builds on these foundational concepts to explore more complex statistical modeling techniques. Finally, look at our Machine Learning in Production in Python skill track to bring your machine learning skills to production level and start deploying advanced models. Build Machine Learning Skills Elevate your machine learning skills to production level. Start Learning for Free Author Vinod Chugani As an adept professional in Data Science, Machine Learning, and Generative AI, Vinod dedicates himself to sharing knowledge and empowering aspiring data scientists to succeed in this dynamic field. Frequently Asked Questions What is a Bernoulli distribution? A Bernoulli distribution is a discrete probability distribution for a random variable that has only two possible outcomes: success (usually denoted as 1) or failure (usually denoted as 0). It is characterized by a single parameter p, which represents the probability of success. What's the difference between a Bernoulli and a Binomial distribution? A Bernoulli distribution models a single trial with two possible outcomes, while a Binomial distribution models the number of successes in a fixed number of independent Bernoulli trials. You can think of a Binomial distribution as a sum of multiple Bernoulli distributions. Can Bernoulli distributions be used for hypothesis testing? Yes, Bernoulli distributions are commonly used in hypothesis testing, especially for scenarios involving proportions or success rates. They're particularly useful in A/B testing, where you're comparing the success rates of two different strategies or treatments. How do Bernoulli distributions relate to odds and log-odds? The odds of success in a Bernoulli trial are p/(1-p), where p is the probability of success. The log-odds, also known as the logit function, is the natural logarithm of the odds. These concepts are commonly used in logistic regression and other statistical analyses. How can I estimate the parameter p of a Bernoulli distribution from data? The most common method to estimate p is the maximum likelihood estimation (MLE). For a Bernoulli distribution, the MLE of p is simply the proportion of successes in your sample. For example, if you observe 7 successes out of 10 trials, your estimate of p would be 0.7. Topics Data ScienceData Analysis Vinod Chugani As an adept professional in Data Science, Machine Learning, and Generative AI, Vinod dedicates himself to sharing knowledge and empowering aspiring data scientists to succeed in this dynamic field. 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1253	https://www.quora.com/What-must-be-the-power-of-a-correcting-lens-for-a-far-sighted-person-who-can-see-near-point-at-100cm	What must be the power of a correcting lens for a far sighted person who can see near point at 100cm? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Vision, Sight Near Point of Convergence Dioptric Corrector Lens Vision Correction Long-sighted Optometry Accurate Optics NearPlace Vision Condition 5 What must be the power of a correcting lens for a far sighted person who can see near point at 100cm? All related (33) Sort Recommended Assistant Bot · 1y To determine the power of a correcting lens for a farsighted person (hyperopia), we can use the lens formula and the concept of the near point. Understanding the Near Point: A normal near point for a young adult is typically around 25 cm. However, in this case, the person's near point is at 100 cm. Calculating the Focal Length: The near point (D) of the person is at 100 cm (1 m). To correct their vision, we want to bring the near point to the normal distance of 25 cm (0.25 m). We can use the lens formula: 1 f=1 v−1 u 1 f=1 v−1 u Here, v v is the image dist Continue Reading To determine the power of a correcting lens for a farsighted person (hyperopia), we can use the lens formula and the concept of the near point. Understanding the Near Point: A normal near point for a young adult is typically around 25 cm. However, in this case, the person's near point is at 100 cm. Calculating the Focal Length: The near point (D) of the person is at 100 cm (1 m). To correct their vision, we want to bring the near point to the normal distance of 25 cm (0.25 m). We can use the lens formula: 1 f=1 v−1 u 1 f=1 v−1 u Here, v v is the image distance (the desired near point, which is -0.25 m because it's a virtual image), and u u is the object distance (the current near point, which is -1.0 m). Substituting the Values: Substituting these values into the lens formula gives: 1 f=1−0.25−1−1.0 1 f=1−0.25−1−1.0 1 f=−4+1=−3 m−1 1 f=−4+1=−3 m−1 Thus, the focal length f f is: f=−1 3 m≈−0.333 m f=−1 3 m≈−0.333 m 4. Calculating the Power of the Lens: The power P P of a lens is given by: P=1 f(in meters)P=1 f(in meters) Therefore: P=1−0.333≈−3 D P=1−0.333≈−3 D Conclusion The power of the correcting lens needed for a farsighted person who can see their near point at 100 cm is approximately -3 diopters. Upvote · Sandhya Uma Shekar Optimistic Optometrist · Author has 450 answers and 1.3M answer views ·7y A 3.00 D lens. Here is how I arrived: Power of the lens = 1/ focal length We now need to calculate the focal length using the near point which is 100cm. f= uv/u+v u= object size v= image size and since the image is virtual it is usually prefixed with a - sign. The near point of a normal eye(u) is 25 cm. Using this, f= 25-100/25+(-100) = 33.3 cm P=1/f and 1 is in meter P=100cm/33.3cm = 3.00 D Upvote · 9 5 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 207 Ed Averbukh Retina surgeon · Upvoted by Mitul Mehta , Vitreoretinal (Eye) surgeon at Univ of California, Irvine · Author has 1.8K answers and 4.8M answer views ·7y Originally Answered: The near point of the hypermetropic eye is 100cm. What is the power of lens required to correct this? · Near point is a useful concept to evaluate accommodation. Without it, it is the same as the far point. There's no near point for hypermetropia, unless you have accommodation. To evaluate hypermetropia in a child we need cycloplegic refraction where accommodation is pharmacologically negated. Only then we can prescribe glasses for hypermetropia. Uncorrected hypermetropia in a child may induce strabismus. If you need to read, but your near point is at one meter you have to add 2 diopters for reading comfortably at 1/3 meter. Assuming hypermetropia and accommodation, you don't need glasses for far Continue Reading Near point is a useful concept to evaluate accommodation. Without it, it is the same as the far point. There's no near point for hypermetropia, unless you have accommodation. To evaluate hypermetropia in a child we need cycloplegic refraction where accommodation is pharmacologically negated. Only then we can prescribe glasses for hypermetropia. Uncorrected hypermetropia in a child may induce strabismus. If you need to read, but your near point is at one meter you have to add 2 diopters for reading comfortably at 1/3 meter. Assuming hypermetropia and accommodation, you don't need glasses for far yet, your accommodation compensates for hypermetropia. Upvote · 9 2 Related questions More answers below A nearsighted woman has a near point of 10 cm and a far point of 50 cm. What power of lens is required to correct her defect? A nearsighted person has a near point of 12 cm and a far point of 17 cm. If the corrective lens is 2.0 cm from his eye, what lens power will enable this person to see distant objects clearly (answer should be in diopters)? A far-sighted person has a near point at 100 cm. What must be the power of the correcting lens? Are concave lenses for near-sighted or far-sighted? What is the far point and near point of the human eye with normal vision? Regina Reyes Works at Self-Employment ·3y Related What is the focal length of the lens required to correct the defect of the eyes of a woman whose near point is 50cm and least distance of vision of the normal eye is 25cm? Since the NP of the eye is greater than the standard NP of 25 cm; the eye defect is hyperopia which is also known as farsightedness. To correct hyperoptic vision, convex lenses are recommended. I'm 56 years old, and I've had to replace my spectacles frequently due to the rapid deterioration of my vision. It got to the point where I couldn't even read the words on a menu board, let alone the writing on a book a few inches away from my face. These last several years have been stressful due to the possibility of going blind very soon, but thanks to this product, I no longer have to be concerned. My Continue Reading Since the NP of the eye is greater than the standard NP of 25 cm; the eye defect is hyperopia which is also known as farsightedness. To correct hyperoptic vision, convex lenses are recommended. I'm 56 years old, and I've had to replace my spectacles frequently due to the rapid deterioration of my vision. It got to the point where I couldn't even read the words on a menu board, let alone the writing on a book a few inches away from my face. These last several years have been stressful due to the possibility of going blind very soon, but thanks to this product, I no longer have to be concerned. My daily activities have been transformed into squinting at everything around me as my eyesight deteriorated from seeing solid objects to hazy smudges. It wasn’t until my doctor recommended me this medicine that I’ve been able to pick myself up again. It's been two weeks, and it's like if I've never had any eye problems in the previous two years! Although my vision is still slightly blurry, it’s better than when I’m using my glasses that have almost half an inch of glass. Ever since I’ve been taking these meds, I’ve only been praising it! Here is how a convex lens works in correcting hyperopia: the user can place the object at the standard NP (25 cm) and the corrective lens would project the image of the object to the user’s NP (in this case, 50 cm). The image has to be ERECT relative to the object (else the eye will have a difficult time adjusting to seeing inverted images!) Erect images are virtual. Using the lens equation, we can calculate the focal length: 1/f = 1/25 - 1/50 → f = 50 cm. This would be a + 2 Diopter lens (reciprocal of f in meters). Incidentally, it is the same type of lens that I am wearing at this very instant that I am writing this! Addendum 1 (bonus): the magnification of this situation is 2 X (Di/Do). Addendum 2 (second bonus): the Far Point (FP) of the woman’s eyes is still infinity. The rule of thumb for measuring infinity is 400 f (where f is the focal length of the eye, which on average is, 1.5 cm). 400 f = 600 cm = 6 meters = 20 ft (which is the number most know about “true” vision). Upvote · 9 2 Inex Es Former Optician (retired) · Author has 257 answers and 2.7M answer views ·2y Related A near sighted woman needed to see at 100 cm away from her eyes. What type of corrective lens and focal length must she use? 100 cm is a bit over 3 ft. There are several possibilities. How old is she? What is her distance Rx? She may want a multifocal lens, she may want a dedicated pair of glasses for within 100 cms or…she may just take them off to see that far and be perfectly fine. Upvote · Sponsored by RedHat Customize AI for your needs, with simpler model alignment tools. Your AI needs context, not common knowledge. Learn More 9 7 Leta Rosetree Former Former Psychotherapist on AIDS & cancer units. · Author has 14.7K answers and 6.8M answer views ·5y Related What power lens is needed to correct for farsightedness where the uncorrected near point distance is 75 cm? You need a licensed Optician or an Optometrist, maybe an Ophthalmologist, to test you to get an accurate answer, not a bunch of strangers online. You should probably be checked for a few other critical factors such at glaucoma as well. Do not mess around with your vision - it has to last the rest of your life. Ask your doctor for a referral for the correct one. Upvote · 9 1 Related questions More answers below Why can people be both near-sighted and far-sighted simultaneously? My left eye is near-sighted while the other is far-sighted, is there any way to correct them? I'm only 16. Can you get glasses with one lens correcting for near-sightedness, and the other for far-sightedness? Can far-sighted people see faraway with their glasses on? Which lens can be used to correct both long and short-sightedness? Ben Wilkinson Former Ophthalmologist (1972–2003) · Author has 19.1K answers and 13.8M answer views ·2y Related A near sighted woman needed to see at 100 cm away from her eyes. What type of corrective lens and focal length must she use? It says that there is an answer, but I cannot find it. It depends on her age and distance correction. With no accommodation, she will need her distance correction + 1.00D to focus at 1.00 metre. She may need no correction at all. Upvote · Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 619 Malcolm Maciver Optometrist & Academic · Author has 429 answers and 1.1M answer views ·5y Related A nearsighted woman wears glasses (3 cm in front of her eyes) to correct for a far point that is 2.0 m from her eyes. What power should her glasses be? 3cm is an awfully long way from the front of the eye… the furthest BVD id typically see would be 14mm. But ill bite:- Ocular far point of 2m would begin by giving a refraction of (F=1/-f=1/-2) -0.50D. Now from a realistic perspective this is as much as you need to know as when prescribing we work in 0.25D steps, from experience (or calculation above) I know that the next higher step (-0.75) has a focal length of 1.33m, while the figure is 3cm in front of the eye - its not far enough to justify altering the correction from -0.50 to -0.75. However if you want from an optics perspective: 2m - 3cm = 1. Continue Reading 3cm is an awfully long way from the front of the eye… the furthest BVD id typically see would be 14mm. But ill bite:- Ocular far point of 2m would begin by giving a refraction of (F=1/-f=1/-2) -0.50D. Now from a realistic perspective this is as much as you need to know as when prescribing we work in 0.25D steps, from experience (or calculation above) I know that the next higher step (-0.75) has a focal length of 1.33m, while the figure is 3cm in front of the eye - its not far enough to justify altering the correction from -0.50 to -0.75. However if you want from an optics perspective: 2m - 3cm = 1.97m F = 1/-1.97 = -0.5076D (so abt 1.5% diff) so picking between -0.50 and -0.75 ill be going with the former Hope this helps PS if you want to make it interesting use examples of far points around 15–18cm in front of the eye - you’ll see a big difference even at 12–14mm Upvote · 9 2 9 1 Jayanta Mukherjee B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990) · Author has 43.6K answers and 11.1M answer views ·4y Related What is the focal length of the corrective lens that will allow the person to see very far away? From this question, it appears that the person is having nearsightedness or myopic vision; which makes it hard for him / her to see distant objects clearly. While the person can have a clear view of objects lying nearer; but will get a blurred vision of the objects lying far-off. Light getting reflected from any object enters our eye to get incident on the ocular lens, which varies its focal length as per need to bend the light rays getting refracted through the lens and making them meet on retina, the light sensitive part of the eye; thus ensuring formation of a real image of the object on ret Continue Reading From this question, it appears that the person is having nearsightedness or myopic vision; which makes it hard for him / her to see distant objects clearly. While the person can have a clear view of objects lying nearer; but will get a blurred vision of the objects lying far-off. Light getting reflected from any object enters our eye to get incident on the ocular lens, which varies its focal length as per need to bend the light rays getting refracted through the lens and making them meet on retina, the light sensitive part of the eye; thus ensuring formation of a real image of the object on retina, enabling us to view the object. When the eyeball is too long or the cornea (protective layer of the eye) is too curved; then the light rays entering the eye from far-off objects won't be focussed properly by the ocular lens and the refracted rays instead of meeting on the retina, actually end up meeting at a point in front of retina. So, the eye senses a hazy image of the far-off object formed on the retina. This defect in vision can be corrected by applying a concave (diverging) lens of suitable focal length; which will make the parallel light rays coming from the object sufficiently divergent before hitting the ocular lens and ensure meeting point of the refracted rays getting shifted slightly backward and on the retina - to enable the eye to sense a clear image of the distant object. So, all it needed is a corrective concave lens of appropriate focal length. Now, to determine the value of focal length of the corrective concave lens; it needs to be calculated from the facts that how far the person can see without any vision aid and upto what distance the person intends to see. Upvote · 9 2 9 1 Sponsored by LPU Online Career Ka Turning Point with LPU Online. 100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support. Apply Now 999 262 Malcolm Maciver Optometrist & Academic · Author has 429 answers and 1.1M answer views ·5y Related A nearsighted person has a near point of 12 cm and a far point of 17 cm. If the corrective lens is 2.0 cm from his eye, what lens power will enable this person to see distant objects clearly (answer should be in diopters)? ok, so the first thing is that the near point isn't relevant to the question at hand (confounder? or subsequent question?) what matters here is the far point. so if we were fitting with a contact lens we’d look at far point focal power of (F=1/-f = 1/-0.17) -5.88D (half way between -5.75 and -5.50), however as we’re discussing spectacles there are two approaches: either the step forward formula F’= K’/(1+dK’) or the differential far point 0.17–0.02 = 0.15, giving a power of F= -6.66D (half way between -6.50 and -6.75D) as for the other info, well far point 17cm and near point 12cm gives an anticip Continue Reading ok, so the first thing is that the near point isn't relevant to the question at hand (confounder? or subsequent question?) what matters here is the far point. so if we were fitting with a contact lens we’d look at far point focal power of (F=1/-f = 1/-0.17) -5.88D (half way between -5.75 and -5.50), however as we’re discussing spectacles there are two approaches: either the step forward formula F’= K’/(1+dK’) or the differential far point 0.17–0.02 = 0.15, giving a power of F= -6.66D (half way between -6.50 and -6.75D) as for the other info, well far point 17cm and near point 12cm gives an anticipated ocular accommodation of 2.45D available. for a follow on question of what add should be added for a near vision point of 40cm, add = dioptric dist - 1/2accomm rounds nicely to a +1.25 add. Hope this helps Upvote · 9 4 9 1 Jim Seymour Former Optometrist-retired at National Health Service (NHS) (1978–2019) · Author has 993 answers and 1.1M answer views ·5y Related What power lens is needed to correct for farsightedness where the uncorrected near point distance is 75 cm? If, by ‘farsightedness’ you mean hypermetropia (i.e. the opposite of myopia) there is insufficient information to calculate the required positive lens power. If, you mean ‘…to correct for distance vision when far point of focus of the eye is 75cm’ … for an eye that is myopic, all you need to do is express the far point distance in metres and calculate the reciprocal and your result will be in dioptres. But why should I deprive you of the satisfaction of doing it for yourself? If one has a answer of more than -6.00, (which you won’t) one should make an allowance for the distance of the trail lens Continue Reading If, by ‘farsightedness’ you mean hypermetropia (i.e. the opposite of myopia) there is insufficient information to calculate the required positive lens power. If, you mean ‘…to correct for distance vision when far point of focus of the eye is 75cm’ … for an eye that is myopic, all you need to do is express the far point distance in metres and calculate the reciprocal and your result will be in dioptres. But why should I deprive you of the satisfaction of doing it for yourself? If one has a answer of more than -6.00, (which you won’t) one should make an allowance for the distance of the trail lens from the cornea vs. the distance of the spectacle lens from the cornea. But this is getting a bit exotic for Quora, an optometry textbook would explain it. Upvote · Ben Wilkinson Former Ophthalmologist (1972–2003) · Author has 19.1K answers and 13.8M answer views ·1y Related Which lens is used for short-sightedness? Minus lenses. Myopia By Ben Wilkinson FRCSC 2020 Dec. Myopia, shortsightedness, is a condition in which the eyeballs grow too (optically) long. It appears to be a largely genetic (hereditary) condition, which develops with growth, and stabilises at the end of growth, around age 17-21. It is becoming more common in recent years, and there is debate about the reasons. One explanation, the one I favour, is that the development of myopia is associated with growth. People are growing taller, and faster than ever before. This can be seen most easily in what has happened in Japan. Before WWI, Japanese p Continue Reading Minus lenses. Myopia By Ben Wilkinson FRCSC 2020 Dec. Myopia, shortsightedness, is a condition in which the eyeballs grow too (optically) long. It appears to be a largely genetic (hereditary) condition, which develops with growth, and stabilises at the end of growth, around age 17-21. It is becoming more common in recent years, and there is debate about the reasons. One explanation, the one I favour, is that the development of myopia is associated with growth. People are growing taller, and faster than ever before. This can be seen most easily in what has happened in Japan. Before WWI, Japanese people were characterised as small, and not needing glasses. They were rarely photographed, or illustrated as wearing glasses. Very quickly, their diet of mainly rice, with flavourings, changed to a much more Western diet, and they grew taller, and far more myopic. Since WWII, they have become taller still, and less slim. Now, it seems, almost all Japanese are myopic. One, unlikely, explanation is that they have always been myopic, but glasses were not available. This is unlikely, as good distance vision did not appear to be a prized skill in warrior times. At present, there is nothing which can be done to control the growth of the eyes, so myopia is untreatable. Once growth has ceased, surgical correction (LASIK at the present time) becomes an option. As myopia begins, it is important to make sure that the change in vision is not due to pseudomyopia. This is a condition of cramping of the focusing muscles, so that they cannot relax, to allow clear distance vision. It is easily diagnosed by remeasuring the patient after the dilating drops have had the effect of temporarily paralysing the focusing muscles. Some ‘treatments’ for myopia have been devised, citing successes which are probably due to treating pseudomyopia, and confusing it with true myopia. High myopia, with a prescription of -6.00D or more, carries with it a risk of certain complications which increase with the stronger prescriptions. The complications are explained as due to a stress between the size of the sclera (the outer, tough layer of the eye) and the retina, the light sensitive ‘film’ in the eye, because they are derived from different kinds of tissue in the embryo. The refractive error caused by myopia is corrected by glasses, contact lenses, or by ortho-keratology. Orthokeratology is a system in which the patient wears contact lenses at night, which bend the cornea so that correction does not need to be worn in the daytime. There is also refractive surgery, in which the curvature of the cornea is altered. The most common procedure is LASIK. Once you need contacts or glasses, there is debate as to whether you should wear them all the time or not. Glasses change the optics to normal. This means that, with your eyes relaxed, you are focused in the far distance (infinity); when an object comes near, you use muscular accommodation to keep the object in focus, this is normal. So, wearing the glasses enables your eyes to behave normally. This has no effect on the myopia, neither making it better or worse. Taking your glasses off is similar to putting reading glasses on, if your prescription is moderate. My advice is that, if you need your glasses frequently, you are wiser to wear them constantly. If you are wearing them, it is very hard to lose them! It is also more difficult to sit on them, or scratch them. So, in general, you should wear them except when swimming, bathing, or in bed. Upvote · 9 2 Ben Wilkinson FRCSC in Ophthalmology, The University of British Columbia (Graduated 1972) · Author has 19.1K answers and 13.8M answer views ·1y Related Is it possible to correct both far-sightedness and near-sightedness with one pair of glasses, or is it necessary to use two pairs of glasses at different distances from the eyes? Presbyopia By Ben Wilkinson 2021 May In the human embryo, the lens of the eye starts as a piece of ectoderm (skin). It is attracted by a bud from the brain, which eventually produces the retina, and sinks below the surface to form the lens. Like other skin structures, it continues to produce cells throughout life. However, it cannot be allowed to increase beyond slightly, so it increases in hardness. The lens hardens at a remarkably predictable rate. In the ‘normal’ (emmetropic) eye, the stiffening of the lens leads to loss of accommodation, so that the near point retreats to about 30 cm by about Continue Reading Presbyopia By Ben Wilkinson 2021 May In the human embryo, the lens of the eye starts as a piece of ectoderm (skin). It is attracted by a bud from the brain, which eventually produces the retina, and sinks below the surface to form the lens. Like other skin structures, it continues to produce cells throughout life. However, it cannot be allowed to increase beyond slightly, so it increases in hardness. The lens hardens at a remarkably predictable rate. In the ‘normal’ (emmetropic) eye, the stiffening of the lens leads to loss of accommodation, so that the near point retreats to about 30 cm by about 45 years of age, and reading glasses (+ lenses) start to be needed. This loss of near vision is termed ‘presbyopia’. The lens continues to harden, and the near point to retreat. Eventually, the lens becomes so hard that cracks and splits begin to appear. When these interfere with vision, it is termed ‘cataract’. Upvote · 9 2 Related questions A nearsighted woman has a near point of 10 cm and a far point of 50 cm. What power of lens is required to correct her defect? A nearsighted person has a near point of 12 cm and a far point of 17 cm. If the corrective lens is 2.0 cm from his eye, what lens power will enable this person to see distant objects clearly (answer should be in diopters)? A far-sighted person has a near point at 100 cm. What must be the power of the correcting lens? Are concave lenses for near-sighted or far-sighted? What is the far point and near point of the human eye with normal vision? Why can people be both near-sighted and far-sighted simultaneously? My left eye is near-sighted while the other is far-sighted, is there any way to correct them? I'm only 16. Can you get glasses with one lens correcting for near-sightedness, and the other for far-sightedness? Can far-sighted people see faraway with their glasses on? Which lens can be used to correct both long and short-sightedness? I have become far-sighted after putting on contact lenses. I was originally near-sighted. Is that normal? Which one is worse: being nearsighted or farsighted? What type of lens is used to correct long sightedness? What happens to the power of the eye lens on myopia? Why can one get far-sighted correction glasses at the pharmacy but not near sighted? Related questions A nearsighted woman has a near point of 10 cm and a far point of 50 cm. What power of lens is required to correct her defect? A nearsighted person has a near point of 12 cm and a far point of 17 cm. If the corrective lens is 2.0 cm from his eye, what lens power will enable this person to see distant objects clearly (answer should be in diopters)? A far-sighted person has a near point at 100 cm. What must be the power of the correcting lens? Are concave lenses for near-sighted or far-sighted? What is the far point and near point of the human eye with normal vision? Why can people be both near-sighted and far-sighted simultaneously? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1254	https://artofproblemsolving.com/wiki/index.php/Ball-and-urn?srsltid=AfmBOorjz6Rnd1Wdzju4G-hClC8E8XCl4ISibA5baNtIbZyAvMKVH7lj	Art of Problem Solving Ball-and-urn - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Ball-and-urn Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Ball-and-urn The ball-and-urn technique, also known as stars-and-bars, sticks-and-stones, or dots-and-dividers, is a commonly used technique in combinatorics. It is used to solve problems of the form: how many ways can one distribute indistinguishable objects into distinguishable bins? We can imagine this as finding the number of ways to drop balls into urns, or equivalently to arrange balls and dividers. For example, represent the ways to put objects in bins. The number of ways to do such is . More generally, the number of ways to put objects into bins is or . The number of ways to put objects into bins, where each bin must have at least 1 object in it, is . Contents 1 Restrictions 2 Reasoning 1 3 Reasoning 2 4 Problems Restrictions Main article: Distinguishability Let's say that we want to put objects in bins, but there must be at least objects in each bin. Without the restriction, we can set the following equation up: . We know that each (the bins) must have at least objects in them, so we can subtract from , since that's how many objects are left. This means that there are ways to distribute the objects. Well, what if we can have at most objects in each bin? We can use the following formula to find this: This can be derived using the Principle of Inclusion-Exclusion. Reasoning 1 Arranging 's and \|'s is the same as saying there are positions: and you want to fill of them with 's and the rest of them with \|'s. Thus you are choosing positions out of total positions, or equivalently, positions for the dividers out of total spots, resulting in a total of ways. Reasoning 2 If you could only put one ball in each urn, then there would be possibilities; the problem is that you can repeat urns, so this does not work. and then you also have urns labeled "repeat 1st", "repeat 2nd", ..., and "repeat -th". After the balls are in urns you can imagine that any balls in the "repeat" urns are moved on top of the correct balls in the first urns, moving from left to right. There is a one-to-one correspondence between the non-repeating arrangements in these new urns and the repeats-allowed arrangements in the original urns. For a simple example, consider balls and urns. The one to one correspondence between several of the possibilities and the "repeated urns" version is shown. Since there are 4 balls, these examples will have three possible "repeat" urns. For simplicity, I am listing the numbers of the urns with balls in them, so "1,1,2,4" means balls in urn in urn and in urn The same is true for the "repeat" urns options but I use the notation etc. (no repeats). . , . , . Since the re-framed version of the problem has urns, and balls that can each only go in one urn, the number of possible scenarios is simply Note: Due to the principle that , we can say that . Problems 1986 AIME Problems/Problem 13 2001 AMC 10 Problems/Problem 19 2003 AMC 10A Problems/Problem 21 2004 AMC 8 Problems/Problem 17 Mock AIME 3 Pre 2005 Problems/Problem 2 2007 AIME I Problems/Problem 10 2018 AMC 8 Problems/Problem 23 2019 AMC 8 Problems/Problem 25 2018 AMC 10A Problems/Problem 11 2020 AMC 10B Problems/Problem 25 Retrieved from " Category: Combinatorics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1255	https://www.jmap.org/LessonPlans/A.CED.A.1b.pdf	H – Quadratics, Lesson 4, Geometric Applications of Quadratics (r. 2018) QUADRATICS Geometric Applications of Quadratics Common Core Standard A-CED.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic func-tions, and simple rational and exponential functions. PARCC: Tasks are limited to linear, quadratic, or exponential equations with integer exponents. Next Generation Standard AI-A.CED.1 Create equations and inequalities in one variable to represent a real-world context. (Shared standard with Algebra II) Notes: • This is strictly the development of the model (equa-tion/inequality). • Limit equations to linear, quadratic, and exponen-tials of the form f(x) = a(b)x where a > 0 and b > 0 (b ≠ 1). • Work with geometric sequences may involve an ex-ponential equation/formula of the form an = arn-1, where a is the first term and r is the common ratio. • Inequalities are limited to linear inequalities. • Algebra I tasks do not involve compound inequalities. LEARNING OBJECTIVES Students will be able to: 1) model quadratic equations that reflect real-world contexts involving the area and dimensions of two-dimensional geometric figures. Overview of Lesson Teacher Centered Introduction Overview of Lesson - activate students’ prior knowledge - vocabulary - learning objective(s) - big ideas: direct instruction - modeling Student Centered Activities guided practice Teacher: anticipates, monitors, selects, sequences, and connects student work - developing essential skills - Regents exam questions - formative assessment assignment (exit slip, explain the math, or journal entry) VOCABULARY area area formulas length width BIG IDEAS Geometric Area Problems: Quadratics are frequently used to model problems involving geometric area. The keys to solving geometric area problems are to use a geometric area formula and draw a sketch to represent the problem. Typical Problem in Context Mathematical Translation Hints and Strategies The area of the rectangular playground enclosure at South School is 500 square meters. The length of the playground is 5 meters longer than the width. Find the dimensions of the playground, in meters. width = 20 and length = 25 A=lw Let A=500 Let width = w Let length = w+5 Write: { } 2 2 500 ( 5) 500 5 0 5 500 0 ( 25)( 20) 25,20 Reject -25 as a solution because width cannot be negative. A lw w w w w w w w w w = = + = + = + − = + − = − Start with a formula. Define variables. Substitute known information into the formula. Sketching a Diagram Can Help to Understand and Solve a Problem The general strategy for solving problems that involve geometric applications of quadratics is to substitute terms with a common variable for length and width in common area formulas. Drawing a picture can also help. For example: A rectangular garden has length of x+2 and width of 2x-3, and the area of the garden is 72 square feet. What are the dimensions of the garden? Start by drawing a picture to help understand the problem. Then, use the formula for finding the area of a rectangle, which is: Substitute information about the length and width of the garden into the area formula for a rectangle, then write: 2x-3 X+2 72 square feet ( )( ) 72 2 2 3 A lw x x = = + − The area of the garden is 72 square feet, so we can write: Solve for x, then for x+2 and 2x-3. The length is 8 feet and the width is 9 feet. DEVELOPING ESSENTIAL SKILLS 1) A contractor needs 54 square feet of brick to construct a rectangular walkway. The length of the walkway is 15 feet more than the width. Write an equation that could be used to determine the dimensions of the walkway. Solve this equation to find the length and width, in feet, of the walkway. 2) A rectangle has an area of 24 square units. The width is 5 units less than the length. What is the length, in units, of the rectangle? 3) Jack is building a rectangular dog pen that he wishes to enclose. The width of the pen is 2 yards less than the length. If the area of the dog pen is 15 square yards, how many yards of fencing would he need to completely enclose the pen? 4) A rectangular park is three blocks longer than it is wide. The area of the park is 40 square blocks. If w represents the width, write an equation in terms of w for the area of the park. Find the length and the width of the park. 5) What is the length of one side of the square whose perimeter has the same numerical value as its area? Answers 1) The formula for the area of a rectangle is A lw = Let 54 represent A. Let w represent the width of the rectangle. Let w+15 represent the length of the rectangle. Write: ( ) ( )( ) { } 2 2 54 15 54 15 0 15 54 0 18 3 18,3 A lw w w w w w w w w w = = + = + = + − = + − = − Reject the negative solution. The width of the sidewalk is 3 feet. The length of the sidewalk is 18 feet. 2) The formula for the area of a rectangle is A lw = Let 24 represent A. Let l represent the length of the rectangle. Let l-5 represent the width of the rectangle. Write: ( ) ( )( ) { } 2 2 24 5 24 5 0 5 24 0 8 3 3,8 l l l l l l l l l = − = − = − − = − + = − Reject the negative solution. The length of the rectangle is 8 units. 3) The formula for the area of a rectangle is A lw = Let 15 represent A. Let l represent the length of the rectangle. Let l-2 represent the width of the rectangle. Write: ( ) ( )( ) { } 2 2 15 2 15 2 0 2 15 0 5 3 3,5 A lw l l l l l l l l l = = − = − = − − = − + = − Reject the negative solution. If the length is 5, the width is 3. The formula for the perimeter of a rectangle is 2 2 P l w = + , so the length of fence needed is ( ) ( ) 2 2 2 5 2 3 16 P l w P P = + = + = 16 yards of fencing are needed. 4) The formula for the area of a rectangle is A lw = The units in this problem are blocks. Let 40 represent A. Let w represent the width of the rectangle. Let w+3 represent the length of the rectangle. Write: ( ) ( )( ) { } 2 2 40 3 40 3 0 3 40 0 8 5 8,5 w w w w w w w w w = + = + = + − = + − = − Reject the negative solution. The park is 5 blocks wide and 8 blocks long. 5) The formula for the area of a square is 2 A s = The formula for the perimeter of a square is 4 P s = Write: ( ) { } 2 2 4 0 4 0 4 0,4 s s s s s s s = = − = − = Reject the zero solution. The length of one side of the square is 4 units. REGENTS EXAM QUESTIONS (through June 2018) A.CED.A.1: Geometric Applications of Quadratics 210) The length of the shortest side of a right triangle is 8 inches. The lengths of the other two sides are represented by consecutive odd integers. Which equation could be used to find the lengths of the other sides of the triangle? 1) 3) 2) 4) 211) New Clarendon Park is undergoing renovations to its gardens. One garden that was originally a square is being adjusted so that one side is doubled in length, while the other side is decreased by three meters. The new rectangular garden will have an area that is 25% more than the original square garden. Write an equation that could be used to determine the length of a side of the original square garden. Explain how your equation models the situation. Determine the area, in square meters, of the new rectangular garden. 212) A rectangular garden measuring 12 meters by 16 meters is to have a walkway installed around it with a width of x meters, as shown in the diagram below. Together, the walkway and the garden have an area of 396 square meters. Write an equation that can be used to find x, the width of the walkway. Describe how your equation models the situation. Determine and state the width of the walkway, in meters. 213) A school is building a rectangular soccer field that has an area of 6000 square yards. The soccer field must be 40 yards longer than its width. Determine algebraically the dimensions of the soccer field, in yards. 214) A landscaper is creating a rectangular flower bed such that the width is half of the length. The area of the flower bed is 34 square feet. Write and solve an equation to determine the width of the flower bed, to the nearest tenth of a foot. 215) A rectangular picture measures 6 inches by 8 inches. Simon wants to build a wooden frame for the picture so that the framed picture takes up a maximum area of 100 square inches on his wall. The pieces of wood that he uses to build the frame all have the same width. Write an equation or inequality that could be used to determine the maximum width of the pieces of wood for the frame Simon could create. Explain how your equation or inequality models the situation. Solve the equation or inequality to determine the maximum width of the pieces of wood used for the frame to the nearest tenth of an inch. 216) Joe has a rectangular patio that measures 10 feet by 12 feet. He wants to increase the area by 50% and plans to increase each dimension by equal lengths, x. Which equation could be used to determine x? 1) 3) 2) 4) 217) A contractor has 48 meters of fencing that he is going to use as the perimeter of a rectangular garden. The length of one side of the garden is represented by x, and the area of the garden is 108 square meters. Determine, algebraically, the dimensions of the garden in meters. SOLUTIONS 210) ANS: 4 Strategy: Use the Pythagorean Theorem, the sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse. The shortest side must be one of the legs, since the longest side is always the hypotenuse. Substitute 8 for a in the equation. The lengths of sides b and c are consecutive odd integers. Let x represent the smaller odd integer and let represent the larger consecutive odd integer. Side c must be represented by because side c represents the hypotenuse, which is always the longest side of a right triangle. Therefore, side b is represented by x and side c is represented by . Substitute these values into the equation. By using the commutative property to rearrange the two terms in the right expression, we obtain the same equation as answer choice d. DIMS? Does It Make Sense? Yes. Transorm the equation for input into a graphing calculator as follows: and we find that the other two sides of the right triangle are 15 and 17. By the Pythagorean Theorem, Everything checks! PTS: 2 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 211) ANS: a) b) Because the original garden is a square, represents the original area, represents the side decreased by 3 meters, represents the doubled side, and represents the new garden with an area 25% larger. c) The length of a side of the original square garden was 8 meters. The area of the new rectangular garden is 80 square meters. Strategy: Draw two pictures: one picture of the garden as it was in the past and one picture of the garden as it will be in the future. Then, write and solve an equation to determine the length of a side of the original garden. STEP 1. Draw 2 pictures. Area of original garden is . Area of new garden is . STEP 2: Use the area formula, , to write an equation for the area of the new garden. STEP 3: Transform the equation for input into a graphing calculator and solve. x x Original Garden New Garden 2x X-3 The length on a side of the original square garden was 8 meters. The area of the new garden is square meters. DIMS? Does It Make Sense? Yes. The dimensions of the original square garden are 8 meters on each side and the area was 64 square meters. The dimensions of the new rectangular garden are 16 meters length and 5 meters width. The new garden will have area of 80 meters. The area of the new garden is 1.25 times the area of the original garden. PTS: 6 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 212) ANS: a) . b) The length, , and the width, , are multiplied and set equal to the area. c) The width of the walkway is 3 meters. Strategy: Use the picture, the area formula ( ), and information from the problem to write an equation, then solve the equation. STEP 1. Use the area formula, the picture, and information from the problem to write an equation. STEP 2. Solve the equation. The width of the walkway is 3 meters. DIMS? Does It Make Sense? Yes. The garden plus walkway is meters long and meters wide. , which fits the information in the problem. PTS: 4 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 213) ANS: The soccer field is 60 yards wide and 100 yards long. Strategy: Draw and label a picture, then use the picture to write and solve an equation based on the area formula: STEP 1: Draw and label a picture. STEP 2: Write and solve an equation based on the area formula: DIMS? Does It Make Sense? Yes. If the width of the soccer field is 60 yards and the length of the soccer field is 100 yards, then the area of the soccer field will be 6,000 square yards, as required by the problem. PTS: 4 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 214) ANS: Width (w) Length (w + 40) Area 6000 a) Equation b) The width of the flower bed is approximately 4.1 feet. Strategy: Draw a picture, then write and solve an equation based on the area formula, . STEP 1. Draw a picture. STEP 2: Write and solve an equation based on the area formula. . PTS: 2 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 215) ANS: The maximum width of the frame should be 1.5 inches. Strategy: Write an inequality, then solve it. STEP 1: Write the inequality. The picture is 6 inches by 8 inches. The area of the picture is . The width of the frame is an unknown variable represented by x. Two widths of the frame (2x) must be added to the length and width of the picture. Therefore, the area of the picture with frame is The area of the picture with frame, , must be less than or equal to 100. Write STEP 2: Solve the inequality. Area = 34 Length (l) Width (l/2) Notes Left Expression Sign Right Expression Given Use Distributive Property to Clear Parentheses 100 Commutative Property 100 Combine Like Terms 100 Subtract 100 from both expressions 0 Use the Quadratic Formula: a=4, b=28, c=-52 DIMS? Does It Make Sense? Yes. If the frame is 1.5 inches wide, then the total picture with frame will be PTS: 6 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 216) ANS: 2 Strategy: STEP 1. First, determine the area of the current rectangular patio and increase its size by 50%, which will be the size of the new patio. STEP 2. Then, increase each dimension of the current rectangular patio by x, as follows: STEP 1. The new patio will have an area of 180 square feet. Eliminate choice (a). STEP 2. Choose answer b. PTS: 2 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics 217) ANS: The garden is a rectangle that measures 18 meters by 6 meters. Strategy: . Solve as a system of two equations, because the question requires solving for two variables: length and width. STEP 1. Draw a picture that illustrates the information in the problem. STEP 2. Using the picture, write two equations using length and area formulas for rectangles. Let l represent the unknown length of the garden and let w represent the unknown width of the garden. The first equation, , is based on the formula for the perimeter of a rectangle, which is . The second equation, , is based on the area formula for rectangles, which is STEP 2. Isolate the length variable Area = 108 Width (w) Length (l) Perimeter equals 2w+2l STEP 3. Solve and as a system using substitution, as follows: The garden is 6 meters wide. The length of the garden can be found using . PTS: 4 NAT: A.CED.A.1 TOP: Geometric Applications of Quadratics
1256	https://www.biointeractive.org/classroom-resources/translation-basic-detail	Skip to main content Explore Related Content Other Resources About the Central Dogma Showing 5 of 14 Animations RNA Splicing Lessons What Causes Different Fur Colors? Card Activities Central Dogma Card Sorting Activity Click & Learn Central Dogma and Genetic Medicine Short Films The Double Helix Labs & Demos Building a Paper Model of CRISPR-Cas9 Animations DNA Transcription (Advanced Detail) Animations Coding Sequences in DNA Animations DNA Transcription (Basic Detail) Animations The Chemical Structure of DNA Animations Triplet Code Animations Paired DNA Strands Animations Building Blocks of DNA Animations Translation (Advanced Detail) Animations RNA Splicing Lessons What Causes Different Fur Colors? Card Activities Central Dogma Card Sorting Activity Click & Learn Central Dogma and Genetic Medicine Short Films The Double Helix Labs & Demos Building a Paper Model of CRISPR-Cas9 Animations DNA Transcription (Advanced Detail) Animations Coding Sequences in DNA Animations DNA Transcription (Basic Detail) Animations The Chemical Structure of DNA Animations Triplet Code Animations Paired DNA Strands Animations Building Blocks of DNA Animations Translation (Advanced Detail) Animations RNA Splicing Lessons What Causes Different Fur Colors? Card Activities Central Dogma Card Sorting Activity Click & Learn Central Dogma and Genetic Medicine Short Films The Double Helix Other Related Resources Showing 5 of 10 Animations DNA Transcription (Advanced Detail) Click & Learn Structure and Function of Telomeres Animations DNA Transcription (Basic Detail) Animations The Chemical Structure of DNA Animations DNA Replication (Advanced Detail) Animations How DNA Is Packaged Animations Translation (Advanced Detail) Click & Learn The p53 Gene and Cancer Animations Triplet Code Labs & Demos Modeling the Structure of DNA Animations DNA Transcription (Advanced Detail) Click & Learn Structure and Function of Telomeres Animations DNA Transcription (Basic Detail) Animations The Chemical Structure of DNA Animations DNA Replication (Advanced Detail) Animations How DNA Is Packaged Animations Translation (Advanced Detail) Click & Learn The p53 Gene and Cancer Animations Triplet Code Labs & Demos Modeling the Structure of DNA Animations DNA Transcription (Advanced Detail) Click & Learn Structure and Function of Telomeres Animations DNA Transcription (Basic Detail) Animations The Chemical Structure of DNA Animations DNA Replication (Advanced Detail) This Resource Appears in the Following Playlists Showing 2 of 2 Teaching Gene Expression Using Genetic Medicine 7 Resources Teaching the Central Dogma Using the Rock Pocket Mouse and Genetic Medicine 7 Resources By: BioInteractive
1257	https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas?srsltid=AfmBOoq-xQG487eYe3KuozErEepA0fL-nd5xHqqw-EMbpDnNQ9GxBEtT	Art of Problem Solving Vieta's Formulas - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Vieta's Formulas Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Vieta's Formulas In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients. It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments. Contents 1 Statement 2 Proof 3 Problems 3.1 Introductory 3.2 Intermediate 4 Advanced 5 See also Statement Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots. Vieta’s formulas then state that This can be compactly summarized as for some such that . Proof Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant . Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof. Problems Here are some problems with solutions that utilize Vieta's quadratic formulas: Introductory 2005 AMC 12B Problem 12 2007 AMC 12A Problem 21 2010 AMC 10A Problem 21 2003 AMC 10A Problem 18 2021 AMC 12A Problem 12 Intermediate 2017 AMC 12A Problem 23 2003 AIME II Problem 9 2008 AIME II Problem 7 2021 Fall AMC 12A Problem 23 2019 AIME I Problem 10 Advanced 2020 AIME I Problem 14 See also Polynomial Retrieved from " Categories: Algebra Polynomials Theorems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1258	https://math-gpt.org/problems/question/a-rectangle-is-inscribed-in-a-circle-such-that-each-vertex-of-the-rectangle	To solve this problem, let's denote the sides of the rectang \| MathGPT To solve this problem, let's denote the sides of the rectangle as a and b where a is the shortest side. According ... Question: To solve this problem, let's denote the sides of the rectangle as a and b where a is the shortest side. According to the problem, the diagonal d of the rectangle can be expressed using the Pythagorean theorem as:d = a^2 + b^2We are also given that the diagonal of the rectangle is twice the length of the shortest side, which gives us the equation:d = 2aThus, we can equate the two expressions for the diagonal:a^2 + b^2 = 2aSquaring both sides, we have:a^2 + b^2 = 4a^2This simplifies to:b^2 = 3a^2b = 3aThe area A of the rectangle is given by:A = a b = a (3a) = 3a^2We know that the area is 1,0893 , so we can set up the equation:3a^2 = 1,0893Dividing both sides by 3 yields:a^2 = 1,089Taking the square root gives:a = 1,089 = 33Now, using the relationship we found earlier for b :b = 3a = 3 33 = 333Next, we can find the length of the diagonal d :d = a^2 + b^2 = 33^2 + (333)^2Calculating this yields:d = 1,089 + 3 1,089 = 1,089(1 + 3) = 1,089 4 = 4,356 = 66Since the diameter of the circle is the same as the length of the diagonal of the rectangle, we find that the diameter is:Diameter = d = 66Thus, the final answer is:66 Answer: Let the shortest side of the rectangle be and the longer side be . Given that the diagonal is twice the length of the shortest side, we have the diagonal as . Since the rectangle is inscribed in the circle, the diagonal of the rectangle is the diameter of the circle. From the Pythagorean theorem, we have: ... Unlock with MathGPT [a^2 + b^2 = 4a^2.] The area of the rectangle is given by . Substituting , we get: Since , we find as: The diameter of the circle, which is the diagonal of the rectangle, is : Thus, the length of the diameter of the circle is units. MathGPTNext Problem ;
1259	https://www.turito.com/ask-a-doubt/Mathematics-which-related-addition-fact-helps-to-solve-15-9-15-11-4-15-6-8-15-5-10-15-9-6-q714894fe	Which related addition fact helps to solve 15 9?-Turito Turito Campus AI Academy Login Register Now+14708451137Login Register Now Are you sure you want to logout? Yes No Dashboard Subjects Tests Report Quiz Zone Bookmarks Help Tutor Connect Have a query? Ask a Tutor!! Access to best educators and exclusive paper discussions Offsite CampusTurito Championship Can’t find what you’re looking for? Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere. Homework- One to One Solutions Understand concepts to solve better Get your doubts solved instantly Please update the Email address in the profile section, to refer a friend Update Later Ask a doubt Back × OK Mathematics Grade-1 Easy Question Which related addition fact helps to solve ‘15 – 9’? 15 = 9 + 6 15 = 5 + 10 15 = 6 + 8 15 = 11 + 4 Answer The correct answer is: 15 = 9 + 6 Solution 9 + 6 = 15 addition fact helps to solve 15 – 9. Related Questions to study With Turito Academy. Subscribe Now With Turito Foundation. Get a Free Demo Get an Expert Advice From Turito. Get a Free Demo Turito Academy With Turito Academy. Subscribe Now Test Prep With Turito Foundation. Get a Free Demo About Turito Turito is a revolutionary learning platform for quality online and on-campus education from the most experienced faculty. We offer two years integrated classroom coaching for IIT-JEE and NEET aspirants at our Hyderabad campus. We provide online coaching for all kinds of competitive exams such as IIT-JEE, NEET, SAT, ACT, PSAT, Subjective SAT, AP & Olympiads, and Foundation Courses for students of Class 6 to 10 (CBSE, ICSE & State Syllabus). Turito's faculty has 30+ years of experience with a proven track record in teaching English, Maths, Physics, Chemistry, Botany, Zoology, Reading Comprehension, and so on. Turito is driven to provide the best online and on-campus programs to students across India, US, Singapore, and Middle East regions. Follow Us at Support help@turito.com +14708451137 1800-599-0009 Download App Quick Links About Us Privacy Policy Terms & Conditions Teachers @ turito Turito Blog Live TV Privacy Policy Terms & Conditions © 2025 turito.com All Right Reserved © 2025 turito.com All Right Reserved Privacy PolicyTerms & Conditions callback button
1260	https://uroweb.org/guidelines/urolithiasis/chapter/guidelines	Guidelines Urolithiasis Guidelines Introduction Methods Guidelines Metabolic Evaluation And Recurrence Prevention Follow Up Of Urinary Stones Bladder Stones References Conflict Of Interest Citation Information No elements found. Consider changing the search query. List is empty. 3. GUIDELINES 3.1. Prevalence, aetiology, risk of recurrence 3.1.1. Introduction Stone incidence depends on geographical, climatic, ethnic, dietary, and genetic factors. The recurrence risk is basically determined by the disease or disorder causing the stone formation. Accordingly, the prevalence rates for urinary stones vary from 1% to 20% . In countries with a high standard of life such as Sweden, Canada or the USA, renal stone prevalence is notably high (> 10%). For some areas, an increase of more than 37% over the last 20 years has been reported [7-9]. There is emerging evidence linking nephrolithiasis to the risk of chronic kidney disease (CKD) . Stones can be stratified into those caused by: infections, non-infectious causes, genetic defects [11,12]; or adverse drug effects (drug stones) (Table 3.1). See also section 3.2. Table 3.1: Stones classified by aetiology \| \| \| \| --- \| Stones classified by aetiology \| \| \| \| Non-infection stones \| \| \| \| • Calcium oxalate \| • Calcium phosphate \| • Uric acid • Ammonium urate \| \| Infection stones \| \| \| \| • Magnesium ammonium phosphate \| • Highly carbonated apatite \| • Ammonium urate \| \| Genetic causes \| \| \| \| • Cystine \| • Xanthine \| • 2,8-Dihydroxyadenine \| \| Drug stones \| \| \| In children in developing countries; in patients with anorexia or laxative-abuse. 3.1.2. Stone composition Stone composition is the basis for further diagnostic and management decisions. Stones are often formed from a mixture of substances. Table 3.2 lists the most clinically relevant substances and their mineral components. Table 3.2: Stone composition \| \| \| \| --- \| Chemical name \| Mineral name \| Chemical formula \| \| Calcium oxalate monohydrate \| Whewellite \| CaC2O4.H2O \| \| Calcium oxalate dihydrate \| Weddelite \| CaC2O4.2H2O \| \| Basic calcium phosphate \| Apatite \| Ca10(PO4)6.(OH)2 \| \| Calcium hydroxyl phosphate \| Carbonate apatite \| Ca5(PO4)3(OH) \| \| b-tricalcium phosphate \| Whitlockite \| Ca3(PO4)2 \| \| Carbonate apatite phosphate \| Dahllite \| Ca5(PO4)3OH \| \| Calcium hydrogen phosphate dihydrate \| Brushite \| CaHPO4.2H2O \| \| Calcium carbonate \| Aragonite \| CaCO3 \| \| Octacalcium phosphate Ca8H2(PO4)6.5H2O \| \| Uric acid \| Uricite \| C5H4N4O3 \| \| Uric acid dihydrate \| Uricite \| C5H4O3.2H20 \| \| Ammonium urate NH4C5H3N4O3 \| \| Sodium acid urate monohydrate NaC5H3N4O3.H2O \| \| Magnesium ammonium phosphate hexahydrate \| Struvite \| MgNH4PO4.6H2O \| \| Magnesium acid phosphate trihydrate \| Newberyite \| \| \| Magnesium ammonium phosphate monohydrate \| Dittmarite \| \| \| Cystine \| \| Xanthine - \| \| 2,8-Dihydroxyadenine - \| \| Proteins - \| \| Cholesterol - \| \| Calcite - \| \| Potassium urate - \| \| Trimagnesium phosphate - \| \| Melamine - \| \| Matrix - \| \| Drug stones \| Active compounds crystallising in urine \| Foreign body calculi - \| 3.1.3. Risk groups for stone formation Determination of the risk for stone formation is imperative for pharmacological treatment. Previous stone history (recurrence, regrowth, stone surgeries) is a fundamental element in determining risk for stone formation. About 50% of recurrent stone formers have just one-lifetime recurrence [9,14]. A review of first-time stone formers calculated a recurrence rate of 26% in five years’ time . Highly recurrent disease is observed in slightly more than 10% of patients. Stone type and disease severity determine low- or high-risk stone formers (Table 3.3) [16-32]. However, the risk status of stone formers should be determined in a holistic way taking into consideration not only the probability of stone recurrence or regrowth, but also the risk of chronic kidney disease (CKD), end-stage kidney disease (ESKD), and metabolic bone disorder (MBD) [33,34]. A comprehensive evaluation of stone risk in patients should also include the risk of developing CKD, ESKD, and MBD (Tables 3.4, 3.5, and 3.6) . Urolithiasis can compromise renal function because of the renal stone (obstruction, infection), renal tissue damage due to the primary condition causing stone formation (some genetic diseases, nephrocalcinosis, enteric hyperoxaluria, etc.), or urological treatments for the condition . Certain risk factors have been shown to be associated with such a risk in stone formers, as shown below. Table 3.3: High-risk stone formers \| \| \| General factors \| \| Early onset of urolithiasis (especially children and teenagers) \| \| Familial stone formation \| \| Recurrent stone formers \| \| Short time since last stone episode \| \| Brushite-containing stones (CaHPO4.2H2O) \| \| Uric acid and urate-containing stones \| \| Infection stones \| \| Solitary kidney (the kidney itself does not particularly increase the risk of stone formation, but prevention of stone recurrence is of crucial importance to avoid acute renal failure) \| \| Chronic Kidney Disease (CKD) \| \| Diseases associated with stone formation \| \| Hyperparathyroidism \| \| Metabolic syndrome \| \| Mineral Bone Disorder (MBD) \| \| Nephrocalcinosis \| \| Polycystic kidney disease (PKD) \| \| Gastrointestinal diseases (i.e., enteric hyperoxaluria due to jejuno-ileal bypass, intestinal resection, Crohn’s disease, malabsorptive conditions, urinary diversion, exocrine pancreatic insufficiency, and bariatric surgery). \| \| Increased levels of vitamin D \| \| Sarcoidosis \| \| Spinal cord injury, neurogenic bladder \| \| Genetically determined stone formation \| \| Cystinuria (type A, B and AB) \| \| Primary hyperoxaluria (PH) \| \| Renal tubular acidosis (RTA) type I \| \| 2,8-Dihydroxyadeninuria \| \| Xanthinuria \| \| Lesch-Nyhan syndrome \| \| Cystic fibrosis \| \| Drug-induced stone formation (see Table 4.11) \| \| Anatomical abnormalities associated with stone formation \| \| Medullary sponge kidney (tubular ectasia) \| \| Ureteropelvic junction (UPJ) obstruction \| \| Calyceal diverticulum, calyceal cyst \| \| Ureteral stricture \| \| Vesico-uretero-renal reflux \| \| Horseshoe kidney \| \| Ureterocele \| \| Environmental and professional factors \| \| High ambient temperatures \| \| Chronic lead and cadmium exposure \| Table 3.4 Risk factors for CKD and ESKD in stone formerss \| \| \| Risk factors for CKD/ESKD in stone formers \| \| Female gender \| \| Overweight \| \| Frequent UTI \| \| Struvite stones \| \| Acquired single kidney \| \| Neurogenic bladder \| \| Previous obstructive nephropathy \| \| Ileal conduit \| Furthermore, some specific kinds of urolithiasis also carry a particular risk of developing CKD/ESKD as shown below. Table 3.5 Risk factors for CKD and renal stones \| \| \| Risk of chronic kidney disease and renal stones \| \| Possible risk of CKD \| \| - Xanthine stones \| \| - Indinavir stones \| \| - Distal renal tubular acidosis (incomplete) \| \| - Primary hyperparathyroidism \| \| - Eating disorders and laxative abuse \| \| - Medullary sponge kidney \| \| Moderate risk of CKD \| \| - Brushite stones \| \| - 2,8-Dihydroxyadenine stones \| \| - Sarcoidosis \| \| - Pyelo-ureteral or ureteral strictures \| \| High risk of CKD \| \| - Cystine stones \| \| - Struvite stones \| \| - Stones in a single kidney \| \| - Distal renal tubular acidosis (complete) \| \| - Secondary hyperoxaluria (bariatric surgery, inflammatory bowel disease, bowel resection and malabsorptive syndromes) \| \| - Other forms of nephrocalcinosis (often associated with genetic conditions with hypercalciuria) \| \| - Anatomical abnormalities of the kidney and urinary tract (for example, horseshoe kidney, ureterocele and vesicoureteral reflux) \| \| - Neurological bladder \| \| Very high risk of CKD \| \| - Primary hyperoxaluria \| \| - Autosomal dominant polycystic kidney \| Table 3.6 Risk factors for metabolic bone disease and calcium renal stones \| \| \| Risk of metabolic bone disease and calcium renal stones \| \| Distal renal tubular acidosis (complete or incomplete) \| \| Medullary sponge kidney \| \| Primary hyperparathyroidism \| \| Malabsorptive syndromes \| \| Fasting hypercalciuria \| \| Genetic disorders \| 3.2. Classification of stones Urinary stones can be classified according to size, location, X-ray characteristics, aetiology of formation, composition, and risk of recurrence [2,9,32]. 3.2.1. Stone size Stone size can be reported in a single, two or three dimensions. Currently, the guidelines still use the linear measurement of cumulative stone diameter to stratify stones in < 5 mm, 5-10 mm, 10-20 mm, and > 20 mm for use in the treatment algorithm. 3.2.2. Stone location Stones can be classified according to anatomical position: upper, middle, or lower calyx; renal pelvis; upper, middle, or distal ureter; and urinary bladder. 3.2.3. X-ray characteristics Stones can be classified according to plain X-ray appearance [kidney-ureter-bladder (KUB) radiography] (Table 3.7), which varies according to mineral composition . Non-contrast-enhanced computed tomography (NCCT) can be used to classify stones according to density, inner structure, and composition, which can affect treatment decisions (Section 3.3) [35,36]. Table 3.8: X-ray characteristics \| \| \| \| --- \| Radiopaque \| Poor radiopacity \| Radiolucent \| \| Calcium oxalate dihydrate \| Magnesium ammonium phosphate \| Uric acid \| \| Calcium oxalate monohydrate \| Cystine \| Ammonium urate \| \| Calcium phosphate \| \| Xanthine \| \| 2,8-Dihydroxyadenine \| \| Drug-stones (Section 4.11) \| 3.3. Diagnostic evaluation 3.3.1. Diagnostic imaging The most appropriate imaging modality will be determined by the clinical situation, which will differ depending on if a ureteral or a renal stone is suspected. Standard evaluation includes a detailed medical history and physical examination. Patients with ureteral stones usually present with loin pain, vomiting, and sometimes fever, but may also be asymptomatic . Immediate evaluation is indicated in patients with solitary kidney, fever or when there is doubt regarding a diagnosis of renal colic. Ultrasound (US) should be used as the primary diagnostic imaging tool, although pain relief, or any other emergency measures, should not be delayed by imaging assessments. Ultrasound is safe (no risk of radiation), reproducible and inexpensive. It can identify stones located in the calyces, pelvis, and pyeloureteric and vesico-ureteral junctions (US with filled bladder), as well as in patients with upper urinary tract (UUT) dilatation. Ultrasound has a sensitivity of 45% and specificity of 94% for ureteral stones and a sensitivity of 45% and specificity of 88% for renal stones [38,39]. The sensitivity and specificity of KUB is 44-77% . Kidney-ureter-bladder radiography is helpful in differentiating between radiolucent and radiopaque stones and could be used for comparison during follow-up. 3.3.1.1. Evaluation of patients with acute flank pain/suspected ureteral stones Non-contrast-enhanced computed tomography has become the standard for diagnosing acute flank pain and has replaced intravenous urography (IVU). Non-contrast-enhanced CT can determine stone location, burden, and density. When stones are absent, the cause of abdominal pain should be identified. In evaluating patients with suspected acute urolithiasis, NCCT is significantly more accurate than IVU or US [42,43]. Non-contrast-enhanced CT can detect uric acid and xanthine stones, which are radiolucent on plain films, but not indinavir stones . Non-contrast-enhanced CT can determine stone density, the inner structure of the stone, skin-to-stone distance, and surrounding anatomy; all of which affect the selection of treatment modality . The advantage of non-contrast imaging must be balanced against the loss of information on renal function and urinary collecting system anatomy, as well as higher radiation dose [46-49]. Radiation risk can be reduced by low-dose CT, which may, however, be difficult to introduce in standard clinical practice [50-53]. In patients with a body mass index (BMI) < 30, low-dose CT has been shown to have a sensitivity of 86% for detecting ureteral stones < 3 mm and 100% for calculi > 3 mm . A meta-analysis (MA) of prospective studies has shown that low-dose CT diagnosed urolithiasis with a pooled sensitivity of 93.1% (95% CI: 91.5-94.4), and a specificity of 96.6% (95% CI: 95.1-97.7%). Dual-energy CT can differentiate uric acid containing stones from calcium-containing stones . \| \| \| --- \| \| Summary of evidence \| LE \| \| Non-contrast-enhanced CT is used to confirm stone diagnosis in patients with acute flank pain, as it is superior to IVU. \| 1a \| \| Computed tomography imaging enables 3D reconstruction of the collecting system, as well as measurement of stone density and skin-to-stone distance. \| 2a \| \| Consider a contrast study if stone removal is planned and the anatomy of the renal collecting system needs to be assessed. \| 3 \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Immediate imaging is indicated with fever or solitary kidney, and when diagnosis is doubtful. \| Strong \| \| Use non-contrast-enhanced computed tomography to confirm stone diagnosis in patients with acute flank pain following initial ultrasound assessment. \| Strong \| 3.3.2. Diagnostics - metabolism-related Besides imaging, each emergency patient with urolithiasis needs a succinct biochemical work-up of urine and blood test. At this point, no distinction is made between high- and low-risk patients for stone formation. 3.3.2.1. Basic laboratory analysis - non-emergency urolithiasis patients Biochemical work-up is similar for all stone patients (see 3.3.2.3). However, if no intervention is planned, examination of sodium, potassium, C-reactive protein (CRP), and blood coagulation time can be omitted. Only patients at high risk for stone recurrence should undergo a more specific analytical programme . Stone-specific metabolic evaluation is described in Chapter 4. The easiest method for diagnosing stones is by analysis of a passed stone using a validated method as listed in section 3.3.2.3. Once the mineral composition is known, a potential metabolic disorder can be identified. 3.3.2.2. Analysis of stone composition Stone analysis should be performed on all first-time stone formers. In clinical practice, repeat stone analysis is needed in the case of: recurrence under pharmacological prevention; early recurrence after interventional therapy with complete stone clearance; late recurrence after a prolonged stone-free period [56,57]. Patients should be instructed to filter their urine to retrieve a concrement for analysis. Stone passage and restoration of baseline renal function should be confirmed. The preferred analytical procedures are infrared spectroscopy (IRS) or X-ray diffraction (XRD) [58,59]. Equivalent results can be obtained by polarisation microscopy. Chemical analysis (wet chemistry) is generally deemed to be obsolete [58,60]. 3.3.2.3. Recommendations for laboratory examinations and stone analysis [17,23,61-63] \| \| \| --- \| \| Recommendations \| Strength rating \| \| Urine \| \| \| Dipstick test of spot urine sample: • red cells; • white cells; • nitrites; • approximate urine pH; • urine microscopy and/or culture. \| Weak \| \| Blood \| \| \| Serum blood sample: • creatinine; • uric acid; • (ionised) calcium; • sodium; • potassium; • blood cell count; • C-reactive protein. \| Strong \| \| Perform a coagulation test (partial thromboplastin time and international normalised ratio) if intervention is likely or planned. \| Strong \| \| Perform stone analysis in first-time formers using a valid procedure (X-ray diffraction or infrared spectroscopy). \| Strong \| \| Repeat stone analysis in patients presenting with: • recurrent stones despite drug therapy; • early recurrence after complete stone clearance; • late recurrence after a long stone-free period because stone composition may change. \| Strong \| 3.3.3. Diagnosis in special groups and conditions 3.3.3.1. Diagnostic imaging during pregnancy In pregnant women, radiation exposure may cause non-stochastic (teratogenesis) or stochastic (carcinogenesis, mutagenesis) effects. Teratogenic effects are cumulative with increasing doses and require a threshold dose (< 50 mGy are considered as safe) and depend on the gestation age (minimum risk prior to the 8th week and after the 23rd week). Carcinogenesis (dose even < 10 mGy present a risk) and mutagenesis (500-1000 mGy doses are required, far in excess of the doses in common radiographic studies) get worse with increasing dose but they do not require a dose threshold and are not dependent on the gestational age . There is no imaging modality that should be routinely repeated in pregnant women. Scientific societies and organizations agree on the safety of the diagnostic evaluation when the US , X-ray imaging [66,67], and MRI are used as and when indicated [69-76]. A radiographic procedure should not be withheld from a pregnant woman if the procedure is clearly indicated and doing so will affect her medical care. It is generally recommended that an investigation resulting in an absorbed dose to the foetus of greater than 0.5 mGy requires justification. Ultrasound (when necessary, using changes in the renal resistive index and transvaginal/transabdominal US with a full bladder) has become the primary radiological diagnostic tool when evaluating pregnant patients suspected of renal colic. However, normal physiological changes in pregnancy can mimic ureteral obstruction [72-74]. Magnetic resonance imaging can be used, as a second-line option , to define the level of urinary tract obstruction, and to visualise stones as a filling defect . The use of gadolinium is not routinely recommended in pregnancy to avoid toxic effects on the embryo . For the detection of urolithiasis during pregnancy, low-dose CT is associated with a higher positive predictive value (95.8%), compared to MRI (80%) and US (77%). As per White et al., low-dose CT offers improved diagnostic accuracy that can avoid negative interventions such as ureteroscopy . Although low-dose CT protocols reduce radiation exposure, judicious use is currently recommended in pregnant women as a last-line option . \| \| \| --- \| \| Summary of evidence \| LE \| \| Only low-level data exist for imaging in pregnant women supporting US and MRI. \| 3 \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Use ultrasound as the preferred method of imaging in pregnant women. \| Strong \| \| Use magnetic resonance imaging as a second-line imaging modality in pregnant women. \| Strong \| \| Use low-dose computed tomography as a last-line option in pregnant women. \| Strong \| 3.3.3.2. Diagnostic imaging in children Children with urinary stones have a high risk of recurrence; therefore, standard diagnostic procedures for high-risk patients apply, including a valid stone analysis (Section 3.1.3 and Chapter 4). The most common nonmetabolic disorders facilitating stone formation are vesicoureteral reflux (VUR), UPJ obstruction, neurogenic bladder, and other voiding difficulties . When selecting diagnostic procedures to identify urolithiasis in children, it should be remembered that these patients might be uncooperative, require anaesthesia, and may be sensitive to ionising radiation. Again, the principle of ALARA (As Low As Reasonably Achievable) should be observed [80-84]. Ultrasound Ultrasound is the primary imaging technique in children. Its advantages are the absence of radiation and no need for anaesthesia. Imaging should include both the fluid-filled bladder with adjoining portion of the ureters, as well as the upper ureter . Colour Doppler US shows differences in the ureteral jet and resistive index of the arciform arteries of both kidneys, which are indicative of the grade of obstruction . Nevertheless, the US may fail to identify ureteral stones and provides limited information on renal function . Plain films (KUB radiography) Kidney-ureter-bladder radiography can help to identify stones and their radiopacity and facilitate follow-up. Intravenous urography The radiation dose for IVU is comparable to that for voiding cystourethrography (0.33 mSV) . However, the need for contrast medium injection is a major drawback. Non-contrast-enhanced computed tomography Low-dose CT protocols have been shown to significantly reduce radiation exposure [90-92]. Sedation or anaesthesia is rarely needed with modern high-speed CT equipment. Magnetic resonance urography Magnetic resonance urography (MRU) cannot be used to detect urinary stones. However, it might provide detailed anatomical information about the urinary collecting system, the location of an obstruction or stenosis in the ureter, and renal parenchymal morphology . 3.3.3.2.1. Summary of evidence and recommendations for diagnostic imaging in children \| \| \| --- \| \| Summary of evidence \| LE \| \| Ultrasound is the first-line imaging modality in children when a stone is suspected; it should include the kidney, fluid-filled bladder, and the ureter next to the kidney and the (filled) bladder. \| 2b \| \| A kidney-ureter-bladder radiography (or low-dose NCCT) is an alternative investigation if US will not provide the required information. \| 2b \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Complete a metabolic evaluation based on stone analysis in all children. \| Strong \| \| Collect stone material for analysis to classify the stone type. \| Strong \| \| Perform ultrasound as first-line imaging modality in children when a stone is suspected; it should include the kidney, fluid-filled bladder, and the ureter. \| Strong \| \| Perform a kidney-ureter-bladder radiography (or low-dose non-contrast-enhanced computed tomography) if ultrasound will not provide the required information. \| Strong \| 3.4. Disease Management The treatment of urolithiasis is based on many parameters and is individualised for each patient. Parameters such as the size, number, location, and constitution of the stones are the cornerstones for deciding the treatment. In addition, the morphology, shape, volume, mobility, and hardness of the stone should be considered. Finally, the anatomy and compliance of the entire pelvic-calyceal system should be assessed for each patient. The design of therapeutic algorithms including all the above parameters is difficult mainly due to the great diversity of lithiasis disease per patient. Furthermore, there is a significant lack of comparative clinical studies to support development of algorithms using parameters other than stone size and composition. 3.4.1. Renal colic Pain relief Non-steroidal anti-inflammatory drugs (NSAIDs) (including metamizole dipyrone), and paracetamol are effective in patients with acute stone colic and have better analgesic efficacy than opioids . Ibuprofen compared to ketorolac is a more rapid-acting drug in controlling pain caused by renal colic with a similar side effect profile . Pain relief from intramuscular (i.m.) diclofenac compared favourably with those from intravenous (i.v.) ibuprofen and i.v. ketorolac; however, no recommendation can be given due to the way in which the results have been reported . The addition of antispasmodics to NSAIDs does not result in better pain control. Patients receiving NSAIDs are less likely to require further analgesia in the short term. It should be taken into consideration that the use of diclofenac and ibuprofen increased major coronary events [94,95]. Oral diclofenac in the long-term increases the risk of cardiovascular events and upper GI bleeding . Patients with significant risk factors for cardiovascular events should be treated with diclofenac only after careful consideration. As risks increase with dose and duration, the lowest effective dose should be used for the shortest duration . Non-steroidal anti-inflammatory drugs may affect renal function in those patients with pre-existing decreased GFR. In an RCT including 150 patients, Intradermal sterile water injection (ISWI) and diclofenac (i.m.) were shown equally effective for pain relief in acute renal colic. Intradermal sterile water injection may be an alternative to NSAIDs in pregnant patients or others where NSAIDs are contra-indicated . Opioids, particularly pethidine, are associated with a high rate of vomiting compared to NSAIDs and carry a greater likelihood of further analgesia being needed [94,101]. If an opioid is used, it is recommended that it is not pethidine. Combination of opioids and NSAIDs increase analgetic effect compared to opioids alone . Acupuncture seems to be effective in renal colic alone or in combination with analgetic drugs, but there is limited data [103,104]. Prevention of recurrent renal colic Facilitation of passage of ureteral stones is discussed in Section 3.4.9. For patients with ureteral stones that are expected to pass spontaneously, NSAID tablets or suppositories (e.g., diclofenac sodium, 100-150 mg/day, 3-10 days) may help reduce inflammation and the risk of recurrent pain [105,106]. Although NSAID can affect renal function in patients with already reduced function, it has no functional effect in patients with normal renal function . The systematic review and MA by Hollingsworth et al., addressed pain reduction as a secondary outcome and concluded that medical expulsive therapy (MET) seems efficacious in reducing pain episodes of patients with ureteral stones. If analgesia cannot be achieved medically, drainage, using stenting, percutaneous nephrostomy, or stone removal, is indicated . 3.4.1.1. Summary of evidence and recommendations for the management of renal colic \| \| \| --- \| \| Summary of evidence \| LE \| \| Non-steroidal anti-inflammatory drugs are very effective in treating renal colic and are superior to opioids. \| 1b \| \| For symptomatic ureteral stones, stone removal as first-line treatment is a feasible option in selected patients. \| 1b \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Offer a non-steroidal anti-inflammatory as the first drug of choice; depending on cardiovascular risk factors and side effects. \| Strong \| \| Offer opiates (hydromorphine, pentazocine or tramadol) as a second choice. \| Weak \| \| Offer renal decompression or ureteroscopic stone removal in case of analgesic refractory colic pain. \| Strong \| 3.4.2. Management of sepsis and/or anuria in obstructed kidney The obstructed kidney with all signs of urinary tract infection (UTI) and/or anuria is a urological emergency. Urgent decompression is often necessary to prevent further complications in infected hydronephrosis secondary to stone-induced, unilateral, or bilateral, renal obstruction. Decompression There are two options for urgent decompression of obstructed collecting systems : placement of an indwelling ureteral stent percutaneous placement of a nephrostomy tube. Several systematic reviews on the subject have been published, all of which emphasize that the available literature comparing different drainage modalities for obstructing stones with or without infection is scarce, based on small cohorts and of medium to very low quality . There appears to be no difference in success rate or complication rate of both procedures and there is not a difference in time to defervescence in the population presenting with fever. Both meta-analyses identified patients receiving a nephrostomy tube to have a longer stay in the hospital. Based on the available data, a DJ stent has a more negative impact on the patients’ quality of life in comparison with a nephrostomy tube, which can be explained mainly by the stent-related symptoms that these patients experience [111,112]. Definitive stone removal should be delayed until the infection is cleared following a complete course of antimicrobial therapy. A small RCT showed the feasibility of immediate ureteroscopic stone removal combined with an appropriate antibiotic regimen; however, at the cost of longer hospital stay and higher analgesic requirements . Further measures Along with urgent decompression of the obstructed and infected urinary collecting system, both urine- and blood samples should be sent for culture-antibiogram sensitivity testing and antibiotics should be initiated immediately [114,115]. The regimen should be re-evaluated in the light of the culture-antibiogram results. Although clinically well accepted, the impact of a second antibiogram test on treatment outcome has not yet been evaluated . Intensive care might become necessary. 3.4.2.1. Summary of evidence and recommendations for the management of sepsis and anuria \| \| \| --- \| \| Summary of evidence \| LE \| \| For decompression of the renal collecting system, ureteral stents and percutaneous nephrostomy catheters are equally effective. \| 1b \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Urgently decompress the collecting system in case of sepsis with obstructing stones, using percutaneous drainage or ureteral stenting. \| Strong \| \| Delay definitive treatment of the stone until sepsis is resolved. \| Strong \| \| Collect (again) urine for antibiogram test following decompression. \| Strong \| \| Start antibiotics immediately (+ intensive care, if necessary). \| Strong \| \| Re-evaluate antibiotic regimen following antibiogram findings. \| Strong \| 3.4.3. Medical expulsive therapy Several drug classes including α-blockers, calcium channel inhibitors, and phosphodiesterase type 5 inhibitors (PDEI-5) are used for MET [117-120] A class effect of α-blockers in MET has been demonstrated in MAs although this is an off-label indication [121-123]. However, there is contradictory evidence between these studies and several well-designed, multicentre, placebo-controlled, double-blinded randomised studies showing limited, or no, benefit using α-blockers, besides some advantage for distal ureteral stones > 5 mm [124-128]. Based on studies with a limited number of patients [120,121,129,130], no recommendation for the use of PDEI-5 or corticosteroids in combination with α-blockers in MET can be made. The panel concludes that MET using α-blockers seems efficacious in the treatment of patients with distal ureteral stones > 5 mm who are amenable to conservative management. Medical expulsive therapy in special situations is addressed in the relevant chapters. 3.4.3.1. Summary of evidence and recommendations for medical expulsive therapy \| \| \| --- \| \| Summary of evidence \| LE \| \| Medical expulsive therapy seems to be efficacious for treating patients with ureteral stones who are amenable to conservative management. The greatest benefit might be among those with > 5 mm (distal) ureteral stones. \| 1a \| \| Insufficient data exist to support the use of PDEI-5 or corticosteroids in combination with α-blockers as an accelerating adjunct. \| 2a \| \| Alpha-blockers increase stone expulsion rates in distal ureteral stones > 5 mm. \| 1a \| \| A class effect of α-blockers has been demonstrated. \| 1a \| \| \| \| --- \| \| Recommendation \| Strength rating \| \| Offer α-blockers as medical expulsive therapy as one of the treatment options for (distal) ureteral stones > 5 mm. \| Strong \| Alpha-blockers are an off-label treatment 3.4.4. Chemolysis Percutaneous irrigation chemolysis Percutaneous chemolysis is rarely used nowadays, for practical reasons. Percutaneous irrigation chemolysis may be an option for infection stones and theoretically also for uric acid stones. For dissolution of struvite stones, Suby’s G solution (10% hemiacidrin; pH 3.5-4) can be used. The method has been described in case series and literature reviews . Oral chemolysis Stones composed of uric acid, but not sodium or ammonium urate stones, can be dissolved by oral chemolysis. Prior stone analysis may provide information on stone composition. Urinary pH measurement and X-ray characteristics can provide information on the type of stone. Oral chemolysis is based on alkalinisation of urine by application of alkaline citrate or sodium bicarbonate. The pH should be adjusted to 7.0-7.2. Chemolysis is more effective at a higher pH, which might, however, promote calcium phosphate stone formation. Patients will need to adjust the dosage of alkalising medication by self-monitoring the pH of their urine. A SR shows a complete or partial dissolution in 80.5%, discontinuation rate of 10.2% with 15.7% requiring further intervention . In the case of uric acid obstruction of the collecting system, oral chemolysis in combination with urinary drainage is indicated. A combination of alkalinisation with tamsulosin can increase the frequency of spontaneous passage of distal ureteral uric acid stones as shown in one RCT for stones > 5 mm . Additional SWL might help to improve the results but evidence is weak . 3.4.4.1. Summary of evidence and recommendations for chemolysis \| \| \| --- \| \| Summary of evidence \| LE \| \| Irrigation chemolysis has been used in limited clinical settings to dissolve struvite stones. \| 3 \| \| Uric acid stones > 5mm can be dissolved based on oral alkalinisation of the urine above 7.0. \| 3 \| \| For obstructing uric acid stones, a combination of oral chemolysis with tamsulosin is more effective than each substance alone, particularly in stones > 8 mm. \| 1b \| \| \| \| --- \| \| Recommendations (oral chemolysis of uric acid stones) \| Strength rating \| \| Inform the patient how to monitor urine-pH by dipstick and to modify the dosage of alkalising medication according to urine pH, as changes in urine pH are a direct consequence of such medication. \| Strong \| \| Carefully monitor patients during/after oral chemolysis of uric acid stones. \| Strong \| \| Combine oral chemolysis with tamsulosin in case of (larger) ureteral stones (if active intervention is not indicated). \| Weak \| 3.4.5. Extracorporeal shock wave lithotripsy (SWL) The success of SWL depends on the efficacy of the lithotripter and the following factors: size, location (ureteral, pelvic, or calyceal), and composition (hardness) of the stones (Section 3.4.9.3); patient’s habitus (Section 3.4.10.3); performance of SWL (best practice, see below). Each of these factors significantly influences the retreatment rate and outcome of SWL. Best clinical practiceStenting Routine use of internal stents before SWL does not improve stone-free rates (SFRs), nor lowers the number of auxiliary treatments. It may, however, reduce the formation of steinstrasse [135-138]. Pacemaker Patients with a pacemaker can be treated with SWL, provided that appropriate technical precautions are taken. Patients with implanted cardioverter defibrillators must be managed with special care (firing mode temporarily reprogrammed during SWL treatment). However, this might not be necessary with new-generation lithotripters . Shock wave rate Lowering shock wave frequency from 120 to 60-90 shock waves/min improves SFRs [140-148]. Ultraslow frequency of 30 shock waves/min may increase SFR . Tissue damage increases with shock wave frequency [150-153]. Number of shock waves, energy setting, and repeat treatment sessions The number of shock waves that can be delivered at each session depends on the type of lithotripter and shock wave power. There is no consensus on the maximum number of shock waves . Starting SWL on a lower energy setting with stepwise power (and SWL sequence) ramping can achieve vasoconstriction during treatment , which prevents renal injury [155-157]. Animal studies and a prospective randomised study have shown better SFRs (96% vs. 72%) using stepwise power ramping, but no difference has been found for fragmentation or evidence of complications after SWL, irrespective of whether ramping was used [160,161]. There are no conclusive data on the intervals required between repeated SWL sessions. However, clinical experience indicates that repeat sessions are feasible (within one day for ureteral stones) . Improvement of acoustic coupling Proper acoustic coupling between the cushion of the treatment head and the patient’s skin is important. Defects (air pockets) in the coupling gel deflect 99% of shock waves . Gentle swiping between the coupled therapy head and the patient skin helps remove air bubbles and improves the coupling . Ultrasound gel is probably the most widely-used agent available as a lithotripsy coupling agent . Procedural control Results of treatment are operator-dependent, and experienced clinicians obtain better results. During the procedure, careful imaging control of localisation contributes to outcome quality . Pain Control Careful control of pain during treatment is necessary to limit pain-induced movements and excessive respiratory excursions [167-170]. Antibiotic prophylaxis No standard antibiotic prophylaxis before SWL is recommended. However, prophylaxis is recommended in the case of internal stent placement ahead of anticipated treatments and in the presence of increased bacterial burden (e.g., indwelling catheter, nephrostomy tube, or infectious stones) [62,171,172]. Medical therapy after extracorporeal shock wave lithotripsy Despite conflicting results, most RCTs and several MAs support MET after SWL for ureteral or renal stones as an adjunct to expedite expulsion and increase SFRs. Medical expulsion therapy might also reduce analgesic requirements . Post-treatment management Mechanical percussion and diuretic therapy can significantly improve SFRs and accelerate stone passage after SWL . Complications of extracorporeal shock wave lithotripsy Compared to percutaneous nephrolithotomy (PCNL) and ureteroscopy (URS), there are fewer overall complications with SWL (Table 3.8). In a Meta-Analysis of 115 RCT´s 18.43% of Clavien I–II complications and 2.48% of Clavien III–IV complications occurred . The relationship between SWL and hypertension or diabetes is unclear. Published data are contradictory; however, no evidence exists supporting the hypothesis that SWL may cause long-term adverse effects [176-182]. Table 3.8: Shock wave lithotripsy-related complications \| \| \| \| \| \| --- --- \| Complications \| \| \| % \| Reference \| \| Related to stone fragments \| Steinstrasse \| \| 4 \| [183-185] \| \| Macroscopic haematuria \| 17.2% \| \| \| Pain \| 12.1% \| \| \| Regrowth of residual fragments \| \| 21 – 59 \| [186,187] \| \| \| Auxiliary procedure \| \| 6.9% \| \| \| \| Renal colic \| \| 2 – 4 \| \| \| Infectious \| Bacteriuria in non-infection stones \| \| 7.7 – 23 \| [186-189] \| \| Sepsis \| 0.15% \| \| \| Tissue effect \| Renal \| Haematoma, symptomatic \| 0.21% \| \| \| Haematoma, asymptomatic \| 1.2% \| \| \| Cardiovascular \| Dysrhythmia \| 11 – 59 \| [186,188] \| \| Morbid cardiac events \| Case reports \| [186,188] \| \| Gastrointestinal \| Bowel perforation \| Case reports \| \| \| Liver, spleen haematoma \| Case reports \| [190-193] \| 3.4.5.1. Summary of evidence and recommendations for Shock wave lithotripsy \| \| \| --- \| \| Summary of evidence \| LE \| \| Stepwise power ramping prevents renal injury. \| 1b \| \| Clinical experience has shown that repeat sessions are feasible (within one day for ureteral stones). \| 4 \| \| Optimal shock wave frequency is 1.0 to 1.5 Hz. \| 1a \| \| Proper acoustic coupling between the cushion of the treatment head and the patient’s skin is important. \| 2 \| \| Careful imaging control of localisation of stone contributes to outcome of treatment. \| 2a \| \| Careful control of pain during treatment is necessary to limit pain-induced movements and excessive respiratory excursions. \| 1a \| \| Antibiotic prophylaxis is recommended in the case of internal stent placement, infected stones, or bacteriuria. \| 1a \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Ensure correct use of the coupling agent because this is crucial for effective shock wave transportation. \| Strong \| \| Maintain careful fluoroscopic and/or ultrasonographic monitoring during shock wave lithotripsy (SWL). \| Strong \| \| Use proper analgesia because it improves treatment results by limiting pain-induced movements and excessive respiratory excursions. \| Strong \| \| Prescribe antibiotics prior to SWL in the case of infected stones or bacteriuria. \| Strong \| 3.4.6. Ureteroscopy (retrograde and antegrade) The current standard for rigid ureteroscopes is a tip diameter of < 8 French (F). Rigid URS can be used for the whole ureter . However, technical improvements, as well as the availability of digital scopes, also favour the use of flexible ureteroscopes in the ureter . Percutaneous antegrade removal of ureteral stones is a consideration in selected cases, i.e. large (> 15 mm), impacted proximal ureteral calculi in a dilated renal collecting system [195,196], or when the ureter is not amenable to retrograde manipulation . Ureteroscopy for renal stones: Retrograde Intrarenal Surgery (RIRS) Technical improvements including endoscope miniaturisation, improved deflection mechanism, enhanced optical quality and tools, and introduction of disposables have led to an increased use of URS for both renal and ureteral stones. Major technological progress has been achieved for RIRS. A systematic review addressing renal stones > 2 cm showed a cumulative SFR of 91% with 1.45 procedures/patient; 4.5% of the complications were > Clavien 3 [198,199]. Digital scopes demonstrate shorter operation times due to the improvement in image quality . Stones that cannot be extracted directly must be disintegrated. If it is difficult to access stones within the lower renal pole that need disintegration it may help to displace them into a more accessible calyx . Best clinical practice in ureteroscopy Access to the upper urinary tract Most interventions are performed under general anaesthesia, although local or spinal anaesthesia is possible . Intravenous sedation is suitable for female patients with distal ureteral stones . Smaller caliber (4.5/6 Fr) semi-rigid ureteroscope was associated with significantly higher SFR, lower rates of ureteric injury, and shorter hospital stay . Antegrade URS is an option for large, impacted, proximal ureteral calculi [195,205]. Reduction of flexible ureteroscope diameter may provide similar vision, deflection, and manoeuvrability to standard flexible ureteroscopes potentially with improved ureteric access . Disposable ureteroscopes provide similar safety and clinical effectiveness to reusable scopes. Concerns regarding cost-effectiveness and environmental sustainability remain [204,207-209]. Safety aspects Fluoroscopic equipment must be available in the operating room. The Panel recommends placement of a safety wire, even though some groups have demonstrated that URS can be performed without it [210-214]. Balloon and plastic dilators should be available, if necessary. Prior rigid URS can be helpful for optical dilatation followed by flexible URS, if necessary. If ureteral access is not possible, insertion of a JJ stent followed by URS after seven to fourteen days offers an alternative . Bilateral URS during the same session is feasible resulting in equivalent-to-lower SFRs, but slightly higher overall complication rates (mostly minor, Clavien 1 and 2) [216,217]. Difficult lower pole anatomy such as steep infundibulopelvic angle predisposes to failure during RIRS . A reusable flexible ureteroscope can be more helpful in reaching a difficult lower pole calyx . Prolonged operative times are linked to increased complication rates in ureteroscopy, and efforts must be made to keep it below 90 minutes . Ureteral access sheaths Hydrophilic-coated ureteral access sheaths, which are available in different calibres (inner diameter from 9 F upwards), can be inserted (via a guide wire) with the tip placed in the proximal ureter. Ureteral access sheaths allow easy, multiple, access to the UUT and therefore significantly facilitate URS. The use of ureteral access sheaths improves vision by establishing a continuous outflow, decreases intrarenal pressure, and potentially reduces operating time [221,222]. The insertion of ureteral access sheaths may lead to ureteral damage, the risk is lowest in pre-stented systems . No data on long-term side effects are available [198,223]. Whilst larger cohort series showed no difference in SFRs and ureteral damage (stricture rates of about 1.8%), they did show lower post-operative infectious complications [224,225]. Increasing sheath size directly determines higher grades of ureteral injury rates but there is no difference in long-term stricture rates . The use of a ureteral access sheath is safe and can be useful for large and multiple renal stones or if long procedural time is expected . Stone extraction The aim of URS is complete stone removal. “Dust and go” strategies should be limited to the treatment of large (renal) stones . Stones can be extracted by endoscopic forceps or baskets. Only baskets made of nitinol can be used for flexible URS . Intracorporeal lithotripsy The most effective lithotripsy system is the holmium: yttrium-aluminum-garnet (Ho: YAG) laser, which is currently the optimum standard for URS and flexible nephroscopy (Section 3.4.6), because it is effective in all stone types [230,231]. Compared to low-power lasers, high-power laser reduces procedural time although the reported difference in clinical outcomes was non-significant and based on a low level of evidence . The only RCT to date shows no clinical difference regarding stone-free rate or operative time . Although pulse-modulation in Ho: YAG lasers has demonstrated several in vitro benefits, a systematic review including 8 comparative studies and only one RCT showed no difference in stone-free rate, complication rate, or operative time . The two available RCTs on the subject both found a shorter operative time, without conferring a difference in success rate [235,236]. Thulium fiber laser (TFL) for stone disease has a promising role and offers good clinical outcomes, which seem to be comparable to Ho: YAG laser (holmium) laser [237-239]. With the limited reports of clinical use available to date, a meta-analysis could not demonstrate the superiority of TFL over Ho: YAG, although the operative time to achieve this stone-free rate seems to be shorter with the use of TFL . More comparative clinical studies are however needed between these two modalities. When a laser is not available, pneumatic and US systems can be used with high disintegration efficacy in rigid URS [241,242]. However, stone migration into the kidney is a common problem, which can be prevented by the placement of special anti-migration tools proximal to the stone . Medical expulsion therapy following Ho: YAG laser lithotripsy increases SFRs and reduces colic episodes . Stenting before and after URS Routine stenting is not necessary before URS. Despite a complete lack of RCTs on this subject, a meta-analysis has been performed, demonstrating that pre-stenting may improve the stone-free rate of ureteroscopic treatment of renal stones, but not of ureteral stones . Although it may facilitate ureteroscopic management of stones and increase success in access sheath placement, intra-operative complications were not significantly different [245,246]. One should also consider that pre-stenting also causes the patient to experience stent-related symptoms during the time the stent is indwelling, prior to a procedure. Randomised prospective trials have found that routine stenting after uncomplicated URS (complete stone removal) is not necessary; stenting might be associated with higher postoperative morbidity and costs . Smaller diameter ureteric stents may reduce urinary symptoms and patient reported pain . A ureteral catheter with a shorter indwelling time (one day) may also be used, with similar results . Stents should be inserted in patients who are at increased risk of complications (e.g., ureteral trauma, residual fragments, bleeding, perforation, UTIs, or pregnancy), and in all doubtful cases, to avoid stressful emergencies. The ideal duration of stenting is not known. Most urologists favour one to two weeks after URS. Alpha-blockers reduce the morbidity of ureteral stents and increase tolerability . Medical expulsive therapy before and after ureteroscopy Medical expulsion therapy before URS might reduce the risk for intra-operative ureteral dilatation, protect against ureteral injury when using access sheaths and increase stone-free rates four weeks after URS [251,252]. Medical expulsion therapy following Ho: YAG laser lithotripsy accelerates the spontaneous passage of fragments and reduces episodes of colic . Complications of ureteroscopy The overall complication rate after URS is 4-25% [253,254]. Most complications are minor and do not require intervention. There is evidence suggesting a risk of post-operative urosepsis of up to 5% [255,256]. Ureteral avulsion and strictures are rare (< 1%). Previous perforations, pre-operative positive urine cultures, comorbidities, and longer operation time are the most important risk factor for complications [220,257,258]. Infectious complications following URS can be minimised using prophylactic antibiotics, limiting stent dwell and procedural time, identification and treatment of UTI, and planning in patients with large stone burden and multiple comorbidities . High intrarenal pressure (IRP) predisposes to URS complications, and measures should be used to reduce IRP. Currently, there are no accurate ways to measure the intra-operative IRP . 3.4.6.1. Summary of evidence and recommendations for retrograde URS, RIRS and antegrade ureteroscopy \| \| \| --- \| \| Summary of evidence \| LE \| \| In uncomplicated URS, a post-procedure stent need not be inserted. \| 1a \| \| In URS, pre-stenting has been shown to improve outcomes for renal stones. \| 1a \| \| An α-blocker can reduce stent-related symptoms and colic episodes. \| 1a \| \| The most effective lithotripsy system for flexible ureteroscopy is the Ho: YAG laser. \| 2a \| \| Pneumatic and US systems can be used with high disintegration efficacy in rigid URS. \| 2a \| \| Percutaneous antegrade removal of proximal ureter stones, or laparoscopic ureterolithotomy are feasible alternatives to retrograde ureteroscopy, in selected cases. \| 1b \| \| Pre-treatment of patients undergoing URS with an α-blocker one week prior to the procedure reduces the need for active dilatation and increases the stone free rate. \| 1a \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Use holmium: yttrium-aluminum-garnet (Ho: YAG) or Thulium fiber laser (TFL) laser lithotripsy for (flexible) ureteroscopy (URS). \| Strong \| \| Perform stone extraction only under direct endoscopic visualisation of the stone. \| Strong \| \| Do not insert a stent in uncomplicated cases. \| Strong \| \| Offer medical expulsive therapy for patients suffering from stent-related symptoms and after Ho: YAG laser lithotripsy to facilitate the passage of fragments. \| Strong \| \| Use percutaneous antegrade removal of ureteral stones as an alternative when shock wave lithotripsy (SWL) is not indicated or has failed, and when the upper urinary tract is not amenable to retrograde URS. \| Strong \| \| Use flexible URS (even for stones > 2 cm) in cases where percutaneous nephrolithotomy or SWL are not an option. However, in this case, there is a higher risk that a follow-up procedure and placement of a ureteral stent may be needed. \| Strong \| 3.4.7. Percutaneous nephrolithotomy Percutaneous nephrolithotomy remains the standard procedure for large renal calculi. Different rigid and flexible endoscopes are available, and the selection is mainly based on the surgeon’s own reference. Standard access tracts are 24-30 F. Smaller access sheaths, < 18 F, were initially introduced for paediatric use, but are now increasingly utilized in the adult population [261,262]. Contraindications Patients receiving anticoagulant therapy must be monitored carefully pre-and post-operatively. Anti-coagulant therapy must be discontinued before PCNL . Other important considerations include: untreated UTI; tumour in the presumptive access tract area; potential malignant kidney tumour; pregnancy (Section 3.4.14.1). Best clinical practice Intracorporeal lithotripsy Several methods for intracorporeal lithotripsy during PCNL are available. Ultrasonic, pneumatic, and combined systems are most commonly used for rigid nephroscopy, whilst the laser is increasingly used for miniaturised and flexible instruments . Pre-operative imaging Pre-procedural imaging evaluations are summarised in Section 3.3.1. In particular, US or CT of the kidney and the surrounding structures can provide information regarding interposed organs within the planned percutaneous path (e.g., spleen, liver, large bowel, pleura, and lung). Positioning of the patient Both prone and supine positions are equally safe. A meta-analysis including twelve studies and a total of 1,290 patients treated, showed a similar SFR but a lower operative time for supine PCNL . The supine position allows simultaneous retrograde access to the collecting system, using a flexible ureteroscope . The combination of PCNL and RIRS may be a good alternative for the treatment of complex renal stones compared to standard PCNL; however, the existing evidence is of low-quality [265,267]. Puncture Although fluoroscopy is still the most common intra-operative imaging method, the use of US as an additional or only means of puncture guidance provides advantages according to two meta-analyses including eight randomised controlled trials. Additional to the expected reduced radiation exposure with the use of ultrasound the meta-analyses also demonstrated a lower complication rate [268,269]. Pre-operative CT or intra-operative US allows identification of the tissue between the skin and kidney and lowers the incidence of visceral injury. As an additional aid to increase puncture accuracy, the calyceal puncture may be done under direct visualisation using simultaneous flexible URS [270-272]. Dilatation Dilatation of the percutaneous access tract can be achieved using a metallic telescopic, single (one-shot or serial) dilator, or balloon dilatator. During PCNL, safety and effectiveness are similar for different tract dilatation methods . Although there are papers demonstrating that single-step dilation is equally effective as other methods and that US only can be used for the dilatation, the difference in outcomes is most likely related to surgeon experience rather than to the technology used [273,274]. A meta-analysis of the most commonly used tract dilation methods suggested that one-step dilation would allow for a shorter operative time and reduced complication rate, including haemoglobin loss and transfusion rate . Choice of instruments Several meta-analyses on mini-PCNL (12-22 F) vs. standard PCNL (> 22 F) have identified that both techniques allow for a similar SFR. Patients treated with mini-PCNL had reduced blood loss and transfusion rates, as well as a shorter hospital stay, without a significant difference in overall complication rates [262,276-278]. However, it is important to note that the level of evidence was downgraded due to heterogeneity of data related to tract sizes used and types of stones treated. There is some evidence for using suction during PCNL to reduce intra-renal pressure and increase SFR . Post-operative drainage The decision on whether, or not, to place a nephrostomy tube or a double J stent at the conclusion of the PCNL procedure depends on several factors, including: presence of residual stones; likelihood of a second-look procedure; significant intra-operative blood loss/ bleeding from the percutaneous tract; urine extravasation; ureteral obstruction; potential persistent bacteriuria due to infected stones; solitary kidney; bleeding diathesis; planned percutaneous chemolitholysis. Small-bore nephrostomies seem to have advantages in terms of postoperative pain [262,280,281]. Tubeless PCNL is performed without a nephrostomy tube and is associated with reduced post-operative pain and hospital stay . When neither a nephrostomy tube nor a ureteral stent is introduced, the procedure is known as a totally tubeless PCNL . In uncomplicated cases, the latter procedure results in a shorter hospital stay, with no disadvantages reported . As reported in the above section on the drainage of an infected or obstructed system [110-112] (section 3.4.2), the quality of life may be slightly lower with a DJ stent in comparison to a short-term nephrostomy tube after PCNL. This should be weighed against the shorter hospital stay with a DJ stent . Complications of percutaneous nephrolithotomy A systematic review of almost 12,000 patients shows the incidence of complications associated with PCNL; fever 10.8%, transfusion 7%, thoracic complication 1.5%, sepsis 0.5%, organ injury 0.4%, embolisation 0.4%, urinoma 0.2%, and death 0.05% . Perioperative fever can occur, even with a sterile pre-operative urinary culture and peri-operative antibiotic prophylaxis, because the renal stones themselves may be a source of infection. The evidence demonstrates that a stone culture or urine culture taken directly from the renal pelvis is more predictive of post-operative SIRS or sepsis. Whenever possible a urine culture from the renal pelvis and/or stone culture should be taken at the time of PCNL . Intra-operative renal stone or renal pelvic urine culture may be more indicative of the causative organism for sepsis; therefore, helping to select the most suitable postoperative antibiotics [287-289]. Although this data is weak, there is limited retrospective data indicating that increased pressures during mPCNL may contribute to febrile complications [290-292]. This contrasts with the previously mentioned meta-analyses on mini vs standard PCNL that do not identify a difference in complication rate between the two procedures [262,276-278]. Bleeding after PCNL may be treated by briefly clamping the nephrostomy tube. Super-selective embolic occlusion of the arterial branch may become necessary in the case of severe bleeding. Several meta-analyses have demonstrated that the use of tranexamic acid reduces bleeding complications and the transfusion rate of PCNL [293-295]. However, the transfusion rate in the control group of the meta-analyses was in the range of 10-12%. Depending on the stone burden and the patient’s anatomy, multiple tracts may be necessary to render the patient stone free in one session of PCNL. While this is a generally accepted practice, it should be highlighted that this comes with an increased risk of postoperative complications including pleural damage, infections, and the need for transfusion . To reduce post-operative pain after PCNL, a peripheral nerve block can be performed at the intercostal nerve, paravertebral region, erector spinae, or quadratus lumborum. Such a block may significantly reduce the need for post-operative opioid analgesics [297,298]. Current evidence shows that a quadratus lumborum block or infiltration of a local anaesthetic around the nephrostomy tube may reduce post-operative pain and opioid consumption after PCNL [299,300]. 3.4.7.1. Summary of evidence and recommendations for endourology techniques for renal stone removal \| \| \| --- \| \| Summary of evidence \| LE \| \| Imaging of the kidney with US or CT can provide information regarding inter-positioned organs within the planned percutaneous path (e.g., spleen, liver, large bowel, pleura, and lung). \| 3 \| \| Both prone and supine positions are equally safe with equivalent SFR. \| 1a \| \| Percutaneous nephrolithotomy performed with small instruments tends to be associated with significantly lower blood loss, but the duration of procedure tended to be significantly longer. There are no significant differences in SFR or any other complications. \| 1a \| \| In uncomplicated cases, a totally tubeless PCNL results in a shorter hospital stay, with no increase in complication rate. \| 1a \| \| Peri-operative use of tranexamic acid may reduce bleeding complications and transfusion rates. \| 1a \| \| Urine cultures taken directly from the renal pelvis, or a stone culture are more predictive of post-PCNL sepsis than a pre-operative midstream urine culture. \| 1a \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Perform pre-procedural computed tomography (CT) imaging, including contrast medium when indicated or retrograde study when starting the procedure, to assess stone comprehensiveness and anatomy of the collecting system to ensure safe access to the renal stone. \| Strong \| \| Perform a tubeless (without nephrostomy tube) or totally tubeless (without nephrostomy tube and ureteral stent) percutaneous nephrolithotomy (PCNL) procedure, in uncomplicated cases. \| Strong \| \| Take a stone culture or urine culture directly from the renal pelvis at time of PCNL, if possible. \| Strong \| 3.4.8. General recommendations and precautions for stone removal 3.4.8.1. Antibiotic therapy Urinary tract infections should always be treated if stone removal is planned. In patients with clinically significant infection and obstruction, drainage should be performed for several days before starting stone removal. A urine culture or urinary microscopy should be performed before treatment . Peri-operative antibiotic prophylaxis The available evidence for prevention of infection following URS and percutaneous stone removal, remains limited . Administration of a single dose of prophylactic antibiotics prior to ureteroscopy was found to be sufficient [302-304]. In a review of a large database of patients undergoing PCNL, it was found that in patients with negative baseline urine culture, antibiotic prophylaxis significantly reduced the rate of postoperative fever and other complications . Based on three meta-analyses, pooling data from small series with varying quality an extended course of pre-operative prophylactic antibiotics prior to PCNL compared to a single dose before anaesthesia significantly reduced post-operative sepsis and fever in patients with an a priori increased risk of infection [288,306,307]. In an RCT including only moderate to high-risk infection patients (patients with pre-operative stents/nephrostomy or positive urine culture), a seven-day course of pre-operative antibiotics reduced the risk of post-PCNL sepsis threefold in comparison to a two-day course . In studies that did not specify the risk of the patient population, a single dose of antibiotic prophylaxis administered at induction was equivalent to an extended pre-operative course [307,309]. In contrast to this, a prolonged course of post-operative antibiotics was not superior to a single dose pre-operatively [288,307]. As national and regional antibiotic resistance patterns can differ significantly, the choice of antibiotic prophylaxis should be tailored to institutional or regional antimicrobial susceptibility . \| \| \| --- \| \| Recommendations \| Strength rating \| \| Obtain a urine culture or perform urinary microscopy before any treatment is planned. \| Strong \| \| Exclude or treat urinary tract infections prior to stone removal. \| Strong \| \| Offer peri-operative antibiotic prophylaxis to all patients undergoing endourological treatment. \| Strong \| 3.4.8.2. Antithrombotic therapy and stone treatment Patients with a bleeding disorder, or receiving antithrombotic therapy, should be referred to an internist for appropriate therapeutic measures before deciding on stone management [310-314]. In patients with an uncorrected bleeding disorder, the following are at elevated risk of haemorrhage or perinephric haematoma (PNH) (high-risk procedures): SWL (hazard ratio of PNH up to 4.2 during anti-coagulant/anti-platelet medication) PCNL; percutaneous nephrostomy; laparoscopic surgery; open surgery . Shock wave lithotripsy is feasible and safe after correction of the underlying coagulopathy [316,317]. In the case of an uncorrected bleeding disorder or continued antithrombotic therapy, URS, in contrast to SWL and PCNL, might offer an alternative approach since it is associated with less morbidity [318-320]. Despite the appropriate cessation of anti-platelet agents, following standardised protocols, prolonged haematuria in tube drainage after PCNL has been reported . Only data on flexible URS are available which support the superiority of URS in the treatment of proximal ureteral stones [322,323]. Although URS is safe in patients with bleeding disorders or anticoagulation, an individualised patient approach is necessary . Table 3.9: Risk stratification for bleeding \| \| \| --- \| \| Risk stratification for bleeding \| \| \| Low-risk bleeding procedures \| Cystoscopy Flexible cystoscopy Ureteral catheterisation Extraction of ureteral stent Ureteroscopy \| \| High-risk bleeding procedures \| Shock wave lithotripsy Percutaneous nephrostomy Percutaneous nephrolithotomy \| Table 3.10: Suggested strategy for antithrombotic therapy in stone removal [312-314] (In collaboration with a cardiologist/internist weigh the risks and benefits of discontinuation of therapy, vs. delaying elective surgical procedures). \| \| \| \| \| \| --- --- \| Suggested strategy for antithrombotic therapy in stone removal \| \| \| \| \| \| Medication/Agent \| Bleeding risk of planned procedure \| Risk of thromboembolism \| \| \| \| Low risk \| Intermediate risk \| High risk \| \| Warfarin Dabigatran Rivaroxaban Apixaban \| Low-risk procedure \| May be continued \| Bridging therapy \| Bridging therapy \| \| High-risk procedure \| May be temporarily discontinued at appropriate interval. Bridging therapy is strongly recommended. \| Bridging therapy \| Bridging therapy \| \| Aspirin \| Low-risk procedure \| Continue \| Continue \| Elective surgery: postpone. Non-deferrable surgery: continue. \| \| High-risk procedure \| Discontinue \| Elective surgery: postpone. Non-deferrable surgery: continue, if it is possible. \| Elective surgery: postpone. Non-deferrable surgery: continue. \| \| Thienopyridine agents (P2Y12 receptor inhibitors) \| Low-risk procedure \| Discontinue five days before intervention. Resume within 24-72 hours with a loading dose. \| Continue \| Elective surgery: postpone. Non-deferrable surgery: continue. \| \| High-risk procedure \| Discontinue five days before intervention and resume within 24-72 hours with a loading dose. \| Elective surgery: postpone. Non-deferrable surgery: discontinue five days before procedure and resume within 24-72 hours with a loading dose. Bridging therapy -glycoprotein IIb/IIIa inhibitors if aspirin is discontinued. \| Elective surgery: postpone. Non-deferrable surgery: discontinue five days before procedure and resume within 24-72 hours, with a loading dose. Bridging therapy -glycoprotein IIb/IIIa inhibitors. \| 3.4.8.2.1. Summary of evidence and recommendations for antithrombotic therapy and stone treatment \| \| \| --- \| \| Summary of evidence \| LE \| \| Active surveillance is indicated in patients at high risk for thrombotic complications in the presence of an asymptomatic calyceal stone. \| 4 \| \| The temporary discontinuation, or bridging of antithrombotic therapy in high-risk patients, should be discussed with the internist. \| 3 \| \| Retrograde (flexible) URS stone removal is associated with less morbidity in patients when antithrombotic therapy cannot be discontinued. \| 2a \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Offer active surveillance to patients at high risk of thrombotic complications in the presence of an asymptomatic calyceal stone. \| Weak \| \| Decide on temporary discontinuation, or bridging of antithrombotic therapy in high-risk patients, in consultation with the internist. \| Strong \| \| Retrograde (flexible) URS is the preferred intervention if stone removal is essential and antithrombotic therapy cannot be discontinued since it is associated with less morbidity. \| Strong \| 3.4.8.3. Obesity A high BMI can pose a higher anaesthetic risk and a lower success rate after SWL and PCNL and may influence the choice of treatment . 3.4.8.4. Stone composition Stones composed of brushite, calcium oxalate monohydrate, or cystine are particularly hard, as well as homogeneous stones with a high density on NCCT [326,327]. Percutaneous nephrolithotomy or RIRS and URS are alternatives for removal of large SWL-resistant stones. \| \| \| --- \| \| Recommendations \| Strength rating \| \| Consider the stone composition before deciding on the method of removal, based on patient history, former stone analysis of the patient or Hounsfield unit on unenhanced computed tomography. \| Strong \| \| Attempt to dissolve radiolucent stones. \| Strong \| 3.4.8.5. Contraindications of procedures Contraindications of extracorporeal SWL There are several contraindications to the use of extracorporeal SWL, including: pregnancy, due to the potential effects on the foetus ; bleeding disorders, which should be compensated for at least 24 hours before and 48 hours after treatment ; uncontrolled UTIs; severe skeletal malformations and severe obesity, which prevent targeting of the stone; arterial aneurysm in the vicinity of the stone ; anatomical obstruction distal to the stone. Contraindications of URS Apart from general problems, for example with general anaesthesia or untreated UTIs, URS can be performed in all patients without any specific contraindications. Contraindications of PCNL Patients receiving anti-coagulant therapy must be monitored carefully pre- and post-operatively. Anti-coagulant therapy must be discontinued before PCNL . Other important contraindications include: untreated UTI; tumour in the presumptive access tract area; potential malignant kidney tumour; pregnancy (Section 3.4.14.1). General contraindication for endourological procedures Endourological interventions do not adversely affect renal function although care must be taken in those with poor pre-operative renal function, diabetes and hypertension . However, a meta-analysis, based on low quality evidence, suggests that patients with impaired renal function and stone disease, may in fact benefit from the procedure to preserve or increase their renal function . 3.4.9. Specific stone management of ureteral stones 3.4.9.1. Conservative treatment/observation There are only limited data regarding spontaneous stone passage according to stone size [333,334]. Spontaneous stone passage was reported for 49-52% of upper ureteral stones, 58-70% of mid ureteral stones and 68-83% of distal ureteral stones. Considering stone size almost 75% of stones < 5 mm and 62% of stones ≥ 5 mm passed spontaneously, with an average time to stone expulsion about 17 days (range 6-29 days) [333,335]. Considering both size and location, stones of <5 mm in the distal ureteral have a 89% chance of spontaneous passage, while 71% of stones <5 mm located in the upper ureter still pass spontaneously . The Panel is aware of the fact that spontaneous stone expulsion decreases with increasing stone size and that there are differences between individual patients. 3.4.9.2. Pharmacological treatment, medical expulsive therapy Medical expulsive therapy should only be used in informed patients if active stone removal is not indicated. Treatment should be discontinued if complications develop (infection, refractory pain, deterioration of renal function). In the case of known uric acid stones in the distal ureter, a combination of alkalinisation with tamsulosin can increase the frequency of spontaneous passage. For details see Sections 3.4.3 and 3.4.4. 3.4.9.3. Indications for treatment of ureteral stones Indications for removal of ureteral stones are [176,334,336]: stones with a low likelihood of spontaneous passage; persistent pain despite adequate analgesic medication; persistent obstruction; renal insufficiency (renal failure, bilateral obstruction, or single kidney). 3.4.9.4. Selection of procedure for removal of ureteral stones The selection of the procedure depends on many factors, including stone-related factors, such as size, location, and density, as well as patient-related factors, such as body habitus, urinary anatomy, bleeding disorders, and other potential comorbidities. These and their influence on the outcomes of each of the procedures should be considered when counselling patients. As previously mentioned in this guideline, CT imaging can provide useful information that may influence the choice of treatment. A meta-analysis outlines that increasing stone density, stone burden, skin-to-stone distance, and hydronephrosis can negatively impact the success of the shockwave lithotripsy . Overall, SFRs after URS or SWL for ureteral stones are comparable. However, larger stones achieve earlier stone-free status with URS. A large multi-center non-inferiority trial compared URS to SWL for ureteral stones. When excluding patients that had spontaneously passed their stone prior to treatment, SWL could not be considered non-inferior to URS with only 12% of patients needing further intervention after URS in comparison to 26% in the SWL arm . In contrast to the success of SWL, comparative data on the outcomes of URS depending on patients’ BMI has shown URS to be as effective and safe in obese and morbidly obese patients as in non-obese patients . The Panel performed a systematic review to assess the benefits and harms of URS compared to SWL . Compared with SWL, URS was associated with a significantly greater SFR of up to four weeks, but the difference was not significant at three months in the included studies. Ureteroscopy was associated with fewer retreatments and the need for secondary procedures but with a higher need for adjunctive procedures, higher complication rates, and longer hospital stay. Counterbalancing for URS’s higher SFRs, SWL is associated with lower morbidity. Success rates and complications of URS are not impacted by previous unsuccessful SWL . Clavien-Dindo grade complications were if reported, less frequent in patients treated with SWL . Apart from the treatment modality, the timing of treatment may also be of importance. Primary or emergent ureteroscopy appears to be a safe and feasible procedure for patients presenting with renal colic due to an obstructive ureteral stone , without however increasing the stone-free rate. These results however are based mainly on low level of evidence reports and should be interpreted with caution . Similarly, SWL can be performed in the acute setting or electively allowing a trial of spontaneous passage. In contrast to acute URS, SWL in the acute setting does provide an increased stone-free rate and reduced need for auxiliary procedures . For large proximal ureteral stones, a percutaneous antegrade approach may provide better stone-free results than a retrograde ureteroscopic approach . Bleeding disorder Ureteroscopy can be performed in patients with bleeding disorders, with a moderate increase in complications (see also Section 3.4.8.2) . 3.4.9.4.1. Summary of evidence and recommendations for selection of procedure for active removal of ureteral stones \| \| \| --- \| \| Summary of evidence \| LE \| \| Observation is feasible in informed patients who develop no complications (infection, refractory pain, deterioration of renal function). \| 2a \| \| Medical expulsive therapy seems to be efficacious for treating patients with ureteral stones who are amenable to conservative management. The greatest benefit might be among those with > 5 mm (distal) stones. \| 1a \| \| Compared with SWL, URS was associated with significantly greater SFRs up to four weeks, but the difference was not significant at three months in the included studies. \| 1a \| \| Ureteroscopy was associated with fewer retreatments and need for secondary procedures, but with a higher need for adjunctive procedures, greater complication rates and longer hospital stay. \| 1a \| \| In the case of severe obesity, URS is a more promising therapeutic option than SWL. \| 2b \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| If active removal is not indicated (Section 3.4.9.3) in patients with newly diagnosed small ureteral stones, observe patient initially with periodic evaluation. \| Strong \| \| Offer α-blockers as medical expulsive therapy as one of the treatment options for (distal) ureteral stones > 5 mm. \| Strong \| \| Inform patients that ureteroscopy (URS) has a better chance of achieving stone-free status with a single procedure. \| Strong \| \| Inform patients that URS has higher complication rates when compared to shock wave lithotripsy. \| Strong \| \| Use URS as first-line therapy for ureteral (and renal) stones in cases of severe obesity. \| Strong \| See stratification data Alpha-blockers are an off-label treatment for this indication Figure 3.1: Treatment algorithm for ureteral stones (if stone removal is indicated)SWL = shock wave lithotripsy; URS = Ureteroscopy. 3.4.10. Specific stone management of renal stones The natural history of small, non-obstructing asymptomatic calculi is not well defined, and the risk of progression is unclear. There is still no consensus on the follow-up duration, timing, and type of intervention. In an RCT patients with small asymptomatic renal stones, who were not treated actively, had a higher incidence of relapse . 3.4.10.1. Conservative treatment (observation) Observation of renal stones, especially in calyces, depends on their natural history (Section 3.4.10.3). The recommendations provided are not supported by high-level literature . There is a prospective trial supporting annual observation for asymptomatic inferior calyceal stones, < 10 mm. In case stone growth is detected, the follow-up interval should be lowered . Intervention is advised for growing stones > 5 mm . In a systematic review of patients with asymptomatic renal stones on active surveillance spontaneous stone passage rates varied from 3-29%, symptom development from 7-77%, stone growth from 5-66%, surgical intervention from 7-26% . 3.4.10.2. Pharmacological treatment of renal stones Dissolution therapy seems to be an option for uric acid stones. See sections 3.4.4. and 3.4.8.4. 3.4.10.3. Indications for stone removal of renal stones Indications for the removal of renal stones include: stone growth; stones in high-risk patients for stone formation; obstruction caused by stones; infection; symptomatic stones (e.g., pain or haematuria) ; patient preference; comorbidity; the social situation of the patient (e.g., profession or traveling); 3.4.10.4. Selection of procedure for active removal of renal stones For general recommendations and precautions see Section 3.4.8. 3.4.10.4.1. Stones in the renal pelvis or upper/middle calyces Shock wave lithotripsy, PCNL and RIRS are available treatment modalities for renal calculi. While PCNL efficacy is hardly affected by the stone size, the SFRs after SWL or URS are inversely proportional to stone size [253,349-355]. Although multiple treatments or sessions may be needed shock wave lithotripsy achieves good SFRs for stones up to 20 mm, except for those at the lower pole [351,356,357]. When SWL is considered, stones with density > 1,000 HU (and with high homogeneity) on non-contrast-enhanced CT are less likely to be disintegrated . Endourology is considered an alternative because of the reduced need for repeated procedures and consequently a shorter time until stone-free status is achieved. For stones > 10 mm, mPCNL achieves a higher SFR than RIRS or SWL, but carries a higher risk of bleeding and is associated with a longer hospital stay; however, there is a high degree of heterogeneity among the included studies [353,355]. Stones > 20 mm should be treated primarily by PCNL, because SWL often requires multiple treatments, and is associated with an increased risk of ureteral obstruction (colic or steinstrasse) with a need for adjunctive procedures (Figure 3.2) . Retrograde renal surgery cannot be recommended as first-line treatment for stones > 20 mm in uncomplicated cases as SFRs decrease, and staged procedures will be required [359-361]. However, it may be a first-line option in patients where PCNL is not an option or contraindicated or in selected patients . The combination of PCNL and RIRS may be a good alternative for the treatment of complex renal stones compared to standard PCNL; however, the level of the existing evidence is low . 3.4.10.4.2. Stones in the lower renal pole The stone clearance rate after SWL seems to be lower for stones in the inferior calyx than for other intra-renal locations. Although the disintegration efficacy of SWL is not limited compared to other locations, the fragments often remain in the calyx and cause recurrent stone formation. The reported SFR of SWL for lower pole calculi is 25-95%. The preferential use of endoscopic procedures is supported by some current reports, even for stones < 1 cm [350,352,356,358,361,363-370]. The following can impair successful stone treatment by SWL [371-377]: steep infundibular-pelvic angle; long calyx; long skin-to-stone distance; narrow infundibulum; shock wave-resistant stones (calcium oxalate monohydrate, brushite, or cystine). Further anatomical parameters cannot yet be established. Supportive measures such as inversion, vibration or hydration may facilitate stone clearance (See 3.4.5 SWL) [174,378,379]. If there are negative predictors for SWL, PCNL and RIRS might be reasonable alternatives, even for smaller calculi . Retrograde renal surgery seems to have comparable efficacy to SWL [350,356,358,380]. Clinical experience has suggested a higher SFR of RIRS compared to SWL but at the expense of greater invasiveness. Depending on operator skills, stones up to 3 cm can be treated by RIRS [362,381]. However, staged procedures are frequently required. Although mini-PCNL has the highest success rate for the treatment of lower pole stones up to 2 cm, it comes at the expense of a higher complication rate and longer hospital stay . In complex stone cases, open or laparoscopic approaches are possible alternatives although they are infrequently used. 3.4.10.5. Summary of evidence and recommendations for the management of renal stones \| \| \| --- \| \| Summary of evidence \| LE \| \| It is still debatable whether renal stones should be treated, or whether annual follow-up is sufficient for asymptomatic calyceal stones that have remained stable for six months. \| 4 \| \| Although the question of whether asymptomatic calyceal stones should be treated is still unanswered, stone growth, de novo obstruction, associated infection, and acute and/or chronic pain are indications for treatment. \| 3 \| \| Percutaneous nephrolithotomy is indicated in renal stones > 2 cm as primary option. \| 1a \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Offer active treatment for renal stones in case of stone growth, de novo obstruction, associated infection, and acute and/or chronic pain. \| Weak \| \| Evaluate stone composition before deciding on the method of removal, based on patient history, former stone analysis of the patient or Hounsfield unit (HU) on unenhanced computed tomography. \| Strong \| \| Perform percutaneous nephrolithotomy (PCNL) as first-line treatment of larger stones > 2 cm. \| Strong \| \| Treat larger stones (> 2 cm) with flexible ureteroscopy or shock wave lithotripsy (SWL), in cases where PCNL is not an option. However, in such instances there is a higher risk that a follow-up procedure and placement of a ureteral stent may be needed. \| Strong \| \| Perform PCNL or retrograde intrarenal surgery for the lower pole, even for stones > 1 cm, as the efficacy of SWL is limited (depending on favourable and unfavourable factors for SWL). \| Strong \| \| Perform PCNL or retrograde intrarenal surgery for the lower pole, even for stones > 1 cm, as the efficacy of SWL is limited (depending on favourable and unfavourable factors for SWL). \| Strong \| Figure 3.2: Treatment algorithm for renal stones (if/when active treatment is indicated)The term ‘Endourology’ encompasses all PCNL and URS interventions. PCNL = percutaneous nephrolithotomy; RIRS = retrograde intrarenal surgery; SWL = shock wave lithotripsy; URS = ureteroscopy 3.4.11. Laparoscopy and open surgery Advances in SWL and endourological surgery (URS and PCNL) have significantly decreased the indications for open or laparoscopic stone surgery [382-387]. There is a consensus that most complex stones, including partial and complete staghorn stones, should be approached primarily with PCNL. Additionally, a combined approach with PCNL and RIRS may also be an appropriate alternative. However, if percutaneous approaches are not likely to be successful, or if multiple endourological approaches have been performed unsuccessfully; open or laparoscopic surgery may be a valid treatment option [388-392]. Few studies have reported laparoscopic stone removal. These procedures are usually reserved for special cases. When expertise is available, laparoscopic ureterolithotomy can be performed for large proximal ureteral stones as an alternative to URS or SWL [393,394]. These more invasive procedures have yielded high SFRs and lower auxiliary procedure rates [196,205,389]. A systematic review showed no difference in the post-operative phase for stented or unstented laparoscopic ureterolithotomy . Laparoscopic pyelolithotomy could be offered for solitary stones > 2 cm located in the renal pelvis as an alternative to PCNL . In addition, in selected cases with an extrarenal and dilated pelvis, RLP can be considered as an alternative management of staghorn calculi . A few studies with limited numbers of patients have reported using robotic surgery in the treatment of urinary stones . Open surgery should be considered as the last treatment option after all other possibilities have been explored. Studies on laparoscopy should be interpreted with caution due to their low design and quality of evidence. 3.4.11.1. Recommendation for laparoscopy and open surgery \| \| \| --- \| \| Recommendation \| Strength rating \| \| Offer laparoscopic or open surgical stone removal in rare cases in which shock wave lithotripsy, retrograde or antegrade ureteroscopy and percutaneous nephrolithotomy fail, or are unlikely to be successful. \| Strong \| 3.4.12. Steinstrasse Steinstrasse is an accumulation of stone fragments or stone gravel in the ureter and may interfere with the passage of urine . Steinstrasse occurs in 4% of cases of SWL [175,183], and the major factor in the development of steinstrasse formation is stone size . A major problem of steinstrasse is ureteral obstruction, which may be silent in up to 23% of cases. A MA including eight RCTs (n = 876) suggested a benefit of stenting before SWL in terms of steinstrasse formation but did not result in a benefit on SFRs or less auxiliary treatments . When steinstrasse is asymptomatic, conservative treatment is an initial option. Medical expulsion therapy increases stone expulsion and reduces the need for endoscopic intervention [398,399]. Ureteroscopy and SWL are effective in the treatment of steinstrasse [185,400]. In the event of UTI or fever, the urinary system should be decompressed, preferably by percutaneous nephrostomy [113,401]. 3.4.12.1. Summary of evidence and recommendations for steinstrasse \| \| \| --- \| \| Summary of evidence \| LE \| \| Medical expulsion therapy increases the stone expulsion rate of steinstrasse. \| 1b \| \| Ureteroscopy is effective for the treatment of steinstrasse. \| 3 \| \| Only low-level evidence is available, supporting SWL or URS for the treatment of steinstrasse. \| 4 \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Treat steinstrasse associated with urinary tract infection (UTI)/fever preferably with percutaneous nephrostomy. \| Weak \| \| Treat steinstrasse when large stone fragments are present with shock wave lithotripsy or ureteroscopy (in absence of signs of UTI). \| Weak \| 3.4.13. Management of patients with residual stones Following initial treatment with SWL, URS or PCNL, residual fragments may remain and require additional intervention [347,402-405]. Most of these studies indicate that initial imaging is performed on the first day or the first week after treatment. However, false positive results from dust or residual fragments, that will pass spontaneously without causing any stone-related event, might lead to over-treatment. Therefore, imaging at four weeks seems most appropriate [406-408]. Compared to US, KUB and IVU, NCCT scan has a higher sensitivity to detect small residual fragments after definitive treatment of ureteral or kidney stones [409,410]. A SR/MA examining residual fragments following any treatment has demonstrated that around a third of patients with either dust or fragments </= 4 mm experience disease progression and re-intervention within three years, whilst a third have spontaneous passage within two years regardless of imaging modality follow-up. For fragments > 4 mm, there are fewer studies, but these suggest low spontaneous passage rates and high intervention rates . Although NCCT has the highest sensitivity to detect residual fragments, this must be balanced to the exposure to ionising radiation when compared with KUB and US. Recurrence risk in patients with residual fragments after treatment of infection stones is higher than for other stones . 3.4.13.1. Recommendation for management of patients with residual stones \| \| \| --- \| \| Recommendation \| Strength rating \| \| Treat residual fragments > 4 mm. \| Weak \| 3.4.14. Management of specific patient groups 3.4.14.1. Management of urinary stones and related problems during pregnancy Clinical management of a pregnant patient with urolithiasis is complex and demands close collaboration between the patient, radiologist, obstetrician, and urologist . For diagnostic imaging see Section 3.3.1. Patients with urolithiasis may be at increased risk of developing adverse maternal or neonatal outcomes . Conservative approaches for symptomatic hydronephrosis as well as for ureteric calculi are the preferred initial management option in pregnant patients [414,415]. If spontaneous passage does not occur, or if complications develop (e.g., intractable symptoms, severe hydronephrosis, spontaneous renal fornix rupture or induction of premature labour), placement of a ureteral stent or a percutaneous nephrostomy tube is necessary as it is more effective than conservative treatment for symptom relief [417-419]. In the treatment of renal stones during pregnancy, when a stent is necessary, PCNL versus ureteral stent placement does not confer a significant difference in rates of adverse pregnancy events. However, ureteral stent placement was associated with a lower incidence of hospital admissions, emergency department visits, exchange procedures, and new UTIs or pyelonephritis . Ureteroscopy has become a reasonable alternative in these situations [408,421]. When compared to temporary ureteral JJ stenting until after delivery, ureteroscopy resulted in fewer needs for stent exchanges, less irritative LUTS and better patient satisfaction [422,423]. Non-urgent ureteroscopy in pregnant women is best performed during the second trimester, by an experienced urologist. Counselling of the patient should include access to neonatal and obstetric services . Although feasible, percutaneous removal of renal stones during pregnancy remains an individual decision and should be performed only in experienced centres . Pregnancy remains an absolute contraindication for SWL. 3.4.14.1.1. Summary of evidence and recommendation for the management of urinary stones and related problems during pregnancy \| \| \| --- \| \| Summary of evidence \| LE \| \| Stent insertion seems to be more effective than conservative treatment in the management of symptomatic moderate-to-severe hydronephrosis during pregnancy. \| 1a \| \| Ureteroscopy is a reasonable alternative to avoid long-term stenting/drainage. \| 1b \| \| There is a higher tendency for stent encrustation during pregnancy. \| 3 \| \| \| \| --- \| \| Recommendation \| Strength rating \| \| Treat all uncomplicated cases of urolithiasis in pregnancy conservatively (except when there are clinical indications for intervention). \| Strong \| 3.4.14.2. Management of stones in patients with urinary diversion Aetiology Patients with urinary diversion are at high risk for stone formation in the renal collecting system and ureter or in the conduit or continent reservoir [425,426]. Metabolic factors (hypercalciuria, hyperoxaluria and hypocitraturia), infection with urease-producing bacteria, foreign bodies, mucus secretion, and urinary stasis are responsible for stone formation (section 3.1.3). One study has shown that the risk for recurrent upper tract stones in patients with urinary diversion subjected to PCNL was 63% at five years . Management Smaller upper-tract stones can be treated effectively with SWL [429,430]. In most cases, endourological techniques are necessary to achieve stone-free status . In individuals with long, tortuous conduits or with invisible ureter orifices, a retrograde endoscopic approach might be difficult or impossible . For stones in the conduit, a trans-stomal approach can be used to remove all stone material (along with the foreign body) using standard techniques, including intracorporeal lithotripsy and flexible endoscopes. Trans-stomal manipulations in continent urinary diversion must be performed carefully to avoid disturbance of the continence mechanism . Before considering any percutaneous approach in these cases, CT should be undertaken to assess the presence of overlying bowel, which could make this approach unsafe , and if present, a surgical approach should be considered. Prevention Recurrence risk is high in patients with urinary diversion . Metabolic evaluation and close follow-up are necessary to obtain the risk parameters for effective long-term prevention. Preventive measures include medical management of metabolic abnormalities, appropriate therapy of urinary infections, and hyper-diuresis or regular irrigation of continent reservoirs . 3.4.14.2.1. Summary of evidence and recommendation for the management of stones in patients with urinary diversion \| \| \| --- \| \| Summary of evidence \| LE \| \| The choice of access depends on the feasibility of orifice identification in the conduit or bowel reservoir. Whenever a retrograde approach is impossible, percutaneous access with antegrade ureteroscopy is the alternative. \| 4 \| \| \| \| --- \| \| Recommendation \| Strength rating \| \| Perform percutaneous lithotomy to remove large renal stones in patients with urinary diversion, as well as for ureteral stones that cannot be accessed via a retrograde approach, or that are not amenable to shock wave lithotripsy. \| Strong \| 3.4.14.3. Management of stones in patients with neurogenic bladder Aetiology, clinical presentation, and diagnosis Patients with neurogenic bladder develop urinary calculi because of additional risk factors such as bacteriuria, hydronephrosis, VUR, renal scarring and lower urinary tract reconstruction [436,437]. The most common causes are urinary stasis and infection (Section 3.1.3). Indwelling catheters and surgical interposition of bowel segments for treatment of bladder dysfunction both facilitate UTI. Although calculi can form at any level of the urinary tract, they occur more frequently in the bladder; especially if bladder augmentation has been performed [438,439]. Diagnosis of stones may be difficult and delayed in the absence of clinical symptoms due to sensory impairment and vesicourethral dysfunction. Difficulties in self-catheterisation should lead to suspicion of bladder calculi. Imaging studies are needed (US, CT) to confirm the clinical diagnosis prior to surgical intervention. Management Management of calculi in patients with neurogenic bladder is similar to that described in Section 3.3. Any surgery in these patients must be performed under general anaesthesia because of the impossibility of using spinal anaesthesia. Bone deformities often complicate positioning on the operating table . The risk of stone formation after augmentation cystoplasty in immobile patients with sensory impairment can be significantly reduced by irrigation protocols . For efficient long-term stone prevention in patients with neurogenic bladder, correction of the metabolic disorder, appropriate infection control, and restoration of normal storing/voiding function of the bladder are needed. 3.4.14.3.1. Summary of evidence and recommendation for the management of stones in patients with neurogenic bladder \| \| \| --- \| \| Summary of evidence \| LE \| \| Patients undergoing urinary diversion and/or suffering from neurogenic bladder dysfunction are at risk for recurrent stone formation. \| 3 \| 3.4.14.4. Management of stones in patients with transplanted kidneys Stones in transplanted kidneys can either be transplanted or present de novo allograft stones. Usually, they are detected by routine US examination, followed by NCCT in cases of unclear diagnosis . Aetiology Transplant patients depend on their solitary kidneys for renal function. Impairment causing urinary stasis/obstruction, therefore, requires immediate intervention or drainage of the transplanted kidney. Stones in kidney allografts have an incidence of 2% . Risk factors for de novo stone formation in these patients are multi-fold: Immunosuppression increases the infection risk, resulting in recurrent UTIs. Hyper-filtration, excessively alkaline urine, renal tubular acidosis (RTA), and increased serum calcium caused by persistent tertiary hyperparathyroidism are biochemical risk factors. Management Selecting the appropriate technique for stone removal in a transplanted kidney is difficult, although management principles are like those applied in other single renal units [443-445]. Additional factors such as transplant function, coagulative status, and anatomical alterations due to the iliac position of the organ, directly influence the surgical strategy. For large or ureteral stones, careful percutaneous access and subsequent antegrade endoscopy are more favourable. The introduction of small flexible ureteroscopes and the holmium laser has made URS a valid treatment option for transplant calculi; however, one must be aware of potential injury to adjacent organs [444,446,447]. Retrograde access to transplanted kidneys can be difficult due to the anterior location of the ureteral anastomosis, and ureteral tortuosity [448-450]. Treatment of donor stones may be needed pre-transplant and increases the pool available for renal transplants. Post-transplant stone disease may also need treatment to maintain the allograft function. A systematic review evaluating the outcomes of pre- vs. post-transplant URS demonstrated a 100% SFR with an overall 7.5% complication rate, compared to a SFR of 60-100% with an overall complication rate of 12.9% for post-transplant URS; most complications were Clavien 1 . A SR shows that SWL is also a safe and effective option for de novo stones after transplantation, with an overall SFR of 81% and a complication rate of 17.2% . 3.4.14.4.1. Summary of evidence and recommendation for the management of stones in patients with transplanted kidneys \| \| \| --- \| \| Summary of evidence \| LE \| \| Conservative treatment for small asymptomatic stones is only possible under close surveillance and in absolutely compliant patients. \| 3 \| \| Shock wave lithotripsy for small calyceal stones is an option with minimal risk of complication, but localisation of the stone can be challenging. \| 3 \| \| \| \| --- \| \| Recommendation \| Strength rating \| \| Offer patients with transplanted kidneys any of the contemporary management options, including shock wave lithotripsy, flexible ureteroscopy and percutaneous nephrolithotomy. \| Strong \| 3.4.14.5. Special problems in stone removal Table 3.11: Special problems in stone removal \| \| \| --- \| \| Special problems in stone removal \| \| \| Calyceal diverticulum stones \| SWL, PCNL (if possible) or RIRS [454,455]. Can also be removed using laparoscopic retroperitoneal surgery [456,457]. Patients may become asymptomatic due to stone disintegration (SWL), whilst well-disintegrated stone material remains in the original position due to narrow calyceal neck. \| \| Horseshoe kidneys \| Can be treated in line with the options described above [458-460]. Passage of fragments after SWL might be poor. Acceptable SFRs (up to 76%) with low major complication rates (2.4%) can be achieved with flexible ureteroscopy [458-460]. \| \| Stones in pelvic kidneys \| SWL, RIRS, PCNL or laparoscopic surgery . \| \| Stones formed in a continent reservoir \| Each stone must be considered and treated individually. \| \| Patients with obstruction of the UPJ \| When outflow abnormality requires correction, stones can be removed by PCNL together with percutaneous endopyelotomy or open/laparoscopic reconstructive surgery. URS together with endopyelotomy with Ho:YAG laser . \| 3.4.15. Management of stones in children The true incidence of nephrolithiasis in children remains unclear due to the global lack of large epidemiological studies. Data derived from nationwide epidemiological studies, studies performed in different counties worldwide and large-scale databases [464,465] indicate that the incidence and prevalence of paediatric urinary stone disease have increased over the last few decades. Although boys are most commonly affected in the first decade of life the greatest increase in incidence has been seen in older female adolescents . Stone composition is similar in children as in adults, with a predominance of calcium oxalate stones. Compared to historical data, metabolic abnormalities responsible for stone formation are less commonly identified in children nowadays [467-469]. Hypocitraturia, low urine volume and hypercalciuria predominate [84,467-469]. Age may affect the predominant metabolic abnormality with hypercalciuria and hypocitraturia being the most common disorder present in children < 10 and > 10 years old, respectively . Genetic or systemic diseases (e.g., cystinuria or nephrocalcinosis) contributing to stone formation are relatively frequent in children accounting for less than 17% of the identifying causes [467,470]. The role of diet remains unclear in children, although there is some evidence that children are drinking less water and taking greater daily amounts of sodium than is recommended [471-473]. For diagnostic procedures see Section 3.3.3.2, for acute decompression see Section 3.4.2. and for metabolic evaluation see Chapter 4. 3.4.15.1. Clinical presentation Children with urinary stones can be asymptomatic or present with non-specific symptoms that necessitate a high index of suspicion for proper diagnosis. Symptoms are age-dependent with infants presenting with crying, irritability and vomiting in 40% of cases while in older children flank pain, micro or gross haematuria and recurrent UTIs are more common . 3.4.15.2. Conservative management There is a lack of evidence on conservative management of paediatric stones with evidence for ureteric calculi coming from the placebo arms of medical expulsive trials, while evidence for renal stones comes from small cohort studies, either on primary stones [476,477] or residual fragments remained after SWL, RIRS or PCNL . Expectant management for single, asymptomatic lower-pole renal stones could be the initial approach with increased odds of stone passage, especially in patients with non-struvite, non-cystine stones < 7 mm, with no anatomic abnormalities . Intervention may be needed for stones located elsewhere independently of their size [476-478]. 3.4.15.3. Medical expulsive therapy in children There are limited studies on MET as off-label expulsive therapy for children with ureteral stones up to 10 mm which show conflicting outcomes. Several systematic reviews and meta-analyses, including six RCTs and one conference abstract of an RCT, have been performed, all unanimously reporting that the use of alpha-blockers for distal ureteric stones increases the stone-free or stone expulsion rate [479-481]. The use of alpha-blockers also reduces the stone expulsion time and decreases pain episodes and analgesia demand with the disadvantage of more side-effects such as headache and nasal congestion [480,481]. 3.4.15.4. Extracorporeal shock wave lithotripsy Shock wave lithotripsy is still the first-line treatment for most ureteral stones in children. However, it is less likely to be successful for stones > 10 mm in diameter, impacted stones, calcium oxalate monohydrate or cystine stones, or for stones in children with unfavourable anatomy and in whom localisation is difficult . Studies on extracorporeal SWL in children suggest an overall SFR of 70-90%, retreatment rate of 4-50% and need for auxiliary procedures in 4-12.5% of cases [483-487]. A MA of fourteen studies reporting on 1,842 paediatric patients treated with SWL found significantly higher SFR for stones < 10 mm than for stones > 10 mm and higher retreatment rates as the stone size increased . For best clinical practice see Section 3.4.5. A MA on slow SWL vs. rapid SWL for renal stones revealed very low-quality evidence about the effects of SWL on SFRs, serious adverse events or complications of treatment and secondary procedures for residual fragments . Shock wave lithotripsy is well tolerated; however, good treatment outcomes are more likely to require the administration of general anaesthesia to children. With improvements in modern (second and third-generation) lithotripters, successful treatment using intravenous sedation, patient-controlled analgesia or no medication at all has been increasingly performed in a select population of older, co-operative children . Based on the results of a MA which compared SWL to dissolution therapy for intra-renal stones, and SWL to ureteroscopy with holmium laser or pneumatic lithotripsy for renal and distal ureteric stones, no firm conclusions can be drawn about the effects of SWL on SFR, serious adverse events or complications of treatment and secondary procedures for residual fragments . When SWL was compared to mini-percutaneous nephrolithotomy for lower pole renal stones 1-2 cm in size SWL resulted in lower SFRs (RR: 0.88, 95% CI: 0.80 - 0.97; moderate-quality evidence) and higher rates of secondary procedures (RR: 2.50, 95% CI: 1.01 - 6.20; low-quality evidence); however, SWL showed less severe adverse events (RR: 0.13, 95% CI: 0.02 - 0.98; low-quality evidence) . 3.4.15.5. Endourological procedures Rigid/semi-rigid ureteroscopy In recent years ureteroscopy is increasingly used in children with ureteral stones . Ureteroscopy proved to be effective with SFR of 81-98% [491-493], retreatment rates of 6.3%-10% and complication rates of 1.9-23% [491-493,495]. Similar to adults, routine stenting is not necessary before URS. Pre-stenting may facilitate URS, increase SFR and decrease complication rates [496,497]. Flexible ureteroscopy/retrograde intrarenal surgery Retrograde intra-renal surgery with flexible ureteroscopes (FURS) has become an efficacious treatment modality for paediatric renal stones. Studies report SFRs of 76-100%, retreatment rates of 0-19% and complication rates of 0-28% [498-501]. Younger age, cystine composition , large stone diameter and lack of pre-stenting predispose to FURS failure in children . A large global study across eight centres shows an SFR of 75.5%; although complications were minor, they were higher in patients < 5 years of age . Although high-level evidence is lacking to support a strong recommendation , FURS may be a particularly effective treatment option for lower calyceal stones in the presence of unfavourable factors for SWL [493,499,504]. For large and complex kidney stones RIRS has a significantly lower SFR compared to PCNL (71% vs. 95%), but is associated with less radiation exposure, lower complication rates, and a shorter hospital stay . Similarly, retrospective data indicate that RIRS may achieve lower SFRs compared to micro percutaneous surgery in favour of shorter operative time, shorter fluoroscopy time, and less hospitalisation time [506,507]. A published MA confirmed these results . Percutaneous nephrolithotomy Indications for PCNL in children are like those in adults and include renal stones > 2 cm, or smaller stones resistant to SWL and ureteroscopic treatment. Reported SFRs with paediatric PCNL are 71.4-95% after a single session [505-507,509,510] with an overall complication rate of 20% . A high degree of hydronephrosis, increased number of tracts and operative time , and large tract size [510,513-515] are associated with increased blood loss. Child age and stone burden predispose to the use of larger instruments during PCNL in children. The miniaturisation of equipment increases the opportunity to perform tubeless PCNL in appropriately selected children, which can reduce the length of hospital stay and post-operative pain [516,517]. A systematic review on the role of mini-PCNL showed an initial and overall SFR of 87.9% and 97% respectively, with no conversions to standard PCNL, and a complication rate of 19%, with a mean transfusion rate of 3.3% . Concerns have been raised regarding the possible adverse effects of PCNL on the renal parenchyma of the developing child. However, focal damage is only reported in 5% of cases . Using pre- and post-PCNL dimercaptosuccinic acid (DMSA) scans, Cicekbilek et al. demonstrated that PCNL tracts between 12-24 Charrière in size did not cause significant harm to paediatric kidneys . 3.4.15.6. Open and laparoscopic/robot-assisted stone surgery With the advances in SWL, PCNL and RIRS, very few cases of paediatric urolithiasis require open surgery. Data extracted from the National Inpatient Sample (NIS) databases for 2001-2014 showed that in the USA incisional procedures (mainly nephrolithotomy, pyelolithotomy and ureterotomy) were performed in 2.6% of hospitalised patients (52% aged 15-17 years) who required surgical intervention for urinary stones . Laparoscopy for the management of paediatric renal and ureteric stones is a safe and effective procedure when specific indications are followed. Stone-free rates of 100% were reported when laparoscopic pyelolithotomy was applied for a ≥ 1 cm single stone located in an extra-renal pelvis , or when laparoscopic ureterolithotomy was applied to impacted ureteric stones ≥ 1.5 cm, or to ureteric stones that were refractory to SWL or URS . There are extremely limited data available on the efficacy and complications of robot-assisted laparoscopic management of paediatric urolithiasis . 3.4.15.7. Special considerations on recurrence prevention All paediatric stone formers need metabolic evaluation and recurrence prevention with respect to the detected stone type. Children are in the high-risk group for stone recurrence (See Chapter 4). 3.4.15.8. Summary of evidence and recommendations for the management of stones in children \| \| \| --- \| \| Summary of evidence \| LE \| \| In children, MET could increase the rate of stone expulsion, reduce the stone expulsion time, and decrease pain episodes/analgesia demand, but it has a higher incidence of side effects. \| 1b \| \| In children, the indications for SWL, URS and PCNL are similar to those in adults. \| 1b \| \| Children with renal stones of a diameter up to 20 mm (~300 mm2) are ideal candidates for SWL. \| 1b \| \| Ureteroscopy has become the treatment of choice for larger distal ureteral stones in children. \| 1a \| \| In children, the indications for PCNL are similar to those in adults. \| 1a \| \| Mini-PCNL is safe and effective in children. \| 1b \| \| \| \| --- \| \| Recommendations \| Strength rating \| \| Offer children with single ureteral stones less than 10 mm shock wave lithotripsy (SWL) if localisation is possible or ureteroscopy as first-line option. \| Strong \| \| Ureteroscopy is a feasible alternative for ureteral stones not amenable to SWL. \| Strong \| \| Offer children with renal stones with a diameter of up to 20 mm (~300 mm2) SWL. \| Strong \| \| Offer children with renal pelvic or calyceal stones with a diameter > 20 mm (~300 mm2) percutaneous nephrolithotomy. \| Strong \| \| Retrograde renal surgery is a feasible alternative for renal stones smaller than 20 mm in all locations. \| Weak \| 3.5. Radiation exposure and protection during endourology The diagnosis and treatment of nephrolithiasis are associated with high levels of ionising radiation exposure to patients [523,524]. Currently, there are no studies estimating the lifetime radiation exposure of stone formers or the subsequent risk of malignancy development. The radiation exposure of endourologists has been extensively studied. Still, there are no studies assessing the risk of radiation-induced malignancies in urologists or operating theatre staff members [525-527]. Current evidence from atomic bomb patients [528,529], retrospective epidemiological data on medical exposure [530,531], and modelling studies [532,533] suggest an age and dose-dependent risk of secondary malignancy from ionising radiation. The International Commission on Radiological Protection (ICRP) recommends a maximum annual occupational exposure of 50mSv . However, the risk of radiation-induced malignancy follows a stochastic model having no known safe threshold of exposure. Taking this into consideration as well as the length of a urologist’s career the upper limit of 50mSv is still highly concerning. Table 3.12 shows the EAU Urolithiasis guidelines panel recommended protection methods to reduce radiation exposure to patients, surgical, anaesthesiologic, and nursing staff. Table 3.12: Radiation protection measures \| \| \| Radiation protection measures \| \| Limit studies or intervention involving radiation exposure to those that are strictly medically necessary. \| \| Implement a patient electronic record of medical imaging. \| \| Make use of imaging studies with lower radiation doses (US, KUB, digital tomosynthesis, low-dose and ultra-low dose CT scan). \| \| Create and follow a precise radiation exposure protection protocol in your department. \| \| Act in accordance with the as low as reasonably achievable (ALARA) principle. \| \| Measure and report fluoroscopy time to the operative surgeon (use dosimeters and perform monthly calculations). \| \| Technical measures to reduce radiation exposure include: Reducing fluoroscopy time; Limiting time adjacent to patient; Using low-dose radiation; Irradiating only to observe motion; Intra-operative use of pulsed fluoroscopy; Reduced fluoroscopy pulse rate; Collimated fields; Avoid digital image acquisition and rely on last image hold and instant replay technology. \| \| Use radiation protection instruments (chest, pelvic and thyroid shields, lead or lead-free gloves, protective glasses, lead protection under the operating table between the x-ray source and the surgeon). \| \| The radiation protection instruments must be cared for appropriately as any damage decreases effectiveness and increases exposure risk. They should be monitored and measured regularly to ensure integrity. \| \| Proper surgeon and operating room setup should be observed (follow the inverse square law, use the x-ray source underneath the patient’s body, decrease the x-ray source to patient distance, reduce magnification, avoid field overlap by not turning the C-arm in extreme angles, operate in the standing rather than the seated position). \| The availability of fluoroscopy is mandatory for endourological procedures. There is an increasing interest in fluoroless and fluoroscopy-free operations in urology. Several RCTs have been published showing a good outcome in means of stone-free and complication rates [166,270,535-537]. These trials have been limited to non-complex cases and they were not sufficiently powered to show the non-inferiority of fluoroscopy in PCNL [270,525] or the superiority of ultrasound in URS [213,214].
1261	https://www.statisticshowto.com/calculus-definitions/closed-interval-open-interval/	Statistics How To Statistics How To Closed Interval / Open Interval: Definition Contents (Click to skip to that section): 1. What is a Closed Interval? A closed interval [a, b] includes all of its endpoints and is represented by square brackets. For example, the closed interval [1, 9] includes all the numbers from 1 to 9, including the endpoints 1 and 9. On the other hand, an open interval, (a, b) — which has round parentheses — does not include the endpoints. For example, the interval (1, 9) includes all the possible numbers from 1 to 9 but not 1 and 9. In other words, square brackets mean the endpoint is included, and round parentheses mean it’s excluded. For example, let’s say we have a set of discrete numbers from 1 to 5: With a closed interval, the endpoints are included in the interval. For an open interval, the endpoints are excluded. So if an interval is half-closed, one of the endpoints are included and the other isn’t. Closed and open intervals complement each other in the sense that any real number can be in a closed interval, an open interval, or both. For example, the real number 3 is in the closed interval [1, 4] and the open interval (1, 4). The empty interval 0 and the interval of all real numbers, (∞, -∞), are also both open and closed. More formally, the definition of a closed interval is an interval that includes all of its limits. Notation Intervals are designated by writing the start point and end point as an ordered pair, within brackets. Square brackets show that an endpoint is included in an interval; so the closed interval beginning at x and ending at y would be denoted [x, y]. For example, say you had a set of numbers {1, 2, 3, 4}. The closed interval could be written as [1, 4]. Intervals which begin or end at infinity can’t include the endpoint, so a standard parentheses is used; but since the limit is included in the interval they can still be closed. For example, closed intervals include: More Precise Definition An interval is more precisely defined as a set of real numbers such that, for any two numbers a and b, any number c that lies between them is also included in the set. 2. Open Intervals Open intervals are defined as those which don’t include their endpoints. For example, let’s say you had a number x, which lies somewhere between zero and 100: Closed and opened intervals complement each other, but they aren’t mutually exclusive. The empty interval 0 and the interval containing all the reals, (∞, -∞), are actually both open and closed. 3. Half-Closed and Half-Open An interval which includes one endpoint but not the other is called half-closed or half-open — depending on your preference. It’s similar to a glass being half full — or half empty. However, you might see it defined by some authors as: When working in certain areas of statistical analysis such as topology , it’s preferable to specify which end is half-closed (or half-open). To achieve this, use the terminology left-closed, left-open, right-closed, right-open instead of half-open/half-closed. For example: Half Closed Interval Notation We denote an open interval by (a, b) and a closed interval by [a,b]; We denote a half-closed interval by a mixture of those two notations. Imagine your interval has endpoints a and b: It looks like mismatched brackets, but it’s really just a clever way of denoting which side is in the interval and which side isn’t. There’s one more way of denoting open intervals which you may not have come across yet: with backward square brackets. Using that notation, your half closed intervals become [a,b[ and ]a,b]. Here we use mathematical set notation to define these phrases: What are the differences between an open Interval and closed Interval? 4. Interval Function In simple terms, an interval function deals with inexact data—data that has to be represented with an interval. Interval functions are a part of interval analysis (a specific case of set-valued analysis), which handles uncertainty in intervals that commonly appear in real-world, deterministic phenomena (Chalco-Cano et al., 2013). For example, areas that often deal with inexact data include computer graphics, experimental physics, robotics, and many others (Budak et al. 2019). The most general type of differentiability for interval-valued functions is Hukuhara differentiability. An early example of an interval function can be traced to Archimedes, who gave the following interval for π: The idea of interval-valued functions was however, largely forgotten until it’s resurrection in the latter half of the 20th century (Zhao, 2019). Example: A Class of Interval Function One widely-used interval function class includes interval-valued objective functions in interval (fuzzy) linear or nonlinear programming. The class is defined as follows (Chalco-Cano et al., 2013): F(t) = C · g(t), Where: Software Interval analysis is fairly esoteric, so you won’t find it widely implemented in software. Many programs have add-ons. For example, in Maple, interval functions can be implemented with the intpakX package (Kramer & Geulig, 2001). They include &sqr, &sqrt, &ln, &exp, &∗∗, &intpower (i.e. xn), &sin, &cos, &tan, &arcsin, &arccos, &arctan, &sinh, &cosh und &tanh. 5. Closed and open intervals and inequalities Interval notation — including closed and open intervals — is a way to note the range of values that makes an inequality true. Therefore, every interval can be written as an inequality (and vice versa). Infinity is always represented as an open interval, or a half-open interval. For example: References Budak, H. et al., (2019) Fractional Hermite-Hadamard-Type Inequalities for Interval-Valued Functions. Proceedings of the American Mathematical Society. Volume 148, Number 2, Feb. Pages 705-718. Chalco-Cano, Y. et al., (2013). Calculus for interval-valued functions using generalized Hukuhara derivative and applications. Retrieved May 27, 2020 from: Hunter, John K. Topology of the Real Numbers. From Introduction to Analysis. Retrieved from on June 23, 2018 LaValle, Steven. Topological Spaces: Closed Sets. Planning Algorithms, 2006, Cambridge University Press. Retrieved from on June 23, 2018. Kramer, W. & Geulig, I. (2001). Interval Calculus in Maple. The Extension intpakX to the Package intpak of the Share-Library. Retrieved May 27, 2020 from: Oregon State University. Interval. Retrieved January 5, 2020 from: Ross, K. Elementary Analysis: The Theory of Calculus (Undergraduate Texts in Mathematics) 2nd ed. 2013 Edition. Springer. Zhao, D. (2019). Chebyshev type inequalities for interval-valued functions. Comments? Need to post a correction? Please Contact Us. Latest articles
1262	https://mathematica.stackexchange.com/questions/48138/implementing-kanes-subset-sum-algorithm	number theory - Implementing Kane's Subset Sum Algorithm - Mathematica Stack Exchange Join Mathematica By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematica helpchat Mathematica Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Implementing Kane's Subset Sum Algorithm Ask Question Asked 11 years, 4 months ago Modified6 years, 11 months ago Viewed 602 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. In the following paper, D. Kane describes an algorithm for subset sum that runs in logspace: I am trying to implement it mathematica subsetsum[sum0_, set0_] := Module[{sum = sum0, set = set0}, c = 0; csum = Abs[sum] + Total[set] + 1; p = NextPrime[csum]; n = Length[set]; check = False; While[c <= n, Print["c=" <> ToString[c] <> "p=" <> ToString[p]]; If[IntegerQ[Sum[x^(-sum)Product[(1 + x^(set)), {i, 1, n}], {x, 1, p - 1}]/p], check = True; Break[];]; c = c + Floor[Log[2, p]]; p = NextPrime[p]; ]; check ]; This initial value C C which seems to be the C C th prime or the next prime after C C is given as: C=\|B\|+∑n i=1\|m i\|+1 C=\|B\|+∑i=1 n\|m i\|+1 B B is the sum to which a subset of some set {m 1,m 2,m 3,…,m n}{m 1,m 2,m 3,…,m n} sum to. Furthermore, on a simple instance such as: mathematica set = {2, 3, 5, 7, 8, 9}; sum = 9; subsetsum[sum, set] the program seems to stop after 2 loops returning False... which is wrong. It should be returning true... Anyone has any ideas on this? algorithm number-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Oct 1, 2018 at 21:07 J. M.'s missing motivation 126k 11 11 gold badges 411 411 silver badges 590 590 bronze badges asked May 19, 2014 at 16:13 user13675user13675 967 1 1 gold badge 6 6 silver badges 15 15 bronze badges 2 I don't see anything unary in your inputs, which IIRC is the target of DK's algorithm. Ping him at Stanford if you need clarification on paper, nice guy and responsive.ciao –ciao 2014-05-20 08:52:23 +00:00 Commented May 20, 2014 at 8:52 I think you want the test to be If[!IntegerQ[...]] since that is when you have a "witness" to existence of one or more solutions. Also, strictly speaking, that algorithm calls for more space efficient arithmetic than is being done above.Daniel Lichtblau –Daniel Lichtblau 2014-05-20 15:29:30 +00:00 Commented May 20, 2014 at 15:29 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. [Okay, I realize this isn't an answer but it's a sort'a neat way to get those coefficients.] One can alternatively use the Cauchy Integral Formula to get the appropriate coefficient of the product polynomial that enumerates the possible subset sums. Below is a numerical implementation. A symbolic version can of course be coded, but it would in effect be computing the coefficients explicitly. mathematica residue[ll_, n_] := Module[ {x, logsum, pts, m = 1101, vals}, logsum = Total[Log[1 + x^ll]] - nLog[x]; pts = N[1/2Exp[2PiIRange[0, m - 1]/m], 200]; vals = Exp[logsum /. Transpose[{Thread[x -> pts]}]]; Total[vals]/m ] Here is an example of usage. ```mathematica vals = Union[NextPrime[RandomInteger[{1, 100}, 20]]] ( Out= {7, 11, 17, 29, 37, 41, 43, 53, 59, 67, 79, 89, 101} ) ``` This next function computes the coefficients explicitly so we can compare. ```mathematica coeffpoly[ll_] := Rest[Module[{x}, CoefficientList[Times @@ (1 + x^ll), x]]] coeffpoly[vals] ( Out= 38 ) ``` SO there are 38 ways to form the value 358 as a subset sum of vals. Now we do this using residue. residue[vals, 358] ( Out= \ 38.0000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000 + 0.10^-90 I ) Almost seems like magic. The weaknesses are (1) One cannot get away with m=1101 in general. The integral approximation could need arbitrarily many terms. (2) One cannot rely on numerical precision of 100 in general. This too might have to be ratcheted arbitrarily high. As subset sums are NP complete clearly one or the other will have to go beyond polynomially large in the size of the input in order to make this into a fully reliable method. I do not know if it is possible to get "usually" acceptable approximations in a way that stays in polynomial complexity. I suspect that, at least for the class of subset sums for which lattice reduction methods usually work, the approach above would not require exponential resources in the problem size. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered May 22, 2014 at 3:26 Daniel LichtblauDaniel Lichtblau 61k 2 2 gold badges 107 107 silver badges 206 206 bronze badges 1 Neat indeed! The numerical stability of complex-valued functions is always tricky...user13675 –user13675 2014-05-27 04:11:31 +00:00 Commented May 27, 2014 at 4:11 Add a comment\| Your Answer Thanks for contributing an answer to Mathematica Stack Exchange! 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1263	https://allen.in/dn/qna/34887573	As theta increases from 0^(@) to 90^(@), the value of cos theta is Increases Decreases Remains constant First decrease then increases. To view this video, please enable JavaScript and consider upgrading to a web browser thatsupports HTML5 video Text Solution AI Generated Solution Topper's Solved these Questions BASIC MATHS BASIC MATHS BASIC MATHS AIIMS 2019 CIRCULAR MOTION Similar Questions Explore conceptually related problems Find value of cos(90+theta) In the odjoining figure, angle of plane theta is increased from 0^(@) to 90^(@) . Plot force of friction f veraus theta graph Knowledge Check If (cos^(2) theta)/(cot^(2) theta - cos^(2) theta) = 3 and 0^(@) lt theta lt 90^(@) , then the value of theta is If 4sin^(2)theta-1=0 and angle theta is less than 90^(@) , the value of cos^(2)theta+tan^(2)theta is : (Take 0^(@)ltthetalt90^(@)) If 0^@ lt theta lt 90^@ , then the value of sin theta + cos theta is Similar Questions Explore conceptually related problems If cos theta + sec theta = 2 ( 0^@ letheta le 90^@) , then the value of cos 10 theta + sec 11 theta is If sin theta+cos theta=sqrt(2)cos(90^(@)-theta) then value of cot theta is If sin theta - cos theta =0, 0 le theta le 90^(@) then the value of theta is _____- If tan^(2)theta - 3sectheta + 3=0, 0^(@) lt theta lt 90^(@) , then the value of sintheta + cos theta is: If 5 sin theta - 4 cos theta = 0, 0^(@) lt theta lt 90^(@) , then the value of (5sintheta-2costheta)/(sintheta+3costheta) is : ALLEN-BASIC MATHS-EXERCISE-2 02:30 \| 00:48 \| 01:08 \| 01:19 \| 02:03 \| 00:58 \| 02:07 \| 01:01 \| 01:42 \| 01:47 \| 01:52 \| 01:10 \| 01:30 \| 03:01 \| 01:49 \| 03:27 \| 01:32 \| 01:03 \| 01:38 \| 03:15 \|
1264	http://ui.adsabs.harvard.edu/abs/2022arXiv220805386H/abstract	The cycle of length four is strictly $F$-Turán-good - Astrophysics Data System Skip to main content Now on article abstract page Toggle navigation ads === Feedback Submit Updates General Feedback ORCID Sign in to ORCID to claim papers in the ADS. About About ADS What's New ADS Blog ADS Help Pages ADS Legacy Services Careers@ADS Sign Up Log In 2025-09-26: ADS servers will be undergoing scheduled maintenance starting 4:00PM EDT 2025-09-26. During this period, we will be unable to send any user emails, including new user registration and password reset emails. The outage is expected to last at least 24 hours, and this banner will be updated when email capabilities are restored. Close Search Bar to Enter New Query quick field:Author First Author Abstract Year Fulltext Select a field or operator All Search Terms Your search returned 0 results Your search returned 0 results Show Menu Full Text Sources view AbstractCitations (1)References (3)Co-ReadsSimilar PapersVolume ContentGraphicsMetricsExport Citation The cycle of length four is strictly $F$-Turán-good Show affiliations Loading affiliations affiliations loading Hei, Doudou (0); Hou, Xinmin (1) Abstract Given an $(r+1)$-chromatic graph $F$ and a graph $H$ that does not contain $F$ as a subgraph, we say that $H$ is strictly $F$-Turán-good if the Turán graph $T_{r}(n)$ is the unique graph containing the maximum number of copies of $H$ among all $F$-free graphs on $n$ vertices for every $n$ large enough. Győri, Pach and Simonovits (1991) proved that cycle $C_4$ of length four is strictly $K_{r+1}$-Turán-good for all $r\geq 2$. In this article, we extend this result and show that $C_4$ is strictly $F$-Turán-good, where $F$ is an $(r+1)$-chromatic graph with $r\ge 2$ and a color-critical edge. Moreover, we show that every $n$-vertex $C_4$-free graph $G$ with $N(H,G)=\ex(n,C_4,F)-o(n^4)$ can be obtained by adding or deleting $o(n^2)$ edges from $T_r(n)$. Our proof uses the flag algebra method developed by Razborov (2007). Publication: eprint arXiv:2208.05386 Pub Date:August 2022 DOI: 10.48550/arXiv.2208.05386 arXiv:arXiv:2208.05386Bibcode: 2022arXiv220805386H Copied!Keywords: Mathematics - Combinatorics; 05C35; 05C38 E-Print Comments: 16 pages Feedback/Corrections? full text sources Preprint Preprint PDF \| Preprint article © The SAO Astrophysics Data System adshelp[at]cfa.harvard.edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement 80NSSC25M7105 The material contained in this document is based upon work supported by a National Aeronautics and Space Administration (NASA) grant or cooperative agreement. Any opinions, findings, conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of NASA. Resources About ADS ADS Help System Status What's New Careers@ADS Web Accessibility Policy Social @adsabs ADS Blog Project Switch to basic HTML Privacy Policy Terms of Use Smithsonian Astrophysical Observatory Smithsonian Institution NASA 🌓 × Back How may we help you? #### Missing/Incorrect Record Submit a missing record or correct an existing record.#### Missing References Submit missing references to an existing ADS record.#### Associated Articles Submit associated articles to an existing record (e.g. arXiv / published paper).#### General Feedback Send your comments and suggestions for improvements. Name Email Feedback Submit You can also reach us at adshelp [at] cfa.harvard.edu This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. Close without submitting Alert
1265	https://www.youtube.com/watch?v=5y82N5ApvOg	Trig Simplifying acosx +bsinx Duncan Loweth 1270 subscribers 3 likes Description 37 views Posted: 21 Feb 2021 you can turn acox + bsinx into Rsin(x+a) or Rcos(x+a) if you like. Here is how and some examples 2 comments Transcript: hello and this time looking at um how to simplify when you've got uh two trig functions a sign and c and there's coefficients in front so this might be something like uh four uh sorry let's put cos first 4 cos x + 5 sin x something like this and what we can do is we can um simplify this into something that looks like R of s x plus a number or we could make it as R sorry cos of X plus a number so it's kind of like doing the reverse of the double angle rule it's sort of saying okay well you know if this did magically turn out to be you know if this was a double angle rule answer um then what would the values of R and A need to be so that it could work backwards because if you think about expanding this forwards then um the double angle rule here is going to give you um sin x cos a uh plus cos x sin a and um and also we've got this coefficient out in the front um and what we're saying is that that R and that c a need to oh no need to equal five and that R and that sin a need to equal the four uh and we can do it with the cause as well um because we can look at that and say that this is R cos x cos a is that minus I think that's minus struggling to remember my addition rules um R sin x sin a and again in this situation the r and the COS a going to be four and the r and the sin a I've got yes R sin a needs to be five and um just by looking at the coefficients saying oh well you know if this thing is supposed to equal this thing then R cos a has to equal four and the r sin a has to equal the five and in which case once we know what those values are we can write it in this form now sometimes that's really helpful uh because it allows us to uh solve it in different way or it allows us to uh maybe draw a graph of it because you know if we have it as y equals uh in this state we can see that it's just transformations of graphs we know it's the cos x graph the plus a is going to move it to the left and the r is going to stretch it uh vertically so it might be that it allows us to solve a problem having in this form that we can't have it in its separated form right that's enough of me rambling on uh let's have a go at a uh an example see if I can uh work through it so here we have um some expression cos 2 Theta - 2 sin 2 Theta so I'm a bit worried we've got double angles all of a sudden um that might add a complication and give it in the form of R cos 2 theta plus Alpha R needs to be greater than zero and Alpha needs to be between not and Pi / 2 which is uh means it's acute angle that's 90° in in in degrees so I'm not actually concerned this 2 Theta anymore because although it's here it's also in this bit so I think they've just thrown that in just to try and worry you um so let's give it a go let's start with this thing and have a look at what this would look like if we expanded it so R cos 2 theta plus Alpha put a bracket around that so the addition formula would give this as COS of the 1 COS of the second us s of the first um s of the second and of course we've got an R out in front here so we need an R uh multiplied through there as well okay and what we actually want here we've just got cos 2 Theta which we've got here cos 2 Theta so that's nice but there's nothing else there's just yes it's just a one isn't it so that means that this thing R and cos Alpha R cos Alpha it's got to equal 1 and here we've got - 2 sin 2 Theta so we can write that sin 2 Theta which means that R and sin Theta needs to equal two so do that as well R sin Alpha did I say Theta I'm not sure R sin Alpha has to equal the two there okay and now in order to get R and Theta we've got two equations two unknowns it's simultaneous equations and there's a little trick to to get each of them do theta f done it again Alpha first by dividing them because we know that s over cos is the same as tan so if we do the second one divid by the first we're going to get sin / cos or tan Alpha equal 2 over 1 okay what's happen to the r well if we have r/ R yes R sin Alpha over cos um R cos Alpha the RS are just going to cancel so that's why it's quite neat dividing them the RS cancel those turn into tan and this gives our fraction you got to get the right way round of course so tan Alpha equals 2 so that's nice and easy uh we can just get uh shift inverse tan of 2 in radians just check that you are in radians 1.17 now you could substitute this back in and solve for R but sub optimal I mean um I suppose if you did it in your calculator it would be relatively exact but you can do it it another way you can uh do it somewhat like it's uh Pythagoras we can uh Square them and add them together so this squar plus this squar uh is R 2 sin 2 Alpha plus this squar so R 2 cos s Alpha and of course that will equal this squar plus this squar so that's 2^ 2 + 1 2 which is 1 uh and then we can factorize out the R 2 and say the R 2 braet sin^2 Alpha + cos 2 Alpha well of course we know that sin squ Alpha plus cos squ Alpha equals 1 that's one of the trigonometric identities so all of that can just disappear cuz that's one and we've just said that this equals uh 2^2 + 1 so R = < TK 5 which is greater than zero we don't need R is minus root 5 we don't need the plus or minus cuz they told us that R is greater than zero so we only need the positive value even though mathematically there would be two we're only using the positive one give your value of alpha to 3 decimal places uh okay yeah did that I didn't check that that rounds yes it's a one so that's okay hence or otherwise solve uh for Theta between n and Pi so that's obtuse uh OBT acute and obtuse angles um so anywhere up to 180° if we're thinking in degrees um uh for this okay well this we've got already so we're following on from what we've just done equals minus 1.5 so if we now look and say okay well we've just said that we can write it out uh with these two values so let's make a bit of space having done all of that the whole point was to have it as a single expression here Asun 5 of 2 Theta + 1.17 and this equals - 1.5 and now it's just a simple case of solving divide both sides byun 5 uh we need to inverse CA to get uh yes this and then we're going to end up with the value of 2 Theta + 1.17 and we're going to get two answers from that from our Cass diagram uh I remember all students t crap and uh this gives me a negative answer uh so presumably it will give me an answer either in uh this quadrant or this one uh and we'll need to um work out the various answers now um of course it's only wanting not Pi but because it's 2 theta plus a it we need to go up to three uh sorry 2 pi plus a bit plus 1.1 um so that we can catch them all so we will need both both of these answers uh for here and then we can code it back to get the Theta okay I'm not going to do that bit cuz um if if you can't do that bit then I'm sure I've got another video of how to do cast diagrams uh lurking somewhere that you can look up but hopefully you can see how uh this simplification bit went and now it's just a case of solving the trigonometry uh if I haven't done a video on that and uh and you don't know how to do the rest of this then put something in the comments and I'll I'll do the video okay and then and then we get answers you see that's that's the Clue answers there's several answers uh there's at least two small chance there might be a third but I suspect there'll be two answers okay so now moving on to another example uh a class were asked to solve 3 cos Theta = 2 - sin Theta for not to 360 one student expressed the equation in the form like this that's very sensible with r is greater than n and a is yes so that's good and correctly solve the equation why isn't that good for them find the values of R and Alpha and hence find her Solutions okay so what we notice straight away is well it doesn't look in the normal way uh but I think it's because they've got cos on one side and S on the other so we need to move that round rearrange it and say that sin Theta having added it to the other side plus 3 cos theta equals uh 2 now actually I was almost expecting it to be our s uh because we've got s and uh S Plus cos uh but uh actually it is CA so never mind it just means that we'll get a minus sign cre creeping in to our um formula later but that's okay not a problem so let's expand this Chap and see uh what that becomes so R cos Theta minus Alpha is going to become R COS of the first COS of the second uh minus oh hang on there's your minus aha so there's not a minus cropping in because they had a minus there so that makes that sign a plus doesn't it um which it is so that's good uh it doesn't make that sign a plus but it makes this s of plus plus uh s of the first s of the second and that matches with what we've got there so not as tricky as I was expecting good so this is what we should have which means that um R cos Alpha should equal three and uh oh there's an R there forgot about that one R sin Alpha should equal one okay well we can solve these easily enough we can say that tan Alpha equal 1 over 3 doing this second one divided by this first one and therefore Alpha equals so let's say shift tan of 1/3 equals no because I'm in radians let's try it again 18 435 okay uh and then we can find R either by shoving that back in and doing 3 / C of that answer let's do that 3 / COS of that answer so oops just need the bracket when it's in a fraction so R is 3.16 let's put the two on as well so R is 3162 but let's just check that we can do it the proper way so this squar plus this squar uh will equal this squ plus this squar and of course the sign and the COS are going to uh become one when we add them together so we get that R2 equals what's that going to be 3 S 9 10 and so R equals uh < TK 10 so let's see what Runk 10 looks like if this is different I'll be very sad but there we are 31622 776 so interesting that uh it didn't recognize it uh as um an exact value because often the calculator if it knows it's a a square root it will it'll display it but it didn't spot it uh but doing it this way we got the absolute exact answer so I'm going to say that that that is a far better way of doing it uh barely takes any effort but you get the exact answer no using that with the trig please um because you'll just get decimals and then get things wrong okay so find the values of R and Theta and hence find her Solutions oh KY this is going to be a huge question then I won't do this bit in fact because this is just the solving like we had before so if we get rid of all of this uh we don't need that anymore then we can rewrite this statement here that < TK 10 C finishes there COS of theta minus 18435 um equals 2 and you can divide by the root 10 you can inverse C you do your cast Di and uh and then you can find your Solutions uh multiple solutions for Theta minus 18.4 and then you can find your solutions for Theta so that's a cast diagram question from now on so I'll leave that one with you another student decided to square both sides of the equation and then form a quadratic equation in sin Theta show that the correct quadratic is this all right well that shouldn't be too bad let's give it a go get rid of all of this make a bit of space so Square both sides that's legitimate so Square the left hand side is going to give you 9 cos² Theta okay well there's no cos Square in this so we'll have to use some um angle formula in a minute and then squaring this of course you have to do it in a bracket don't you so we're going to get 4 minus double the two things times together so that's 4 sin Theta and then plus sin s Theta and this should oh no I was going to say equals not but no we've got an equals already so we we'll get that equals by moving it around later okay this looks promising uh we've got a four not a five but we've got Theus 4 sin Theta so that's good and we've got a sin Theta but we really need another uh another nine and look here's nine over here so right this will no doubt be a um uh simple case of swapping this C out for S uh we know know that sin^2 Theta + cos 2 Theta = 1 so 1 uh cos squ which want to get rid of is 1 - sin 2 Theta so we can just replace that and we get 9 - 9 sin^2 Theta so this is looking very good indeed equals all of this and now we can just bring it over have zero on this side if we bring the N sign over we're going to get 10 sin squ and if we bring uh the minus four sign's already here and looking good and then bring the N9 across we're going to get minus5 and okay I've got my equals z on the wrong side but it is shown and we're happy now we can solve this equation okay I'll set this up but I'm not going to do all of the number work I'll let you um finish that off but just to show you uh if you've forgotten how to do this just pretend that uh or not pretend but substitute something like U = sin Theta and then you can just solve 10 u^2 - 4 U - 5al not now that might factorize it might not uh but you can definitely solve that if well presumably you can I'd hope so if you're doing this kind of trigonometry then you can solve that equation and get two values of U and then what you can say is well okay if sin theta equals that value and sin theta equals that value then you can do inverse sign for each of and of course a cast diagram because you're going to get two answers from that within the range not to 360 so four answers in total for Theta two from well two u values from the quadratic and then two trigonometry values uh you Theta values from each of those because of the cast diagram explain why not all of the answers satisfy this thing um I think what you should probably do is try them so you've got four answers for Theta try them and see what happens uh and I think the what you find is that because you squared the answer um you know you uh because we squared it if if this equals this and that's all well and square it when we then come out and square root it and we have you know solve the quadratic it's going to give us negative answers as well or at least two answers uh presumably you know one will be positive negative you'll know this uh but even if it doesn't give one positive one negative it'll give two answers and it's quite possible that this was not true uh before we squared it but because we squared it it introduce that other value to come in uh but the the what you want to do there is is try them and see which ones work and you might find that the answers from the positive U work and the answers from the negative value of U that you've then solved for Theta don't work or or possibly the other way around okay I'm I'm guessing you can tell me if I'm wrong in the comments ments but I suspect that both of these values either will or won't work and opposite for the other appreciate I've left you a reasonable amount of uh solving to do there um if you need help with that and I haven't done videos on that already then let me know and I can do that it's turned into another giant uh length video but it is quite a big topic uh well not a big topic but quite a niggly thing to do so I hope uh it was helpful to see those examples if indeed you're still watching uh and uh and yeah the focus is really on getting uh those S plusus cause um situations into this one thing by expanding this with the addition formula uh obviously you still have to be able to solve quadratics and solve the trig and all that kind of stuff uh but as I say that's hopefully covered in another video right I shall stop rambling and let you go thanks very much and cheerio
1266	https://dictionary.cambridge.org/us/grammar/british-grammar/modal-verbs-and-modality	Cambridge Dictionary +Plus My profile +Plus help Log out {{userName}} Cambridge Dictionary +Plus My profile +Plus help Log out Log in / Sign up English (US) Modal verbs and modality Modality is about a speaker’s or a writer’s attitude towards the world. A speaker or writer can express certainty, possibility, willingness, obligation, necessity and ability by using modal words and expressions. Speakers often have different opinions about the same thing. These speakers are looking at the same thing. Here are the main verbs we use to express modal meanings: Core modal verbs: can, could, may, might, will, shall, would, should, must Semi-modals: dare, need, ought to, used to Other verbs with modal meanings: have (got) to, be going to and be able to Modal words and expressions There are a number of other words and expressions in English, apart from the main modal verbs, which also express modal meanings. Here are some examples: \| modal form \| example \| \| noun \| There is a possibility of snow this evening. \| \| adjective \| It’s clear that the Prime Minister is worried about education. \| \| adverb \| I’m certainly not fed up with it. \| See also: Modality: meanings and uses Modality: other modal words and expressions Be expressions (be able to, be due to) CanCouldCould, may and mightDareHad betterMayMightModality: formsModality: meanings and usesModality: tenseModality: other verbsModality: other modal words and expressionsMustNeedOught toShallShouldWillWouldBe expressions (be able to, be due to) Test your vocabulary with our fun image quizzes Try a quiz now Word of the Day rock something to its foundations to seriously damage, upset, or change an organization or someone's beliefs About this Blog Taking the rough with the smooth: talking about imperfect situations Read More New Words SIM farm More new words has been added to list To top Contents Adjectives and adverbs Easily confused words Nouns, pronouns and determiners Prepositions and particles Using English Verbs Words, sentences and clauses Cambridge Dictionary +Plus My profile +Plus help Log out English (US) Change English (UK) English (US) Español Português 中文 (简体) 正體中文 (繁體) Dansk Deutsch Français Italiano Nederlands Norsk Polski Русский Türkçe Tiếng Việt Svenska Українська 日本語 한국어 ગુજરાતી தமிழ் తెలుగు বাঙ্গালি मराठी हिंदी Follow us Choose a dictionary Recent and Recommended English Grammar English–Spanish Spanish–English Definitions Clear explanations of natural written and spoken English English Learner’s Dictionary Essential British English Essential American English Grammar and thesaurus Usage explanations of natural written and spoken English Grammar Thesaurus Pronunciation British and American pronunciations with audio English Pronunciation Translation Click on the arrows to change the translation direction. 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Especially or specially? Except or except for? Expect, hope or wait? Experience or experiment? Fall or fall down? Far or a long way? Farther, farthest or further, furthest? Fast, quick or quickly? Fell or felt? Female or feminine; male or masculine? Finally, at last, lastly or in the end? First, firstly or at first? Fit or suit? Following or the following? For or since? Forget or leave? Full or filled? Fun or funny? Get or go? Grateful or thankful? Hear or listen (to)? High or tall? Historic or historical? House or home? How is …? or What is … like? If or when? If or whether? Ill or sick? Imply or infer? In the way or on the way? It’s or its? Late or lately? Lay or lie? Lend or borrow? Less or fewer? Look at, see or watch? Low or short? Man, mankind or people? Maybe or may be? Maybe or perhaps? Nearest or next? Never or not … ever? Nice or sympathetic? No doubt or without doubt? No or not? Nowadays, these days or today? Open or opened? Opportunity or possibility? Opposite or in front of? Other, others, the other or another? Out or out of? Permit or permission? Person, persons or people? Pick or pick up? Play or game? Politics, political, politician or policy? Price or prize? Principal or principle? Quiet or quite? Raise or rise? Remember or remind? Right or rightly? Rob or steal? Say or tell? So that or in order that? Sometimes or sometime? Sound or noise? Speak or talk? Such or so? There, their or they’re? Towards or toward? Wait or wait for? Wake, wake up or awaken? Worth or worthwhile? Nouns, pronouns and determiners Determiners A/an and the Determiners (the, my, some, this) Determiners and types of noun Determiners: position and order Determiners: typical errors Determiners used as pronouns Every Possession (John’s car, a friend of mine) Such This, that, these, those Whole Nouns Nouns Nouns: form Nouns and prepositions Nouns: compound nouns Nouns: countable and uncountable Nouns: forming nouns from other words Nouns: singular and plural Uncountable nouns Accommodation Equipment Furniture Information Luck and lucky News Progress Weather Noun phrases Noun phrases: dependent words Noun phrases: order Noun phrases: uses Noun phrases: noun phrases and verbs Noun phrases: two noun phrases together Pronouns Pronouns Each other, one another Everyone, everybody, everything, everywhere It Gender No one, nobody, nothing, nowhere One One and one’s Pronouns: personal (I, me, you, him, it, they, etc.) Pronouns: possessive (my, mine, your, yours, etc.) Pronouns: reflexive (myself, themselves, etc.) Pronouns: indefinite (-body, -one, -thing, -where) Pronouns: one, you, we, they Relative pronouns Questions: interrogative pronouns (what, who) Someone, somebody, something, somewhere That Quantifiers A bit All Any Both Either Enough Least, the least, at least Less Little, a little, few, a few Lots, a lot, plenty Many More Most, the most, mostly Much, many, a lot of, lots of: quantifiers No, none and none of Plenty Some Some and any Question words How What When Where Which Who, whom Whose Why Using nouns Piece words and group words Comparison: nouns (more money, the most points) Nouns and gender Reported speech: reporting nouns Age Half Holiday and holidays Mind Opinion Promise Reason Sort, type and kind Thing and stuff View Way Work (noun) Prepositions and particles Prepositions Prepositional phrases Above After, afterwards Against Among and amongst As At At, in and to (movement) At, on and in (place) At, on and in (time) Below Beneath Beyond By During For For + -ing From In front of In spite of and despite In, into Near and near to Of On, onto Over To Under Until With Within Without Using English Collocation Functions Commands and instructions Commentaries Invitations Offers Requests Greetings and farewells: hello, goodbye, Happy New Year Suggestions Telephoning Warnings Numbers Dates Measurements Number Time People and places Geographical places Names and titles: addressing people Nationalities, languages, countries and regions Place names Place and movement Abroad Away and away from Back Inside Nearby Outside Up Politeness Reported speech Reported speech Reported speech: direct speech Reported speech: indirect speech Sexist language Spoken English Pronunciation Intonation Politeness Interjections (ouch, hooray) Tags Chunks Ellipsis Headers and tails Hyperbole Vague expressions Downtoners Hedges (just) Substitution All right and alright Please and thank you Here and there Just Kind of and sort of Oh So and not with expect, hope, think, etc. So Yes Anyway Discourse markers (so, right, okay) In fact Okay, OK Well You know You see Types of English British and American English Dialect Double negatives and usage Formal and informal language Newspaper headlines Register Slang Standard and non-standard language Swearing and taboo expressions Useful phrases According to Actual and actually Approximations (around four o’clock) At all Else Hear that, see that However, whatever, whichever, whenever, wherever, whoever It’s time May as well and might as well More or less Of course Point of view Writing Apostrophe (’) Apposition Contractions Contrasts Detached impersonal style Internet discourse and text messages It, this and that in paragraphs Paragraphs Punctuation Speech into writing Spelling Such as Verbs Tenses and time Past Past simple (I worked) Past continuous (I was working) Past continuous or past simple? Past simple or present perfect? Used to Past perfect simple (I had worked) Past perfect continuous (I had been working) Past perfect simple or past perfect continuous? Past perfect simple or past simple? Past verb forms referring to the present Past: typical errors Present Present continuous (I am working) Present perfect continuous (I have been working) Present perfect simple (I have worked) Present perfect simple or present perfect continuous? Present perfect: typical errors Present simple (I work) Present simple or present continuous? Present: typical errors Present verb forms referring to the past Future Future: will and shall Future: be going to (I am going to work) Future: other expressions to talk about the future Future continuous (I will be working) Future in the past Future perfect continuous (I will have been working here ten years) Future perfect simple (I will have worked eight hours) Future: present continuous to talk about the future (I’m working tomorrow) Future: present simple to talk about the future (I work tomorrow) Future: typical errors Going to Verb forms Finite and non-finite verbs Imperative clauses (Be quiet!) Infinitives with and without to Infinitive: active or passive? Perfect infinitive with to (to have worked) Verbs: basic forms Verbs: formation Verb patterns Hate, like, love and prefer Hear, see, etc. + object + infinitive or -ing Help somebody (to) do Look forward to Stop + -ing form or to-infinitive Verb patterns: verb + infinitive or verb + -ing? Verb patterns: verb + that-clause Verb patterns: with and without objects Would like Would rather, would sooner Phrasal verbs and multi-word verbs Passive voice Get passive Have something done Passive: forms Passives with and without an agent Passive: uses Passive: other forms Passive: typical errors Modal verbs and modality Can Could Could, may and might Dare Had better May Might Modality: forms Modality: meanings and uses Modality: tense Modality: other verbs Modality: other modal words and expressions Must Need Ought to Shall Should Will Would Conditionals and wishes Conditionals Conditionals: if Conditionals: other expressions (unless, should, as long as) Conditionals: typical errors If only In case (of) Suppose, supposing and what if Wish Using verbs Verbs: types Verb phrases Verbs and verb phrases: typical errors Appear Ask and ask for Be Be expressions (be able to, be due to) Come Do Enable Enjoy Explain Get Go Happen Have Have got and have Hope Know Let, let’s Like Look Made from, made of, made out of, made with Make Marry and divorce Matter Mean Miss Prefer Put See Seem Suggest Take Think Want Table of irregular verbs Words, sentences and clauses Word classes and phrase classes Word formation Word formation Prefixes Suffixes Compounds Abbreviations, initials and acronyms -ish and -y Diminutives (-let, -y and mini-) Hyphens Word order and focus Word order and focus Word order: structures Cleft sentences (It was in June we got married.) Fronting Inversion No sooner Not only … but also Conjunctions and linking words And As if and as though As long as and so long as Because, because of and cos, cos of Before But Conjunctions Conjunctions: adding Conjunctions: causes, reasons, results and purpose Conjunctions: contrasting Conjunctions: time Either … or… If In order to Or Since Unless Whereas Whether While and whilst Yet Clauses and sentences Adjuncts Clauses Clauses: finite and non-finite Clause types Complements Dummy subjects Exclamations Heads Objects Sentences Subjects Subject complements Subject–verb agreement Relative clauses Relative clauses Relative clauses referring to a whole sentence Relative clauses: defining and non-defining Relative clauses: typical errors Negation Negation Neither, neither … nor and not … either Not Neither, neither … nor and not … either Not Forming negative statements, questions and imperatives Negation: two negatives Negative clauses with any, anybody, anyone, anything, anywhere Negation in non-finite clauses Negative prefixes and suffixes Negative adverbs: hardly, seldom, etc. Negation: emphasising Negation of think, believe, suppose, hope Questions Questions: alternative questions (Is it black or grey?) Questions: statement questions (you’re over 18?) Questions: two-step questions Questions: typical errors Questions: wh-questions Questions: yes-no questions (Are you feeling cold?) Questions: follow-up questions Questions: echo and checking questions Questions: short forms My word lists To add ${headword} to a word list please sign up or log in. Sign up or Log in My word lists Add ${headword} to one of your lists below, or create a new one. {{name}} Go to your word lists Tell us about this example sentence: By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts. Privacy and Cookies Policy
1267	https://www.maths.gla.ac.uk/wws/cabripages/hyperbolic/common.html	hyperbolic geometry page perpendiculars and common perpendiculars Our first result is to generalize the euclidean theorem that guarantees a unique line through a given point and perpendicular to a given line. the existence of perpendiculars Suppose that K is a complete hyperbolic line, and that P is a point of E, not asymptotic for K, then there is a unique hyperbolic line through P perpendicular to K. proof Let K have asymptotic points A and B, so P ≠ A,B, and K,C are in B(A,B). We require an i-line through P orthogonal to K and to C. By theorem O2, The family of i-lines orthogonal to K,C is A(A,B). There is exacly one member of this family through P. Notice that this result applies even when P is on C. Also notice that, if K has asymptotic points A,B and H is the perpendicular through P, then as H is orthogonal to K, inversion in H maps K to itself. It follows that this inversion swaps A and B. It also fixes P which lies on H. Thus the inversion maps (A,B,P) to (B,A,P). For P on C, the existence of a hyperbolic transformation with this effect is implicit in the fundamental theorem of weird geometry. What is new is that this hyperbolic transformation is an inversion. In euclidean geometry, two lines have a common perpendicular if and only if they are parallel. Two parallel lines have an infinite family of perpendiculars. In hyperbolic geometry, the situation is quite different, as we shall show. In fact, both results are really consequences of results in inversive geometry. We shall leave the euclidean result as an exercise. the common perpendicular theorem Two complete hyperbolic lines have a common perpendicular if and only if they are ultraparallel. When the lines are ultraparallel, the perpendicular is unique. proof Suppose that K,L are complete hyperbolic lines. In other words, they are arcs of i-lines K,L orthogonal to C. An i-line H will give a hyperbolic line perpendicular to both if and only if H is orthogonal to C (to give a hyperbolic line) as well as to K and L By theorem O3, such an H will exist if and only if K and L are disjoint. This is equivalent to saying that K,L are ultraparallel. The theorem also shows that the common perpendicular is unique when K,Lare ultraparallel. main asymptotic page
1268	https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/slct.202403695	BF3 ⋅ OEt2 as a Versatile Reagent: Applications in Organic Synthesis - Kumar - 2024 - ChemistrySelect - Wiley Online Library Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. 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Search for more papers by this author Aditi Arora, Aditi Arora Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Authors have equal contribution. Search for more papers by this author Riya Chaudhary, Riya Chaudhary Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Search for more papers by this author Shivani Sapra, Shivani Sapra Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Search for more papers by this author Bhawani Shankar, Bhawani Shankar Department of Chemistry, Deshbandhu College, University of Delhi, Delhi, 110019 India Search for more papers by this author Sandeep Kumar, Corresponding Author Sandeep Kumar skbanger388@gmail.com orcid.org/0000-0002-2461-8329 Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Department of Chemistry, Ramjas College, University of Delhi, Delhi, 110007 India Search for more papers by this author Brajendra K. Singh, Corresponding Author Brajendra K. Singh singhbk@chemistry.du.ac.in Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Search for more papers by this author Sumit Kumar, Sumit Kumar Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Authors have equal contribution. Search for more papers by this author Aditi Arora, Aditi Arora Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Authors have equal contribution. Search for more papers by this author Riya Chaudhary, Riya Chaudhary Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Search for more papers by this author Shivani Sapra, Shivani Sapra Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Search for more papers by this author Bhawani Shankar, Bhawani Shankar Department of Chemistry, Deshbandhu College, University of Delhi, Delhi, 110019 India Search for more papers by this author Sandeep Kumar, Corresponding Author Sandeep Kumar skbanger388@gmail.com orcid.org/0000-0002-2461-8329 Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Department of Chemistry, Ramjas College, University of Delhi, Delhi, 110007 India Search for more papers by this author Brajendra K. Singh, Corresponding Author Brajendra K. Singh singhbk@chemistry.du.ac.in Bioorganic Laboratory, Department of Chemistry, University of Delhi, Delhi, 110007 India Search for more papers by this author First published: 29 October 2024 Citations: 1 Read the full text About References ---------- Related ------- Information ----------- PDF PDF Tools Request permission Export citation Add to favorites Track citation ShareShare Give access Share full text access Close modal Share full-text access Please review our Terms and Conditions of Use and check box below to share full-text version of article. [x] I have read and accept the Wiley Online Library Terms and Conditions of Use Shareable Link Use the link below to share a full-text version of this article with your friends and colleagues. Learn more. Copy URL Share a link Share on Email Facebook x LinkedIn Reddit Wechat Bluesky Abstract Boron trifluoride diethyl etherate, commonly referred to as BF 3 ⋅ OEt 2, plays a crucial role as a potent reagent in various organic transformations, particularly in the synthesis of heterocyclic systems through the establishment of carbon-carbon & carbon-heteroatom bonds. Its Lewis acid properties, reactivity, and stability at room temperature make it essential for a wide range of chemical reactions. BF 3 ⋅ OEt 2 demonstrates remarkable efficacy in processes such as hydroxylation of double bonds, epoxide cleavage, acid esterification, cyanation, coupling, and diverse cyclization reactions, surpassing other Lewis acids in performance. Its significant impact on organic synthesis extends to the formation of natural products, highlighting its profound significance in the realm of chemistry. The unique properties and versatility of boron trifluoride diethyl etherate make it a fundamental tool for a diverse array of chemical transformations. The numerous applications of this reagent in organic synthesis have prompted the development of a comprehensive review, focusing on its various roles and contributions in this field. Graphical Abstract BF 3 ⋅ OEt 2 is a highly versatile Lewis acid in organic synthesis, enabling transformations like alkylation, glycosylation, rearrangements, cyanation, coupling, ring-opening, michael addition, cyclization, etc. It plays a crucial role in the synthesis of natural products, showcasing its importance and adaptability in various chemical reactions. Conflict of Interests The authors declare no competing financial interest. Open Research Data Availability Statement Data sharing is not applicable to this article as no new data were created or analyzed in this study. References 1 M. J. O'Neil, P. E. Heckelman, C. B. Koch, K. J. 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Lett. 2014, 16, 1442. 10.1021/ol500238k CASPubMedWeb of Science®Google Scholar Citing Literature Volume 9, Issue 41 November 4, 2024 e202403695 References ---------- Related ------- Information ----------- Recommended Fluorobenziodoxole−BF 3 Reagent for Iodo(III)etherification of Alkynes in Ethereal Solvent Jinkui Chai,Wei Ding,Junliang Wu,Naohiko Yoshikai, Chemistry – An Asian Journal Ynamides: Versatile Tools in Organic Synthesis Gwilherm Evano Dr.,Alexis Coste,Kévin Jouvin, Angewandte Chemie International Edition β‐halovinyl Aldehydes: Multifaceted Versatile Building Blocks In Organic Synthesis Paras Saini,Jyoti,Pawan K. Sharma,Virender Singh, European Journal of Organic Chemistry Boron Trifluoride Etherate Veronica Cornel,Carl J. 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1269	http://stanleyrabinowitz.com/download/reciprocalwsums-revised.pdf	1 Algorithmic Simpliﬁcation of Reciprocal Sums Stanley Rabinowitz 12 Vine Brook Road Westford, MA 01886 USA 1. Introduction. Algorithms for evaluating or simplifying sums of the form N X n=1 1 Fn+a1Fn+a2 · · · Fn+ar where the Fi are Fibonacci numbers and the ai are integers have been discussed in . It is the goal of this paper to generalize these results to arbitrary second-order linear recurrences. Consider the second order linear recurrences deﬁned by un+2 = Pun+1 −Qun, u0 = 0, u1 = 1, (1) vn+2 = Pvn+1 −Qvn, v0 = 2, v1 = P, (2) and wn+2 = Pwn+1 −Qwn, w0, w1 arbitrary. (3) Let r1 and r2 denote the roots of the characteristic equation x2 −Px + Q = 0. Let D = P 2 −4Q and e = w0w2 −w2 1. (4) Throughout this paper, we shall assume that D 6= 0, e 6= 0, Q 6= 0, and wn 6= 0 for n > 0. In the case of the Fibonacci sequence, we showed that all reciprocal sums can be expressed in closed form in terms of FN = N X n=1 1 Fn , GN = N X n=1 (−1)n Fn , and KN = N X n=1 1 FnFn+1 . (5) It is our intent to generalize these results to apply to the sequence hwni. For the following deﬁnitions, let r be a positive integer and let a1, a2,. . . , ar be distinct nonnegative integers. Deﬁnition 1 (Unit Reciprocal Sum). A unit reciprocal sum of order r is a sum of the form N X n=1 1 wn+a1wn+a2 · · · wn+ar . Reprinted from Applications of Fibonacci Numbers, volume 8 (edited by Frederic T. Howard), Kluwer Academic Publishers, Dordrecht, The Netherlands: 1999, pp. 277–292 2 Deﬁnition 2 (Q-Reciprocal Sum). A Q-reciprocal sum of order r is a sum of the form N X n=1 Qkn wn+a1wn+a2 · · · wn+ar where k = br/2c. Deﬁnition 3 (Reciprocal Sum). A reciprocal sum is a unit reciprocal sum or a Q-reciprocal sum. Deﬁnition 4 (Rational Sum). A rational sum of order r is a sum of the form N X n=1 f(x1, x2, . . . , xs) wn+a1wn+a2 · · · wn+ar where f(x1, x2, . . . , xs) is a polynomial with each of the variables xi being of the form wn+ci or Qn. In this paper, we will show that all reciprocal sums of orders 1 and 2 can be expressed in closed form in terms of XN = N X n=1 1 wn , YN = N X n=1 Qn wn , and WN = N X n=1 1 wnwn+1 . (6) We will also show that all Q-reciprocal sums (of any order) can be expressed in closed form in terms of WN, XN, and YN. Finally, we shall show that if Q = ±1, then all rational sums can be expressed in terms of UN, VN, WN, XN, and YN, where UN = N X n=1 wn+1 wn and VN = N X n=1 Qnwn+1 wn . We shall also present mechanical algorithms for ﬁnding these closed forms. We need the following results. Theorem 1 (The Representation Theorem). If a, b, and c are integers and ua−b 6= 0, then wn+c = uc−b ua−b wn+a + uc−a ub−a wn+b. (7) (This expresses wn+c in terms of wn+a and wn+b.) Proof: The identity can be mechanically veriﬁed by using algorithm LucasSimplify from . Theorem 1 can be put into a more symmetrical form: 3 Theorem 2. For all integers a, b, and c, Qcub−cwn+a + Qauc−awn+b + Qbua−bwn+c = 0. (8) (This gives a symmetric connection between wn+a, wn+b, and wn+c.) Proof: This follows from the Representation Theorem by making use of the well-known Negation Formula : u−n = −unQ−n. Theorem 3. If a, b, and c are integers and ua−b 6= 0, then 1 wn+awn+b = A wn+cwn+a + B wn+cwn+b (9) where A = uc−a ub−a and B = uc−b ua−b . (This allows one to convert reciprocal sums of order 2 to those in which wn+c occurs as a factor of the denominator.) Proof: This is an immediate consequence of the Representation Theorem. 2. Reciprocal Sums of Order 1. There are no known elementary forms for the reciprocal sums of order 1, so we shall give them names: XN = N X n=1 1 wn and YN = N X n=1 Qn wn . (10) Strictly speaking, we should write these as XN(w0, w1, P, Q) and YN(w0, w1, P, Q); but we will simply write XN and YN when w0, w1, P, and Q are ﬁxed. If a > 0, we have N X n=1 1 wn+a = XN+a −Xa (11) and N X n=1 Qn wn+a = YN+a −Ya. (12) Thus all reciprocal sums of order 1 can be expressed in terms of XN and YN. 4 3. Q-Reciprocal Sums of Order 2. Theorem 4. If a > 0, then ua N X n=1 Qn wnwn+a = Q e " a X n=1 wn−1 wn − a X n=1 wN+n−1 wN+n # . (13) Proof: We begin with the identity wn+awn−1 −wnwn+a−1 = Qn−1eua (14) which comes from d’Ocagne’s Identity (see ). Thus, we have wn−1 wn −wn+a−1 wn+a = Qn−1eua wnwn+a . Summing from 1 to N yields ua N X n=1 Qn−1 wnwn+a = 1 e " N X n=1 wn−1 wn − N X n=1 wn+a−1 wn+a # = 1 e " a X n=1 wn−1 wn − a X n=1 wN+n−1 wN+n # which is the desired result. This can be put into a more symmetrical form. The following theorem is a general-ization of a result by Good and was proven by Andr´ e-Jeannin . Theorem 5 (Symmetry Property for Reciprocal Sums). If a > 0, then ua N X n=1 Qn wnwn+a = uN a X n=1 Qn wnwn+N . (15) Proof: Again we use d’Ocagne’s identity. Putting a = N in formula (14) gives wn−1wN+n −wnwN+n−1 = Qn−1euN. Combining the two sums on the right-hand side of Theorem 4 and applying this identity yields the desired result. Corollary (letting a = 1). N X n=1 Qn wnwn+1 = QuN w1wN+1 = Q e w0 w1 − wN wN+1 % . (16) 5 4. Unit Reciprocal Sums of Order 2. For non-alternating reciprocal Fibonacci sums, we had to introduce (in ) the sym-bol KN = N X n=1 1 FnFn+1 (17) for a sum with no known simple closed form. In a similar manner, we need to do the same thing for unit reciprocal sums for the sequence hwni. Let WN = N X n=1 1 wnwn+1 (18) with the understanding that W0 = 0. Again, we should really write this as WN(w0, w1, P, Q); but if hwni is a ﬁxed sequence, we will simply write this as WN. Theorem 6. If c > 0 and uc 6= 0, then N X n=1 1 wnwn+c = 1 uc c−1 X i=0 Qi(WN+i −Wi). (19) Proof: Letting a = i, b = i + 1, and c = 0 in Theorem 3, we get ui+1 wnwn+i+1 − ui wnwn+i = Qi wn+iwn+i+1 using the Negation Formula u−n = −unQ−n. Summing as n goes from 1 to N yields ui+1WN(i + 1) −uiWN(i) = Qi(WN+i −Wi) where WN(a) = N X n=1 1 wnwn+a . Now sum as i goes from 0 to c −1. The left side telescopes and we get ucWN(c) = c−1 X i=0 Qi(WN+i −Wi) which gives our desired result. Thus, all reciprocal sums of order 2 can be expressed in terms of WN. 6 5. The Reduction Process. We now show how to simplify certain reciprocal sums with three or more factors in the denominator. Theorem 7 (The Partial Fraction Decomposition Formula for w). For all n, Qn wn+awn+bwn+c = A wn+a + B wn+b + C wn+c (20) where A = −Q−a eub−auc−a , B = −Q−b euc−bua−b , and C = −Q−c eua−cub−c . (21) Proof: This result can be mechanically proven using algorithm LucasSimplify from . Theorem 8 (The Reduction Theorem for w). If r > 2, then any Q-reciprocal sum of order r can be expressed in terms of Q-reciprocal sums of order r −2. Proof: By Theorem 7, we have Qkn wn+awn+bwn+c = AQ(k−1)n wn+a + BQ(k−1)n wn+b + CQ(k−1)n wn+c (22) where A, B, and C are given in display (21). If r > 2 and k = br/2c, then we can take the last three factors in the denominator and apply Theorem 7. This breaks the given sum up into sums with r −2 factors in the denominator. The numerators have terms that are constant multiples of Q(k−1)n where k−1 = b(r −2)/2c, thus making these sums multiples of Q-reciprocal sums of order r −2. Corollary. Any Q-reciprocal sum of order r can be expressed in terms of reciprocal sums of order 1 or 2. If Q = ±1, then any reciprocal sum of order r can be expressed in terms of reciprocal sums of order 1 or 2. Proof: Apply Theorem 8 repeatedly, until the order of the Q-reciprocal sum becomes 1 or 2. If Q = ±1, then formula (20) can be written in the form 1 wn+awn+bwn+c = A(−1)n wn+a + B(−1)n wn+b + C(−1)n wn+c . (23) Applying this repeatedly reduces the order of the reciprocal sum to 1 or 2. By induction, we can state a more general form of the Partial Fraction Decomposition Theorem. 7 Theorem 9 (The Generalized Partial Fraction Decomposition Formula). If r is a positive integer, then 1 wn1wn2wn3 · · · wn2r+1 = 2r+1 X i=1 Ai wni (24) where A−1 i = (−eQni)r Q j6=i unj−ni. 6. The Simpliﬁcation Algorithm. We can also handle sums similar to reciprocal sums, but in which the numerators are polynomials in the w’s. These are called rational sums. We need to add in two new primitives: UN = N X n=1 wn+1 wn and VN = N X n=1 Qnwn+1 wn . (25) Once again, these would more properly be written as UN(w0, w1, P, Q) and VN(w0, w1, P, Q); but UN and VN will suﬃce when the sequence is ﬁxed. We now show how to evaluate a wide class of reciprocal and rational sums in closed form in terms of the quantities UN, VN, WN, XN, and YN. Deﬁnition. A w-polynomial in the variable n is any polynomial f(x1, x2, . . . , xr) with constant coeﬃcients where each xi is of the form wx or Qx, with each x of the form n+cj, where the cj are integer constants. For purposes of this deﬁnition, the quantities P, Q, w0, w1, and e are to be considered constants. Theorem 10 (The Simpliﬁcation Theorem for Q = ±1). Suppose that P, Q, w0, and w1 are ﬁxed constants, thereby determining the sequence hwni. Let f(n) be any w-polynomial in the variable n. For r a positive integer, let cj, j = 1, 2, . . . , r be distinct integers. Assume that wn > 0 and un > 0 for all n > 0. Furthermore, if Q = ±1, then we can ﬁnd N X n=1 f(n) wn+c1wn+c2 · · · wn+cr in closed form in terms of UN, VN, WN, XN, and YN. Proof: As proof, we give the following algorithm. Algorithm WReciprocalSum to evaluate certain reciprocal sums in closed form: INPUT: A rational sum meeting the conditions of Theorem 9. 8 OUTPUT: A closed form for the sum expressed in terms of the quantities UN, VN, WN, XN, and YN. STEP 1: [Reduce the order.] If the denominator consists of three or more terms of the form wx, choose any three of them, say wn+a, wn+b, and wn+c, and make the following substitution: 1 wn+awn+bwn+c = AQ−n wn+a + BQ−n wn+b + CQ−n wn+c (26) where A, B, and C are given by formula (21). Expand out and make the transformation N X n=1 [f(n) + g(n)] = N X n=1 f(n) + N X n=1 g(n) (27) summing each term on the right by this algorithm. STEP 2: [Normalize subscripts in denominator.] If the denominator is of the form wn+a or of the form wn+awn+b with a 6= 0 and a < b, then apply one of the following transformations: N X n=1 f(n) wn+awn+b = a+N X n=a+1 f(n −a) wnwn+b−a (28) N X n=1 f(n) wn+a = a+N X n=a+1 f(n −a) wn . (29) STEP 3: [Normalize index of summation.] If the index of summation does not start at 1, then add or subtract a ﬁnite number of terms to make the index start at 1. Speciﬁcally, if n0 is a constant and n0 6= 1, then apply the transformation N X n=n0 f(n) = 8 > > > > < > > > > : N X n=1 f(n) − n0−1 X n=1 f(n), if n0 > 1, N X n=1 f(n) + 0 X n=n0 f(n), if n0 0. (30) STEP 4: [Break up sums.] Expand out the numerator. If the numerator consists of a sum of terms, then sum each term individually. That is, apply the transformation N X n=1 f(n) + g(n) d = N X n=1 f(n) d + N X n=1 g(n) d . (31) 9 In each fraction, cancel any factors of the form wn+c common to the numerator and denominator. Then evaluate each sum recursively using this algorithm. Return the sum of the results so obtained. STEP 5: [Normalize numerator.] If the denominator is of the form wnwn+a with a > 0 and if the numerator contains a subexpression of the form wn+c where c 6= 0 and c 6= a, then express this subexpression in terms of wn and wn+a by using the Representation Theorem. Speciﬁcally, make the substitution wn+c = Qauc−a ua wn + uc−a ua wn+a. (32) If the numerator contains a subexpression of the form Qn, then express this subexpression in terms of wn and wn+a by using the formula Qn = vawnwn+a −Qaw2 n −w2 n+a eu2 a . (33) Go back to step 4. STEP 6: [Normalize numerator (continued).] If the denominator is of the form wn, and if the numerator contains a subexpression of the form wn+c where c 6= 0 and c 6= 1, then express this subexpression in terms of wn and wn+1 by using the Representation Theorem. Speciﬁcally, make the substitution wn+c = ucwn+1 −Quc−1wn. (34) Go back to step 4. STEP 7: [Evaluate polynomial sums.] If the summand is a w-polynomial in the variable n, evaluate the sum by using algorithm LucasSum from . Exit. STEP 8: [Reduce numerator.] If the denominator is of the form wn and if the numerator contains a subexpression of the form wr n+1 with r > 1, then write wr n+1 as wr−2 n+1w2 n+1 and reduce the exponent by 1 by applying the substitution w2 n+1 = Pwnwn+1 −Qw2 n −eQn. (35) Expand the numerator. If Q = −1, replace any terms of the form Qrn+d by Qd(Qr)n. Repeat this step as long as possible, then go back to step 4. STEP 9: [Pull out constants.] Replace any expressions of the form Qn+b where b is a constant by QbQn. If the numerator is of the form c, cQn, cQnwx, or cwr x, where c is a constant (c 6= 0 and c 6= 1), then apply the transformation N X n=1 cf(n) = c N X n=1 f(n). (36) 10 STEP 10:[Handle sums of order 2.] If the denominator is of the form wnwn+a, then evaluate the sum by using one of the following formulas: N X n=1 Qn wnwn+a = uN ua a X n=1 Qn wnwn+N ; (37) N X n=1 1 wnwn+a = 1 ua a−1 X i=0 Qi(WN+i −Wi). (38) STEP 11:[Handle basic sums.] If the summand is of one of the following forms, make the substitution shown. N X n=1 wn+1 wn = UN N X n=1 Qnwn+1 wn = VN N X n=1 1 wnwn+1 = WN N X n=1 1 wn = XN N X n=1 Qn wn = YN. (39) Proof of Correctness. Step 1 reduces the order to 1 or 2. This step introduces terms of the form Q−n. If Q = ±1, then Q−n = Qn. Thus, the numerator will remain a w-polynomial if Q = ±1. Step 2 guarantees that if there is a denominator, then its ﬁrst factor will be wn. Step 3 ensures that the index of summation begins with 1. The upper limit can be any expression, since N need not be just a variable, but may be any expression. Step 4 guarantees that there will be no sums (or di↵erences) in the numerator. Step 5 is justiﬁed by the Representation Theorem. Formula (33) comes from . At the end of step 5, there will be no terms of the form wx in the numerator of any reciprocal sum of order 2. Step 6 is justiﬁed by the Representation Theorem. After step 4, the numerator consists only of a product of terms. Steps 5 and 6 ensure that these terms only involve w’s that cancel with w’s in the denominator or are of the form wn+1. Thus, by the time we get to step 7, the only w’s left in the numerator are those of the form wn+1. Of course, if the denominator went away, then we are left with a w-polynomial and it is easily summed in step 7. 11 Step 8 reduces the degree of the variable wn+1 to 0 or 1. Step 9 removed any constants from the numerator. Step 10 is justiﬁed by Theorems 5 and 6. None of the previous steps introduce terms of the form Qrn in the numerator (for r > 1). Thus steps 10 and 11 handle all the remaining cases. Note. It should be noted that algorithm WReciprocalSum also works in the cases where r < 3 and deg f(n) < 2 or for any r if f(n) = Qkn where k = br/2c. In step 1, if f(n) = Qkn, the Q−n term introduced changes Qkn into Q(k−1)n and the order of the sum decrements by 1 until it reaches 1 or 2. The degree of Qn will increase only if the degree of f(n) was larger than 1, so if deg f(n) < 2, no terms of the form Qcn are introduced with c > 1. This gives us the following two theorems. Theorem 11 (The Simpliﬁcation Theorem for Q-reciprocal sums). Let r be a positive integer and let k = br/2c. Let cj, j = 1, 2, . . . , r be distinct integers. Then we can ﬁnd N X n=1 Qkn wn+c1wn+c2 · · · wn+cr in closed form in terms of UN, VN, WN, XN, and YN. Theorem 12 (The Simpliﬁcation Theorem for Low-Order Reciprocal Sums). Let f(n) be any w-polynomial in the variable n with deg f(n) < 2. Let a and b be distinct integers. Then we can ﬁnd N X n=1 f(n) wn+a and N X n=1 f(n) wn+awn+b in closed form in terms of UN, VN, WN, XN, and YN. 7. Some General Formulas. We have given an algorithm for evaluating certain reciprocal sums. However, in some special cases, simple explicit formulas can be given. We can take a formula, such as that given by Theorem 4, which involves expressions of the form wn+a and turn it into a valid formula involving expressions of the form wk(n+a) where k is a ﬁxed positive integer. We do this by applying the Dilation Theorem (see ) which says we can transform an identity into another identity by replacing all occurrences of wx by wkx provided that we also replace ux by ukx/uk, vx by vkx, Q by Qk, P by vk, and e by euk. Applying the Dilation Theorem to Theorem 4 gives us the following theorem: 12 Theorem 13. If a > 0, k > 0, uk 6= 0, and uka 6= 0, then N X n=1 Qkn wknwk(n+a) = Qk eukuka " a X n=1 wk(n−1) wkn − a X n=1 wk(N+n−1) wk(N+n) # = 1 eukuka " a X n=1 wk(N+n+1) wk(N+n) − a X n=1 wk(n+1) wkn # . (40) The last equality comes from the identity wn+1 wn = Pwn −Qwn−1 wn = P −Qwn−1 wn (41) which when dilated by k gives wk(n+1) wkn = vk −Qk wk(n−1) wkn . (42) Corollary (letting a = 1). If k > 0 and uk 6= 0, then N X n=1 Qkn wknwk(n+1) = Qk eu2 k w0 wk − wkN wk(N+1) % = 1 eu2 k wk(N+2) wk(N+1) −w2k wk % . (43) This agrees with the result given by Lucas in 1878 for the sequences huni and hvni. Furthermore, when k = 1, we get the result found by Kappus which generalized the result of Hillman in . When wn = Fn, this reduces to the results found by Swamy in . When wn is either the Pell polynomials or the Pell-Lucas polynomials, formula (43) is equivalent to results found by Mahon and Horadam in . In a similar manner, applying the Dilation Theorem to Theorem 5 yields Theorem 14: Theorem 14. If a > 0 and k > 0, then uka N X n=1 Qkn wknwk(n+a) = ukN a X n=1 Qkn wknwk(n+N) . (44) Theorem 15. If a > 0, b > 0, k > 0, and uka 6= 0, then N X n=1 Qkn wkn+bwk(n+a)+b = ukN uka a X n=1 Qkn wkn+bwk(n+N)+b . (45) 13 Proof: Apply the Translation Theorem (see ) to Theorem 6 to convert the sequence hwni into the sequence hwn+bi. If P = x and Q = −1, Theorems 5, 6, and 14 give results about partial sums of Fibonacci polynomials that were found by Bergum and Hoggatt . Corollary (letting a = 1). If b > 0, k > 0, and uk 6= 0, then N X n=1 Qk(n−1) wkn+bwk(n+1)+b = ukN ukwk+bwk(N+1)+b . (46) This is equivalent (with b = a −k) to the results found by Popov for the sequences huni and hvni. Theorem 16. If a < b, k > 0, and uk(b−a) 6= 0, then N X n=1 Qkn wk(n+a)wk(n+b) = ukN uk(b−a) b−a X n=1 Qkn wk(n+a)wk(n+N+a) . (47) Proof: Apply the Translation Theorem (see ) to change the sequence hwmi in Theorem 6 into the sequence hwm+kai. Then let c = b −a. Theorem 17. If k > 0, c > 0, and ukc 6= 0, then N X n=1 1 wknwk(n+c) = uk ukc c−1 X i=0 Qki(Wk,N+i −Wk,i). (48) where Wk,N = N X n=1 1 wknwk(n+1) . Proof: Apply the Dilation Theorem to Theorem 5. Applying the Translation Theorem to Theorem 15 gives us the following result. Theorem 18. If a < b and uk(b−a) 6= 0, then N X n=1 1 wk(n+a)wk(n+b) = uk uk(b−a) b−a−1 X i=0 Qki(Wk,a+N+i −Wk,a+i). (49) 14 8. Special Cases. Although unit reciprocal sums of order 2 cannot in general be evaluated in closed form (without involving terms of the form UN, VN, WN, XN, or YN), a closed form can be found for some important special cases (such as when Q = ±1). Theorem 19. If Q = −1, then N X n=1 1 wn+awn+a+2 = 1 P  1 wa+1wa+2 − 1 wN+a+1wN+a+2 % . (50) Proof: Since Q = −1, we have wm+2 = Pwm+1 + wm. Thus, P wn+awn+a+2 = Pwn+a+1 wn+awn+a+1wn+a+2 = wn+a+2 −wn+a wn+awn+a+1wn+a+2 = 1 wn+awn+a+1 − 1 wn+a+1wn+a+2 . Summing from 1 to N gives the desired result since the right-hand side telescopes. Lemma. If Q = 1, then ck = rk 1wk(n+1) −wkn rk(n+1) 1 (51) is independent of n. In particular, ck = (w1 −w0)r2uk. Proof: Since Q = 1, we have r1r2 = 1. The Binet form for wn is known to be wn = Arn 1 + Brn 2 where A = w1−w0r2 r1−r2 and B = w0r1−w1 r1−r2 . Then ckrk(n+1) 1 = rk 1wk(n+1) −wkn = rk 1 h Ark(n+1) 1 + Brk(n+1) 2 i − ⇥ Arkn 1 + Brkn 2 ⇤ = Arkn+2k 1 −Arkn 1 = Arkn 1 [r2k 1 −1] = Arkn 1 [r2k 1 −(r1r2)k] = Ark(n+1) 1 [rk 1 −rk 2] = Ark(n+1) 1 (r1 −r2)uk. Therefore ck = A(r1 −r2)uk = (w1 −w0r2)uk. 15 Theorem 20. If Q = 1, k 6= 0, and uk 6= 0, then N X n=1 1 wknwk(n+1) = 1 (w1 −w0r2)rk 1uk 1 wk − 1 rkN 1 wk(N+1) % . (52) Proof: Using the Lemma, we have 1 rkn 1 wkn − 1 rk(n+1) 1 wk(n+1) = rk 1wk(n+1) −wkn rk(n+1) 1 wknwk(n+1) = (w1 −w0r2)uk wknwk(n+1) . Summing as n goes from 1 to N, we ﬁnd that the left side telescopes and we reach the stated result. This theorem generalizes the results found by Melham and Shannon . The idea for the proof comes from that paper. Alternatively, we could let Q = 1 in formula (43). Corollary (letting k = 1). If Q = 1, then N X n=1 1 wnwn+1 = 1 (w1 −w0r2)r1 1 w1 − 1 rN 1 wN+1 % . (53) 9. Open Problems. Query 1. Is there a simple closed form for any of the quantities UN, VN, WN, XN, or YN? Query 2. Is there a simple algebraic relationship between any of the quantities UN, VN, WN, XN, and YN? Query 3. Can N X n=1 1 w2nw2n+1 be expressed in terms of UN, VN, WN, XN, and YN? Query 4. Can N X n=1 1 w2 n be expressed in terms of UN, VN, WN, XN, and YN? Query 5. Can N X n=1 1 wnwn+1wn+2 be expressed in terms of UN, VN, WN, XN, and YN? 16 Query 6. Can N X n=1 1 wn+awn+bwn+c be expressed in terms of UN, VN, WN, XN, YN, a, b, and c? References R. Andr´ e-Jeannin, “Summation of Reciprocals in Certain Second-Order Recurring Se-quences”, The Fibonacci Quarterly 35.1(1997)68–74. G. E. Bergum and V. E. Hoggatt, Jr., “Inﬁnite Series with Fibonacci and Lucas Poly-nomials”, The Fibonacci Quarterly 17.2(1979)147–151. Brother Alfred Brousseau, “Fibonacci-Lucas Inﬁnite Series – Research Topic”, The Fibonacci Quarterly 7.2(1969)211–217. I. J. Good, “A Symmetry Property of Alternating Sums of Products of Reciprocals”, The Fibonacci Quarterly 32.3(1994)284–287. A. P. Hillman, “Solution to Problem B-697: A Sum of Quotients”, The Fibonacci Quarterly 30.3(1992)280–281. Hans Kappus, “Solution to Problem B-747: Great Sums from Partial Sums”, The Fi-bonacci Quarterly 33.1(1995)87–88. Edouard Lucas, “Th´ eorie des Fonctions Num´ eriques Simplement P´ eriodiques”, Amer-ican Journal of Mathematics 1(1878)184–240, 289–321. Br. J. M. Mahon and A. F. Horadam, “Inﬁnite Series Summation Involving Recipro-cals of Pell Polynomials” in Fibonacci Numbers and Their Applications, volume 1 (edited by G. E. Bergum, A. F. Horodam, and Andreas N. Philippou), D. Reidel Publishing Company, Dordrecht, The Netherlands: 1986. pp. 163–180. R. S. Melham and A. G. Shannon, “On Reciprocal Sums of Chebyshev Related Se-quences”, The Fibonacci Quarterly 33.3(1995)194–202. Blagoj S. Popov, “Summation of Reciprocal Series of Numerical Functions of Second Order”, The Fibonacci Quarterly 24.1(1986)17–21. Stanley Rabinowitz, “Algorithmic Manipulation of Fibonacci Identities” in Applica-tions of Fibonacci Numbers, volume 6 (edited by G. E. Bergum, A. F. Horodam, and Andreas N. Philippou), Kluwer Academic Publishers, Dordrecht, The Netherlands: 1996. pp. 389–408. Stanley Rabinowitz, “Algorithmic Manipulation of Second-Order Linear Recurrences”, The Fibonacci Quarterly 37.2(1999)162–177. Stanley Rabinowitz, “Algorithmic Summation of Reciprocals of Products of Fibonacci Numbers”, The Fibonacci Quarterly 37.2(1999)122–127. M. N. S. Swamy, “More Fibonacci Identities”, The Fibonacci Quarterly 4.4(1966)369– 372. AMS Classiﬁcation Numbers: 11B37, 11Y16, 11B39. 17 July 7, 2018 Addendum Equation (12) is incorrect. The correct result is the following. If a > 1, then N X n=1 Qn wn+a = 1 Qa [YN+a −Ya] . (120) Proof: Letting k = n + a gives N X n=1 Qn wn+a = a+N X k=a+1 Qk−a wk = Q−a a+N X k=a+1 Qk wk = Q−a "a+N X k=1 Qk wk − a X k=1 Qk wk # = 1 Qa [YN+a −Ya] .
1270	https://www.aafp.org/pubs/afp/issues/2003/0315/p1349a.html	Scheduled maintenance is planned for September 26–29. You may experience brief interruptions during this time. RICHARD SADOVSKY, M.D. Am Fam Physician. 2003;67(6):1349-1353 Accurate identification of periorbital and orbital cellulitis is essential for appropriate treatment. Givner reviewed the characteristics (see accompanying table) and treatment of these two clinical entities. Sources of periorbital infection can be trauma, including insect bites, or bacteremia. The resultant cellulitis is preseptal, or anterior to the orbital septum, involving a fibrous layer beginning at the periosteum of the skull and extending to the eyelids. Periorbital cellulitis does not progress to orbital cellulitis because of this protective fibrous barrier. Orbital cellulitis is postseptal and involves the orbit itself. The most common cause is extension of infection from sinusitis, although penetrating trauma causes some cases. The orbit is surrounded by the sinuses. The ethmoid sinus is the most common source of orbital infection because a very thin septum separates it from the orbit. Generally, orbital cellulitis occurs in older children because sinusitis is more common as children reach the preteen years. Orbital cellulitis can cause bacteremia. The clinical signs help to distinguish periorbital from orbital cellulitis. Periorbital infection causes erythema, induration, and tenderness of the periorbital tissues, with patients rarely showing signs of systemic illness. The latter signs are more likely if the infection results from bacteremia. Orbital cellulitis may have the same signs and symptoms in the periorbital tissue but also results in proptosis, edema of the conjunctiva, ophthalmoplegia, or decreased visual acuity. Because the eyelids usually are swollen shut, computed tomographic (CT) scanning with contrast infusion may be necessary to confirm the diagnosis. Nearby sinusitis generally is evident. Although it is associated with orbital cellulitis, periorbital edema can occur without infection, but this condition can be identified by a lack of induration or tenderness. The causative organism is related to the pathogenesis of infection. Post-traumatic periorbital cellulitis usually is caused by Staphylococcus aureus or Streptococcus pyogenes. Because Streptococcus pneumoniae is the most common cause of bacteremia in young children, it is also the most common cause of periorbital cellulitis. Haemophilus influenzae type b is becoming a rare cause because of the prevalence of H. influenzae vaccinations. Orbital cellulitis results from the pathogens causing sinus infection, often similar to those causing periorbital infection, but may be polymicrobial. Givner concludes with treatment recommendations. Uncomplicated post-traumatic periorbital cellulitis usually responds to oral antibiotics, such as cephalexin, dicloxacillin, or clindamycin, that treat gram-positive microbes. If there is evidence of bacteremia, the risk of meningitis may require lumbar puncture. Children should be hospitalized initially for parenteral treatment with ceftriaxone (if they have normal cerebrospinal fluid findings), and vancomycin and ceftriaxone (if they have abnormal spinal fluid findings). Response to antibiotics in children with periorbital cellulitis usually is rapid, and a 10-day course of treatment generally is sufficient. Orbital cellulitis requires careful examination of the eyes and sinuses. Ampicillin/ sulbactam is a good treatment choice; operative drainage is no longer being done. Patients with a large abscess or significant symptoms may require surgical drainage of the abscess and involved sinuses. Improvement should be noted in all patients within 24 to 36 hours. Those who do not improve should have repeat CT scanning and, possibly, surgery. Three weeks of antibiotic therapy, with at least the first week of therapy being administered parenterally, is required in the treatment of children with orbital cellulitis. Givner LB. Periorbital versus orbital cellulitis. Pediatr Infect Dis J. December 2002;21:1157-8. Continue Reading More in AFP Copyright © 2003 by the American Academy of Family Physicians. This content is owned by the AAFP. A person viewing it online may make one printout of the material and may use that printout only for his or her personal, non-commercial reference. This material may not otherwise be downloaded, copied, printed, stored, transmitted or reproduced in any medium, whether now known or later invented, except as authorized in writing by the AAFP. See permissions for copyright questions and/or permission requests. Copyright © 2025 American Academy of Family Physicians. All Rights Reserved.
1271	https://courses.lumenlearning.com/suny-microbiology/chapter/oxygen-requirements-for-microbial-growth/	Oxygen Requirements for Microbial Growth \| Microbiology Skip to main content Microbiology Microbial Growth Search for: Oxygen Requirements for Microbial Growth Learning Objectives Interpret visual data demonstrating minimum, optimum, and maximum oxygen or carbon dioxide requirements for growth Identify and describe different categories of microbes with requirements for growth with or without oxygen: obligate aerobe, obligate anaerobe, facultative anaerobe, aerotolerant anaerobe, microaerophile, and capnophile Give examples of microorganisms for each category of growth requirements Ask most people “What are the major requirements for life?” and the answers are likely to include water and oxygen. Few would argue about the need for water, but what about oxygen? Can there be life without oxygen? The answer is that molecular oxygen (O 2) is not always needed. The earliest signs of life are dated to a period when conditions on earth were highly reducing and free oxygen gas was essentially nonexistent. Only after cyanobacteria started releasing oxygen as a byproduct of photosynthesis and the capacity of iron in the oceans for taking up oxygen was exhausted did oxygen levels increase in the atmosphere. This event, often referred to as the Great Oxygenation Event or the Oxygen Revolution, caused a massive extinction. Most organisms could not survive the powerful oxidative properties of reactive oxygen species (ROS), highly unstable ions and molecules derived from partial reduction of oxygen that can damage virtually any macromolecule or structure with which they come in contact. Singlet oxygen (O 2•), superoxide (O−2)(O 2−), peroxides (H 2 O 2), hydroxyl radical (OH•), and hypochlorite ion (OCl−), the active ingredient of household bleach, are all examples of ROS. The organisms that were able to detoxify reactive oxygen species harnessed the high electronegativity of oxygen to produce free energy for their metabolism and thrived in the new environment. Oxygen Requirements of Microorganisms Many ecosystems are still free of molecular oxygen. Some are found in extreme locations, such as deep in the ocean or in earth’s crust; others are part of our everyday landscape, such as marshes, bogs, and sewers. Within the bodies of humans and other animals, regions with little or no oxygen provide an anaerobic environment for microorganisms (Figure 1). Figure 1. Anaerobic environments are still common on earth. They include environments like (a) a bog where undisturbed dense sediments are virtually devoid of oxygen, and (b) the rumen (the first compartment of a cow’s stomach), which provides an oxygen-free incubator for methanogens and other obligate anaerobic bacteria. (credit a: modification of work by National Park Service; credit b: modification of work by US Department of Agriculture) We can easily observe different requirements for molecular oxygen by growing bacteria in thioglycolate tube cultures. A test-tube culture starts with autoclaved thioglycolate medium containing a low percentage of agar to allow motile bacteria to move throughout the medium. Thioglycolate has strong reducing properties and autoclaving flushes out most of the oxygen. The tubes are inoculated with the bacterial cultures to be tested and incubated at an appropriate temperature. Over time, oxygen slowly diffuses throughout the thioglycolate tube culture from the top. Bacterial density increases in the area where oxygen concentration is best suited for the growth of that particular organism. The growth of bacteria with varying oxygen requirements in thioglycolate tubes is illustrated in Figure 2. In tube A, all the growth is seen at the top of the tube. The bacteria are obligate (strict) aerobes that cannot grow without an abundant supply of oxygen. Tube B looks like the opposite of tube A. Bacteria grow at the bottom of tube B. Those are obligate anaerobes, which are killed by oxygen. Tube C shows heavy growth at the top of the tube and growth throughout the tube, a typical result with facultative anaerobes. Facultative anaerobes are organisms that thrive in the presence of oxygen but also grow in its absence by relying on fermentation or anaerobic respiration, if there is a suitable electron acceptor other than oxygen and the organism is able to perform anaerobic respiration. The aerotolerant anaerobes in tube D are indifferent to the presence of oxygen. They do not use oxygen because they usually have a fermentative metabolism, but they are not harmed by the presence of oxygen as obligate anaerobes are. Tube E on the right shows a “Goldilocks” culture. The oxygen level has to be just right for growth, not too much and not too little. These microaerophiles are bacteria that require a minimum level of oxygen for growth, about 1%–10%, well below the 21% found in the atmosphere. Examples of obligate aerobes are Mycobacterium tuberculosis, the causative agent of tuberculosis and Micrococcus luteus, a gram-positive bacterium that colonizes the skin. Neisseria meningitidis, the causative agent of severe bacterial meningitis, and N. gonorrheae, the causative agent of sexually transmitted gonorrhea, are also obligate aerobes. Figure 2. Diagram of bacterial cell distribution in thioglycolate tubes. Many obligate anaerobes are found in the environment where anaerobic conditions exist, such as in deep sediments of soil, still waters, and at the bottom of the deep ocean where there is no photosynthetic life. Anaerobic conditions also exist naturally in the intestinal tract of animals. Obligate anaerobes, mainly Bacteroidetes, represent a large fraction of the microbes in the human gut. Transient anaerobic conditions exist when tissues are not supplied with blood circulation; they die and become an ideal breeding ground for obligate anaerobes. Another type of obligate anaerobe encountered in the human body is the gram-positive, rod-shaped Clostridium spp. Their ability to form endospores allows them to survive in the presence of oxygen. One of the major causes of health-acquired infections is C. difficile, known as C. diff. Prolonged use of antibiotics for other infections increases the probability of a patient developing a secondary C. difficile infection. Antibiotic treatment disrupts the balance of microorganisms in the intestine and allows the colonization of the gut by C. difficile, causing a significant inflammation of the colon. Other clostridia responsible for serious infections include C. tetani, the agent of tetanus, and C. perfringens, which causes gas gangrene. In both cases, the infection starts in necrotic tissue (dead tissue that is not supplied with oxygen by blood circulation). This is the reason that deep puncture wounds are associated with tetanus. When tissue death is accompanied by lack of circulation, gangrene is always a danger. The study of obligate anaerobes requires special equipment. Obligate anaerobic bacteria must be grown under conditions devoid of oxygen. The most common approach is culture in an anaerobic jar (Figure 3). Anaerobic jars include chemical packs that remove oxygen and release carbon dioxide (CO 2). An anaerobic chamber is an enclosed box from which all oxygen is removed. Gloves sealed to openings in the box allow handling of the cultures without exposing the culture to air (Figure 3). Figure 3. (a) An anaerobic jar is pictured that is holding nine Petri plates supporting cultures. (b) Openings in the side of an anaerobic box are sealed by glove-like sleeves that allow for the handling of cultures inside the box. (credit a: modification of work by Centers for Disease Control and Prevention; credit b: modification of work by NIST) Staphylococci and Enterobacteriaceae are examples of facultative anaerobes. Staphylococci are found on the skin and upper respiratory tract. Enterobacteriaceae are found primarily in the gut and upper respiratory tract but can sometimes spread to the urinary tract, where they are capable of causing infections. It is not unusual to see mixed bacterial infections in which the facultative anaerobes use up the oxygen, creating an environment for the obligate anaerobes to flourish. Examples of aerotolerant anaerobes include lactobacilli and streptococci, both found in the oral microbiota. Campylobacter jejuni, which causes gastrointestinal infections, is an example of a microaerophile and is grown under low-oxygen conditions. The optimum oxygen concentration, as the name implies, is the ideal concentration of oxygen for a particular microorganism. The lowest concentration of oxygen that allows growth is called the minimum permissive oxygen concentration. The highest tolerated concentration of oxygen is the maximum permissive oxygen concentration. The organism will not grow outside the range of oxygen levels found between the minimum and maximum permissive oxygen concentrations. Think about It Would you expect the oldest bacterial lineages to be aerobic or anaerobic? Which bacteria grow at the top of a thioglycolate tube, and which grow at the bottom of the tube? An Unwelcome Anaerobe Figure 4. This clinical photo depicts ulcers on the foot of a diabetic patient. Dead tissue accumulating in ulcers can provide an ideal growth environment for the anaerobe C. perfringens, a causative agent of gas gangrene. (credit: Shigeo Kono, Reiko Nakagawachi, Jun Arata, Benjamin A Lipsky) Charles is a retired bus driver who developed type 2 diabetes over 10 years ago. Since his retirement, his lifestyle has become very sedentary and he has put on a substantial amount of weight. Although he has felt tingling and numbness in his left foot for a while, he has not been worried because he thought his foot was simply “falling asleep.” Recently, a scratch on his foot does not seem to be healing and is becoming increasingly ugly. Because the sore did not bother him much, Charles figured it could not be serious until his daughter noticed a purplish discoloration spreading on the skin and oozing (Figure 4). When he was finally seen by his physician, Charles was rushed to the operating room. His open sore, or ulcer, is the result of a diabetic foot. The concern here is that gas gangrene may have taken hold in the dead tissue. The most likely agent of gas gangrene is Clostridium perfringens, an endospore-forming, gram-positive bacterium. It is an obligate anaerobe that grows in tissue devoid of oxygen. Since dead tissue is no longer supplied with oxygen by the circulatory system, the dead tissue provides pockets of ideal environment for the growth of C. perfringens. A surgeon examines the ulcer and radiographs of Charles’s foot and determines that the bone is not yet infected. The wound will have to be surgically debrided (debridement refers to the removal of dead and infected tissue) and a sample sent for microbiological lab analysis, but Charles will not have to have his foot amputated. Many diabetic patients are not so lucky. In 2008, nearly 70,000 diabetic patients in the United States lost a foot or limb to amputation, according to statistics from the Centers for Disease Control and Prevention. Which growth conditions would you recommend for the detection of C. perfringens? Detoxification of Reactive Oxygen Species Aerobic respiration constantly generates reactive oxygen species (ROS), byproducts that must be detoxified. Even organisms that do not use aerobic respiration need some way to break down some of the ROS that may form from atmospheric oxygen. Three main enzymes break down those toxic byproducts: superoxide dismutase, peroxidase, and catalase. Each one catalyzes a different reaction. Reactions of type seen in Reaction 1 are catalyzed by peroxidases. (1)X−(2 H+)+H 2 O 2→oxidized-X+2 H 2 O(1)X−(2 H+)+H 2 O 2→oxidized-X+2 H 2 O In these reactions, an electron donor (reduced compound; e.g., reduced nicotinamide adenine dinucleotide [NADH]) oxidizes hydrogen peroxide, or other peroxides, to water. The enzymes play an important role by limiting the damage caused by peroxidation of membrane lipids. Reaction 2 is mediated by the enzyme superoxide dismutase (SOD) and breaks down the powerful superoxide anions generated by aerobic metabolism: (2)2 O 2-+2 H+→H 2 O 2+O 2(2)2 O 2-+2 H+→H 2 O 2+O 2 The enzyme catalase converts hydrogen peroxide to water and oxygen as shown in Reaction 3. (3)2 H 2 O 2→2 H 2 O+O 2(3)2 H 2 O 2→2 H 2 O+O 2 Figure 5. The catalase test detects the presence of the enzyme catalase by noting whether bubbles are released when hydrogen peroxide is added to a culture sample. Compare the positive result (right) with the negative result (left). (credit: Centers for Disease Control and Prevention) Obligate anaerobes usually lack all three enzymes. Aerotolerant anaerobes do have SOD but no catalase. Reaction 3, shown occurring in Figure 5, is the basis of a useful and rapid test to distinguish streptococci, which are aerotolerant and do not possess catalase, from staphylococci, which are facultative anaerobes. A sample of culture rapidly mixed in a drop of 3% hydrogen peroxide will release bubbles if the culture is catalase positive. Bacteria that grow best in a higher concentration of CO 2 and a lower concentration of oxygen than present in the atmosphere are called capnophiles. One common approach to grow capnophiles is to use a candle jar. A candle jar consists of a jar with a tight-fitting lid that can accommodate the cultures and a candle. After the cultures are added to the jar, the candle is lit and the lid closed. As the candle burns, it consumes most of the oxygen present and releases CO 2. Think about It What substance is added to a sample to detect catalase? What is the function of the candle in a candle jar? Clinical Focus: Nataliya, Part 2 This example continues Nataliya’s story that started in How Microbes Grow. The health-care provider who saw Nataliya was concerned primarily because of her pregnancy. Her condition enhances the risk for infections and makes her more vulnerable to those infections. The immune system is downregulated during pregnancy, and pathogens that cross the placenta can be very dangerous for the fetus. A note on the provider’s order to the microbiology lab mentions a suspicion of infection by Listeria monocytogenes, based on the signs and symptoms exhibited by the patient. Nataliya’s blood samples are streaked directly on sheep blood agar, a medium containing tryptic soy agar enriched with 5% sheep blood. (Blood is considered sterile; therefore, competing microorganisms are not expected in the medium.) The inoculated plates are incubated at 37 °C for 24 to 48 hours. Small grayish colonies surrounded by a clear zone emerge. Such colonies are typical of Listeria and other pathogens such as streptococci; the clear zone surrounding the colonies indicates complete lysis of blood in the medium, referred to as beta-hemolysis (Figure 6). When tested for the presence of catalase, the colonies give a positive response, eliminating Streptococcus as a possible cause. Furthermore, a Gram stain shows short gram-positive bacilli. Cells from a broth culture grown at room temperature displayed the tumbling motility characteristic of Listeria (Figure 6). All of these clues lead the lab to positively confirm the presence of Listeria in Nataliya’s blood samples. Figure 6. (a) A sample blood agar test showing beta-hemolysis. (b) A sample motility test showing both positive and negative results. (credit a: modification of work by Centers for Disease Control and Prevention; credit b: modification of work by “VeeDunn”/Flickr) How serious is Nataliya’s condition and what is the appropriate treatment? We’ll return to Nataliya’s example in later pages. Key Concepts and Summary Aerobic and anaerobic environments can be found in diverse niches throughout nature, including different sites within and on the human body. Microorganisms vary in their requirements for molecular oxygen. Obligate aerobes depend on aerobic respiration and use oxygen as a terminal electron acceptor. They cannot grow without oxygen. Obligate anaerobes cannot grow in the presence of oxygen. They depend on fermentation and anaerobic respiration using a final electron acceptor other than oxygen. Facultative anaerobes show better growth in the presence of oxygen but will also grow without it. Although aerotolerant anaerobes do not perform aerobic respiration, they can grow in the presence of oxygen. Most aerotolerant anaerobes test negative for the enzyme catalase. Microaerophiles need oxygen to grow, albeit at a lower concentration than 21% oxygen in air. Optimum oxygen concentration for an organism is the oxygen level that promotes the fastest growth rate. The minimum permissive oxygen concentration and the maximum permissive oxygen concentration are, respectively, the lowest and the highest oxygen levels that the organism will tolerate. Peroxidase, superoxide dismutase, and catalase are the main enzymes involved in the detoxification of the reactive oxygen species. Superoxide dismutase is usually present in a cell that can tolerate oxygen. All three enzymes are usually detectable in cells that perform aerobic respiration and produce more ROS. A capnophile is an organism that requires a higher than atmospheric concentration of CO 2 to grow. Multiple Choice An inoculated thioglycolate medium culture tube shows dense growth at the surface and turbidity throughout the rest of the tube. What is your conclusion? The organisms die in the presence of oxygen The organisms are facultative anaerobes. The organisms should be grown in an anaerobic chamber. The organisms are obligate aerobes. Show Answer Answer b. The organisms are facultative anaerobes. An inoculated thioglycolate medium culture tube is clear throughout the tube except for dense growth at the bottom of the tube. What is your conclusion? The organisms are obligate anaerobes. The organisms are facultative anaerobes. The organisms are aerotolerant. The organisms are obligate aerobes. Show Answer Answer a. The organisms are obligate anaerobes. Pseudomonas aeruginosa is a common pathogen that infects the airways of patients with cystic fibrosis. It does not grow in the absence of oxygen. The bacterium is probably which of the following? an aerotolerant anaerobe an obligate aerobe an obligate anaerobe a facultative anaerobe Show Answer Answer b. The bacterium is probably an obligate aerobe. Streptococcus mutans is a major cause of cavities. It resides in the gum pockets, does not have catalase activity, and can be grown outside of an anaerobic chamber. The bacterium is probably which of the following? a facultative anaerobe an obligate aerobe an obligate anaerobe an aerotolerant anaerobe Show Answer Answer d. The bacterium is probably an aerotolerant anaerobe. Why do the instructions for the growth of Neisseria gonorrheae recommend a CO 2-enriched atmosphere? It uses CO 2 as a final electron acceptor in respiration. It is an obligate anaerobe. It is a capnophile. It fixes CO 2 through photosynthesis. Show Answer Answer c. It is a capnophile. Matching Four tubes are illustrated with cultures grown in a medium that slows oxygen diffusion. Match the culture tube with the correct type of bacteria from the following list: facultative anaerobe, obligate anaerobe, microaerophile, aerotolerant anaerobe, obligate aerobe. Show Answer Tube a is an obligate anaerobe. Tube b is an obligate aerobe. Tube c is a microaerophile. Tube d is a facultative anaerobe Think about It Why are some obligate anaerobes able to grow in tissues (e.g., gum pockets) that are not completely free of oxygen? Why should Haemophilus influenzae be grown in a candle jar? In terms of oxygen requirements, what type of organism would most likely be responsible for a foodborne illness associated with canned foods? A microbiology instructor prepares cultures for a gram-staining practical laboratory by inoculating growth medium with a gram-positive coccus (nonmotile) and a gram-negative rod (motile). The goal is to demonstrate staining of a mixed culture. The flask is incubated at 35 °C for 24 hours without aeration. A sample is stained and reveals only gram-negative rods. Both cultures are known facultative anaerobes. Give a likely reason for success of the gram-negative rod. Assume that the cultures have comparable intrinsic growth rates. Candela Citations CC licensed content, Shared previously OpenStax Microbiology. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at Centers for Disease Control and Prevention. "Living With Diabetes: Keep Your Feet Healthy." ↵ Licenses and Attributions CC licensed content, Shared previously OpenStax Microbiology. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at PreviousNext Privacy Policy
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Generating m distinct random numbers in the range [0..n-1] Ask Question Asked 14 years, 1 month ago Modified6 years, 9 months ago Viewed 36k times This question shows research effort; it is useful and clear 33 Save this question. Show activity on this post. I have two methods of generating m distinct random numbers in the range [0..n-1] Method 1: //C++-ish pseudocode int result[m]; for(i = 0; i < m; ++i) { int r; do { r = rand()%n; }while(r is found in result array at indices from 0 to i) result[i] = r; } Method 2: //C++-ish pseudocode int arr[n]; for(int i = 0; i < n; ++i) arr[i] = i; random_shuffle(arr, arr+n); result = first m elements in arr; The first method is more efficient when n is much larger than m, whereas the second is more efficient otherwise. But "much larger" isn't that strict a notion, is it? :) Question: What formula of n and m should I use to determine whether method1 or method2 will be more efficient? (in terms of mathematical expectation of the running time) c++ algorithm random performance Share Improve this question Follow Follow this question to receive notifications edited Aug 4, 2011 at 19:51 Veger 38k 11 11 gold badges 110 110 silver badges 118 118 bronze badges asked Aug 4, 2011 at 19:42 Armen TsirunyanArmen Tsirunyan 134k 61 61 gold badges 335 335 silver badges 442 442 bronze badges 3 If m is really small, does efficiency matter all that much? Optimize for the case that's more likely to cause problems.Mark Ransom –Mark Ransom 2011-08-04 19:48:11 +00:00 Commented Aug 4, 2011 at 19:48 @Mark: I will dynamically get n and m. I will have to determine runtime which method to use Armen Tsirunyan –Armen Tsirunyan 2011-08-04 19:49:58 +00:00 Commented Aug 4, 2011 at 19:49 Have you tested the two methods with a couple of different parameters? Just to get a rough feel of how long they take.biziclop –biziclop 2011-08-04 20:00:30 +00:00 Commented Aug 4, 2011 at 20:00 Add a comment\| 11 Answers 11 Sorted by: Reset to default This answer is useful 17 Save this answer. Show activity on this post. Pure mathematics: Let's calculate the quantity of rand() function calls in both cases and compare the results: Case 1: let's see the mathematical expectation of calls on step i = k, when you already have k numbers chosen. The probability to get a number with one rand() call is equal to p = (n-k)/n. We need to know the mathematical expectation of such calls quantity which leads to obtaining a number we don't have yet. The probability to get it using 1 call is p. Using 2 calls - q p, where q = 1 - p. In general case, the probability to get it exactly after n calls is (q^(n-1))p. Thus, the mathematical expectation is Sum[ n q^(n-1) p ], n = 1 --> INF. This sum is equal to 1/p (proved by wolfram alpha). So, on the step i = k you will perform 1/p = n/(n-k) calls of the rand() function. Now let's sum it overall: Sum[ n/(n - k) ], k = 0 --> m - 1 = n T - the number of rand calls in method 1. Here T = Sum[ 1/(n - k) ], k = 0 --> m - 1 Case 2: Here rand() is called inside random_shuffle``n - 1 times (in most implementations). Now, to choose the method, we have to compare these two values: n T ? n - 1. So, to choose the appropriate method, calculate T as described above. If T < (n - 1)/n it's better to use the first method. Use the second method otherwise. Share Improve this answer Follow Follow this answer to receive notifications answered Aug 5, 2011 at 9:11 Grigor GevorgyanGrigor Gevorgyan 6,893 4 4 gold badges 38 38 silver badges 64 64 bronze badges 2 Comments Add a comment abksrv abksrvOver a year ago Would be great if you could please expand - " Using 2 calls - q p, where q = 1 - p." I don't really understand it. 2017-02-14T11:14:17.4Z+00:00 0 Reply Copy link abksrv abksrvOver a year ago I would also consider the search effort(which is 0 for case 2) in each iteration of while loop, not only the times of random calls 2017-02-14T11:36:19.397Z+00:00 1 Reply Copy link This answer is useful 11 Save this answer. Show activity on this post. Check the Wikipedia description of the original Fisher-Yates algorithm. It advocates using essentially your method 1 for up to n/2, and your method 2 for the remainder. Share Improve this answer Follow Follow this answer to receive notifications answered Aug 4, 2011 at 19:53 Mark RansomMark Ransom 310k 44 44 gold badges 422 422 silver badges 659 659 bronze badges 3 Comments Add a comment Nobody moving away from SE Nobody moving away from SEOver a year ago I think it is a tweaking issue. He will just have to measure it on the target system. 2011-08-04T20:00:21.527Z+00:00 0 Reply Copy link biziclop biziclopOver a year ago This sounds about right, certainly from a memory efficiency perspective. Using method 2 can consume an awful lot of memory unnecessarily with smaller m values. 2011-08-04T20:02:31.807Z+00:00 0 Reply Copy link Markus MarkusOver a year ago At visualstudiomagazine.com/articles/2013/07/01/… you can find a good explained c# implementation of the Fisher method. The post also shows also an alternative Approach with The Reservoir Method and the differences between them. 2016-10-26T07:22:35.007Z+00:00 0 Reply Copy link Add a comment This answer is useful 7 Save this answer. Show activity on this post. Personally, I would use Method 1, and then if M > N/2, choose N-M values, and then invert the array (return the numbers that were not picked). So for example, if N is 1000 and you want 950 of them, chose 50 values using Method 1, and then return the other 950. Edit: Though, if consistent performance is your goal, I would use a modified method 2, which doesn't do the full shuffle, but only shuffles the first M elements of your N length array. ``` int arr[n]; for(int i = 0; i < n; ++i) arr[i] = i; for (int i =0; i < m; ++i) { int j = rand(n-i); // Pick random number from 0 <= r < n-i. Pick favorite method // j == 0 means don't swap, otherwise swap with the element j away if (j != 0) { std::swap(arr[i], arr[i+j]); } } result = first m elements in arr; ``` Share Improve this answer Follow Follow this answer to receive notifications edited Aug 4, 2011 at 20:47 answered Aug 4, 2011 at 20:38 Dave SDave S 21.2k 3 3 gold badges 52 52 silver badges 69 69 bronze badges Comments Add a comment This answer is useful 6 Save this answer. Show activity on this post. Here's an algorithm that will work in O(n) memory and O(n) time (where n is the number of returned results, not the size of the set you're selecting from) for any result set. It's in Python for convenience because it uses a hashtable: def random_elements(num_elements, set_size): state = {} for i in range(num_elements): # Swap state[i] with a random element swap_with = random.randint(i, set_size - 1) state[i], state[swap_with] = state.get(swap_with, swap_with), state.get(i, i) return [state[i] for i in range(num_elements) # effectively state[:num_elements] if it were a list/array. This is just a partial fisher-yates shuffle, with the array being shuffled implemented as a sparse hashtable - any element that is not present is equal to its index. We shuffle the first num_elements indices, and return those values. In the case that set_size = 1, this is equivalent to picking a random number in the range, and in the case that num_elements = set_size, this is equivalent to a standard fisher-yates shuffle. It's trivial to observe that this is O(n) time, and because each iteration of the loop initializes at most two new indices in the hashtable, it's O(n) space, too. Share Improve this answer Follow Follow this answer to receive notifications edited Feb 25, 2016 at 17:27 answered Aug 8, 2011 at 5:34 Nick JohnsonNick Johnson 101k 17 17 gold badges 131 131 silver badges 198 198 bronze badges 4 Comments Add a comment astraujums astraujumsOver a year ago Hashtable/dictionary access is not constant time O(1) but rather logarithmic time O(log(n)) so the overall complexity will be O(n log(n)). 2013-11-19T22:07:52.73Z+00:00 0 Reply Copy link ghord ghordOver a year ago @astraujums It is constant amortized actually 2015-02-13T09:11:48.453Z+00:00 1 Reply Copy link cidermole cidermoleOver a year ago I believe it should be swap_with = random.randint(i, set_size-1) since randint() uses an inclusive range? @nick-johnson 2016-02-25T12:48:42.987Z+00:00 0 Reply Copy link Nick Johnson Nick JohnsonOver a year ago @cidermole Wow, you're right. Nice Python library gotcha there that I'd never noticed before. Fixed. 2016-02-25T17:27:17.613Z+00:00 0 Reply Copy link Add a comment This answer is useful 3 Save this answer. Show activity on this post. What about a third method? int result[m]; for(i = 0; i < m; ++i) { int r; r = rand()%(n-i); r += (number of items in result <= r) result[i] = r; } Edit it should be <=. and it would actually additional logic to avoid collisions. This is better, an example using the Modern Method from Fisher-Yates ``` //C++-ish pseudocode int arr[n]; for(int i = 0; i < n; ++i) arr[i] = i; for(i = 0; i < m; ++i) swap(arr, n-i, rand()%(n-i) ); result = last m elements in arr; ``` Share Improve this answer Follow Follow this answer to receive notifications edited Aug 4, 2011 at 21:29 answered Aug 4, 2011 at 20:03 Jacob EggersJacob Eggers 9,330 2 2 gold badges 27 27 silver badges 43 43 bronze badges 5 Comments Add a comment Nobody moving away from SE Nobody moving away from SEOver a year ago What do you mean by (number of items in result < r)? Should this mean that 1 is to be added if r is bigger than the number of items in result? 2011-08-04T20:06:32.23Z+00:00 0 Reply Copy link biziclop biziclopOver a year ago +1 Once again, we were missing the obvious. It'd consume a lot of space though because you'd have to keep your results in a sorted list too to perform the (number of items in result < r) check efficiently. 2011-08-04T20:07:47.303Z+00:00 0 Reply Copy link Jacob Eggers Jacob EggersOver a year ago @Nobody I mean +1 for every item in result <= r. This is to compensate for the fact that we are getting rand() to n-i 2011-08-04T20:14:03.353Z+00:00 0 Reply Copy link Armen Tsirunyan Armen TsirunyanOver a year ago @Jacob: I mean, even if we assume that rand() returns a uniformly distributed number, your algorithm will not generate uniformly distributes sequences 2011-08-04T20:23:40.043Z+00:00 0 Reply Copy link Jacob Eggers Jacob EggersOver a year ago @Armen Why not? Isn't it essentially saying pick a random integer in [0,n-1], then pick a random one from the n-1 remaining...finally pick one from the n-m+1 remaining. It's the start of the Fisher-Yates shuffle, but stopping at n-m instead of 2. 2011-08-04T20:30:44.007Z+00:00 0 Reply Copy link Add a comment This answer is useful 2 Save this answer. Show activity on this post. Talking about mathematical expectation, it's pretty useless but I will post it anyway :D Shuffle is simple O(m). Now the other algorithm is a bit more complex. The number of steps needed to generate the next number is the expected value of the number of trials, and the probability of the trial length is a geomtric distribution. So... p=1 E[X1]=1 = 1 = 1 p=1-1/n E[x2]=1/(1-1/n) = 1 + 1/(n-1) = 1 + 1/(n-1) p=1-2/n E[x3]=1/(1-1/n) = 1 + 2/(n-2) = 1 + 1/(n-2) + 1/(n-2) p=1-3/n E[X4]=1/(1-2/n) = 1 + 3/(n-3) = 1 + 1/(n-3) + 1/(n-3) + 1(n-3) .... p=1-(m-1)/n) E[Xm]=1/(1-(m-1)/n)) Note that the sum can be split up into a triangle shape, see right hand side. Let's use the formula for the harmonic series: H_n = Sum k=0->n (1/k) = approx ln(k) Sum(E[Xk]) = m + ln(n-1)-ln(n-m-1) + ln(n-2)-ln(n-m-1) + ... = m + ln(n-1) + ln(n-2) + ... - (m-1)ln(n-m-1) .. And there is some forumla for the sum of harmonic series, if you are still interested I will look it up... Update: actually it's quite nice formula (thanks to the brilliant Concrete Mathematics book) Sum(H_k) k=0->n = nH_n - n So the expected number of steps: Sum(E[Xk]) = m + (n-1)ln(n-1) - (n-1) - (n-m-1)ln(n-m-1) - (n-m-1)) - (m-1)ln(n-m-1). Note: I haven't verified it. Share Improve this answer Follow Follow this answer to receive notifications edited Aug 4, 2011 at 20:24 answered Aug 4, 2011 at 20:13 Karoly HorvathKaroly Horvath 96.7k 11 11 gold badges 123 123 silver badges 181 181 bronze badges Comments Add a comment This answer is useful 2 Save this answer. Show activity on this post. What about using set instead of array, i think it is much easier than array set<int> Numbers; while (Numbers.size() < m) { Numbers.insert(rand() % n); } Share Improve this answer Follow Follow this answer to receive notifications answered Dec 20, 2018 at 13:25 Hani ShamsHani Shams 381 3 3 silver badges 5 5 bronze badges Comments Add a comment This answer is useful 1 Save this answer. Show activity on this post. This is a bit of a long shot, but it could work, depending on your system. Start with some reasonable ratio, like 0.5. When a request comes in, process it with whichever method you get from the current value of the threshold ratio. Record the time it takes and when you have "empty" time, perform the same task with the other method. If the alternative solution is much faster than the original one, adjust the threshold up or down. The obvious flaw in this method is that on highly variable load systems your "offline" test won't be too reliable. Share Improve this answer Follow Follow this answer to receive notifications answered Aug 4, 2011 at 20:05 biziclopbiziclop 49.9k 12 12 gold badges 80 80 silver badges 108 108 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. There was suggested Fisher-Yates shuffle. Don't know if the next code generates equally distributed integers, but it is at least compact and one-pass: std::random_device rd; std::mt19937 g(rd()); for (size_type i = 1; i < std::size(v); ++i) { v[i] = std::exchange(v[g() % i], i); } Share Improve this answer Follow Follow this answer to receive notifications answered Feb 8, 2018 at 15:48 Tomilov AnatoliyTomilov Anatoliy 16.9k 11 11 gold badges 77 77 silver badges 187 187 bronze badges Comments Add a comment This answer is useful -1 Save this answer. Show activity on this post. Quite possibly it would be more simple to start it in debug mode (and keep one method as a note) for a couple of times to get an average, then use the other method to get an average from that Share Improve this answer Follow Follow this answer to receive notifications answered Apr 15, 2016 at 21:42 JustinJustin 121 1 1 silver badge 4 4 bronze badges Comments Add a comment This answer is useful -1 Save this answer. Show activity on this post. I don't advise this method but it works ``` include include include using namespace std; int randArray; int index = 0; bool unique(int rand) { for (int i = 0; i < index; i++) if (rand == randArray[i]) return false; index++; return true; } int main() { srand(time(NULL)); for (int i = 1; i < 26; i++) randArray[i] = -1; for (int i = 0; i < 26; i++) { randArray[i] = rand() % 26; while (!unique(randArray[i])) { randArray[i] = rand() % 26; } } for (int i = 0; i < 26; i++) { cout << randArray[i] << " "; } cout << "\n" << index << endl; return 0; } ``` Share Improve this answer Follow Follow this answer to receive notifications answered Aug 3, 2018 at 16:41 joejoe 128 9 9 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 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1273	https://onlinelibrary.wiley.com/doi/10.1002/ca.22778	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Volume 30, Issue 1 pp. 61-70 Computerized tomography of the transverse pericardial sinus: Normal or pathologic? Anna Żurada, Corresponding Author Anna Żurada anna.zurada@gmail.com Department of Radiology, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Correspondence to: Anna Żurada MD, Ph.D. Department of Radiology, University Hospital in Olsztyn, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland, ul. Warszawska 30; 10-182 Olsztyn. E-mail: anna.zurada@gmail.comSearch for more papers by this author Andrzej Ustymowicz, Andrzej Ustymowicz Department of Radiology, Medical University of Bialystok, Poland Search for more papers by this author Marios Loukas, Marios Loukas Department of Anatomical Sciences, St. George's University, School of Medicine, Grenada, West Indies Department of Anatomy, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Search for more papers by this author Maciej Michalak, Maciej Michalak Department of Radiology, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Search for more papers by this author Dorota Czyżewska, Dorota Czyżewska Department of Diagnostic Imaging, Independent Public Provincial Hospital of J. Sniadecki in Bialystok, Poland Search for more papers by this author Jerzy Gielecki, Jerzy Gielecki Department of Anatomy, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Search for more papers by this author Anna Żurada, Corresponding Author Anna Żurada anna.zurada@gmail.com Department of Radiology, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Correspondence to: Anna Żurada MD, Ph.D. Department of Radiology, University Hospital in Olsztyn, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland, ul. Warszawska 30; 10-182 Olsztyn. E-mail: anna.zurada@gmail.comSearch for more papers by this author Andrzej Ustymowicz, Andrzej Ustymowicz Department of Radiology, Medical University of Bialystok, Poland Search for more papers by this author Marios Loukas, Marios Loukas Department of Anatomical Sciences, St. George's University, School of Medicine, Grenada, West Indies Department of Anatomy, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Search for more papers by this author Maciej Michalak, Maciej Michalak Department of Radiology, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Search for more papers by this author Dorota Czyżewska, Dorota Czyżewska Department of Diagnostic Imaging, Independent Public Provincial Hospital of J. Sniadecki in Bialystok, Poland Search for more papers by this author Jerzy Gielecki, Jerzy Gielecki Department of Anatomy, Faculty of Medicine, University of Warmia and Mazury in Olsztyn, Poland Search for more papers by this author First published: 06 October 2016 Citations: 8 Abstract The transverse pericardial sinus is a uniquely located structure subdivided into many parts. However, discrepancies still exist on the nomenclature and divisions. As noninvasive diagnostic technology such as CT and MR imaging improve, the transverse pericardial sinus and constituent recesses are visualized with more clarity, increasing the risk for misinterpretation. In this review, we will explore the anatomy of the transverse pericardial sinus and associated recesses with the goal of heightening awareness regarding the differential diagnosis between normal and pathological states as seen on CT. In addition, the inconsistencies of the right lateral superior aortic recess are also addressed. Last, we describe the clinical and surgical significance of the transverse pericardial sinus. Clin. Anat. 30:61–70, 2017. © 2016 Wiley Periodicals, Inc. REFERENCES Arey LB. 1974. Developmental Anatomy: A Textbook and Laboratory Manual of Embryology. Philadelphia and London: WB Saunders. p 389. Google Scholar Aribandi M. 2007. High-riding superior pericardial recess: Temporal change can help distinguish from mediastinal pathology. South Med J 100: 407–408. 10.1097/01.smj.0000242788.28442.8d PubMed Web of Science® Google Scholar Aronberg DJ, Peterson RR, Glazer HS, Sagel S. 1984. The superior sinus of the pericardium: CT appearance. Radiology 153: 489–492. 10.1148/radiology.153.2.6484182 CAS PubMed Web of Science® Google Scholar Basile A, Bisceglie P, Giulietti G, Calcara G, Figuera M, Mundo E, Granata A, Runza G, Privitera C, Privitera G, Patti MT. 2006. Prevalence of “high-riding” superior pericardial recesses on thin-section 16-MDCT scans. 59: 265–269. 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Surg Radiol Anat 8: 221–227. 10.1007/BF02425071 PubMed Web of Science® Google Scholar Citing Literature Volume30, Issue1 January 2017 Pages 61-70 ## References ## Related ## Information Metrics 8 Details © 2016 Wiley Periodicals, Inc. Keywords transverse pericardial sinus divisions anatomy superior aortic recess pulmonic recess high riding superior pericardial sinus computerized tomography Publication History 12 December 2016 06 October 2016 26 August 2016 24 August 2016 Close Figure Viewer Previous FigureNext Figure Caption Download PDF Log in to Wiley Online Library NEW USER > INSTITUTIONAL LOGIN > Change Password Password Changed Successfully Your password has been changed Create a new account The full text of this article hosted at iucr.org is unavailable due to technical difficulties.
1274	https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_7?srsltid=AfmBOoplxAF1vqSyL4ZoRMqOJcLD8KxBeOina5eFpVG4hlixlK1VMdVL	Art of Problem Solving 2002 AIME II Problems/Problem 7 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2002 AIME II Problems/Problem 7 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2002 AIME II Problems/Problem 7 Problem It is known that, for all positive integers , . Find the smallest positive integer such that is a multiple of . Solution Using the given formula, we require to be a multiple of , i.e. to be a multiple of . This is equivalent to being divisible by each of these factors (, , and ) separately, since they are coprime. Thus, for divisibility by , we observe that is necessarily odd, while exactly one of and is even. It follows that the (at least) factors of must be either all contained in , or all contained in , yielding or respectively. This reduces to . Next, if , then obviously will be divisible by 3; otherwise, we will have either , giving , or , giving . Hence is in fact always divisible by , regardless of the value of . Similarly, considering the cases in turn, the residues modulo of , , and respectively are ; ; ; ; and . As never appears more than once in any of these cases, we deduce that at most one of , , and is divisible by 5. Analogously to above, it follows that the (at least) factors of must be either all contained in , all contained in , or all contained in . These respectively give , which reduces to . Accordingly, as there are possible residues for modulo and possible residues modulo , we obtain a total of pairs of simultaneous congruences. By the Chinese remainder theorem, as and are coprime, each pair has a unique solution modulo , which we can easily calculate by simply writing out the solutions of the first congruence, then checking whether each of them also satisfies the second congruence. For instance, to solve , , we can write out the positive multiples of (to satisfy the first congruence, together with the fact that ), then subtract from each, until we find a multiple of for which this subtraction gives a multiple of . As it turns out, and , so the solution of this pair is precisely . By applying this algorithm to each of the remaining pairs of congruences, we eventually obtain the complete solution , so the smallest possible positive value of is . See also 2002 AIME II (Problems • Answer Key • Resources) Preceded by Problem 6Followed by Problem 8 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Intermediate Number Theory Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1275	https://2012books.lardbucket.org/pdfs/principles-of-general-chemistry-v1.0/s31-appendix-c-dissociation-consta.pdf	This is “Appendix C: Dissociation Constants and pKa Values for Acids at 25°C”, appendix 3 from the book Principles of General Chemistry (index.html) (v. 1.0). This book is licensed under a Creative Commons by-nc-sa 3.0 ( 3.0/) license. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. This content was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz ( in an effort to preserve the availability of this book. Normally, the author and publisher would be credited here. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally, per the publisher's request, their name has been removed in some passages. More information is available on this project's attribution page ( For more information on the source of this book, or why it is available for free, please see the project's home page ( You can browse or download additional books there. i Chapter 27 Appendix C: Dissociation Constants and pKa Values for Acids at 25°C Name Formula K a1 pKa1 K a2 pKa2 K a3 pKa3 Acetic acid CH3CO2H 1.75 × 10−5 4.756 Arsenic acid H3AsO4 5.5 × 10−3 2.26 1.7 × 10−7 6.76 5.1 × 10−12 11.29 Benzoic acid C6H5CO2H 6.25 × 10−5 4.204 Boric acid H3BO3 5.4 × 10−10 9.27 >1 × 10−14 >14 Bromoacetic acid CH2BrCO2H 1.3 × 10−3 2.90 Carbonic acid H2CO3 4.5 × 10−7 6.35 4.7 × 10−11 10.33 Chloroacetic acid CH2ClCO2H 1.3 × 10−3 2.87 Chlorous acid HClO2 1.1 × 10−2 1.94 Chromic acid H2CrO4 1.8 × 10−1 0.74 3.2 × 10−7 6.49 Citric acid C6H8O7 7.4 × 10−4 3.13 1.7 × 10−5 4.76 4.0 × 10−7 6.40 Cyanic acid HCNO 3.5 × 10−4 3.46 Dichloroacetic acid CHCl2CO2H 4.5 × 10−2 1.35 Fluoroacetic acid CH2FCO2H 2.6 × 10−3 2.59 Formic acid CH2O2 1.8 × 10−4 3.75 Hydrazoic acid HN3 2.5 × 10−5 4.6 Hydrocyanic acid HCN 6.2 × 10−10 9.21 Hydrofluoric acid HF 6.3 × 10−4 3.20 Measured at 20°C, not 25°C. ‡ Measured at 18°C, not 25°C. 2990 Name Formula K a1 pKa1 K a2 pKa2 K a3 pKa3 Hydrogen selenide H2Se 1.3 × 10−4 3.89 1.0× 10−11 11.0 Hydrogen sulfide H2S 8.9 × 10−8 7.05 1 × 10−19 19 Hydrogen telluride H2Te 2.5 × 10−3‡ 2.6‡ 1 × 10−11 11 Hypobromous acid HBrO 2.8 × 10−9 8.55 Hypochlorous acid HClO 4.0 × 10−8 7.40 Hypoiodous acid HIO 3.2 × 10−11 10.5 Iodic acid HIO3 1.7 × 10−1 0.78 Iodoacetic acid CH2ICO2H 6.6 × 10−4 3.18 Nitrous acid HNO2 5.6 × 10−4 3.25 Oxalic acid C2H2O4 5.6 × 10−2 1.25 1.5 × 10−4 3.81 Periodic acid HIO4 2.3 × 10−2 1.64 Phenol C6H5OH 1.0 × 10−10 9.99 Phosphoric acid H3PO4 6.9 × 10−3 2.16 6.2 × 10−8 7.21 4.8 × 10−13 12.32 Phosphorous acid H3PO3 5.0 × 10−2 1.3 2.0 × 10−7 6.70 Pyrophosphoric acid H4P2O7 1.2 × 10−1 0.91 7.9 × 10−3 2.10 2.0 × 10−7 6.70 4. Resorcinol C6H4(OH)2 4.8 × 10−10 9.32 7.9 × 10−12 11.1 Selenic acid H2SeO4 Strong Strong 2.0 × 10−2 1.7 Selenious acid H2SeO3 2.4 × 10−3 2.62 4.8 × 10−9 8.32 Sulfuric acid H2SO4 Strong Strong 1.0 × 10−2 1.99 Sulfurous acid H2SO3 1.4 × 10−2 1.85 6.3 × 10−8 7.2 meso-Tartaric acid C4H6O6 6.8 × 10−4 3.17 1.2 × 10−5 4.91 Measured at 20°C, not 25°C. ‡ Measured at 18°C, not 25°C. Chapter 27 Appendix C: Dissociation Constants and pKa Values for Acids at 25°C 2991 Name Formula K a1 pKa1 K a2 pKa2 K a3 pKa3 Telluric acid H2TeO4 2.1 × 10−8‡ 7.68‡ 1.0 × 10−11‡ 11.0‡ Tellurous acid H2TeO3 5.4 × 10−7 6.27 3.7 × 10−9 8.43 Trichloroacetic acid CCl3CO2H 2.2 × 10−1 0.66 Trifluoroacetic acid CF3CO2H 3.0 × 10−1 0.52 Measured at 20°C, not 25°C. ‡ Measured at 18°C, not 25°C. Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004). Chapter 27 Appendix C: Dissociation Constants and pKa Values for Acids at 25°C 2992
1276	https://www.cs.cmu.edu/~15859n/RelatedWork/Goemans-LP-notes.pdf	18.415/6.854 Advanced Algorithms October 1994 Linear Programming Lecturer: Michel X. Goemans 1 An Introduction to Linear Programming Linear programming is a very important class of problems, both algorithmically and combinatorially. Linear programming has many applications. From an algorithmic point-of-view, the simplex was proposed in the forties (soon after the war, and was motivated by military applications) and, although it has performed very well in prac-tice, is known to run in exponential time in the worst-case. On the other hand, since the early seventies when the classes P and NP were deﬁned, it was observed that linear programming is in NP∩co-NP although no polynomial-time algorithm was known at that time. The ﬁrst polynomial-time algorithm, the ellipsoid algorithm, was only dis-covered at the end of the seventies. Karmarkar’s algorithm in the mid-eighties lead to very active research in the area of interior-point methods for linear programming. We shall present one of the numerous variations of interior-point methods in class. From a combinatorial perspective, systems of linear inequalities were already studied at the end of the last century by Farkas and Minkovsky. Linear programming, and especially the notion of duality, is very important as a proof technique. We shall illustrate its power when discussing approximation algorithms. We shall also talk about network ﬂow algorithms where linear programming plays a crucial role both algorithmically and combinatorially. For a more in-depth coverage of linear programming, we refer the reader to [1, 4, 7, 8, 5]. A linear program is the problem of optimizing a linear objective function in the decision variables, x1 . . . xn, subject to linear equality or inequality constraints on the xi’s. In standard form, it is expressed as: Min n X j=1 cjxj (objective function) subject to: n X j=1 aijxj = bi, i = 1 . . .m (constraints) xj ≥0, j = 1 . . .n (non-negativity constraints) where {aij, bi, cj} are given. A linear program is expressed more conveniently using matrices: min cTx subject to ( Ax = b x ≥ 0 LP-1 where x =     x1 . . . xn     ∈Rn×1 b =     b1 . . . bm     ∈Rm×1 c =     c1 . . . cn     ∈Rn×1 A =     a11 ... amn     ∈Rm×n 2 Basic Terminology Deﬁnition 1 If x satisﬁes Ax = b, x ≥0, then x is feasible. Deﬁnition 2 A linear program (LP) is feasible if there exists a feasible solution, otherwise it is said to be infeasible. Deﬁnition 3 An optimal solution x∗is a feasible solution s.t. cTx∗= min{cTx : Ax = b, x ≥0}. Deﬁnition 4 LP is unbounded (from below) if ∀λ ∈R, ∃a feasible x∗s.t. cTx∗≤λ. 3 Equivalent Forms A linear program can take on several forms. We might be maximizing instead of minimizing. We might have a combination of equality and inequality contraints. Some variables may be restricted to be non-positive instead of non-negative, or be unrestricted in sign. Two forms are said to be equivalent if they have the same set of optimal solutions or are both infeasible or both unbounded. 1. A maximization problem can be expressed as a minimization problem. max cTx ⇔min −cTx 2. An equality can be represented as a pair of inequalities. aT i x = bi ⇔ ( aT i x ≤ bi aT i x ≥ bi ⇔ ( aT i x ≤ bi −aT i x ≤ −bi LP-2 3. By adding a slack variable, an inequality can be represented as a combination of equality and non-negativity constraints. aT i x ≤bi ⇔aT i x + si = bi, si ≥0. 4. Non-positivity constraints can be expressed as non-negativity constraints. To express xj ≤0, replace xj everywhere with −yj and impose the condition yj ≥0. 5. x may be unrestricted in sign. If x is unrestricted in sign, i.e. non-positive or non-negative, everywhre replace xj by x+ j −x− j , adding the constraints x+ j , x− j ≥0. In general, an inequality can be represented using a combination of equality and non-negativity constraints, and vice versa. Using these rules, min n cTx s.t. Ax ≥b} can be transformed into min n cTx+ −cTx− s.t. Ax+ −Ax−−Is = b, x+, x−, s ≥0}. The former LP is said to be in canonical form, the latter in standard form. Conversely, an LP in standard form may be written in canonical form. min n cTx s.t. Ax = b, x ≥0} is equivalent to min n cTx s.t. Ax ≥b, −Ax ≥−b, Ix ≥0}. This may be rewritten as A ′x ≥b ′, where A ′ =    A -A I   and b ′ =    b -b 0   . 4 Example Consider the following linear program: min x2 subject to          x1 ≥ 2 3x1 − x2 ≥ 0 x1 + x2 ≥ 6 −x1 + 2x2 ≥ 0 The optimal solution is (4, 2) of cost 2 (see Figure 1). If we were maximizing x2 instead of minimizing under the same feasible region, the resulting linear program would be unbounded since x2 can increase arbitrarily. From this picture, the reader should be convinced that, for any objective function for which the linear program is bounded, there exists an optimal solution which is a “corner” of the feasible region. We shall formalize this notion in the next section. LP-3 (2,1) (2,4) x1 x2 3x − x > 0 2 1 1 1 1 x + x > 6 x > 2 2 2 −x + 2x > 0 Figure 1: Graph representing primal in example. An example of an infeasible linear program can be obtained by reversing some of the inequalities of the above LP: x1 ≤ 2 3x1 − x2 ≥ 0 x1 + x2 ≥ 6 −x1 + 2x2 ≤ 0. 5 The Geometry of LP Let P = {x : Ax = b, x ≥0} ⊆Rn. Deﬁnition 5 x is a vertex of P if ̸ ∃y ̸= 0 s.t. x + y, x −y ∈P. Theorem 1 Assume min{cTx : x ∈P} is ﬁnite, then ∀x ∈P, ∃a vertex x ′ such that cTx ′ ≤cTx. Proof: If x is a vertex, then take x ′ = x. If x is not a vertex, then, by deﬁnition, ∃y ̸= 0 s.t. x + y, x −y ∈P. Since A(x + y) = b and A(x −y) = b, Ay = 0. LP-4 WLOG, assume cTy ≤0 (take either y or −y). If cTy = 0, choose y such that ∃j s.t. yj < 0. Since y ̸= 0 and cTy = cT(−y) = 0, this must be true for either y or −y. Consider x + λy, λ > 0. cT(x + λy) = cTx + λcTy ≤cTx, since cTy is assumed non-positive. Case 1 ∃j such that yj < 0 As λ increases, component j decreases until x + λy is no longer feasible. Choose λ = min{j:yj<0}{xj/−yj} = xk/−yk. This is the largest λ such that x + λy ≥0. Since Ay = 0, A(x + λy) = Ax + λAy = Ax = b. So x + λy ∈P, and moreover x + λy has one more zero component, (x + λy)k, than x. Replace x by x + λy. Case 2 yj ≥0 ∀j By assumption, cTy < 0 and x + λy is feasible for all λ ≥0, since A(x + λy) = Ax+ λAy = Ax = b, and x+ λy ≥x ≥0. But cT(x+ λy) = cTx+ λcTy →−∞ as λ →∞, implying LP is unbounded, a contradiction. Case 1 can happen at most n times, since x has n components. By induction on the number of non-zero components of x, we obtain a vertex x ′. ✷ Remark: The theorem was described in terms of the polyhedral set P = {x : Ax = b : x ≥0}. Strictly speaking, the theorem is not true for P = {x : Ax ≥ b}. Indeed, such a set P might not have any vertex. For example, consider P = {(x1, x2) : 0 ≤x2 ≤1} (see Figure 2). This polyhedron has no vertex, since for any x ∈P, we have x + y, x −y ∈P, where y = (1, 0). It can be shown that P has a vertex iﬀRank(A) = n. Note that, if we transform a program in canonical form into standard form, the non-negativity constraints imply that the resulting matrix A has full column rank, since Rank    A -A I   = n. Corollary 2 If min{cTx : Ax = b, x ≥0} is ﬁnite, There exists an optimal solution, x∗, which is a vertex. Proof: Suppose not. Take an optimal solution. By Theorem 1 there exists a vertex costing no more and this vertex must be optimal as well. ✷ Corollary 3 If P = {x : Ax = b, x ≥0} ̸= ∅, then P has a vertex. LP-5 (0,1) (0,0) x x 1 2 Figure 2: A polyhedron with no vertex. Theorem 4 Let P = {x : Ax = b, x ≥0}. For x ∈P, let Ax be a submatrix of A corresponding to j s.t. xj > 0. Then x is a vertex iﬀAx has linearly independent columns. (i.e. Ax has full column rank.) Example A =    2 1 3 0 7 3 2 1 0 0 0 5   x =      2 0 1 0     Ax =    2 3 7 2 0 0   , and x is a vertex. Proof: Show ¬i →¬ii. Assume x is not a vertex. Then, by deﬁnition, ∃y ̸= 0 s.t. x + y, x −y ∈P. Let Ay be submatrix corresponding to non-zero components of y. As in the proof of Theorem 1, Ax + Ay = b Ax − Ay = b ) ⇒Ay = 0. Therefore, Ay has dependent columns since y ̸= 0. Moreover, x + y ≥ 0 x − y ≥ 0 ) ⇒yj = 0 whenever xj = 0. Therefore Ay is a submatrix of Ax. Since Ay is a submatrix of Ax, Ax has linearly dependent columns. LP-6 Show ¬ii →¬i. Suppose Ax has linearly dependent columns. Then ∃y s.t. Axy = 0, y ̸= 0. Extend y to Rn by adding 0 components. Then ∃y ∈Rn s.t. Ay = 0, y ̸= 0 and yj = 0 wherever xj = 0. Consider y ′ = λy for small λ ≥0. Claim that x + y ′, x −y ′ ∈P, by argument analogous to that in Case 1 of the proof of Theorem 1, above. Hence, x is not a vertex. ✷ 6 Bases Let x be a vertex of P = {x : Ax = b, x ≥0}. Suppose ﬁrst that \|{j : xj > 0}\| = m (where A is m × n). In this case we denote B = {j : xj > 0}. Also let AB = Ax; we use this notation not only for A and B, but also for x and for other sets of indices. Then AB is a square matrix whose columns are linearly independent (by Theorem 4), so it is non-singular. Therefore we can express x as xj = 0 if j ̸∈B, and since ABxB = b, it follows that xB = A−1 B b. The variables corresponding to B will be called basic. The others will be referred to as nonbasic. The set of indices corresponding to nonbasic variables is denoted by N = {1, . . . , n} −B. Thus, we can write the above as xB = A−1 B b and xN = 0. Without loss of generality we will assume that A has full row rank, rank(A) = m. Otherwise either there is a redundant constraint in the system Ax = b (and we can remove it), or the system has no solution at all. If \|{j : xj > 0}\| < m, we can augment Ax with additional linearly independent columns, until it is an m × m submatrix of A of full rank, which we will denote AB. In other words, although there may be less than m positive components in x, it is convenient to always have a basis B such that \|B\| = m and AB is non-singular. This enables us to always express x as we did before, xN = 0, xB = A−1 B b. Summary x is a vertex of P iﬀthere is B ⊆{1, . . . , n} such that \|B\| = m and 1. xN = 0 for N = {1, . . . , n} −B 2. AB is non-singular 3. xB = A−1 B b ≥0 In this case we say that x is a basic feasible solution. Note that a vertex can have several basic feasible solution corresponding to it (by augmenting {j : xj > 0} in diﬀerent ways). A basis might not lead to any basic feasible solution since A−1 B b is not necessarily nonnegative. LP-7 Example: x1 + x2 + x3 = 5 2x1 −x2 + 2x3 = 1 x1, x2, x3 ≥0 We can select as a basis B = {1, 2}. Thus, N = {3} and AB = 1 1 2 −1 ! A−1 B = 1 3 1 3 2 3 −1 3 ! A−1 B b = 2 3 ! x = (2, 3, 0) Remark. A crude upper bound on the number of vertices of P is n m . This number is exponential (it is upper bounded by nm). We can come up with a tighter approx-imation of n−m 2 m 2 , though this is still exponential. The reason why the number is much smaller is that most basic solutions to the system Ax = b (which we counted) are not feasible, that is, they do not satisfy x ≥0. 7 The Simplex Method The Simplex algorithm [Dantzig,1947] solves linear programming problems by focusing on basic feasible solutions. The basic idea is to start from some vertex v and look at the adjacent vertices. If an improvement in cost is possible by moving to one of the adjacent vertices, then we do so. Thus, we will start with a bfs corresponding to a basis B and, at each iteration, try to improve the cost of the solution by removing one variable from the basis and replacing it by another. We begin the Simplex algorithm by ﬁrst rewriting our LP in the form: min cBxB + cNxN s.t. ABxB + ANxN = b xB, xN ≥0 Here B is the basis corresponding to the bfs we are starting from. Note that, for any solution x, xB = A−1 B b −A−1 B ANxN and that its total cost, cTx can be speciﬁed as follows: LP-8 cTx = cBxB + cNxN = cB(A−1 B b −A−1 B ANxN) + cNxN = cBA−1 B b + (cN −cBA−1 B AN)xN We denote the reduced cost of the non-basic variables by ˜ cN, ˜ cN = cN −cBA−1 B AN, i.e. the quantity which is the coeﬃcient of xN above. If there is a j ∈N such that ˜ cj < 0, then by increasing xj (up from zero) we will decrease the cost (the value of the objective function). Of course xB depends on xN, and we can increase xj only as long as all the components of xB remain positive. So in a step of the Simplex method, we ﬁnd a j ∈N such that ˜ cj < 0, and increase it as much as possible while keeping xB ≥0. It is not possible any more to increase xj, when one of the components of xB is zero. What happened is that a non-basic variable is now positive and we include it in the basis, and one variable which was basic is now zero, so we remove it from the basis. If, on the other hand, there is no j ∈N such that ˜ cj < 0, then we stop, and the current basic feasible solution is an optimal solution. This follows from the new expression for cTx since xN is nonnegative. Remarks: 1. Note that some of the basic variables may be zero to begin with, and in this case it is possible that we cannot increase xj at all. In this case we can replace say j by k in the basis, but without moving from the vertex corresponding to the basis. In the next step we might replace k by j, and be stuck in a loop. Thus, we need to specify a “pivoting rule” to determine which index should enter the basis, and which index should be removed from the basis. 2. While many pivoting rules (including those that are used in practice) can lead to inﬁnite loops, there is a pivoting rule which will not (known as the minimal index rule - choose the minimal j and k possible [Bland, 1977]). This fact was discovered by Bland in 1977. There are other methods of “breaking ties” which eliminate inﬁnite loops. 3. There is no known pivoting rule for which the number of pivots in the worst case is better than exponential. 4. The question of the complexity of the Simplex algorithm and the last remark leads to the question of what is the length of the shortest path between two vertices of a convex polyhedron, where the path is along edges, and the length of the path in measured in terms of the number of vertices visited. LP-9 Hirsch Conjecture: For m hyperplanes in d dimensions the length of the shortest path between any two vertices of the arrangement is at most m −d. This is a very open question — there is not even a polynomial bound proven on this length. On the other hand, one should note that even if the Hirsch Conjecture is true, it doesn’t say much about the Simplex Algorithm, because Simplex generates paths which are monotone with respect to the objective function, whereas the shortest path need not be monotone. Recently, Kalai (and others) has considered a randomized pivoting rule. The idea is to randomly permute the index columns of A and to apply the Simplex method, always choosing the smallest j possible. In this way, it is possible to show a subexponential bound on the expected number of pivots. This leads to a subexponential bound for the diameter of any convex polytope deﬁned by m hyperplanes in a d dimension space. The question of the existence of a polynomial pivoting scheme is still open though. We will see later a completely diﬀerent algorithm which is polynomial, although not strongly polynomial (the existence of a strongly polynomial algo-rithm for linear programming is also open). That algorithm will not move from one vertex of the feasible domain to another like the Simplex, but will conﬁne its interest to points in the interior of the feasible domain. A visualization of the geometry of the Simplex algorithm can be obtained from considering the algorithm in 3 dimensions (see Figure 3). For a problem in the form min{cTx : Ax ≤b} the feasible domain is a polyhedron in R3, and the algorithm moves from vertex to vertex in each step (or does not move at all). 8 When is a Linear Program Feasible ? We now turn to another question which will lead us to important properties of linear programming. Let us begin with some examples. We consider linear programs of the form Ax = b, x ≥0. As the objective function has no eﬀect on the feasibility of the program, we ignore it. We ﬁrst restrict our attention to systems of equations (i.e. we neglect the non-negativity constraints). Example: Consider the system of equations: x1 + x2 + x3 = 6 2x1 + 3x2 + x3 = 8 2x1 + x2 + 3x3 = 0 and the linear combination LP-10 Objective function Figure 3: Traversing the vertices of a convex body (here a polyhedron in R3). −4 × x1 + x2 + x3 = 6 1 × 2x1 + 3x2 + x3 = 8 1 × 2x1 + x2 + 3x3 = 0 The linear combination results in the equation 0x1 + 0x2 + 0x3 = −16 which means of course that the system of equations has no feasible solution. In fact, an elementary theorem of linear algebra says that if a system has no solution, there is always a vector y such as in our example (y = (−4, 1, 1)) which proves that the system has no solution. Theorem 5 Exactly one of the following is true for the system Ax = b: 1. There is x such that Ax = b. 2. There is y such that ATy = 0 but yTb = 1. This is not quite enough for our purposes, because a system can be feasible, but still have no non-negative solutions x ≥0. Fortunately, the following lemma establishes the equivalent results for our system Ax = b, x ≥0. Theorem 6 (Farkas’ Lemma) Exactly one of the following is true for the system Ax = b, x ≥0: LP-11 1. There is x such that Ax = b, x ≥0. 2. There is y such that ATy ≥0 but bTy < 0. Proof: We will ﬁrst show that the two conditions cannot happen together, and then than at least one of them must happen. Suppose we do have both x and y as in the statement of the theorem. Ax = b = ⇒yTAx = yTb = ⇒xTATy = yTb but this is a contradiction, because yTb < 0, and since x ≥0 and ATy ≥0, so xTATy ≥0. The other direction is less trivial, and usually shown using properties of the Sim-plex algorithm, mainly duality. We will use another tool, and later use Farkas’ Lemma to prove properties about duality in linear programming. The tool we shall use is the Projection theorem, which we state without proof: Theorem 7 (Projection Theorem) Let K be a closed convex (see Figure 4) non-empty set in Rn, and let b be any point in Rn. The projection of b onto K is a point p ∈K that minimizes the Euclidean distance ∥b −p∥. Then p has the property that for all z ∈K, (z −p)T(b −p) ≤0 (see Figure 5) non-empty set. not convex convex Figure 4: Convex and non-convex sets in R2. We are now ready to prove the other direction of Farkas’ Lemma. Assume that there is no x such that Ax = b, x ≥0; we will show that there is y such that ATy ≥0 but yTb < 0. Let K = {Ax : x ≥0} ⊆Rm (A is an m × n matrix). K is a cone in Rm and it is convex, non-empty and closed. According to our assumption, Ax = b, x ≥0 has no solution, so b does not belong to K. Let p be the projection of b onto K. LP-12 b p z Figure 5: The Projection Theorem. Since p ∈K, there is a w ≥0 such that Aw = p. According to the Projection Theorem, for all z ∈K, (z −p)T(b−p) ≤0 That is, for all x ≥0 (Ax−p)T(b−p) ≤0 We deﬁne y = p−b, which implies (Ax−p)Ty ≥0. Since Aw = p, (Ax−Aw)Ty ≥ 0. (x −w)T(ATy) ≥0 for all x ≥0 (remember that w was ﬁxed by choosing b). Set x = w +             0 0 . . . 1 . . . 0             (w plus a unit vector with a 1 in the i-th row). Note that x is non-negative, because w ≥0. This will extract the i-th column of A, so we conclude that the i-th component of ATy is non-negative (ATy)i ≥0, and since this is true for all i, ATy ≥0. Now it only remains to show that yTb < 0. ytb = (p−y)Ty = pTy−yTy Since (Ax−p)Ty ≥0 for all x ≥0, taking x to be zero shows that pTy ≤0. Since b ̸∈K, y = p−b ̸= 0, so yTy > 0. So yTb = pTy −yTy < 0. ✷ Using a very similar proof one can show the same for the canonical form: Theorem 8 Exactly one of the following is true for the system Ax ≤b: 1. There is x such that Ax ≤b. 2. There is y ≥0 such that ATy = 0 but yTb < 0. LP-13 The intuition behind the precise form for 2. in the previous theorem lies in the proof that both cannot happen. The contradiction 0 = 0x = (yTA)x = yT(Ax) = yTb < 0 is obtained if ATy = 0 and yTb < 0. 9 Duality Duality is the most important concept in linear programming. Duality allows to provide a proof of optimality. This is not only important algorithmically but also it leads to beautiful combinatorial statements. For example, consider the statement In a graph, the smallest number of edges in a path between two spec-iﬁed vertices s and t is equal to the maximum number of s −t cuts (i.e. subsets of edges whose removal disconnects s and t). This result is a direct consequence of duality for linear programming. Duality can be motivated by the problem of trying to ﬁnd lower bounds on the value of the optimal solution to a linear programming problem (if the problem is a maximization problem, then we would like to ﬁnd upper bounds). We consider problems in standard form: min cTx s.t. Ax = b x ≥0 Suppose we wanted to obtain the best possible upper bound on the cost function. By multiplying each equation Amx = bm by some number ym and summing up the resulting equations, we obtain that yTAx = bTy. if we impose that the coeﬃcient of xj in the resulting inequality is less or equal to cj then bTy must be a lower bound on the optimal value since xj is constrained to be nonnegative. To get the best possible lower bound, we want to solve the following problem: max bTy s.t. ATy ≤c This is another linear program. We call this one the dual of the original one, called the primal. As we just argued, solving this dual LP will give us a lower bound on the optimum value of the primal problem. Weak duality says precisely this: if we denote the optimum value of the primal by z, z = min cTx, and the optimum value of the dual by w, then w ≤z. We will use Farkas’ lemma to prove strong duality which says that these quantities are in fact equal. We will also see that, in general, the dual of the dual is the problem. LP-14 Example: z = min x1 + 2x2 + 4x3 x1 + x2 + 2x3 = 5 2x1 + x2 + 3x3 = 8 The ﬁrst equality gives a lower bound of 5 on the optimum value z, since x1 + 2x2 + 4x3 ≥x1 + x2 + 2x3 = 5 because of nonnegativity of the xi. We can get an even better lower bound by taking 3 times the ﬁrst equality minus the second one. This gives x1 + 2x2 + 3x3 = 7 ≤x1 + 2x2 + 4x3, implying a lower bound of 7 on z. For x =    3 2 0   , the objective function is precisely 7, implying optimality. The mechanism of generating lower bounds is formalized by the dual linear program: max 5y1 + 8y2 y1 + 2y2 ≤ 1 y1 + y2 ≤ 2 2y1 + 3y2 ≤ 4 y1 represents the multiplier for the ﬁrst constraint and y2 the multiplier for the second constraint, This LP’s objective function also achieves a maximum value of 7 at y = 3 −1 ! . We now formalize the notion of duality. Let P and D be the following pair of dual linear programs: (P) z = min{cTx : Ax = b, x ≥0} (D) w = max{bTy : ATy ≤c}. (P) is called the primal linear program and (D) the dual linear program. In the proof below, we show that the dual of the dual is the primal. In other words, if one formulates (D) as a linear program in standard form (i.e. in the same form as (P)), its dual D(D) can be seen to be equivalent to the original primal (P). In any statement, we may thus replace the roles of primal and dual without aﬀecting the statement. Proof: The dual problem D is equivalent to min{−bTy : ATy + Is = c, s ≥0}. Changing forms we get min{−bTy+ +bTy−: ATy+ −ATy−+Is = c, and y+, y−, s ≥0}. Taking the dual of this we obtain: max{−cTx : A(−x) ≤−b, −A(−x) ≤b, I(−x) ≤0}. But this is the same as min{cTx : Ax = b, x ≥0} and we are done. ✷ We have the following results relating w and z. Lemma 9 (Weak Duality) z ≥w. LP-15 Proof: Suppose x is primal feasible and y is dual feasible. Then, cTx ≥yTAx = yTb, thus z = min{cTx : Ax = b, x ≥0} ≥max{bTy : ATy ≤c} = w. ✷ From the preceding lemma we conclude that the following cases are not possible (these are dual statements): 1. P is feasible and unbounded and D feasible. 2. P is feasible and D is feasible and unbounded. We should point out however that both the primal and the dual might be infeasible. To prove a stronger version of the weak duality lemma, let’s recall the following corollary of Farkas’ Lemma (Theorem 8): Corollary 10 Exactly one of the following is true: 1. ∃x′ : A′x′ ≤b′. 2. ∃y′ ≥0 : (A′)Ty′ = 0 and (b′)Ty′ < 0. Theorem 11 (Strong Duality) If P or D is feasible then z = w. Proof: We only need to show that z ≤w. Assume without loss of generality (by duality) that P is feasible. If P is unbounded, then by Weak Duality, we have that z = w = −∞. Suppose P is bounded, and let x∗be an optimal solution, i.e. Ax∗= b, x∗≥0 and cTx∗= z. We claim that ∃y s.t. ATy ≤c and bTy ≥z. If so we are done. Suppose no such y exists. Then, by the preceding corollary, with A′ = AT −bT ! , b′ = c −z ! , x′ = y, y′ = x λ ! , ∃x ≥0, λ ≥0 such that Ax = λb and cTx < λz. We have two cases • Case 1: λ ̸= 0. Since we can normalize by λ we can assume that λ = 1. This means that ∃x ≥0 such that Ax = b and cTx < z. But this is a contradiction with the optimality of x∗. • Case 2: λ = 0. This means that ∃x ≥0 such that Ax = 0 and cTx < 0. If this is the case then ∀µ ≥0, x∗+ µx is feasible for P and its cost is cT(x∗+ µx) = cTx∗+ µ(cTx) < z, which is a contradiction. ✷ LP-16 9.1 Rules for Taking Dual Problems If P is a minimization problem then D is a maximization problem. If P is a maxi-mization problem then D is a minimization problem. In general, using the rules for transforming a linear program into standard form, we have that the dual of (P): z = min cT 1 x1 + cT 2 x2 + cT 3 x3 s.t. A11x1 + A12x2 + A13x3 = b1 A21x1 + A22x2 + A23x3 ≥ b2 A31x1 + A32x2 + A33x3 ≤ b3 x1 ≥0 , x2 ≤0 , x3 UIS (where UIS means “unrestricted in sign” to emphasize that no constraint is on the variable) is (D) w = max bT 1 y1 + bT 2 y2 + bT 3 y3 s.t. AT 11y1 + AT 21y2 + AT 31y3 ≤ c1 AT 12y1 + AT 22y2 + AT 32y3 ≥ c2 AT 13y1 + AT 23y2 + AT 33y3 = c3 y1 UIS , y2 ≥0 , y3 ≤0 10 Complementary Slackness Let P and D be (P) z = min{cTx : Ax = b, x ≥0} (D) w = max{bTy : ATy ≤c}, and let x be feasible in P, and y be fesible in D. Then, by weak duality, we know that cTx ≥bTy. We call the diﬀerence cTx −bTy the duality gap. Then we have that the duality gap is zero iﬀx is optimal in P, and y is optimal in D. That is, the duality gap can serve as a good measure of how close a feasible x and y are to the optimal solutions for P and D. The duality gap will be used in the description of the interior point method to monitor the progress towards optimality. It is convenient to write the dual of a linear program as (D) w = max{bTy : ATy + s = c for some s ≥0} Then we can write the duality gap as follows: cTx −bTy = cTx −xTATy = xT(c −ATy) = xTs LP-17 since ATy + s = c. The following theorem allows to check optimality of a primal and/or a dual solu-tion. Theorem 12 (Complementary Slackness) Let x∗, (y∗, s∗) be feasible for (P), (D) respectively. The following are equivalent: 1. x∗is an optimal solution to (P) and (y∗, s∗) is an optimal solution to (D). 2. (s∗)Tx∗= 0. 3. x∗ js∗ j = 0, ∀j = 1, . . . , n. 4. If s∗ j > 0 then x∗ j = 0. Proof: Suppose (1) holds, then, by strong duality, cTx∗= bTy∗. Since c = ATy∗+ s∗and Ax∗= b, we get that (y∗)TAx∗+ (s∗)Tx∗= (x∗)TATy∗, and thus, (s∗)Tx∗= 0 (i.e (2) holds). It follows, since x∗ j, s∗ j ≥0, that x∗ js∗ j = 0, ∀j = 1, . . . , n (i.e. (3) holds). Hence, if s∗ j > 0 then x∗ j = 0, ∀j = 1, . . . , n (i.e. (4) holds). The converse also holds, and thus the proof is complete. ✷ In the example of section 9, the complementary slackness equations corresponding to the primal solution x = (3, 2, 0)T would be: y1 + 2y2 = 1 y1 + y2 = 2 Note that this implies that y1 = 3 and y2 = −1. Since this solution satisﬁes the other constraint of the dual, y is dual feasible, proving that x is an optimum solution to the primal (and therefore y is an optimum solution to the dual). 11 Size of a Linear Program 11.1 Size of the Input If we want to solve a Linear Program in polynomial time, we need to know what would that mean, i.e. what would the size of the input be. To this end we introduce two notions of the size of the input with respect to which the algorithm we present will run in polynomial time. The ﬁrst measure of the input size will be the size of a LP, but we will introduce a new measure L of a LP that will be easier to work with. Moreover, we have that L ≤size(LP), so that any algorithm running in time polynomial in L will also run in time polynomial in size(LP). LP-18 Let’s consider the linear program of the form: min cTx s.t. Ax = b x ≥0 where we are given as inputs the coeﬃcients of A (an m × n matrix), b (an m × 1 vector), and c (an n × 1 vector), whith rationial entries. We can further assume, without loss of generality, that the given coeﬃcients are all integers, since any LP with rational coeﬃcients can be easily transformed into an equivalent one with integer coeﬃcients (just multiply everything by l.c.d.). In the rest of these notes, we assume that A, b, c have integer coeﬃcients. For any integer n, we deﬁne its size as follows: size(n) △ = 1 + ⌈log2(\|n\| + 1)⌉ where the ﬁrst 1 stands for the fact that we need one bit to store the sign of n, size(n) represents the number of bits needed to encode n in binary. Analogously, we deﬁne the size of a p × 1 vector d, and of a p × l matrix M as follows: size(v) △ = Pp i=1 size(vi) size(M) △ = Pp i=1 Pl j=1 size(mij) We are then ready to talk about the size of a LP. Deﬁnition 6 (Size of a linear program) size(LP) △ = size(A) + size(b) + size(c). A more convenient deﬁnition of the size of a linear program is given next. Deﬁnition 7 L △ = size(detmax) + size(bmax) + size(cmax) + m + n where detmax △ = max A′ (\| det(A′)\|) bmax △ = max i (\|bi\|) cmax △ = max j (\|cj\|) and A′ is any square submatrix of A. LP-19 Proposition 13 L < size(LP), ∀A, b, c. Before proving this result, we ﬁrst need the following lemma: Lemma 14 1. If n ∈Z then \|n\| ≤2size(n)−1 −1. 2. If v ∈Zn then ∥v∥≤∥v∥1 ≤2size(v)−n −1. 3. If A ∈Zn×n then \|det(A)\| ≤2size(A)−n2 −1. Proof: 1. By deﬁnition. 2. 1+∥v∥≤1+∥v∥1 = 1+ n X i=1 \|vi\| ≤ n Y i=1 (1+\|vi\|) ≤ n Y i=1 2size(vi)−1 = 2size(v)−n where we have used 1. 3. Let a1, . . . , an be the columns of A. Since \|det(A)\| represents the volume of the parallelepiped spanned by a1, . . . , an, we have \|det(A)\| ≤ n Y i=1 ∥ai∥. Hence, by 2, 1 + \|det(A)\| ≤1 + n Y i=1 ∥ai∥≤ n Y i=1 (1 + ∥ai∥) ≤ n Y i=1 2size(ai)−n = 2size(A)−n2. ✷ We now prove Proposition 13. Proof: If B is a square submatrix of A then, by deﬁnition, size(B) ≤size(A). Moreover, by lemma 14, 1 + \|det(B)\| ≤2size(B)−1. Hence, ⌈log(1 + \|det(B)\|)⌉≤size(B) −1 < size(B) ≤size(A). (1) Let v ∈Zp. Then size(v) ≥size(maxj \|vj\|) + p −1 = ⌈log(1 + maxj \|vj\|)⌉+ p. Hence, size(b) + size(c) ≥⌈log(1 + max j \|cj\|)⌉+ ⌈log(1 + max i \|bi\|)⌉+ m + n. (2) Combining equations (1) and (2), we obtain the desired result. ✷ Remark 1 detmax ∗bmax ∗cmax ∗2m+n < 2L, since for any integer n, 2size(n) > \|n\|. In what follows we will work with L as the size of the input to our algorithm. LP-20 11.2 Size of the Output In order to even hope to solve a linear program in polynomial time, we better make sure that the solution is representable in size polynomial in L. We know already that if the LP is feasible, there is at least one vertex which is an optimal solution. Thus, when ﬁnding an optimal solution to the LP, it makes sense to restrict our attention to vertices only. The following theorem makes sure that vertices have a compact representation. Theorem 15 Let x be a vertex of the polyhedron deﬁned by Ax = b, x ≥0. Then, xT = p1 q p2 q . . . pn q ! , where pi (i = 1, . . . , n), q ∈N, and 0 ≤pi < 2L 1 ≤q < 2L. Proof: Since x is a basic feasible solution, ∃a basis B such that xB = A−1 B b and xN = 0. Thus, we can set pj = 0, ∀j ∈N, and focus our attention on the xj’s such that j ∈B. We know by linear algebra that xB = A−1 B b = 1 det(AB)cof(AB)b where cof(AB) is the cofactor matrix of AB. Every entry of AB consists of a deter-minant of some submatrix of A. Let q = \|det(AB)\|, then q is an integer since AB has integer components, q ≥1 since AB is invertible, and q ≤detmax < 2L. Finally, note that pB = qxB = \|cof(AB)b\|, thus pi ≤ Pm j=1 \|cof(AB)ij\|\|bj\| ≤m detmax bmax < 2L. ✷ 12 Complexity of linear programming In this section, we show that linear programming is in NP∩co-NP. This will follow from duality and the estimates on the size of any vertex given in the previous section. Let us deﬁne the following decision problem: Deﬁnition 8 (LP) Input: Integral A, b, c, and a rational number λ, Question: Is min{cTx : Ax = b, x ≥0} ≤λ? LP-21 Theorem 16 LP ∈NP ∩co-NP Proof: First, we prove that LP ∈NP. If the linear program is feasible and bounded, the “certiﬁcate” for veriﬁcation of instances for which min{cTx : Ax = b, x ≥0} ≤λ is a vertex x′ of {Ax = b, x ≥0} s.t. cTx′ ≤λ. This vertex x′ always exists since by assumption the minimum is ﬁnite. Given x′, it is easy to check in polynomial time whether Ax′ = b and x′ ≥0. We also need to show that the size of such a certiﬁcate is polynomially bounded by the size of the input. This was shown in section 11.2. If the linear program is feasible and unbounded, then, by strong duality, the dual is infeasible. Using Farkas’ lemma on the dual, we obtain the existence of ˜ x: A˜ x = 0, ˜ x ≥0 and cT ˜ x = −1 < 0. Our certiﬁcate in this case consists of both a vertex of {Ax = b, x ≥0} (to show feasiblity) and a vertex of {Ax = 0, x ≥0, cTx = −1} (to show unboundedness if feasible). By choosing a vertex x′ of {Ax = 0, x ≥0, cTx = −1}, we insure that x′ has polynomial size (again, see Section 11.2). This proves that LP ∈NP. (Notice that when the linear program is infeasible, the answer to LP is “no”, but we are not responsible to oﬀer such an answer in order to show LP ∈NP). Secondly, we show that LP ∈co-NP, i.e. LP ∈NP, where LP is deﬁned as: Input: A, b, c, and a rational number λ, Question: Is min{cTx : Ax = b, x ≥0} > λ? If {x : Ax = b, x ≥0} is nonempty, we can use strong duality to show that LP is indeed equivalent to: Input: A, b, c, and a rational number λ, Question: Is max{bTy : ATy ≤c} > λ? which is also in NP, for the same reason as LP is. If the primal is infeasible, by Farkas’ lemma we know the existence of a y s.t. ATy ≥0 and bTy = −1 < 0. This completes the proof of the theorem. ✷ 13 Solving a Liner Program in Polynomial Time The ﬁrst polynomial-time algorithm for linear programming is the so-called ellipsoid algorithm which was proposed by Khachian in 1979 . The ellipsoid algorithm was in fact ﬁrst developed for convex programming (of which linear programming is a special case) in a series of papers by the russian mathematicians A.Ju. Levin and, D.B. Judin and A.S. Nemirovskii, and is related to work of N.Z. Shor. Though of polynomial running time, the algorithm is impractical for linear programming. Nevertheless it has extensive theoretical applications in combinatorial optimization. For example, the stable set problem on the so-called perfect graphs can be solved in polynomial time using the ellipsoid algorithm. This is however a non-trivial non-combinatorial algorithm. LP-22 In 1984, Karmarkar presented another polynomial-time algorithm for linear pro-gramming. His algorithm avoids the combinatorial complexity (inherent in the sim-plex algorithm) of the vertices, edges and faces of the polyhedron by staying well inside the polyhedron (see Figure 13). His algorithm lead to many other algorithms for linear programming based on similar ideas. These algorithms are known as interior point methods. Figure 6: Exploring the interior of a convex body. It still remains an open question whether there exists a strongly polynomial algo-rithm for linear programming, i.e. an algorithm whose running time depends on m and n and not on the size of any of the entries of A, b or c. In the rest of these notes, we discuss an interior-point method for linear program-ming and show its polynomiality. High-level description of an interior-point algorithm: 1. If x (current solution) is close to the boundary, then map the polyhedron onto another one s.t. x is well in the interior of the new polyhedron (see Figure 7). 2. Make a step in the transformed space. 3. Repeat (a) and(b) until we are close enough to an optimal solution. Before we give description of the algorithm we give a theorem, the corollary of which will be a key tool used in determinig when we have reached an optimal solution. LP-23 Theorem 17 Let x1, x2 be vertices of Ax = b, x ≥0. If cTx1 ̸= cTx2 then \|cTx1 −cTx2\| > 2−2L. Proof: By Theorem 15, ∃qi, q2, such that 1 ≤q1, q2 < 2L, and q1x1, q2x2 ∈Nn. Further-more, \|cTx1 −cTx2\| = q1cTx1 q1 −q2cTx2 q2 = q1q2(cTx1 −cTx2) q1q2 ≥ 1 q1q2 since cTx1 −cTx2 ̸= 0, q1, q2 ≥1 > 1 2L2L = 2−2L since q1, q2 < 2L. ✷ Corollary 18 Assume z = min{cTx : Ax = b, x ≥0 \| {z } polyhedron P }. Assume x is feasible to P, and such that cTx ≤z + 2−2L. Then, any vertex x′ such that cTx′ ≤cTx is an optimal solution of the LP. Proof: Suppose x′ is not optimal. Then, ∃x∗, an optimal vertex, such that cTx∗= z. Since x′ is not optimal, cTx′ ̸= cTx∗, and by Theorem 17 ⇒ cTx′ −cTx∗ > 2−2L ⇒ cTx′ > cTx∗+ 2−2L = Z + 2−2L ≥ cTx by deﬁnition of x ≥ cTx′ by deﬁnition of x′ ⇒ cTx′ > cTx′, a contradiction. ✷ What this corollary tells us is that we do not need to be very precise when choosing an optimal vertex. More precisely we only need to compute the objective function with error less than 2−2L. If we ﬁnd a vertex that is within that margin of error, then it will be optimal. LP-24 x P x’ P’ Figure 7: A centering mapping. If x is close to the boundary, we map the polyhedron P onto another one P ′, s.t. the image x′ of x is closer to the center of P ′. 13.1 Ye’s Interior Point Algorithm In the rest of these notes we present Ye’s interior point algorithm for linear pro-gramming. Ye’s algorithm (among several others) achieves the best known asymptotic running time in the literature, and our presentation incorporates some simpliﬁcations made by Freund . We are going to consider the following linear programming problem: (P)      minimize Z = cTx subject to Ax = b, x ≥0 and its dual (D)      maximize W = bTy subject to ATy + s = c, s ≥0. The algorithm is primal-dual, meaning that it simultaneously solves both the primal and dual problems. It keeps track of a primal solution x and a vector of dual slacks s (i.e. ∃y : ATy = c −s) such that x > 0 and s > 0. The basic idea of this algorithm is to stay away from the boundaries of the polyhedron (the hyperplanes xj ≥0 and sj ≥0, j = 1, 2, . . . , n) while approaching optimality. In other words, we want to make the duality gap cTx −bTy = xTs > 0 LP-25 very small but stay away from the boundaries. Two tools will be used to achieve this goal in polynomial time. Tool 1: Scaling (see Figure 7) Scaling is a crucial ingredient in interior point methods. The two types of scaling commonly used are projective scaling (the one used by Karmarkar) and aﬃne scaling (the one we are going to use). Suppose the current iterate is x > 0 and s > 0, where x = (x1, x2, . . . , xn)T, then the aﬃne scaling maps x to x′ as follows. x =           x1 x2 . . . xn           − →x′ =            x1 x1 x2 x2 . . . xn xn            . Notice this transformation maps x to e = (1, . . . , 1)T. We can express the scaling transformation in matrix form as x′ = X −1x or x = Xx′, where X =          x1 0 0 . . . 0 0 x2 0 . . . 0 . . . . . . . . . 0 0 . . . xn−1 0 0 0 . . . 0 xn          . Using matrix notation we can rewrite the linear program (P) in terms of the trans-formed variables as: minimize Z = cTXx′ subject to AXx′ = b, x′ ≥0. If we deﬁne c = Xc (note that X = X T) and A = AX we can get a linear program in the original form as follows. minimize Z = cTx′ subject to Ax′ = b, x′ ≥0. We can also write the dual problem (D) as: maximize W = bTy subject to (AX)Ty + Xs = c, Xs ≥0 LP-26 or, equivalently, maximize W = bTy subject to A Ty + s′ = c, s′ ≥0 where s′ = Xs, i.e. s′ =         s1x1 s2x2 . . snxn         . One can easily see that xjsj = x′ js′ j ∀j ∈{1, . . . , n} (3) and, therefore, the duality gap xTs = P j xjsj remains unchanged under aﬃne scaling. As a consequence, we will see later that one can always work equivalently in the transformed space. Tool 2: Potential Function Our potential function is designed to measure how small the duality gap is and how far the current iterate is away from the boundaries. In fact we are going to use the following “logarithmic barrier function”. Deﬁnition 9 (Potential Function, G(x, s)) G(x, s) △ = q ln(xTs) − n X j=1 ln(xjsj), for some q, where q is a parameter that must be chosen appropriately. Note that the ﬁrst term goes to −∞as the duality gap tends to 0, and the second term goes to +∞as xi →0 or si →0 for some i. Two questions arise immediately concerning this potential function. Question 1: How do we choose q? Lemma 19 Let x, s > 0 be vectors in Rn×1. Then n ln xTs − n X j=1 ln xjsj ≥n ln n. LP-27 Proof: Given any n positive numbers t1, . . . , tn, we know that their geometric mean does not exceed their arithmetic mean, i.e.   n Y j=1 tj   1/n ≤1 n   n X j=1 tj   . Taking the logarithms of both sides we have 1 n   n X j=1 ln tj  ≤ln   n X j=1 tj  −ln n. Rearranging this inequality we get n ln   n X j=1 tj  −   n X j=1 ln tj  ≥n ln n. (In fact the last inequality can be derived directly from the concavity of the logarith-mic function). The lemma follows if we set tj = xjsj. ✷ Since our objective is that G →−∞as xTs →0 (since our primary goal is to get close to optimality), according to Lemma 19, we should choose some q > n (notice that ln xTs →−∞as xTs →0) . In particular, if we choose q = n + 1, the algorithm will terminate after O(nL) iterations. In fact we are going to set q = n + √n, which gives us the smallest number — O(√nL) — of iterations by this method. Question 2: When can we stop? Suppose that xTs ≤2−2L, then cTx −Z ≤cTx −bTy = xTs ≤2−2L, where Z is the optimum value to the primal problem. From Corollary 18, the following claim follows immediately. Claim 20 If xTs ≤2−2L, then any vertex x∗satisfying cTx∗≤cTx is optimal. In order to ﬁnd x∗from x, two methods can be used. One is based on purely algebraic techniques (but is a bit cumbersome to describe), while the other (the cleanest one in literature) is based upon basis reduction for lattices. We shall not elaborate on this topic, although we’ll get back to this issue when discussing basis reduction in lattices. Lemma 21 Let x, s be feasible primal-dual vectors such that G(x, s) ≤−k√nL for some constant k. Then xTs < e−kL. LP-28 Proof: By the deﬁnition of G(x, s) and the previous theorem we have: −k√nL ≥ G(x, s) = (n + √n) ln xTs − n X j=1 ln xjsj ≥ √n ln xTs + n ln n. Rearranging we obtain ln xTs ≤ −kL −√n ln n < −kL. Therefore xTs < e−kL. ✷ The previous lemma and claim tell us that we can stop whenever G(x, s) ≤ −2√nL. In practice, the algorithm can terminate even earlier, so it is a good idea to check from time to time if we can get the optimal solution right away. Please notice that according to Equation (3) the aﬃne transformation does not change the value of the potential function. Hence we can work either in the original space or in the transformed space when we talk about the potential function. 14 Description of Ye’s Interior Point Algorithm Initialization: Set i = 0. Choose x0 > 0, s0 > 0, and y0 such that Ax0 = b, ATy0 + s0 = c and G(x0, s0) = O(√nL). (Details are not covered in class but can be found in the appendix. The general idea is as follows. By augmenting the linear program with additional variables, it is easy to obtain a feasible solution. Moreover, by carefully choosing the augmented linear program, it is possible to have feasible primal and dual solutions x and s such that all xj’s and sj’s are large (say 2L). This can be seen to result in a potential of O(√nL).) Iteration: while G(xi, si) > −2√nL do      either a primal step (changing xi only) or a dual step (changing si only) ) to get (xi+1, si+1) i := i + 1 The iterative step is as follows. Aﬃne scaling maps (xi, si) to (e, s′). In this transformed space, the point is far away from the boundaries. Either a dual or LP-29 a a g g−d d 1 2 Null space of A {x: Ax = 0} Figure 8: Null space of A and gradient direction g. primal step occurs, giving (˜ x, ˜ s) and reducing the potential function. The point is then mapped back to the original space, resulting in (xi+1, si+1). Next, we are going to describe precisely how the primal or dual step is made such that G(xi+1, si+1) −G(xi, si) ≤−7 120 < 0 holds for either a primal or dual step, yielding an O(√nL) total number of iterations. In order to ﬁnd the new point (˜ x, ˜ s) given the current iterate (e, s′) (remember we are working in the transformed space), we compute the gradient of the potential function. This is the direction along which the value of the potential function changes at the highest rate. Let g denote the gradient. Recall that (e, s′) is the map of the current iterate, we obtain g = ∇xG(x, s)\|(e,s′) = q xTss −     1/x1 . . . 1/xn     (e,s′) = q eTs′s′ −e (4) We would like to maximize the change in G, so we would like to move in the direction of −g. However, we must insure the new point is still feasible (i.e. A˜ x = b). Let d be the projection of g onto the null space {x : Ax = 0} of A. Thus, we will move in the direction of −d. Claim 22 d = (I −A(A A T)−1A)g. LP-30 Proof: Since g −d is orthogonal to the null space of A, it must be the combination of some row vectors of A. Hence we have ( Ad = 0 ∃w, s.t. A Tw = g −d. This implies    A Tw = g −d (A A T)w = Ag (normal equations). Solving the normal equations, we get w = (A A T)−1Ag and d = g −A T(A A T)−1Ag = (I −A T(A A T)−1A)g. ✷ A potential problem arises if g is nearly perpendicular to the null space of A. In this case, \|\|d\|\| will be very small, and each primal step will not reduce the potential greatly. Instead, we will perform a dual step. In particular, if \|\|d\|\| = \|\|d\|\|2 = √ dTd ≥0.4, we make a primal step as follows. ˜ x = e − 1 4\|\|d\|\|d ˜ s = s′. Claim 23 ˜ x > 0. Proof: ˜ xj = 1 −1 4 dj \|\|d\|\| ≥3 4 > 0. ✷ This claim insures that the new iterate is still an interior point. For the similar reason, we will see that ˜ s > 0 when we make a dual step. Proposition 24 When a primal step is made, G(˜ x, ˜ s) −G(e, s′) ≤−7 120. If \|\|d\|\| < 0.4, we make a dual step. Again, we calculate the gradient h = ∇sG(x, s)\|(e,s′) = q eTs′e −     1/s′ 1 . . . 1/s′ n     (5) LP-31 Notice that hj = gj/sj, thus h and g can be seen to be approximately in the same direction. Suppose the current dual feasible solution is y′, s′ such that A Ty′ + s′ = c. Again, we restrict the solution to be feasible, so A Ty + ˜ s = c ˜ s −s′ = A T(y′ −y) Thus, in the dual space, we move perpendicular to the null space and in the direction of −(g −d). Thus, we have ˜ s = s′ −(g −d)µ For any µ, ∃y A Ty + ˜ s = c So, we can choose µ = eT s′ q and get A T(y′ + µw) + ˜ s = c. Therefore, ˜ s = s′ −eTs′ q (g −d) = s′ −eTs′ q (q s′ eTs′ −e −d) = eTs′ q (d + e) ˜ x = x′ = e. One can show that ˜ s > 0 as we did in Claim 23. So such move is legal. Proposition 25 When a dual step is made, G(˜ x, ˜ s) −G(e, s′) ≤−1 6 According to these two propositions, the potential function decreases by a con-stant amount at each step. So if we start from an initial interior point (x0, s0) with G(x0, s0) = O(√nL), then after O(√nL) iterations we will obtain another interior point (xj, sj) with G(xj, sj) ≤−k√nL. From Lemma 21, we know that the duality gap (xj)Tsj satisﬁes (xj)Tsj ≤2−kL, and the algorithm terminates by that time. Moreover, each iteration requires O(n3) operations. Indeed, in each iteration, the only non-trivial task is the computation of the projected gradient d. This can be done by solving the linear system ( ¯ A ¯ AT)w = ¯ Ag in O(n3) time using Gaussian elimination. Therefore, the overall time complexity of this algorithm is O(n3.5L). By using approximate solutions to the linear systems, we can obtain O(n2.5) time per iteration, and total time O(n3L). LP-32 15 Analysis of the Potential Function In this section, we prove the two propositions of the previous section, which concludes the analysis of Ye’s algorithm. Proof of Proposition 24: G(˜ x, ˜ s) −G(e, s′) = G(e − 1 4\|\|d\|\|d, ˜ s) −G(e, s′) = q ln eTs′ −dTs′ 4\|\|d\|\| ! − n X j=1 ln 1 − dj 4\|\|d\|\| ! − n X j=1 ln s′ j − −q ln eTs′ + n X j=1 ln 1 + n X j=1 ln s′ j = q ln 1 − dTs′ 4\|\|d\|\|eTs′ ! − n X j=1 ln 1 − dj 4\|\|d\|\| ! . Using the relation −x − x2 2(1 −a) ≤ln(1 −x) ≤−x (6) which holds for \|x\| ≤a < 1, we get: G(˜ x, ˜ s) −G(e, s′) ≤ −q dTs′ 4\|\|d\|\|eTs′ + n X j=1 dj 4\|\|d\|\| + n X j=1 d2 j 16\|\|d\|\|2 2(3/4) for a = 1/4 = −q dTs′ 4\|\|d\|\|eTs′ + eTd 4\|\|d\|\| + 1 24 = 1 4\|\|d\|\|(e − q eTs′s′)Td + 1 24 = 1 4\|\|d\|\|(−g)Td + 1 24 = −\|\|d\|\|2 4\|\|d\|\| + 1 24 = −\|\|d\|\| 4 + 1 24 ≤ −1 10 + 1 24 = −7 120. Note that gTd = \|\|d\|\|2, since d is the projection of g. (This is where we use the fact that d is the projected gradient!) ✷ Before proving Proposition 25, we need the following lemma. LP-33 Lemma 26 n X j=1 ln(˜ sj) −n ln(eT ˜ s n ) ≥−2 15 . Proof: Using the equality ˜ s = ∆ q (e + d) and Equation 6, which holds for \|x\| ≤a < 1, we see that Pn j=1 ln(˜ sj) −n ln( eT ˜ s n ) = Pn j=1 ln( ∆ q (1 + dj)) −n ln( ∆ q (1 + eT d n )) ≥ Pn j=1(dj − dj2 2(3/5)) −neT d n ≥−\|\|d\|\|2 6/5 ≥−2 15 ✷ Proof of Proposition 25: Using Lemma 26 and the inequality n X j=1 ln(sj) ≤n ln(eTs n ), which follows from the concavity of the logarithm function, we have G(e, ˜ s) −G(e, s′) = q ln( eT ˜ s eT s′) −Pn j=1 ln( ˜ sj) + Pn j=1 ln(s′ j) ≤q ln( eT ˜ s eT s′) + 2 15 −n ln( eT ˜ s n ) + n ln( eT s′ n ) = 2 15 + √n ln( eT ˜ s eT s′) On the other hand, eT ˜ s = ∆ q (n + eTd) and recall that ∆= eTs′, eT ˜ s eTs′ = 1 q(n + eTd) ≤ 1 n + √n(n + 0.4√n), since, by Cauchy-Schwartz inequality, \|eTd\| ≤\|\|e\|\| \|\|d\|\| = √n\|\|d\|\|. Combining the above inequalities yields G(e, ˜ s) −G(e, s′) ≤ 2 15 + √n ln(1 −0.6√n n+√n) ≤ 2 15 − 0.6n n+√n ≤ 2 15 −3 10 = −1 6 LP-34 since n + √n ≤2n. ✷ This completes the analysis of Ye’s algorithm. 16 Bit Complexity Throughout the presentation of the algorithm, we assumed that all operations can be performed exactly. This is a fairly unrealistic assumption. For example, notice that ∥d∥might be irrational since it involves a square root. However, none of the thresholds we set were crucial. We could for example test whether ∥d∥≥0.4 or ∥d∥≤0.399. To test this, we need to compute only a few bits of ∥d∥. Also, if we perform a primal step (i.e. ∥d∥≥0.4) and compute the ﬁrst few bits of ∥d∥so that the resulting approximation ∥d∥ap satisﬁes (4/5)∥d∥≤∥d∥ap ≤∥d∥then if we go through the analysis of the primal step performed in Proposition 1, we obtain that the reduction in the potential function is at least 19/352 instead of the previous 7/120. Hence, by rounding ∥d∥we can still maintain a constant decrease in the potential function. Another potential problem is when using Gaussian elimination to compute the pro-jected gradient. We mentioned that Gaussian elimination requires O(n3) arithmetic operations but we need to show that, during the computation, the numbers involved have polynomial size. For that purpose, consider the use of Gaussian elimination to solve a system Ax = b where A = A(1) =        a(1) 11 a(1) 12 . . . a(1) 1n a(1) 21 a(1) 22 . . . a(1) 2n . . . . . . ... . . . a(1) m1 a(1) m2 . . . a(1) mn        . Assume that a11 ̸= 0 (otherwise, we can permute rows or columns). In the ﬁrst iteration, we substract a(1) i1 /a(1) 11 times the ﬁrst row from row i where i = 2, . . . , m, resulting in the following matrix: A(2) =        a(2) 11 a(2) 12 . . . a(2) 1n 0 a(2) 22 . . . a(2) 2n . . . . . . ... . . . 0 a(2) m2 . . . a(2) mn        . In general, A(i+1) is obtained by subtracting a(i) ji /a(i) ii times row i from row j of A(i) for j = i + 1, . . . , m. Theorem 27 For all i ≤j, k, a(i) jk can be written in the form det(B)/ det(C) where B and C are some submatrices of A. LP-35 Proof: Let Bi denote the i×i submatrix of A(i) consisting of the ﬁrst i entries of the ﬁrst i rows. Let B(i) jk denote the i × i submatrix of A(i) consisting of the ﬁrst i −1 rows and row j, and the ﬁrst i −1 columns and column k. Since Bi and B(i) jk are upper triangular matrices, their determinants are the products of the entries along the main diagonal and, as a result, we have: a(i) ii = det(Bi) det(Bi−1) and a(i) jk = det(B(i) jk ) det(Bi−1). Moreover, remember that row operations do not aﬀect the determinants and, hence, the determinants of B(i) jk and Bi−1 are also determinants of submatrices of the original matrix A. ✷ Using the fact that the size of the determinant of any submatrix of A is at most the size of the matrix A, we obtain that all numbers occuring during Gaussian elimination require only O(L) bits. Finally, we need to round the current iterates x, y and s to O(L) bits. Otherwise, these vectors would require a constantly increasing number of bits as we iterate. By rounding up x and s, we insure that these vectors are still strictly positive. It is fairly easy to check that this rounding does not change the potential function by a signiﬁcant amount and so the analysis of the algorithm is still valid. Notice that now the primal and dual constraints might be slightly violated but this can be taken care of in the rounding step. A Transformation for the Interior Point Algorithm In this appendix, we show how a pair of dual linear programs Min cTx Max bTy (P) s.t. Ax = b (D) s.t. ATy + s = c x ≥0 s ≥ 0 can be transformed so that we know a strictly feasible primal solution x0 and a strictly feasible vector of dual slacks s0 such that G(x0; s0) = O(√nL) where G(x; s) = q ln(xTs) − n X j=1 ln(xjsj) and q = n + √n. LP-36 Consider the pair of dual linear programs: Min cTx + kcxn+1 (P ′) s.t. Ax + (b −22LAe)xn+1 = b (24Le −c)Tx + 24Lxn+2 = kb x ≥0 xn+1 ≥0 xn+2 ≥0 and Min bTy + kbym+1 (D′) s.t. ATy + (24Le −c)ym+1 + s = c (b −22LAe)Ty + sn+1 = kc 24Lym+1 + sn+2 = 0 s, sn+1, sn+2 ≥ 0 where kb = 26L(n + 1) −22LcTe is chosen in such a way that x′ = (x, xn+1, xn+2) = (22Le, 1, 22L) is a (strict) feasible solution to (P ′) and kc = 26L. Notice that (y′, s′) = (y, ym+1, s, sn+1, sn+2) = (0, −1, 24Le, kc, 24L) is a feasible solution to (D′) with s′ > 0. x′ and (y′, s′) serve as our initial feasible solutions. We have to show: 1. G(x′; s′) = O( √ n′L) where n′ = n + 2, 2. the pair (P ′) −(D′) is equivalent to (P) −(D), 3. the input size L′ for (P ′) as deﬁned in the lecture notes does not increase too much. The proofs of these statements are simple but heavily use the deﬁnition of L and the fact that vertices have all components bounded by 2L. We ﬁrst show 1. Notice ﬁrst that x′ js′ j = 26L for all j, implying that G(x′; s′) = (n′ + √ n′) ln(x′Ts′) − n′ X j=1 ln(x′ js′ j) = (n′ + √ n′) ln(26Ln′) −n′ ln(26L) = √ n′ ln(26L) + (n′ + √ n′) ln(n′) = O( √ n′L) In order to show that (P ′) −(D′) are equivalent to (P) −(D), we consider an optimal solution x∗to (P) and an optimal solution (y∗, s∗) to (D) (the case where (P) or (D) is infeasible is considered in the problem set). Without loss of generality, we can assume that x∗and (y∗, s∗) are vertices of the corresponding polyhedra. In particular, this means that x∗ j, \|y∗ j\|, s∗ j < 2L. LP-37 Proposition 28 Let x′ = (x∗, 0, (kb−(24Le−c)Tx∗)/24L) and let (y′, s′) = (y∗, 0, , s∗, kc− (b −22LAe)Ty∗, 0). Then 1. x′ is a feasible solution to (P ′) with x′ n+2 > 0, 2. (y′, s′) is a feasible solution to (D′) with s′ n+1 > 0, 3. x′ and (y′, s′) satisfy complementary slackness, i.e. they constitute a pair of optimal solutions for (P ′) −(D′). Proof: To show that x′ is a feasible solution to (P ′) with x′ n+2 > 0, we only need to show that kb −(24Le −c)Tx∗> 0 (the reader can easily verify that x′ satisfy all the equalities deﬁning the feasible region of (P ′)). This follows from the fact that (24Le −c)Tx∗≤n(24L + 2L)2L = n(25L + 22L) < n26L and kb = 26L(n + 1) −22LcTe ≥26L(n + 1) −22Ln max j \|cj\| ≥26Ln + 26L −23L > n26L where we have used the deﬁnition of L and the fact that vertices have all their entries bounded by 2L. To show that (y′, s′) is a feasible solution to (D′) with s′ n+1 > 0, we only need to show that kc −(b −22LAe)Ty∗> 0. This is true since (b −22LAe)Ty∗ ≤ bTy∗−22LeTATy∗ ≤ m max i \|bi\|2L + 22Lnm max i,j \|aij\|2L = 22L + 24L < 26L = kc. x′ and (y′, s′) satisfy complementary slackness since • x∗Ts∗= 0 by optimality of x∗and (y∗, s∗) for (P) and (D) • x′ n+1s′ n+1 = 0 and • x′ n+2s′ n+2 = 0. ✷ This proposition shows that, from an optimal solution to (P)−(D), we can easily construct an optimal solution to (P ′) −(D′) of the same cost. Since this solution has s′ n+1 > 0, any optimal solution ˆ x to (P ′) must have ˆ xn+1 = 0. Moreover, since x′ n+2 > 0, any optimal solution (ˆ y, ˆ s) to (D′) must satisfy ˆ sn+2 = 0 and, as a result, ˆ ym+1 = 0. Hence, from any optimal solution to (P ′) −(D′), we can easily deduce an optimal solution to (P) −(D). This shows the equivalence between (P) −(D) and (P ′) −(D′). By some tedious but straightforward calculations, it is possible to show that L′ (corresponding to (P ′)−(D′)) is at most 24L. In other words, (P)−(D) and (P ′)−(D′) have equivalent sizes. LP-38 References V. Chvatal. Linear Programming. W.H. Freeman and Company, 1983. G. Dantzig. Maximization of a linear function of variables subject to linear in-equalities. In T. Koopmans, editor, Activity Analysis of Production and Allocation, pages 339–347. John Wiley & Sons, Inc., 1951. R. M. Freund. Polynomial-time algorithms for linear programming based only on primal scaling and project gradients of a potential function. Mathematical Programming, 51:203–222, 1991. D. Goldfarb and M. Todd. Linear programming. In Handbook in Operations Re-search and Management Science, volume 1, pages 73–170. Elsevier Science Pub-lishers B.V., 1989. C. C. Gonzaga. Path-following methods for linear programming. SIAM Review, 34:167–224, 1992. L. Khachian. A polynomial algorithm for linear programming. Doklady Akad. Nauk USSR, 244(5):1093–1096, 1979. K. Murty. Linear Programming. John Wiley & Sons, 1983. A. Schrijver. Theory of Linear and Integer Programming. John Wiley & Sons, 1986. Y. Ye. An O(n3L) potential reduction algorithm for linear programming. Math-ematical Programming, 50:239–258, 1991. LP-39 LP-40
1277	https://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Field-and-the-Movement-of-Charge	Physics Classroom - Home Electric Field and the Movement of Charge Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers. Perhaps one of the most useful yet taken-for-granted accomplishments of the recent centuries is the development of electric circuits. The flow of charge through wires allows us to cook our food, light our homes, air-condition our work and living space, entertain us with movies and music and even allows us to drive to work or school safely. In this unit of The Physics Classroom, we will explore the reasons for why charge flows through wires of electric circuits and the variables that affect the rate at which it flows. The means by which moving charge delivers electrical energy to appliances in order to operate them will be discussed in detail. One of the fundamental principles that must be understood in order to grasp electric circuits pertains to the concept of how an electric field can influence charge within a circuit as it moves from one location to another. The concept of electric field was first introduced in the unit on Static Electricity. In that unit, electric force was described as a non-contact force. A charged balloon can have an attractive effect upon an oppositely charged balloon even when they are not in contact. The electric force acts over the distance separating the two objects. Electric force is an action-at-a-distance force. Action-at-a-distance forces are sometimes referred to as field forces. The concept of a field force is utilized by scientists to explain this rather unusual force phenomenon that occurs in the absence of physical contact. The space surrounding a charged object is affected by the presence of the charge; an electric field is established in that space. A charged object creates an electric field - an alteration of the space or field in the region that surrounds it. Other charges in that field would feel the unusual alteration of the space. Whether a charged object enters that space or not, the electric field exists. Space is altered by the presence of a charged object; other objects in that space experience the strange and mysterious qualities of the space. As another charged object enters the space and moves deeper and deeper into the field, the effect of the field becomes more and more noticeable. Electric field is a vector quantity whose direction is defined as the direction that a positive test charge would be pushed when placed in the field. Thus, the electric field direction about a positive source charge is always directed away from the positive source. And the electric field direction about a negative source charge is always directed toward the negative source. Electric Field, Work, and Potential Energy Electric fields are similar to gravitational fields - both involve action-at-a-distance forces. In the case of gravitational fields, the source of the field is a massive object and the action-at-a-distance forces are exerted upon other masses. When the concept of the force of gravity and energy was discussed in Unit 5 of the Physics Classroom, it was mentioned that the force of gravity is an internal or conservative force. When gravity does work upon an object to move it from a high location to a lower location, the object's total amount of mechanical energy is conserved. However, during the course of the falling motion, there was a loss of potential energy (and a gain of kinetic energy). When gravity does work upon an object to move it in the direction of the gravitational field, then the object loses potential energy. The potential energy originally stored within the object as a result of its vertical position is lost as the object moves under the influence of the gravitational field. On the other hand, energy would be required to move a massive object against its gravitational field. A stationary object would not naturally move against the field and gain potential energy. Energy in the form of work would have to be imparted to the object by an external force in order for it to gain this height and the corresponding potential energy. The important point to be made by this gravitational analogy is that work must be done by an external force to move an object against nature - from low potential energy to high potential energy. On the other hand, objects naturally move from high potential energy to low potential energy under the influence of the field force. It is simply natural for objects to move from high energy to low energy; but work is required to move an object from low energy to high energy. In a similar manner, to move a charge in an electric field against its natural direction of motion would require work. The exertion of work by an external force would in turn add potential energy to the object. The natural direction of motion of an object is from high energy to low energy; but work must be done to move the object against nature. On the other hand, work would not be required to move an object from a high potential energy location to a low potential energy location. When this principle is logically extended to the movement of charge within an electric field, the relationship between work, energy and the direction that a charge moves becomes more obvious. Consider the diagram above in which a positive source charge is creating an electric field and a positive test charge being moved against and with the field. In Diagram A, the positive test charge is being moved against the field from location A to location B. Moving the charge in this direction would be like going against nature. Thus, work would be required to move the object from location A to location B and the positive test charge would be gaining potential energy in the process. This would be analogous to moving a mass in the uphill direction; work would be required to cause such an increase in gravitational potential energy. In Diagram B, the positive test charge is being moved with the field from location B to location A. This motion would be natural and not require work from an external force. The positive test charge would be losing energy in moving from location B to location A. This would be analogous to a mass falling downward; it would occur naturally and be accompanied by a loss of gravitational potential energy. One can conclude from this discussion that the high energy location for a positive test charge is a location nearest the positive source charge; and the low energy location is furthest away. The above discussion pertained to moving a positive test charge within the electric field created by a positive source charge. Now we will consider the motion of the same positive test charge within the electric field created by a negative source charge. The same principle regarding work and potential energy will be used to identify the locations of high and low energy. In Diagram C, the positive test charge is moving from location A to location B in the direction of the electric field. This movement would be natural - like a mass falling towards Earth. Work would not be required to cause such a motion and it would be accompanied by a loss of potential energy. In Diagram D, the positive test charge is moving from location B to location A against the electric field. Work would be required to cause this motion; it would be analogous to raising a mass within Earth's gravitational field. Since energy is imparted to the test charge in the form of work, the positive test charge would be gaining potential energy as the result of the motion. One can conclude from this discussion that the low energy location for a positive test charge is a location nearest a negative source charge and the high energy location is a location furthest away from a negative source charge. As we begin to discuss circuits, we will apply these principles regarding work and potential energy to the movement of charge about a circuit. Just as we reasoned here, moving a positive test charge against the electric field will require work and result in a gain in potential energy. On the other hand, a positive test charge will naturally move in the direction of the field without the need for work being done on it; this movement will result in the loss of potential energy. Before making this application to electric circuits, we need to first explore the meaning of the concept electric potential.
1278	https://www.geeksforgeeks.org/dsa/prims-minimum-spanning-tree-mst-greedy-algo-5/	Prim’s Algorithm for Minimum Spanning Tree (MST) Last Updated : 29 Aug, 2025 Suggest changes 411 Likes Prim’s algorithm is a Greedy algorithm like Kruskal's algorithm. This algorithm always starts with a single node and moves through several adjacent nodes, in order to explore all of the connected edges along the way. The algorithm starts with an empty spanning tree. The idea is to maintain two sets of vertices. The first set contains the vertices already included in the MST, and the other set contains the vertices not yet included. At every step, it considers all the edges that connect the two sets and picks the minimum weight edge from these edges. After picking the edge, it moves the other endpoint of the edge to the set containing MST. A group of edges that connects two sets of vertices in a graph is called Articulation Points (or Cut Vertices) in graph theory. So, at every step of Prim’s algorithm, find a cut, pick the minimum weight edge from the cut, and include this vertex in MST Set (the set that contains already included vertices). Working of the Prim's Algorithm Step 1: Determine an arbitrary vertex as the starting vertex of the MST. We pick 0 in the below diagram. Step 2: Follow steps 3 to 5 till there are vertices that are not included in the MST (known as fringe vertex). Step 3: Find edges connecting any tree vertex with the fringe vertices. Step 4: Find the minimum among these edges. Step 5: Add the chosen edge to the MST. Since we consider only the edges that connect fringe vertices with the rest, we never get a cycle. Step 6: Return the MST and exit 1 / 12 How to implement Prim's Algorithm? Follow the given steps to utilize the Prim's Algorithm mentioned above for finding MST of a graph: Create a set mstSet that keeps track of vertices already included in MST. Assign a key value to all vertices in the input graph. Initialize all key values as INFINITE. Assign the key value as 0 for the first vertex so that it is picked first. While mstSet doesn't include all vertices Pick a vertex u that is not there in mstSetand has a minimum key value. Include u in the mstSet. Update the key value of all adjacent vertices of u. To update the key values, iterate through all adjacent vertices. For every adjacent vertex v, if the weight of edge u-v is less than the previous key value of v, update the key value as the weight of u-v. The idea of using key values is to pick the minimum weight edge from the cut. The key values are used only for vertices that are not yet included in MST, the key value for these vertices indicates the minimum weight edges connecting them to the set of vertices included in MST. Below is the implementation of the approach: Try it on GfG Practice C++ ```` // A C++ program for Prim's Minimum // Spanning Tree (MST) algorithm. The program is // for adjacency matrix representation of the graph include using namespace std; // A utility function to find the vertex with // minimum key value, from the set of vertices // not yet included in MST int minKey(vector &key, vector &mstSet) { // Initialize min value int min = INT_MAX, min_index; for (int v = 0; v < mstSet.size(); v++) if (mstSet[v] == false && key[v] < min) min = key[v], min_index = v; return min_index; } // A utility function to print the // constructed MST stored in parent[] void printMST(vector &parent, vector> &graph) { cout << "Edge \tWeight\n"; for (int i = 1; i < graph.size(); i++) cout << parent[i] << " - " << i << " \t" << graph[parent[i]][i] << " \n"; } // Function to construct and print MST for // a graph represented using adjacency // matrix representation void primMST(vector> &graph) { int V = graph.size(); // Array to store constructed MST vector<int> parent(V); // Key values used to pick minimum weight edge in cut vector<int> key(V); // To represent set of vertices included in MST vector<bool> mstSet(V); // Initialize all keys as INFINITE for (int i = 0; i < V; i++) key[i] = INT_MAX, mstSet[i] = false; // Always include first 1st vertex in MST. // Make key 0 so that this vertex is picked as first // vertex. key = 0; // First node is always root of MST parent = -1; // The MST will have V vertices for (int count = 0; count < V - 1; count++) { // Pick the minimum key vertex from the // set of vertices not yet included in MST int u = minKey(key, mstSet); // Add the picked vertex to the MST Set mstSet[u] = true; // Update key value and parent index of // the adjacent vertices of the picked vertex. // Consider only those vertices which are not // yet included in MST for (int v = 0; v < V; v++) // graph[u][v] is non zero only for adjacent // vertices of m mstSet[v] is false for vertices // not yet included in MST Update the key only // if graph[u][v] is smaller than key[v] if (graph[u][v] && mstSet[v] == false && graph[u][v] < key[v]) parent[v] = u, key[v] = graph[u][v]; } // Print the constructed MST printMST(parent, graph); } // Driver's code int main() { vector> graph = { { 0, 2, 0, 6, 0 }, { 2, 0, 3, 8, 5 }, { 0, 3, 0, 0, 7 }, { 6, 8, 0, 0, 9 }, { 0, 5, 7, 9, 0 } }; // Print the solution primMST(graph); return 0; } ```` // A C++ program for Prim's Minimum // A C++ program for Prim's Minimum // Spanning Tree (MST) algorithm. The program is // Spanning Tree (MST) algorithm. The program is // for adjacency matrix representation of the graph // for adjacency matrix representation of the graph ``` include #include ``` using namespace std; using namespace std // A utility function to find the vertex with // A utility function to find the vertex with // minimum key value, from the set of vertices // minimum key value, from the set of vertices // not yet included in MST // not yet included in MST int minKey(vector<int> &key, vector<bool> &mstSet) {int minKey vector< int>& key vector< bool>& mstSet ``` ``` // Initialize min value // Initialize min value int min = INT_MAX, min_index; int min = INT_MAX min_index for (int v = 0; v < mstSet.size(); v++) for int v = 0 v< mstSet size v ++ if (mstSet[v] == false && key[v] < min) if mstSet v == false&& key v< min min = key[v], min_index = v; min = key v min_index = v return min_index; return min_index } // A utility function to print the // A utility function to print the // constructed MST stored in parent[] // constructed MST stored in parent[] void printMST(vector<int> &parent, vector<vector<int>> &graph) {void printMST vector< int>& parent vector< vector< int>>& graph cout << "Edge \tWeight\n"; cout<< "Edge \tWeight\n" for (int i = 1; i < graph.size(); i++) for int i = 1 i< graph size i ++ cout << parent[i] << " - " << i << " \t" cout<< parent i<<" - "<< i<< " \t" << graph[parent[i]][i] << " \n";<< graph parent i i<< " \n" } // Function to construct and print MST for // Function to construct and print MST for // a graph represented using adjacency // a graph represented using adjacency // matrix representation // matrix representation void primMST(vector<vector<int>> &graph) {void primMST vector< vector< int>>& graph ``` ``` int V = graph.size(); int V = graph size ``` ``` // Array to store constructed MST // Array to store constructed MST vector<int> parent(V); vector< int> parent V // Key values used to pick minimum weight edge in cut // Key values used to pick minimum weight edge in cut vector<int> key(V); vector< int> key V // To represent set of vertices included in MST // To represent set of vertices included in MST vector<bool> mstSet(V); vector< bool> mstSet V // Initialize all keys as INFINITE // Initialize all keys as INFINITE for (int i = 0; i < V; i++) for int i = 0 i< V i ++ key[i] = INT_MAX, mstSet[i] = false; key i = INT_MAX mstSet i = false // Always include first 1st vertex in MST. // Always include first 1st vertex in MST. // Make key 0 so that this vertex is picked as first // Make key 0 so that this vertex is picked as first // vertex. // vertex. key = 0; key 0 = 0 ``` ``` // First node is always root of MST // First node is always root of MST parent = -1; parent 0 = - 1 // The MST will have V vertices // The MST will have V vertices for (int count = 0; count < V - 1; count++) {for int count = 0 count< V - 1 count ++ ``` ``` // Pick the minimum key vertex from the // Pick the minimum key vertex from the // set of vertices not yet included in MST // set of vertices not yet included in MST int u = minKey(key, mstSet); int u = minKey key mstSet // Add the picked vertex to the MST Set // Add the picked vertex to the MST Set mstSet[u] = true; mstSet u = true // Update key value and parent index of // Update key value and parent index of // the adjacent vertices of the picked vertex. // the adjacent vertices of the picked vertex. // Consider only those vertices which are not // Consider only those vertices which are not // yet included in MST // yet included in MST for (int v = 0; v < V; v++) for int v = 0 v< V v ++ // graph[u][v] is non zero only for adjacent // graph[u][v] is non zero only for adjacent // vertices of m mstSet[v] is false for vertices // vertices of m mstSet[v] is false for vertices // not yet included in MST Update the key only // not yet included in MST Update the key only // if graph[u][v] is smaller than key[v] // if graph[u][v] is smaller than key[v] if (graph[u][v] && mstSet[v] == false if graph u v&& mstSet v == false && graph[u][v] < key[v])&& graph u v< key v parent[v] = u, key[v] = graph[u][v]; parent v = u key v = graph u v } // Print the constructed MST // Print the constructed MST printMST(parent, graph); printMST parent graph } // Driver's code // Driver's code int main() {int main vector<vector<int>> graph = { { 0, 2, 0, 6, 0 }, vector< vector< int>> graph = 0 2 0 6 0 { 2, 0, 3, 8, 5 }, 2 0 3 8 5 { 0, 3, 0, 0, 7 }, 0 3 0 0 7 { 6, 8, 0, 0, 9 }, 6 8 0 0 9 { 0, 5, 7, 9, 0 } }; 0 5 7 9 0 // Print the solution // Print the solution primMST(graph); primMST graph return 0; return 0 } C ```` // A C program for Prim's Minimum // Spanning Tree (MST) algorithm. The program is // for adjacency matrix representation of the graph include include include // Number of vertices in the graph define V 5 // A utility function to find the vertex with // minimum key value, from the set of vertices // not yet included in MST int minKey(int key[], bool mstSet[]) { // Initialize min value int min = INT_MAX, min_index; for (int v = 0; v < V; v++) if (mstSet[v] == false && key[v] < min) min = key[v], min_index = v; return min_index; } // A utility function to print the // constructed MST stored in parent[] int printMST(int parent[], int graph[V][V]) { printf("Edge \tWeight\n"); for (int i = 1; i < V; i++) printf("%d - %d \t%d \n", parent[i], i, graph[parent[i]][i]); } // Function to construct and print MST for // a graph represented using adjacency // matrix representation void primMST(int graph[V][V]) { // Array to store constructed MST int parent[V]; // Key values used to pick minimum weight edge in cut int key[V]; // To represent set of vertices included in MST bool mstSet[V]; // Initialize all keys as INFINITE for (int i = 0; i < V; i++) key[i] = INT_MAX, mstSet[i] = false; // Always include first 1st vertex in MST. // Make key 0 so that this vertex is picked as first // vertex. key = 0; // First node is always root of MST parent = -1; // The MST will have V vertices for (int count = 0; count < V - 1; count++) { // Pick the minimum key vertex from the // set of vertices not yet included in MST int u = minKey(key, mstSet); // Add the picked vertex to the MST Set mstSet[u] = true; // Update key value and parent index of // the adjacent vertices of the picked vertex. // Consider only those vertices which are not // yet included in MST for (int v = 0; v < V; v++) // graph[u][v] is non zero only for adjacent // vertices of m mstSet[v] is false for vertices // not yet included in MST Update the key only // if graph[u][v] is smaller than key[v] if (graph[u][v] && mstSet[v] == false && graph[u][v] < key[v]) parent[v] = u, key[v] = graph[u][v]; } // print the constructed MST printMST(parent, graph); } // Driver's code int main() { int graph[V][V] = { { 0, 2, 0, 6, 0 }, { 2, 0, 3, 8, 5 }, { 0, 3, 0, 0, 7 }, { 6, 8, 0, 0, 9 }, { 0, 5, 7, 9, 0 } }; // Print the solution primMST(graph); return 0; } ```` Java ```` // A Java program for Prim's Minimum Spanning Tree (MST) // algorithm. The program is for adjacency matrix // representation of the graph import java.io.; import java.lang.; import java.util.; class MST { // A utility function to find the vertex with minimum // key value, from the set of vertices not yet included // in MST int minKey(int key[], Boolean mstSet[]) { // Initialize min value int min = Integer.MAX_VALUE, min_index = -1; for (int v = 0; v < mstSet.length; v++) if (mstSet[v] == false && key[v] < min) { min = key[v]; min_index = v; } return min_index; } // A utility function to print the constructed MST // stored in parent[] void printMST(int parent[], int graph[][]) { System.out.println("Edge \tWeight"); for (int i = 1; i < graph.length; i++) System.out.println(parent[i] + " - " + i + "\t" + graph[parent[i]][i]); } // Function to construct and print MST for a graph // represented using adjacency matrix representation void primMST(int graph[][]) { int V = graph.length; // Array to store constructed MST int parent[] = new int[V]; // Key values used to pick minimum weight edge in // cut int key[] = new int[V]; // To represent set of vertices included in MST Boolean mstSet[] = new Boolean[V]; // Initialize all keys as INFINITE for (int i = 0; i < V; i++) { key[i] = Integer.MAX_VALUE; mstSet[i] = false; } // Always include first 1st vertex in MST. // Make key 0 so that this vertex is // picked as first vertex key = 0; // First node is always root of MST parent = -1; // The MST will have V vertices for (int count = 0; count < V - 1; count++) { // Pick the minimum key vertex from the set of // vertices not yet included in MST int u = minKey(key, mstSet); // Add the picked vertex to the MST Set mstSet[u] = true; // Update key value and parent index of the // adjacent vertices of the picked vertex. // Consider only those vertices which are not // yet included in MST for (int v = 0; v < V; v++) // graph[u][v] is non zero only for adjacent // vertices of m mstSet[v] is false for // vertices not yet included in MST Update // the key only if graph[u][v] is smaller // than key[v] if (graph[u][v] != 0 && mstSet[v] == false && graph[u][v] < key[v]) { parent[v] = u; key[v] = graph[u][v]; } } // Print the constructed MST printMST(parent, graph); } public static void main(String[] args) { MST t = new MST(); int graph[][] = new int[][] { { 0, 2, 0, 6, 0 }, { 2, 0, 3, 8, 5 }, { 0, 3, 0, 0, 7 }, { 6, 8, 0, 0, 9 }, { 0, 5, 7, 9, 0 } }; // Print the solution t.primMST(graph); } } ```` Python ```` A Python3 program for Prim's Minimum Spanning Tree (MST) algorithm. The program is for adjacency matrix representation of the graph Library for INT_MAX import sys class Graph(): def init(self, vertices): self.V = vertices self.graph = [[0 for column in range(vertices)] for row in range(vertices)] # A utility function to print # the constructed MST stored in parent[] def printMST(self, parent): print("Edge \tWeight") for i in range(1, self.V): print(parent[i], "-", i, "\t", self.graph[parent[i]][i]) # A utility function to find the vertex with # minimum distance value, from the set of vertices # not yet included in shortest path tree def minKey(self, key, mstSet): # Initialize min value min = sys.maxsize for v in range(self.V): if key[v] < min and mstSet[v] == False: min = key[v] min_index = v return min_index # Function to construct and print MST for a graph # represented using adjacency matrix representation def primMST(self): # Key values used to pick minimum weight edge in cut key = [sys.maxsize] self.V parent = [None] self.V # Array to store constructed MST # Make key 0 so that this vertex is picked as first vertex key = 0 mstSet = [False] self.V parent = -1 # First node is always the root of for cout in range(self.V): # Pick the minimum distance vertex from # the set of vertices not yet processed. # u is always equal to src in first iteration u = self.minKey(key, mstSet) # Put the minimum distance vertex in # the shortest path tree mstSet[u] = True # Update dist value of the adjacent vertices # of the picked vertex only if the current # distance is greater than new distance and # the vertex in not in the shortest path tree for v in range(self.V): # graph[u][v] is non zero only for adjacent vertices of m # mstSet[v] is false for vertices not yet included in MST # Update the key only if graph[u][v] is smaller than key[v] if self.graph[u][v] > 0 and mstSet[v] == False \ and key[v] > self.graph[u][v]: key[v] = self.graph[u][v] parent[v] = u self.printMST(parent) Driver's code if name == 'main': g = Graph(5) g.graph = [[0, 2, 0, 6, 0], [2, 0, 3, 8, 5], [0, 3, 0, 0, 7], [6, 8, 0, 0, 9], [0, 5, 7, 9, 0]] g.primMST() ```` C# ```` // A C# program for Prim's Minimum Spanning Tree (MST) // algorithm. The program is for adjacency matrix // representation of the graph using System; using System.Collections.Generic; class MST { // A utility function to find the vertex with minimum // key value, from the set of vertices not yet included // in MST int MinKey(int[] key, bool[] mstSet) { // Initialize min value int min = int.MaxValue, minIndex = -1; for (int v = 0; v < mstSet.Length; v++) if (!mstSet[v] && key[v] < min) { min = key[v]; minIndex = v; } return minIndex; } // A utility function to print the constructed MST // stored in parent[] void PrintMST(int[] parent, int[,] graph) { Console.WriteLine("Edge \tWeight"); for (int i = 1; i < graph.GetLength(0); i++) Console.WriteLine(parent[i] + " - " + i + "\t" + graph[parent[i], i]); } // Function to construct and print MST for a graph // represented using adjacency matrix representation public void PrimMST(int[,] graph) { int V = graph.GetLength(0); // Array to store constructed MST int[] parent = new int[V]; // Key values used to pick minimum weight edge in // cut int[] key = new int[V]; // To represent set of vertices included in MST bool[] mstSet = new bool[V]; // Initialize all keys as INFINITE for (int i = 0; i < V; i++) { key[i] = int.MaxValue; mstSet[i] = false; } // Always include first 1st vertex in MST. // Make key 0 so that this vertex is // picked as first vertex key = 0; // First node is always root of MST parent = -1; // The MST will have V vertices for (int count = 0; count < V - 1; count++) { // Pick the minimum key vertex from the set of // vertices not yet included in MST int u = MinKey(key, mstSet); // Add the picked vertex to the MST Set mstSet[u] = true; // Update key value and parent index of the // adjacent vertices of the picked vertex. // Consider only those vertices which are not // yet included in MST for (int v = 0; v < V; v++) // graph[u][v] is non zero only for adjacent // vertices of m mstSet[v] is false for // vertices not yet included in MST Update // the key only if graph[u][v] is smaller // than key[v] if (graph[u, v] != 0 && !mstSet[v] && graph[u, v] < key[v]) { parent[v] = u; key[v] = graph[u, v]; } } // Print the constructed MST PrintMST(parent, graph); } public static void Main(string[] args) { MST t = new MST(); int[,] graph = new int[,] { { 0, 2, 0, 6, 0 }, { 2, 0, 3, 8, 5 }, { 0, 3, 0, 0, 7 }, { 6, 8, 0, 0, 9 }, { 0, 5, 7, 9, 0 } }; // Print the solution t.PrimMST(graph); } } ```` JavaScript ```` // Number of vertices in the graph let V = 5; // A utility function to find the vertex with // minimum key value, from the set of vertices // not yet included in MST function minKey(key, mstSet) { // Initialize min value let min = Number.MAX_VALUE, min_index = -1; for (let v = 0; v < V; v++) if (!mstSet[v] && key[v] < min) { min = key[v]; min_index = v; } return min_index; } // A utility function to print the // constructed MST stored in parent[] function printMST(parent, graph) { console.log("Edge Weight"); for (let i = 1; i < V; i++) console.log(parent[i] + " - " + i + " " + graph[parent[i]][i]); } // Function to construct and print MST for // a graph represented using adjacency matrix function primMST(graph) { // Array to store constructed MST let parent = new Array(V); // Key values used to pick minimum weight edge in cut let key = new Array(V); // To represent set of vertices included in MST let mstSet = new Array(V); // Initialize all keys as INFINITE for (let i = 0; i < V; i++) { key[i] = Number.MAX_VALUE; mstSet[i] = false; } // Always include first vertex in MST. key = 0; parent = -1; // First node is always root of MST // The MST will have V vertices for (let count = 0; count < V - 1; count++) { // Pick the minimum key vertex from the set of vertices not yet included in MST let u = minKey(key, mstSet); // Add the picked vertex to the MST Set mstSet[u] = true; // Update key value and parent index of the adjacent vertices of the picked vertex. for (let v = 0; v < V; v++) { // graph[u][v] is non-zero only for adjacent vertices of u // mstSet[v] is false for vertices not yet included in MST // Update the key only if graph[u][v] is smaller than key[v] if (graph[u][v] && !mstSet[v] && graph[u][v] < key[v]) { parent[v] = u; key[v] = graph[u][v]; } } } // Print the constructed MST printMST(parent, graph); } // Driver code let graph = [ [ 0, 2, 0, 6, 0 ], [ 2, 0, 3, 8, 5 ], [ 0, 3, 0, 0, 7 ], [ 6, 8, 0, 0, 9 ], [ 0, 5, 7, 9, 0 ] ]; // Print the solution primMST(graph); ```` Output ``` Edge Weight 0 - 1 2 1 - 2 3 0 - 3 6 1 - 4 5 ``` Time Complexity: O(V2), As, we are using adjacency matrix, if the input graph is represented using an adjacency list, then the time complexity of Prim's algorithm can be reduced to O((E+V) logV) with the help of a binary heap. Auxiliary Space: O(V) Optimized Implementation using Adjacency List Representation (of Graph) and Priority Queue We transform the adjacency matrix into adjacency list using ArrayList>. in Java, list of list in Python and array of vectors in C++. Then we create a Pair class to store the vertex and its weight . We sort the list on the basis of lowest weight. We create priority queue and push the first vertex and its weight in the queue Then we just traverse through its edges and store the least weight in a variable called ans. At last after all the vertex we return the ans. C++ ```` include using namespace std; // Function to find sum of weights of edges of the Minimum Spanning Tree. int spanningTree(int V, int E, vector> &edges) { // Create an adjacency list representation of the graph vector<vector<int>> adj[V]; // Fill the adjacency list with edges and their weights for (int i = 0; i < E; i++) { int u = edges[i]; int v = edges[i]; int wt = edges[i]; adj[u].push_back({v, wt}); adj[v].push_back({u, wt}); } // Create a priority queue to store edges with their weights priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq; // Create a visited array to keep track of visited vertices vector<bool> visited(V, false); // Variable to store the result (sum of edge weights) int res = 0; // Start with vertex 0 pq.push({0, 0}); // Perform Prim's algorithm to find the Minimum Spanning Tree while(!pq.empty()){ auto p = pq.top(); pq.pop(); int wt = p.first; // Weight of the edge int u = p.second; // Vertex connected to the edge if(visited[u] == true){ continue; // Skip if the vertex is already visited } res += wt; // Add the edge weight to the result visited[u] = true; // Mark the vertex as visited // Explore the adjacent vertices for(auto v : adj[u]){ // v represents the vertex and v represents the edge weight if(visited[v] == false){ pq.push({v, v}); // Add the adjacent edge to the priority queue } } } return res; // Return the sum of edge weights of the Minimum Spanning Tree } int main() { vector> graph = {{0, 1, 5}, {1, 2, 3}, {0, 2, 1}}; cout << spanningTree(3, 3, graph) << endl; return 0; } ```` ``` include#include ``` using namespace std; using namespace std // Function to find sum of weights of edges of the Minimum Spanning Tree. // Function to find sum of weights of edges of the Minimum Spanning Tree. int spanningTree(int V, int E, vector<vector<int>> &edges) {int spanningTree int V int E vector< vector< int>>& edges ``` ``` // Create an adjacency list representation of the graph // Create an adjacency list representation of the graph vector<vector<int>> adj[V]; vector< vector< int>> adj V ``` ``` // Fill the adjacency list with edges and their weights // Fill the adjacency list with edges and their weights for (int i = 0; i < E; i++) {for int i = 0 i< E i ++ int u = edges[i]; int u = edges i 0 int v = edges[i]; int v = edges i 1 int wt = edges[i]; int wt = edges i 2 adj[u].push_back({v, wt}); adj u push_back v wt adj[v].push_back({u, wt}); adj v push_back u wt } ``` ``` // Create a priority queue to store edges with their weights // Create a priority queue to store edges with their weights priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq; priority_queue< pair< int int> vector< pair< int int>> greater< pair< int int>>> pq ``` ``` // Create a visited array to keep track of visited vertices // Create a visited array to keep track of visited vertices vector<bool> visited(V, false); vector< bool> visited V false ``` ``` // Variable to store the result (sum of edge weights) // Variable to store the result (sum of edge weights) int res = 0; int res = 0 ``` ``` // Start with vertex 0 // Start with vertex 0 pq.push({0, 0}); pq push 0 0 ``` ``` // Perform Prim's algorithm to find the Minimum Spanning Tree // Perform Prim's algorithm to find the Minimum Spanning Tree while(!pq.empty()){while! pq empty auto p = pq.top(); auto p = pq top pq.pop(); pq pop ``` ``` int wt = p.first; // Weight of the edge int wt = p first // Weight of the edge int u = p.second; // Vertex connected to the edge int u = p second // Vertex connected to the edge ``` ``` if(visited[u] == true){if visited u == true continue; // Skip if the vertex is already visited continue // Skip if the vertex is already visited } ``` ``` res += wt; // Add the edge weight to the result res += wt // Add the edge weight to the result visited[u] = true; // Mark the vertex as visited visited u = true // Mark the vertex as visited ``` ``` // Explore the adjacent vertices // Explore the adjacent vertices for(auto v : adj[u]){for auto v adj u // v represents the vertex and v represents the edge weight // v represents the vertex and v represents the edge weight if(visited[v] == false){if visited v 0 == false pq.push({v, v}); // Add the adjacent edge to the priority queue pq push v 1 v 0 // Add the adjacent edge to the priority queue } } } ``` ``` return res; // Return the sum of edge weights of the Minimum Spanning Tree return res // Return the sum of edge weights of the Minimum Spanning Tree } int main() {int main vector<vector<int>> graph = {{0, 1, 5}, vector< vector< int>> graph = 0 1 5 {1, 2, 3}, 1 2 3 {0, 2, 1}}; 0 2 1 cout << spanningTree(3, 3, graph) << endl; cout<< spanningTree 3 3 graph<< endl return 0; return 0 } Java ```` // A Java program for Prim's Minimum Spanning Tree (MST) // algorithm. The program is for adjacency list // representation of the graph import java.io.; import java.util.; // Class to form pair class Pair implements Comparable { int v; int wt; Pair(int v,int wt) { this.v=v; this.wt=wt; } public int compareTo(Pair that) { return this.wt-that.wt; } } class GFG { // Function of spanning tree static int spanningTree(int V, int E, int edges[][]) { ArrayList<ArrayList<Pair>> adj=new ArrayList<>(); for(int i=0;i<V;i++) { adj.add(new ArrayList<Pair>()); } for(int i=0;i<edges.length;i++) { int u=edges[i]; int v=edges[i]; int wt=edges[i]; adj.get(u).add(new Pair(v,wt)); adj.get(v).add(new Pair(u,wt)); } PriorityQueue<Pair> pq = new PriorityQueue<Pair>(); pq.add(new Pair(0,0)); int[] vis=new int[V]; int s=0; while(!pq.isEmpty()) { Pair node=pq.poll(); int v=node.v; int wt=node.wt; if(vis[v]==1) continue; s+=wt; vis[v]=1; for(Pair it:adj.get(v)) { if(vis[it.v]==0) { pq.add(new Pair(it.v,it.wt)); } } } return s; } // Driver code public static void main (String[] args) { int graph[][] = new int[][] {{0,1,5}, {1,2,3}, {0,2,1}}; // Function call System.out.println(spanningTree(3,3,graph)); } } ```` Python ```` import heapq def tree(V, E, edges): # Create an adjacency list representation of the graph adj = [[] for _ in range(V)] # Fill the adjacency list with edges and their weights for i in range(E): u, v, wt = edges[i] adj[u].append((v, wt)) adj[v].append((u, wt)) # Create a priority queue to store edges with their weights pq = [] # Create a visited array to keep track of visited vertices visited = [False] V # Variable to store the result (sum of edge weights) res = 0 # Start with vertex 0 heapq.heappush(pq, (0, 0)) # Perform Prim's algorithm to find the Minimum Spanning Tree while pq: wt, u = heapq.heappop(pq) if visited[u]: continue # Skip if the vertex is already visited res += wt # Add the edge weight to the result visited[u] = True # Mark the vertex as visited # Explore the adjacent vertices for v, weight in adj[u]: if not visited[v]: heapq.heappush(pq, (weight, v)) # Add the adjacent edge to the priority queue return res # Return the sum of edge weights of the Minimum Spanning Tree if name == "main": graph = [[0, 1, 5], [1, 2, 3], [0, 2, 1]] # Function call print(tree(3, 3, graph)) ```` C# ```` using System; using System.Collections.Generic; public class MinimumSpanningTree { // Function to find sum of weights of edges of the Minimum Spanning Tree. public static int SpanningTree(int V, int E, int[,] edges) { // Create an adjacency list representation of the graph List> adj = new List>(); for (int i = 0; i < V; i++) { adj.Add(new List()); } // Fill the adjacency list with edges and their weights for (int i = 0; i < E; i++) { int u = edges[i, 0]; int v = edges[i, 1]; int wt = edges[i, 2]; adj[u].Add(new int[] { v, wt }); adj[v].Add(new int[] { u, wt }); } // Create a priority queue to store edges with their weights PriorityQueue<(int, int)> pq = new PriorityQueue<(int, int)>(); // Create a visited array to keep track of visited vertices bool[] visited = new bool[V]; // Variable to store the result (sum of edge weights) int res = 0; // Start with vertex 0 pq.Enqueue((0, 0)); // Perform Prim's algorithm to find the Minimum Spanning Tree while (pq.Count > 0) { var p = pq.Dequeue(); int wt = p.Item1; // Weight of the edge int u = p.Item2; // Vertex connected to the edge if (visited[u]) { continue; // Skip if the vertex is already visited } res += wt; // Add the edge weight to the result visited[u] = true; // Mark the vertex as visited // Explore the adjacent vertices foreach (var v in adj[u]) { // v represents the vertex and v represents the edge weight if (!visited[v]) { pq.Enqueue((v, v)); // Add the adjacent edge to the priority queue } } } return res; // Return the sum of edge weights of the Minimum Spanning Tree } public static void Main() { int[,] graph = { { 0, 1, 5 }, { 1, 2, 3 }, { 0, 2, 1 } }; // Function call Console.WriteLine(SpanningTree(3, 3, graph)); } } // PriorityQueue implementation for C# public class PriorityQueue where T : IComparable { private List heap = new List(); public int Count => heap.Count; public void Enqueue(T item) { heap.Add(item); int i = heap.Count - 1; while (i > 0) { int parent = (i - 1) / 2; if (heap[parent].CompareTo(heap[i]) <= 0) break; Swap(parent, i); i = parent; } } public T Dequeue() { int lastIndex = heap.Count - 1; T frontItem = heap; heap = heap[lastIndex]; heap.RemoveAt(lastIndex); --lastIndex; int parent = 0; while (true) { int leftChild = parent 2 + 1; if (leftChild > lastIndex) break; int rightChild = leftChild + 1; if (rightChild <= lastIndex && heap[leftChild].CompareTo(heap[rightChild]) > 0) leftChild = rightChild; if (heap[parent].CompareTo(heap[leftChild]) <= 0) break; Swap(parent, leftChild); parent = leftChild; } return frontItem; } private void Swap(int i, int j) { T temp = heap[i]; heap[i] = heap[j]; heap[j] = temp; } } ```` JavaScript ```` class PriorityQueue { constructor() { this.heap = []; } enqueue(value) { this.heap.push(value); let i = this.heap.length - 1; while (i > 0) { let j = Math.floor((i - 1) / 2); if (this.heap[i] >= this.heap[j]) { break; } [this.heap[i], this.heap[j]] = [this.heap[j], this.heap[i]]; i = j; } } dequeue() { if (this.heap.length === 0) { throw new Error("Queue is empty"); } let i = this.heap.length - 1; const result = this.heap; this.heap = this.heap[i]; this.heap.pop(); i--; let j = 0; while (true) { const left = j 2 + 1; if (left > i) { break; } const right = left + 1; let k = left; if (right <= i && this.heap[right] < this.heap[left]) { k = right; } if (this.heap[j] <= this.heap[k]) { break; } [this.heap[j], this.heap[k]] = [this.heap[k], this.heap[j]]; j = k; } return result; } get count() { return this.heap.length; } } function spanningTree(V, E, edges) { // Create an adjacency list representation of the graph const adj = new Array(V).fill(null).map(() => []); // Fill the adjacency list with edges and their weights for (let i = 0; i < E; i++) { const [u, v, wt] = edges[i]; adj[u].push([v, wt]); adj[v].push([u, wt]); } // Create a priority queue to store edges with their weights const pq = new PriorityQueue(); // Create a visited array to keep track of visited vertices const visited = new Array(V).fill(false); // Variable to store the result (sum of edge weights) let res = 0; // Start with vertex 0 pq.enqueue([0, 0]); // Perform Prim's algorithm to find the Minimum Spanning Tree while (pq.count > 0) { const p = pq.dequeue(); const wt = p; // Weight of the edge const u = p; // Vertex connected to the edge if (visited[u]) { continue; // Skip if the vertex is already visited } res += wt; // Add the edge weight to the result visited[u] = true; // Mark the vertex as visited // Explore the adjacent vertices for (const v of adj[u]) { // v represents the vertex and v represents the edge weight if (!visited[v]) { pq.enqueue([v, v]); // Add the adjacent edge to the priority queue } } } return res; // Return the sum of edge weights of the Minimum Spanning Tree } // Example usage const graph = ; // Function call console.log(spanningTree(3, 3, graph)); ```` Output ``` 4 ``` Time Complexity: O((E+V)log(V)) where V is the number of vertex and E is the number of edges Auxiliary Space: O(E+V) where V is the number of vertex and E is the number of edges Advantages and Disadvantages of Prim's algorithm Advantages: Prim's algorithm is guaranteed to find the MST in a connected, weighted graph. It has a time complexity of O((E+V)log(V)) using a binary heap or Fibonacci heap, where E is the number of edges and V is the number of vertices. It is a relatively simple algorithm to understand and implement compared to some other MST algorithms. Disadvantages: Like Kruskal's algorithm, Prim's algorithm can be slow on dense graphs with many edges, as it requires iterating over all edges at least once. Prim's algorithm relies on a priority queue, which can take up extra memory and slow down the algorithm on very large graphs. The choice of starting node can affect the MST output, which may not be desirable in some applications. 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1279	https://math.stackexchange.com/questions/446923/binomial-co-efficients-as-a-basis-for-polynomials	generating functions - Binomial Co-efficients as a basis for Polynomials - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Binomial Co-efficients as a basis for Polynomials Ask Question Asked 12 years, 2 months ago Modified12 years, 2 months ago Viewed 283 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Well, I was just doing some random reading, and came across this: Let P(n)P(n) be a polynomial of degree d d, further, define, p k=Δ k(P(n))\|n=0 p k=Δ k(P(n))\|n=0. (Here, Δ Δ indicates the forward difference operator.) Then, P(n)=∑i=0 d p i(n i)P(n)=∑i=0 d p i(n i), where, (n i)(n i) is the generalised binomial co-efficient. This seems like a pretty marvellous result, and I notice the similarity to Maclaurin or Taylor series expansions, but, well, does anybody know where I can find a proof? And where could I read more about results like this? polynomials generating-functions Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jul 18, 2013 at 22:13 AlpArslanAlpArslan 633 5 5 silver badges 14 14 bronze badges 2 The keyword here is "calculus of finite differences." The proof is by induction on the degree.Qiaochu Yuan –Qiaochu Yuan 2013-07-18 22:20:25 +00:00 Commented Jul 18, 2013 at 22:20 @QiaochuYuan Thanks for the heads up. It's curiously fortuitous that you, of all people, commented. I was, after all, reading your notes at the time. As for the induction, well, I feel sheepish that I didn't try it beforehand. Regardless, I tried it myself, but it seems incredibly messy. Would you mind handing me some direction, man?AlpArslan –AlpArslan 2013-07-19 19:08:22 +00:00 Commented Jul 19, 2013 at 19:08 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Apply the operators Δ k\|n=0 Δ k\|n=0 on both sides for all k=0,1,...k=0,1,... and check you get the same every time. The operators applied on the binomial coefficients act nicely. Use the basic property of the binomial coefficients. Of course, you also need to prove that if a polynomial gives you zero after applying all these operators then you get zero. But this clearly gives you that the polynomial is constant at too many points. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jul 19, 2013 at 1:28 answered Jul 19, 2013 at 1:23 OR.OR. 6,099 2 2 gold badges 25 25 silver badges 43 43 bronze badges 2 Ha, I tried that, and yes they do act rather prettily, indeed. I was wondering though if it were possible to derive this sort of formula analytically? Otherwise, it just seems sort of marvellously contrived, I suppose.AlpArslan –AlpArslan 2013-07-19 19:19:17 +00:00 Commented Jul 19, 2013 at 19:19 What would it be analytically? For example: How would it be a proof of P(x)=∑d k=0 P(k)(a)(x−a)k P(x)=∑k=0 d P(k)(a)(x−a)k analytically? Here we are proving the same type of formula but for finite differences instead of derivatives. It is likely that any proof of one can also be adapted to the other.OR. –OR. 2013-07-19 23:30:19 +00:00 Commented Jul 19, 2013 at 23:30 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polynomials generating-functions See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 3basis for space of polynomials 6Find a n,i,j a n,i,j in the expansion (x+D)n=∑i,j a n,i,j x i D j.(x+D)n=∑i,j a n,i,j x i D j. 1Change of basis for polynomials 3Schur polynomials form a basis for the space of symmetric polynomials 3Hermite polynomials as a basis for generating function 0Basis for polynomials of degree k or lower 0Showing a basis for polynomials 4Binomial theorem-respecting polynomials 0Basis for polynomial ring modulo polynomials 3Some identities for polynomials defined by their generating function. 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1280	https://www.youtube.com/watch?v=wRYBxd36AXM	A common mistake in geometry, especially in triangle proofs! #maths #teacher #geometry #education Mr. Allen Math 53900 subscribers 4 likes Description 323 views Posted: 4 Dec 2024 Transcript:
1281	https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_9?srsltid=AfmBOoo29zPw3KQQbnFdfc9poKwqLwLxsazX-lb8SHJrxXRwB03avyQi	Art of Problem Solving 1960 AHSME Problems/Problem 9 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1960 AHSME Problems/Problem 9 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1960 AHSME Problems/Problem 9 Problem The fraction is (with suitable restrictions of the values of a, b, and c): Solution Use the commutative property to get Factor perfect square trinomials to get Factor difference of squares to get Cancel out like terms (with suitable restrictions of a, b, c) to get The answer is . See Also 1960 AHSC (Problems • Answer Key • Resources) Preceded by Problem 8Followed by Problem 10 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25•26•27•28•29•30•31•32•33•34•35•36•37•38•39•40 All AHSME Problems and Solutions Retrieved from " Category: Introductory Algebra Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1282	https://www.nist.gov/si-redefinition	An official website of the United States government Here’s how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. Redefinition SI stands for the International System of Units, also known as the metric system. On November 16, 2018, in Versailles, France, a group of 60 countries made history. With a unanimous vote, they dramatically transformed the international system that underpins global science and trade. This single action finally realized scientists’ 150-year dream of a measurement system based entirely on unchanging fundamental properties of nature. On that day, the International System of Units, informally known as the metric system — the way in which the world measures everything from coffee to the cosmos — changed in a way that is more profound than anything since its establishment following the French Revolution. It was a turning point for humanity. READ MORE THE SI The SI — the modern metric system — has seven base units from which all other measurement units can be derived. On May 20, 2019, four of them — the kilogram, kelvin, ampere and mole — were redefined in terms of constants of nature. The remaining three — the second, meter, and candela — are already based on universal constants. Click on the SI symbols below for more information. Scientists Voted on Metric Makeover After decades of groundbreaking laboratory work, the world’s scientific and technical community has redefined four of the seven base units for the International System of Units (SI). A vote to adopt the change happened on November 16, 2018, at Versailles, France, and the change went into effect on May 20, 2019. Scientists Voted on Metric Makeover - English THE CONSTANTS Credit: Stoughton/NIST The units in the revised SI are based completely on seven unchanging quantities or “universal constants,” including the speed of light, the amount of electric charge in an electron, and the Planck constant. Learn more about each of these “invariants of nature” and how they come into play in the revised SI. Meet The Constants Was this page helpful?
1283	http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/nucnot.html	\| \| \| \| \| \| \| \| --- --- --- \| Nuclear Notation Standard nuclear notation shows the chemical symbol, the mass number and the atomic number of the isotope. \| \| \| \| \| --- --- \| \| \| \| \| \| --- \| \| Example: the isotopes of carbon. The element is determined by the atomic number 6. Carbon-12 is the common isotope, with carbon-13 as another stable isotope which makes up about 1%. Carbon 14 is radioactive and the basis for carbon dating. \| \| \| The mass of an element that is numerically equal to the atomic mass A in grams is called a mole and will contain Avogadro's number NA of nuclei. If the density ρ of the material is known, then the number of nuclei per unit volume n can be calculated from n =ρNA/A. This is useful in calculating the cross section for nuclear scattering. \| \| \| Atoms and elements \| \| Index \| \| \| \| \| --- \| \| HyperPhysics Nuclear \| R Nave \| \| Go Back \| \| \| \| \| \| \| --- --- \| Isotopes The different isotopes of a given element have the same atomic number but different mass numbers since they have different numbers of neutrons. The chemical properties of the different isotopes of an element are identical, but they will often have great differences in nuclear stability. For stable isotopes of light elements, the number of neutrons will be almost equal to the number of protons, but a growing neutron excess is characteristic of stable heavy elements. The element tin (Sn) has the most stable isotopes with 10, the average being about 2.6 stable isotopes per element. Information about the isotopes of each element and their abundances can be found by going to the periodic table and choosing an element. Then take the link to nuclear data. Isotopes are (almost) Chemically Identical It is significant to note that the three isotopes of hydrogen change in mass by a factor of three, but their chemical properties are virtually identical. A tiny difference in the spectral frequencies of hydrogen and deuterium comes from an essentially mechanical source, the slight change in the "reduced mass" associated with the orbiting electron. But for practical purposes the chemical behavior of the isotopes of any element are identical. The dominant contributer to the interactions between atoms and their environment is the electromagnetic force. It should not be surprising that an extra neutron or two in the nucleus has almost no effect on that interaction with the world. Examination of a scale model of the atom makes it evident that the nucleus is extremely tiny compared ot the size of the atom. The nuclear radius of carbon-12 is 2.7 x 10-15 m while the size of the atom from the periodic table is about 0.9 x 10-10 m, about 33,000 times larger! \| \| \| --- \| \| Nuclear notation \| Heavy element synthesis \| \| \| \| Example: isotopic abundances of krypton \| \| Index \| \| \| \| \| --- \| \| HyperPhysics Nuclear \| R Nave \| \| Go Back \| \| \| \| \| --- \| Nuclear Forces Within the incredibly small nuclear size, the two strongest forces in nature are pitted against each other. When the balance is broken, the resultant radioactivity yields particles of enormous energy. Click on any of the text for more detail The electron in a hydrogen atom is attracted to the proton nucleus with a force so strong that gravity and all other forces are negligible by comparison. But two protons touching each other would feel a repulsive force over 100 million times stronger!! So how can such protons stay in such close proximity? This may give you some feeling for the enormity of the nuclear strong force which holds the nuclei together. \| \| \| Nuclear scale model \| \| Index \| \| \| \| \| --- \| \| HyperPhysics Nuclear \| R Nave \| \| Go Back \| \| \| \| --- \| \| Nuclear Size The size of the nucleus compared to the size of the atom in which it resides is so small that it has invited a number of interesting comparisons. For example, the space inside an atom can be compared to the space in the solar system in a scale model. Scaling the gold nucleus suggests that the atomic radius is some 18,000 times the size of the nucleus. This great disparity in size was first discovered by Rutherford scattering of alpha particles off a thin gold foil. The extremity of this space comparison is highlighted by the fact that for an atom with equal numbers of neutrons and protons, the nucleus comprises about 99.97% of the mass of the atom! Experimental evidence suggests that nuclear matter is almost of uniform density, so that the size of a nucleus can be estimated from its mass number. \| Index \| \| \| \| \| --- \| \| HyperPhysics Nuclear \| R Nave \| \| Go Back \|
1284	https://www.youtube.com/watch?v=Px-Kio0mNgc	MIT Integration Bee: How to Handle the Floor Function Like a Pro! Michael Penn 326000 subscribers 1010 likes Description 14739 views Posted: 26 Jul 2023 🌟Support the channel🌟 Patreon: Channel Membership: Merch: My amazon shop: 🟢 Discord: 🌟my other channels🌟 mathmajor: pennpav podcast: 🌟My Links🌟 Personal Website: Instagram: Twitter: Randolph College Math: Research Gate profile: Google Scholar profile: 🌟How I make Thumbnails🌟 Canva: Color Pallet: 🌟Suggest a problem🌟 41 comments Transcript: today I've got a really nice integral that comes from an MIT integration B I think the year was around 2015. and what I like about this integral is it mixes The Continuous and discrete so integrals are inherently continuous objects whereas this has the floor function which turns this into a discrete operation okay so let's see what we have we'd like to calculate the integral from 0 to 1 of DX over the floor of 1 minus the log base 2 of X that's another thing that I think makes this interesting the fact that we have a log base 2 of there although I think we could probably very easily generate this to the log base a where a is really any positive number I'll let you guys think about how you might do that and post it in the comments now before we start our real calculation let's notice that this has a discontinuity at zero the limit as X Goes To Zero from above of any logarithm is negative Infinity so that means to properly do this I need to write this in terms of a limit and that's exactly what I'll do so I'll write this as the limit as we'll call it t Goes To Zero from above of the integral from T to 1 of well it's the same integrand or the same function so it's essentially one over the four of one minus the log base 2 of x but now we can really put anything in here for this T that we want as long as it approaches 0 from above so we might as well put something in here that interacts with the log base 2 well just like we talked about before so I think that would be a Power of Two so let's maybe write it as follows so let's write this as the limit as I'm going to write this as n goes to infinity and then the integral of 1 over 2 to the N notice that encodes exactly what we have right here T approaching 0 from above as n goes to Infinity 1 over 2 to the N approaches 0 from above okay so anyway up to one of DX over the floor of 1 minus the log base 2 of X great and now we're going to split this up into a bunch of different integrals and so what integrals will those be well we're going to start way down here at the bottom 1 over 2 to the n and work all the way up to one so in other words we'll have the limit as n goes to Infinity of I'm going to write it like this at first the integral from of 1 over 2 to the N to 1 over 2 to the N minus 1 plus the integral of 1 over 2 to the N minus 1 to 1 over 2 to the N minus 2 all the way down to the integral from one quarter to one half plus the integral from one half to one so I think that encodes everything but I can smush that all together into one object and that's going to look like this so I'll just bring my limit down as n goes to infinity and now I'll have the sum as little n goes from let's see It'll be zero up to capital n so n equals 0 is going to be this thing right here this will be 1 over 2 to the zero and n equals capital N will be this term right here and now that I look at it I think I need this to be capital N minus 1. and then this will be the integral from 1 over 2 to the N plus one to one over two to the N of our function so DX over the floor of 1 minus log base 2 of X but now the whole reason we split this up into pieces based off of powers of 2 is that so we could do some simplification and that's exactly what we'll do over here so let's maybe put this as a little bit of an observation or a side calculation so if we have X between 1 over 2 to the N plus 1 and 1 over 2 to the n then that means that the log base 2 of x is between two things that are nice well the log base 2 of X is an increasing function so that means it will preserve our inequality okay so let's put our log base 2 of X in the middle and then we can pay take the log base 2 of this we'll get minus n take the log base 2 of this we'll get minus n plus one now let's maybe negate the whole thing that'll Swap all the inequalities we'll have n is less than negative log base 2 of X Which is less than n plus one and then we'll add 1 to everything here and we'll have n plus 1 is less than 1 minus log base 2 of X Which is less than n Plus 2. but notice that puts 1 minus log base 2 of X between two consecutive integers so if it's between two consecutive integers and we take its floor we get the smaller integer so in other words we have the floor of 1 minus log base 2 of X is simply equal to n plus 1 in this case so let's put a box around that and then maybe point out that that will be inserted right here okay so let's see where that leaves us so we'll still have this limit as capital N goes to Infinity we'll still have this sum as little n goes from zero to capital N minus one and then we'll have the integral from well it's the same bounds of integration so 1 over 2N plus one to one over two n but now we're integrating a constant because we can switch out this floor of 1 minus log base 2 of X with just 1 over n plus one so we have 1 over n plus 1 DX but if we're integrating a constant over you know like an interval we simply get a constant times the length of that interval well essentially it's just the area of a rectangle so that's how you want to think about that okay so that means we can calculate this as 1 over n plus 1 times 1 over 2 to the N minus 1 over 2 to the N Plus 1. again it's the constant times the length of that interval but it's pretty easy to calculate this difference here and we'll see that it's 1 over 2 to the N Plus 1. so that gives us the following object we have the limit as capital N goes to Infinity of the sum as little n goes from 0 to capital N minus 1 of let's see 1 over n plus 1 times 1 half to the N plus one I'm going to write it like that but look at that that's the limit of some partial sums and in fact it's pretty easy to check that this series converges notice it's less than the series if we left out the 1 over n plus one and that's simply a geometric series which we know converges so since this converges we can rewrite this as an infinite sum so we have this is the sum as little n goes from 0 to Infinity of well it's going to be the same thing 1 over n plus 1 and then one half to the N plus one okay so let's bring that up so this is where we just ended so our integral was the following sum but now I'm going to take the sum and rewrite it so that it looks like well I'll call it a sum of zeroth integrals so this is going to be the sum as n goes from 0 to Infinity of 1 over n plus one times x to the N plus 1 evaluated from 0 to 1 half that's what I mean by a zeroth integral it's like just the evaluation so clearly if we evaluate that at x equals zero we get zero evaluating at x equals half we retrieve our sum but now I'm going to apply the fundamental theorem of calculus to turn this zeroth integral to a first integral I can do that by taking the derivative of this and then switch that whole thing out with an integral so now we've got the sum n goes from 0 up to Infinity well taking the derivative of that we actually get something nice it's just x to the N so we have the integral from 0 up to one half of x to the N DX but now by the dominated convergence theorem we can switch the order of summation and integration to give us the integral from 0 to 1 half of the sum as n goes from 0 up to Infinity of x to the N DX but check it out if x is between 0 and 1 half this is simply a geometric Series where we are within the interval of convergence for a geometric Series so that's nice and so we'll take the well-known value of this geometric series which is one over one minus X great so if there's like one series that you should remember from calculus two it's a geometric series if there's two probably the Taylor expansion for e to the X okay so that's going to leave us with the integral from 0 to half of one over one minus X DX the antiderivative of that is quite easy it is minus natural log of 1 minus X we'll evaluate that from 0 to 1 half so evaluating it at one half we get minus natural log of one-half evaluating it at zero we get natural log of 1 which is zero but now using logarithms we can very simply reduce this to the natural log of two and that's a good place to stop thanks for watching and sticking around until the end of the video and since you're here don't forget to gently press that like button subscribe ring the bell and select all notifications to never miss a video if you want to get your name in the credits like you see here access the live seminar series review videos before release and more go to patreon.com michaelpinmath and become a patreon member today if you want full ad free course content subscribe to my second Channel math major I've got courses on linear algebra complex analysis and proof writing among several others and that's everything bye
1285	https://bestpractice.bmj.com/topics/en-us/734	Skip to main content Skip to search Periorbital and orbital cellulitis  Menu  Close Overview  Theory  Diagnosis  Management  Follow up  Resources  Log in or subscribe to access all of BMJ Best Practice Last reviewed: 7 Aug 2025 Last updated: 17 Sep 2024 Summary Periorbital cellulitis (also known as preseptal cellulitis) is an infective process occurring in the eyelid tissues superficial to (anterior to) the orbital septum. It is usually due to superficial tissue injury (e.g., insect bite or chalazion) or adjacent sinusitis. Orbital cellulitis is an infective process affecting soft tissue within the orbit, posterior or deep to the orbital septum, not involving the globe. It is usually due to underlying bacterial sinusitis. Periorbital cellulitis is of concern in children because it may be secondary to occult underlying bacterial sinusitis or, rarely, due to bacteremic spread from a primary infection (e.g., pneumonia), and may rapidly progress to orbital cellulitis in children. Complications include subperiosteal abscess, cavernous sinus thrombosis, intracranial abscess, and subsequent loss of vision and/or death. Definition Periorbital (also known as preseptal) cellulitis is inflammation and infection of the superficial eyelid. The inflammation remains confined to the soft tissue layers superficial to the orbital septum and ocular function remains intact. Orbital cellulitis is an infection within the orbital soft tissues. It may be associated with ocular dysfunction and is usually due to underlying bacterial sinusitis. Orbital cellulitis is a far more serious condition and warrants hospital admission. It has much higher morbidity than periorbital cellulitis, and warrants urgent imaging and surgical evaluation by oculoplastic as well as a head-and-neck specialist. History and exam Key diagnostic factors recent sinus infection recent eyelid injury redness and swelling of eye ocular pain decreased vision proptosis eyelid edema insect bite on eyelid stye or chalazion ear or facial infection ophthalmoplegia and diplopia Full details Other diagnostic factors orbital trauma skin infection chemosis tenderness around eye fever eyelid erythema elevated intraocular pressure headache malaise previous dental infection or dental work orbital fracture foreign body in eye or orbit drowsiness nausea/vomiting nasal discharge Full details Risk factors sinusitis young age male sex lack of Hib vaccine in children Full details Log in or subscribe to access all of BMJ Best Practice Diagnostic tests 1st tests to order clinical exam CT sinus and orbits with contrast medium WBC count Full details Tests to consider blood culture microbiology swabs (conjunctiva, nasopharnyx, external wounds) MRI head and orbits with contrast medium Orbital ultrasonography lumbar puncture Full details Log in or subscribe to access all of BMJ Best Practice Treatment algorithm ACUTE periorbital cellulitis: causative organism not identified periorbital cellulitis: causative organism identified orbital cellulitis: causative organism not identified orbital cellulitis: causative organism identified Log in or subscribe to access all of BMJ Best Practice Contributors Authors Sudarshan Srivatsan, MD Oculoplastics Fellow Moran Eye Center University of Utah Salt Lake City UT Disclosures SS declares that he has no competing interests. Robert Kersten, MD, FACS, FASOPRS Professor of Clinical Ophthalmology Division Chief, Oculoplastics Moran Eye Center University of Utah Salt Lake City UT Disclosures RK declares that he has no competing interests. Acknowledgements Dr Sudarshan Srivatsan and Dr Robert Kersten would like to gratefully acknowledge Dr Sandra Lora Cremers, Dr Sarosh Janjua, and Dr H. Jane Kim, previous contributors to this topic. SLC, SJ and HJK declare that they have no competing interests. Peer reviewers David M. Ozog, MD Director of Cosmetic Dermatology Department of Dermatology Cosmetic and Procedural Dermatology Henry Ford Health System Detroit MI Disclosures DMO declares that he has no competing interests. I-Hui (Elaine) Wu, MD Resident Wilmer Eye Institute Baltimore MD Disclosures IW declares that she has no competing interests. Cristine Radojicic, MD Staff Physician Cleveland Clinic Cleveland OH Disclosures CR declares that she has no competing interests. Jonathan Smith, MD Specialist Registrar in Ophthalmology Royal Victoria Infirmary Newcastle UK Disclosures JS declares that he has no competing interests. Peer reviewer acknowledgements BMJ Best Practice topics are updated on a rolling basis in line with developments in evidence and guidance. The peer reviewers listed here have reviewed the content at least once during the history of the topic. Disclosures Peer reviewer affiliations and disclosures pertain to the time of the review. References Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible. Key articles Robinson A, Beech T, McDermott AL, et al. Investigation and management of adult periorbital or orbital cellulitis. J Laryngol Otol. 2007;121:545-7. Abstract Chandler JR, Langenbrunner DJ, Stevens ER. The pathogenesis of orbital complications in acute sinusitis. Laryngoscope. 1970;80:1414-1428. Abstract Hamed-Azzam S, AlHashash I, Briscoe D, et al. Common orbital infections ~ state of the art ~ Part I. J Ophthalmic Vis Res. 2018 Apr-Jun;13(2):175-82. Full text Abstract American Academy of Ophthalmology. Orbital cellulitis. Apr 2024 [internet publication]. Full text Reference articles A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice. Differentials Orbital pseudotumor (idiopathic orbital inflammation) Thyroid eye disease Panophthalmitis More Differentials #### Guidelines Guide to utilization of the microbiology laboratory for diagnosis of infectious diseases: 2024 update by the Infectious Diseases Society of America (IDSA) and the American Society for Microbiology (ASM) Appropriateness criteria: orbital imaging and visual loss - child More Guidelines #### Patient information Cellulitis and erysipelas Sinusitis More Patient information Log in or subscribe to access all of BMJ Best Practice Use of this content is subject to our disclaimer Log in or subscribe to access all of BMJ Best Practice Log in or subscribe to access all of BMJ Best Practice Log in to access all of BMJ Best Practice person personal subscription or user profile Access through your institution OR SUBSCRIPTION OPTIONS Cookies and privacy We and our 225 partners store and access personal data, like browsing data or unique identifiers, on your device. 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1286	https://www.tsc.fl.edu/media/divisions/learning-commons/resources-by-subject/math/foundational-math/number-sense/Properties-of-Real-Numbers.pdf	Properties of Real Numbers We will introduce some properties of real numbers which will be very important throughout your algebra courses. It is important that you understand each property and how it works. These properties apply to all numbers in the real number system. REAL NUMBER SYSTEM Rational numbers are real numbers that can be written as the ratio of two integers. Irrational numbers cannot. As all irrational numbers are non-repeating and non-terminating decimals, and all rational numbers are repeating or terminating decimals, an easy way to think of the real number system is as the set of all decimal numbers. These properties then apply to any number which can be written as a decimal. We will separate the properties into properties of addition and properties of multiplication. The Commutative Property of Addition If a and b are real numbers, then a + b = b + a This property of addition allows us to change the order in which we add two numbers without changing the answer For example, we know that 3 + 5 = 8 and 5 + 3 = 8 also. a + b = b + a 3+ 5 = 5 + 3 8 = 8 a + b = b + a −9+ 4 = 4 + (-9) -5 = -5 NOTE that a and b can stand for any two numbers. Int egers W hole Nu mbers Natural Numbers Ra tional Nu mbers Irrational Numbers The Associative Property of Addition If a, b and c are real numbers, then (a + b) + c = a + (b + c). This property of addition allows us to change the way we group the numbers that are being added, without changing the answer. (a +b) +c = a +(b+c) (3+ 2)+ 4 = 3+(2+ 4) 5+ 4 = 3+6 9 = 9 (a +b) +c = a +(b+c) [3 + (- 7)] + −( 2) = 3 + [-7 + (- 2)] −4 + (- 2) = 3 + (- 9) -6 = -6 NOTE that a, b and c can be any three numbers. Be careful that you can distinguish between these two properties of addition. It is not always easy. EXAMPLE: (3 + 2) + 7 = 7 + (3 + 2) Is this an example of the Associative Property of Addition or the Commutative Property of Addition? Look carefully! What has changed? At first glance you may think it is the Associative Property, but don’t let those parentheses fool you! If you look carefully, you will see that the group HAS NOT CHANGED. The 3 and the 2 are still grouped together. What HAS changed is the order. It was 3, 2, 7, and it became 7, 3, 2. This is an example of the Commutative Property of Addition! a + b = b + a (3 + 2) + 7 = 7+ (3 + 2) The Addition Property of Zero If a is a real number, then a + 0 = 0 + a = a This property means that we can add zero to any number without changing the value of that number. Zero is also called the Identity for addition. This is because zero is the only number which can be added without changing the value of the other addend. Adding zero to a number does not change that number. NOTE that a can be any real number. The Inverse Property of Addition If a is a real number, then, a + (−a) = −a + a = 0 This property means that whenever we add a number and its opposite, the sum will always be zero. The opposite of a number is also called the additive inverse of that number. a + −( a) = -a + a = 0 5 + (- 5) = −5 + 5= 0 a + −( a) = -a + a = 0 -3 + [-(-3)] = [-(-3)] + (-3) = 0 NOTE that a can be any real number. If you have a good understanding of the properties of addition, the Commutative and Associative Laws of Multiplication will be easy for you. They are the same as the laws for addition except that they apply to the operation of multiplication. The Commutative Property of Multiplication If a and b are real numbers, then 𝑎∙𝑏= 𝑏∙𝑎 ⋅ This property allows us to change the order of two numbers when we are multiplying without changing the answer. 𝑎∙𝑏= 𝑏∙𝑎 3 ∙5 = 5 ∙3 15 =15 𝑎∙𝑏= 𝑏∙𝑎 −6 ∙2 = 2 ∙(−6) −12 = −12 NOTE that a and b can be any two numbers. The Associative Property of Multiplication If a, b, and c are real numbers, then (𝑎∙𝑏) ∙𝑐= 𝑎∙(𝑏∙𝑐) This property allows us to change the way factors are grouped in a multiplication problem. (3 ∙2) ∙8 = 3 ∙(2 ∙8) (−3 ∙4) ∙2 = −3 ∙(4 ∙2) 6 ∙8 = 3 ∙16 −12 ∙2 = −3 ∙8 ⋅ 48 = 48 −24 = −24 NOTE that a, b and c can be any three numbers. As with addition, be sure that you can distinguish between the two laws of multiplication. (3 ∙2) ∙5 = 5 ∙(3 ∙2) What has changed? Order or grouping? The grouping is still the same. It is the order that has changed, so this is an example of the Commutative Property of Multiplication. The Multiplication Property of Zero If a is a real number, then 𝑎∙0 = 0 ∙𝑎 This means that when any number is multiplied by zero, the product is always zero. 6 ∙0 = 0 ∙6 = 0 : note that a can be any number. The Multiplication Property of One If a is a real number, then 𝑎∙1 = 1 ∙𝑎= 𝑎 This property means that we can multiply any number by one without changing the value of that number. 2 ∙1 = 1 ∙2 = 2 One is also called the identity for multiplication. This is because “one” is the only number that you can multiply by so that the value of the other number does not change. Multiplying a number by one does not change that number. NOTE that a can be any number. The Inverse Property of Multiplication If a is a real number and a ≠ 0, 𝑎∙ 1 𝑎= 1 𝑎∙𝑎= 1 This means that the product of any number and its reciprocal (also called the multiplicative inverse) will always be the number one. 𝑎∙1 𝑎= 1 𝑎∙𝑎= 1 2 ∙1 2 = 1 2 ∙2 = 1 NOTE that a can be any number except zero. We cannot have zero in the denominator of a fraction because division by zero is undefined. The last property we will look at involves both multiplication and addition. The Distributive Property If a, b and c are real numbers, then 𝑎(𝑏+ 𝑐) = 𝑎𝑏+ 𝑎𝑐 or (𝑏+ 𝑐)𝑎= 𝑏𝑎+ 𝑐𝑎 This property allows us to either add inside the parentheses first and then multiply that sum, or to multiply each number inside the parentheses by the number on the outside, and then add the products. The answer will be the same. 𝑎(𝑏+ 𝑐) = 𝑎𝑏+ 𝑎𝑐 3(2 + 4) = 3(2) + 3(4) 3(6) = 6 + 12 18 = 18 NOTE that a, b and c can be any three numbers. This property will be particularly helpful when we need to remove parentheses from expressions containing variables. III. EXERCISES: Identify the property which justifies each statement. a. -5 + 5 = 0 b. 3(2 + 6) = 3(2) + 3(6) c. – 4 + 10 = 10 + (-4) d. (-2 + 7) + 4 = -2 + (7 + 4) e. 0 ∙6 = 0 f. 1 3 ∙3 = 1 g. (-2)(-4) = (-4)(-2) i. (3 ∙7) ∙9 = 3 ∙(7 ∙9) h. -8 + 0 = -8 j. 12 ∙1 = 12 KEY: b. Distributive Property c. Commutative Property of Addition d. Associative Property of Addition e. Multiplication Property of Zero f. Inverse Property of Multiplication g. Commutative Property of Multiplication h. Addition Property of Zero i. Associative Property of Multiplication j. Multiplication Property of One
1287	https://www.sprachcaffe.com/espanol/cursos-espanol-extranjero/aprende-espanol/gramatica-espanol/articulos-y-genero-sustantivos.htm	Crea tu presupuesto Los artículos y el género de los sustantivos en español Los artículos, tanto definidos como indefinidos, desempeñan un papel crucial en la lengua española al proporcionar información esencial sobre los sustantivos que acompañan. Existen dos tipos de artículos: los artículos definidos y los artículos indefinidos. En español los sustantivos tienen género gramatical, es decir, se clasifican como masculinos o femeninos, aunque es fundamental que tengas en cuenta que el género del sustantivo no siempre está relacionado con el sexo biológico del objeto o ser al que hace referencia. Contenido de la página » Los artículos definidos e indefinidos en español» El género de los sustantivos en español» Conclusión Los artículos definidos e indefinidos en español Los artículos en español se clasifican en dos tipos: 1. Los artículos definidos: Determinan a un sustantivo de manera precisa, indicando que se está hablando de algo específico o conocido para el hablante y el oyente. La elección del artículo definido depende del género (masculino o femenino) y el número (singular o plural) del sustantivo al que acompaña. Los artículos definidos en español son "el", "la", "los" y "las". \| \| \| --- \| \| "El" y "los" \| Se utilizan con sustantivos masculinos, tanto en singular como en plural. + Ejemplo singular: "el libro" (the book) + Ejemplo plural: "los libros" (the books) \| \| "La" y "las" \| Se utilizan con sustantivos femeninos, tanto en singular como en plural. + Ejemplo singular: "la casa" (the house) + Ejemplo plural: "las casas" (the houses) \| 2. Los artículos indefinidos: Se utilizan para referirse a algo de manera no específica o desconocida, es decir, para expresar que no se está hablando de un objeto o conjunto de objetos en particular. Los artículos indefinidos en español son "un", "una", "unos" "unas". \| \| \| --- \| \| "Un" y "unos": \| Se utilizan con sustantivos masculinos, tanto en singular como en plural. + Ejemplo singular: "un libro" (a book) + Ejemplo plural: "unos libros" (some books) \| \| "Una" y "unas" \| Se utilizan con sustantivos femeninos, tanto en singular como en plural. + Ejemplo singular: "una casa" (a house) + Ejemplo plural: "unas casas" (some houses) \| Viajes de idiomas en español ¡Mejora tu español y disfruta de tus vacaciones! Comienza tu curso de español El género de los sustantivos en español A diferencia de otras lenguas como el alemán, en español no existe el género neutro. Los sustantivos se clasifican en dos géneros: masculinos y femeninos. Los sustantivos masculinos: Se utilizan para referirse a seres, objetos o conceptos de género masculino. Ejemplos: el libro, el perro, el árbol, el día. Los sustantivos femeninos: Se utilizan para referirse a eres, objetos o conceptos de género femenino. Ejemplos: la casa, la mesa, la silla, la noche. Reglas para determinar el género de los sustantivos A menudo, los sustantivos que terminan en -o son masculinos, mientras que aquellos que terminan en -a son femeninos. Ejemplos: el gato (masculino), la ventana (femenino) Sin embargo, esta regla no es absoluta y hay muchas excepciones. Algunos de los sustantivos más comunes que rompen esta regla son: Sustantivos masculinos que terminan en "a": El día (the day) El mapa (the map) El problema (the problem) El poema (the poem) Sustantivos femeninos que terminan en "o": La mano (the hand) La radio (the radio) La foto (the photo) La moto (the motorcycle) Conclusión \| \| \| Los artículos definidos ("el", "la", "los", "las") permiten identificar a qué objeto o conjunto de objetos nos referimos, indicando si son conocidos o específicos en el contexto. Por otro lado, los artículos indefinidos ("un", "una", "unos", "unas") señalan un objeto o conjunto de objetos de manera no específica, introduciendo algo nuevo o desconocido en la conversación. Los artículos definidos e indefinidos concuerdan en género y número con los sustantivos a los que acompañan. En español no existen los sustantivos neutros. Generalmente los sustantivos que acaban en "-o" son masculinos y los que acaban en "-a" son femeninos. \| Con estos consejos aprenderás más fácilmente cómo se expresa el género de los sustantivos y artículos. ¡No te olvides de concordar el artículo con el sustantivo al que acompaña! Memoriza las reglas y apunta en una lista las palabras que no siguen esta regla. Pon en práctica tus conocimientos y viaja a España o Cuba con Sprachcaffe para mejorar tu español. Aprende más español Los pronombres Aprende español con Sprachcaffe Viajes de idiomas en español Realiza un viaje de idiomas en España o Cuba con Sprachcaffe. Campamentos de idiomas Disfruta de un campamento de verano mientras aprendes un idioma nuevo. Curso de español online Aprende español desde la comodidad de tu casa con un curso online. Sprachcaffe / Idiomas / Español / Gramática español / artículos y género sustantivos
1288	https://www.ngsslifescience.com/science/punnett-square-worksheets	NGSS Life Science Punnett Square Worksheets Punnett square worksheets and genetics lesson plans that are designed for high school, middle school, and elementary school life science teachers are downloadable here for free. The Punnett square worksheets lets students practice the probability of inheritance. Mendel pea plant worksheets help students learn the history of genetics, the vocabulary of genetics and introduced to the probabilities of genetics. NGSS Life Science offers many great lessons. Free Lesson Plan (PDF) link below or become a member to get access to the answer key and editable file. Free Punnett square curriculum includes: \| \| \| Punnett Square Worksheet – Pea Plant Genetics \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Worksheet Students will learn about the genotypes, phenotypes, and probabilities by completing this Punnett square worksheet for the different characteristics of pea plants. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Dragon Genetics - Online Game \| \| Free Lesson Plan \| \| \| High School Project Students learn heredity, meiosis, proteins, and more by playing a free online game that models the genetics of virtual dragons. Highly visual and interactive, students love playing this game! You can run the game for 5 days up to two weeks. \| \| NGSS Standard HS-LS1-1 HS-LS3-3 \| \| \| Published by The Concord Consortium, Inc \| \| \| \| Genetics Test Question Bank \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Test Genetics question bank to help you build assessments. An active membership is required to view questions. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Genetics with SpongeBob - Dihybrid Crosses \| \| Free Lesson Plan \| \| \| High School Worksheet Students complete dihybird crosses about SpongeBob. SpongeBob SquarePants and all related characters are trademarks of Viacom International Inc. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Andrea Stonebraker \| \| \| \| Blood Type Genetics Lab \| \| Free Student Handout Free Teacher Notes \| \| \| High School Lab In this blood typing lab, students pretend to investigate blood samples (actually simple chemicals) and use their analysis of the experiment to solve a crime and determine if two babies were switched in the hospital. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Ingrid Waldron and Jennifer Doherty \| \| \| \| Independent Assortment and Gene Linkage Lab - Dragon Genetics \| \| Free Student Handout Free Teacher Notes \| \| \| High School Lab Students learn the principles of independent assortment and gene linkage in activities that analyze the inheritance of multiple genes on the same or different chromosomes in hypothetical dragons. The concepts of independent assortment of chromosomes during meiosis and random fertilization are modeled with linked genes. \| \| NGSS Standard HS-LS1-4 HS-LS3-3 \| \| \| Published by Ingrid Waldron and Jennifer Doherty \| \| \| \| Demystifying Punnett Squares Simulation \| \| Free Multimedia \| \| \| High School Lab In the simulation, students breed pairs of mice with different fur colors and observe the phenotypic ratios in the resulting offspring. Students select one breeding pair from the nest to investigate the ratios of fur color phenotypes, genotypes, and sex. Students then click “breed”. After each parent pair breeding, the resulting litter of offspring is shown, along with a pie chart, which updates to display the ratios in the total number of offspring for those parent mice. Example questions to ask students:Why do some of the pie charts change as you breed the mice?Why do the pie charts stabilize?Where are the numbers different from the ratios in Punnett Square? Where are they the same? \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Concord Consortium \| \| \| \| Punnett Square Worksheet - Fruit Fly Genetics \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Worksheet In this Punnet square worksheet, students will learn about the genotypes, phenotypes, and probabilities of genetics for different characteristics of fruit flies. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Autosomal Pedigrees Worksheet \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Worksheet Students will learn how to use pedigrees to trace a autosomal genes as they pass from generation to the next generation. Students will also learn about genetic disorders. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Genetics with SpongeBob - Incomplete Dominance \| \| Free Lesson Plan \| \| \| High School Worksheet This worksheet provides several practice problems related to incomplete dominance with Poofkins (a special kind of flower) and Goobers (a special breed of jellyfish!) An answer key is provided. SpongeBob SquarePants and all related characters are trademarks of Viacom International Inc. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Tracy Trimpe \| \| \| \| SpongeBob Genetics Quiz \| \| Free Lesson Plan \| \| \| High School Test This SpongeBob genetics quiz includes heterozygous, homozygous, dominant, recessive, phenotype, genotype, and Punnett squares. SpongeBob SquarePants and all related characters are trademarks of Viacom International Inc. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Tracy Trimpe \| \| \| \| Punnett Square Genetics with SpongeBob #2 \| \| Free Lesson Plan \| \| \| High School Worksheet A second helping of Bikini Bottom fun exploring the genetics of SpongeBob and his pals! An answer key is provided. SpongeBob SquarePants and all related characters are trademarks of Viacom International Inc. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Tracy Trimpe \| \| \| \| Punnett Square Genetics with SpongeBob #1 \| \| Free Lesson Plan \| \| \| High School Worksheet What happens when SpongeBob SquarePants marries SpongeSusie RoundPants? Explore the genetics of Sponge Bob and his pals with this fun worksheet! An answer key is provided! SpongeBob SquarePants and all related characters are trademarks of Viacom International Inc. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Tracy Trimpe \| \| \| \| Genetics Lab - Introduction \| \| Free Student Handout Free Teacher Notes \| \| \| High School Lab In this genetics lab, students are introduced to genetics vocabulary, including meiosis, fertilization, Punnett squares, and pedigree analysis. Students flip coins to mimic the probability of genetics and use the coin flip results to answer inheritance questions. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by Ingrid Waldron and Jennifer Doherty \| \| \| \| Dihybrid Cross Worksheet \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Worksheet In this dihybrid cross worksheet, students will practice performing two trait crosses. Students will cross different parental genotypes and determine the offspring phenotype, and probabilities of having each trait. The last question asks students to make the connection between multiple traits and Mendel’s law of independent assortment. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Punnett Square Worksheet - Human Genetic Disorders \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Worksheet In this Punnett square worksheet, students will learn about human genetic disorders for autosomal, sex-linked, pleiotropy, and incomplete dominant traits. Students will cross different parental genotypes and determine the offspring genotype, phenotype, and probabilities of having the genetic disorder and being a carrier of the genetic disorder. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Punnett Square Worksheet - Incomplete Dominant and Codominant Traits \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Worksheet Students will learn the difference between incomplete dominance and codominant traits by completing Punnett squares. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Heredity Lab \| \| Free Lesson Plan (PDF) Lesson Plan (DOCX) & Answer Key with Membership \| \| \| High School Lab Students will model how DNA is passed from parent to offspring to cause the physical traits of a fictitious baby. \| \| NGSS Standard HS-LS3-3 \| \| \| Published by NGSS Life Science \| \| \| \| Mendelian Dragon Genetics Lab \| \| Free Student Handout Free Teacher Notes \| \| \| High School Lab In this genetics lab, students use Popsicle sticks to represent a pair of homologous chromosomes with multiple genetic traits. Teams of students use their sets of Popsicle sticks to model fertilization and then identify the genetic makeup and phenotypic traits, resulting in a fictitious baby dragon. \| \| NGSS Standard HS-LS3-3 MS-LS3-2 (Types of Reproduction) \| \| \| Published by Ingrid Waldron and Jennifer Doherty \| > ## MembershipIncludes: All Lesson Plans + Editable Files + Answer Keys + Test Question Banks Discover How Punnett Square Concepts Classical Genetics - Gregor Mendel, inheritance, heredity of traits, pea plants experiments. Punnett Squares - genotype, phenotype, genotypic ratios, phenotypic ratios Monohybrid Cross Punnett Squares - pea plant genetics, genetic disorders, genetic probability with a single trait. Dihybird Cross Punnett Squares - fruit fly genetics, genetic probability with two traits. Advanced Punnett Squares - incomplete dominance, pleiotropy, autosomal traits, sex-linked traits. Mendel’s Law of Dominance - dominant, recesive, carrier of a trait, dominant allele, recessive allele. Law of Segregation - Gene, allele, homozygous, heterozygous, gametes. Law of Independent Assortment - multi-trait crosses, 9:3:3:1 ratio, chromosomes assort independently. Genetic Disorders - cystic fibrosis, albinism (albino), neurofibromatosis, sickle-cell anemia, hemophilia A, color blindness. Gametes - sperm, egg, ovum, gonads. Pedigrees - autosomal and sex-linked. Lessons Organized by NGSS Standard LS1 From Molecules to Organisms LS2 Ecosystems LS3 Heredity LS4 Biological Evolution
1289	https://www.youtube.com/watch?v=UqKiSjVXLIo	Ring Opening of Epoxides Under Acidic Conditions Michael Evans 29700 subscribers 8 likes Description 603 views Posted: 18 Jul 2023 00:00 Ring Opening in Acid 02:25 Stereochemistry 04:13 Regiochemistry 06:57 Tertiary Epoxides 09:00 Example 1 11:01 Example 2 Transcript: Ring Opening in Acid because of the strain built into the three-membered epoxide ring these cyclic ethers are susceptible to nucleophilic substitution reactions and strong nucleophiles will open epoxides through a pretty much pure sn2 mechanism with concerted formation of the nucleophile carbon Bond and cleavage of the carbon oxygen Bond under acidic conditions with much weaker nucleophiles in the presence of an acid nucleophilic substitution of epoxides can also occur so ring opening can be done under acidic conditions as well but in this case protonation of the epoxide oxygen comes before nucleophilic substitution and so we're dealing with an electrophile that is not a neutral epoxide but a protonated epoxide and that protonated epoxide is highly electrophilic so neutral El uh nucleophiles neutral nucleophiles like alcohols and water can then open the epoxide and so as a general mechanism here the first thing we're going to do is protonate the epoxide now you might imagine particularly when the epoxide is highly substituted imagine I had two other groups here you might imagine that this could actually open to form an open carbocation with the hydroxy group linked to the carbon next to the cationic center um however this generally does not occur in ring opening reactions of epoxides even when we have a highly substituted carbon like this we'll discuss the evidence for this actually when we get down to the stereochemistry CH istry but for the time being very briefly just take my word for it that no open carbocation forms in this reaction mechanism this can be proven using the stereochemical outcome and so in general what we observe here as well is an sn2 type mechanism involving the protonated epoxide being attacked by the nucleophile with cleavage of the carbon oxygen Bond occurring at the same time so for example if we use a hydrohalic acid HX the x minus anion the halide anion attacks at the electrophilic carbon and the co Bond breaks all at the same time and we get this Halo hydron interesting way to make Halo hydrogens we could also use alcohols and then we get these beta or two Al coxy alcohol products and other nucleophiles work in this context as well as long as they're weak right we're looking at Water neutral alcohols neutral carboxilic acids that kind of Stereochemistry thing now the stereo chemistry here is very interesting because again we could imagine an open carbocation intermediate a kind of S in1 type of reactivity for proteinated epoxides under acidic conditions this jives with what we've seen for alkal halides where when we use weak nucleophiles it's quite common for an sn1 mechanism to be involved however the stereochemical outcome of the reaction argues against this for example if we take this epoxide and we hit it with HX hbr HCL what we observe is selective substitution at the more substituted carbon and we'll talk about why that is on the next slide but at the same time while this suggests maybe there's a tertiary carbocation involved we observe exclusive inversion of configuration at the electrophilic carbon so this carbon highlighted in purple is the one that reacts and we go from an S configuration to an R configuration and the key thing to notice is that the carbon oxygen bond is sort of up and to the left and the new carbon X spond is down to the right so an inversion of configuration has occurred a kind of umbrella flip right of these three groups has taken place in going from the reactant to the product so there's an inversion of configuration that takes place here this argues against an open carbocation since an open carbo cation would lead to a mixture of inversion and retention of configuration and this inversion is observed regardless of the configuration of the starting material right so if we start with the epoxide we get get R product if we switch up the configuration in the reactant start with the r epoxide we get the S product so this is a stereos specific inversion of configuration great evidence for an sn2 mechanism the regiochemistry of ring Regiochemistry opening of epoxides under acidic conditions is probably the most complicated regiochemical situation that you'll encounter maybe in both organic chemistry 1 and organic chemistry 2 uh because it gets quite subtle since there's a competition between two different things going on here we saw in the last slide with the stereochemistry that the mechanism of this reaction is likely sn2 even when the epoxide is protonated there's no ring opening to a carbo canion this suggests the uh reaction should be sensitive to steric effects and that steric hindrance should play a role in determining the regiochemical outcome this is observed in some circumstances however in other circumstances we observe regiochemistry that suggests that it's actually the partial positive charge in the epoxide and where that's living where most of the partial positive charge is living that drives the Regio selectivity and these are two different carbons in the epoxide and what happens actually depends on the substitution patterns of both of the carbons of the epoxide ring so we're going to look at two cases where the outcomes are different in the first case we've got a primary carbon versus a secondary carbon so here for example we have a primary carbon just a ch2 and a secondary carbon a CH with another alkal substituent in this case reaction occurs selectively at the less hindered primary position and the main reason for this is the typ the classic sn2 factor of steric hindrance the secondary carbon is more sterically hindered than the primary carbon and so nucleophilic attack occurs here we get an X link to the less substituted position the second case we're going to look at is a tertiary epoxide carbon versus a primary or secondary so this is any epoxide that has a tertiary carbon in it essentially so here for example we have a primary carbon versus a tertiary carbon under these circumstances reaction occurs selectively at the tertiary carbon so there's a switch in the Regio selectivity now where apparently sterix is not the main driving Factor right if sterix were the main driving Factor reaction would occur at the primary carbon not the tertiary one even when that carbon is secondary we still get selective reaction at the tertiary carbon even though the secondary carbon again is going to be less hindered than the tertiary carbon so it's not sterix that's driving this effect in fact it's an electronic effect having to do with where positive charge really lives in the protonated epoxide intermediate so Tertiary Epoxides here for example in the first case and this result generalizes to really any epoxide containing a fully substituted carbon like this we can actually draw resonance forms of this epoxide where we break the carbon oxygen bonds toward oxygen showing that the carbons are sharing some of the positive charge and the proton epoxide this isn't exactly a street legal version of resonance right since we're breaking a sigma Bond but it's illustrative and to me it's legal as long as we keep that in mind we're not actually moving the atoms we can do that to generate this tertiary carbocation or we can break the other Co bond to generate a hypothetical primary carbocation but this doesn't look right right this is a primary carbocation and those are a no no so this is a minimal contributor to the true structure of the protonated epoxide and what it shows us is that most of the partial positive charge is actually living on this more substituted carbon in the proted epoxide intermediate and so nucleophiles are drawn to this partial positive charge right there's an electrostatic attraction between the x minus anion right the Hali anion and this partial positive charge this explains why x minus ends up linking to that more substituted carbon in the protonated epoxide so it's all about the electronics of the protonated epoxide intermediate it's really only this tertiary carbon that can support enough positive charge to make this an important effect when all we have quote unquote is a secondary carbon in the epoxide well then we default back to the it's an sn2 reaction uh idea and we're going to attack at the less substituted position in that case so this is complicated it's a pretty complicated regiochemical situation where a tertiary carbon leads to a different outcome than a a secondary carbon and attack at tertiary carbons kind of wins the day due to this electronic effect so you want to be really careful when predicting the products of ring opening of epoxides under acidic conditions when the epoxide has two carbons with different subst stitution patterns this Example 1 is a skill that I think really requires a lot of practice and so on this slide we're going to predict the products of these acidic ring opening of ring openings of epoxides thinking mechanistically throughout this process and so we know we're in acidic conditions for example in the first reaction because we're dealing with sulfuric acid h2so4 so the first step is going to be protonation of the epoxide oxygen the other thing we want to notice here is that we have a tertiary carbon and a secondary carbon in the epoxide so a difference in substitution pattern meaning we're going to have a regiochemical issue so here's the proteinated epoxide and based on our kind of regiochemical ideas we know that this is where the positive charge is really located that tertiary carbon and this is where nucleophilic attack is is going to occur the finel ring actually plays a role in this as well since it can stabilize positive charge even more than just a plain vanilla alkal group right so the nucleophile is going to attack here what is the nucleophile well ethanol is around in solvent quantities and the ethanol oxygen is a decent nucleophile so we're going to attack there and open the ring and an sn2 step this is important for the stereochemical outcome right we want to make sure to depict inversion of configuration at this electrophilic Center in this step and it's worth pausing the video now and verifying that in fact this does show an inversion of configuration again there's some kind of mental gymnastics where you want to do here where you imagine the eth coming from behind this ethyl group pushing the ethyl group forward such that the ethanol group ends up in the back and then we're going to deprotonate that ethoxy group to generate the neutral product which has the ethoxy group linked to the more substituted carbon and the O group derived from the oxygen of the epoxide still link to the less substituted carbon and notice that this stereo Center the secondary carbon that was just along for the ride the whole time its configuration stays the same throughout the mechanism in this second case we again Example 2 have a difference in substitution pattern with a tertiary or fully substituted carbon and a secondary carbon and we've got hbr now as the acid it's a strong acid we're going to protonate the epoxide first this generates the proteinated epoxide and BR minus with a tertiary carbon that's where we're going to do the attack it's an sn2 ring opening and so we're going to imagine an inversion of configuration although interestingly enough this electrophilic carbon is not a stereo Center so it actually does not matter that inversion of configuration occurs in this particular case and the opening occurs to give this Halo hydron in fact and again notice this carbon wasn't directly involved this secondary carbon was not directly involved in the ring opening and so its configuration Remains the Same throughout the reaction
1290	https://www.youtube.com/watch?v=l6lI0mfciUU	Truth Tables (Introduction) Houston Math Prep 51600 subscribers 27 likes Description 2960 views Posted: 28 Aug 2020 This truth table tutorial video teaches you the very basics of building truth tables and explains how to make truth tables for negation conjunction and disjunction. We start with the smallest truth table that involves only one statement. Then we move on to the standard arrangement for a truth table that involves two statements (p and q). We then show you the truth tables for the conjunction and disjunction, followed by two additional examples of building truth tables that involve negation conjunction and disjunction. Truth tables and logic are a common topic found in a discrete mathematics course. 0:00 Introduction to truth tables 2:32 Conjunctions in truth tables (and) 4:18 Disjunctions in truth tables (or) 5:53 Making a truth table (Example 1) 7:30 Making a truth table (Example 2) Houston Math Prep Foundation of Math Playlist: Houston Math Prep Finite Math Playlist: Transcript:
1291	https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_6?srsltid=AfmBOopehkuISMRydJZjiUpADdqwVkt-zY-I5tBboDVNCrIu2weXbEnR	Art of Problem Solving 1983 IMO Problems/Problem 6 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1983 IMO Problems/Problem 6 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1983 IMO Problems/Problem 6 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 5 Solution4 Problem Let , and be the lengths of the sides of a triangle. Prove that Determine when equality occurs. Solution 1 By Ravi substitution, let , , . Then, the triangle condition becomes . After some manipulation, the inequality becomes: . By Cauchy, we have: with equality if and only if . So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral. Solution 2 Without loss of generality, let . By Muirhead or by AM-GM, we see that . If we can show that , we are done, since then , and we can divide by . We first see that, , so . Factoring, this becomes . This is the same as: . Expanding and refactoring, this is equal to . (This step makes more sense going backwards.) Expanding this out, we have , which is the desired result. Solution 3 Let be the semiperimeter, , of the triangle. Then, , , and . We let , and (Note that are all positive, since all sides must be shorter than the semiperimeter.) Then, we have , , and . Note that , so Plugging this into and doing some expanding and cancellation, we get The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by , which we know is positive from earlier so we can maintain the sign of the inequality. This gives We move the negative terms to the right, giving We rewrite this as where is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of and on the right sum to 1 the right side will be . Solution4 "Solution from 111 problems in Algebra and Number Theory" Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1292	https://d43fweuh3sg51.cloudfront.net/media/alfresco/u/pr/WPSU/Math%20Interactive%20%26%20Lesson%20Plan%20What%20are%20the%20Odds_c46ebb52-e58e-4ca6-9e3b-c0c0a5d3c70b/What%20are%20the%20Odds_LP.pdf	TITLE: WHAT ARE THE ODDS? SUBJECT: Mathematics/Algebra 2 Grade level: High school ALIGNMENTS: Standards: 2.7.A2.A, 2.7.A2.C ANCHOR: A2.2.3.2.2 ELIGIBLE CONTENT: M11.E.3.1.2 VOCABULARY: • Probability: A number from 0 to 1 that indicates how likely something is to happen. Can be expressed as a decimal, fraction, or percent. • Odds: The odds of an event occurring is the ratio of the number of ways the event can occur (successes) to the number of ways the event cannot occur (failures). • Favorable outcomes: the outcomes for an event you wish to have happen OBJECTIVES: • Students will find the probability and/or odds of an event. • Students will convert probability to odds and/or odds to probability. • Students will determine which event has the better chance of occurring based on the odds of one event and probability of another event. ESSENTIAL QUESTIONS: • How do we calculate probability and odds? • How do we convert probability to odds and odds to probability? • How can we determine which event has a better chance of occurring based on the odds of one event and the probability of the other event? DURATION: 45-60 min MATERIALS: • Board/Interactive white board • Partner/Team Worksheet • Internet access • Calculators INSTRUCTIONAL PROCEDURES: W: For a warm up, have decimal numbers that students have to convert to a percent, percent that they have to convert to decimal, decimal they have to convert to fraction, fraction they have to convert to percent. Here are some examples you could have on the board/interactive white board as a warm up. 1. Convert the decimal 0.52 to a fraction. A: 13/25 2. Convert the fraction 16/82 to a percent. A: 19.51% 3. Convert the fraction 4/9 to a decimal. A: 0.44 4. Convert the percent 59% to a decimal. A: 0.59 5. Convert the percent 32% to a fraction. A: 8/25 Check answers before moving on to lesson. “What does probability mean?” Allow students time to think and respond. Write their responses on board/interactive white board. Have students write down responses on their notes. The students should come up with something like: the likelihood that an event will occur. “What does odds mean?” Allow students time to think and respond. This one might be more difficult for students. Write their responses on the board/interactive board. Have students write down responses on their notes. Students will probably say probability or give examples of how they have heard odds being used, like: the chances of winning the lottery are 1 to 100,000. “What are some other words that can be used for probability?” Allow students time to think and respond. Write their responses on board/interactive white board. Have students write down responses on their notes. The students should come up with something like: chances, percentage, long shot, proportion, likelihood, maybe even odds. “Where do you see/hear probability/odds being used in every day life?” Allow students time to think and respond. Write their responses on board/interactive white board. Have students write down responses on their notes. The students should come up with something like: Weather – 30% chance of rain, lottery – less than 10% chance of winning, horse races – horse has a 10 to 2 chance of winning, and buying a vehicle – the chances that you will have to repair it. “Good, today we are going to be finding the probability and odds of events. Then we are going to be given the odds of an event and convert it to probability and vice versa. We are going to do a few guided problems together, then you and your partner will do 8 problems, and when you are done with that, we will use the internet game, “What are the Odds?” to evaluate your understanding.” H: “We know what probability means but how do we find it? Probability is the favorable outcomes divided by the total number of outcomes.” Write this on the board/interactive whiteboard for students. Have them write it in their notes. “What is a favorable outcome?” Allow students time to think/talk with partner/team. Students should come up with something like ‘what we want to happen’. “Yes, a favorable outcome is what we want to happen. And when we divide that by total number of outcomes, what is that?” Allow students time to think/talk with partner/team. Students should come up with ‘the total outcomes possible’ or ‘everything that could happen’. “Let’s say an algebra class has 17 males and 16 females. One student is chosen at random from the class. What is the probability that the student chosen is a female?” Allow students time to respond – talk with their partner/team to come up with an answer. Students should come up with 16/33 or 48.48%, or 0.4848. Write all responses on board/interactive board. “Probability can be expressed as a decimal, fraction, or percent. Remember when we reviewed that earlier in the warm up? That is why we reviewed that because all of these answers, 16/33, 48.48% or 0.4848, are correct.” “Now let’s take that same example -- an algebra class has 17 males and 16 females -- and find the odds of selecting a male. Odds are computed by the number of favorable outcomes: number of unfavorable outcomes.” Write this on board for students. Have them write it on their notes. “The colon is where you would say “to”. For example, if the odds of an event were 5:3, you could say “5 to 3”. So let’s find the odds of selecting a male from the algebra class.” Allow students time to respond – talk with partner/team to come up with an answer. Students should come up with 16:17. “The answer is 16:17. Let’s say we found odds of an event to be 2:6. Do we leave it that way or could we simplify that?” Allow students time to think or talk with partner/team. Students should come up with an answer of 1:3. “The answer is 1:3. You do need to simplify the ratio.” “Now let’s get into converting probability to odds. Let’s say you have the probability of an event is 2/5.” (Write on board.) “What would the odds be for the same event?” Allow students time to respond – talk with partner/team to come up with an answer. Student should come up with an answer of 2:3. If a person/team/partner got the right answer, have him/her/them explain how they got the answer. “The odds of the same event are 2:3. The odds have to add up to the total, which if you look at the probability, the total is 5. So to find the odds from probability, you keep the first number, the 2, then subtract the two numbers, 5-2, to get the second number for the odds. 2:3.” (Write on board.) “Now let’s convert the odds, 4:23 (write on board) of an event to probability.” Allow students time to respond – talk with partner/team to come up with answer. Student should come up with an answer 4/27. If a person/team/partner got the right answer, have him/her/team explain how they got the answer. “The probability of the same event is 4/27. The probability needs the total so you have to add up the two numbers to get the total, 27. Then it is the first number, 4, over 27. 4/27.” (Write on board.) E: “Okay, now we are going to do 8 problems with your partner. When you are done, let me check them and then you can get on the computer and play “What are the Odds?” Hand out Partner/Team worksheet. You could have students work in teams or partners, however things work in your classroom. The main thing here is to make sure students check their work with someone before moving onto the next problem. Answers: 1. 3/17 2. 2/10 or 1/5 3. 10: 50 or 1:5 4. 6: 3 or 2:1 5. 1:24 6. 4: 9 7. 5/17 8. 7/26 R: As students are working in partners, monitor student performance. Visit each group and have students explain their thinking and clarify any misunderstandings. Once they have show mastery, have students get on the internet for activity below. E: For individual practice, have students go to SUGGESTED INSTRUCTIONAL STRATEGIES: T: For those students who are showing strong proficiency of the concepts, you can have one of the partners make up their own probability (or odds) word problem and the other partner will then work it out. Then they have to check the work. For those still struggling with the concept after the partner activity, pull together a small group to continue practicing. Then have them get on the Internet. O: The goal of this lesson is to find probability and odds of events and, to convert probability to odds and vice versa. The activities in this lesson are intended to move the student from the teacher guided items to independent rigorous problems for converting probability to odds and odds to probability. FORMATIVE ASSESSMENT: • Ongoing teacher observation during partner work work, student interaction, and computer activity • Partner worksheet KEY WORDS: probability, odds, converting probability to odds, converting odds to probability, favorable outcomes, RELATED MATERIALS & RESOURCES: Interactive: Student Handout: See next page. PROBABILITY & ODDS Partner/Team Worksheet. Name: ______ 1. There are 3 blue marbles, 8 red marbles, and 6 orange marbles in a jar. What is the probability that Caleb will draw a blue marble at random? 2. What is the probability a randomly selected letter in the word METROPOLIS is an "O"? 3. A machine randomly selects a stuffed animal. There are 12 giraffes, 3 panda bears, 10 elephants, 5 horses, 21 teddy bears, 4 dogs, and 5 cats. What are the odds that you get an elephant? 4. You randomly choose an integer from 0 through 9. What are the odds that the integer is 4 or more? 5. The probability of a person winning a door prize is 1 25 . What are the odds that the person will win the door prize? 6. The probability of an event is 4 13 . What are the odds of the same event occurring? 7. The odds of an event occurring are 5:12. What is the probability of the same event occurring? 8. The odds of an event occurring are 7:19. What is the probability of the same event occurring?
1293	https://hull-awe.org.uk/index.php/Repel_-_repulse	Repel - repulse - Hull AWE Repel - repulse From Hull AWE Jump to: navigation, search Repel and repulse have a common history. Both are old; but to repel is an older word (first recorded in 1450, according to OED). Although the noun 'a repulse' goes back to 1533, the verb 'to repulse' is as old as 1489 in one sense at least. They mean the same. To some extent, OED views them as interchangeable. So the feeling among British academics that 'to repel' is the more traditional form of the word is not supported by the evidence; there is nothing to choose between them. Still, AWE's choice (prejudice) is for 'to repel' as the verb: we prefer to use repulse only as a noun. Style always depends on the taste of the speaker. Retrieved from " Categories: Usage Disambig Navigation menu Personal tools Log in Namespaces Page Talk Variants Views Read View source View history Actions Search Navigation Main Page Help Feedback Categories Bibliography editors Editors' Portal Recent changes Random page Category list Tools What links here Related changes Special pages Printable version Permanent link Page information This page was last modified on 17 February 2017, at 17:26. This page has been accessed 2,830 times. Privacy policy About Hull AWE Disclaimers
1294	https://stats.libretexts.org/Courses/Citrus_College/Statistics_C1000%3A_Introduction_to_Statistics/04%3A_Probability_and_Combinatorics/4.07%3A_Chapter_4_Formulas	4.7: Chapter 4 Formulas - Statistics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 4: Probability and Combinatorics Statistics C1000 - Introduction to Statistics { } { "4.01:Empirical_Probability" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.02:_Theoretical_Probability" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.03:_Addition_Rules_for_Probabilities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.04:_Multiplication_Rules_and_Conditional_Probability" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.05:_At_Least_One_Rule_and_Tree_Diagrams" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.06:_Counting_Rules" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.07:_Chapter_4_Formulas" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.08:_Chapter_4-_Key_Terms_and_Symbols" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Fundamentals_of_Statistics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Frequency_Distributions_and_Graphs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Descriptive_Statistics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Probability_and_Combinatorics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Discrete_Probability_Distributions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Continuous_Probability_Distribution" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Confidence_Interval_for_One_Sample" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Hypothesis_Testing_for_One_Sample" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Hypothesis_Testing_for_Two_Samples" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Correlation_and_Linear_Regression" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Chi-Square_Test" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Analysis_of_Variance(ANOVA)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Statistical_Tables" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Wed, 13 Aug 2025 01:13:57 GMT 4.7: Chapter 4 Formulas 56444 56444 Joshua Halpern { } Anonymous Anonymous 2 false false [ "article:topic", "showtoc:no", "license:ccby", "licenseversion:40", "authorname:tberberyan-tnguyen-aswa" ] [ "article:topic", "showtoc:no", "license:ccby", "licenseversion:40", "authorname:tberberyan-tnguyen-aswa" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Citrus College 4. Statistics C1000 - Introduction to Statistics 5. 4: Probability and Combinatorics 6. 4.7: Chapter 4 Formulas Expand/collapse global location Statistics C1000 - Introduction to Statistics Front Matter 1: Fundamentals of Statistics 2: Frequency Distributions, and Graphs 3: Descriptive Statistics 4: Probability and Combinatorics 5: Discrete Probability Distributions 6: Continuous Probability Distribution 7: Confidence Interval for One Sample 8: Hypothesis Testing for One Sample 9: Hypothesis Testing for Two Samples 10: Correlation and Linear Regression 11: Chi-Square Test 12: Analysis of Variance (ANOVA) 13: Statistical Tables Back Matter 4.7: Chapter 4 Formulas Last updated Aug 13, 2025 Save as PDF 4.6: Counting Rules 4.8: Chapter 4 - Key Terms and Symbols picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report View on CommonsDonate Page ID 56444 Toros Berberyan, Tracy Nguyen, and Alfie Swan Citrus College ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Chapter 4 Formulas 1. General Formula for the Probability of an Event 2. Addition Rule #I for Probability for Mutually Exclusive Events 3. Addition Rule #II for Probability for Non-Mutually Exclusive Events 4. At Least One Probability 5. Complement Rule for Probability 6. Conditional Probability 7. Multiplication Rule I for Probability for Independent Events 8. Multiplication Rule II for Probability for Dependent Events 9. Combination Rule 10. Fundamental Counting Rule 11. n-Factorial 12. Permutation of n Distinct Objects 13. Permutation with Repetition of Identical Items 14. Permutation Rule for Selecting r Objects from n Distinct Items Authors Chapter 4 Formulas General Formula for the Probability of an Event The probability that an event E will occur is the ratio of the number of outcomes in the event to the total number of outcomes in the sample space. It gives a numerical measure of the likelihood that the event will happen. P⁡(E)=number of outcomes in the event total number of outcomes Addition Rule #I for Probability for Mutually Exclusive Events The Addition Rule I in probability theory is used to find the probability that one of two events occurs when the events are mutually exclusive. Mutually exclusive events cannot occur at the same time. This means that if events A and B cannot happen at the same time, the probability of either A or B happening is the sum of their probabilities. P⁡(A or B)=P⁡(A)+P⁡(B) Addition Rule #II for Probability for Non-Mutually Exclusive Events The Addition Rule II in probability theory is used to find the probability that one of two events occurs, even if the events are not mutually exclusive. When events can occur at the same time, we must subtract the probability of both events occurring together to avoid double-counting. P⁡(A or B)=P⁡(A)+P⁡(B)−P⁡(A and B) At Least One Probability The "at least one" rule in probability is used to calculate the probability that an event occurs at least once in multiple trials. Rather than computing all possible ways the event can happen, it is often simpler to use the complement rule: subtract the probability that the event does not occur at all from 1. P⁡(at least one)=1−P⁡(none) Complement Rule for Probability The complement rule in probability is used to find the probability that an event does not occur. It is based on the fact that the total probability of all possible outcomes is 1. Therefore, the probability that event A does not happen is equal to 1 minus the probability that it does happen. P⁡(Not A)=1−P⁡(A) Conditional Probability Conditional probability is the probability that one event occurs given that another event has already occurred. It helps us understand how the occurrence of one event affects the likelihood of another. This means the probability of B given A is equal to the probability of both A and B occurring divided by the probability of A. P⁡(B∣A)=P⁡(A and B)P⁡(A) Multiplication Rule I for Probability for Independent Events The Multiplication Rule I in probability theory is used to find the probability that two events occur in sequence, assuming the events are independent. Independent events are those where the outcome of one event does not affect the outcome of the other. P⁡(A and B)=P⁡(A)⋅P⁡(B) Multiplication Rule II for Probability for Dependent Events The Multiplication Rule II in probability theory is used to find the probability that two events occur in sequence when the events are dependent. Dependent events are those where the outcome of the first event affects the outcome of the second event. This means the probability of both A and B occurring is equal to the probability of A times the probability of B, given that A has occurred. P⁡(A and B)=P⁡(A)⋅P⁡(B∣A) Combination Rule The combination rule is used to determine the number of ways to choose r items from a group of n items when the order of selection does not matter. This is useful in situations where the arrangement of selected items is irrelevant. n⁢C r=n!r!(n−r)! Fundamental Counting Rule The fundamental counting rule is used to determine the total number of possible outcomes when a sequence of events occurs and each event has a specific number of possible outcomes. If the first event can occur in m 1 ways, the second in m 2 ways, and so on, then the total number of combined outcomes is the product of all these values. Total outcomes=m 1⋅m 2⋅m 3⋯m k n-Factorial The term "n factorial" (written as n!) represents the product of all positive integers from 1 to n. It is used frequently in counting and probability problems, especially in permutations and combinations. n!=n⋅(n−1)⋅(n−2)⋅⋯⋅2⋅1 Permutation of n Distinct Objects The permutation of n items is an arrangement of those items in a specific order. If all n items are used, the total number of different permutations is computed using the formula below. n! Permutation with Repetition of Identical Items The permutation rule with objects that repeat is used when some objects are identical (i.e., indistinguishable from each other); the total number of distinct permutations is reduced. The formula accounts for repeated items by dividing by the factorial of each group of repeated objects, where there are n total items, with r 1 indistinguishable of one kind, r 2 of another, and so on is provided below. n!r 1!⋅r 2!⋅⋯⋅r k! Permutation Rule for Selecting r Objects from n Distinct Items The permutation rule for n choose r gives the number of ways to arrange r items from a set of n distinct items, where the order matters and no repetitions are allowed. The number of such permutations is computed using the formula below. n⁢P r=n!(n−r)! Authors "4.7: Chapter 4 Formulas" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY 4.0 This page titled 4.7: Chapter 4 Formulas is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Toros Berberyan, Tracy Nguyen, and Alfie Swan. Back to top 4.6: Counting Rules 4.8: Chapter 4 - Key Terms and Symbols Was this article helpful? Yes No Recommended articles 4.1: Empirical ProbabilityEmpirical probability is based on observed outcomes from experiments rather than theoretical calculations. It uses the ratio of favorable outcomes to ... 4.2: Theoretical ProbabilityTheoretical probability is based on known possible outcomes, assuming each is equally likely. Probabilities range from 0 to 1, where 0 means impossibl... 4.4: Multiplication Rules and Conditional ProbabilityThe multiplication rules help find the probability of two events happening together. If the events are independent, meaning one does not affect the ot... 4.6: Counting RulesCounting rules help determine how many ways events can occur. The fundamental counting rule multiplies the number of choices for each step in a sequen... 4.3: Addition Rules for ProbabilitiesThe two addition rules for probability determine the chance of either one event or another occurring. If the events are mutually exclusive, meaning th... Article typeSection or PageAuthorToros Berberyan, Tracy Nguyen, and Alfie SwanLicenseCC BYLicense Version4.0Show TOCno Tags This page has no tags. © Copyright 2025 Statistics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 4.6: Counting Rules 4.8: Chapter 4 - Key Terms and Symbols Complete your gift to make an impact
1295	https://habitaware.com/blogs/habitaware-blog/what-does-excoriate-mean?srsltid=AfmBOopnABl7BD9LnoG3Rh3Xq77xrMwG3nzL2qIPDGuAcicAEpIuEuS3	Change your life at the Great BFRB Get-Together Virtual Conference! Item added to your cart What Does Excoriate Mean? Excoriate, as a medical term, refers to the action of damaging or removing part of the skin's surface. This behavior can result from various underlying causes, and understanding its implications is essential. In this blog, we will learn what does excoriate mean and identify its medical and psychological causes. Excoriation Definition in Medical Terminology In medical practice, excoriate is used to describe the act of damaging the skin, which can include scratching, picking, or scraping the skin's surface. This behavior can lead to abrasions, sores, and wounds, potentially exposing the body to infections and other complications. Causes and Triggers of Excoriation Excoriation can arise from a range of physical and psychological factors. In some cases, skin conditions like eczema, psoriasis, or dermatitis can lead to itching and subsequent scratching, resulting in excoriated skin. Allergic reactions to certain substances may also provoke the urge to scratch and damage the skin. Psychological triggers can also play a significant role in compulsive skin picking, which might lead to excoriation. Anxiety, stress, and obsessive-compulsive tendencies can drive individuals to engage in this behavior to cope with emotional turmoil. Excoriate Definition by HabitAware Excoriated meaning signifies the repetitive behavior of skin picking, stemming from emotions such as anxiety. This viewpoint acknowledges it as a coping mechanism, providing momentary solace while revealing concealed distress. The motive isn't self-harm; rather, it serves as a means to navigate overpowering sentiments. Psychological treatments prioritize understanding self-kindness, confronting triggers, and nurturing more positive methods. With empathy, those grappling with excoriation can discover comfort in comprehending its emotional origin, thereby nurturing recovery and self-approval. Approaches to Managing Excoriation Managing excoriation involves addressing both the physical and psychological aspects of the behavior. For those with underlying skin conditions, dermatological treatment can help alleviate itching and reduce the urge to scratch. Moisturizers, topical medications, and antihistamines may be prescribed to soothe the skin and prevent further damage. For individuals experiencing excoriation as a result of psychological triggers, therapy and counseling can play a vital role in identifying the root causes and developing healthier coping strategies. Cognitive-behavioral therapy (CBT) is particularly effective in helping patients recognize negative thought patterns and modify behaviors associated with skin picking. HabitAware’s Approach for Managing Excoriation HabitAware, specializing in mental health and behavioral awareness, is directly related to the phenomenon of excoriation and dermatillomania which are both medical terms for compulsive skin picking. Our tools and approaches are relevant to helping individuals struggling with this condition. Our excoriated definition has allowed us to provide the following management tips: Wrapping Up Excoriating definition refers to the act of damaging or removing part of the skin's surface. Understanding the potential causes and consequences of this behavior is essential in providing appropriate care and support for affected individuals. By addressing both the physical and psychological aspects of excoriation, medical professionals can help patients regain control over their skin health and overall well-being. If excoriation affects you or someone you care about, it's important to note that HabitAware's reputation extends beyond dermatillomania and excoriation disorder. Our team understands the gamut of Body Focused Repetitive Behaviors - which include compulsive skin picking, hair pulling, nail biting, lip picking, cheek biting, finger/feet picking and more. Through enhancing consciousness, furnishing tools for behavior adjustment, and fostering a caring community, HabitAware significantly aids individuals in learning excoriations’ definition, managing excoriation, promoting behavioral control, and enhancing their mental well-being. "The first step is being aware and bringing it to your attention... [Keen2] really is a brilliant idea." –Howie Mandel on the Marcus Lemonis podcast Life-changing wisdom, in your inbox. Get 10% off for signing up. Need Help? We're Here for You: About HabitAware HabitAware services, products and content are not substitutes for medical advice. By continuing to engage with this content you are acknowledging and agreeing to our medical disclaimer. HabitAware is protected under U.S. Pat. No. 9,721,450 B2. Country/region Country/region
1296	https://www.statisticshowto.com/additive-interval-property/	Skip to content Additive Interval Property Integrals > The additive interval property (sometimes called the additive integral property) tells us that we can add up parts of an integral to get a whole. Formally, the rule is stated as : All this is saying is that a definite integral (i.e. an area under a curve) from point a to point b can be split at a point “c“. The two pieces or “step integrals” can be added together to get the total area. More specifically, a definite integral evaluated over the closed interval [a, b] is the same as the sum of parts of the integrals (split at point c) over [a, c] and [c, b] when a ≤ c ≤ b. The integral can be split at any point, and you can have more than two pieces. Example of the Additive Interval Property The above image shows the definite integral of f(x) = 2x on the interval [0, 3]. Calculating this definite integral using the integral power rule, we get: We can split the integral at x = 2 and get the same result: Using the Additive Property for Integrals The addition problem can be useful to work backwards, finding intervals that have a certain area. For example, if you knew that the whole integral had an area of 9, and one part of the integral had an area of 3, you could conclude that the “missing” interval had an area of 6. From there, you could work out the missing bounds of integration. Ex: Property of Definite Integral Addition References Larson, R. & Edwards, B. (2017). Calculus, 11th Edition. Cengage Learning. Comments? Need to post a correction? Please Contact Us. Leave a Comment You must be logged in to post a comment.
1297	https://www.quora.com/What-is-a-bipartite-graph	Something went wrong. Wait a moment and try again. Graph Theory Terminology Computer Science Bipartite Graphs Discrete Mathmatics Computational Graph Theor... Mathematical Sciences Math (Discrete) Graph Theory 5 What is a bipartite graph? · Sep 5 A bipartite graph is a graph whose vertices can be divided into two disjoint sets such that every edge connects a vertex in one set with a vertex in the other set. No edge has both endpoints in the same set. Key properties and characterizations Partition description: G = (U, V, E) with U ∩ V = ∅, U ∪ V = V(G), and E ⊆ { {u,v} : u ∈ U, v ∈ V }. Equivalent condition (odd-cycle characterization): A graph is bipartite ⇔ it contains no cycles of odd length. 2-colorability: A graph is bipartite ⇔ its vertices can be colored with two colors so that every edge joins differently colored vertices. Components A bipartite graph is a graph whose vertices can be divided into two disjoint sets such that every edge connects a vertex in one set with a vertex in the other set. No edge has both endpoints in the same set. Key properties and characterizations Partition description: G = (U, V, E) with U ∩ V = ∅, U ∪ V = V(G), and E ⊆ { {u,v} : u ∈ U, v ∈ V }. Equivalent condition (odd-cycle characterization): A graph is bipartite ⇔ it contains no cycles of odd length. 2-colorability: A graph is bipartite ⇔ its vertices can be colored with two colors so that every edge joins differently colored vertices. Components: Each connected component of a bipartite graph is bipartite independently; bipartiteness can be checked component-wise. Common examples Even cycles (C4, C6, ...), trees, and complete bipartite graphs K_{m,n} (every vertex in U is connected to every vertex in V). Matching problems: bipartite graphs model assignments, job-worker matching, and network flow bipartite formulations. Standard algorithms and results BFS/DFS 2-coloring: Use BFS (or DFS) to attempt 2-coloring; conflict implies non-bipartite. Runs in O(\|V\|+\|E\|). Maximum matching: Hopcroft–Karp algorithm finds a maximum matching in O(√V · E) for bipartite graphs. Konig’s theorem: In bipartite graphs, size of maximum matching = size of minimum vertex cover. Hall’s marriage theorem: Characterizes existence of a perfect matching from one part to the other: every subset S of U must satisfy \|N(S)\| ≥ \|S\| (N(S) = neighborhood). Representations and applications Representation: adjacency lists, incidence between U and V, biadjacency matrix (rows = U, columns = V). Applications: assignment problems, scheduling, recommendation systems, bipartite network analysis (user–item), chemical reaction networks, and reduced formulations for constraints. Testing and construction tips To test bipartiteness: run BFS from any unvisited vertex assigning alternate colors; detect conflicts on edges joining same-colored vertices. To construct K_{m,n}: choose two sets of sizes m and n, draw all cross edges. To avoid odd cycles, ensure no edges inside a part. Concise summary A bipartite graph is exactly a 2-colorable graph, equivalent to having no odd cycles, and is the canonical structure for modeling two-type relations (e.g., people ↔ jobs, users ↔ items) where edges only cross between types. Related questions How is a tree a bipartite graph? What is the smallest bipartite graph? What are regular bipartite graphs? How would you explain a bipartite graph to someone who is younger? Is the Petersen graph bipartite? Debo Olaosebikan Physics PhD. Candidate at Cornell · Author has 102 answers and 959.4K answer views · 14y A visual way to look at it: It is a bunch of dots which can be either of two colors and lines connecting the dots where no two dots of the same color are connected by a line. The figure below makes it clear, "dots" are more technically referred to as "vertices" and lines are called "edges" The bipartite graph structure can be used to capture a relationship between two types of objects where the distinction between the types of objects is important. A student-applied-to-college relationship can (for example) be stored in the bipartite graph depicted above. Blue dots could be students (e.g. Mary, A visual way to look at it: It is a bunch of dots which can be either of two colors and lines connecting the dots where no two dots of the same color are connected by a line. The figure below makes it clear, "dots" are more technically referred to as "vertices" and lines are called "edges" The bipartite graph structure can be used to capture a relationship between two types of objects where the distinction between the types of objects is important. A student-applied-to-college relationship can (for example) be stored in the bipartite graph depicted above. Blue dots could be students (e.g. Mary, Jane, Joe) and Red dots could be colleges they applied to (e.g. Princeton, Cornell, MIT) Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. 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Anonymous Upvoted by Timothy Johnson , PhD Computer Science, University of California, Irvine (2018) · 14y There are plenty of technical definitions of bipartite graphs all over the web like this one from (ppt): But an easy way to understand and reason about bipartite graphs is to think of two groups of people being paired in some activity. For example, boys and girls dating. Assuming only heterosexual couples, even without monogamy, you get a bipartite graph where the vertices are people, edges represent relationships between them, and the two sets are the set of boys and the set of girls. There are plenty of technical definitions of bipartite graphs all over the web like this one from (ppt): But an easy way to understand and reason about bipartite graphs is to think of two groups of people being paired in some activity. For example, boys and girls dating. Assuming only heterosexual couples, even without monogamy, you get a bipartite graph where the vertices are people, edges represent relationships between them, and the two sets are the set of boys and the set of girls. Ohad Assulin Chief Data Scientist at HP Softwre · 14y A graph (V,E) Whereas the vertices(V) can be split into two sets V1 and V2 s.t. every edge e is a pair {v1,v2} s.t v1 belong to V1 and v2 belong to V2; And V1,V2 are disjoint sets Pay attention that such a graph does not contain odd length cycles. Moreover a graph without odd length cycles is bipartite E.g. tree is always bipartite (it doesn't contain cycles and esp' not odd length cycles), In order to split some tree's vertices into two sets you can run the BFS algorithm and mark all the vertices that you found on any odd step as V1 and all the other vertices as V2 Related questions What is an example of a bipartite graph? Is every complete bipartite graph complete? What are the limitations of the bipartite graph? Which complete bipartite graphs are trees? What is the difference between bipartite and complete bipartite graph? Sourav Mishra MTech in Computational and Data Sciences (CDS), Indian Institute of Science, Bangalore (IISc) (Graduated 2020) · 7y A bipartite graph is a graph whose vertices can be partitioned into two disjoint sets such that all the edges connect vertices from one set to vertices in the other set. There are no edges joining two vertices from the same set. To put things in a more clear and slightly less mathematical perspective, consider the example of a match the following problem which you might have come across in textbooks. In this case the two disjoint sets of vertices are obvious - the two columns which need to be matched. All the edges which represent the matchings connect elements from one column to elements in th A bipartite graph is a graph whose vertices can be partitioned into two disjoint sets such that all the edges connect vertices from one set to vertices in the other set. There are no edges joining two vertices from the same set. To put things in a more clear and slightly less mathematical perspective, consider the example of a match the following problem which you might have come across in textbooks. In this case the two disjoint sets of vertices are obvious - the two columns which need to be matched. All the edges which represent the matchings connect elements from one column to elements in the other column. There is no matching/edge between the elements of the same column. Hence a match the following type problem can be represented as a bipartite graph. Some other facts about bipartite graphs - A connected bipartite graph can be represented as a tree and vice versa. 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The Relation or edges in these graphs exists only between two vertices which are not from the same set of vertices. Famous example : Assigning n jobs to m persons can be modelled using a Bipartite Graph.. The edge exists if a job i can be allotted to a person p. Obviously this requires a Bipartite model of graph, Since an edge here can not exist between a job to another job or a person to another person. As there are few explanations in here before I write... I just will go with a simple answer.. Bipartite Graphs are those which have two sets of vertices which are mutually disjoint . The Relation or edges in these graphs exists only between two vertices which are not from the same set of vertices. Famous example : Assigning n jobs to m persons can be modelled using a Bipartite Graph.. The edge exists if a job i can be allotted to a person p. Obviously this requires a Bipartite model of graph, Since an edge here can not exist between a job to another job or a person to another person. Andrew Morgan previous CEO of a data science and engineering company · 11y The descriptions above outline the properties, but not the utility or reason you should care. What is it? I think of it as a search criteria over graph data, where I am looking for a very useful special case. I am looking for a data structure called an Adjacency Matrix, aka: a 2-mode matrix. A two mode matrix maps together instances of entities of two types. That means for example your row headers are schools, and your column headers are students. In databases these are resolver tables that enable many-to-many relations. Prathyush Kashyap's answer illustrates this nicely. Sponsored by CDW Corporation What would an evergreen data storage architecture do for you? Get uninterrupted access to data that is update-proof when you invest in storage from CDW & Pure Storage. Jason Wang Avid learner, climber and home chef · 12y Think Bi-partite (2 parts). It is a graph G=(V,E) with a set V split into V1 and V2 and set E for edges. What makes a bipartite graph special that every edge (i,j) that exists in E has (i in V1 )and (j in V2). Simply, all edges in the bipartite graph must have one end in V1 and the other end in V2 (so edges connecting nodes in V1 or nodes in V2 are allowed). Daniel R. Page Theoretical Computer Scientist, CS PhD · Author has 3K answers and 5.3M answer views · 4y Originally Answered: Is a graph bipartite? · Some graphs are bipartite, some are not. For example, all trees are bipartite. Bipartite graphs can also have cycles in them; however, if there is an odd-length cycle in your graph, it cannot be bipartite (see Graph is Bipartite iff No Odd Cycles for a proof). A minimal example of a graph that is not bipartite is simply a triangle (three vertices, all adjacent to one another, otherwise known as the complete graph on three vertices, K3, or a cycle graph on three vertices, C3): Try to lump three vertices all adjacent to each other into two disjoint sets in this fashion, you cannot because of th Some graphs are bipartite, some are not. For example, all trees are bipartite. Bipartite graphs can also have cycles in them; however, if there is an odd-length cycle in your graph, it cannot be bipartite (see Graph is Bipartite iff No Odd Cycles for a proof). A minimal example of a graph that is not bipartite is simply a triangle (three vertices, all adjacent to one another, otherwise known as the complete graph on three vertices, K3, or a cycle graph on three vertices, C3): Try to lump three vertices all adjacent to each other into two disjoint sets in this fashion, you cannot because of the existence of that one cycle of odd length consisting of all the graph’s edges. Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Daniel R. Page Ph.D. in Computer Science, University of Western Ontario (Graduated 2019) · Author has 3K answers and 5.3M answer views · 9y Let A and B be sets of vertices, such that they share no common elements (i.e., they are disjoint). A bipartite graph G = ( A ∪ B , E ) is where every edge { u , v } ∈ E such that u ∈ A and v ∈ B , but there is no edge that connects any two vertices in A or any two vertices in B . That is, edges only go between A and B . Luis Argerich College Professor, Amateur Astronomer and Astrophotographer. · Author has 850 answers and 4.6M answer views · 10y A graph that can be split in two sets in a way that there's no link between members of a set. All links go from set A to B or from B to A. Chris McCoy Upvoted by Kiran Kannar , Master's Computer Science, University of California, San Diego (2018) · Author has 117 answers and 192.4K answer views · 10y Sports competitions are a bipartite graph: Two teams playing a game. Vertices are the teams, edges are the relationships of both teams in previous games/matches, and the two sets are players on team 1 and players on team 2. Radhey Shyam Pandey Former Software Developer at SRM CARE · Author has 245 answers and 479.7K answer views · 7y Bipartite is the graph which is divided into two sets of graph such that in which no two adjacent nodes are come in the same set. It is the powerful algorithm technique such that it is reduced the time complexity and also it needs only small memory.However,the time complexity of the bipartite is much less than the simple graph. Related questions How is a tree a bipartite graph? What is the smallest bipartite graph? What are regular bipartite graphs? How would you explain a bipartite graph to someone who is younger? Is the Petersen graph bipartite? What is an example of a bipartite graph? Is every complete bipartite graph complete? What are the limitations of the bipartite graph? Which complete bipartite graphs are trees? What is the difference between bipartite and complete bipartite graph? Is a graph with no edges bipartite? Is a graph bipartite? What is the bipartite graph matching problem? What is a simple or bipartite graph? How is a subgraph of a bipartite graph is bipartite? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1298	https://www.purplemath.com/modules/solvrtnl2.htm	Home Lessons HW Guidelines Study Skills Quiz Find Local Tutors Demo MathHelp.com Join MathHelp.com Login Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Solving Rational Equations IntroGraphs Purplemath Solve the following equation: The rational expressions in this equation have variables in the denominators. So my first step is to check for which x-values are not allowed, because they'd cause division by zero. Setting each denominator equal to zero and solving, I get: x ≠ −4, −1 Looking at the equation, I notice that this equation is a proportion; that is, the equation is of the form "(one fraction) equals (another fraction)". So all I need to do here is "cross-multiply"; that is, I can use Method 3: Content Continues Below MathHelp.com Solving Rational Equations 10(4(x + 1)) = 15(x + 4) 40x + 40 = 15x + 60 25x + 40 = 60 25x = 20 Since this solution won't cause any division-by-zero problems, it is a valid solution to the equation, and my answer is: Affiliate Solve the following equation: There is only one fraction, so the common denominator is the only denominator; namely, x. Also, setting that denominator equal to zero, I see that the solution to this equation cannot be x = 0. Method 1: To solve, I can convert everything to this common denominator, and then solve the numerators: x2 + x = 72 x2 + x − 72 = 0 (x + 9)(x − 8) = 0 x = −9 or x = 8 Method 2: To solve, I can start by multiplying through on both sides by x: x2 + x = 72 x2 + x − 72 = 0 (x + 9)(x − 8) = 0 x = −9 or x = 8 Either way, the solution is the same. Since neither solution causes a division-by-zero problem in the original equation, both solution values are valid. x = −9, 8 Content Continues Below Solve the following equation: First, I'll solve the denominators. Setting each equal to zero to find the disallowed values, I get: x ≠ 0, 2 The lowest common denominator of these fractions will be x(x − 2). Method 1: I can solve the equation by converting all of the rational expressions to the common denominator, and then solving the numerators: 10 + (4x − 8) = 5x 10 + 4x − 8 = 5x 4x + 2 = 5x 2 = x Method 2: I can also solve this equation by multiplying through on both sides of the equation by the denominator. (The equation becomes a bit of a mess. I've used color below to highlight which parts cancel off. Use caution!) 10 + 4(x − 2) = 5(x) 10 + 4x − 8 = 5x 4x + 2 = 5x 2 = x Using either method, I get the same answer; namely, x = 2. However, checking back to the beginning, where I first noted the disallowed values for the original equation, I see that x ≠ 2. In other words, my only solution value would actually cause division by zero. Since the only possible solution causes division by zero, then this equation really has no solution. My answer then is: no solution It doesn't look like I did anything wrong, mathematically, in the previous exercise. So how did I end up with an entirely invalid solution? Affiliate Advertisement When dealing with rational expressions and equations, we're not allowed to divide by zero. When there are variables in denominators, we then will have certain values which can cause division by zero. Whichever method one uses to solve a given rational equation, one will, at some point, get rid of those denominators. In other words, at some point, one will, in effect, make those division-by-zero problems magically disappear. But they aren't actually gone; they've merely been ignored at some stage. In the end, one must go back to the beginning and check one's solution against those original disallowed values. And it is perfectly possible that a given equation will have no solution at all. Whenever you solve a rational equation, always check your (interim) solution against the denominators (and their disallowed values) from the original equation. It is entirely possible that a problem will have an invalid (that is, an "extraneous") solution. This is especially true on tests. So always check! URL: Page 1Page 3 Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Share This Page Terms of Use Privacy Contact About Purplemath About the Author Tutoring from PM Advertising Linking to PM Site licencing Visit Our Profiles © 2024 Purplemath, Inc. All right reserved. Web Design by
1299	https://sites.lsa.umich.edu/kesmith/wp-content/uploads/sites/1309/2024/06/OrthonormalityW2017.pdf	217 Orthonormal basis Professor Karen E. Smith (c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Definition: A set of vectors {⃗ v1, . . . ,⃗ vm} in Rn is orthonormal if for all indices i, j between 1 and m, we have ⃗ vi · ⃗ vj = 0 for all i ̸= j AND ⃗ vi · ⃗ vi = 1. A. Orthonormal Coordinates. 1. Discuss the geometric meaning of the deﬁnition above. Be sure you discuss what BOTH ⃗ vi · ⃗ vj = 0 AND ⃗ vi · ⃗ vi = 1 mean. Use a theorem in the book to explain why n orthonormal vectors in Rn always form a basis of Rn. 2. Is the standard basis orthonormal? Find1 an orthonormal basis B of R2 that includes the vector 3 5 4 5 . Draw a picture. 3. Suppose that {⃗ v1, . . . ,⃗ vn} is an orthonormal set. Let ⃗ x = c1⃗ v1 + · · · + cm⃗ vm. Compute ⃗ x · ⃗ vi for any i = 1, 2, . . . , m. 4. Suppose that B = (⃗ v1, . . . ,⃗ vm) is an orthonormal basis for Rn. For an arbitrary vector ⃗ x ∈Rn, prove that [⃗ x]B =      ⃗ x · ⃗ v1 ⃗ x · ⃗ v2 . . . ⃗ x · ⃗ vn     . Why might this make working in an orthonormal basis advantageous? 5. Use your formula in (5) to ﬁnd the coordinates of the vector 5 −10 in your basis from (2). 6. For your basis B in (2), ﬁnd the change of coordinate matrix SB→E where E is the standard basis. Use it to ﬁnd (the standard coordinates of) the vector ⃗ v ∈R2 whose B-coordinates are 5 5 . Also use it to ﬁnd what vector has B-coordinates 1 0 ? Also use it to ﬁnd ⃗ w such that [⃗ w]B = 2 1 ? 7. Find the change of coordinates matrix SE→B. Verify, using the three change of coordinates you did in (6) that it really takes the standard coordinates of each vector in (6) back the B-coordinates. 8. Let R2 T →R2 be left multiplication by A = −31/5 −36/5 −36/5 −34/5 . Discuss two diﬀerent ways to ﬁnd the matrix B representing T in the basis B. Use one of these methods to ﬁnd B. 1There is more than one correct answer. To make your answer the same as the answer key’s, choose the answer in which the ﬁrst coordinate is negative. Solution note: 1. Vectors (⃗ v1, . . . ,⃗ vm) in Rn are orthonormal means that they are all perpendicular to ea-chother and have length 1. A theorem in the book says that orthonormal vectors are linearly independent. Hence if there are n of them, they will be a basis for Rn, since Rn has dimension n. 2. Yes, the standard basis satisﬁes ⃗ ei · ⃗ ej = 0 if i ̸= j and ⃗ ei · ⃗ ei = 1. 3 5 4 5 , −4 5 3 5 is one of two possible answers. The other multiplies the second vector by −1. 3. Write ⃗ x = c1⃗ v1 + · · · + cn⃗ vn. Dot with ⃗ vi and use the distributive property of dot product: ⃗ x⃗ vi = c1⃗ v1 · ⃗ vi + · · · + cn⃗ vn · ⃗ vi. Now use the orthonormality: most of the ⃗ vj · ⃗ vi = 0 so this reduces to ⃗ x · ⃗ vi = ci. 4. Write ⃗ x = c1⃗ v1 + · · · + cn⃗ vn be an arbitrary ⃗ x expressed in the basis B. We use the previous problem to compute the scalar ci as ⃗ x · ⃗ vi. This means that the B-coordinate column is      ⃗ x · ⃗ v1 ⃗ x · ⃗ v2 . . . ⃗ x · ⃗ vn     . 5. For B = ( 3 5 4 5 , −4 5 3 5 ), we have 5 −10 B = −5 −10 . We got this by computing the two dot products of 5 −10 with each of our two basis elements. 6. Using the Handy Technique to ﬁnd the S-matrix: we compute each column separately. To go FROM B TO E, for the ﬁrst column we take the ﬁrst vector in basis B and write in E-coordinates. This gives us 3 5 4 5 for the ﬁrst column. For the second column we take the second vector in basis B and write in E-coordinates. This gives us −4 5 3 5 for the second column. So SB→E = 3 5 −4 5 4 5 3 5 . This means that the standard coordinates of the vector whose B-coordinates are [5 5]T is SB→E 5 5 = −1 7 . Also 3 5 4 5 has B-coordinates 1 0 . The ⃗ w that satisﬁes [⃗ w]B = 2 1 is 2⃗ v1 + 1⃗ v2 = 2/5 11/5 . 7. SE→B = S−1 B→E = 3 5 4 5 −4 5 3 5 . 8. We have [T]E = A. So [T]B = SE→B[T]ESB→E. So [T]E = 3 5 4 5 −4 5 3 5 −31/5 −36/5 −36/5 −34/5 3 5 −4 5 4 5 3 5 . An alternate technique would be to use the handy method. Compute column by column. For the ﬁrst, we take the ﬁrst basis element of B and see where it goes under T (ie, multiply by A). Then we have to rewrite in B-coordinates again, which is relatively easy because we can ﬁnd them by dotting with our orthonormal basis. Same for second column. B. Prove that if the set {⃗ v1, . . . ,⃗ vm} is an orthonormal set in Rn, then it is linearly independent. Solution note: Suppose we have a relation c1⃗ v1 + · · · + cm⃗ vm = 0. Dot both sides with ⃗ vi. This gives (c1⃗ v1 + · · · + cm⃗ vm) · ⃗ vi = 0 · ⃗ vi = 0 c1(⃗ v1 · ⃗ vi) + · · · + cm(⃗ vm · ⃗ vi) = 0 so using the fact that the ⃗ vi are orthonormal to each other, we have ci = 0, for each i. QED. C. Definition: Fix a subspace V of Rn. A vector ⃗ w ∈Rn is orthogonal to V if ⃗ w · ⃗ v = 0 for all ⃗ v ∈V . The orthogonal complement V ⊥is the set of all vectors orthogonal to V : V ⊥= {⃗ w ∈Rn \| ⃗ w · ⃗ v = 0 for all ⃗ v ∈V }. 1. Let L be the subspace of R2 spanned by 1 4 . Draw a picture showing L and L⊥. Find a basis for L⊥. 2. Prove that ⃗ w ∈Rn is in V ⊥if and only if ⃗ w · ⃗ vi = 0 for each ⃗ vi in a basis B for V . 3. Prove that V ⊥is a subspace of Rn. 4. Consider the map R2 πL →R2 projecting orthogonally onto L (with L as in (1)). Use a geometric argument to ﬁnd the kernel and image of πL in terms of terms of L and L⊥. What does rank nullity for πL tell us about dim L⊥+ dim L? 5. Now let W ⊂R3 be a 2 dimensional subspace of R3. Consider the orthogonal projection R3 πW →R3 onto W. Use a geometric argument to ﬁnd the kernel and image of πW in terms of terms of W and W ⊥. If W is the plane deﬁned by ax + by + cz = 0, can you ﬁnd a basis for W ⊥? 6. For any subspace V ⊂Rn, what can you say about dim V ⊥+ dim V ? Make sure your answer agrees with the special cases you did in (4) and (6). 7. Prove that V ∩V ⊥= 0. Solution note: 1. You should see two perpendicular lines, L and L⊥, crossing at the origin. A basis for L⊥is [−4 1]T . 2. We have two directions to prove: (a) If ⃗ w ∈Rn is in V ⊥, then ⃗ w · ⃗ vi = 0 for each ⃗ vi in a basis B for V . (b) If ⃗ w · ⃗ vi = 0 for each ⃗ vi in a basis B for V , then ⃗ w ∈V ⊥. Note that (a) is obvious, since the basis elements ⃗ vi are in V , so by deﬁnition of ⃗ w ∈V ⊥we know ⃗ w · ⃗ vi = 0. For (b), suppose ⃗ w dots with each ⃗ vi ∈B. To show ⃗ w ∈V ⊥we need to show ⃗ w · ⃗ v = 0 for all ⃗ v ∈V . Take an arbitrary element ⃗ v ∈V . Write ⃗ v = a1⃗ v1 +· · ·+ad⃗ vd where ⃗ vi ∈V. Compute ⃗ w · ⃗ v = ⃗ w · (a1⃗ v1 + · · · + ad⃗ vd) = a1(⃗ w · ⃗ v1) + · · · + ad(⃗ w · ⃗ vd) using basic properties of dot product. Since every ⃗ w ·⃗ vi = 0, so also ⃗ w ·⃗ v = 0. Since this argument applies for every ⃗ v ∈V , we conclude that v ∈V ⊥. 3. There are three things to check: (a) ⃗ 0 ∈V ⊥. (b) If ⃗ x, ⃗ y ∈V ⊥, then ⃗ x + ⃗ y ∈V ⊥. (c) If ⃗ x ∈V ⊥and c ∈R, then c⃗ x ∈V ⊥. For this, we take arbitrary ⃗ v ∈V . W (a) ⃗ 0 · v = 0. This is clear since dotting with 0 always gives 0. (b) If ⃗ x, ⃗ y ∈V ⊥, then ⃗ x + ⃗ y ∈V ⊥. (c) If ⃗ x, ⃗ y ∈V ⊥, we need (⃗ x + ⃗ y) · ⃗ v = 0. But (⃗ x + ⃗ y) · ⃗ v = ⃗ x · ⃗ v + ⃗ y · ⃗ v = 0 + 0 = 0, since both ⃗ x and ⃗ y are in V ⊥. (d) Take ⃗ x ∈V ⊥and scalar c. Check (c⃗ x) · ⃗ v = c(⃗ x · ⃗ v) = c0 = 0 since ⃗ x ∈V ⊥. 4. Thinking about the projection geometrically, we see every point is mapped to something in L and the points on L are mapped to themselves. So the image is L. The vectors in the kernel are those that project to the origin: this is exactly L⊥. 5. dim L⊥+ dim L = 1 + 1 = 2. It is rank nullity applied to the map πL, since the kernel of π2 is L⊥, the image is L, and the source is R2. 6. Here the image is W and the kernel is W ⊥. By rank nullity, since W has dimension 2, we know W ⊥is dimension 1. A basis for W ⊥is [a b c]T , since this vector has dot product zero with every point in the plane ax + by + cz = 0. 7. In generally, dim V + dim V ⊥= n where n is the dimension of the ambient space where V lies. We can get this from rank-nullity: the projection onto V has image V and kernel V ⊥, so their dimensions add to n. 8. Let ⃗ x ∈V ∩V ⊥. Since ⃗ x ∈V ⊥, we know its dot product with any element in V is zero. But ⃗ x ∈V , which means ⃗ x·⃗ x = 0. But ⃗ x·⃗ x = \|\|x\|\|2 can be zero only if x = 0. So the only element of V ∩V ⊥is 0. QED. D. How many diﬀerent ways can we chose a vector in R3 to complete the vectors   0 0 1  ,    − √ 2 2 √ 2 2 0   to an orthonormal basis for R3? Find2 one such explicit way to complete the basis, and call it A. Compute the A-coordinates of   √ 6 √ 6 −3  . Solution note: There are two vectors: we can take either unit vector perpendicular to the plane spanned by the given vectors. So to get A, we just add in    √ 2 2 √ 2 2 0   . For A = (   0 0 1  ,    √ 2 2 − √ 2 2 0   ,    √ 2 2 √ 2 2 0   ), we have   √ 6 √ 6 −3   A =   −3 0 2 √ 3  . E. Let V ⊂R3 be the plane deﬁned by x + y −z = 0. 1. Find an orthonormal basis B for V . 2. Consider the element ⃗ v =   1 1 2  . Find [⃗ v]B. 3. Extend your basis B for V to an orthonormal basis A for R3. 4. Let R3 π →R3 be the projection onto V . Describe the source, target, image, kernel, rank and nullity using geometric language. 5. Compute the A-matrix for π. 6. Use your A-matrix for π to compute π(⃗ v) where ⃗ v is the sum of the vectors in your basis A. 7. Compute the change of basis matrices SA→E and SE→A where E denotes the standard basis for R3. 8. Express [π]E as a product of other matrices that are easier to ﬁnd. What is another way to ﬁnd [π]E? 2There is more than one correct answer. To make your answer the same as the answer key’s, choose the answer in which the ﬁrst coordinate is positive. Solution note: 1. We can pick any unit vector ⃗ u1 = [ √ 2/2 0 √ 2/2]T in V for the ﬁrst element in the basis. Now we need to ﬁnd just one more element in the plane V which is perpendicular to this one (and then scale it down to be a unit vector). We need to ﬁnd ⃗ w = [a b c]T such that ⃗ w ·⃗ u1 = 0. Clearly this means that a = −c. But it also has to lie on V , so b must be −2a. For example, ⃗ w = [1 −2 −1]T is in V , so ⃗ u1, ⃗ w is a basis for V . This is not quite orthonormal: the vectors are perpendicular, but the second one is not a unit vector. We ﬁx this by dividing the second vector by its length. So an orthonormal basis is (   √ 2/2 0 √ 2/2  ,   1/ √ 6 −2/ √ 6 −1/ √ 6  ). 2. Coordinates for an orthonormal basis are easy to ﬁnd using dot product! Since ⃗ v · ⃗ u1 =   1 1 2  ·   √ 2/2 0 √ 2/2  = √ 2/2+2 √ 2/2 = 3 √ 2/2 and ⃗ v· ⃗ w =   1 1 2  ·   1/ √ 6 −2/ √ 6 −1/ √ 6  = 1/ √ 6−2/ √ 6−2/ √ 6 = −3/ √ 6 = − √ 6/2, we have   1 1 2   B = 3 √ 2/2 − √ 6/2 . 3. We need a unit vector perpendicular to both the vectors in B. This is the same as being perpendicular to the plane V . A normal vector to the plane is   1 1 −1  , so we only need to scale it to length 1. Thus we can take A = (   √ 2/2 0 √ 2/2  ,   1/ √ 6 −2/ √ 6 −1/ √ 6  ,   √ 3/3 √ 3/3 − √ 3/3  ). 4. The source and target are both R3, the image is V , the kernel is the line through the origin perpendicular to V . The rank is 2 (dimension of V ) and the nullity is 1 (dimension of kernel). 5. [π]A =   1 0 0 0 1 0 0 0 0  . 6. [⃗ v]A =   1 1 1  . So [π(⃗ v)]A =   1 0 0 0 1 0 0 0 0     1 1 1  =   1 1 0  . This output is the A-coordinate vector of our desired vector. To write it in standard coordinates, we note it is representing ⃗ v1 + ⃗ v2, or   √ 2/2 0 √ 2/2  +   1/ √ 6 −2/ √ 6 −1/ √ 6  =   √ 2/2 + 1/ √ 6 −2/ √ 6 √ 2/2 −1/ √ 6  . 7. SA→E =   √ 2/2 1/ √ 6 √ 3/3 0 −2/ √ 6 √ 3/3 √ 2/2 −1/ √ 6 − √ 3/3  . Continued next page. Solution note: To get SE→A we could use row reduction to get the inverse of SA→E. Or, since A is orthonormal, it is not too bad to get it by direct computation (using the dot-product to ﬁnd coordinates.) We get SE→A =   √ 2/2 0 √ 2/2 1/ √ 6 −2/ √ 6 −1/ √ 6 √ 3/3 √ 3/3 − √ 3/3  . Note that SE→A is the TRANSPOSE of SA→E. Why is this? The ﬁrst student who can explain to me why change of basis matrices between any two orthonormal bases also satisfy S−1 A→B = ST A→B will get an extra credit point on ”quiz 11”. I would like to see if anyone is actually reading these solutions! Finally, we have [π]E = SA→E[π]ASE→A. This is   √ 2/2 1/ √ 6 √ 3/3 0 −2/ √ 6 √ 3/3 √ 2/2 −1/ √ 6 − √ 3/3     1 0 0 0 1 0 0 0 0     √ 2/2 0 √ 2/2 1/ √ 6 −2/ √ 6 −1/ √ 6 √ 3/3 √ 3/3 − √ 3/3   =   √ 2/2 1/ √ 6 √ 3/3 0 −2/ √ 6 √ 3/3 √ 2/2 −1/ √ 6 − √ 3/3     √ 2/2 0 √ 2/2 1/ √ 6 −2/ √ 6 −1/ √ 6 0 0 0  .

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