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17500	https://journals.lww.com/ccmjournal/fulltext/2023/01001/823__standard_versus_low_dose_levetiracetam_for.788.aspx	Critical Care Medicine SCCM Critical Care Medicine Pediatric Critical Care Medicine Critical Care Explorations Visit our other sites SCCM Critical Care Medicine Pediatric Critical Care Medicine Critical Care Explorations SCCM Critical Care Medicine Pediatric Critical Care Medicine Critical Care Explorations Log in or Register Subscribe to journal Subscribe Get new issue alerts Get alerts;;) Subscribe to eTOC;;) ### Secondary Logo Enter your Email address: Privacy Policy ### Journal Logo Articles Advanced Search Toggle navigation SubscribeRegisterLogin Browsing History Articles & Issues Current Issue Previous Issues Latest Articles Latest Articles Collections Top 10 Cited Articles Over the Last 10 Years 2022 Top Cited Articles on CCM SCCM Guidelines Concise Definitive Reviews Infographics SCCM COVID-19 Articles Methods and Analysis Neurologic Critical Care Pediatric Critical Care Viewpoints Featured Supplements SCCM 50th Anniversary Articles Chinese Translations Japanese Translations Podcasts For Authors Submit a Manuscript Information for Authors Language Editing Services Author Permissions Open Access Journal Info About the Journal Open Access Editorial Board Affiliated Society Advertising Subscription Services Reprints Rights and Permissions Become a Member Articles Advanced Search January 2023 - Volume 51 - Issue 1 Previous Abstract Next Abstract Cite Copy Export to RIS Export to EndNote Share Email Facebook X LinkedIn Favorites Permissions More Cite Permissions Article as EPUB Export All Images to PowerPoint FileAdd to My Favorites Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. Export to End Note Procite Reference Manager [x] Save my selection Research Snapshot Theater: Pharmacology/Trauma 823: STANDARD VERSUS LOW-DOSE LEVETIRACETAM FOR SEIZURE PROPHYLAXIS IN ELDERLY TRAUMATIC BRAIN INJURIES Merkel, Alison 1; McMullen, Lisanne 2; Jenniches, Daniel 1; Hanna, Katherine 1; Chung, Eunice 1 Author Information 1 Allegheny General Hospital 2 Rochester General Hospital, Rochester, NY Critical Care Medicine 51(1):p 403, January 2023. \| DOI: 10.1097/01.ccm.0000909020.35463.f7 Buy Copyright © 2022 by the Society of Critical Care Medicine and Wolters Kluwer Health, Inc. All Rights Reserved. Full Text Access for Subscribers: ##### Individual Subscribers Log in for access Log in with your SCCM credentials. ##### Institutional Users Access through Ovid® Not a Subscriber? Buy Subscribe Request Permissions Become a Society Member You can read the full text of this article if you: Log InAccess through Ovid Source 823: STANDARD VERSUS LOW-DOSE LEVETIRACETAM FOR SEIZURE PROPHYLAXIS IN ELDERLY TRAUMATIC BRAIN INJURIES Critical Care Medicine51(1):403, January 2023. Full-Size Email Favorites Export View in Gallery Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. 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17501	https://brightchamps.com/en-us/math/algebra/square-root-of-17	Our Programs Learn More About us Login Table Of Contents What Is the Square Root of 17? Finding the Square Root of 17 Square Root of 17 By Prime Factorization Method Square Root of 17 by Long Division Method Square Root of 11 by Approximation Method Common Mistakes and How to Avoid Them in the Square Root of 17 Square Root of 17 Examples FAQs on 17 Square Root Important Glossaries for Square Root of 17 Explore More algebra Summarize this article: 406 Learners Last updated on August 5, 2025 Square root of 17 The square root of 17 is the inverse operation of squaring a value “y” such that when “y” is multiplied by itself → y ⤫ y, the result is 17. It contains both positive and a negative root, where the positive root is called the principal square root. Square root of 17forUSStudents What Is the Square Root of 17? The square root of 17 is ±4.12310562562. The positive value,4.12310562562 is the solution of the equation x2 = 17. As defined, the square root is just the inverse of squaring a number, so, squaring 4.12310562562 will result in 17. The square root of 17 is expressed as √17 in radical form, where the ‘√’ sign is called “radical” sign. In exponential form, it is written as (17)1/2 Finding the Square Root of 17 We can find the square root of 17 through various methods. They are: i) Prime factorization method ii) Long division method iii) Approximation/Estimation method Square Root of 17 By Prime Factorization Method The prime factorization of 17 involves breaking down a number into its factors. Divide 17 by prime numbers, and continue to divide the quotients until they can’t be separated anymore. After factorizing 17, make pairs out of the factors to get the square root. If there exists numbers which cannot be made pairs further, we place those numbers with a “radical” sign along with the obtained pairs So, Prime factorization of 17 =17 × 1 for 17, no pairs of factors are obtained, but a single 17 is obtained. So, it can be expressed as √17 = √(17 × 1) = √17 √17 is the simplest radical form of √17 Square Root of 17 by Long Division Method This is a method used for obtaining the square root for non-perfect squares, mainly. It usually involves the division of the dividend by the divisor, getting a quotient and a remainder too sometimes. Follow the steps to calculate the square root of 17: Step 1 : Write the number 17, and draw a bar above the pair of digits from right to left. Step 2 : Now, find the greatest number whose square is less than or equal to 17. Here, it is 4, Because 42=16 < 17 Step 3 : Now divide 17 by 4 (the number we got from Step 2) such that we get 4 as quotient, and we get a remainder. Double the divisor 4, we get 8 and then the largest possible number A1=1 is chosen such that when 1 is written beside the new divisor, 8, a 2-digit number is formed →81 and multiplying 1 with 81 gives 81 which is less than 100. Repeat the process until you reach remainder 0 We are left with the remainder, 871 (refer to the picture), after some iterations and keeping the division till here, at this point Step 4 : The quotient obtained is the square root. In this case, it is 4.123… Square Root of 11 by Approximation Method Approximation or estimation of square root is not the exact square root, but it is an estimate. Here, through this method, an approximate value of square root is found by guessing. Follow the steps below: Step 1 : Identify the square roots of the perfect squares above and below 17 Below : 16→ square root of 16 = 4 ……..(i) Above : 25 →square root of 25= 5 ……..(ii) Step 2 : Divide 17 with one of 4 or 5 If we choose 4, and divide 17 by 4, we get 4.25 …….(iii) Step 3: Find the average of 4 (from (i)) and 4.25 (from (iii)) (4+4.25)/2 = 4.125 Hence, 4.125 is the approximate square root of 17 Common Mistakes and How to Avoid Them in the Square Root of 17 When we find the square root of 17, we often make some key mistakes, especially when we solve problems related to that. So, let’s see some common mistakes and their solutions. Mistake 1 Incorrectly applying the square root property Square root do not distribute over additions. For example, √17 ≠ √10 + √7 Mistake 2 Ignoring precision Provide an approximate answer without acknowledging the extent of precision is not expected Mistake 3 Assuming √17 as a perfect square firstly find the approximate value of the square root before assuming it as a perfect square Mistake 4 Incorrect formula for finding square roots through different methods The formula should be applied accurately for iterative methods Mistake 5 Rounding √17 too early and not able to get precision Regulate decimal places as needed. For example, if needed to have correct up to 2 decimal places, we can keep the value as 4.12. Hey! Square Root of 17 Examples Problem 1 Simplify 5√17? Okay, lets begin 5√17 = 5⤬√17 = 5⤬4.123 = 20.615 Answer : 20.615 Explanation √17= 4.123, so multiplying the square root value with 5 Well explained 👍 Problem 2 What is √11 + √17 ? Okay, lets begin √11+ √17= 3.316+ 4.123 = 7.439 Answer: 7.439 Explanation adding the square root value of 11 with that of square root value of 17. Well explained 👍 Problem 3 Find the value of√17 /√16? Okay, lets begin √17/√16 = 4.123 / 4 = 1.03075 Answer: 1.03075 Explanation we divide √17 by the value of √16 Well explained 👍 Problem 4 If y=√17, find y² Okay, lets begin firstly, y=√17= 4.123 Now, squaring y, we get, y2= (4.123)2=17 or, y2=17 Answer : 17 Explanation squaring “y” which is same as squaring the value of √17 resulted to 17 Well explained 👍 Problem 5 Find √17 - √9 Okay, lets begin √17-√9 = 4.123–3 = 1.123 Answer : 1.123 Explanation subtracting the square root value of 9 from square root value of 17 Well explained 👍 FAQs on 17 Square Root 1.Is 17 a perfect cube? No, 17 is not a perfect cube, that is, there is no whole number exists which when multiplied by itself thrice returns 17 as a product. 2.Is √17 a complex number? No, √17 is not a complex number, it is a real numbe 3.Is 17 a perfect square or non-perfect square? 17 is a non-perfect square, since 17 =(4.12310562562) 2. 4.Is the square root of 17 a rational or irrational number? The square root of 17 is ±4.12310562562. So, 4.12310562562 is an irrational number, since it cannot be obtained by dividing two integers and cannot be written in the form p/q, where p and q are integers. 5. How would you represent √17 on a number line? we can locate √17 on a number line. It is between 4 and 5 but more close to 4, precisely between 4.0 and 4.2 6.How does learning Algebra help students in United States make better decisions in daily life? Algebra teaches kids in United States to analyze information and predict outcomes, helping them in decisions like saving money, planning schedules, or solving problems. 7.How can cultural or local activities in United States support learning Algebra topics such as Square root of 17? Traditional games, sports, or market activities popular in United States can be used to demonstrate Algebra concepts like Square root of 17, linking learning with familiar experiences. 8.How do technology and digital tools in United States support learning Algebra and Square root of 17? At BrightChamps in United States, we encourage students to use apps and interactive software to demonstrate Algebra’s Square root of 17, allowing students to experiment with problems and see instant feedback for better understanding. 9.Does learning Algebra support future career opportunities for students in United States? Yes, understanding Algebra helps students in United States develop critical thinking and problem-solving skills, which are essential in careers like engineering, finance, data science, and more. Important Glossaries for Square Root of 17 Exponential form:An algebraic expression that includes an exponent. It is a way of expressing the numbers raised to some power of their factors. It includes continuous multiplication involving base and exponent.Ex: 3 ⤬ 3 ⤬ 3 ⤬ 3 = 81 Or, 3 4 = 81, where 3 is the base, 4 is the exponent Factorization:Expressing the given expression as a product of its factors Ex: 52=2 ⤬ 2 ⤬ 13 Prime Numbers :Numbers which are greater than 1, having only 2 factors as →1 and Itself. Ex: 1,3,5,7,.... Rational numbers and Irrational numbers:The Number which can be expressed as p/q, where p and q are integers and q not equal to 0 are called Rational numbers. Numbers which cannot be expressed as p/q, where p and q are integers and q not equal to 0 are called Irrational numbers. Perfect and non-perfect square numbers:Perfect square numbers are those numbers whose square roots do not include decimal places. Ex: 4,9,25 Non-perfect square numbers are those numbers whose square roots comprise decimal places. Ex :2, 8, 18 Explore More algebra Previous to Square root of 17 Square root of 193\|Square root of 194\| Square Root of 196\|Square root of 200\|Square Root of 225\|Square Root of 250\|Square Root of 256\| Square root of 260\|Square root of 288\|Square Root of 289\|Square root of 300\|Square root of 320\| Square root of 324\|Square root of 325\| Square root of 360\|Square root of 361 \|Square root of 400\|Square root of 900.\|Square root of 1600\|Square root of 61 Next to Square root of 17 Square root of 39 \|Square root of 41 \|Square root of 43\|Square root of 44\|Square root of 441 \|Square Root of 138\|Square Root of 142\|Square root of 168\|Square Root of 0\|Square Root of 0.0004\|Square Root of 0.0008\|Square Root of 0.001\|Square Root of 0.0025\|Square Root of 0.01\|Square Root of 0.015\|Square Root of 0.03\|Square Root of 0.04\|Square Root of 0.05\|Square Root of 0.0625\|Square Root of 0.09 About BrightChamps in United States At BrightCHAMPS, we understand algebra is more than just symbols it’s a gateway to endless possibilities! Our goal is to empower kids throughout the United States to master key math skills, like today’s topic on the Square root of 17, with a special emphasis on understanding square roots in an engaging, fun, and easy-to-grasp manner. Whether your child is calculating how fast a roller coaster zooms through Disney World, keeping track of scores during a Little League game, or budgeting their allowance for the latest gadgets, mastering algebra boosts their confidence to tackle everyday problems. Our hands-on lessons make learning both accessible and exciting. Since kids in the USA learn in diverse ways, we customize our methods to suit each learner’s style. From the lively streets of New York City to the sunny beaches of California, BrightCHAMPS brings math alive, making it meaningful and enjoyable all across America. Let’s make square roots an exciting part of every child’s math adventure! 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17502	https://www.cs.umd.edu/~samir/grant/gas-j.pdf	To Fill or not to Fill: The Gas Station Problem. SAMIR KHULLER Dept. of Computer Science and Institute for Advanced Computer Studies University of Maryland, College Park AZARAKHSH MALEKIAN Dept. of Computer Science University of Maryland, College Park and JULI´ AN MESTRE Max-Planck-Institut f¨ ur Informatik, Saarbr¨ ucken In this paper we study several routing problems that generalize shortest paths and the Traveling Salesman Problem. We consider a more general model that incorporates the actual cost in terms of gas prices. We have a vehicle with a given tank capacity. We assume that at each vertex gas may be purchased at a certain price. The objective is to ﬁnd the cheapest route to go from s to t, or the cheapest tour visiting a given set of locations. Surprisingly, the problem of ﬁnd the cheapest way to go from s to t can be solved in polynomial time and is not NP-complete. For most other versions however, the problem is NP-complete and we develop polynomial time approximation algorithms for these versions. Categories and Subject Descriptors: F.2.2 [Analysis of Algorithms and Problem Complex-ity]: Nonnumerical Alg orithms and Problems General Terms: Graph Theory, Algorithms Additional Key Words and Phrases: Approximation Algorithms, Shortest Paths, Vehicle Routing 1. INTRODUCTION Optimization problems related to computing the shortest (or cheapest) tour vis-iting a set of locations, or that of computing the shortest path between a pair of locations are pervasive in Computer Science and Operations Research. Typically, the measures that we optimize are in terms of “distance” traveled, or time spent (or in some cases, a combination of the two). There are literally thousands of papers dealing with problems related to shortest-path and tour problems. In this paper, we consider a more general model that incorporates the actual cost This research was supported by NSF grant CCF-0430650, while J. Mestre was at the University of Maryland, College Park. Contact Information: S. Khuller and A. Malekian: Email: {samir,malekian}@cs.umd.edu. J. Mestre: Email: jmestre@mpi-inf.mpg.de Permission to make digital/hard copy of all or part of this material without fee for personal or classroom use provided that the copies are not made or distributed for proﬁt or commercial advantage, the ACM copyright/server notice, the title of the publication, and its date appear, and notice is given that copying is by permission of the ACM, Inc. To copy otherwise, to republish, to post on servers, or to redistribute to lists requires prior speciﬁc permission and/or a fee. c ⃝20YY ACM 0000-0000/20YY/0000-0001 $5.00 ACM Journal Name, Vol. V, No. N, Month 20YY, Pages 1–0??. 2 · Khuller, Malekian, Mestre in terms of gas prices. We have a vehicle with a given tank capacity of U. In fact, we will assume that U is the distance the vehicle may travel on a full tank of gas (this can easily be obtained by taking the product of the tank size and the mileage per gas unit of the vehicle). Moreover, we may assume that we start with some given amount of gas µ (≤U) in the tank. We assume that at each vertex v gas may be purchased at a price of c(v). This price is the cost of gas per mile. For example if gas costs $3.40 per gallon and the vehicle can travel 17 miles per gallon, then the cost per mile is 20 cents. At each gas station we may ﬁll up some amount of gas to “extend” the range of the vehicle by a certain amount. Moreover, since gas prices vary, the cost depends on where we purchase gas from. In addition to ﬂuctuating gas prices, there is signiﬁcant variance in the price of gas between gas stations in diﬀerent areas. For example, in the Washington DC area alone, the variance in gas prices between gas stations in diﬀerent areas (on the same day) can be by as much as 20%. Due to diﬀerent state taxes, gas prices in adjacent states also vary. Finally, one may ask: why do we expect such information to be available? In fact, there are a collection of web sites [gas ; ] that currently list gas prices in an area speciﬁed by zip code. So it is reasonable to assume that information about gas prices is available. What we are interested in are algorithms that will let us compute solutions to some basic problems, given this information. In this general framework, we are interested in a collection of basic questions. (1) (The gas station problem) Given a start node s and a target node t, how do we go from s to t in the cheapest possible way if we start at s with µs amount of gas? In addition we consider the variation in which we are willing to stop to get gas at most ∆times1. Another generalization we study is the sequence gas station problem. Here, we want to ﬁnd the cheapest route that visits a set of p locations in a speciﬁed order (for example by a delivery vehicle). (2) (The ﬁxed-path gas station problem) An interesting special case is when we ﬁx the path along which we would like to travel. Our goal is to ﬁnd an optimal set of reﬁll stops along the path. (3) (The uniform cost tour gas station problem) Given a collection of cities T , and a set of gas stations S at which we are willing to purchase gas, ﬁnd the shortest tour that visits T . We have to ensure that we never run out of gas. Clearly this problem generalizes the Traveling Salesman Problem. The problem gets more interesting when S ̸= T , and we address this case. This models the situation when a large transportation company has a deal with a certain gas company, and their vehicles may ﬁll up gas at any station of this company at a pre-negotiated price. Here we assume that gas prices are the same at each gas station. This could also model a situation where some gas stations with very high prices are simply dropped from consideration, and the set S is simply the set of gas stations that we are willing to use. 1This restriction makes sense, because in some situations where the gas prices are decreasing as we approach our destination, the cheapest solution may involve an arbitrarily large number of stops, since we only ﬁll up enough gas to make it to a cheaper station further down the path. ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 3 (4) (The tour gas station problem) This is the same as the previous problem, except that the prices at diﬀerent stations can vary. Of all the above problems, only the tour problems are NP-hard. For the ﬁrst two we develop polynomial time algorithms, and for the tour problems we develop approximation algorithms. We now give a short summary of the results in the paper: (1) (The gas station problem) For the basic gas station problem, our algorithm runs in time O(∆n2 log n) and computes an optimal solution. If we want to visit a sequence of p cities we can ﬁnd an optimal solution in time O(∆(np)2 log(np)). In addition, we develop a second algorithm for the all-pairs version that runs in time O(n3∆2). This method is better than repeating the ﬁxed-destination algorithm n times when ∆< log n. (2) (The ﬁxed-path gas station problem) For the ﬁxed-path version with an un-bounded number of stops, we develop a fast O(n log n) time algorithm. (3) (The uniform cost tour gas station problem) Since this problem is NP-hard, we focus on polynomial time approximation algorithms. We assume that every city has a gas station within a distance of α U 2 for some α < 1. This assumption is reasonable since in any case, every city has to have a gas station within distance U 2 , otherwise there is no way to visit it. A similar assumption is made in the work on distance constrained vehicle routing problem [Li et al. 1992]. We develop an approximation algorithm with an approximation factor of 3 2( 1+α 1−α). We also consider a special case, namely when there is only one gas station. This is the same as having a central depot, and requiring the vehicle to return to the depot after traveling a maximum distance of U. For this special case, we develop an algorithm with factor O(ln 1 1−α) and this improves the bound of 3 2(1−α) given by Li et al. [Li et al. 1992] for the distance constrained vehicle routing problem. (4) (The tour gas station problem) For the tour problem with arbitrary prices, we can use the following scheme: sort all the gas prices in non-decreasing order c1 ≤c2 ≤. . . cn. Now guess a range of prices [ci . . . cj] one is willing to pay, and let βij = cj ci . Let Sij include all the gas stations v such that ci ≤c(v) ≤cj. We can run the algorithm for the uniform cost tour gas station problem with set Sij and cities T . This will yield a tour T [i,j]. We observe that the cost of the tour T [i,j] is at most O( βij 1−α) times the cost of an optimal solution, since its possible that we always pay a factor βij more than the optimal solution, at each station where we ﬁll gas. Taking the best solution over all O(n2) possible choices gives a valid solution to the tour gas station problem. 1.1 Related Work The problems of computing shortest paths and the shortest TSP tour are clearly the most relevant ones here and are widely studied, and discussed in several books [Lawler et al. 1985; Papadimitriou and Steiglitz 1998]. One closely related problem is the Orienteering problem [Arkin et al. 1998; Awer-buch et al. 1998; Golden et al. 1987; Blum et al. 2003]. In this problem the goal is to compute a path of a ﬁxed length L that visits as many locations as possible, ACM Journal Name, Vol. V, No. N, Month 20YY. 4 · Khuller, Malekian, Mestre starting from a speciﬁed vertex. For this problem, a factor 3 approximation has been given recently by Bansal et al. [Bansal et al. 2004]. (In fact, they can ﬁx the starting and ending vertices.) This algorithm is used as subroutine for developing a bicriteria bound for Deadline TSP. By using the 3 approximation for the Orien-teering problem, we develop an O(log \|T \|) approximation for the single gas station tour problem. This is not surprising, since we would like to cover all the locations by ﬁnding walks of length at most U. There has been some recent work by Nagarajan and Ravi [Nagarajan and Ravi 2006] on minimum vehicle routing that is closely related to the single gas station tour problem. In this problem, a designated root vertex (depot) and a deadline D are given and the goal is to use the minimum number of vehicles from the root so that each location is met by at least one of the vehicles, and each vehicle traverses length at most D. (In their deﬁnition, vehicles do not have to go back to the root.) They give a 4-approximation for the case where locations are in a tree and an O(log D) approximation for graphs with integer weights. Another closely related piece of work is by Arkin et al. [Arkin et al. 2006] where tree and tour covers of bounded length are computed. What makes their problem easier is that there is no speciﬁed root node, or a set of gas stations one of which should be included in any bounded length tree or tour. Several pieces of work deal with vehicle routing problems [M. Haimovich 1985; 1988; Frederickson et al. 1978] with multiple vehicles, where the objective is to bound the total cost of the solution, or to minimize the longest tour. However these problems are signiﬁcantly easier to develop approximation algorithms for. 2. THE GAS STATION PROBLEM The input to our problem consists of a complete graph G = (V, E) with edge lengths d : E →R+, gas costs c : V →R+ and a tank capacity U. (Equivalently, if we are not given a complete graph we can deﬁne duv to be the distance between u and v in G.) Our goal is to go from a source s to a destination t in the cheapest possible way using at most ∆stops to ﬁll gas. For ease of exposition we concentrate on the case where we start from s with an empty tank. The case in which we start with µs units of gas can be reduced to the former as follows. Add a new node s′ such that ds′s = U −µs and c(s′) = 0. The problem of starting from s with µs units of gas and that of starting from s′ with an empty tank using one additional stop are equivalent. We would also like to note that our strategy yields a solution where the gas tank will be empty when one reaches a location where gas can be ﬁlled cheaply. In practice, this is not safe and one might run out of gas (for example if one gets stuck in traﬃc). For that reason we suggest deﬁning U to be smaller than the actual tank capacity so that we always have some “reserve” capacity. In this section we develop an O(∆n2 log n) time algorithm for the gas station problem. In addition, when ∆= n we show how to solve the problem in O(n3) time for general graphs, and O(n log n) time for the case where G is a ﬁxed path. One interesting generalization of the problem is the sequence gas station problem where we are given a sequence s1, s2, . . . , sp of vertices that we must visit in the speciﬁed order. This variant can be reduced to the s-t version in an appropriately ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 5 deﬁned graph. 2.1 The gas station problem using ∆stops We will solve the gas station problem using the following dynamic program (DP) formulation: A(u, q, g) = Minimum cost of going from u to t using q reﬁll stops, starting with g units of gas. We consider u to be one of the q stops. The main diﬃculty in dealing with the problem stems from the fact that, in principle, we need to consider every value of g ∈[0, U]. One way to avoid this is to discretize the values g can take. Unfortunately this only yields a pseudo-polynomial time algorithm. To get around this we need to take a closer look at the structure of the optimal solution. Lemma 2.1. Let s = u1, u2, . . . , ul be the reﬁll stops of an optimal solution using at most ∆stops. The following is an optimal strategy for deciding how much gas to ﬁll at each stop: At ul ﬁll just enough to reach t with an empty tank; for j < l i) If c(uj) < c(uj+1) then at uj ﬁll up the tank. ii) If c(uj) ≥c(uj+1) then at uj ﬁll just enough gas to reach uj+1. Proof. If c(uj) < c(uj+1) and the optimal solution does not ﬁll up at uj then we can increase the amount ﬁlled at uj and decrease the amount ﬁlled at uj+1. This improves the cost of the solution, which contradicts the optimality assump-tion. Similarly, if c(uj) ≥c(uj+1) then we can decrease the amount ﬁlled at uj and increase the amount ﬁlled at uj+1 (without increasing the overall cost of the solution) until the condition is met. Consider a reﬁll stop u ̸= s in the optimal solution. Let w be the stop right before u. Lemma 2.1 implies that if c(w) ≥c(u), we reach u with an empty tank, otherwise we reach u with U −dwu gas. Therefore, in our DP formulation we need to keep track of at most n diﬀerent values of gas for u. Let GV (u) be the set of such values, namely GV (u) = {U −dwu \| w ∈V and c(w) < c(u) and dwu ≤U} ∪{0} The following recurrence allows us to compute A(u, q, g) for any g ∈GV (u): A(u, 1, g) = (dut −g) c(u) if g ≤dut ≤U ∞ otherwise A(u, q, g) = min v s.t. duv≤U A(v, q−1, 0) + (duv −g) c(u) \| c(v) ≤c(u) ∧g ≤duv A(v, q−1, U −duv) + (U −g) c(u) \| c(v) > c(u) The cost of the optimal solution is min1≤l≤∆A(s, l, 0). The naive way of ﬁlling the table takes O(∆n3) time. However, this can be done more eﬃciently. Theorem 2.2. There is an O(∆n2 log n) time algorithm for the gas station prob-lem with ∆stops. Instead of spending O(n) time computing a single entry of the table, we spend O(log n) amortized time per entry. More precisely, for ﬁxed u ∈V and 1 < q ≤∆ ACM Journal Name, Vol. V, No. N, Month 20YY. 6 · Khuller, Malekian, Mestre we show how to compute all entries of the form A(u, q, ∗) in O(n log n) time using entries of the form A(∗, q−1, ∗). Theorem 2.2 follows immediately. The DP recursion for A(u, q, g) ﬁnds the minimum, over all v such that duv ≤ U, of terms that corresponds to the cost of going from u to t through v. Split each of these terms into two parts based on whether they depend on g or not. Thus we have an independent part, which is either A(v, q −1, 0) + duv c(u) or A(v, q −1, U −duv) + Uc(u); and a dependent part, −g c(u). fill-row(u, q) 1 R ←{v ∈V \| duv ≤U} 2 for v ∈R do 3 if c(v) ≤c(u) 4 then indep(v) ←C[v, q −1, 0] + duvc(u) 5 else indep(v) ←C[v, q −1, U −duv] + Uc(u) 6 sort R in increasing indep(·) value 7 let v ∈R be ﬁrst in sorted order 8 for g ∈GV (u) in increasing value do 9 while g > duv do 10 let v ∈R be next vertex in sorted order 11 C[u, q, g] ←indep(v) −gc(u) Fig. 1. An O(n log n) time procedure for computing C[u, q, ∗]. Our procedure begins by sorting the independent part of every term. Note that the minimum of these corresponds to the entry for g = 0. As we increase g, the terms decrease uniformly. Thus, to compute the table entry for g > 0 just subtract g c(u) from the smallest independent part available. The only caveat is that the term corresponding to a vertex v such that c(v) ≤c(u) should not be considered any more once g > duv, we say such a term expires after g > duv. Since the independent terms are sorted, once the smallest independent term expires we can walk down the sorted list to ﬁnd the next vertex which has not yet expired. The procedure is dominated by the time spent sorting the independent terms which takes O(n log n) time. Its pseudocode is given in Figure 1. Theorem 2.3. When ∆= n the problem can be solved in O(n3) time. We can reduce the problem to a shortest path question on a new graph H. The vertices of H are pairs (u, g), where u ∈V and g ∈GV (u). The edges of H and their weight w(·) are deﬁned by the DP recurrence. Namely, for every u, v ∈V and g ∈GV (u) such that duv ≤U we have w (u, q), (v, 0) = (duv−g) c(u) if c(v) ≤c(u) and g ≤duv, or w (u, q), (v, U −duv = (U −g) c(u) if c(v) > c(u). Our objective is to ﬁnd a shortest path from (s, 0) to (t, 0). Note that H has at most n2 vertices and at most n3 edges. Using Dijkstra’s algorithm [Cormen et al. 2001] the theorem follows. 2.2 Faster algorithm for the all-pairs version Consider the case in which we wish to solve the problem for all starting nodes i, with µi amount of gas in the tank initially. Using the method described in the ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 7 previous section, we get a running time of O(n3∆log n) since we run the algorithm for each possible destination. We will show that for ∆< log n we can improve this and get a bound of O(n3∆2). Add new nodes i′ such that di′i = U −µi and c(i′) = 0. If we start at i with µi units of gas, it is the same as starting from i′ where gas is free. We ﬁll up the tank to capacity U, and then by the time we reach i we will have exactly µi units of gas in the tank. (Since gas is free at any node i′ in any optimal solution we ﬁll up the tank to capacity U). This will use one extra stop. We deﬁne B(i, h, p) as the minimum cost solution to go from i to h (destination), with p stops to get gas, given that we start with an empty tank at i. Since we start with an empty tank, we have to ﬁll up gas at the starting point (and this is included as one of the stops). Clearly, we will also reach h (destination) with an empty tank, assuming that there is no trivial solution, such as one that arrives at the destination with no ﬁll-ups on the way. Our goal is to compute B(i′, h, ∆+ 1) which is a minimum cost solution to go from i′ to h with at most ∆stops in-between. Note that the ﬁrst ﬁll-up is the one that takes place at node i′, after that we stop at most ∆times. We will now show how to compute B(i, h, p). There are two options: —If the gas price at the ﬁrst stop after i (e.g. k) is cheaper than c(i) then we will reach that station with an empty tank after ﬁlling dik units of gas at i (as long as dik ≤U): B(i, h, p) = B(k, h, p −1) + dikc(i) —If the ﬁrst place where the cost of gas decreases from the previous stop is the q + 1st stop and the price is in increasing order in the ﬁrst q stops then B(i, h, p) = C(i, k, q) + B(k, h, p −q) We deﬁne C(i, k, q) as the minimum cost way of going from i to k with at most q stops to get gas, such that we start at i with an empty tank (and get gas at i, which counts as a stop) and ﬁnally reach k with an empty tank. In addition, the price of gas in intermediate stations is in increasing order except for the last stop. We deﬁne B(h, h, p) = 0. For i ̸= h let B(i, h, 1) = c(i) dih if dih ≤U, and B(i, h, 1) = ∞otherwise. In general: B(i, h, p) = min ( min 1≤k≤n 1 c(iq+1). In fact, at i1 we will get U amount of gas. When we reach ij for 1 < j < q, we will get dij−1ij units of gas (the amount that we consumed since the previous ﬁll-up) at a cost of c(ij) per unit of ACM Journal Name, Vol. V, No. N, Month 20YY. 8 · Khuller, Malekian, Mestre Refill stop Start with empty tank Reach with Cost of gas empty tank i2 i = i1 i3 i4 k = i5 Fig. 2. Example to show C(i, k, q) for q = 4. gas. The amount of gas we will get at iq is just enough to reach k with an empty tank. Now we can see that the total cost is equal to Uc(i1) + di1i2c(i2) + . . . + diq−2iq−1c(iq−1) + (diq−1iq + diqk −U)c(iq). Note that the last term is not negative, since we could not reach k from iq−1 even with a full tank at iq−1, without stopping to get a small amount of gas. We compute C(i, k, q) as follows. First note that if dik ≤U then the answer is dikc(i). Otherwise we build a directed graph G′ = (V ∪VD, E ∪ED), where V is the set of vertices, and VD = {i′\|i ∈V }. We deﬁne E: add a directed edge from i ∈V to j for each vertex j ∈V \ {i} such that dij ≤U and c(i) ≤c(j). The weight of this edge is dijc(j). We deﬁne ED as follows: add a directed edge from each j ∈V to k′ for each vertex k′ ∈VD \ {j′} such that U < djk ≤2U. The weight of this edge is min (djz + dzk −U)c(z) \| c(j), c(k) < c(z) and djz, dzk ≤U Now we can express C(i, k, q) as Sp(i, k′, q)+Uc(i) where Sp(i, k′, q) is the short-est path from i to k′ in the graph G′ using at most q edges. To see why it is true, we can see that for any given order of stops between i and k (where the gas price is in increasing order in consecutive stops), the minimum cost is equal to the weight of the path in G′ that starts from i, goes to the second stop in the given order (e.g., i2) and then traverses the vertices of V in the same order and from the second last stop goes to k′. It is also possible that q = 2 and the path goes directly from i = i1 to k in this case, and i2 is the choice for z that achieves the minimum cost for the edge (i, k′). For any given path P in G′ between i and k′, if the weight of the path is WP we can ﬁnd a feasible plan for ﬁlling the tank at the stations so that the cost is equal to WP + Uc(i). It is enough to ﬁll up the tank at the stations that are in the path, except the last one in which the tank is ﬁlled to only the required level to reach k. We can conclude that C(i, k, q) is equal to Sp(i, k′, q) + Uc(i). The running time for ﬁnding the shortest path between all pairs of nodes with diﬀerent number of stops (at most ∆) can be computed in O(n3∆) by dynamic programming [Lawler 2001]. If we precompute C(i, k, q) the running time for com-puting B(i′, h, ∆+ 1) is O(n3∆2) assuming we start at i with µi amount of gas. So in general the running time is O(n3∆2). ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 9 2.3 The sequence gas station problem Suppose instead of a given source and destination, we are asked to ﬁnd the cheapest way to start from a given location, visit some a set of locations in a given order during the trip and then reach the ﬁnal destination. We deﬁne the problem in a formal way as follows: Given an edge weighted graph G = (V, E) and a list of vertices s0, . . . , sp, we wish to ﬁnd the cheapest way to start from s0, visit s1, . . . , sp−1 in this order and then reach sp. Note that we cannot reduce this problem to p separate source-destination sub-problems and combine the solutions directly. To see why, consider the case where the gas price is very high at some station si and on the way from si−1 to si there is a very cheap gas station near si. If we want to use the solution for the separate subproblems and then combine them, we will reach si with an empty tank so we have to ﬁll the tank at si since we are out of gas; but the optimal solution is to reach si with some gas in the tank to make it possible to reach next station after si without ﬁlling the tank at si. between some node sj and sj+1 is not an optimal way that would be chosen in the To solve this problem, we will make a new graph as follows: Make p −1 new copies of the current graph G and call them G1, . . . , Gp−1. G will become G0. Call vi in Gj as vi,j. Now connect Gi and Gi+1 by merging si+1,i and si+1,i+1 into one node. The solution to the original problem is to ﬁnd the cheapest way to go from s1,0 to sp,p−1 in the new graph. we can see that any path in this graph that goes from s1,0 to sp,p−1 will pass through si+1,i ∀i 0 ≤i ≤p −1. 2.4 Fixed-path Number the nodes along the path from 1 to n, so that we start at 1 and want to reach n. Without loss of generality assume we start with an empty tank. We present a fast, yet simple, exact algorithm for the case where the number stops is unbounded. Theorem 2.4. There is an O(n log n) time algorithm for the ﬁxed-path gas sta-tion problem with an unbounded number of stops. The ﬁrst step consists in ﬁnding, for each gas station i, its previous and next station. Deﬁne prev(i) as the station j ≤i with the cheapest gas among those that satisfy dji ≤U. Similarly let next(i) be the station j > i with the cheapest gas such that dij ≤U. Any eventual tie is broken by favoring the station closest to n. To compute these two values we keep a priority queue on the stations that lie on a moving window of length U. Starting at 1, we slide the window toward n inserting and removing stations as we go along. Right after inserting into (removing from) the queue some station i, asking for the minimum in the queue gives us prev(i) (next(i)). The whole procedure takes O(n log n) time. Station i is said to be a break point if prev(i) = i. Identifying such stations is important because we can break our problem into smaller subproblems (to go from one break point to the next) and then paste these solutions to get a global optimal solution. Lemma 2.5. Let i be a break point. There is an optimal solution that reaches i ACM Journal Name, Vol. V, No. N, Month 20YY. 10 · Khuller, Malekian, Mestre with an empty tank. Proof. Let j < i be the last station we stopped to get gas before reaching i. Since i is a break point, we have c(i) ≤c(j). Therefore at j we ﬁll just enough gas to reach i with an empty tank. Now consider the subproblem of going from i to k starting and ending with an empty tank, such that there is no break point in (i, k). The following algorithm solves our subproblem optimally. drive-to-next(i, k) 1 Let x be i. 2 If dxk ≤U then just ﬁll enough gas to go k. 3 Otherwise, ﬁll up and drive to next(x). Let x be next(x), go to step 2. The key observation is that for every station x considered by the algorithm, if dxk > U then c(x) ≤c(next(x)). Since all stations in a range of U after x oﬀer gas at cost at least c(x), an optimal solution ﬁlls up at x and drives up to the next cheapest station, i.e., next(x). Remark: even though drive-to-next solves our special subproblem optimally, the strategy does not work in general. To see why consider an instance where c(i) > c(i + 1) and d1n = U. While the optimum stops on every station, drive-to-next will tell us to go straight from 1 to n. 3. THE UNIFORM COST TOUR GAS STATION PROBLEM In this section we study a variant of the gas station problem where we must visit a set of cities T in arbitrary order. We consider the case where gas costs the same at every gas station, but some cities may not have a gas station. More formally, the input to our problem consists of a complete undirected graph G = (V, E) with edge lengths d : E →R+, a set of cities T ⊆V , a set of gas stations S ⊆V , and tank capacity U for our vehicle. The objective is to ﬁnd a minimum length tour that visits all cities in T , and possibly some gas stations in S. We are allowed to visit a location multiple times if necessary. We require any segment of the tour of length U to contain at least one gas station, this ensures we never run out of gas. We call this the uniform cost tour gas station problem. We assume that we start with an empty tank at a gas-station. The problem is NP-hard as it generalizes the well-known traveling salesman problem: just set the tank capacity to the largest distance between any two cities and let T = S. In fact, there is a closer connection between the two problems: If every city has a gas station, i.e., T ⊆S, we can reduce the gas station problem to the TSP. Consider a TSP instance on T under metric ℓ: T × T →R+, where ℓxy is the minimum cost of going between cities x and y starting with an empty tank (this can be computed by standard techniques). Since the cost of gas is the same everywhere, a TSP tour can be turned into a driving plan that visits all cities with the same cost and vice-versa. Let OPT denote an optimal solution, and c(OPT ) its cost. ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 11 As mentioned earlier, we can use the algorithm for the uniform cost case to derive an approximation algorithm for the general case by paying a factor β in the approximation ratio. Here β is the ratio of the maximum price that an optimal solution pays for buying a unit of gas, to the minimum price it pays for buying a unit of gas (in practice this ranges from 1 to 1.2). Unfortunately this reduction to the TSP breaks down when cities are not guar-anteed to have a gas station. Consider going from x to y, where x does not have a gas station. The distance between x and y will depend on how much gas we have at x, which in turn depends on which city was visited before x and what route we took to get there. An interesting case of the tour gas station problem is that of an instance with a single gas station. This is also known as the distance constrained vehicle routing problem and was studied by Li et al. [Li et al. 1992] who gave a 3 2(1−α) approx-imation algorithm, where the distance from the gas station to the most distant city is α U 2 , for some α < 1. We improve this by providing an O(log 1 1−α) approxi-mation algorithm. Without making any assumptions on α we show that a greedy algorithm that ﬁnds bounded length tours visiting the most cities at a time is a O(log \|T \|)-factor approximation. For the general case we make the assumption that every city has a gas station at distance at most α U 2 . This assumption is reasonable, because if a city has no gas station within distance U 2 , there is no way to visit it. We show a 3(1+α) 2(1−α) approximation for this problem. Note that when α = 0, this gives the same bound as the Christoﬁdes method for the TSP. 3.1 The tour gas station problem For each city x ∈T let g(x) ∈S be the closest gas station to x, and let dx be the distance from x to g(x). We assume that every city has a gas station at distance at most α U 2 ; in order words, dx ≤α U 2 for all x ∈T . Recall that it is assumed that the price of the gas is the same at all the gas stations. We deﬁne a new distance function for the distance between each pair of cities. The distance ℓis deﬁned as follows: For each pair of cities x and y, ℓxy is the length of the shortest traversal to go from x to y starting with U −dx amount of gas and reaching y with dy amount of gas. If dxy ≤U −dx −dy then we can go directly from x to y, and ℓxy = dxy. Otherwise, we can compute this as follows. Create a graph whose vertex set is S, the set of gas stations. To this graph add x and y. We now add edges from x to all gas stations within distance U −dx from x. Similarly we add edges from y to all gas stations within distance U −dy to y. Between all pairs of gas stations, we add an edge if the distance between the pair of gas stations is at most U. All edges have length equal to the distance between their end points. The length of the shortest path in this graph from x to y will be ℓxy. Note that the shortest path (in general) will start at x and then go through a series of gas stations before reaching y. This path yields a valid plan to drive from x to y without running out of gas, once we reach x with U −dx units of gas. When we reach y, we have enough gas to go to gy. Also note that ℓxy = ℓyx since the path is essentially “reversible”. In Fig. 3 we illustrate the deﬁnition of function ℓxy. We assume here that all ACM Journal Name, Vol. V, No. N, Month 20YY. 12 · Khuller, Malekian, Mestre F C A B D E y U −dx gx x dx gy U −dy dy Fig. 3. Function ℓxy. The path shown is the shortest valid path from x to y. distances are Euclidean. Note that from x, we can only go to B and not A since we start from x with U −dx units of gas. From B, we cannot go to D since the distance between B and D is more than U, even though the path through D to y would be shorter. From C we go to E since going through F will give a longer path, since from F we cannot go to y directly. Note that the function ℓmay not satisfy triangle inequality. To see this, suppose we have three cities x, y, z. Let dxy = dyz = U 2 . Let dx = dy = dz = U 4 and dxz = U. We ﬁrst observe that ℓxy = ℓyz = U 2 . However, if we compute ℓxz, we cannot go from x to z directly since we only have 3 4U units of gas when we start at x and need to reach z with U 4 units of gas. So we have to visit gy along the way, and thus ℓxz = 3 2U. The algorithm is as follows: (1) Create a new graph G′, with a vertex for each city. For each pair of cities x, y compute ℓxy as shown earlier. (2) Find the minimum spanning tree in (G′, ℓ). Also ﬁnd a minimum weight perfect matching M on the odd degree vertices in the MST. Combine the MST and M to ﬁnd an Euler tour T . (3) Start traversing the Eulerian tour. Add reﬁll trips whenever needed. (Details on this follow.) It can be shown that the total length of the MST is less than the optimal solution cost. Suppose x1, . . . , xn is the order in which the optimal solution visits the cities. Clearly, the cost of going from xi to xi+1 in the optimal solution is at least ℓxixi+1. Since the collection of edges (xi, xi+1) forms a spanning tree, we can be conclude that the weight of the ℓ(MST) ≤c(OPT ). Next we show that the cost of M is at most c(OPT ) 2 . Suppose the odd degree vertices are in the optimal solution in the order o1, . . . , ok. We can see that ℓoioi+1 is at most equal to the distance we travel in the optimal solution to go from oi to oi+1. So the cost of minimum weighted matching on the odd degree vertices is at most c(OPT ) 2 . So the total cost of the Eulerian tour T is at most 3c(OPT ) 2 . Now we need to transform the Eulerian tour into a feasible plan. First, every ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 13 indirect edge refill trip direct edge city gas station x1 i x2 i x3 i x4 i xk+1 i x0 i xk i . . . Fig. 4. Decomposition of the solution into strands. edge (x, y) in T is replaced with the actual plan to drive from x to y that we found when computing ℓxy. If dxy ≤U −dx −dy the plan is simply to go straight from x to y, we call these direct edges. Otherwise the plan must involve stopping along the way in one or more gas stations, we call these indirect edges. Notice that the cost of this plan is exactly that of the Eulerian tour T . Unfortunately, as we will see below this plan need not be feasible. Deﬁne a strand, to be a sequence of consecutive cities in the tour connected by direct edges. If a city is connected with two indirect edges, then it forms a strand by itself. Suppose the ith strand has cities x1 i , . . . , xk i . To this we add x0 i (xk+1 i ), the last (ﬁrst) gas station in the indirect edge connecting x1 i (xk i ) with the rest of the tour. Each strand now starts and ends with a gas station. We can view the tour as a decomposition into strands as shown in Fig. 4. Note that if the distance between x0 i and xk+1 i is more than U the overall plan is not feasible. To ﬁx this we add for every city a reﬁll trip to its closest gas station and then greedily try to remove them, while maintaining feasibility, until we get a minimal set of reﬁll trips. Let us bound the extra cost these trips incur. Lemma 3.1. Let Li be the length of the ith strand. Then the total distance traveled on the reﬁll trips of cities in the strand is at most 2α 1−αLi. Proof. Assume there are qi reﬁll trips in this strand. Label the cities with reﬁll trips to their nearest gas stations xj1 i , . . . , x jqi i . Also label x0 i as xj0 i and xk i as x jqi+1 i . Note that ℓ(T (xjp i , xjp+2 i )) ≥(1 −α)U (otherwise the reﬁll trip at xjp+1 i can be dropped). This gives us: 2Li > X 0≤p≤qi−1 ℓ(T (xjp i , xjp+2 i )) ≥qi(1 −α)U = ⇒qi ≤ 2Li (1 −α)U The length of each reﬁll trip no more than αU. Therefore, the total length of the reﬁll trips is at most αUqi, and the lemma follows. The cost of the solution is the total length of the strands (which is the length of the tour) plus the total cost of the reﬁll trips. (Note that without loss of generality we can assume that our tour always starts from a gas station. For the case with only direct edges, there is exactly one strand, starting and ending at the ﬁrst city with the gas station). In other words, the total cost of the solution is: ℓ(T ) + X i αUqi ≤ 1 + 2α 1 −α ℓ(T ) ≤ 1 + α 1 −α 3 2 c(OPT). ACM Journal Name, Vol. V, No. N, Month 20YY. 14 · Khuller, Malekian, Mestre Theorem 3.2. There is a 3 (1+α) 2 (1−α)-approximation for the tour gas station prob-lem. 3.2 Single Gas Station In this version, there is a single gas station and our vehicle starts there. It must return to the gas station before it runs out of gas after traveling a distance of at most U from the previous ﬁll-up. Fix constants (ρ1, ρ2, . . . , ρl). Our algorithm ﬁrst visits cities at distance ρ1 U 2 from the gas station (we refer to these cities as C0). Beyond ρ1 U 2 we work in iterations. In the ith iteration we visit cities (Ci) that lie at distance U 2 ρi, U 2 ρi+1 from the gas station. If we make 1−ρi 1−ρi+1 = γ a constant, after l logγ 1−ρ1 1−α m iterations we will have visited all cities. We will argue that in each iteration we travel O(c(OPT )) distance, which gives us the desired result. The ρi values will be chosen to minimize the constants involved to get the following theorem. Theorem 3.3. There is a 6.362 ln 1 1−α −1.534 factor approximation for the uni-form cost tour gas station problem with a single station, for α ≥0.5. Notice that that for α ≥0.5 the above approximation ratio is ≥1. First we consider the cities C0 at distance ρ1 U 2 or less from the gas station. Find a TSP tour on the gas station and C0 and chop it into segments of length (1−ρ1)U. The distance from the gas station to any location is at most ρ1 U 2 and so the segments can be traversed with loops of length at most U. In fact we can start chopping the TSP tour at the gas station and make the ﬁrst and the last segment be of length (1 −ρ1 2 )U. The total length of these tours will be: cost(C0) ≤ cost(TSP) −ρ1U (1 −ρ1)U U ≤cost(TSP) (1 −ρ1) ≤ 3 2(1 −ρ1) · OPT The second inequality holds if we assume ρ1 ≥.5. The third comes from using Christoﬁdes [Christoﬁdes 1976] algorithm [Christoﬁdes 1976] to ﬁnd the TSP tour and the fact that OPT is a valid TSP tour. Notice that it does not work well when cities are far away from the gas station (α ≈1). In our scheme those far away cities will be visited in a diﬀerent fashion. In the ith iteration we visit cities Ci at distance (ρi U 2 , ρi+1 U 2 ] by ﬁnding a collection of paths of length at most (1−ρi+1) U spanning Ci and then turning these segments into loops. Suppose we knew that in the optimal solution there are ki loops that span some city in Ci—this quantity can be guessed. First we run Kruskal’s algorithm but stop once the number of components becomes ki, let Ri be the resulting forest. Each tree is doubled to form a loop and then chopped into segments of length (1 −ρi+1) U. Let k′ i be the number of such segments. The cost of the these loops is therefore, cost(Ci) ≤2 cost(Ri) + k′ i ρi+1U Lemma 3.4. The number of segments k′ i is at most (2γ + 1)ki. Proof. The edges in Ri form a minimum weight forest with ki components, we can relate this to the cost of OPT. Consider turning each loop in OPT into a path ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 15 by keeping the stretch between the ﬁrst and the last city in Ci. The set P of such paths is a forest with ki components, therefore cost(Ri) ≤cost(P) ≤(1 −ρi)Uki Using this we can bound the number of segments we get after doubling and chopping Ri: k′ i ≤ 2 cost(Ri) (1 −ρi+1)U + ki ≤ 2 (1 −ρi)Uki (1 −ρi+1)U + ki ≤(2 γ + 1) ki We now bound the cost of visiting the cities in Ci. cost(Ci) ≤2 cost(Ri) + k′ i ρi+1U ≤2 cost(OPT) −2kiρiU + (2 γ + 1) ki ρi+1U ≤2 cost(OPT) + (2 γ −1) (cost(OPT) −kiρiU) −2kiρiU + (2 γ + 1) ki ρi+1U ≤(2 γ + 1) cost(OPT) + (2 γ + 1) ki (ρi+1 −ρi)U Let k be the number of loops in the optimal solution whose length is greater than ρ1U, notice that loops spanning cities beyond ρ1 U 2 must be at least this long, therefore k ≥ki for all i. Adding up over all iterations we get: l X i=1 cost(Ci) ≤(2 γ + 1) (l cost(OPT) + k(ρl −ρ1)U) ≤(2 γ + 1) l + 1 −ρ1 ρ1 cost(OPT) After l = l logγ 1−ρ1 1−α m iterations we will have visited all cities at a cost of: 3 2(1 −ρ1) + (2γ + 1) logγ 1 −ρ1 1 −α + 1 + 1 ρ1 −1 cost(OPT) We can use numerical optimization to minimize the approximation ratio in the expression from above. The values ρ1 = 0.7771 and γ = 3.1811 gives us Theo-rem 3.3. 3.3 A Greedy Algorithm In this case we do not make any assumption on the maximum distance from a city to its closest gas station. We will use the Point-to-Point Orienteering path as the basis of the greedy scheme. In the Point-to-Point Orienteering problem, each vertex in the graph has a prize. The goal is to ﬁnd a path P of maximum length d (predeﬁned) between two given vertices s and t so that the total prize of P is maximized. A 3-approximation algorithm for this problem is described in [Bansal et al. 2004]. The greedy algorithm works as follows: At the beginning the prize of all the cities are initialized to 1. As the algorithm proceeds whenever we visit a city in a tour, we reset its prize to 0. The greedy algorithm will repeatedly choose the Point-to-Point Orienteering path that begins and ends at s with maximum length U, until the prize of all the vertices are reset to zero. Using an argument similar to that in set-cover it can be shown that both the total cost and the number of cycles given by this approach is at most O(log \|T \|) times the optimum cost. ACM Journal Name, Vol. V, No. N, Month 20YY. 16 · Khuller, Malekian, Mestre Theorem 3.5. The greedy method gives an O(log \|T \|) approximation guarantee for both the total cost and the number of the cycles in the single gas station problem. Proof. Observe that if an algorithm approximates the number of cycles in the optimal solution, it also approximates the total length of the tour over the optimal solution. For any given solution, we can merge each two cycles of length less than U 2 together. The new tour is still feasible and of length less or equal the initial tour. Thus, there exists a minimum-length solution in which the sum of the lengths of any two cycles is at least U . Consider the solution with minimum length and with the property that we can merge no more cycles. If the number of cycles in this tour is Nc and the total length traversed is L, by the above argument we conclude that L ≥⌈J 2 ⌉U. Now, suppose we give an algorithm which cover all the points in aOPTc cycles where OPTc is the optimal number of cycles to cover all the points.We can conclude that the length of the tour is at most 2aT . From now on we try to ﬁnd the approximation factor for the number of cycles in our solution. Suppose the optimal number of cycles is J. The total length of the tour will be at most U × J. Let ui and si denote the number of elements covered in round i and the total number of elements covered from the beginning till this round, respectively. Therefore, con-sidering the way we choose the cycles we can assert that u1 ≥ n 3J (where n in the number of cities), and also for each ui ui ≥n−si−1 3J holds. The algorithm continues until si ≥n. We deﬁne si, by the following recursion: si ≥ n 3J i = 1 si−1 + n−si−1 3J i > 1 After solving the above recursion, we see that si ≥n(1 − 1 3J )i. Our goal is to ﬁnd the smallest i so that si > n −1. Hence, if iterate for i > J × O(log n), all the cities would be covered. So the greedy method will give us an O(log n) approximate solution for both length and number of cycles. 4. CONCLUSIONS Current problems of interest are to explore improvements in the approximation factors for the special cases of Euclidean metrics, and planar graphs. In addition we would also like to develop faster algorithms for the single source and destination case, perhaps at the cost of sacriﬁcing optimality of the solution. REFERENCES Arkin, E. M., Hassin, R., and Levin, A. 2006. Approximations for minimum and min-max vehicle routing problems. Journal of Algorithms 59, 1, 1–18. Arkin, E. M., Mitchell, J. S. B., and Narasimhan, G. 1998. Resource-constrained geomet-ric network optimization. In Proceedings of the 14th Annual Symposium on Computational Geometry (SoCG). 307–316. Awerbuch, B., Azar, Y., Blum, A., and Vempala, S. 1998. New approximation guarantees for minimum-weight k-trees and prize-collecting salesmen. SIAM Journal on Computing 28, 1, 254–262. ACM Journal Name, Vol. V, No. N, Month 20YY. The Gas Station Problem · 17 Bansal, N., Blum, A., Chawla, S., and Meyerson, A. 2004. Approximation algorithms for deadline-TSP and vehicle routing with time-windows. In Proceedings of the 36th annual ACM symposium on Theory of computing (STOC). 166–174. Blum, A., Chawla, S., Karger, D. R., Lane, T., Meyerson, A., and Minkoff, M. 2003. Approximation algorithms for orienteering and discounted-reward TSP. In Proceedings of the 44rd Annual IEEE Symposium on Foundations of Computer Science (FOCS). 46. Christofides, N. 1976. Worst-case analysis of a new heuristic for the traveling salesman problem. Tech. rep., Graduate School of Industrial Administration, Carnegie-Mellon University. Cormen, T. H., Leiserson, C. E., Rivest, R. L., and Stein, C. 2001. Introduction to Algorithms. M.I.T. Press and McGraw-Hill. Frederickson, G. N., Hecht, M. S., and Kim, C. E. 1978. Approximation algorithms for some routing problems. SIAM Journal on Computing 7, 2, 178–193. Golden, B. L., Levy, L., and Vohra, R. 1987. The orienteering problem. Naval Research Logistics 34, 307–318. Lawler, E. L. 2001. Combinatorial Optimization: Networks and Matroids. Dover Publications. Lawler, E. L., Lenstra, J. K., Kan, A. H. G. R., and Shmoys, D. B. 1985. The Traveling Salesman Problem : A Guided Tour of Combinatorial Optimization. John Wiley & Sons. Li, C.-L., Simchi-Levi, D., and Desrochers, M. 1992. On the distance constrained vehicle routing problem. Operations Research 40, 4, 790–799. M. Haimovich, A. R. K. 1985. Bounds and heuristics for capacitated routing problems. Mathe-matics of Operations Research 10, 4, 527–542. M. Haimovich, A.G. Rinnoooy Kan, L. S. 1988. Analysis of heuristics for vehicle routing problems. Vehicle Routing: Methods and Studies, 47–61. Nagarajan, V. and Ravi, R. 2006. Minimum vehicle routing with a common deadline. In Pro-ceedings of the 9th International Workshop on Approximation Algorithms for Combinatorial Optimization Problems (APPROX). 212–223. Papadimitriou, C. H. and Steiglitz, K. 1998. Combinatorial Optimization. Dover Publications, Inc. ACM Journal Name, Vol. V, No. N, Month 20YY.
17503	https://math.stackexchange.com/questions/2476613/counting-number-of-plates-with-no-restrictions	combinatorics - counting number of plates with no restrictions - 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 counting number of plates with no restrictions Ask Question Asked 7 years, 11 months ago Modified7 years, 11 months ago Viewed 158 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. If there are no restrictions on where the digits and letters are placed, how many 8-character license plates consisting of 5 letters and 3 digits are possible if no repetitions of letters or digits are allowed? my ans: For the five letters there 26⋅25⋅24⋅23⋅22 26⋅25⋅24⋅23⋅22 and for the digits 10⋅9⋅8 10⋅9⋅8 so we have 26⋅25⋅24⋅23⋅22⋅10⋅9⋅8 26⋅25⋅24⋅23⋅22⋅10⋅9⋅8 such a plates. But, my answer key says the answer is (26 5)(10 3)⋅8! (26 5)(10 3)⋅8! I dont understand why this is the answer. combinatorics Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Oct 17, 2017 at 11:06 ILoveMathILoveMath 11.1k 9 9 gold badges 57 57 silver badges 128 128 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. I will break the key answer bit by bit, to try and give an intuition of why it is correct: (26 5)(26 5) is the number of ways you have to pick 5 5 letters from the alphabet without repetition; (10 3)(10 3) is the number of ways you have to pick 3 3 digits without repetitions. At this point you have a bag with 5 5 letters and 3 3 digits. To assemble a plate, you need to order them in some way. There are 8!8! ways of ordering the contents of your bag, thus for each bag there are 8!8! plates. If there are (26 5)(10 3)(26 5)(10 3) bags, then there are (26 5)(10 3)8! (26 5)(10 3)8! plates. Maybe we can also see why your answer doesn't quite work: The product 26⋅25⋅24⋅23⋅22⋅10⋅9⋅8 26⋅25⋅24⋅23⋅22⋅10⋅9⋅8 is accounting for the number of ways you have to create a sequence of 5 5 letters and 3 3 numbers. You are not taking into account the fact that the numbers and letters can have their relative positions changed. Take a smaller alphabet as a practical example: assume A,B A,B are the only letters and 1,2,3 1,2,3 the only digits, and that a plate is a letter and two numbers. Your product was 2⋅3⋅2=12 2⋅3⋅2=12, which would correspond to the following plates: A 12,A 13,A 21,A 23,A 31,A 32,B 12,B 13,B 21,B 23,B 31,B 32 A 12,A 13,A 21,A 23,A 31,A 32,B 12,B 13,B 21,B 23,B 31,B 32 But there are many more plates; namely the ones obtained by permutating the contents of the above plates. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 17, 2017 at 11:16 answered Oct 17, 2017 at 11:10 RGSRGS 9,883 2 2 gold badges 21 21 silver badges 35 35 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics See similar questions with these tags. 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17504	https://www.collinsdictionary.com/us/dictionary/english-thesaurus/lazy	Synonyms of LAZY \| Collins American English Thesaurus - [x] - [x] TRANSLATOR LANGUAGE GAMES SCHOOLS BLOG RESOURCES More [x] English Thesaurus [x] English English Dictionary English Thesaurus English Word Lists COBUILD English Usage [x] English Grammar Easy Learning Grammar COBUILD Grammar Patterns English Conjugations English Sentences [x] English ⇄ French English-French Dictionary French-English Dictionary Easy Learning French Grammar French Pronunciation Guide French Conjugations French Sentences [x] English ⇄ German English-German Dictionary German-English Dictionary Easy Learning German Grammar German Conjugations German Sentences [x] English ⇄ Italian English-Italian Dictionary Italian-English Dictionary Easy Learning Italian Grammar Italian Conjugations Italian Sentences [x] English ⇄ Spanish English-Spanish Dictionary Spanish-English Dictionary Easy Learning Spanish Grammar Easy Learning English Grammar in Spanish Spanish Pronunciation Guide Spanish Conjugations Spanish Sentences [x] English ⇄ Portuguese English-Portuguese Dictionary Portuguese-English Dictionary Easy Learning Portuguese Grammar Portuguese Conjugations [x] English ⇄ Hindi English-Hindi Dictionary Hindi-English Dictionary [x] English ⇄ Chinese English-Simplified Dictionary Simplified-English Dictionary English-Traditional Dictionary Chinese-Traditional Dictionary [x] English ⇄ Korean English-Korean Dictionary Korean-English Dictionary [x] English ⇄ Japanese English-Japanese Dictionary Japanese-English Dictionary English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese More [x] English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese DefinitionsSummarySynonymsSentencesPronunciationCollocationsConjugationsGrammar Credits × Synonyms of 'lazy' in American English lazy 1(adjective)in the sense of idle Synonyms idle inactive indolent inert slack slothful slow workshy 2(adjective)in the sense of lethargic Synonyms lethargic drowsy languid languorous sleepy slow-moving sluggish somnolent torpid Synonyms of 'lazy' in British English lazy 1(adjective)in the sense of idle Definition not inclined to work or exert oneself I was too lazy to learn how to read music. Synonyms idle I've never met such an idle bunch of workers! inactive They certainly were not politically inactive. indolent indolent teenagers who won't lift a finger to help slack Many publishers have simply become far too slack. negligent The jury ruled that the Council had acted in a negligent manner. inert He covered the inert body with a blanket remiss (formal) I would be remiss if I did not do something about it. workshy slothful (formal) He was not slothful: he had been busy all night. shiftless a shiftless husband See examples for synonyms Opposites active, stimulated, energetic, diligent, industrious, assiduous 2(adjective)in the sense of lethargic Definition moving in a sluggish manner We would have a lazy lunch and then lie on the beach in the sun. Synonyms lethargic He felt too miserable and lethargic to get dressed. languorous slow-moving languid He's a large languid man with a round and impassive face. sleepy I was beginning to feel amazingly sleepy. sluggish feeling sluggish and lethargic after a big meal drowsy He felt pleasantly drowsy. somnolent The sedative makes people very somnolent. torpid He led a lazy, torpid life at the weekends. See examples for synonyms Opposites quick Copyright © 2016 by HarperCollins Publishers. All rights reserved. Additional synonyms in the sense of drowsy Definition feeling sleepy He felt pleasantly drowsy. Synonyms sleepy, tired, lethargic, heavy, nodding, dazed, dozy, comatose, dopey (slang), half asleep, somnolent, torpid in the sense of inactive They certainly were not politically inactive. Synonyms lazy, passive, slow, quiet, dull, low-key (informal), sluggish, lethargic, sedentary, indolent, somnolent, torpid, slothful (formal) in the sense of indolent Definition lazy indolent teenagers who won't lift a finger to help Synonyms lazy, slack, idle, slow, sluggish, inactive, inert, languid, lethargic, listless, lackadaisical, torpid, good-for-nothing, workshy, slothful (formal), lumpish, fainéant You may also like English Quiz Confusables English Word lists Latest Word Submissions English Grammar Grammar Patterns Language Lover's Blog Collins Scrabble The Paul Noble Method Browse alphabetically lazy lazily laziness lazing lazy lazybones leach lead All ENGLISH synonyms that begin with 'L' 123 Wordle Helper ------------- Scrabble Tools -------------- Quick word challenge Quiz Review Question: 1 Score: 0 / 5 SYNONYMS Select the synonym for: enormous straightforward amiable effulgent prodigious SYNONYMS Select the synonym for: liberty solution freedom reproduction goal SYNONYMS Select the synonym for: mountainous amiable goofy unsure towering SYNONYMS Select the synonym for: environment gossip setting guarantee acclaim SYNONYMS Select the synonym for: frantically sardonically feverishly secretly purposely Your score: Check See the answer Next Next quiz Review Study guides for every stage of your learning journey Whether you're in search of a crossword puzzle, a detailed guide to tying knots, or tips on writing the perfect college essay, Harper Reference has you covered for all your study needs. 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17505	https://stjoesemresidency.com/wp-content/uploads/2018/08/Ch-48-Chest-Pain.pdf	Tintinalli’s Emergency Medicine: A Comprehensive Study Guide, 8e > Chapter 48: Chest Pain A. Mahler Simon INTRODUCTION AND EPIDEMIOLOGY About 8 million patients with chest pain present to a U.S. ED each year.1 Of these, 50% to 70% are placed into an observation unit or admitted to the hospital, yet only about 10% are eventually diagnosed with an acute coronary syndrome.2,3,4,5 Still, 2% to 5% of patients with acute myocardial infarctions are missed on initial presentation and discharged from the ED.2 We discuss the features and approach that help differentiate acute coronary syndrome from other causes of chest pain. The chapters titled "Acute Coronary Syndromes" and "Low Probability Acute Coronary Syndromes" discuss management of these specific syndromes. Acute chest pain is the recent onset of pain, pressure, or tightness in the anterior thorax between the xiphoid, suprasternal notch, and both midaxillary lines. Acute coronary syndrome includes acute myocardial infarction and acute ischemia (unstable angina). Acute myocardial infarction is defined by myocardial necrosis with elevation of cardiac biomarkers and is classified by ECG findings as ST-segment elevation myocardial infarction or non-ST-segment elevation myocardial infarction. Unstable angina is a clinical diagnosis defined by chest pain or an equivalent (neck or upper extremity pain) from inadequate myocardial perfusion that is new, occurring with greater frequency, less activity, or at rest. Patients with unstable angina do not have pathologic ST-segment elevation on ECG or cardiac biomarker elevation, but they are at risk of eventual myocardial damage absent recognition and treatment. PATHOPHYSIOLOGY The chest wall, from the dermis to the parietal pleura, is innervated by somatic pain fibers. Neurons enter the spinal cord at specific levels corresponding to the skin dermatomes. Visceral pain fibers are found in internal organs, such as the heart, blood vessels, esophagus, and visceral pleura. Visceral pain fibers enter the spinal cord and map to areas on the parietal cortex corresponding to cord levels shared with somatic fibers. Stimulation of visceral or somatic afferent pain fibers results in two distinct pain syndromes. Pain from somatic fibers is usually easily described, precisely located, and often experienced as a sharp sensation. Pain from visceral fibers is generally more difficult to describe and imprecisely localized. Patients with visceral pain are more likely to use terms such as discomfort, heaviness, pressure, tightness, or aching. Visceral pain is often referred to an area of the body corresponding to adjacent somatic nerves, which explains why pain from an acute myocardial infarction may radiate to the neck, jaw, or arms. Factors such as age, sex, comorbid illnesses, medications, drugs, and alcohol may interact with psychological and cultural factors to alter pain perception and communication. CLINICAL FEATURES RISK ASSESSMENT Patients with abnormal vital signs, concerning ECG findings (if available initially), a history of prior coronary artery disease, multiple atherosclerotic risk factors, or any abrupt, new, or severe chest pain or dyspnea should be quickly placed into a treatment bed. Initiate cardiac monitoring and IV access, and obtain an ECG, ideally within 10 minutes of arrival. Identify and treat immediate life needs like supporting the airway, breathing, and circulation. Measure vital signs promptly and at regular intervals. Administer oxygen if ambient saturation is <95%. Once the patient is stable, focus on history, physical exam, and laboratory findings associated with cardiac (acute coronary syndrome) versus noncardiac chest pain causes. Obtain a focused history that includes symptoms, abridged past medical history, and review of systems, seeking features of life-threatening causes of chest pain, such as acute coronary syndrome, aortic dissection, pulmonary embolism, severe pneumonia, and esophageal rupture. Ask about the onset, timing, severity, radiation, and character of the chest pain; alleviating and exacerbating factors; and presence of associated symptoms, such as diaphoresis, dyspnea, nausea, vomiting, palpitations, and dizziness. Focus the physical examination on findings pertinent to life-threatening causes of chest pain. Inspect the thorax for prior surgical incisions, chest wall deformities, and symmetric rise and fall with respiration. Palpate the chest wall for tenderness, masses, or crepitus. Auscultate to identify chest consolidation or pneumothorax, murmurs, gallops, or friction rubs, and obtain a chest x-ray immediately to identify immediate life-threatening processes. HISTORY Patients with serious and life-threatening intrathoracic disorders, including acute coronary syndrome, may report pain outside the chest, such as in the epigastrium, neck, jaw, shoulders, or arms. Some patients never experience chest pain or have migratory pain that is no longer in the chest at the time of medical evaluation. Patients with acute myocardial infarction who present without chest pain have diagnostic and treatment delays and have an in-hospital mortality rate more than twice that of acute myocardial infarction patients with chest pain.6 Classic Chest Pain Terms such as "typical" and "atypical" symptoms are misleading because symptoms among patients with acute coronary syndrome vary and may not include classic findings. Classic cardiac chest pain is retrosternal left anterior chest crushing, squeezing, tightness, or pressure. Cardiac chest pain is often brought on or exacerbated by exertion and relieved by rest. Traditional teaching is that anginal pain lasts 2 to 10 minutes, unstable angina pain lasts 10 to 30 minutes, and pain from acute myocardial infarction often lasts longer than 30 minutes, but great overlap exists. Other classic features of acute coronary syndrome presentation include radiation of the pain to the arms, neck, or jaw; diaphoresis; dyspnea; and nausea or vomiting.7 Nonclassic Chest Pain Patients with acute coronary syndrome frequently present without a "classic" chest pain story. The absence of classic symptoms contributes to delays in seeking care and in evaluation once they reach the ED. Nonclassic presentations include chest pain lasting for seconds, constant pains lasting for 12 to 24 hours or more without waxing and waning intensity, or pain worsened by specific body movements or positions, such as twisting and turning of the thorax. Reports of stabbing, well-localized, positional, or pleuritic chest pain are uncommon with acute coronary syndrome but do not exclude it with certainty. The Multicenter Chest Pain Study reported that 22% of patients with acute myocardial infarction described their chest pain as sharp or stabbing.8 Nonclassic presentations of acute coronary syndrome occur more frequently in women, racial minorities, diabetics, the elderly, and patients with psychiatric disease or altered mental status than in other patient groups.6,9,10 Multiple prescription medications, drugs, alcohol, patient or provider sex, and cultural differences can impact the pain perception or reporting of symptoms.11,12,13 For example, the term "sharp" in some cultures is interpreted to mean "severe," rather than knife-like.14 Premenopausal and early menopausal women with acute coronary syndrome are more likely to present with pain unrelated to exercise, pain not relieved by rest or nitroglycerin, pain relieved by antacids, palpitations without chest pain, or a chief complaint of fatigue.15 Associated symptoms of nausea, emesis, jaw pain, neck pain, and back pain are more common in women with acute coronary syndrome, while diaphoresis is more common among men.15 Anginal Equivalents One large public hospital reported that 47% of 721 consecutive patients with myocardial infarction presented complaining of symptoms other than chest pain.10 This means ED physicians must consider potential anginal-equivalent symptoms like dyspnea at rest or with exertion, nausea, light-headedness, generalized weakness, acute changes in mental status, diaphoresis, or shoulder, arm, or jaw discomfort. Patients with dyspnea alone have a fourfold increased risk of sudden death from cardiac causes compared with asymptomatic patients, and a twofold increased risk compared with patients with classic angina.16 Epigastric or upper abdominal discomfort, even when relieved with antacids, should raise suspicion for acute coronary syndrome, especially for patients >50 years old and those with known coronary artery disease. In these two high-risk groups, include an ECG in routine evaluation of abdominal pain. Consider acute coronary syndrome in patients presenting with palpitations, because myocardial ischemia may increase automaticity and irritability, leading to dysrhythmias. Furthermore, tachycardia can cause an increase in myocardial oxygen demand, triggering myocardial ischemia. Risk Factors Major risk factors for coronary artery disease include age >40 years old, male or postmenopausal female, hypertension, tobacco use, hypercholesterolemia, diabetes, truncal obesity, family history, and a sedentary lifestyle.17,18 Cocaine use is associated with acute myocardial infarction even in young people with minimal or no coronary artery disease. Chronic cocaine use may accelerate atherosclerosis and severe coronary artery disease,19 although some suggest no relationship once controlling for other cardiovascular risk factors.20 Human immunodeficiency virus infection and treatment with highly active antiretroviral therapy can accelerate atherosclerosis.21 Although cardiac risk factors are useful in predicting coronary artery disease risk within a given population, they are less useful for diagnosing the presence or absence of acute coronary syndrome in an individual patient.7,22,23 Patients with known coronary artery disease and prior acute coronary syndrome are at risk for another acute coronary syndrome event. So, identify previous episodes of chest pain, prior echocardiography, stress testing or coronary angiography, or prior revascularization (stent placement or coronary artery bypass graft surgery). ACUTE MYOCARDIAL INFARCTION SIGNS AND SYMPTOMS: LIKELIHOOD RATIOS There are no historical features with sufficient sensitivity and specificity to either diagnose or exclude acute coronary syndrome. Radiation to the arms and shoulders, particularly to the right arm or both arms, is the historical feature most strongly associated with acute coronary syndrome (likelihood ratio range of 2.3– 4.7).22,24,25 Chest pain with exertion or associated symptoms of dyspnea, diaphoresis, nausea, or vomiting are associated with twofold likelihood of acute coronary syndrome.24,25 Pressure-like chest sensation has limited value in the prediction of acute coronary syndrome.22 Sharp, pleuritic, positional chest pain is associated with a decreased likelihood of acute coronary syndrome but cannot eliminate the diagnosis.22 Lack of exertional pain or pain radiation has no diagnostic value for exclusion of acute coronary syndrome.22,25 Since classic cardiac ischemic pain is not universal and men and women both present with nonclassic symptoms, the diagnostic utility of specific chest pain descriptions does not differ significantly between men and women.26 Tables 48–1 and 48–2 summarize the chest pain characteristics associated with increased or decreased likelihood ratios of acute myocardial infarction. Table 48–1 Acute Myocardial Infarction Symptoms: Positive Likelihood Ratios22,24,25 Pain Descriptor Study No. of Patients Studied Positive Likelihood Ratio (95% Confidence Interval) Radiation to right arm or shoulder Chun et al. 770 4.7 (1.9–12.0) Radiation to both arms or shoulders Goodacre et al. 893 4.1 (2.5–6.5) Associated with exertion Goodacre et al. 893 2.4 (1.5–3.8) Radiation to left arm Panju et al. 278 2.3 (1.7–3.1) Associated with diaphoresis Panju et al. 8426 2.0 (1.9–2.2) Associated with nausea or vomiting Panju et al. 970 1.9 (1.7–2.3) Worse than previous angina or similar to previous myocardial infarction Chun et al. 7734 1.8 (1.6–2.0) Described as pressure Chun et al. 11,504 1.3 (1.2–1.5) Table 48–2 AMI Symptoms: Negative Likelihood Ratios22,25,27 Pain Descriptor Study No. of Patients Studied Positive Likelihood Ratio (95% Confidence Interval) Described as pleuritic Chun et al. 8822 0.2 (0.1–0.3) Described as positional Chun et al. 8330 0.3 (0.2–0.5) Described as sharp Chun et al. 1088 0.3 (0.2–0.5) Reproducible with palpation Chun et al. 8822 0.3 (0.2–0.4) Reproducible with positioning Chun et al. 8330 0.3 (0.2–0.5) Inframammary location Everts et al. 903 0.8 (0.7–0.9) Not associated with exertion Goodacre et al. 893 0.8 (0.6–0.9) PHYSICAL EXAMINATION The examination of patients with acute coronary syndrome is often normal, and there are no exam findings that are sensitive or specific enough to exclude or diagnose acute coronary syndrome. Use the exam in conjunction with history to identify or exclude other causes of chest pain and to guide therapy. Vital sign abnormalities from acute coronary syndrome may include hyper- or hypotension, tachycardia, or bradycardia. Tachycardia may result from increased sympathetic tone and decreased left ventricular stroke volume. Bradycardia may occur due to ischemia or infarction involving the conduction system or alterations in sympathetic and parasympathetic activation of the sinoatrial or atrioventricular nodes. Patients with acute myocardial ischemia or infarction may have abnormal heart sounds due to changes in ventricular function or compliance, such as an S3 or S4 gallop, diminished S1, or a paradoxically split S2. New murmurs in patients with chest pain may be associated with acute myocardial infarction with chordae tendineae rupture or aortic root dissection. Ischemia-induced congestive heart failure may produce crackles on auscultation of the lungs. Physical examination findings most strongly associated with acute myocardial infarction in patients presenting with acute chest pain are hypotension, S3 gallop, and diaphoresis, although the frequency, interrater reliability, and added diagnostic value are limited.22 Reproducible chest wall tenderness is suggestive of a musculoskeletal etiology but is reported in up to 15% of patients with confirmed acute myocardial infarction and cannot alone exclude the diagnosis of acute coronary syndrome.28 RESPONSE TO THERAPY Response to medications poorly discriminates between cardiac and noncardiac chest pain. While nitroglycerin reduces anginal pain, it may also relieve the pain from noncardiac conditions such as esophageal spasm.22,29,30,31 Similarly, relief from antacid or combination "GI cocktail" therapy does not represent a noncardiac cause of chest pain.32,33 Combine the above responses with other features to best assess the likely presence or absence of acute coronary syndrome. DIAGNOSIS Life-threatening concerns in acute chest pain are acute coronary syndrome, aortic dissection, pulmonary embolism, pneumonia, tension pneumothorax, and esophageal rupture. Other diagnoses with the potential for morbidity and mortality include simple pneumothorax, myocarditis, pericarditis, aortic stenosis, perforated ulcer, and cholecystitis. Benign causes of chest pain include anxiety, musculoskeletal pain, esophagitis, and gastritis. Common causes of chest pain are listed in Table 48–3. Table 48–4 summarizes the classic symptoms of the life-threatening causes of acute chest pain. Table 48–3 Common Causes of Acute Chest Pain Visceral Pain Pleuritic Pain Chest Wall Pain Typical angina Pulmonary embolism Costosternal syndrome Unstable angina Pneumonia Costochondritis (Tietze's syndrome) Acute myocardial infarction Spontaneous pneumothorax Precordial catch syndrome Aortic dissection Pericarditis Xiphodynia Esophageal rupture Pleurisy Radicular syndromes Esophageal reflux or spasm Intercostal nerve syndromes Mitral valve prolapse Fibromyalgia Abbreviations: L = left; R = right. Atypical presentations are common. Table 48–4 Classic Symptoms of Potentially Life-Threatening Causes of Chest Pain Disorder Pain Location Pain Character Radiation Associated Signs and Symptoms Acute coronary syndrome Retrosternal, L chest, or epigastric Crushing, tightness, squeezing, pressure R or L shoulder, R or L arm/hand, jaw Dyspnea, diaphoresis, nausea Pulmonary embolism Focal chest Pleuritic None Tachycardia, tachypnea, hypoxia, may have hemoptysis Aortic dissection Midline, substernal Ripping, tearing Intrascapular area of back Secondary arterial branch occlusion Pneumonia Focal chest Sharp, pleuritic None Fever, hypoxia, may see signs of sepsis Esophageal rupture Substernal Sudden, sharp, after forceful vomiting Back Dyspnea, diaphoresis, may see signs of sepsis Pneumothorax One side of chest Sudden, sharp, lancinating, pleuritic Shoulder, back Dyspnea Pericarditis Substernal Sharp, constant or pleuritic Back, neck, shoulder Fever, pericardial friction rub Perforated peptic ulcer Epigastric Severe, sharp Back, up into chest Acute distress, diaphoresis PULMONARY EMBOLISM Symptoms of pulmonary embolism include sharp chest pain (may worsen with inspiration, called "pleuritic"), dyspnea, hypoxemia, syncope, or shock. There may be associated cough or hemoptysis. Patients with pulmonary embolism may be febrile and have leg swelling or pain, and some patients will report chest wall tenderness. Common physical examination findings include tachypnea, tachycardia, and hypoxemia. Pulmonary embolism risk factors include recent surgery, trauma, prolonged immobility, active cancer, estrogens from birth control pills or hormone replacement therapy (particularly when combined with smoking), procoagulant syndromes, or a history of prior pulmonary embolism or deep venous thrombosis.34,35 Clinical decision aids, such as the Wells and Revised Geneva Scores, can risk stratify patients with possible pulmonary embolism.36,37 The Pulmonary Embolism Rule-Out Criteria exclude pulmonary embolism in patients with a low pretest probability without further diagnostic testing.38 Normal D-dimer testing, measured by a sensitive enzyme-linked immunosorbent assay, in a hemodynamically stable low- to intermediate-risk patient (with a Revised Geneva Criteria Score of 0 to 10) makes pulmonary embolism exceptionally unlikely; in those with higher risk assessment, a negative D-dimer has limited value.39,40 In patients with pulmonary embolism, elevated cardiac troponin (cTn) indicates ventricular dysfunction and identifies patients with an elevated risk of death and complications.41 In pulmonary embolism, ECG findings are nonspecific, with the most common finding being sinus tachycardia. Chest radiographs are usually normal, but in rare cases may show signs of pulmonary infarction. CT pulmonary angiography is the test of choice and is highly sensitive for the detection of large to medium-sized pulmonary emboli. See more details on pulmonary embolism in the chapter 56, "Venous Thromboembolism." AORTIC DISSECTION Pain from aortic dissection is classically described as a ripping or tearing sensation radiating to the interscapular area of the back. The pain is often sudden in onset, maximal at the time of symptom onset, and may migrate or be noted above and below the diaphragm. Lack of sudden-onset pain decreases the probability of aortic dissection but cannot exclude it.42 Secondary symptoms of aortic dissection result from arterial branch occlusions and include stroke, acute myocardial infarction, or limb ischemia. Risk factors include male sex, age over 50 years, poorly controlled hypertension, cocaine or amphetamine use, a bicuspid aortic valve or prior aortic valve replacement, connective tissue disorders (Marfan's syndrome and Ehlers-Danlos syndrome), and pregnancy.43 Physical exam findings for aortic dissection lack sensitivity and specificity. A unilateral pulse deficit of the carotid, radial, or femoral arteries is suggestive of aortic dissection (likelihood ratio 5.7; 95% confidence interval, 1.4–23).42 Focal neurologic deficits are rare, occurring in only 17% of patients with aortic dissection, but the combination of chest pain and a focal neurologic deficit greatly increase the likelihood of aortic dissection.42 While a completely normal chest radiograph lowers the likelihood of aortic dissection being present, it does not exclude dissection. A negative D-dimer lowers the probability of aortic dissection (detecting the clotting/declotting expected), but it also cannot exclude the disease.44 ECG changes are common among patients with aortic dissection, with up to 40% to 50% presenting with ST-segment or T-wave changes.45,46 Elevated cTn among patients with aortic dissection is associated with increased mortality.47 If aortic dissection is suspected, obtain a CT aortogram or transesophageal echocardiogram. See chapter 59, "Aortic Dissection and Related Aortic Syndromes." PNEUMONIA Pneumonia is potentially life threatening in the elderly, immunocompromised, or patients with multiple comorbid conditions. Chest pain from pneumonia is usually described as sharp, pleuritic, and associated with fever, cough, sputum production, and possibly hypoxemia.48 Auscultation may reveal decreased breath sounds, rales, or bronchial breath sounds over the affected areas of consolidation. A chest radiograph usually confirms the diagnosis. See chapter 65, "Pneumonia and Pulmonary Infiltrates." ESOPHAGEAL RUPTURE (BOERHAAVE'S SYNDROME) Patients classically present with a history of sudden-onset, sharp, substernal chest pain following forceful vomiting. Patients with esophageal rupture are usually ill-appearing and may be tachycardic, febrile, dyspneic, or diaphoretic. Physical examination may reveal crepitus in the neck or chest from subcutaneous emphysema. Hamman's crunch, audible crepitus that varies with the heartbeat on auscultation of the precordium, is a rare finding associated with pneumomediastinum. Chest radiography may demonstrate a pleural effusion (left more common than right), pneumothorax, pneumomediastinum, pneumoperitoneum, or subcutaneous air, although a normal x-ray cannot exclude esophageal rupture. If esophageal rupture is suspected, obtain a CT with oral water-soluble contrast. See chapter 77, "Esophageal Emergencies." SPONTANEOUS PNEUMOTHORAX The symptoms of spontaneous pneumothorax are sudden-onset, sharp, pleuritic chest pain with dyspnea. Classically, spontaneous pneumothorax occurs in tall, slender males. Risk factors for spontaneous pneumothorax include smoking and chronic lung diseases such as asthma and chronic obstructive pulmonary disease. Approximately 1% to 3% of patients with a spontaneous pneumothorax progress to develop a tension pneumothorax.49 Auscultation may reveal decreased breath sounds and hyperresonance to percussion on the ipsilateral side. However, the physical exam findings of a simple pneumothorax are inconstant and cannot be used to exclude presence, with the diagnosis made by chest radiography. See chapter 68, "Pneumothorax." ACUTE PERICARDITIS Pain from acute pericarditis is classically described as a sharp, severe, constant pain with a substernal location. The pain may radiate to the back, neck, or shoulders; worsens by lying flat and by inspiration; and is relieved by sitting up and leaning forward. A pericardial friction rub is the most specific physical exam finding but is not always evident. The classic ECG findings are diffuse ST-segment elevation with PR depression.50 See chapter 55, "Cardiomyopathies and Pericardial Disease." MITRAL VALVE PROLAPSE Symptoms attributed to mitral valve prolapse include sharp chest pain, palpitations, fatigue, anxiety, and dyspnea unrelated to exertion. A midsystolic click may be heard on auscultation. However, most patients are asymptomatic and have no consistent association of chest pain, dyspnea, or anxiety with the disorder.51,52 See chapter 54, "Valvular Emergencies." CHEST WALL PAIN Musculoskeletal or chest wall pain is characterized by sharp, highly localized, and positional pain. The pain should be completely reproducible by light to moderate palpation or by specific movements and may be increased by inspiration or coughing. However, chest wall tenderness is also reported by some patients with acute coronary syndrome and pulmonary embolism. Costochondritis (Tietze's syndrome) is an inflammation of the costal cartilages or their sternal articulations and causes chest pain that is variably sharp, dull, and often increased with respirations. Xiphodynia is inflammation of the xiphoid process that causes sharp, pleuritic chest pain reproduced by light palpation. Precordial catch syndrome is a short, lancinating chest pain occurring in bunches lasting 1 to 2 minutes near the cardiac apex and is associated with inspiration, poor posture, and inactivity. Pleurisy is inflammation of the parietal pleura resulting in sharp pleuritic chest pain. GI PAIN GI disorders often cannot be reliably differentiated from acute coronary syndrome by history and physical examination alone.25,53 Gastritis and esophageal reflux typically produce burning or gnawing pain in the lower half of the chest, with a brackish or acidic taste in the back of the mouth. The pain may be lessened with antacids and exacerbated by recumbency. Peptic ulcer disease is classically described as a postprandial, dull, boring pain in the epigastric region. Patients often describe being awakened from sleep by discomfort. Duodenal ulcer pain may be relieved after eating food, whereas gastric ulcer pain is often exacerbated by eating. Antacid medications usually provide symptomatic relief. Acute pancreatitis and biliary disease typically present with right upper quadrant or epigastric pain and tenderness but can also cause chest pain. Esophageal spasm is often associated with reflux disease and is characterized by a sudden onset of dull or tight substernal chest pain. The pain is frequently precipitated by consumption of hot or cold liquids or a large food bolus and may be relieved by nitroglycerin. See chapter 77, "Esophageal Emergencies" and 78, "Peptic Ulcer Disease and Gastritis." PANIC DISORDER Panic disorder is characterized by recurrent, unexpected, and discrete periods of intense fear or discomfort (panic attacks) with at least four of the following symptoms: chest pain, dyspnea, palpitations, diaphoresis, nausea, tremor, choking, dizziness, fear of losing control or dying, paresthesias, chills, or hot flashes. In one study, 25% of ED patients with chest pain met diagnostic criteria for panic disorder. Conversely, 9% of the patients identified as having panic disorder were ultimately diagnosed with acute coronary syndrome on hospital discharge.54 This means panic disorder is at best a diagnosis of exclusion or a co-diagnosis with acute coronary syndrome (or another cause). Do not assume panic disorder in a patient with chest pain in the ED until further testing allows better risk stratification. See chapter 289, "Mood and Anxiety Disorders." DIAGNOSTIC TESTING Focus initial diagnostic testing for patients with chest pain on the exclusion or confirmation of serious pathology based on the differential diagnosis drawn from the history and physical examination. When history and exam make acute coronary syndrome a potential cause, testing commonly includes an ECG, chest x-ray, and cardiac biomarkers. Stress testing, advanced cardiac imaging, serial or continuous ECG monitoring, and serial cardiac biomarker measurements are discussed in chapter 49, "Acute Coronary Syndromes." IMAGING Chest radiography is commonly performed in the evaluation of ED patients with chest pain. Most patients with acute coronary syndrome have a normal chest x-ray, but the images are useful to diagnose or exclude other conditions such as pneumonia and pneumothorax.55 Other imaging modalities such as CT help evaluate conditions such as aortic dissection or pulmonary embolism. ECG Guidelines recommend a screening ECG within 10 minutes of ED arrival on patients with chest pain or other symptoms concerning for acute coronary syndrome.56 Rapid ECG screening is essential because delay in identification of an ST-segment elevation myocardial infarction is associated with increased mortality.57 Routine triage ECG testing and prehospital ECG transmission reduce delays to ST-segment elevation myocardial infarction identification, decrease door-to-balloon or door-to-needle time, and improve patient outcomes.58,59,60,61 Less than 5% of patients presenting to the ED with chest pain have evidence of an ST-segment elevation myocardial infarction on ECG.62,63 However, new ST-segment elevation of ≥1 mm in at least two contiguous leads represents an acute myocardial infarction that will benefit from rapid reperfusion interventions.22,64 ST-segment elevation also occurs in patients with pericarditis, myocarditis, early repolarization, left ventricular hypertrophy, and ventricular aneurysms. ST-segment depression and T-wave inversions are also associated with an increase in risk of acute myocardial infarction.22,65 A normal ECG lacks the sensitivity to exclude acute coronary syndrome, notably unstable angina, or non-ST-segment elevation myocardial infarction. In a large multicenter observational study of 391,208 patients with an evaluable ECG and diagnosis of acute myocardial infarction, 57% had "diagnostic" ECG changes, 35% had nonspecific changes, and 8% had normal ECGs. Diagnostic changes were defined as ST-segment elevation, ST-segment depression, or left bundle-branch block.66 Other studies document normal or near normal ECGs in 5% to 10% of patients with acute myocardial infarction.67,68,69,70 A normal ECG is also an independent risk factor for missed acute myocardial infarction and inappropriate ED discharge (odds ratio 7.7; 95% confidence interval, 2.9–20.2).2 However, among young patients (<40 years old) without known coronary artery disease, a normal ECG is associated with a cardiovascular event rate of less than 1% at 30 days.71 Misinterpretation of ECGs (i.e., failure to detect ischemic changes that are present) occurs in up to 40% of missed acute myocardial infarction cases.70 In addition, the initial ECG represents only a single time point in a dynamic pathophysiologic process; the diagnostic value of an ECG is improved by comparing it to a prior ECG or repeating it.72 CARDIAC BIOMARKERS Cardiac Troponins Cardiac cTns are proteins essential to cardiac muscle contraction, which are complexed with actin and myosin filaments within cardiac myofibrils and are present within cardiac myocyte cytoplasm.73 Myocardial injury resulting in the disruption of myocyte cell membrane integrity or myofibril destruction results in extracellular cTn leak, which can be detected in the patient's peripheral blood and used to identify and quantify myocardial damage.74 Due to its high sensitivity and nearly complete cardiac specificity, cTn is the biomarker of choice for the detection of myocardial injury.75 Although cTn elevation is specific for myocardial necrosis, elevation does not indicate the mechanism of injury, nor does it necessarily indicate acute myocardial infarction. There are numerous nonischemic causes of cTn elevations, which are summarized in Table 48–5. Acute myocardial infarction can be differentiated from nonischemic cTn elevations based on the pattern of cTn elevation and the clinical context. The diagnostic criteria for acute myocardial infarction include a gradual rise and fall of cTn with a maximum value above the 99th percentile of a reference population (the upper reference limit), combined with any of the following: symptoms consistent with ischemia, characteristic acute ECG changes (ST- and T-wave changes, new left bundle-branch block, or new Q waves), or imaging evidence of a new regional wall motion abnormality or new loss of viable myocardium.76 Table 48–5 Conditions Associated with Elevated Cardiac Troponin Levels inthe Absence of Ischemic Heart Disease Cardiac contusion Cardiac procedures (surgery, ablation, pacing, stenting) Acute or chronic congestive heart failure Aortic dissection Aortic valve disease Hypertrophic cardiomyopathy Arrhythmias (tachyarrhythmia or bradyarrhythmia) Apical ballooning syndrome Rhabdomyolysis with cardiac injury Pulmonary hypertension Pulmonary embolism Acute neurologic disease (e.g., stroke, subarachnoid hemorrhage) Myocardial infiltrative diseases (amyloid, sarcoid, hemochromatosis, scleroderma) Inflammatory cardiac diseases (myocarditis, endocarditis, pericarditis) Drug toxicity Respiratory failure Sepsis Burns Extreme exertion (e.g., endurance athletes) Immunoassays have been developed for the isoforms of cTnI and cTnT. Isoforms I and T provide nearly identical information, and selection between them is driven mainly by central laboratory vendor and equipment preference.77 A single manufacturer produces the cTnT assay; however, multiple manufacturers produce cTnI assays, which differ in their upper reference limits (the cTn value above the 99th percentile of a reference population), coefficients of variability, and lower limits of detection. Over the past 25 years, cTn assays have become more analytically sensitive, pushing down the upper reference limits (URLs) and limits of detection. For example, a "first-generation" assay, available in 1995, had a URL of 0.4 nanograms/mL, whereas a commonly used contemporary (current-generation) assay has a 10-fold lower URL (0.04 nanograms/mL).78,79 Current commercially available contemporary cTn assays have URLs ranging from 0.023 to 0.20 nanograms/mL and lower limits of detection ranging from 0.006 to 0.15 nanograms/mL.80 Point-of-care assays offer a shorter turnaround time but with slightly lower analytic sensitivities than conventional assays.76,81 With contemporary assays, cTn is detected in serum as early as 2 hours after symptom onset of an acute myocardial infarction, but elevations are not reliably present until 6 hours or more.82 Elevations peak at approximately 48 hours from symptom onset unless repeat injury occurs, and cTns remain elevated for up to 10 days (Figure 48–1). This persistence makes cTn a good tool in diagnosing acute myocardial infarction in patients with delayed presentations. However, in patients with intermittent symptoms over a period of days, an elevated cTn could represent a remote or new infarct. In this rare setting, the concomitant use of creatine kinase-MB fraction, which returns to normal sooner, can help differentiate acute from remote infarction. FIGURE 48–1. Typical pattern of contemporary serum marker elevation after acute myocardial infarction (AMI). CK-MB = MB fraction of creatine kinase; cTnI = cardiac troponin I; cTnT = cardiac troponin T; LD1 = lactate dehydrogenase isoenzyme 1; MLC = myosin light chain. Obtain cardiac cTn levels in all patients with suspected acute coronary syndrome.55 Contemporary cTn assays will identify most patients (approximately 80%) with acute myocardial infarction within 2 to 3 hours of ED arrival.82,83,84 Patients with early presentations (within 6 hours of symptom onset) or those with intermittent symptoms should have serial measurements of cTn over time. In patients with constant symptoms for >8 to 12 hours, a single cTn may be sufficient to exclude acute myocardial infarction.56,85 Measurement of cTn at short time intervals, such as 2 to 4 hours, to evaluate for serial change (delta cTn) is more sensitive for acute myocardial infarction than a single cTn approach.55 Newer high-sensitivity cTn assays have a 10-fold higher analytical sensitivity compared to contemporary assays; these are currently pending U.S. Food and Drug Administration approval.80,87,88 Compared with contemporary assays, high-sensitivity cTn assays are more sensitive for the detection of acute myocardial infarction in all patients (94% to 96% vs. 85% to 90%) and increase the early detection of myocardial injury.83,84 Among patients presenting within 3 hours of chest pain onset, high-sensitivity cTn assays are 92% to 94% sensitive for acute myocardial infarction compared to 76% for a contemporary assay.83 However, the increased sensitivity of high-sensitivity cTn assays for acute myocardial infarction is balanced by the detection of more patients with non–acute myocardial infarction cTn elevations.83,88 The overall impact on ED decision making and ultimate outcome is not yet defined or shown to be clearly superior to previous assay use. An elevated cTn is associated with an increased risk of cardiac death or acute myocardial infarction at 30 days (odds ratio 3.4; 95% confidence interval, 2.9–4.0).89 This elevated risk of death or cardiovascular complications is independent of ECG findings or creatine kinase-MB levels.90 Higher cTn elevations also are associated with more adverse events, even with minimal elevations.91 Patients with renal disease often have an elevated cTnT (15% to 50%), whereas cTnI elevations are less common (<10%). After dialysis, serum levels of cTnT generally increase while cTnI levels decrease.92 Despite these features, cTnT and cTnI assays remain highly sensitive for acute myocardial infarction in patients with renal failure, particularly when new measures can be compared with baseline measures. Furthermore, renal failure patients with elevated cTn levels are at higher risk for death and adverse events than patients with normal cTn levels.93 Creatine Kinase-MB and Myoglobin Troponin testing has made these markers almost obsolete in acute coronary syndrome care. Creatine kinase-MB fraction levels elevate within 4 to 8 hours after acute myocardial infarction, peak between 12 and 24 hours, and return to normal between 36 and 72 hours (Figure 48–1). When used with cTn, creatine kinase-MB provides little additional information.94 Creatine kinase-MB testing may be useful in a small subset of patients in whom the timing of infarction is unclear. Elevated creatine kinase-MB and cTn indicate an acute infarct, whereas a negative creatine kinase-MB with an elevated cTn suggests a remote or subacute infarction. Myoglobin is a small heme-containing protein found in skeletal and cardiac muscle. After acute myocardial infarction, serum myoglobin levels rise within 3 hours of symptoms, peak at 4 to 9 hours, and return to baseline within 24 hours (Figure 48–1). False-positive results are common, and false-negative results may occur in patients with delayed presentations. Due to the improved sensitivity of contemporary cTn assays, myoglobin does not appear to have added value in the early detection of acute myocardial infarction.95 B-Type Natriuretic Peptide Natriuretic peptide elevations are not specific to myocardial ischemia or infarction and will rise with any ventricular dysfunction. Patients with acute coronary syndrome and an elevated natriuretic peptide level have higher short-term mortality, although the lab test does not aid in specific patient management actions. Other Biomarkers High-sensitivity C-reactive protein aids long-term cardiac event prediction, but this test is not recommended for ED care.54 A variety of other assays have been studied as cardiac biomarkers, such as ischemia-modified albumin, interleukin-6, vascular cell adhesion molecule, intercellular adhesion molecule, E-selectin, P-selectin, pregnancy-associated plasma protein A, and myeloperoxidase. Current evidence does not support the use of these novel biomarkers for ED chest pain evaluations. CLINICAL RISK SCORES AND DECISION AIDS The Thrombosis in Myocardial Infarction risk score or Global Registry of Acute Coronary Events score can aid acute coronary syndrome risk stratification (Figure 48–2).56 The Thrombosis in Myocardial Infarction and Global Registry of Acute Coronary Events scores were drawn from groups with acute coronary syndrome present or strongly suspected, and then were applied to a wider population. Each stratifies patients into low-, intermediate-, or high-risk groups for acute coronary syndrome. However, a low-risk score is not sensitive enough to exclude acute coronary syndrome or identify patients for early discharge without further evaluation.96,97 FIGURE 48–2. Thrombosis in Myocardial Infarction (TIMI) score and the Global Registry of Acute Coronary Events (GRACE) score. ACS = acute coronary syndrome; Cr = creatinine; HR = heart rate; SBP = systolic blood pressure. The HEART score, ADAPT, and the North American Chest Pain Rule (Figure 48–3) combine clinical information to risk stratify patients and guide key decisions, notably discharge with follow-up for the lowest risk patients or observation/admission for the remaining patients. Although these decision support tools may improve the quality and efficiency of chest pain care in the ED, they require further impact assessment before routine use.98,99,100 1. 2. FIGURE 48–3. ADAPT, the North American Chest Pain Rule (NACPR), and the HEART score. With ADAPT and NACPR, a patient is considered low risk if the patient has none of the high-risk criteria. For ADAPT, risk factors include family history of coronary disease, hypertension, hypercholesterolemia, diabetes mellitus, and current smoker. With the HEART score, low risk is a score of 0 to 3, and high risk is a score of 4 or greater. Risk factors include currently treated diabetes mellitus, current or recent (<90 days) smoker, diagnosed and/or treated hypertension, diagnosed hypercholesterolemia, family history of coronary artery disease, obesity (body mass index >30), or a history of significant atherosclerosis (coronary revascularization, myocardial infarction, stroke, or peripheral arterial disease). ACS = acute coronary syndrome; TIMI = Thrombosis in Myocardial Infarction score. 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17506	https://caddellprep.com/common-core-algebra-assignments/verbal-to-algebraic-expressions/	Skip to content Give us a call (917) 722-0677 You can excel with Caddell! Verbal To Algebraic Expressions & Equations In this video, we are going to learn how to turn verbal expressions and equations into algebraic expressions and equations. When turning expressions into mathematical expressions, we use keywords to determine the operation that should be used. Specific keywords or phrases in the verbal expression allows us to also determine the order of numbers and variables, “greater than”, “less than”, and “difference of”. After you finish this lesson, view all of our Pre-Algebra and Algebra 1 lessons and practice problems. Verbal Expressions to Algebraic Expressions x increased by 5 x decreased by 5 x less than 5 Verbal Equations to Algebraic Equations Examples of Verbal To Algebraic Expressions & Equations Example 1 The product of and decreased by The product of and would be which we know is . Decreased means we subtract. Therefore, the algebraic expression is, Example 2 The quotient of and is less than Since quotient is the answer to division, it would be The less than sign is Therefore, Video-Lesson Transcript Let’s go over changing verbal expression into algebraic expressions. For example: increased by this will be . decreased by since it’s a decrease, we are taking away from . So this becomes . greater than , here we have more than . We have and we’re increasing it by . Here we have . less than . So think about this, if we have less than . We have and we’re taking away from it. So it’s . The sum of and is just . The difference of and , difference is the answer to subtraction so we have is . Then the product of and would be which we know is . The order isn’t important in multiplication and addition. The quotient of and , since quotient is the answer to division our answer would be . You should get familiar with these different expressions. Some may memorize these. But it’s nice to look at these expressions and make sense of it. Now, let’s look at changing verbal equations to algebraic equations. Verbal equations are expressions that have equal signs. So, we’re going to do this in equality. Like equal to, less than, greater than, less than or equal to, or greater than or equal to. Let’s have some examples. greater than is . This becomes . The product of and is less than . The difference of and is greater than . more than the product of and is . Let me point out the difference between these two: less than andis less than less than is While is less than is . “Is” in the sentence is an equality sign. That’s why we come up with the less than sign. About Contact Us Blog Online Test Prep for Schools Terms of Service Privacy Policy SHSAT SAT TACHS Vocabulary Past Regents Exams Use of the Caddell Prep service and this website constitutes acceptance of our Terms of Use and Privacy Policy. Disclaimer: Use of this website does not guarantee an increase in school grades, test performance, etc., unless otherwise noted. SAT is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. Test names are the trademarks of their respective owners, who are not affiliated with Caddell Prep LLC or Caddell Prep Inc. As seen on: Want to apply for a tutoring position? Apply here Caddell Prep Email: [email protected] Phone: (917) 722-0677 Url: cash, check, credit card, invoice, paypal 91 Guyon Ave Staten Island, NY 10306
17507	https://www.youtube.com/watch?v=X9K8LT7SCZ0	30.1 Introduction to Torque and Rotational Dynamics MIT OpenCourseWare 5940000 subscribers 302 likes Description 25924 views Posted: 2 Jun 2017 MIT 8.01 Classical Mechanics, Fall 2016 View the complete course: Instructor: Prof. Deepto Chakrabarty License: Creative Commons BY-NC-SA More information at More courses at 17 comments Transcript: consider a rod of mass m and suppose I apply a force to this rod let's say of course like that so we know that as a result of that applied force the center of mass of the rod which we can imagine is right at the center of the rod will translate with some acceleration such that the vector F is equal to the mass of the rod times the acceleration of the center of mass now recall that for a rigid body this equation will be true regardless of where on the rod I apply the force so for example if I draw the rod again over here and I apply the same force vector F but I apply it let's say on the right hand side I'll still have the same F equals MA all I specify with F is its magnitude and direction but for a rigid body no matter where on the body I apply that force the translation the the acceleration of the center of mass will be the same now we know from experience however that the motion of the rod is different if I push it at the center or at the at one end or at the other end the point is though that the motion the overall motion the overall translation of the object doesn't just involve the translation in the center of mass but it also involves some rotation in general so in fact there is a theorem called Sholes as theorem which tells us that the most general displacement of a rigid body can be split into the translation of the center of mass and a rotation about the center of mass our F equals MA relation tells us how an applied force affects the first part the translation of the center of mass but how does that apply force affect the rotation of the rigid body will see that this depends not only on the magnitude and direction of the force but also on where the force is applied okay that's different than the translation center of mass which only depends upon what the force is the rotation about the center of mass depends both upon the force itself and where on the object that it's applied in fact in the next few lessons we'll see that there's a special equation of motion for describing rotational motion that's analogous to Newton's second law of motion analogous tentacles MA applying two rotational motion and we write that relationship this rotational equation of motion is written as tau which is a vector is equal to I times alpha which is a vector so tau is a new quantity we haven't talked about before called the torque it's a vector I is the moment of inertia that we've already encountered and alpha is the angular acceleration of the rotation of motion now Tao is a sort of rotational analog to the force and it depends upon not just the force itself but on where the force is applied computing the torque Tao depends upon something called a vector product or cross product it's a way of multiplying two vectors together to produce a third vector in that way it's different than the dot product or scalar product that we've discussed earlier where we multiply two vectors together and get a scalar or a pure number there's this is a different operation the vector product or cross product torque is the first physical quantity we've encountered but it won't be the last that involves a cross product in its definition and so before giving you a formal definition of torque will first review the mathematics of cross products and we'll do that in the next lesson now the angular acceleration alpha tells us about how the rotational motion changes again just like in F equals MA where we could divide the two sides of this equation into dynamics and kinematics kinematics is telling us about a geometrical description of the motion of the translational motion and the dynamics F is telling us about how the applied forces cause changes in the motion changes in the kinematics likewise for rotational motion the angular acceleration alpha tells us about the geometry of the rotation of motion and specifically how the rotational motion is changing and the torque is telling us about the application of forces and how that causes changes in the rotational motion so that leaves one other quantity which is the moment of inertia and I want to talk about what that means for just a moment so the term inertia term inertia and physics represents a resistance to an applied force it tells us how difficult it is to change an object's motion so for example suppose I have two blocks one of mass and another that's ten times as massive so 10 and Newton's second law F equals MA tells us that if I want to accelerate the heavier mass to the same rate that I do with a smaller mass I'll need to apply a force that's ten times larger so for translation of motion the mass M represents the notion of inertia the resistance to a force it tells us how much force we have to apply to achieve a certain change in motion now for rotational motion that role is played by the moment of inertia I that tells us how difficult it is to change the rotation of an object if I increase the moment of inertia by a factor of 10 then I'll need a torque that is 10 times larger in order to achieve the same change in rotational motion the same angular acceleration but recall that our definition of moment of inertia for some rigid body is if I break up the rigid body into a bunch of little pieces and for each piece I take the product of the mass of that piece Delta M J times the perpendicular distance of that piece from the rotation axis I'll call that R sub J and if I sum that over the entire object the entire body that gives me my moment of inertia well so I'm sorry this is R squared so it's the the mass of each element times the distance of the axis squared Delta M J times are J squared now I can increase the moment of inertia by a factor of 10 by increasing the total mass of the object but you can see from this equation that I can also do it with this same mass by changing the location of the mass if I increase the ars if I move the mass the same mass but i but I move it to be farther away from the rotation axis that also achieves an increase in the moment of inertia so what that tells us the for rotation of motion it's not just the amount of mass that matters but also how that mass is distributed okay so in that sense rotational inertia is different than translational inertia for translational motion all that matters is the mass if I increase the mass by a factor of 10 then it will become a factor of 10 more difficult to get that object to accelerate I need a factor of 10 larger force but with rotational motion I can increase the notion of inertia I can make it more difficult to change the rotation of motion either by changing the mass by increasing and or by Inc by making the distribution mass be further away from the rotation axis so as an example imagine if you were rolling a wheel up a hill it's different if I have a wheel whose mass is distributed evenly over the whole disk or if I have all of the mass in the rim if all of the masses in the rim then from this equation we see the moment of inertia is larger for the same amount of total mass and so it's much more difficult to roll a wheel up the hill if all of the mass is in the rim than it is if the mass is distributed evenly over the wheel so our rotational equivalent to Newton's second law is tau equals I alpha the torque equals the moment of inertia times the angular acceleration in the next few lessons we'll see how torque is defined and we'll derive this expression for the dynamics of rotational motion and see how to apply it
17508	https://www.youtube.com/watch?v=ctYOVjyMR0s	Create NaCl Crystal structure and generate XRD data using VESAT PhDzzz 3650 subscribers 16 likes Description 1634 views Posted: 11 Dec 2023 In this short tutorial, learn how to create a unit cell or the crystal structure of sodium chloride (NaCl) and generate its X-ray diffraction (XRD) data. First, download the CIF file of NaCl from the COD website and then drag it to the VESTA software. Then follow the tutorial and learn to visualize crystal structure, edit the atoms and bond, and generate XRD of NaCl structure. nacl #crystalstructure #xrd #crystallography #unitcell #vestasoftware #cif #cod 4 comments Transcript: in this video I will show you how to draw sodium chloride NL crystal structure by using Vista software so first go to the Google and type crystallography open database go to this website click on the search over here type the atom name is a sodium and the chloride and then number of Minima and maximum atom and click on the same over here you will find the CIF file for sodium chloride with fmus 3m space group that is a cubic structure click on the CF file the CF file will be downloaded now go to the Google again and type for Vista software click on the download Vista over here you will find the Vista software for Windows Mac and the Linux so depending on your OS select anyone zf5 Z file and download it I have downloaded Vista software go to the download folder and drag the CIF file okay so this is the unit s of NSL if you click on the objective you can find out the color for sodium is yellow and for chlorine is green click over here you can change the color so let's select blue and for this one maybe iron okay so then you can go and click on the edit or here you can find the edit data phase so this is sodium chloride title if you click on the unit cell is a cubic unit cell with a space group of fmus 3M space group is 2: 5 lce constant is 5.62 and Alpha Beta is a cubic so is 909090 if you click on the structure parameter then you can find the cartisian coordinate for sodium is 0 0 0 and for cloting is .55 and .5 so close this one again you can go and click on the edit you can find out detail about the bond Vector if you want to draw some lce plane so let's select this and uh let's click on the new and over here this is lce plane of one0 so you can see the pink color is 1 0 0 if you change if you want to change this one to say 11 one 1 one then it can change to 111 if you want to move it position so this is Multiplied D and this is distance from the origin is 5.62 so if you change this number it will change the position of this plan so let's consider the spacing of one so plane will be over here and click okay so this way the plane is located at this place okay and and uh here you may be wondering about this extra chlorine atom so how to delete those uh maybe you can select the orientation along a axis and then select this option and select the item and just PR press the delete button on the keyboard so just repeat it for all the other atems similarly you can click around the B and just oh sorry select this atem delete it to the same part here and the C that's all so now this is your unit cell sorry so this is the unit cell for here you can find the option for Zoom plus and minus and also this one it will help you to move unit cell along the right and left this one help you to rotate the unit cell and uh I think what we can do if you want to generate the xrd of this unit cell what you can do is go to the utility and click on the powder deflection pattern so it will go to the new window and over here click on the condition and you can reduce number of wavelength to one uh you can Define the wavelength so this is by default copper key Alpha which is 1.54 and then click on the calculator so it will help you to generate the reflection list over here you can find the 2 Theta values as well as the D spacing and the hql plane and if you go to the plot you'll find the xrd plot this plot these values can be compared with the experimental xrd if you have any and then from that you can know about the lce constant and also if you want to know the impurity you can compare the peak and their intensity if you want to know how to perform R refinement you can check out my other video do share and subscribe and don't forget to press the Bell icon button
17509	https://math.stackexchange.com/questions/116011/proof-by-induction-of-summation-inequality-1-frac12-frac13-frac1	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Ask Question Asked Modified 8 years, 10 months ago Viewed 8k times 1 $\begingroup$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$. We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers. I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used. This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$ Let $P(n)$ be summation shown above. Base case for $n=1$, the first positive integer, $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$ so base case is true. Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$ This can be written as $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$ I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes. inequality summation induction harmonic-numbers Share edited Nov 5, 2016 at 14:07 Martin Sleziak 56.3k2020 gold badges211211 silver badges391391 bronze badges asked Mar 3, 2012 at 17:00 JaredJared 39511 gold badge77 silver badges1717 bronze badges $\endgroup$ 6 $\begingroup$ Please check that I didnt make any inadvertent changes when I added $\LaTeX$ to your post. $\endgroup$ Brian M. Scott – Brian M. Scott 2012-03-03 17:17:44 +00:00 Commented Mar 3, 2012 at 17:17 $\begingroup$ Youre not trying to reach a contradiction: youre trying to show that if $P(n)$ is true, then $P(n+1)$ is true. $\endgroup$ Brian M. Scott – Brian M. Scott 2012-03-03 17:19:59 +00:00 Commented Mar 3, 2012 at 17:19 $\begingroup$ Thanks for adding the LaTex. I know I should be proving the summation is true but my math always results in a contradiction. I guess I'm not manipulating the terms correctly. Got any tips? $\endgroup$ Jared – Jared 2012-03-03 17:27:43 +00:00 Commented Mar 3, 2012 at 17:27 $\begingroup$ 1+1/2+1/4=7/4 but 1+2/2 = 2 <7/4, thus, what you are trying to prove is false for n=2. $\endgroup$ Per Alexandersson – Per Alexandersson 2012-03-03 17:35:10 +00:00 Commented Mar 3, 2012 at 17:35 $\begingroup$ @Pax: $2 < 7/4$? Indeed, it is false, but not because $2 < 7/4$. $\endgroup$ TMM – TMM 2012-03-03 19:31:36 +00:00 Commented Mar 3, 2012 at 19:31 \| Show 1 more comment 3 Answers 3 Reset to default 6 $\begingroup$ I think that your notation is rather badly confused: I strongly suspect that youre supposed to be showing that $$\sum_{k=1}^{2^n}\frac1k\ge 1+\frac{n}2\;,\tag{1}$$ from which one can conclude that the harmonic series diverges. The basis step for your induction should then be to check that $(1)$ is true for $n=0$, which it is: $$\sum_{k=1}^{2^n}\frac1k=\frac11\ge 1+\frac02\;.$$ Now your induction hypothesis, $P(n)$, should be equation $(1)$, and you want to show that this implies $P(n+1)$, which is the inequality $$\sum_{k=1}^{2^{n+1}}\frac1k\ge 1+\frac{n+1}2\tag{2}\;.$$ You had the right idea when you broke up the bigger sum into the old part and the new part, but the details are way off: $$\begin{align}\sum_{k=1}^{2^{n+1}}\frac1k&=\sum_{k=1}^{2^n}\frac1k+\sum_{k=2^n+1}^{2^{n+1}}\frac1k\ &\ge 1+\frac{n}2+\sum_{k=2^n+1}^{2^{n+1}}\frac1k\tag{3} \end{align}$$ by the induction hypothesis $P(n)$. Now look at that last summation in $(3)$: it has $2^{n+1}-2^n=2^n$ terms, and the smallest of those terms is $\dfrac1{2^{n+1}}$, so $$\sum_{k=2^n+1}^{2^{n+1}}\frac1k\ge 2^n\cdot\frac1{2^{n+1}}=\frac12\;.$$ If you plug this into $(3)$, you find that $$\sum_{k=1}^{2^{n+1}}\frac1k\ge 1+\frac{n}2+\frac12=1+\frac{n+1}2\;,$$ which is exactly $P(n+1)$, the statement that you were trying to prove. Youve now checked the basis step and carried out the induction step, so you can conclude that $(1)$ is true for all $n\ge 0$. Share answered Mar 3, 2012 at 17:33 Brian M. ScottBrian M. Scott 633k5757 gold badges824824 silver badges1.4k1.4k bronze badges $\endgroup$ 3 $\begingroup$ I understand now that my summation notation was incorrect. However, the base case here is invalid because 0 is not a positive integer. Therefore the base case should start at n=1. $\endgroup$ Jared – Jared 2012-03-04 18:44:38 +00:00 Commented Mar 4, 2012 at 18:44 $\begingroup$ @Izzy: It doesnt really matter: the result is true for $n=0$, so theres no harm starting there. $\endgroup$ Brian M. Scott – Brian M. Scott 2012-03-05 05:33:28 +00:00 Commented Mar 5, 2012 at 5:33 $\begingroup$ What does the "smallest of those terms" mean? Where does 2^(n+1) - 2^n = 2^n come from? $\endgroup$ Zip – Zip 2017-05-11 03:31:59 +00:00 Commented May 11, 2017 at 3:31 Add a comment \| 0 $\begingroup$ The base case looks okay. For the inductive step, you want to assume $P(n)$ is true, and you need to show that $P(n) \rightarrow P(n+1)$. Your wording suggests that you are assuming that implication. So, you assume for all $k \geq 1$ $$\sum\limits_{i=0}^{k} \frac{1}{2^i} \geq 1 + \frac{k}{2}.$$ Then we have the following when $n = k + 1$ $$\sum\limits_{i=0}^{k+1} \frac{1}{2^i} = \frac{1}{2^{k+1}} + \sum\limits_{i=0}^{k} \frac{1}{2^i} \geq \frac{1}{2^{k+1}} + 1 + \frac{k}{2}.$$ We know that $\frac{1}{2^{k+1}} > \frac{1}{2}$, so we have $$\sum\limits_{i=0}^{k+1} \frac{1}{2^i} \geq \frac{1}{2^{k+1}} + 1 + \frac{k}{2} > 1 + \frac{k+1}{2}.$$ Thus, for all $n \geq 1$, $P(n) \rightarrow P(n+1)$, so the hypothesis holds. Share answered Mar 3, 2012 at 17:26 Kurtis ZimmermanKurtis Zimmerman 79344 silver badges1111 bronze badges $\endgroup$ 2 $\begingroup$ $\sum_{i\ge 0}\frac1{2^i}=2$, and the stated result is false for all sufficiently large $k$. $\endgroup$ Brian M. Scott – Brian M. Scott 2012-03-03 17:34:49 +00:00 Commented Mar 3, 2012 at 17:34 $\begingroup$ Oh I don't know where my mind went. I made a terrible mistake in saying that $\frac{1}{2^{k+1}} > \frac{1}{2}$. Just a stupid error. $\endgroup$ Kurtis Zimmerman – Kurtis Zimmerman 2012-03-03 18:14:11 +00:00 Commented Mar 3, 2012 at 18:14 Add a comment \| -1 $\begingroup$ Hint: $$\sum_{k=0}^n\frac1{2^k}=\frac{2^{n+1}-1}{2^n}$$ Share answered Mar 3, 2012 at 17:28 Salech AlhasovSalech Alhasov 6,98422 gold badges3131 silver badges4747 bronze badges $\endgroup$ 1 1 $\begingroup$ True, but it wont help the OP, since the result that he stated is false (and almost certainly isnt the one that he was actually supposed to prove). $\endgroup$ Brian M. Scott – Brian M. Scott 2012-03-03 17:36:16 +00:00 Commented Mar 3, 2012 at 17:36 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 inequality summation induction harmonic-numbers 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 Linked How to show that $ \sum_{k=1}^{2^n}\frac{1}{k}\geq 1+\frac{n}{2}$? 1 Prove $1 + \frac{n}{2} \leq 1+ \frac{1}{2} +\frac{1}{3} +\cdots + \frac{1}{2^n}$ for all natural numbers $n$ Related 2 Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction 1 Induction on summation inequality stuck on Induction step 0 tricky summation proof by induction 0 Induction proof with summation 1 Proving a Summation using induction 1 Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{3^n}\ge 1+\frac{2n}3$ Hot Network Questions Why include unadjusted estimates in a study when reporting adjusted estimates? Is it safe to route top layer traces under header pins, SMD IC? 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17510	https://arxiv.org/html/2403.02030v2	Rational distances from given points in the plane Back to arXiv Back to arXiv This is experimental HTML to improve accessibility. We invite you to report rendering errors. Use Alt+Y to toggle on accessible reporting links and Alt+Shift+Y to toggle off. Learn more about this project and help improve conversions. Why HTML?Report IssueBack to AbstractDownload PDF Table of Contents Abstract 1 Introduction 2 Preliminary geometrical considerations 3 Rational distances from two points 3.1 A third approach for distances from two points 4 Squared Distances from Three Points 4.1 The collinear case 4.2 The non-collinear case. Proof of Theorem 1.2 5 Rational distances from three points 5.1 The collinear case 5.2 The non-collinear case 5.2.1 Some sections for the cubic family 5.2.2 Good reduction 5.3 An alternative construction 6 The Geometry of the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT 6.1 The surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT as a ramified cover of ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT 6.2 Alternative models for 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT 6.3 The Kummer model for the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT 7 Distance from four or more points 8 Integral Distances from two points 8.1 Integral points with coordinates in a real quadratic number field 8.2 Points with coordinates in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] References License: CC BY-NC-SA 4.0 arXiv:2403.02030v2 [math.NT] 10 Dec 2024 Rational distances from given points in the plane Report issue for preceding element Pietro Corvaja Dipartimento di Scienze Matematiche, Informatiche e Fisiche Università di Udine Via delle Scienze, 206 33100 Udine – Italy pietro.corvaja@dimi.uniud.it,Amos Turchet Dipartimento di Matematica e Fisica Università di Roma Tre Largo San L. Murialdo, 1 00146 Roma – Italy amos.turchet@uniroma3.itand Umberto Zannier Scuola Normale Superiore Piazza dei Cavalieri, 7 56126 Pisa – Italy umberto.zannier@sns.it Report issue for preceding element (Date: December 10, 2024) Abstract. Report issue for preceding element In this paper we study sets of points in the plane with rational distances from r 𝑟 r italic_r prescribed points P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P_{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT. A crucial case arises for r=3 𝑟 3 r=3 italic_r = 3, where we provide simple necessary and sufficient conditions for the density of this set in the real topology. We show in Theorem 1.1 that these conditions can be checked effectively (via congruences), proving that a related class of K⁢3 𝐾 3 K3 italic_K 3 surfaces satisfies the local-global principle. In particular, these conditions are always satisfied when P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT are rational. This result completes and goes beyond the analysis of Berry , who worked under stronger assumptions, not always fulfilled for instance in all the cases where P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT are rational. On the other hand, for r≥4 𝑟 4 r\geq 4 italic_r ≥ 4, we show that points with rational distances correspond to rational points in a surface of general type, hence conjecturally not Zariski dense. However, at the present, we lack methods to prove this, given the fact that the surface is simply-connected, as we shall show. We give explicit proofs as well as describe in detail the geometry of the surfaces involved. In addition we discuss certain analogues for points with distances in certain ring of integers. Report issue for preceding element Key words and phrases: Report issue for preceding element rational points, rational distances, algebraic surfaces, elliptic fibrations 2020 Mathematics Subject Classification: Report issue for preceding element 14G05,14J27,51N35 1. Introduction Report issue for preceding element In this paper we continue the investigation started by the third author in on sets of lattice points in the plane with integral distance to each point in a prescribed finite set. We focus here mainly on points having rational distances rather than integral distances, though we shall also consider some aspects of integrality. We shall further consider variants of the problem where we allow the points and the distances to take value in different rings or fields, such as number fields and their ring of integers, e.g. the ring ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] of Gaussian integers. Report issue for preceding element The investigation of subsets of the plane with integral distances is a fascinating historical subject that appears already in a paper of Anning and Erdős and motivated the so-called Erdős-Ulam problem that is still an open question; see [2, 27, 28] for discussions and further references. We also refer to Guy’s collection for related problems, raised by several authors. Report issue for preceding element We denote by d⁢(P,Q)𝑑 𝑃 𝑄 d(P,Q)italic_d ( italic_P , italic_Q ) the Euclidean distance between the points P 𝑃 P italic_P and Q 𝑄 Q italic_Q in ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Let P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P_{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT be fixed points on the Euclidean plane ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT; we shall usually suppose they are in general position, i.e. pairwise distinct and three-by-three non aligned; however, we shall treat in detail also the case r=3 𝑟 3 r=3 italic_r = 3 and P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT aligned. The main focus for the present article is the set of points with rational distances to P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P_{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT. Not surprisingly, this set becomes sparser when r 𝑟 r italic_r increases; conjecturally, it is not Zariski-dense whenever r≥4 𝑟 4 r\geq 4 italic_r ≥ 4. Report issue for preceding element The case r=2 𝑟 2 r=2 italic_r = 2 is easy; the set of points of the plane having rational distances from two given points P 1,P 2 subscript 𝑃 1 subscript 𝑃 2 P_{1},P_{2}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is dense in the real (hence also in the Zariski) topology. Report issue for preceding element In the case r=3 𝑟 3 r=3 italic_r = 3, T. G. Berry proved, under certain conditions, that the set is dense (in the real topology): he required that all square distances d⁢(P i,P j)2 𝑑 superscript subscript 𝑃 𝑖 subscript 𝑃 𝑗 2 d(P_{i},P_{j})^{2}italic_d ( italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are rational and at least one of the distances is rational. Completing his work, we provide an ‘if and only if’ criterion for density: Report issue for preceding element Theorem 1.1. Report issue for preceding element Let P 1,P 2,P 3∈ℝ 2 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 superscript ℝ 2 P_{1},P_{2},P_{3}\in{\mathbb{R}}^{2}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT be three non-aligned points. The following are equivalent: Report issue for preceding element (i)the set of points Q∈ℝ 2 𝑄 superscript ℝ 2 Q\in{\mathbb{R}}^{2}italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT having rational distance from P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is dense in ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT; Report issue for preceding element (ii)the quadratic form Report issue for preceding element (x,y)↦‖x⁢P 1⁢P 3→+y⁢P 2⁢P 3→‖2 maps-to 𝑥 𝑦 superscript norm 𝑥→subscript 𝑃 1 subscript 𝑃 3 𝑦→subscript 𝑃 2 subscript 𝑃 3 2(x,y)\mapsto\\|x\overrightarrow{P_{1}P_{3}}+y\overrightarrow{P_{2}P_{3}}\\|^{2}( italic_x , italic_y ) ↦ ∥ italic_x over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG + italic_y over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(1.1) is defined over ℚ ℚ{\mathbb{Q}}blackboard_Q and represents a square. Report issue for preceding element (iii)the square distances d⁢(P i,P j)2 𝑑 superscript subscript 𝑃 𝑖 subscript 𝑃 𝑗 2 d(P_{i},P_{j})^{2}italic_d ( italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, i,j=1,2,3 formulae-sequence 𝑖 𝑗 1 2 3 i,j=1,2,3 italic_i , italic_j = 1 , 2 , 3, are all rational and there exists at least one point Q∈ℝ 2 𝑄 superscript ℝ 2 Q\in{\mathbb{R}}^{2}italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT having rational distance from P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT; Report issue for preceding element Thanks to the Hasse-Minkowski theorem for quadratic forms, our condition (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ) can always be checked by verifying the solvability of finitely many congruences. Also, it is plainly satisfied when P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT are rational points, unlike Berry’s one. Report issue for preceding element The following statement provides a further condition equivalent to (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ), and hence to (i)𝑖(i)( italic_i ) and (i⁢i⁢i)𝑖 𝑖 𝑖(iii)( italic_i italic_i italic_i ). We first recall the Iwasawa decomposition in GL 2+⁡(ℝ)superscript subscript GL 2 ℝ\operatorname{GL}{2}^{+}({\mathbb{R}})roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT + end_POSTSUPERSCRIPT ( blackboard_R ), the connected component of the identity of the group of invertible 2×2 2 2 2\times 2 2 × 2 matrices with real coefficients: every matrix T∈GL 2+⁡(ℝ)𝑇 superscript subscript GL 2 ℝ T\in\operatorname{GL}{2}^{+}({\mathbb{R}})italic_T ∈ roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT + end_POSTSUPERSCRIPT ( blackboard_R ) can be uniquely decomposed as a product Report issue for preceding element T=A⋅D⋅U 𝑇⋅𝐴 𝐷 𝑈 T=A\cdot D\cdot U italic_T = italic_A ⋅ italic_D ⋅ italic_U where A∈O⁢(2)𝐴 O 2 A\in\mathrm{O}(2)italic_A ∈ roman_O ( 2 ) is orthogonal, D 𝐷 D italic_D is diagonal with positive entries on the diagonal and U 𝑈 U italic_U is upper triangular unipotent. Report issue for preceding element Addendum to Theorem 1.1. Let P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT be three non-aligned points on the plane. Choose an ordering such that the matrix T=(P 1⁢P 3→,P 2⁢P 3→)𝑇→subscript 𝑃 1 subscript 𝑃 3→subscript 𝑃 2 subscript 𝑃 3 T=(\overrightarrow{P_{1}P_{3}},\overrightarrow{P_{2}P_{3}})italic_T = ( over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG , over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ) has positive determinant. Then (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ) is equivalent to the condition Report issue for preceding element (iv)D=(r 0 0 s)𝐷 matrix 𝑟 0 0 𝑠 D=\left(\begin{matrix}\sqrt{r}&0\ 0&\sqrt{s}\end{matrix}\right)italic_D = ( start_ARG start_ROW start_CELL square-root start_ARG italic_r end_ARG end_CELL start_CELL 0 end_CELL end_ROW start_ROW start_CELL 0 end_CELL start_CELL square-root start_ARG italic_s end_ARG end_CELL end_ROW end_ARG ) with r,s 𝑟 𝑠 r,s italic_r , italic_s positive rational numbers such that the quadratic form r⁢x 2+s⁢y 2 𝑟 superscript 𝑥 2 𝑠 superscript 𝑦 2 rx^{2}+sy^{2}italic_r italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_s italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT represents 1 1 1 1 and U 𝑈 U italic_U has rational entries. Report issue for preceding element Let us call admissible a triangle (P 1,P 2,P 3)subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3(P_{1},P_{2},P_{3})( italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ) for which condition (ii) (hence also condition (iv)) is satisfied. The family of admissible triangles is acted on by a natural group of affine transformations. Of course, this set is invariant by all translations by vectors of the group ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. We find it remarkable that it is also invariant by the affine transformations whose differential lies in the group GL 2⁢(ℚ)subscript GL 2 ℚ\mathrm{GL}_{2}({\mathbb{Q}})roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ); moreover, it is also invariant by the homotheties of a factor λ 𝜆\lambda italic_λ, where λ 2 superscript 𝜆 2\lambda^{2}italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT sum of two squares in ℚ ℚ{\mathbb{Q}}blackboard_Q. It turns out that the group generated by these three kinds of transformations is the maximal group of affine transformations leaving invariant the set of admissible triangles; it can be characterised as the group of affine transformations Report issue for preceding element (x y)↦T⋅(x y)+(u v),maps-to binomial 𝑥 𝑦⋅𝑇 binomial 𝑥 𝑦 binomial 𝑢 𝑣\binom{x}{y}\mapsto T\cdot\binom{x}{y}+\binom{u}{v},( FRACOP start_ARG italic_x end_ARG start_ARG italic_y end_ARG ) ↦ italic_T ⋅ ( FRACOP start_ARG italic_x end_ARG start_ARG italic_y end_ARG ) + ( FRACOP start_ARG italic_u end_ARG start_ARG italic_v end_ARG ) , where T∈GL 2⁡(ℝ)𝑇 subscript GL 2 ℝ T\in\operatorname{GL}{2}({\mathbb{R}})italic_T ∈ roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_R ), (u,v)∈ℝ 2 𝑢 𝑣 superscript ℝ 2(u,v)\in{\mathbb{R}}^{2}( italic_u , italic_v ) ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and T 𝑇 T italic_T is a product λ⋅A⋅𝜆 𝐴\lambda\cdot A italic_λ ⋅ italic_A, with A∈GL 2⁡(ℚ)𝐴 subscript GL 2 ℚ A\in\operatorname{GL}{2}({\mathbb{Q}})italic_A ∈ roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) and λ∈ℝ∗𝜆 superscript ℝ\lambda\in{\mathbb{R}}^{}italic_λ ∈ blackboard_R start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT such that λ 2 superscript 𝜆 2\lambda^{2}italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is a sum of two rational squares. Let us call admissible such an affine transformation. The action of the group of admissible transformations on the set of admissible triangles is not transitive, as shown by the following example: Report issue for preceding element Example: the triangles (0,(1,0),(0,1))0 1 0 0 1(0,(1,0),(0,1))( 0 , ( 1 , 0 ) , ( 0 , 1 ) ) and (0,(1,0),(0,3)(0,(1,0),(0,\sqrt{3})( 0 , ( 1 , 0 ) , ( 0 , square-root start_ARG 3 end_ARG ), where the symbol 0 0 denotes the origin of the plane, are both admissible, but no admissible affine transformation sends one triangle into the other. The first triangle is admissible since it has rational vertices; the second one leads to the quadratic form x 2+3⁢y 2 superscript 𝑥 2 3 superscript 𝑦 2 x^{2}+3y^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 3 italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT which represents a square. To see that the two triangles do not belong to the same orbit under the mentioned group of affine transformations, it suffices to check one by one the six affine transformations of the plane sending the first triangle to the second one. Since the group of admissible transformations acts transitively on the vertices of the first triangle, it suffices to check only one transformation sending the first triangle onto the second one, e.g. the affine transformation fixing the points 0,(1,0)0 1 0 0,(1,0)0 , ( 1 , 0 ) and sending (0,1)0 1(0,1)( 0 , 1 ) to 0,3)0,\sqrt{3})0 , square-root start_ARG 3 end_ARG ). This transformation is not admissible. Report issue for preceding element One can look at the triangles P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT satisfying the first part of condition (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ), namely the rationality of the quadratic form (1.1). Choosing for simplicity P 3 subscript 𝑃 3 P_{3}italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT equal to the origin, the triple P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is identified by a non-singular 2×2 2 2 2\times 2 2 × 2 matrix T∈GL 2⁡(ℝ)𝑇 subscript GL 2 ℝ T\in\operatorname{GL}{2}({\mathbb{R}})italic_T ∈ roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_R ), whose column are the vectors P 1,P 3→,P 2,P 3→→subscript 𝑃 1 subscript 𝑃 3→subscript 𝑃 2 subscript 𝑃 3\overrightarrow{P{1},P_{3}},\overrightarrow{P_{2},P_{3}}over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG , over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG. The rationality condition (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ) amounts to the rationality of the symmetric matrix T t⋅T⋅superscript 𝑇 𝑡 𝑇{}^{t}T\cdot T start_FLOATSUPERSCRIPT italic_t end_FLOATSUPERSCRIPT italic_T ⋅ italic_T. A set of matrices of this form is given by those of the form A⋅T 0⋅𝐴 subscript 𝑇 0 A\cdot T_{0}italic_A ⋅ italic_T start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT, with T 0∈GL 2⁡(ℚ)subscript 𝑇 0 subscript GL 2 ℚ T_{0}\in\operatorname{GL}_{2}({\mathbb{Q}})italic_T start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT ∈ roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) and A 𝐴 A italic_A in the orthogonal group O⁢(2)O 2\mathrm{O}(2)roman_O ( 2 ). Report issue for preceding element In turn one could search for a complete characterization of such matrices (which will be given below): for instance, if the points are defined over a number field K 𝐾 K italic_K (say Galois over ℚ ℚ{\mathbb{Q}}blackboard_Q with Galois group G 𝐺 G italic_G), the condition amounts to Report issue for preceding element T t⋅T=T σ t⋅T σ⋅superscript 𝑇 𝑡 𝑇⋅superscript superscript 𝑇 𝜎 𝑡 superscript 𝑇 𝜎{}^{t}T\cdot T={}^{t}T^{\sigma}\cdot T^{\sigma}start_FLOATSUPERSCRIPT italic_t end_FLOATSUPERSCRIPT italic_T ⋅ italic_T = start_FLOATSUPERSCRIPT italic_t end_FLOATSUPERSCRIPT italic_T start_POSTSUPERSCRIPT italic_σ end_POSTSUPERSCRIPT ⋅ italic_T start_POSTSUPERSCRIPT italic_σ end_POSTSUPERSCRIPT(1.2) for each σ∈G 𝜎 𝐺\sigma\in G italic_σ ∈ italic_G, i.e. T⋅T σ−1⋅𝑇 superscript superscript 𝑇 𝜎 1 T\cdot{T^{\sigma}}^{-1}italic_T ⋅ italic_T start_POSTSUPERSCRIPT italic_σ end_POSTSUPERSCRIPT start_POSTSUPERSCRIPT - 1 end_POSTSUPERSCRIPT belongs to the orthogonal group O⁢(2)O 2\mathrm{O}(2)roman_O ( 2 ). So the map σ↦T σ⁢T−1 maps-to 𝜎 superscript 𝑇 𝜎 superscript 𝑇 1\sigma\mapsto T^{\sigma}T^{-1}italic_σ ↦ italic_T start_POSTSUPERSCRIPT italic_σ end_POSTSUPERSCRIPT italic_T start_POSTSUPERSCRIPT - 1 end_POSTSUPERSCRIPT is a 1 1 1 1-cocycle of G 𝐺 G italic_G with value in the orthogonal group. If the first cohomology group H 1⁢(G,O⁢(2,K))superscript H 1 𝐺 O 2 𝐾\mathrm{H}^{1}(G,\mathrm{O}(2,{K}))roman_H start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT ( italic_G , roman_O ( 2 , italic_K ) ) would vanish, each such matrix would be of the above mentioned shape A⁢T 0 𝐴 subscript 𝑇 0 AT_{0}italic_A italic_T start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT (i.e. the triangle P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT would be isometric to a rational one). Report issue for preceding element However, the above example with the triangle 0,(1,0),(0,3)0 1 0 0 3 0,(1,0),(0,\sqrt{3})0 , ( 1 , 0 ) , ( 0 , square-root start_ARG 3 end_ARG ) shows that this is not the case. Report issue for preceding element See below for a connection with this fact with the field of moduli for the surfaces 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT associated to the triangles P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. Report issue for preceding element An important ingredient in the proof of Theorem 1.1 will be the following result, of easier nature, about rationality of the squares of the distances: Report issue for preceding element Theorem 1.2. Report issue for preceding element Given three non-aligned points P 1,P 2 subscript 𝑃 1 subscript 𝑃 2 P_{1},P_{2}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT and P 3 subscript 𝑃 3 P_{3}italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT, the following are equivalent: Report issue for preceding element (i)the set of points Q∈ℝ 2 𝑄 superscript ℝ 2 Q\in{\mathbb{R}}^{2}italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT having rational squared-distance from P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is dense; Report issue for preceding element (ii)the quadratic form Report issue for preceding element (x,y)↦‖x⁢P 1⁢P 3→+y⁢P 2⁢P 3→‖2 maps-to 𝑥 𝑦 superscript norm 𝑥→subscript 𝑃 1 subscript 𝑃 3 𝑦→subscript 𝑃 2 subscript 𝑃 3 2(x,y)\mapsto\\|x\overrightarrow{P_{1}P_{3}}+y\overrightarrow{P_{2}P_{3}}\\|^{2}( italic_x , italic_y ) ↦ ∥ italic_x over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG + italic_y over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(1.3) is defined over ℚ ℚ{\mathbb{Q}}blackboard_Q. Report issue for preceding element (iii)the square distances d⁢(P i,P j)2 𝑑 superscript subscript 𝑃 𝑖 subscript 𝑃 𝑗 2 d(P_{i},P_{j})^{2}italic_d ( italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, i,j=1,2,3 formulae-sequence 𝑖 𝑗 1 2 3 i,j=1,2,3 italic_i , italic_j = 1 , 2 , 3, are all rational. Report issue for preceding element Moreover, under the above conditions, the set of points having rational squared-distance from P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is equal to the set of points of the form P 3+x⁢P 1⁢P 3→+y⁢P 2⁢P 3→subscript 𝑃 3 𝑥→subscript 𝑃 1 subscript 𝑃 3 𝑦→subscript 𝑃 2 subscript 𝑃 3 P_{3}+x\overrightarrow{P_{1}P_{3}}+y\overrightarrow{P_{2}P_{3}}italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT + italic_x over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG + italic_y over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG, with x,y∈ℚ 𝑥 𝑦 ℚ x,y\in{\mathbb{Q}}italic_x , italic_y ∈ blackboard_Q. Report issue for preceding element This result will be proved in section 4.2. Report issue for preceding element Arguing as in Theorem 1.1, we can express (ii) above in terms of Iwasawa decomposition: Report issue for preceding element Addendum to Theorem 1.2. Setting T=(P 1⁢P 3→,P 2⁢P 3→)∈GL 2⁡(ℝ)𝑇→subscript 𝑃 1 subscript 𝑃 3→subscript 𝑃 2 subscript 𝑃 3 subscript GL 2 ℝ T=(\overrightarrow{P_{1}P_{3}},\overrightarrow{P_{2}P_{3}})\in\operatorname{GL% }_{2}({\mathbb{R}})italic_T = ( over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG , over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ) ∈ roman_GL start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_R ), and T=A⋅D⋅U 𝑇⋅𝐴 𝐷 𝑈 T=A\cdot D\cdot U italic_T = italic_A ⋅ italic_D ⋅ italic_U be its Iwasawa decomposition, the above conditions are equivalent to Report issue for preceding element (iv)D=(r 0 0 s)𝐷 matrix 𝑟 0 0 𝑠 D=\left(\begin{matrix}\sqrt{r}&0\ 0&\sqrt{s}\end{matrix}\right)italic_D = ( start_ARG start_ROW start_CELL square-root start_ARG italic_r end_ARG end_CELL start_CELL 0 end_CELL end_ROW start_ROW start_CELL 0 end_CELL start_CELL square-root start_ARG italic_s end_ARG end_CELL end_ROW end_ARG ) with r,s 𝑟 𝑠 r,s italic_r , italic_s positive rational numbers and U 𝑈 U italic_U has rational entries. Report issue for preceding element Going back to our original problem concerning rationality of the distances, the case r=4 𝑟 4 r=4 italic_r = 4 becomes much harder: for instance no point Q 𝑄 Q italic_Q is known having rational distances from all the vertices of the unit square. We shall see that the set of points having rational distances from four given points in general position corresponds to the set of rational points on a certain algebraic surface of general type. According to the Bombieri-Lang conjecture, such rational points should be contained in a finite union of algebraic curves. Report issue for preceding element Note that if we replace the unit square by the unit pentagon (meaning a regular pentagon with sides of length 1 1 1 1) our condition (ii) of Theorem 1.2 is not satisfied for any choice of three of the five vertices, hence even the set of points with rational square distances from the vertices is degenerate (actually it can be proven to be empty). The same holds for n 𝑛 n italic_n-sided regular unit polygon for n≥7 𝑛 7 n\geq 7 italic_n ≥ 7 (see for a complete proof). Report issue for preceding element To see the link between rational points on algebraic surfaces and the solutions to our problem, let us introduce the following notation. Report issue for preceding element Given r≥3 𝑟 3 r\geq 3 italic_r ≥ 3 points P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P_{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT in general position, let 𝒮 P 1,…,P r=𝒮 r subscript 𝒮 subscript 𝑃 1…subscript 𝑃 𝑟 subscript 𝒮 𝑟{\mathcal{S}}{P{1},\ldots,P_{r}}={\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT end_POSTSUBSCRIPT = caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT be the surface in 𝔸 r superscript 𝔸 𝑟{\mathbb{A}}^{r}blackboard_A start_POSTSUPERSCRIPT italic_r end_POSTSUPERSCRIPT consisting in the set of elements (z 1,…,z r)∈𝔸 r subscript 𝑧 1…subscript 𝑧 𝑟 superscript 𝔸 𝑟(z{1},\ldots,z_{r})\in{\mathbb{A}}^{r}( italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_z start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT ) ∈ blackboard_A start_POSTSUPERSCRIPT italic_r end_POSTSUPERSCRIPT such that there exists a point Q∈𝔸 2 𝑄 superscript 𝔸 2 Q\in{\mathbb{A}}^{2}italic_Q ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT such that Report issue for preceding element (d⁢(P i,Q))2=z i 2 i=1,…,r.formulae-sequence superscript 𝑑 subscript 𝑃 𝑖 𝑄 2 superscript subscript 𝑧 𝑖 2 𝑖 1…𝑟(d(P_{i},Q))^{2}=z_{i}^{2}\qquad i=1,\ldots,r.( italic_d ( italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_Q ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_i = 1 , … , italic_r . Clearly, such a point Q 𝑄 Q italic_Q is unique, so the surface 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT is endowed with a projection 𝒮 r→𝔸 2→subscript 𝒮 𝑟 superscript 𝔸 2{\mathcal{S}}{r}\to{\mathbb{A}}^{2}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT → blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT sending (z 1,…,z r)subscript 𝑧 1…subscript 𝑧 𝑟(z_{1},\ldots,z_{r})( italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_z start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT ) to the corresponding point Q 𝑄 Q italic_Q. Report issue for preceding element Working over the complex number field, or even over the reals, the surface 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT could also be defined as the subsets of 𝔸 2+r superscript 𝔸 2 𝑟{\mathbb{A}}^{2+r}blackboard_A start_POSTSUPERSCRIPT 2 + italic_r end_POSTSUPERSCRIPT formed by the points (Q,z 1,…,z r)𝑄 subscript 𝑧 1…subscript 𝑧 𝑟(Q,z{1},\ldots,z_{r})( italic_Q , italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_z start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT ) satisfying the above equation. However, our arithmetic problem consists in finding points with (z 1,…,z n)∈ℚ r subscript 𝑧 1…subscript 𝑧 𝑛 superscript ℚ 𝑟(z_{1},\ldots,z_{n})\in{\mathbb{Q}}^{r}( italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_z start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ∈ blackboard_Q start_POSTSUPERSCRIPT italic_r end_POSTSUPERSCRIPT, regardless of the rationality of the point Q 𝑄 Q italic_Q. Hence our definition. Report issue for preceding element Our main results can be summarized as follows: Report issue for preceding element (1)The surfaces 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT have smooth models which are elliptic K 𝐾 K italic_K 3 surfaces. Under the necessary and sufficient condition (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ) above, their rational points are dense in the real topology (proven in section 5). Report issue for preceding element 2. (2)Smooth models for the surfaces 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}_{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT, for r≥4 𝑟 4 r\geq 4 italic_r ≥ 4, are simply connected and of general type (proven in section 7). According to the Bombieri-Lang conjecture, their set of rational points are degenerate, that is, not dense in the Zariski topology. Report issue for preceding element We remark at once that no example is known to date of a simply connected surface of general type when one can prove this fact. Report issue for preceding element 3. (3)Concerning integral distances: the set of points in the plane with coordinates in a ring R 𝑅 R italic_R of S 𝑆 S italic_S-integers in a number field, and having distances from two fixed points that belongs to R 𝑅 R italic_R, is Zariski dense for instance when R 𝑅 R italic_R is a real or imaginary quadratic ring. This fits with the Vojta’s conjectures (see ). Report issue for preceding element The condition that the quadratic form (1.1) is defined over ℚ ℚ{\mathbb{Q}}blackboard_Q will be proved to be equivalent to the condition that the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is defined over ℚ ℚ{\mathbb{Q}}blackboard_Q; hence its necessity for the Zariski-density of its rational points. Report issue for preceding element The condition that the quadratic form (x,y)↦‖x⁢P 1⁢P 3→+y⁢P 2⁢P 3→‖2 maps-to 𝑥 𝑦 superscript norm 𝑥→subscript 𝑃 1 subscript 𝑃 3 𝑦→subscript 𝑃 2 subscript 𝑃 3 2(x,y)\mapsto\\|x\overrightarrow{P_{1}P_{3}}+y\overrightarrow{P_{2}P_{3}}\\|^{2}( italic_x , italic_y ) ↦ ∥ italic_x over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG + italic_y over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT represents a square is equivalent to the existence of at least one point Q 𝑄 Q italic_Q having rational distance from P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. Report issue for preceding element It is worth noticing that varying P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT among triples with mutual rational square distances one obtains a family of K⁢3 𝐾 3 K3 italic_K 3 surfaces with the following arithmetic property: whenever one such a surface contains a rational point outside certain given curves “at infinity”, since condition (i⁢i⁢i)𝑖 𝑖 𝑖(iii)( italic_i italic_i italic_i ) of Theorem 1.1 is satisfied this surface contains a Zariski-dense set of rational points. This phenomenon, which occurs for Brauer-Severi varieties and for certain Del Pezzo surfaces, is rather unusual in the case of K⁢3 𝐾 3 K3 italic_K 3 surfaces. Report issue for preceding element We note that the surfaces 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT depend on the r 𝑟 r italic_r points P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT only up to the action of the group of similitudes 1 1 1 By similitude we mean the composition of a homothety with an isometry. on the affine plane; hence over the complex number field ℂ ℂ{\mathbb{C}}blackboard_C, the surfaces 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT form an algebraic family of dimension 2⁢r−4 2 𝑟 4 2r-4 2 italic_r - 4. In particular, the surfaces 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT form a two-dimensional family (of Kummer K 𝐾 K italic_K 3 surfaces). Report issue for preceding element We shall show (see the last Remark in section 7) that the conjectured degeneracy the rational points of 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}_{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT, for r≥4 𝑟 4 r\geq 4 italic_r ≥ 4, would answer the famous Erdős-Ulam problem in the negative, i.e. every infinite set of points having mutual rational distances would be contained in a line or a circle. In particular we obtain, a different proof of the result proved by Tao in , namely that the Bombieri-Lang conjecture implies a negative answer to the Erdős-Ulam problem. Report issue for preceding element The conditional result hinges on the fact that the set of points under consideration corresponds to rational points in a simply connected surface of general type (see Proposition 7.2). This is relevant since no example is known when the rational points can be proved (unconditionally) to be degenerate (over every number field) for such a surface of general type. Report issue for preceding element Concerning the case r≤2 𝑟 2 r\leq 2 italic_r ≤ 2, while the original problem has a trivial solution (the set of points having rational distances from P 1,P 2 subscript 𝑃 1 subscript 𝑃 2 P_{1},P_{2}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is always dense) an interesting problem arises by requiring that Q 𝑄 Q italic_Q is also a rational points. We are then led to studying the surface 𝒳 2 subscript 𝒳 2{\mathcal{X}}_{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT defined as Report issue for preceding element 𝒳 2={(Q,z 1,z 2)∈𝔸 2×𝔸 2:d⁢(P i,Q)=z i,i=1,2},subscript 𝒳 2 conditional-set 𝑄 subscript 𝑧 1 subscript 𝑧 2 superscript 𝔸 2 superscript 𝔸 2 formulae-sequence 𝑑 subscript 𝑃 𝑖 𝑄 subscript 𝑧 𝑖 𝑖 1 2{\mathcal{X}}{2}={(Q,z{1},z_{2})\in{\mathbb{A}}^{2}\times{\mathbb{A}}^{2}\,% :\,d(P_{i},Q)=z_{i},\,i=1,2},caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = { ( italic_Q , italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_z start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT × blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT : italic_d ( italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_Q ) = italic_z start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_i = 1 , 2 } , and investigating the distribution of its rational points. Report issue for preceding element We shall prove the following Report issue for preceding element Theorem 1.1. Report issue for preceding element The surface 𝒳 2 subscript 𝒳 2{\mathcal{X}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is a Del Pezzo surface of degree 4 4 4 4. Whenever P 1,P 2 subscript 𝑃 1 subscript 𝑃 2 P{1},P_{2}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT are rational, it is rational over ℚ ℚ{\mathbb{Q}}blackboard_Q, so in particular its set of rational points is Zariski-dense. Report issue for preceding element Acknowledgments. We thank Fabrizio Catanese for several discussions about the geometry and the topology of the surface 𝒮 4 subscript 𝒮 4{\mathcal{S}}_{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT, and to provide us with the argument (and the reference) for Proposition 7.2; thanks to Francesco Veneziano for pointing out reference . We thank the anonymous referees for several comments that greatly improved our presentation, and in particular to point out the results of . AT is partially supported by the PRIN 2022 project 2022HPSNCR: Semiabelian varieties, Galois representations and related Diophantine problems and the PRIN 2020 project 2020KKWT53: Curves, Ricci flat Varieties and their Interactions and is a member of the INDAM group GNSAGA. PC is a member of the INdAM group GNSAGA and is supported by the Advanced grant “Hyperbolicity in Diophantine Geometry”. Report issue for preceding element 2. Preliminary geometrical considerations Report issue for preceding element Following the notation introduced above, given points P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P_{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT on the plane 𝔸 2⁢(ℝ)=ℝ 2 superscript 𝔸 2 ℝ superscript ℝ 2{\mathbb{A}}^{2}({\mathbb{R}})={\mathbb{R}}^{2}blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( blackboard_R ) = blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, we define the algebraic surfaces 𝒳 r subscript 𝒳 𝑟{\mathcal{X}}{r}caligraphic_X start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT and 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT by letting Report issue for preceding element 𝒳 r:{(Q,z 1,…,z n)∈𝔸 r+2(d(Q,P i))2=z i 2}⊂𝔸 r+2{\mathcal{X}}{r}:\quad{(Q,z{1},\ldots,z_{n})\in{\mathbb{A}}^{r+2}\,>\,(d(Q% ,P_{i}))^{2}=z_{i}^{2}}\subset{\mathbb{A}}^{r+2}caligraphic_X start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT : { ( italic_Q , italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_z start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ∈ blackboard_A start_POSTSUPERSCRIPT italic_r + 2 end_POSTSUPERSCRIPT ( italic_d ( italic_Q , italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT } ⊂ blackboard_A start_POSTSUPERSCRIPT italic_r + 2 end_POSTSUPERSCRIPT and, for r≥3 𝑟 3 r\geq 3 italic_r ≥ 3, Report issue for preceding element 𝒮 r:{(z 1,…,z n)∈𝔸 r∃Q∈ℚ 2,(d(Q,P i))2=z i 2}.{\mathcal{S}}{r}:\quad{(z{1},\ldots,z_{n})\in{\mathbb{A}}^{r}\,>\,\exists Q% \in{\mathbb{Q}}^{2},\,(d(Q,P_{i}))^{2}=z_{i}^{2}}.caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT : { ( italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_z start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ∈ blackboard_A start_POSTSUPERSCRIPT italic_r end_POSTSUPERSCRIPT ∃ italic_Q ∈ blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_d ( italic_Q , italic_P start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT } . We shall frequently denote by the same symbols different projective or affine models of such surfaces. We note that for r≥3 𝑟 3 r\geq 3 italic_r ≥ 3 and P 1,…,P r subscript 𝑃 1…subscript 𝑃 𝑟 P_{1},\ldots,P_{r}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT in general position the surfaces 𝒳 r subscript 𝒳 𝑟{\mathcal{X}}{r}caligraphic_X start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT and 𝒮 r subscript 𝒮 𝑟{\mathcal{S}}{r}caligraphic_S start_POSTSUBSCRIPT italic_r end_POSTSUBSCRIPT are isomorphic over ℂ ℂ{\mathbb{C}}blackboard_C. Report issue for preceding element About the surface 𝒳 1 subscript 𝒳 1{\mathcal{X}}_{1}caligraphic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. Let us consider the points in the plane with rational distance to the origin. This leads to the Pythagorean equation x 2+y 2=z 2 superscript 𝑥 2 superscript 𝑦 2 superscript 𝑧 2 x^{2}+y^{2}=z^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, which we view as the relation between the coordinates of a point Q=(x,y)∈𝔸 2 𝑄 𝑥 𝑦 superscript 𝔸 2 Q=(x,y)\in{\mathbb{A}}^{2}italic_Q = ( italic_x , italic_y ) ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and its distance z=d⁢(O,Q)𝑧 𝑑 𝑂 𝑄 z=d(O,Q)italic_z = italic_d ( italic_O , italic_Q ) from the origin. A rational solution corresponds to a rational point with rational distance from the origin. Such equation defines a cone in 𝔸 3 superscript 𝔸 3{\mathbb{A}}^{3}blackboard_A start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT, hence a singular affine surface; it is the quadratic cover of 𝔸 2 superscript 𝔸 2{\mathbb{A}}^{2}blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ramified over the pair of complex lines defined by the equation x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0. Of course, it is a rational surface, indeed also rational over ℚ ℚ{\mathbb{Q}}blackboard_Q, and since the ancient times we know the rational parametrization given by formulas Report issue for preceding element x=2⁢λ⁢p y=λ⁢(p 2−1)z=λ⁢(p 2+1).formulae-sequence 𝑥 2 𝜆 𝑝 formulae-sequence 𝑦 𝜆 superscript 𝑝 2 1 𝑧 𝜆 superscript 𝑝 2 1 x=2\lambda p\qquad y=\lambda(p^{2}-1)\qquad z=\lambda(p^{2}+1).italic_x = 2 italic_λ italic_p italic_y = italic_λ ( italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 1 ) italic_z = italic_λ ( italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 ) . However, in view of the study of the problem of points with rational distance from three given points, we would like to study also a different model. Report issue for preceding element A non-singular projective model, which will be denoted by 𝒳 1 subscript 𝒳 1{\mathcal{X}}{1}caligraphic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, can be obtained as follows: noticing that the two aforementioned complex lines intersect at the origin, let us consider the blow-up of the projective plane over the origin, i.e. the point of coordinates (x:y:t)=(0:0:1)(x:y:t)=(0:0:1)( italic_x : italic_y : italic_t ) = ( 0 : 0 : 1 ); the strict transform of the reducible conic x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 is the union of two disjoint curves on such a surface, so a smooth divisor. The quadratic cover ramified over these curves is our surface 𝒮 1 subscript 𝒮 1{\mathcal{S}}{1}caligraphic_S start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. Report issue for preceding element Let us now look for explicit equations: the blow-up of the plane can be defined inside ℙ 2×ℙ 1 subscript ℙ 2 subscript ℙ 1{\mathbb{P}}{2}\times{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT as the set of solutions (x:y:t),(ξ:η)(x:y:t),(\xi:\eta)( italic_x : italic_y : italic_t ) , ( italic_ξ : italic_η ) to the linear equation Report issue for preceding element (x+i⁢y)⁢η=(x−i⁢y)⁢ξ.𝑥 𝑖 𝑦 𝜂 𝑥 𝑖 𝑦 𝜉(x+iy)\eta=(x-iy)\xi.( italic_x + italic_i italic_y ) italic_η = ( italic_x - italic_i italic_y ) italic_ξ . Adjoining to the function field the square-root of the rational function (x 2+y 2)/t 2 superscript 𝑥 2 superscript 𝑦 2 superscript 𝑡 2(x^{2}+y^{2})/t^{2}( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) / italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT amounts to extracting the square root of ξ⁢η 𝜉 𝜂\xi\eta italic_ξ italic_η, or of ξ/η 𝜉 𝜂\xi/\eta italic_ξ / italic_η. Hence we can define our surface inside ℙ 2×ℙ 1×ℙ 1 subscript ℙ 2 subscript ℙ 1 subscript ℙ 1{\mathbb{P}}{2}\times{\mathbb{P}}{1}\times{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT with coordinates (x:y:z),(ξ:η),(z 0:z 1)(x:y:z),(\xi:\eta),(z{0}:z_{1})( italic_x : italic_y : italic_z ) , ( italic_ξ : italic_η ) , ( italic_z start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT : italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) by the equation above combined with the new equation Report issue for preceding element η⁢z 0 2=ξ⁢z 1 2.𝜂 superscript subscript 𝑧 0 2 𝜉 superscript subscript 𝑧 1 2\eta z_{0}^{2}=\xi z_{1}^{2}.italic_η italic_z start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_ξ italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . Substituting in the above equation, i.e. projecting from ℙ 2×ℙ 1×ℙ 1→ℙ 2×ℙ 1→subscript ℙ 2 subscript ℙ 1 subscript ℙ 1 subscript ℙ 2 subscript ℙ 1{\mathbb{P}}{2}\times{\mathbb{P}}{1}\times{\mathbb{P}}{1}\to{\mathbb{P}}{2% }\times{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT by forgetting the (ξ:η):𝜉 𝜂(\xi:\eta)( italic_ξ : italic_η ) projective line, we obtain the single equation in ℙ 2×ℙ 1 subscript ℙ 2 subscript ℙ 1{\mathbb{P}}{2}\times{\mathbb{P}}_{1}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT Report issue for preceding element (x+i⁢y)⁢z 1 2=(x−i⁢y)⁢z 0 2.𝑥 𝑖 𝑦 superscript subscript 𝑧 1 2 𝑥 𝑖 𝑦 superscript subscript 𝑧 0 2(x+iy)z_{1}^{2}=(x-iy)z_{0}^{2}.( italic_x + italic_i italic_y ) italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_x - italic_i italic_y ) italic_z start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . This is the equation of the surface 𝒳 1 subscript 𝒳 1{\mathcal{X}}{1}caligraphic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. Note that the projection to (z 0:z 1)∈ℙ 1(z{0}:z_{1})\in{\mathbb{P}}{1}( italic_z start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT : italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) ∈ blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT endows 𝒮 1 subscript 𝒮 1{\mathcal{S}}{1}caligraphic_S start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT with a structure of ℙ 1 subscript ℙ 1{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT-bundle over ℙ 1 subscript ℙ 1{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT: indeed the surface 𝒮 1 subscript 𝒮 1{\mathcal{S}}{1}caligraphic_S start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT is the so-called Hirzebruch surface of degree two, abstractly defined as ℙ⁢(𝒪 ℙ 1⊕𝒪 ℙ 1⁢(2))ℙ direct-sum subscript 𝒪 subscript ℙ 1 subscript 𝒪 subscript ℙ 1 2{\mathbb{P}}({{\mathcal{O}}}{{\mathbb{P}}{1}}\oplus{{\mathcal{O}}}{{\mathbb% {P}}{1}}(2))blackboard_P ( caligraphic_O start_POSTSUBSCRIPT blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT ⊕ caligraphic_O start_POSTSUBSCRIPT blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT ( 2 ) ), i.e. the completion of the total space of a line bundle of degree two over ℙ 1 subscript ℙ 1{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, obtained by adding a single point to each fiber. Report issue for preceding element About the surface 𝒳 2 subscript 𝒳 2{\mathcal{X}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT The above construction can be mimicked to study the surface 𝒳 2 subscript 𝒳 2\mathcal{X}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. In this setting we are interested in points Q∈𝔸 2 𝑄 superscript 𝔸 2 Q\in{\mathbb{A}}^{2}italic_Q ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT with rational distance to two fixed points. Without loss of generality we can assume that one of the points is the origin O=(0,0)𝑂 0 0 O=(0,0)italic_O = ( 0 , 0 ) and the other point is P=(a,b)𝑃 𝑎 𝑏 P=(a,b)italic_P = ( italic_a , italic_b ), with a,b∈ℚ 𝑎 𝑏 ℚ a,b\in{\mathbb{Q}}italic_a , italic_b ∈ blackboard_Q. Then, as above, we consider a bi-quadratic cover of ℙ 2 subscript ℙ 2{\mathbb{P}}_{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ramified over the pair of singular conics defined by the equations x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 and (x−a)2+(y−b)2=0 superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 0(x-a)^{2}+(y-b)^{2}=0( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0. Again, after blowing up the singular points of these conics, i.e. the points O 𝑂 O italic_O, P 𝑃 P italic_P in our original problem, we can replace the singular conics by a smooth divisor; the bi-quadratic covering ramified over such a divisor turns out to be a smooth surface, and indeed is a Del Pezzo surface of degree four (see the explicit computation in the next section). Report issue for preceding element Integral points in 𝒳 2 subscript 𝒳 2{\mathcal{X}}_{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. The distribution of integral points on such affine surfaces often represents a deep problem: it was implicitly considered by M. Davis in in his investigations on the Diophantine representation of the exponential functions (see also for recent developments). Report issue for preceding element In our setting, the integral points on the plane having integral distances from the two given points can be viewed as the rational points on the (complete) surface 𝒳 2 subscript 𝒳 2{\mathcal{X}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT which are integral with respect to a certain divisor, namely the pull-back of the divisor at infinity of ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT in the covering 𝒳 2→ℙ 2→subscript 𝒳 2 subscript ℙ 2{{\mathcal{X}}}{2}\to{\mathbb{P}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. Denoting by D 𝐷 D italic_D this divisor, we note that D 𝐷 D italic_D lies in the anti-canonical class of 𝒮 2 subscript 𝒮 2{{\mathcal{S}}}{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, so the open surface 𝒮 2∖D subscript 𝒮 2 𝐷{{\mathcal{S}}}{2}\setminus D caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∖ italic_D is a so-called log–K⁢3 𝐾 3 K3 italic_K 3 surface. By a result of B. Hassett and Yu. Tschinkel, the integral points on surfaces obtained by removing an anti-canonical divisor from a Del Pezzo surface are potentially dense, i.e. the set of points with coordinates in a suitable ring of S 𝑆 S italic_S-integers is dense (actually, in this was proved under the assumption that the divisor at infinity is smooth, but S. Coccia removed this condition in the recent work ). Report issue for preceding element However, restricting to the ring ℤ ℤ{\mathbb{Z}}blackboard_Z of rational integers makes the situation different: the third author, in proved that under suitable simple necessary conditions the Zariski-closure of the set of points with coordinates in ℤ ℤ{\mathbb{Z}}blackboard_Z of 𝒳 2 subscript 𝒳 2{\mathcal{X}}_{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT projects to the union of at most two lines and a non-empty finite union of hyperbolas. Report issue for preceding element On the other hand, the set of integral points having integral distances from three given points is expected to be degenerate over any number field; this would follow from Vojta’s Conjecture, applied to an affine open set of a K3 surface. Report issue for preceding element 3. Rational distances from two points Report issue for preceding element The goal of this section is proving by explicit computation the following result: Report issue for preceding element Theorem 3.1. Report issue for preceding element The set of rational points on the plane having rational distances from two given rational points is Zariski-dense on the plane, and actually also dense in the real topology. Report issue for preceding element As before, without loss of generality, we can assume that one of the points is the origin O 𝑂 O italic_O and we will denote by P=(a,b)𝑃 𝑎 𝑏 P=(a,b)italic_P = ( italic_a , italic_b ) the second point, for rational numbers a,b∈ℚ 𝑎 𝑏 ℚ a,b\in{\mathbb{Q}}italic_a , italic_b ∈ blackboard_Q. We are interested in studying the affine set 𝒮 2 subscript 𝒮 2{\mathcal{S}}_{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT of points Q=(x,y)∈𝔸 2 𝑄 𝑥 𝑦 superscript 𝔸 2 Q=(x,y)\in{\mathbb{A}}^{2}italic_Q = ( italic_x , italic_y ) ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT such that the distances d⁢(O,Q)𝑑 𝑂 𝑄 d(O,Q)italic_d ( italic_O , italic_Q ) and d⁢(P,Q)𝑑 𝑃 𝑄 d(P,Q)italic_d ( italic_P , italic_Q ) are rational, i.e. the set Report issue for preceding element 𝒳 2=𝒮 2⁢(ℚ)={Q∈𝔸 2⁢(ℚ):d⁢(O,Q)∈ℚ⁢and⁢d⁢(P,Q)∈ℚ}.subscript 𝒳 2 subscript 𝒮 2 ℚ conditional-set 𝑄 superscript 𝔸 2 ℚ 𝑑 𝑂 𝑄 ℚ and 𝑑 𝑃 𝑄 ℚ{\mathcal{X}}{2}={\mathcal{S}}{2}({\mathbb{Q}})={Q\in{\mathbb{A}}^{2}({% \mathbb{Q}}):d(O,Q)\in{\mathbb{Q}}\text{ and }d(P,Q)\in{\mathbb{Q}}}.caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) = { italic_Q ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( blackboard_Q ) : italic_d ( italic_O , italic_Q ) ∈ blackboard_Q and italic_d ( italic_P , italic_Q ) ∈ blackboard_Q } . We stress here that we do not ask that the points of 𝒮 2 subscript 𝒮 2{\mathcal{S}}{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT have mutual rational distance, i.e. given Q 1,Q 2∈𝒮 2⁢(ℚ)subscript 𝑄 1 subscript 𝑄 2 subscript 𝒮 2 ℚ Q{1},Q_{2}\in{\mathcal{S}}{2}({\mathbb{Q}})italic_Q start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_Q start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∈ caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) the distance d⁢(Q 1,Q 2)𝑑 subscript 𝑄 1 subscript 𝑄 2 d(Q{1},Q_{2})italic_d ( italic_Q start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_Q start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) might be, in general, an irrational number (so in particular 𝒮 2 subscript 𝒮 2{\mathcal{S}}_{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is not a “rational distance set” in the sense of [2, 27]). Report issue for preceding element A point Q∈𝒳 2⁢(ℚ)𝑄 subscript 𝒳 2 ℚ Q\in{\mathcal{X}}_{2}({\mathbb{Q}})italic_Q ∈ caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) corresponds to a rational solution to the system Report issue for preceding element x 2+y 2=z 2,superscript 𝑥 2 superscript 𝑦 2 superscript 𝑧 2\displaystyle x^{2}+y^{2}=z^{2},italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(3.1) (x−a)2+(y−b)2=(z−k)2,superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 superscript 𝑧 𝑘 2\displaystyle(x-a)^{2}+(y-b)^{2}=(z-k)^{2},( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_z - italic_k ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(3.2) where z=d⁢(O,Q)𝑧 𝑑 𝑂 𝑄 z=d(O,Q)italic_z = italic_d ( italic_O , italic_Q ), and z−k=d⁢(P,Q)𝑧 𝑘 𝑑 𝑃 𝑄 z-k=d(P,Q)italic_z - italic_k = italic_d ( italic_P , italic_Q ). The above presentation shows that the set 𝒮 2 subscript 𝒮 2{\mathcal{S}}{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is the set of ℚ ℚ{\mathbb{Q}}blackboard_Q-rational points of the intersection of two quadrics in 𝔸 4 superscript 𝔸 4{\mathbb{A}}^{4}blackboard_A start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT, namely the solutions of equations (3.1) and (3.2) respectively, where x,y,z 𝑥 𝑦 𝑧 x,y,z italic_x , italic_y , italic_z and k 𝑘 k italic_k are the coordinates of 𝔸 4 superscript 𝔸 4{\mathbb{A}}^{4}blackboard_A start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT. The (projective completion of the) variety 𝒮 2 subscript 𝒮 2{\mathcal{S}}{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT defined by these equations is a Del Pezzo surface of degree four, hence a geometrically rational surface. Following , one can show that this variety is rational over ℚ ℚ{\mathbb{Q}}blackboard_Q, so its rational points are Zariski dense. Report issue for preceding element We can eliminate z 𝑧 z italic_z in the two equations as follows. First, one can subtract (3.2) from (3.1) to get Report issue for preceding element 2⁢k⁢z=2⁢a⁢x+2⁢b⁢y−δ,2 𝑘 𝑧 2 𝑎 𝑥 2 𝑏 𝑦 𝛿 2kz=2ax+2by-\delta,2 italic_k italic_z = 2 italic_a italic_x + 2 italic_b italic_y - italic_δ ,(3.3) where δ=a 2+b 2−k 2 𝛿 superscript 𝑎 2 superscript 𝑏 2 superscript 𝑘 2\delta=a^{2}+b^{2}-k^{2}italic_δ = italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Squaring (3.3) and using (3.1) we get Report issue for preceding element 𝒞 k:(2 a x+2 b y−δ)2=4 k 2(x 2+y 2).{\mathcal{C}}_{k}:\quad(2ax+2by-\delta)^{2}=4k^{2}(x^{2}+y^{2}).caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT : ( 2 italic_a italic_x + 2 italic_b italic_y - italic_δ ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4 italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) .(3.4) The latter equation represents a family of affine conics 𝒞→B≃𝔸 1→𝒞 𝐵 similar-to-or-equals superscript 𝔸 1{\mathcal{C}}\to B\simeq\mathbb{A}^{1}caligraphic_C → italic_B ≃ blackboard_A start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT indexed by k 𝑘 k italic_k. We note that (at least for k≠0 𝑘 0 k\neq 0 italic_k ≠ 0) rational points on a fiber 𝒞 k subscript 𝒞 𝑘{\mathcal{C}}{k}caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT correspond to points Q 𝑄 Q italic_Q with rational distances from O 𝑂 O italic_O and P 𝑃 P italic_P where the difference d⁢(O,Q)−d⁢(P,Q)=k 𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 𝑘 d(O,Q)-d(P,Q)=k italic_d ( italic_O , italic_Q ) - italic_d ( italic_P , italic_Q ) = italic_k is fixed. In particular, if 𝒞 𝒞{\mathcal{C}}caligraphic_C has a dense set of rational points then 𝒮 2⁢(ℚ)subscript 𝒮 2 ℚ{\mathcal{S}}{2}({\mathbb{Q}})caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) will be Zariski-dense too; actually it will turn out that 𝒮 2⁢(ℚ)subscript 𝒮 2 ℚ{\mathcal{S}}{2}({\mathbb{Q}})caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) is also dense in 𝒮 2⁢(ℝ)subscript 𝒮 2 ℝ{\mathcal{S}}{2}({\mathbb{R}})caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_R ) in the real topology. We observe that the projective completion of 𝒞 𝒞{\mathcal{C}}caligraphic_C is a conic bundle 𝒞¯→B¯→¯𝒞¯𝐵\overline{{\mathcal{C}}}\to\overline{B}over¯ start_ARG caligraphic_C end_ARG → over¯ start_ARG italic_B end_ARG over a projective curve B¯≃ℙ 1 similar-to-or-equals¯𝐵 subscript ℙ 1\overline{B}\simeq{\mathbb{P}}_{1}over¯ start_ARG italic_B end_ARG ≃ blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. Moreover, to show that 𝒞 𝒞{\mathcal{C}}caligraphic_C (and hence 𝒞¯¯𝒞\overline{{\mathcal{C}}}over¯ start_ARG caligraphic_C end_ARG) has a dense set of rational points, it is sufficient to produce a rational section B→𝒞→𝐵 𝒞 B\to{\mathcal{C}}italic_B → caligraphic_C. Report issue for preceding element To continue with our explicit computation, we start with a rational parametrization of (3.1) with parameters λ 𝜆\lambda italic_λ and p 𝑝 p italic_p such as Report issue for preceding element x=2⁢λ⁢p y=λ⁢(p 2−1)z=λ⁢(p 2+1).formulae-sequence 𝑥 2 𝜆 𝑝 formulae-sequence 𝑦 𝜆 superscript 𝑝 2 1 𝑧 𝜆 superscript 𝑝 2 1 x=2\lambda p\qquad y=\lambda(p^{2}-1)\qquad z=\lambda(p^{2}+1).italic_x = 2 italic_λ italic_p italic_y = italic_λ ( italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 1 ) italic_z = italic_λ ( italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 ) .(3.5) Then we use (3.2) with x=a 𝑥 𝑎 x=a italic_x = italic_a to get k=z−y+b 𝑘 𝑧 𝑦 𝑏 k=z-y+b italic_k = italic_z - italic_y + italic_b. Finally we can express λ=a 2⁢p 𝜆 𝑎 2 𝑝\lambda=\frac{a}{2p}italic_λ = divide start_ARG italic_a end_ARG start_ARG 2 italic_p end_ARG and p=a k−b 𝑝 𝑎 𝑘 𝑏 p=\frac{a}{k}-b italic_p = divide start_ARG italic_a end_ARG start_ARG italic_k end_ARG - italic_b thus expressing x 𝑥 x italic_x and y 𝑦 y italic_y (and z 𝑧 z italic_z) in terms of a,b 𝑎 𝑏 a,b italic_a , italic_b and k 𝑘 k italic_k as wanted. This defines a rational section B→𝒞→𝐵 𝒞 B\to{\mathcal{C}}italic_B → caligraphic_C; since B⁢(ℚ)𝐵 ℚ B({\mathbb{Q}})italic_B ( blackboard_Q ) is dense in B⁢(ℝ)𝐵 ℝ B({\mathbb{R}})italic_B ( blackboard_R ) in the real topology, and almost every fiber of a rational point of B 𝐵 B italic_B has a dense (in the real topology) set of rational point, we obtain that 𝒞 𝒞{\mathcal{C}}caligraphic_C has a dense set of rational points. Actually, the section shows that the surface is rational: for the section meets every conic in a rational point, and that conic can be parametrized rationally in terms of such rational point. Report issue for preceding element This proves Theorem 3.1 that the set 𝒳 2 subscript 𝒳 2{\mathcal{X}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, of rational points in the plane with rational distance from two fixed points O 𝑂 O italic_O and P 𝑃 P italic_P with rational coordinates is, in particular, dense in 𝔸 2 superscript 𝔸 2{\mathbb{A}}^{2}blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, or, what amounts to the same, the set of rational points 𝒳 2⁢(ℚ)subscript 𝒳 2 ℚ{\mathcal{X}}{2}({\mathbb{Q}})caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) is dense on the surface 𝒳 2 subscript 𝒳 2{\mathcal{X}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT: the density holds actually simultaneously with respect to any finite set of places; in other words, we can prescribe that our point with rational distance from both P 1,P 2 subscript 𝑃 1 subscript 𝑃 2 P{1},P_{2}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT lies near to any prescribed point of the plane, and satisfies every prescribed congruence modulo any given integer. Report issue for preceding element This is an explicit description parallel to the more algebro-geometric viewpoint that we described in section 2, where we started from a geometrical description of the involved algebraic surface. Report issue for preceding element 3.1. A third approach for distances from two points Report issue for preceding element The aim of this section is showing the geometrical meaning of the explicit calculations made in the previous section; we take the opportunity to reprove in an alternative and more synthetic way the density of the set of rational points having rational distances from two given rational points (i.e. the Zariski-density of 𝒳 2⁢(ℚ)subscript 𝒳 2 ℚ{\mathcal{X}}_{2}({\mathbb{Q}})caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q )). Report issue for preceding element Confocal families of conics. Given two points O,P 𝑂 𝑃 O,P italic_O , italic_P on the plane 𝔸 2⁢(ℝ)=ℝ 2 superscript 𝔸 2 ℝ superscript ℝ 2{\mathbb{A}}^{2}({\mathbb{R}})={\mathbb{R}}^{2}blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( blackboard_R ) = blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, let us consider the algebraic family of conics (either ellipses or hyperbolae) having foci on O 𝑂 O italic_O and P 𝑃 P italic_P. They form a 1 1 1 1-parameter family, but not a linear system: every generic point of the plane belongs to two such conics. A confocal family can always be defined as the set of conics which are tangent to four fixed given lines; hence it is the dual family of a pencil of conics in the dual projective plane. Report issue for preceding element The four (complex) lines tangent to every conic of the family are the tangent drawn from the foci: those passing through O 𝑂 O italic_O are given by the equation x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 (i.e. set of (complex) points with “distance zero” from O 𝑂 O italic_O) and analogously for the other pair of lines, those passing through P 𝑃 P italic_P). Report issue for preceding element Curiously, such considerations on families of confocal conics arose recently in a work by the first and third authors on finiteness results on elliptical billiards . Report issue for preceding element The ellipses in a confocal family are defined as the set points Q∈ℝ 2 𝑄 superscript ℝ 2 Q\in{\mathbb{R}}^{2}italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT such that d⁢(O,Q)+d⁢(P,Q)=k 𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 𝑘 d(O,Q)+d(P,Q)=k italic_d ( italic_O , italic_Q ) + italic_d ( italic_P , italic_Q ) = italic_k, where k 𝑘 k italic_k is a given real number >d⁢(O,P)absent 𝑑 𝑂 𝑃>d(O,P)> italic_d ( italic_O , italic_P ). The hyperbolae in the family are defined by the relation \|d⁢(Q,O)−d⁢(Q,P)\|=k 𝑑 𝑄 𝑂 𝑑 𝑄 𝑃 𝑘\|d(Q,O)-d(Q,P)\|=k\| italic_d ( italic_Q , italic_O ) - italic_d ( italic_Q , italic_P ) \| = italic_k, where now k 𝑘 k italic_k is a real number with 0<k<d⁢(O,P)0 𝑘 𝑑 𝑂 𝑃 0<k<d(O,P)0 < italic_k < italic_d ( italic_O , italic_P ). Report issue for preceding element On the proof of Theorem 3.1. We note that the conics 𝒞 k subscript 𝒞 𝑘\mathcal{C}_{k}caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT, defined by equation (3.4) and used in the previous section to prove our density statement, are precisely the conics having foci at O 𝑂 O italic_O and P 𝑃 P italic_P. Report issue for preceding element The fact that their rational points are solutions to our initial problem, i.e. they have rational distance from both O 𝑂 O italic_O and P 𝑃 P italic_P can be seen as follows: first of all the squared distances (d⁢(O,Q))2,(d⁢(P;Q))2 superscript 𝑑 𝑂 𝑄 2 superscript 𝑑 𝑃 𝑄 2(d(O,Q))^{2},(d(P;Q))^{2}( italic_d ( italic_O , italic_Q ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_d ( italic_P ; italic_Q ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are rational numbers, and this happens for every rational point Q 𝑄 Q italic_Q on the plane, so is the difference Report issue for preceding element (d⁢(O,Q))2−(d⁢(P,Q))2=(d⁢(O,Q)+d⁢(P,Q))⋅(d⁢(O,Q)−d⁢(P,Q)).superscript 𝑑 𝑂 𝑄 2 superscript 𝑑 𝑃 𝑄 2⋅𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 𝑑 𝑂 𝑄 𝑑 𝑃 𝑄(d(O,Q))^{2}-(d(P,Q))^{2}=(d(O,Q)+d(P,Q))\cdot(d(O,Q)-d(P,Q)).( italic_d ( italic_O , italic_Q ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ( italic_d ( italic_P , italic_Q ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_d ( italic_O , italic_Q ) + italic_d ( italic_P , italic_Q ) ) ⋅ ( italic_d ( italic_O , italic_Q ) - italic_d ( italic_P , italic_Q ) ) . Hence if one between the numbers d⁢(O,Q)+d⁢(P,Q)𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 d(O,Q)+d(P,Q)italic_d ( italic_O , italic_Q ) + italic_d ( italic_P , italic_Q ) and d⁢(O,Q)−d⁢(P,Q)𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 d(O,Q)-d(P,Q)italic_d ( italic_O , italic_Q ) - italic_d ( italic_P , italic_Q ) is rational, so is the other (provided neither is zero) and then both numbers d⁢(O,Q)𝑑 𝑂 𝑄 d(O,Q)italic_d ( italic_O , italic_Q ) and d⁢(P,Q)𝑑 𝑃 𝑄 d(P,Q)italic_d ( italic_P , italic_Q ) are rational. In geometric terms, if the point Q 𝑄 Q italic_Q belongs to either a hyperbola or an ellipse corresponding to a rational parameter k 𝑘 k italic_k as before, then d⁢(O,Q)𝑑 𝑂 𝑄 d(O,Q)italic_d ( italic_O , italic_Q ) and d⁢(P,Q)𝑑 𝑃 𝑄 d(P,Q)italic_d ( italic_P , italic_Q ) are rational numbers; also, if Q 𝑄 Q italic_Q belongs to a hyperbola of equation \|d⁢(Q,O)−d⁢(Q,P)\|=k 𝑑 𝑄 𝑂 𝑑 𝑄 𝑃 𝑘\|d(Q,O)-d(Q,P)\|=k\| italic_d ( italic_Q , italic_O ) - italic_d ( italic_Q , italic_P ) \| = italic_k for some rational k 𝑘 k italic_k, then it also belongs to a confocal ellipse of equation d⁢(Q,O)+d⁢(Q,P)=k′𝑑 𝑄 𝑂 𝑑 𝑄 𝑃 superscript 𝑘′d(Q,O)+d(Q,P)=k^{\prime}italic_d ( italic_Q , italic_O ) + italic_d ( italic_Q , italic_P ) = italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, where k′superscript 𝑘′k^{\prime}italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT is again a rational number. Report issue for preceding element We could conclude the proof of Theorem 3.1 as follows: first we choose a single rational point Q 0 subscript 𝑄 0 Q_{0}italic_Q start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT having rational distances from both O 𝑂 O italic_O and P 𝑃 P italic_P, and note that we can choose Q 0 subscript 𝑄 0 Q_{0}italic_Q start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT outside the line segment joining O 𝑂 O italic_O and P 𝑃 P italic_P. (This first step is easy; see e.g. the lemma below for a much stronger result.) Then, consider the ellipse having foci at O 𝑂 O italic_O and P 𝑃 P italic_P passing through Q 0 subscript 𝑄 0 Q_{0}italic_Q start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT. This conic must contain infinitely many rational points, since it is smooth, defined over ℚ ℚ{\mathbb{Q}}blackboard_Q and contains at least one rational point by construction; for almost every point on this ellipse, so in particular for almost every rational point, there exists also a confocal hyperbola passing through it. By what we have just said, each rational point on each such hyperbola is a point with rational distance from both O 𝑂 O italic_O and P 𝑃 P italic_P. This reproves Theorem 3.1. Report issue for preceding element In our proof in §3 we followed a different pattern, implicitly based on the following lemma: Report issue for preceding element Lemma 3.2. Report issue for preceding element Let O∈𝔸 2⁢(ℚ)=ℚ 2 𝑂 superscript 𝔸 2 ℚ superscript ℚ 2 O\in{\mathbb{A}}^{2}({\mathbb{Q}})={\mathbb{Q}}^{2}italic_O ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( blackboard_Q ) = blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT be a rational point and l⊂𝔸 2 𝑙 superscript 𝔸 2 l\subset{\mathbb{A}}^{2}italic_l ⊂ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT a line defined over ℚ ℚ{\mathbb{Q}}blackboard_Q not passing through O 𝑂 O italic_O. Then there exist infinitely many rational points Q∈l⁢(ℚ)𝑄 𝑙 ℚ Q\in l({\mathbb{Q}})italic_Q ∈ italic_l ( blackboard_Q ) having rational distance from O 𝑂 O italic_O. Report issue for preceding element Proof. Report issue for preceding element We choose coordinates in such a way that O 𝑂 O italic_O is the origin. We may assume that l 𝑙 l italic_l is defined by y=a⁢x+b 𝑦 𝑎 𝑥 𝑏 y=ax+b italic_y = italic_a italic_x + italic_b, with rational a,b 𝑎 𝑏 a,b italic_a , italic_b, where b≠0 𝑏 0 b\neq 0 italic_b ≠ 0. The squared distance from O 𝑂 O italic_O of a point (x,y)∈l 𝑥 𝑦 𝑙(x,y)\in l( italic_x , italic_y ) ∈ italic_l is given by x 2+(a⁢x+b)2=(a 2+1)⁢x 2+2⁢a⁢b⁢x+b 2 superscript 𝑥 2 superscript 𝑎 𝑥 𝑏 2 superscript 𝑎 2 1 superscript 𝑥 2 2 𝑎 𝑏 𝑥 superscript 𝑏 2 x^{2}+(ax+b)^{2}=(a^{2}+1)x^{2}+2abx+b^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_a italic_x + italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 ) italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_a italic_b italic_x + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. For x=0 𝑥 0 x=0 italic_x = 0 such distance is rational, hence the conic z 2=(a 2+1)⁢x 2+2⁢a⁢b⁢x+b 2 superscript 𝑧 2 superscript 𝑎 2 1 superscript 𝑥 2 2 𝑎 𝑏 𝑥 superscript 𝑏 2 z^{2}=(a^{2}+1)x^{2}+2abx+b^{2}italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 ) italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_a italic_b italic_x + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT has a rational point, and so it has infinitely many such points. (For instance set z=b+t⁢x 𝑧 𝑏 𝑡 𝑥 z=b+tx italic_z = italic_b + italic_t italic_x; after dividing by x≠0 𝑥 0 x\neq 0 italic_x ≠ 0, we get t 2⁢x+2⁢b⁢t=(a 2+1)⁢x+2⁢a⁢b superscript 𝑡 2 𝑥 2 𝑏 𝑡 superscript 𝑎 2 1 𝑥 2 𝑎 𝑏 t^{2}x+2bt=(a^{2}+1)x+2ab italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_x + 2 italic_b italic_t = ( italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 ) italic_x + 2 italic_a italic_b hence x=2⁢b⁢(a−t)⁢(t 2−a 2−1)−1 𝑥 2 𝑏 𝑎 𝑡 superscript superscript 𝑡 2 superscript 𝑎 2 1 1 x=2b(a-t)(t^{2}-a^{2}-1)^{-1}italic_x = 2 italic_b ( italic_a - italic_t ) ( italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 1 ) start_POSTSUPERSCRIPT - 1 end_POSTSUPERSCRIPT is a parametrization in t 𝑡 t italic_t). Report issue for preceding element ∎ Report issue for preceding element To finish the proof of Theorem 3.1, we choose a point Q 0≠P subscript 𝑄 0 𝑃 Q_{0}\neq P italic_Q start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT ≠ italic_P having rational distance from P 𝑃 P italic_P and such that the line l 𝑙 l italic_l joining P 𝑃 P italic_P and Q 0 subscript 𝑄 0 Q_{0}italic_Q start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT does not contain Q 𝑄 Q italic_Q. Note that every rational point in l 𝑙 l italic_l has rational distance from P 𝑃 P italic_P; by the above lemma, infinitely many rational points of l 𝑙 l italic_l have rational distance also from O 𝑂 O italic_O. For each such point Q 𝑄 Q italic_Q, consider the hyperbola having foci in O,P 𝑂 𝑃 O,P italic_O , italic_P passing through Q 𝑄 Q italic_Q. Such hyperbola contains infinitely many rational points, each of them having rational distance from both O 𝑂 O italic_O and P 𝑃 P italic_P. This is the strategy that we followed in the previous paragraph. Report issue for preceding element Yet another variant. We note that we could have exploited Lemma 3.2, working with an infinite family of lines passing through P 𝑃 P italic_P, each of them possessing a rational point with rational distance from P 𝑃 P italic_P (for instance, take the unit circle centered on P 𝑃 P italic_P, which has infinitely many rational point; each of them provides a suitable line); then, by Lemma 3.2 each such line contains infinitely many rational points having rational distance from O 𝑂 O italic_O, producing a dense set of rational points having rational distances from both O 𝑂 O italic_O and P 𝑃 P italic_P. Report issue for preceding element A final geometrical remark. The algebraic surface 𝒳 2 subscript 𝒳 2{\mathcal{X}}{2}caligraphic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is obtained as a degree-4 cover of the (x,y)𝑥 𝑦(x,y)( italic_x , italic_y )-plane, a Galois cover of type (2,2)2 2(2,2)( 2 , 2 ), ramified over the two degenerate conics x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 and (x−a)2+(y−b)2=0 superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 0(x-a)^{2}+(y-b)^{2}=0( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0, (a,b)𝑎 𝑏(a,b)( italic_a , italic_b ) being the coordinates of P 𝑃 P italic_P. A generic conic on the plane lifts to a genus-one curve of the surface 𝒮 2 subscript 𝒮 2{\mathcal{S}}{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. However, the conics of the confocal family 𝒞 k subscript 𝒞 𝑘{\mathcal{C}}{k}caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT (see equation (3.4)) considered in the above proof lift to rational curves on 𝒮 2 subscript 𝒮 2{\mathcal{S}}{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT; this is due to the fact that this family coincides precisely with the family of conics tangent to every component of the ramification divisor, which consists, as we said, in the union of four lines. Due to this tangency, the pre-image of each curve 𝒞 k subscript 𝒞 𝑘{\mathcal{C}}_{k}caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT totally splits into four components, each one being a rational curve. Report issue for preceding element 4. Squared Distances from Three Points Report issue for preceding element In this section we begin the study of the set of points Q∈ℝ 2 𝑄 superscript ℝ 2 Q\in{\mathbb{R}}^{2}italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT that have rational distance from three fixed points, proving Theorem 1.2. Report issue for preceding element Notation. Without loss of generality, we can assume that one of the three fixed points is the origin O 𝑂 O italic_O. We will denote by P=(a,b)𝑃 𝑎 𝑏 P=(a,b)italic_P = ( italic_a , italic_b ) and P′=(a′,b′)superscript 𝑃′superscript 𝑎′superscript 𝑏′P^{\prime}=(a^{\prime},b^{\prime})italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = ( italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) the second and third point. For Q,Q′∈ℝ 2 𝑄 superscript 𝑄′superscript ℝ 2 Q,Q^{\prime}\in{\mathbb{R}}^{2}italic_Q , italic_Q start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT we shall denote by ‖Q‖norm 𝑄\\|Q\\|∥ italic_Q ∥ the usual euclidean length, i.e. ‖Q‖=d⁢(Q,0)norm 𝑄 𝑑 𝑄 0\\|Q\\|=d(Q,0)∥ italic_Q ∥ = italic_d ( italic_Q , 0 ) and by Q.Q′formulae-sequence 𝑄 superscript 𝑄′Q.Q^{\prime}italic_Q . italic_Q start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, or (Q.Q′)formulae-sequence 𝑄 superscript 𝑄′(Q.Q^{\prime})( italic_Q . italic_Q start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ), the associated scalar product. Moreover, we will denote the three distances as follows d 0=‖Q‖=d⁢(O,Q)subscript 𝑑 0 norm 𝑄 𝑑 𝑂 𝑄 d_{0}=\\|Q\\|=d(O,Q)italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT = ∥ italic_Q ∥ = italic_d ( italic_O , italic_Q ), d=‖Q−P‖=d⁢(P,Q)𝑑 norm 𝑄 𝑃 𝑑 𝑃 𝑄 d=\\|Q-P\\|=d(P,Q)italic_d = ∥ italic_Q - italic_P ∥ = italic_d ( italic_P , italic_Q ) and d′=‖Q−P′‖=d⁢(P′,Q)superscript 𝑑′norm 𝑄 superscript 𝑃′𝑑 superscript 𝑃′𝑄 d^{\prime}=\\|Q-P^{\prime}\\|=d(P^{\prime},Q)italic_d start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = ∥ italic_Q - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ = italic_d ( italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_Q ). Report issue for preceding element In order to identify necessary and sufficient conditions for the density of the set of points with rational distances from O,P 𝑂 𝑃 O,P italic_O , italic_P and P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, we consider the set of points Q∈ℝ 2 𝑄 superscript ℝ 2 Q\in{\mathbb{R}}^{2}italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT such that the squared distances d 0 2=‖Q‖2,d 2=‖Q−P‖2,d′⁣2=‖Q−P′‖2 formulae-sequence superscript subscript 𝑑 0 2 superscript norm 𝑄 2 formulae-sequence superscript 𝑑 2 superscript norm 𝑄 𝑃 2 superscript 𝑑′2 superscript norm 𝑄 superscript 𝑃′2 d_{0}^{2}=\\|Q\\|^{2},d^{2}=\\|Q-P\\|^{2},d^{\prime 2}=\\|Q-P^{\prime}\\|^{2}italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ∥ italic_Q ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ∥ italic_Q - italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT = ∥ italic_Q - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are all rational. Report issue for preceding element In particular we suppose neither that P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, or Q 𝑄 Q italic_Q, are rational points, nor that the distances between these points are rational. The output of this analysis, which is simple but apparently missing from previous papers in the topic, will lead to results holding essentially without restrictions. Report issue for preceding element We shall study conditions which ensure that the set Report issue for preceding element J:={Q∈ℝ 2:d 0 2,d 2,d′⁣2∈ℚ}.assign 𝐽 conditional-set 𝑄 superscript ℝ 2 superscript subscript 𝑑 0 2 superscript 𝑑 2 superscript 𝑑′2 ℚ J:={Q\in{\mathbb{R}}^{2}:d_{0}^{2},d^{2},d^{\prime 2}\in{\mathbb{Q}}}.italic_J := { italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT : italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ∈ blackboard_Q } .(4.1) is Zariski-dense in the plane ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. These are necessary conditions to hold in order for the set of points with rational distance from O,P 𝑂 𝑃 O,P italic_O , italic_P and P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT to be Zariski dense. Report issue for preceding element Remark 4.1. Report issue for preceding element It will follow from the analysis below that it suffices to assume that J 𝐽 J italic_J is not contained in a finite union of conics, in order to achieve the same conclusions. Report issue for preceding element Setting Q:=(x,y)assign 𝑄 𝑥 𝑦 Q:=(x,y)italic_Q := ( italic_x , italic_y ) we have that a point Q∈J 𝑄 𝐽 Q\in J italic_Q ∈ italic_J satisfies Report issue for preceding element x 2+y 2=d 0 2 superscript 𝑥 2 superscript 𝑦 2 superscript subscript 𝑑 0 2\displaystyle x^{2}+y^{2}=d_{0}^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(4.2) (x−a)2+(y−b)2=d 2 superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 superscript 𝑑 2\displaystyle(x-a)^{2}+(y-b)^{2}=d^{2}( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_d start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(4.3) (x−a′)2+(y−b′)2=d′⁣2,superscript 𝑥 superscript 𝑎′2 superscript 𝑦 superscript 𝑏′2 superscript 𝑑′2\displaystyle(x-a^{\prime})^{2}+(y-b^{\prime})^{2}=d^{\prime 2},( italic_x - italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_d start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ,(4.4) and, by construction of J 𝐽 J italic_J, we presently assume that the right hand sides belong to ℚ ℚ{\mathbb{Q}}blackboard_Q. Report issue for preceding element By subtraction, setting g:=d 0 2−d 2 assign 𝑔 superscript subscript 𝑑 0 2 superscript 𝑑 2 g:=d_{0}^{2}-d^{2}italic_g := italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_d start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, g′=d 0 2−d′⁣2 superscript 𝑔′superscript subscript 𝑑 0 2 superscript 𝑑′2 g^{\prime}=d_{0}^{2}-d^{\prime 2}italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_d start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT, we get Report issue for preceding element 2⁢a⁢x+2⁢b⁢y=g+\|P\|2,2⁢a′⁢x+2⁢b′⁢y=g′+\|P′\|2.formulae-sequence 2 𝑎 𝑥 2 𝑏 𝑦 𝑔 superscript 𝑃 2 2 superscript 𝑎′𝑥 2 superscript 𝑏′𝑦 superscript 𝑔′superscript superscript 𝑃′2 2ax+2by=g+\|P\|^{2},\qquad 2a^{\prime}x+2b^{\prime}y=g^{\prime}+\|P^{\prime}\|^{2}.2 italic_a italic_x + 2 italic_b italic_y = italic_g + \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , 2 italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_x + 2 italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_y = italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT + \| italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT .(4.5) 4.1. The collinear case Report issue for preceding element Suppose first that O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are collinear, so P′=q⁢P superscript 𝑃′𝑞 𝑃 P^{\prime}=qP italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_q italic_P for a q∈ℝ∗∖{1}𝑞 superscript ℝ 1 q\in{\mathbb{R}}^{}\setminus{1}italic_q ∈ blackboard_R start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ∖ { 1 }. Then Report issue for preceding element g′=q⁢g+(q⁢‖P‖2−‖P′‖2)=q⁢g+q⁢(1−q)⁢‖P‖2.superscript 𝑔′𝑞 𝑔 𝑞 superscript norm 𝑃 2 superscript norm superscript 𝑃′2 𝑞 𝑔 𝑞 1 𝑞 superscript norm 𝑃 2 g^{\prime}=qg+(q\\|P\\|^{2}-\\|P^{\prime}\\|^{2})=qg+q(1-q)\\|P\\|^{2}.italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_q italic_g + ( italic_q ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) = italic_q italic_g + italic_q ( 1 - italic_q ) ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . Hence (g,g′)𝑔 superscript 𝑔′(g,g^{\prime})( italic_g , italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) lies on a certain fixed line, not parallel to any of the axes. Since P≠O 𝑃 𝑂 P\neq O italic_P ≠ italic_O, if J 𝐽 J italic_J is Zariski-dense the set of such (g,g′)𝑔 superscript 𝑔′(g,g^{\prime})( italic_g , italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) is Zariski-dense on the line. Hence the line is a rational line, which amounts to q∈ℚ 𝑞 ℚ q\in{\mathbb{Q}}italic_q ∈ blackboard_Q, \|P\|2∈ℚ superscript 𝑃 2 ℚ\|P\|^{2}\in{\mathbb{Q}}\| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q. Report issue for preceding element Furthermore we can show that these conditions are also sufficient. Report issue for preceding element Proposition 4.2. Report issue for preceding element Suppose that O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are collinear: P′=q⁢P superscript 𝑃′𝑞 𝑃 P^{\prime}=qP italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_q italic_P, with q≠0,1 𝑞 0 1 q\neq 0,1 italic_q ≠ 0 , 1. Then J 𝐽 J italic_J is Zariski-dense if and only if p:=\|P\|2∈ℚ assign 𝑝 superscript 𝑃 2 ℚ p:=\|P\|^{2}\in{\mathbb{Q}}italic_p := \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q and q∈ℚ 𝑞 ℚ q\in{\mathbb{Q}}italic_q ∈ blackboard_Q. Under these assumptions, if R=(−b,a)𝑅 𝑏 𝑎 R=(-b,a)italic_R = ( - italic_b , italic_a ) and Q=t⁢P+u⁢R 𝑄 𝑡 𝑃 𝑢 𝑅 Q=tP+uR italic_Q = italic_t italic_P + italic_u italic_R, where t=(P.Q)/p t=(P.Q)/p italic_t = ( italic_P . italic_Q ) / italic_p and u=(R.Q)/p u=(R.Q)/p italic_u = ( italic_R . italic_Q ) / italic_p, the necessary and sufficient conditions for Q 𝑄 Q italic_Q to be in J 𝐽 J italic_J are that t∈ℚ 𝑡 ℚ t\in{\mathbb{Q}}italic_t ∈ blackboard_Q and u 2∈ℚ superscript 𝑢 2 ℚ u^{2}\in{\mathbb{Q}}italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q. Report issue for preceding element Proof. Report issue for preceding element The above discussion already proved the necessity. Report issue for preceding element Conversely, let us assume that q,‖P‖2∈ℚ 𝑞 superscript norm 𝑃 2 ℚ q,\\|P\\|^{2}\in{\mathbb{Q}}italic_q , ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q (where P′=q⁢P superscript 𝑃′𝑞 𝑃 P^{\prime}=qP italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_q italic_P). First of all if d 0 2∈ℚ superscript subscript 𝑑 0 2 ℚ d_{0}^{2}\in{\mathbb{Q}}italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q and if the first of 4.5 holds for a g∈ℚ 𝑔 ℚ g\in{\mathbb{Q}}italic_g ∈ blackboard_Q, then the g′superscript 𝑔′g^{\prime}italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT obtained by the second of (4.5) will be rational as well, and (x,y)𝑥 𝑦(x,y)( italic_x , italic_y ) will belong to J 𝐽 J italic_J. Hence we need only study (4.2) and (4.3) in rationals d 0 2,d 2 superscript subscript 𝑑 0 2 superscript 𝑑 2 d_{0}^{2},d^{2}italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. In turn, this is like imposing that both x 2+y 2=d 0 2 superscript 𝑥 2 superscript 𝑦 2 superscript subscript 𝑑 0 2 x^{2}+y^{2}=d_{0}^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and a x+b y=:σ ax+by=:\sigma italic_a italic_x + italic_b italic_y = : italic_σ are rational, where we are assuming that a 2+b 2 superscript 𝑎 2 superscript 𝑏 2 a^{2}+b^{2}italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is rational. Now, for given (d 0 2,σ)∈ℚ 2 superscript subscript 𝑑 0 2 𝜎 superscript ℚ 2(d_{0}^{2},\sigma)\in{\mathbb{Q}}^{2}( italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_σ ) ∈ blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT we can solve for x,y 𝑥 𝑦 x,y italic_x , italic_y, so the set J 𝐽 J italic_J is indeed Zariski-dense. Specifically, we obtain the equations for x,y 𝑥 𝑦 x,y italic_x , italic_y given by Report issue for preceding element \|P\|2⁢x 2−2⁢a⁢σ⁢x+σ 2−b 2⁢d 0 2=0,\|P\|2⁢y 2−2⁢b⁢σ⁢y+σ 2−a 2⁢d 0 2=0.formulae-sequence superscript 𝑃 2 superscript 𝑥 2 2 𝑎 𝜎 𝑥 superscript 𝜎 2 superscript 𝑏 2 superscript subscript 𝑑 0 2 0 superscript 𝑃 2 superscript 𝑦 2 2 𝑏 𝜎 𝑦 superscript 𝜎 2 superscript 𝑎 2 superscript subscript 𝑑 0 2 0\|P\|^{2}x^{2}-2a\sigma x+\sigma^{2}-b^{2}d_{0}^{2}=0,\qquad\|P\|^{2}y^{2}-2b% \sigma y+\sigma^{2}-a^{2}d_{0}^{2}=0.\| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 italic_a italic_σ italic_x + italic_σ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 , \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 italic_b italic_σ italic_y + italic_σ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 . Finally, if we let R=(−b,a)𝑅 𝑏 𝑎 R=(-b,a)italic_R = ( - italic_b , italic_a ), we have that ‖R‖2=‖P‖2 superscript norm 𝑅 2 superscript norm 𝑃 2\\|R\\|^{2}=\\|P\\|^{2}∥ italic_R ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is rational and (P.R)=0(P.R)=0( italic_P . italic_R ) = 0. Given Q=t⁢P+u⁢R 𝑄 𝑡 𝑃 𝑢 𝑅 Q=tP+uR italic_Q = italic_t italic_P + italic_u italic_R for t,u∈ℝ 𝑡 𝑢 ℝ t,u\in{\mathbb{R}}italic_t , italic_u ∈ blackboard_R, we have that \|Q\|2 superscript 𝑄 2\|Q\|^{2}\| italic_Q \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and (P.Q)formulae-sequence 𝑃 𝑄(P.Q)( italic_P . italic_Q ) have both to be rational. In this case, we have \|Q\|2=(t 2+u 2)⁢\|P\|2 superscript 𝑄 2 superscript 𝑡 2 superscript 𝑢 2 superscript 𝑃 2\|Q\|^{2}=(t^{2}+u^{2})\|P\|^{2}\| italic_Q \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and (P.Q)=t∥P∥2(P.Q)=t\\|P\\|^{2}( italic_P . italic_Q ) = italic_t ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, thus obtaining the desired result. ∎ Report issue for preceding element Remark 4.3. Report issue for preceding element If we require that for instance the points in J∩ℚ 2 𝐽 superscript ℚ 2 J\cap{\mathbb{Q}}^{2}italic_J ∩ blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are still Zariski-dense then we find that P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT must be rational points as well; and this condition is also sufficient. Report issue for preceding element 4.2. The non-collinear case. Proof of Theorem 1.2 Report issue for preceding element Suppose now that 0,P,P′0 𝑃 superscript 𝑃′0,P,P^{\prime}0 , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are not collinear, so Δ:=a⁢b′−a′⁢b≠0 assign Δ 𝑎 superscript 𝑏′superscript 𝑎′𝑏 0\Delta:=ab^{\prime}-a^{\prime}b\neq 0 roman_Δ := italic_a italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_b ≠ 0. Report issue for preceding element Now, if J 𝐽 J italic_J is Zariski-dense we deduce from (4.5) that the set Γ⊂ℚ 2 Γ superscript ℚ 2\Gamma\subset{\mathbb{Q}}^{2}roman_Γ ⊂ blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT of corresponding pairs (g,g′)𝑔 superscript 𝑔′(g,g^{\prime})( italic_g , italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) is also Zariski-dense. Report issue for preceding element Solving the equations (4.5) for x,y 𝑥 𝑦 x,y italic_x , italic_y and inserting the values so obtained in (4.2) we obtain a condition of the form Φ⁢(g,g′)∈ℚ Φ 𝑔 superscript 𝑔′ℚ\Phi(g,g^{\prime})\in{\mathbb{Q}}roman_Φ ( italic_g , italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) ∈ blackboard_Q, where Φ Φ\Phi roman_Φ is a certain nonzero quadratic polynomial depending only on P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. More precisely we have Report issue for preceding element x=A⁢(g+‖P‖2)+B⁢(g′+‖P′‖2),y=C⁢(g+‖P‖2)+D⁢(g′+‖P′‖2),formulae-sequence 𝑥 𝐴 𝑔 superscript norm 𝑃 2 𝐵 superscript 𝑔′superscript norm superscript 𝑃′2 𝑦 𝐶 𝑔 superscript norm 𝑃 2 𝐷 superscript 𝑔′superscript norm superscript 𝑃′2 x=A(g+\\|P\\|^{2})+B(g^{\prime}+\\|P^{\prime}\\|^{2}),\qquad y=C(g+\\|P\\|^{2})+D(g^% {\prime}+\\|P^{\prime}\\|^{2}),italic_x = italic_A ( italic_g + ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_B ( italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT + ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) , italic_y = italic_C ( italic_g + ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + italic_D ( italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT + ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ,(4.6) where Report issue for preceding element 2⁢Δ⁢A=b′,2⁢Δ⁢B=b,2⁢Δ⁢C=−a′,2⁢Δ⁢D=−a.formulae-sequence 2 Δ 𝐴 superscript 𝑏′formulae-sequence 2 Δ 𝐵 𝑏 formulae-sequence 2 Δ 𝐶 superscript 𝑎′2 Δ 𝐷 𝑎 2\Delta A=b^{\prime},\quad 2\Delta B=b,\quad 2\Delta C=-a^{\prime},\quad 2% \Delta D=-a.2 roman_Δ italic_A = italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , 2 roman_Δ italic_B = italic_b , 2 roman_Δ italic_C = - italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , 2 roman_Δ italic_D = - italic_a .(4.7) Setting 𝐯:=(A,C)assign 𝐯 𝐴 𝐶{\bf v}:=(A,C)bold_v := ( italic_A , italic_C ), 𝐰:=(B,D)assign 𝐰 𝐵 𝐷{\bf w}:=(B,D)bold_w := ( italic_B , italic_D ), we then have Report issue for preceding element Φ(u,v)=∥𝐯∥2 u 2+2(𝐯.𝐰)u v+∥𝐰∥2 v 2+E u+F v+G,\Phi(u,v)=\\|{\bf v}\\|^{2}u^{2}+2({\bf v}.{\bf w})uv+\\|{\bf w}\\|^{2}v^{2}+Eu+Fv% +G,roman_Φ ( italic_u , italic_v ) = ∥ bold_v ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 ( bold_v . bold_w ) italic_u italic_v + ∥ bold_w ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_E italic_u + italic_F italic_v + italic_G ,(4.8) where Report issue for preceding element E=2∥𝐯∥2∥P∥2+2(𝐯.𝐰)∥P′∥2,F=2∥𝐰∥2∥P′∥2+2(𝐯.𝐰)∥P∥2,G=∥𝐯∥2∥P∥2+∥𝐰∥2∥P′∥2.E=2\\|{\bf v}\\|^{2}\\|P\\|^{2}+2({\bf v}.{\bf w})\\|P^{\prime}\\|^{2},\quad F=2\\|{% \bf w}\\|^{2}\\|P^{\prime}\\|^{2}+2({\bf v}.{\bf w})\\|P\\|^{2},\quad G=\\|{\bf v}\\|% ^{2}\\|P\\|^{2}+\\|{\bf w}\\|^{2}\\|P^{\prime}\\|^{2}.italic_E = 2 ∥ bold_v ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 ( bold_v . bold_w ) ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_F = 2 ∥ bold_w ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 ( bold_v . bold_w ) ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_G = ∥ bold_v ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ∥ bold_w ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . Now, if the vectors (g 2,g⁢g′,g′⁣2,g,g′,1)∈ℚ 6 superscript 𝑔 2 𝑔 superscript 𝑔′superscript 𝑔′2 𝑔 superscript 𝑔′1 superscript ℚ 6(g^{2},gg^{\prime},g^{\prime 2},g,g^{\prime},1)\in{\mathbb{Q}}^{6}( italic_g start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_g italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_g start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT , italic_g , italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , 1 ) ∈ blackboard_Q start_POSTSUPERSCRIPT 6 end_POSTSUPERSCRIPT lie in a proper vector subspace of ℚ 6 superscript ℚ 6{\mathbb{Q}}^{6}blackboard_Q start_POSTSUPERSCRIPT 6 end_POSTSUPERSCRIPT (for varying Q 𝑄 Q italic_Q in J 𝐽 J italic_J) then a fixed equation Ψ⁢(g,g′)=0 Ψ 𝑔 superscript 𝑔′0\Psi(g,g^{\prime})=0 roman_Ψ ( italic_g , italic_g start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = 0 would hold, Ψ Ψ\Psi roman_Ψ being some nonzero quadratic polynomial, and, by (4.5), J 𝐽 J italic_J would not be Zariski-dense. Report issue for preceding element Therefore the vectors span ℚ 6 superscript ℚ 6{\mathbb{Q}}^{6}blackboard_Q start_POSTSUPERSCRIPT 6 end_POSTSUPERSCRIPT, and we deduce that Φ Φ\Phi roman_Φ must have rational coefficients. In the first place this implies that the three quantities ∥P∥2/Δ 2,(P.P′)/Δ 2,∥P′∥2/Δ 2\\|P\\|^{2}/\Delta^{2},(P.P^{\prime})/\Delta^{2},\\|P^{\prime}\\|^{2}/\Delta^{2}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) / roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT (which are essentially the coefficients of the homogeneous quadratic part of Φ Φ\Phi roman_Φ) are rational. Now, let us look at E,F 𝐸 𝐹 E,F italic_E , italic_F, which may be seen as linear forms in ‖P‖2,‖P′‖2 superscript norm 𝑃 2 superscript norm superscript 𝑃′2\\|P\\|^{2},\\|P^{\prime}\\|^{2}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT with rational coefficients. The matrix of these linear forms up to a factor 2 2 2 2 has rows (∥𝐯∥2\|,𝐯.𝐰)(\\|{\bf v}\\|^{2}\|,{\bf v}.{\bf w})( ∥ bold_v ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT \| , bold_v . bold_w ) and (‖𝐰‖2,𝐯.𝐰)formulae-sequence superscript norm 𝐰 2 𝐯 𝐰(\\|{\bf w}\\|^{2},{\bf v}.{\bf w})( ∥ bold_w ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , bold_v . bold_w ), which is the product of the matrix with rows 𝐯,𝐰 𝐯 𝐰{\bf v},{\bf w}bold_v , bold_w with its transpose (𝐯 t,𝐰 t)superscript 𝐯 𝑡 superscript 𝐰 𝑡({\bf v}^{t},{\bf w}^{t})( bold_v start_POSTSUPERSCRIPT italic_t end_POSTSUPERSCRIPT , bold_w start_POSTSUPERSCRIPT italic_t end_POSTSUPERSCRIPT ). But this product matrix is rational, and it is also nonsingular, the determinant of (𝐯 t,𝐰 t)superscript 𝐯 𝑡 superscript 𝐰 𝑡({\bf v}^{t},{\bf w}^{t})( bold_v start_POSTSUPERSCRIPT italic_t end_POSTSUPERSCRIPT , bold_w start_POSTSUPERSCRIPT italic_t end_POSTSUPERSCRIPT ) being (4⁢Δ)−1 superscript 4 Δ 1(4\Delta)^{-1}( 4 roman_Δ ) start_POSTSUPERSCRIPT - 1 end_POSTSUPERSCRIPT. Hence \|P\|2,\|P′\|2 superscript 𝑃 2 superscript superscript 𝑃′2\|P\|^{2},\|P^{\prime}\|^{2}\| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , \| italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are rational, whence Δ 2 superscript Δ 2\Delta^{2}roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is rational, and therefore (P.P′)formulae-sequence 𝑃 superscript 𝑃′(P.P^{\prime})( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) is rational as well. Report issue for preceding element Note that these conditions amount simply to P,P′∈J 𝑃 superscript 𝑃′𝐽 P,P^{\prime}\in J italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ italic_J; in fact, this is equivalent to ‖P‖2,‖P′‖2,‖P−P′‖2 superscript norm 𝑃 2 superscript norm superscript 𝑃′2 superscript norm 𝑃 superscript 𝑃′2\\|P\\|^{2},\\|P^{\prime}\\|^{2},\\|P-P^{\prime}\\|^{2}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ∥ italic_P - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT to be all rational, and the claim follows in view of the equality ∥P−P′∥2=∥P∥2−2(P.P′)+∥P′∥2\\|P-P^{\prime}\\|^{2}=\\|P\\|^{2}-2(P.P^{\prime})+\\|P^{\prime}\\|^{2}∥ italic_P - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) + ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Report issue for preceding element Remark 4.4. Report issue for preceding element We note in passing that the rationality of ∥P∥2,(P.P′),∥P′∥2\\|P\\|^{2},(P.P^{\prime}),\\|P^{\prime}\\|^{2}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT implies itself that of Δ 2 superscript Δ 2\Delta^{2}roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. In fact is α 𝛼\alpha italic_α is the angle P⁢O⁢P′𝑃 𝑂 superscript 𝑃′POP^{\prime}italic_P italic_O italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT we have (P.P′)=∥P∥∥P′∥cos α(P.P^{\prime})=\\|P\\|\\|P^{\prime}\\|\cos\alpha( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = ∥ italic_P ∥ ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ roman_cos italic_α, Δ 2=∥P∥2∥P′∥2 sin 2 α=∥P∥2∥P′∥2−(P.P′)2\Delta^{2}=\\|P\\|^{2}\\|P^{\prime}\\|^{2}\sin^{2}\alpha=\\|P\\|^{2}\\|P^{\prime}\\|^{% 2}-(P.P^{\prime})^{2}roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT roman_sin start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_α = ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Report issue for preceding element Conversely, suppose that ∥P∥2,(P.P′),∥P′∥2∈ℚ\\|P\\|^{2},(P.P^{\prime}),\\|P^{\prime}\\|^{2}\in{\mathbb{Q}}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q. Then by Remark 4.4 necessarily Δ 2∈ℚ superscript Δ 2 ℚ\Delta^{2}\in{\mathbb{Q}}roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q and the above argument can be reversed, giving a Zariski-dense set of points Q 𝑄 Q italic_Q with rational squared-distance from O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. Report issue for preceding element We now summarize these conclusions and describe in a simple way the set J 𝐽 J italic_J. Report issue for preceding element Proposition 4.5. Report issue for preceding element Suppose that 0,P,P′0 𝑃 superscript 𝑃′0,P,P^{\prime}0 , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are not collinear. If J 𝐽 J italic_J is Zariski-dense, then ∥P∥2,(P.P′),∥P′∥2\\|P\\|^{2},(P.P^{\prime}),\\|P^{\prime}\\|^{2}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are rational, or equivalently P,P′∈J 𝑃 superscript 𝑃′𝐽 P,P^{\prime}\in J italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ italic_J, and this condition is also sufficient. Also, if this holds we have J=ℚ⁢P+ℚ⁢P′𝐽 ℚ 𝑃 ℚ superscript 𝑃′J={\mathbb{Q}}P+{\mathbb{Q}}P^{\prime}italic_J = blackboard_Q italic_P + blackboard_Q italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, and actually J 𝐽 J italic_J is an orthogonal sum J=ℚ⁢P+ℚ⁢R 𝐽 ℚ 𝑃 ℚ 𝑅 J={\mathbb{Q}}P+{\mathbb{Q}}R italic_J = blackboard_Q italic_P + blackboard_Q italic_R for a suitable R∈J 𝑅 𝐽 R\in J italic_R ∈ italic_J with (P.R)=0(P.R)=0( italic_P . italic_R ) = 0, ‖R‖2∈ℚ superscript norm 𝑅 2 ℚ\\|R\\|^{2}\in{\mathbb{Q}}∥ italic_R ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q. Finally, the squared-distance between any two points in J 𝐽 J italic_J is rational. Report issue for preceding element Proof. Report issue for preceding element Let L:=ℚ⁢P+ℚ⁢P′assign 𝐿 ℚ 𝑃 ℚ superscript 𝑃′L:={\mathbb{Q}}P+{\mathbb{Q}}P^{\prime}italic_L := blackboard_Q italic_P + blackboard_Q italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, a vector space of dimension 2 2 2 2 over ℚ ℚ{\mathbb{Q}}blackboard_Q, contained in ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Since ∥P∥2,(P.P′)∈ℚ\\|P\\|^{2},(P.P^{\prime})\in{\mathbb{Q}}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) ∈ blackboard_Q, we may find a nonzero vector R∈L 𝑅 𝐿 R\in L italic_R ∈ italic_L with (P.R)=0(P.R)=0( italic_P . italic_R ) = 0: it suffices to put R:=P′−(P.P′)‖P‖2⁢P assign 𝑅 superscript 𝑃′formulae-sequence 𝑃 superscript 𝑃′superscript norm 𝑃 2 𝑃 R:=P^{\prime}-\frac{(P.P^{\prime})}{\\|P\\|^{2}}P italic_R := italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - divide start_ARG ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG start_ARG ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG italic_P. Then L=ℚ⁢P+ℚ⁢R 𝐿 ℚ 𝑃 ℚ 𝑅 L={\mathbb{Q}}P+{\mathbb{Q}}R italic_L = blackboard_Q italic_P + blackboard_Q italic_R. Now, if Q∈L 𝑄 𝐿 Q\in L italic_Q ∈ italic_L, we have Q=t⁢P+u⁢R 𝑄 𝑡 𝑃 𝑢 𝑅 Q=tP+uR italic_Q = italic_t italic_P + italic_u italic_R with rational t,u 𝑡 𝑢 t,u italic_t , italic_u and then \|Q\|2=t 2⁢\|P\|2+u 2⁢\|R\|2∈ℚ superscript 𝑄 2 superscript 𝑡 2 superscript 𝑃 2 superscript 𝑢 2 superscript 𝑅 2 ℚ\|Q\|^{2}=t^{2}\|P\|^{2}+u^{2}\|R\|^{2}\in{\mathbb{Q}}\| italic_Q \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT \| italic_R \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q, and similarly \|Q−S\|2∈ℚ superscript 𝑄 𝑆 2 ℚ\|Q-S\|^{2}\in{\mathbb{Q}}\| italic_Q - italic_S \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q for every S∈L 𝑆 𝐿 S\in L italic_S ∈ italic_L. Also, if Q∈J 𝑄 𝐽 Q\in J italic_Q ∈ italic_J, namely if ‖Q‖2,‖Q−P‖2,‖Q−P′‖2∈ℚ superscript norm 𝑄 2 superscript norm 𝑄 𝑃 2 superscript norm 𝑄 superscript 𝑃′2 ℚ\\|Q\\|^{2},\\|Q-P\\|^{2},\\|Q-P^{\prime}\\|^{2}\in{\mathbb{Q}}∥ italic_Q ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ∥ italic_Q - italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ∥ italic_Q - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q, we have that (Q.P),(Q.P′)∈ℚ(Q.P),(Q.P^{\prime})\in{\mathbb{Q}}( italic_Q . italic_P ) , ( italic_Q . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) ∈ blackboard_Q, whence t,u∈ℚ 𝑡 𝑢 ℚ t,u\in{\mathbb{Q}}italic_t , italic_u ∈ blackboard_Q, so Q∈L 𝑄 𝐿 Q\in L italic_Q ∈ italic_L as wanted. The last assertion also follows. ∎ Report issue for preceding element Corollary 4.6. Report issue for preceding element If J 𝐽 J italic_J is Zariski-dense and contains two non collinear points in ℚ 2 superscript ℚ 2{\mathbb{Q}}^{2}blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, then J=ℚ 2 𝐽 superscript ℚ 2 J={\mathbb{Q}}^{2}italic_J = blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Report issue for preceding element Proof. Report issue for preceding element Indeed, under these assumptions P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT cannot be collinear with the origin, and we may apply the proposition. Then J 𝐽 J italic_J has a ℚ ℚ{\mathbb{Q}}blackboard_Q-basis formed of rational points, whence the conclusion. ∎ Report issue for preceding element Proof of the addendum to Theorem1.2. Let P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT be a triangle satisfying condition (ii) of Theorem1.2. Consider the vectors P 1⁢P 3→,P 2⁢P 3→→subscript 𝑃 1 subscript 𝑃 3→subscript 𝑃 2 subscript 𝑃 3\overrightarrow{P_{1}P_{3}},\overrightarrow{P_{2}P_{3}}over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG , over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG and suppose they form an oriented basis of ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Consider the quadratic form (x,y)↦‖x⁢P 1⁢P 3→+y⁢P 2⁢P 3→‖2 maps-to 𝑥 𝑦 superscript norm 𝑥→subscript 𝑃 1 subscript 𝑃 3 𝑦→subscript 𝑃 2 subscript 𝑃 3 2(x,y)\mapsto\\|x\overrightarrow{P_{1}P_{3}}+y\overrightarrow{P_{2}P_{3}}\\|^{2}( italic_x , italic_y ) ↦ ∥ italic_x over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG + italic_y over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. We want to prove that it is defined over ℚ ℚ{\mathbb{Q}}blackboard_Q if and only if condition (iv) of the Addendum is satisfied. Report issue for preceding element Since the euclidean quadratic form is invariant under the orthogonal group O⁢(2)O 2\mathrm{O}(2)roman_O ( 2 ), we can act with a matrix A∈O⁢(2)𝐴 O 2 A\in\mathrm{O}(2)italic_A ∈ roman_O ( 2 ) in such a way that the vector P 1⁢P 3→→subscript 𝑃 1 subscript 𝑃 3\overrightarrow{P_{1}P_{3}}over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG takes the form (a,0)t{}^{t}(a,0)start_FLOATSUPERSCRIPT italic_t end_FLOATSUPERSCRIPT ( italic_a , 0 ), where a>0 𝑎 0 a>0 italic_a > 0, without changing the quadratic form (x,y)↦‖x⁢P 1⁢P 3→+y⁢P 2⁢P 3→‖2 maps-to 𝑥 𝑦 superscript norm 𝑥→subscript 𝑃 1 subscript 𝑃 3 𝑦→subscript 𝑃 2 subscript 𝑃 3 2(x,y)\mapsto\\|x\overrightarrow{P_{1}P_{3}}+y\overrightarrow{P_{2}P_{3}}\\|^{2}( italic_x , italic_y ) ↦ ∥ italic_x over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG + italic_y over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. The second vector P 2⁢P 3→→subscript 𝑃 2 subscript 𝑃 3\overrightarrow{P_{2}P_{3}}over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG will be of the form (u,b)t{}^{t}(u,b)start_FLOATSUPERSCRIPT italic_t end_FLOATSUPERSCRIPT ( italic_u , italic_b ) with b>0 𝑏 0 b>0 italic_b > 0. The quadratic form reads Report issue for preceding element a 2⁢x 2+2⁢a⁢u⁢x⁢y+(u 2+b 2)⁢y 2.superscript 𝑎 2 superscript 𝑥 2 2 𝑎 𝑢 𝑥 𝑦 superscript 𝑢 2 superscript 𝑏 2 superscript 𝑦 2 a^{2}x^{2}+2auxy+(u^{2}+b^{2})y^{2}.italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_a italic_u italic_x italic_y + ( italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . Its rationality implies that a 𝑎 a italic_a is of the form a=r 𝑎 𝑟 a=\sqrt{r}italic_a = square-root start_ARG italic_r end_ARG, for some rational r>0 𝑟 0 r>0 italic_r > 0 and u=r⋅ξ 𝑢⋅𝑟 𝜉 u=\sqrt{r}\cdot\xi italic_u = square-root start_ARG italic_r end_ARG ⋅ italic_ξ for some ξ∈ℚ 𝜉 ℚ\xi\in{\mathbb{Q}}italic_ξ ∈ blackboard_Q; then b 2∈ℚ superscript 𝑏 2 ℚ b^{2}\in{\mathbb{Q}}italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q, so we can write b=s 𝑏 𝑠 b=\sqrt{s}italic_b = square-root start_ARG italic_s end_ARG for some rational number s>0 𝑠 0 s>0 italic_s > 0; so the new matrix (P 1⁢P 3→,P 2⁢P 3→)→subscript 𝑃 1 subscript 𝑃 3→subscript 𝑃 2 subscript 𝑃 3(\overrightarrow{P_{1}P_{3}},\overrightarrow{P_{2}P_{3}})( over→ start_ARG italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG , over→ start_ARG italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_ARG ) takes the form (r r⁢ξ 0 s)matrix 𝑟 𝑟 𝜉 0 𝑠\left(\begin{matrix}\sqrt{r}&\sqrt{r}\xi\ 0&\sqrt{s}\end{matrix}\right)( start_ARG start_ROW start_CELL square-root start_ARG italic_r end_ARG end_CELL start_CELL square-root start_ARG italic_r end_ARG italic_ξ end_CELL end_ROW start_ROW start_CELL 0 end_CELL start_CELL square-root start_ARG italic_s end_ARG end_CELL end_ROW end_ARG ). After multiplying this matrix by the inverse of the diagonal matrix D=(r 0 0 s)𝐷 matrix 𝑟 0 0 𝑠 D=\left(\begin{matrix}\sqrt{r}&0\ 0&\sqrt{s}\end{matrix}\right)italic_D = ( start_ARG start_ROW start_CELL square-root start_ARG italic_r end_ARG end_CELL start_CELL 0 end_CELL end_ROW start_ROW start_CELL 0 end_CELL start_CELL square-root start_ARG italic_s end_ARG end_CELL end_ROW end_ARG ) we obtain a unipotent upper triangular matrix whose upper-right entry is the rational number ξ 𝜉\xi italic_ξ as wanted. Report issue for preceding element 5. Rational distances from three points Report issue for preceding element Using the analysis from the previous section we consider now the set Report issue for preceding element J∗={Q∈ℝ 2:d 0:=‖Q‖∈ℚ,d:=‖Q−P‖∈ℚ,d′:=‖Q−P′‖∈ℚ}.superscript 𝐽 conditional-set 𝑄 superscript ℝ 2 formulae-sequence assign subscript 𝑑 0 norm 𝑄 ℚ assign 𝑑 norm 𝑄 𝑃 ℚ assign superscript 𝑑′norm 𝑄 superscript 𝑃′ℚ J^{}={Q\in{\mathbb{R}}^{2}:d_{0}:=\\|Q\\|\in{\mathbb{Q}},d:=\\|Q-P\\|\in{\mathbb% {Q}},d^{\prime}:=\\|Q-P^{\prime}\\|\in{\mathbb{Q}}}.italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT = { italic_Q ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT : italic_d start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT := ∥ italic_Q ∥ ∈ blackboard_Q , italic_d := ∥ italic_Q - italic_P ∥ ∈ blackboard_Q , italic_d start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT := ∥ italic_Q - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ ∈ blackboard_Q } . As before, the points P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are distinct points in the plane, which are also distinct from the origin O 𝑂 O italic_O. Report issue for preceding element Of course J∗⊂J superscript 𝐽 𝐽 J^{}\subset J italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ⊂ italic_J (see Definition 4.1), and in fact a priori J∗superscript 𝐽 J^{}italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT is a much smaller set. While the density of J 𝐽 J italic_J is a necessary condition for the density of J∗superscript 𝐽 J^{}italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT we will show that if J 𝐽 J italic_J is Zariski-dense then J∗superscript 𝐽 J^{}italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT too is Zariski-dense, provided certain very simple necessary conditions hold. In particular this will extend the main theorem of by providing the optimal set of necessary conditions that guarantee the density of points with rational distances from the given points O,P 𝑂 𝑃 O,P italic_O , italic_P and P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. Report issue for preceding element 5.1. The collinear case Report issue for preceding element Let us start again with the collinear case. We put p:=‖P‖2 assign 𝑝 superscript norm 𝑃 2 p:=\\|P\\|^{2}italic_p := ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, P′=q⁢P superscript 𝑃′𝑞 𝑃 P^{\prime}=qP italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_q italic_P, and we assume that p,q∈ℚ∗𝑝 𝑞 superscript ℚ p,q\in{\mathbb{Q}}^{}italic_p , italic_q ∈ blackboard_Q start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT which, as we have seen, amounts to J 𝐽 J italic_J being Zariski-dense. Report issue for preceding element As above, we set P=(a,b)𝑃 𝑎 𝑏 P=(a,b)italic_P = ( italic_a , italic_b ), R=(−b,a)𝑅 𝑏 𝑎 R=(-b,a)italic_R = ( - italic_b , italic_a ) so ‖R‖2=p superscript norm 𝑅 2 𝑝\\|R\\|^{2}=p∥ italic_R ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p and (P.R)=0(P.R)=0( italic_P . italic_R ) = 0. Then we have verified that J 𝐽 J italic_J consists of the points t⁢P+u⁢R 𝑡 𝑃 𝑢 𝑅 tP+uR italic_t italic_P + italic_u italic_R such that t,u 2∈ℚ 𝑡 superscript 𝑢 2 ℚ t,u^{2}\in{\mathbb{Q}}italic_t , italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q. Now for a point Q=t⁢P+u⁢R 𝑄 𝑡 𝑃 𝑢 𝑅 Q=tP+uR italic_Q = italic_t italic_P + italic_u italic_R we have Report issue for preceding element ‖Q‖2=(t 2+u 2)⁢p,‖Q−P‖2=((t−1)2+u 2)⁢p,‖Q−P′‖2=((t−q)2+u 2)⁢p.formulae-sequence superscript norm 𝑄 2 superscript 𝑡 2 superscript 𝑢 2 𝑝 formulae-sequence superscript norm 𝑄 𝑃 2 superscript 𝑡 1 2 superscript 𝑢 2 𝑝 superscript norm 𝑄 superscript 𝑃′2 superscript 𝑡 𝑞 2 superscript 𝑢 2 𝑝\\|Q\\|^{2}=(t^{2}+u^{2})p,\qquad\\|Q-P\\|^{2}=((t-1)^{2}+u^{2})p,\qquad\\|Q-P^{% \prime}\\|^{2}=((t-q)^{2}+u^{2})p.∥ italic_Q ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) italic_p , ∥ italic_Q - italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( ( italic_t - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) italic_p , ∥ italic_Q - italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( ( italic_t - italic_q ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) italic_p . Hence by multiplying by p 𝑝 p italic_p we see that Q 𝑄 Q italic_Q lies in J∗superscript 𝐽 J^{}italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT if and only if t 2+u 2,(t−1)2+u 2,(t−q)2+u 2 superscript 𝑡 2 superscript 𝑢 2 superscript 𝑡 1 2 superscript 𝑢 2 superscript 𝑡 𝑞 2 superscript 𝑢 2 t^{2}+u^{2},(t-1)^{2}+u^{2},(t-q)^{2}+u^{2}italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_t - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_t - italic_q ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are of the shape p 𝑝 p italic_p times a rational square. Putting v:=u 2∈ℚ assign 𝑣 superscript 𝑢 2 ℚ v:=u^{2}\in{\mathbb{Q}}italic_v := italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q we arrive at the system Report issue for preceding element t 2+v=p⁢z 2,(t−1)2+v=p⁢(z+k)2,(t−q)2+v=p⁢(z+k′)2,formulae-sequence superscript 𝑡 2 𝑣 𝑝 superscript 𝑧 2 formulae-sequence superscript 𝑡 1 2 𝑣 𝑝 superscript 𝑧 𝑘 2 superscript 𝑡 𝑞 2 𝑣 𝑝 superscript 𝑧 superscript 𝑘′2 t^{2}+v=pz^{2},\qquad(t-1)^{2}+v=p(z+k)^{2},\qquad(t-q)^{2}+v=p(z+k^{\prime})^% {2},italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_v = italic_p italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_t - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_v = italic_p ( italic_z + italic_k ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_t - italic_q ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_v = italic_p ( italic_z + italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , where t,v,z,k,k′∈ℚ 𝑡 𝑣 𝑧 𝑘 superscript 𝑘′ℚ t,v,z,k,k^{\prime}\in{\mathbb{Q}}italic_t , italic_v , italic_z , italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_Q. By subtraction, on putting v:=p⁢z 2−t 2 assign 𝑣 𝑝 superscript 𝑧 2 superscript 𝑡 2 v:=pz^{2}-t^{2}italic_v := italic_p italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_t start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, this amounts to solving Report issue for preceding element −2⁢t+1=p⁢k 2+2⁢p⁢z⁢k,−2⁢q⁢t+q 2=p⁢k′⁣2+2⁢p⁢z⁢k′,t,z,k,k′∈ℚ.formulae-sequence 2 𝑡 1 𝑝 superscript 𝑘 2 2 𝑝 𝑧 𝑘 formulae-sequence 2 𝑞 𝑡 superscript 𝑞 2 𝑝 superscript 𝑘′2 2 𝑝 𝑧 superscript 𝑘′𝑡 𝑧 𝑘 superscript 𝑘′ℚ-2t+1=pk^{2}+2pzk,\qquad-2qt+q^{2}=pk^{\prime 2}+2pzk^{\prime},\qquad t,z,k,k^% {\prime}\in{\mathbb{Q}}.- 2 italic_t + 1 = italic_p italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_p italic_z italic_k , - 2 italic_q italic_t + italic_q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT + 2 italic_p italic_z italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_t , italic_z , italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_Q . In turn, given k,z 𝑘 𝑧 k,z italic_k , italic_z we may define t 𝑡 t italic_t by the equation on the left, and we remain with the single equation Report issue for preceding element p⁢q⁢k 2+2⁢p⁢q⁢z⁢k+q 2−q=p⁢k′⁣2+2⁢p⁢z⁢k′,z,k,k′∈ℚ.formulae-sequence 𝑝 𝑞 superscript 𝑘 2 2 𝑝 𝑞 𝑧 𝑘 superscript 𝑞 2 𝑞 𝑝 superscript 𝑘′2 2 𝑝 𝑧 superscript 𝑘′𝑧 𝑘 superscript 𝑘′ℚ pqk^{2}+2pqzk+q^{2}-q=pk^{\prime 2}+2pzk^{\prime},\qquad z,k,k^{\prime}\in{% \mathbb{Q}}.italic_p italic_q italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_p italic_q italic_z italic_k + italic_q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_q = italic_p italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT + 2 italic_p italic_z italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_z , italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_Q . This defines a family of conics parametrized by z 𝑧 z italic_z. Since we may solve rationally for z 𝑧 z italic_z, this defines a rational surface over ℚ ℚ{\mathbb{Q}}blackboard_Q, whence in particular its rational points are Zariski-dense (and satisfy weak approximation). Report issue for preceding element 5.2. The non-collinear case Report issue for preceding element We now turn to the non-collinear case, which is less evident and involves a condition which is not always verified. In fact, this is not surprising since the rationality in question amounts to study rational points on a surface which is not rational: indeed, it is a Kummer surface (see Theorem6.1). Report issue for preceding element So let now P,P′∈ℝ 2 𝑃 superscript 𝑃′superscript ℝ 2 P,P^{\prime}\in{\mathbb{R}}^{2}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT be non-collinear. Report issue for preceding element In the previous section we have found a simple necessary and sufficient condition for J 𝐽 J italic_J to be Zariski-dense, given by Proposition 4.5 therein. Using that result we may restrict our attention to points Q 𝑄 Q italic_Q in the vector space J=ℚ⁢P+ℚ⁢R 𝐽 ℚ 𝑃 ℚ 𝑅 J={\mathbb{Q}}P+{\mathbb{Q}}R italic_J = blackboard_Q italic_P + blackboard_Q italic_R, where (R.P)=0(R.P)=0( italic_R . italic_P ) = 0 and r:=‖R‖2∈ℚ assign 𝑟 superscript norm 𝑅 2 ℚ r:=\\|R\\|^{2}\in{\mathbb{Q}}italic_r := ∥ italic_R ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q. It follows from the proof of Proposition4.5 that we may take Report issue for preceding element r=‖P′−(P.P′)‖P‖2⁢P‖2=‖P′‖2−((P.P′)‖P‖)2=Δ 2‖P‖2 𝑟 superscript norm superscript 𝑃′formulae-sequence 𝑃 superscript 𝑃′superscript norm 𝑃 2 𝑃 2 superscript norm superscript 𝑃′2 superscript formulae-sequence 𝑃 superscript 𝑃′norm 𝑃 2 superscript Δ 2 superscript norm 𝑃 2 r=\\|P^{\prime}-\frac{(P.P^{\prime})}{\\|P\\|^{2}}P\\|^{2}=\\|P^{\prime}\\|^{2}-(% \frac{(P.P^{\prime})}{\\|P\\|})^{2}=\frac{\Delta^{2}}{\\|P\\|^{2}}italic_r = ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - divide start_ARG ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG start_ARG ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ( divide start_ARG ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG start_ARG ∥ italic_P ∥ end_ARG ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = divide start_ARG roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG start_ARG ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG . Report issue for preceding element Setting Report issue for preceding element p:=∥P∥2,p′:=∥P′∥2,s:=(P.P′),p:=\\|P\\|^{2},\qquad p^{\prime}:=\\|P^{\prime}\\|^{2},\qquad s:=(P.P^{\prime}),italic_p := ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT := ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_s := ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) ,(5.1) the quadratic form ‖Q‖2 superscript norm 𝑄 2\\|Q\\|^{2}∥ italic_Q ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT with respect to the basis P,R 𝑃 𝑅 P,R italic_P , italic_R of J 𝐽 J italic_J takes the shape Report issue for preceding element Φ⁢(x,y)=p⁢x 2+r⁢y 2,Φ 𝑥 𝑦 𝑝 superscript 𝑥 2 𝑟 superscript 𝑦 2\Phi(x,y)=px^{2}+ry^{2},roman_Φ ( italic_x , italic_y ) = italic_p italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , if x,y∈ℚ 𝑥 𝑦 ℚ x,y\in{\mathbb{Q}}italic_x , italic_y ∈ blackboard_Q are the coordinates of Q 𝑄 Q italic_Q in the said basis. As before, we may take r=p′−(s 2/p)𝑟 superscript 𝑝′superscript 𝑠 2 𝑝 r=p^{\prime}-(s^{2}/p)italic_r = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - ( italic_s start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / italic_p ). Report issue for preceding element Note that \|⋅\|\|\cdot\|\| ⋅ \| may attain a nonzero rational value on ℚ⁢P+ℚ⁢P′ℚ 𝑃 ℚ superscript 𝑃′{\mathbb{Q}}P+{\mathbb{Q}}P^{\prime}blackboard_Q italic_P + blackboard_Q italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT only if the quadratic form Φ Φ\Phi roman_Φ represents a square over ℚ ℚ{\mathbb{Q}}blackboard_Q, or (by definition) if the so-called Hilbert symbol (p,r)=1 𝑝 𝑟 1(p,r)=1( italic_p , italic_r ) = 1. (Note also that (p,r)=(p,p⁢Δ 2)=(p,−Δ 2)𝑝 𝑟 𝑝 𝑝 superscript Δ 2 𝑝 superscript Δ 2(p,r)=(p,p\Delta^{2})=(p,-\Delta^{2})( italic_p , italic_r ) = ( italic_p , italic_p roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) = ( italic_p , - roman_Δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ).) This also amounts to the fact that the ternary form Φ⁢(x,y)−z 2 Φ 𝑥 𝑦 superscript 𝑧 2\Phi(x,y)-z^{2}roman_Φ ( italic_x , italic_y ) - italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is isotropic (i.e. represents 0 0) over ℚ ℚ{\mathbb{Q}}blackboard_Q. We shall prove that this condition is sufficient. Report issue for preceding element Remark 5.1. Report issue for preceding element Note that a priori, given the condition, it does not seem obvious even that there exists some point Q 𝑄 Q italic_Q with rational distance from O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, the condition stating only the existence of a point Q∈ℚ⁢P+ℚ⁢P′𝑄 ℚ 𝑃 ℚ superscript 𝑃′Q\in{\mathbb{Q}}P+{\mathbb{Q}}P^{\prime}italic_Q ∈ blackboard_Q italic_P + blackboard_Q italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT with rational distance from O 𝑂 O italic_O. Note also that the condition is obvious when for instance P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are rational points, in which case J=ℚ 2 𝐽 superscript ℚ 2 J={\mathbb{Q}}^{2}italic_J = blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. See equation (5.11) for explicit formulae. giving rational solutions. Report issue for preceding element Theorem 5.2. Report issue for preceding element Suppose that P,P′∈ℝ 2 𝑃 superscript 𝑃′superscript ℝ 2 P,P^{\prime}\in{\mathbb{R}}^{2}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT are two non collinear points. If ∥P∥2,(P.P′),∥P′∥2∈ℚ\\|P\\|^{2},(P.P^{\prime}),\\|P^{\prime}\\|^{2}\in{\mathbb{Q}}∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) , ∥ italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∈ blackboard_Q, the set J∗superscript 𝐽 J^{}italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT is Zariski-dense if and only if there exists a point ℚ⁢P+ℚ⁢P′ℚ 𝑃 ℚ superscript 𝑃′{\mathbb{Q}}P+{\mathbb{Q}}P^{\prime}blackboard_Q italic_P + blackboard_Q italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT with nonzero rational distance from the origin. On the other hand, if the opening assumption does not hold, then already J 𝐽 J italic_J is not Zariski-dense. Report issue for preceding element To prove the theorem it will be convenient to work inside ℚ 2 superscript ℚ 2{\mathbb{Q}}^{2}blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT equipped with the form Φ Φ\Phi roman_Φ, which corresponds to work in J 𝐽 J italic_J with the basis P,R 𝑃 𝑅 P,R italic_P , italic_R. Report issue for preceding element In the said basis P 𝑃 P italic_P has coordinates (1,0)1 0(1,0)( 1 , 0 ) whereas we may choose R 𝑅 R italic_R as in the proof of Proposition4.5, so that P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT has coordinates (c,1)𝑐 1(c,1)( italic_c , 1 ) in this basis, for some c∈ℚ 𝑐 ℚ c\in{\mathbb{Q}}italic_c ∈ blackboard_Q. We shall work throughout with these coordinates. However now the quadratic form x 2+y 2 superscript 𝑥 2 superscript 𝑦 2 x^{2}+y^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is replaced by the new one. Report issue for preceding element Note that with this notation we have s=(P.P′)=(c P+R.P)=c∥P∥2=c p s=(P.P^{\prime})=(cP+R.P)=c\\|P\\|^{2}=cp italic_s = ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = ( italic_c italic_P + italic_R . italic_P ) = italic_c ∥ italic_P ∥ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_c italic_p. Report issue for preceding element Let us write down the relevant equations for a point Q=(x,y)∈J∗𝑄 𝑥 𝑦 superscript 𝐽 Q=(x,y)\in J^{}italic_Q = ( italic_x , italic_y ) ∈ italic_J start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT. Report issue for preceding element p⁢x 2+r⁢y 2=z 2 𝑝 superscript 𝑥 2 𝑟 superscript 𝑦 2 superscript 𝑧 2\displaystyle px^{2}+ry^{2}=z^{2}italic_p italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(5.2) p⁢(x−1)2+r⁢y 2=(z−k)2 𝑝 superscript 𝑥 1 2 𝑟 superscript 𝑦 2 superscript 𝑧 𝑘 2\displaystyle p(x-1)^{2}+ry^{2}=(z-k)^{2}italic_p ( italic_x - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_z - italic_k ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(5.3) p⁢(x−c)2+r⁢(y−1)2=(z−k′)2,𝑝 superscript 𝑥 𝑐 2 𝑟 superscript 𝑦 1 2 superscript 𝑧 superscript 𝑘′2\displaystyle p(x-c)^{2}+r(y-1)^{2}=(z-k^{\prime})^{2},italic_p ( italic_x - italic_c ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r ( italic_y - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_z - italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(5.4) Similarly to the above, on subtracting the second equation from the first, we obtain a family of affine conics with parameter k 𝑘 k italic_k with fibers given by Report issue for preceding element 𝒞 k:2 p x−δ=2 k z,δ=p−k 2.{\mathcal{C}}_{k}:\quad 2px-\delta=2kz,\qquad\delta=p-k^{2}.caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT : 2 italic_p italic_x - italic_δ = 2 italic_k italic_z , italic_δ = italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT .(5.5) Using the first and third equation we also obtain a second family, with parameter k′superscript 𝑘′k^{\prime}italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT: Report issue for preceding element 𝒞 k′′:2 p c x+2 r y−δ′=2 k′z,δ′=p c 2+r−k′⁣2.{\mathcal{C}}_{k^{\prime}}^{\prime}:\quad 2pcx+2ry-\delta^{\prime}=2k^{\prime}% z,\qquad\delta^{\prime}=pc^{2}+r-k^{\prime 2}.caligraphic_C start_POSTSUBSCRIPT italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT : 2 italic_p italic_c italic_x + 2 italic_r italic_y - italic_δ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = 2 italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_z , italic_δ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT .(5.6) Let us write down explicitly one possible cubic fibration. Setting λ=y/x 𝜆 𝑦 𝑥\lambda=y/x italic_λ = italic_y / italic_x we obtain y=λ⁢x 𝑦 𝜆 𝑥 y=\lambda x italic_y = italic_λ italic_x, z=p+r⁢λ 2⁢x 𝑧 𝑝 𝑟 superscript 𝜆 2 𝑥 z=\sqrt{p+r\lambda^{2}}x italic_z = square-root start_ARG italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG italic_x (for some choice of the square root) and so the equations for 𝒞 k,𝒞 k′′subscript 𝒞 𝑘 superscript subscript 𝒞 superscript 𝑘′′{\mathcal{C}}{k},{\mathcal{C}}{k^{\prime}}^{\prime}caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT , caligraphic_C start_POSTSUBSCRIPT italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT become Report issue for preceding element {2⁢p⁢x−p=−k 2+2⁢k⁢p+r⁢λ 2⁢x(2⁢p⁢c+2⁢r⁢λ)⁢x−p⁢c 2−r=−k′⁣2+2⁢k′⁢p+r⁢λ 2⁢x\left{\begin{matrix}2px-p&=&-k^{2}+2k\sqrt{p+r\lambda^{2}}\,x\ (2pc+2r\lambda)x-pc^{2}-r&=&-k^{\prime 2}+2k^{\prime}\sqrt{p+r\lambda^{2}}\,x% \end{matrix}\right.{ start_ARG start_ROW start_CELL 2 italic_p italic_x - italic_p end_CELL start_CELL = end_CELL start_CELL - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_k square-root start_ARG italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG italic_x end_CELL end_ROW start_ROW start_CELL ( 2 italic_p italic_c + 2 italic_r italic_λ ) italic_x - italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_r end_CELL start_CELL = end_CELL start_CELL - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT + 2 italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT square-root start_ARG italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG italic_x end_CELL end_ROW end_ARG(5.7) which correspond to the compositum of two quadratic covers of the x 𝑥 x italic_x-line, each one ramified over two points. Such a curve has genus 1 1 1 1. Note that we can choose λ 𝜆\lambda italic_λ to be rational and such that p+r⁢λ 2 𝑝 𝑟 superscript 𝜆 2 p+r\lambda^{2}italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is a rational square in infinitely many ways. With such choices the coordinate z=p+r⁢λ 2⁢x 𝑧 𝑝 𝑟 superscript 𝜆 2 𝑥 z=\sqrt{p+r\lambda^{2}}\,x italic_z = square-root start_ARG italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG italic_x also becomes rational; moreover both curves 𝒞 k subscript 𝒞 𝑘{\mathcal{C}}{k}caligraphic_C start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT and 𝒞 k′′superscript subscript 𝒞 superscript 𝑘′′{\mathcal{C}}{k^{\prime}}^{\prime}caligraphic_C start_POSTSUBSCRIPT italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT will have a rational point above x=∞𝑥 x=\infty italic_x = ∞, so that the genus-one curve defined by the system of equations (5.7) will have a marked rational point (actually four rational points), turning that curve into an elliptic curve (with rational 2 2 2 2-torsion). Report issue for preceding element Let us derive a cubic curve out of these equations. Report issue for preceding element Write μ 2=p+r⁢λ 2 superscript 𝜇 2 𝑝 𝑟 superscript 𝜆 2\mu^{2}=p+r\lambda^{2}italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. This is a conic, denoted S 𝑆 S italic_S, which by assumption has a rational point, hence can be parametrized over ℚ ℚ{\mathbb{Q}}blackboard_Q.2 2 2 We have used affine coordinates, but of course we may refer to points at infinity on the conic, given by (1:±r:0):1 plus-or-minus 𝑟:0(1:\pm\sqrt{r}:0)( 1 : ± square-root start_ARG italic_r end_ARG : 0 ) in homogeneous coordinates. In terms of these coordinates the equations may be rewritten as Report issue for preceding element {2⁢(p−k⁢μ)⁢x=p−k 2 2⁢(p⁢c+r⁢λ−k′⁢μ)⁢x=p⁢c 2+r−k′⁣2.\left{\begin{matrix}2(p-k\mu)x&=&p-k^{2}\ 2(pc+r\lambda-k^{\prime}\mu)x&=&pc^{2}+r-k^{\prime 2}.\end{matrix}\right.{ start_ARG start_ROW start_CELL 2 ( italic_p - italic_k italic_μ ) italic_x end_CELL start_CELL = end_CELL start_CELL italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL end_ROW start_ROW start_CELL 2 ( italic_p italic_c + italic_r italic_λ - italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_μ ) italic_x end_CELL start_CELL = end_CELL start_CELL italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT . end_CELL end_ROW end_ARG Eliminating x 𝑥 x italic_x we obtain a family of cubic equations, depending on the point (λ,μ)𝜆 𝜇(\lambda,\mu)( italic_λ , italic_μ ) on our conic S 𝑆 S italic_S: Report issue for preceding element (p−k 2)⁢(p⁢c+r⁢λ−μ⁢k′)=(p⁢c 2+r−k′⁣2)⁢(p−μ⁢k).𝑝 superscript 𝑘 2 𝑝 𝑐 𝑟 𝜆 𝜇 superscript 𝑘′𝑝 superscript 𝑐 2 𝑟 superscript 𝑘′2 𝑝 𝜇 𝑘(p-k^{2})(pc+r\lambda-\mu k^{\prime})=(pc^{2}+r-k^{\prime 2})(p-\mu k).( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_p italic_c + italic_r italic_λ - italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = ( italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) .(5.8) Note that if for rational (λ,μ)𝜆 𝜇(\lambda,\mu)( italic_λ , italic_μ ) in our conic we have a rational point (k,k′)𝑘 superscript 𝑘′(k,k^{\prime})( italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) in the corresponding cubic, then, provided p≠μ⁢k 𝑝 𝜇 𝑘 p\neq\mu k italic_p ≠ italic_μ italic_k and p⁢c+r⁢λ≠μ⁢k′𝑝 𝑐 𝑟 𝜆 𝜇 superscript 𝑘′pc+r\lambda\neq\mu k^{\prime}italic_p italic_c + italic_r italic_λ ≠ italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, we can use the previous equations so as to obtain rational values for x,y,z 𝑥 𝑦 𝑧 x,y,z italic_x , italic_y , italic_z, hence a rational point Q 𝑄 Q italic_Q with the sought properties. Report issue for preceding element 5.2.1. Some sections for the cubic family Report issue for preceding element We also abbreviate p′:=p⁢c 2+r=\|P′\|2 assign superscript 𝑝′𝑝 superscript 𝑐 2 𝑟 superscript superscript 𝑃′2 p^{\prime}:=pc^{2}+r=\|P^{\prime}\|^{2}italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT := italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r = \| italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, η:=p⁢c+r⁢λ assign 𝜂 𝑝 𝑐 𝑟 𝜆\eta:=pc+r\lambda italic_η := italic_p italic_c + italic_r italic_λ. Report issue for preceding element A homogeneous equation for the cubic family: In homogeneous coordinates (K:K′:H):𝐾 superscript 𝐾′:𝐻(K:K^{\prime}:H)( italic_K : italic_K start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT : italic_H ) the above equation reads Report issue for preceding element 𝒞:(p H 2−K 2)(η H−μ K′)=(p′H 2−K′⁣2)(p H−μ K).\mathcal{C}:\quad(pH^{2}-K^{2})(\eta H-\mu K^{\prime})=(p^{\prime}H^{2}-K^{% \prime 2})(pH-\mu K).caligraphic_C : ( italic_p italic_H start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_K start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_η italic_H - italic_μ italic_K start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = ( italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_H start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_K start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) ( italic_p italic_H - italic_μ italic_K ) .(5.9) It is the equation of a cubic curve depending on a parameter which is a point of the conic S 𝑆 S italic_S of equation μ 2=p+r⁢λ 2 superscript 𝜇 2 𝑝 𝑟 superscript 𝜆 2\mu^{2}=p+r\lambda^{2}italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. The cubic is generically smooth because it is birationally equivalent to the former curve, which has genus 1 1 1 1. (One can also check this by differentiation: one finds that possible singular points would have k=k′𝑘 superscript 𝑘′k=k^{\prime}italic_k = italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, which is easy to exclude. See also Lemma 5.7 below.) Report issue for preceding element The affine coordinates k,k′𝑘 superscript 𝑘′k,k^{\prime}italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT are obtained from the homogeneous ones by k=K/H,k′=K′/H formulae-sequence 𝑘 𝐾 𝐻 superscript 𝑘′superscript 𝐾′𝐻 k=K/H,\,k^{\prime}=K^{\prime}/H italic_k = italic_K / italic_H , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_K start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT / italic_H. Report issue for preceding element Four rational sections. There are four rational sections, namely Report issue for preceding element A:=(1:0:0),B:=(0:1:0),C:=(1:1:0),N:=(p:η:μ).A:=(1:0:0),\,B:=(0:1:0),\,C:=(1:1:0),\,N:=(p:\eta:\mu).italic_A := ( 1 : 0 : 0 ) , italic_B := ( 0 : 1 : 0 ) , italic_C := ( 1 : 1 : 0 ) , italic_N := ( italic_p : italic_η : italic_μ ) . 3 3 3 Here A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C are independent of (λ,μ)𝜆 𝜇(\lambda,\mu)( italic_λ , italic_μ ) but we view them as points with coordinates which are constant functions on S 𝑆 S italic_S. We note that, as is easy to check, N 𝑁 N italic_N does not coincide with any of A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C, no matter the point (λ,μ)𝜆 𝜇(\lambda,\mu)( italic_λ , italic_μ ) on the conic (not even at infinity): this follows easily from the positivity of \|⋅\|2\|\cdot\|^{2}\| ⋅ \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT on ℝ 2∖{O}superscript ℝ 2 𝑂{\mathbb{R}}^{2}\setminus{O}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ∖ { italic_O }. Note also that two of the four factors defining the cubic vanish at N 𝑁 N italic_N. Report issue for preceding element Further, it is easily checked that these points are smooth on 𝒞 𝒞\mathcal{C}caligraphic_C for every point on the conic. Report issue for preceding element It will be useful to note that Report issue for preceding element μ 2⁢p−p 2=(p+r⁢λ 2)⁢p−p 2=p⁢r⁢λ 2 μ 2⁢p′−η 2=(p+r⁢λ 2)⁢(p⁢c 2+r)−(p⁢c+r⁢λ)2=p⁢r⁢(c⁢λ−1)2 matrix superscript 𝜇 2 𝑝 superscript 𝑝 2 𝑝 𝑟 superscript 𝜆 2 𝑝 superscript 𝑝 2 𝑝 𝑟 superscript 𝜆 2 superscript 𝜇 2 superscript 𝑝′superscript 𝜂 2 𝑝 𝑟 superscript 𝜆 2 𝑝 superscript 𝑐 2 𝑟 superscript 𝑝 𝑐 𝑟 𝜆 2 𝑝 𝑟 superscript 𝑐 𝜆 1 2\begin{matrix}\mu^{2}p-p^{2}&=&(p+r\lambda^{2})p-p^{2}&=&pr\lambda^{2}\ \mu^{2}p^{\prime}-\eta^{2}&=&(p+r\lambda^{2})(pc^{2}+r)-(pc+r\lambda)^{2}&=&pr% (c\lambda-1)^{2}\ \end{matrix}start_ARG start_ROW start_CELL italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_p - italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL start_CELL = end_CELL start_CELL ( italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) italic_p - italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL start_CELL = end_CELL start_CELL italic_p italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL end_ROW start_ROW start_CELL italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_η start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL start_CELL = end_CELL start_CELL ( italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r ) - ( italic_p italic_c + italic_r italic_λ ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL start_CELL = end_CELL start_CELL italic_p italic_r ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_CELL end_ROW end_ARG(5.10) Lemma 5.3. Report issue for preceding element The tangents drawn from the points A,B 𝐴 𝐵 A,B italic_A , italic_B intersect on the fourth point. Report issue for preceding element Proof. Report issue for preceding element Let us prove that the tangent to the point A=(1:0:0)A=(1:0:0)italic_A = ( 1 : 0 : 0 ) passes through the fourth point. De-homogenizing with H=1 𝐻 1 H=1 italic_H = 1 the equation becomes Report issue for preceding element (p−k 2)⁢(η−μ⁢k′)−(p′−k′⁣2)⁢(p−μ⁢k)=0.𝑝 superscript 𝑘 2 𝜂 𝜇 superscript 𝑘′superscript 𝑝′superscript 𝑘′2 𝑝 𝜇 𝑘 0(p-k^{2})(\eta-\mu k^{\prime})-(p^{\prime}-k^{\prime 2})(p-\mu k)=0.( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_η - italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) - ( italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) = 0 . The line joining (1:0:0):1 0:0(1:0:0)( 1 : 0 : 0 ) with (p:η:μ):𝑝 𝜂:𝜇(p:\eta:\mu)( italic_p : italic_η : italic_μ ) has equation K′=H⁢η/μ superscript 𝐾′𝐻 𝜂 𝜇 K^{\prime}=H\eta/\mu italic_K start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_H italic_η / italic_μ, i.e. k′=η/μ superscript 𝑘′𝜂 𝜇 k^{\prime}=\eta/\mu italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_η / italic_μ in affine coordinates. Substituting we obtain Report issue for preceding element (p′−η 2/μ 2)⁢(p−μ⁢k)=0 superscript 𝑝′superscript 𝜂 2 superscript 𝜇 2 𝑝 𝜇 𝑘 0(p^{\prime}-\eta^{2}/\mu^{2})(p-\mu k)=0( italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_η start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) = 0 which has the only solution k=p/μ 𝑘 𝑝 𝜇 k=p/\mu italic_k = italic_p / italic_μ. Since a line has a unique point at infinity, this means that there is a double solution at infinity on the cubic. Report issue for preceding element The verification for the point (0:1:0):0 1:0(0:1:0)( 0 : 1 : 0 ) is symmetric. ∎ Report issue for preceding element Corollary 5.4. Report issue for preceding element Taking the point N 𝑁 N italic_N as origin for the group law on the cubic curve, supposed to be smooth, we have identically on S 𝑆 S italic_S Report issue for preceding element A+C=B,B+C=A formulae-sequence 𝐴 𝐶 𝐵 𝐵 𝐶 𝐴 A+C=B,\qquad B+C=A italic_A + italic_C = italic_B , italic_B + italic_C = italic_A and so Report issue for preceding element 2⁢C=0.2 𝐶 0 2C=0.2 italic_C = 0 . This follows immediately from the linear equivalences provided by the lemma, together with the fact that A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C are collinear points on the cubic: A+B+C∼2⁢A+N∼2⁢B+N similar-to 𝐴 𝐵 𝐶 2 𝐴 𝑁 similar-to 2 𝐵 𝑁 A+B+C\sim 2A+N\sim 2B+N italic_A + italic_B + italic_C ∼ 2 italic_A + italic_N ∼ 2 italic_B + italic_N. Report issue for preceding element Let us now consider the point (1:1:0):1 1:0(1:1:0)( 1 : 1 : 0 ). We have the following lemma. Report issue for preceding element Lemma 5.5. Report issue for preceding element Suppose μ≠0 𝜇 0\mu\neq 0 italic_μ ≠ 0. The line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT joining C=(1:1:0)C=(1:1:0)italic_C = ( 1 : 1 : 0 ) to N=(p:η:μ)N=(p:\eta:\mu)italic_N = ( italic_p : italic_η : italic_μ ) is tangent to the cubic at the point (1:1:0):1 1:0(1:1:0)( 1 : 1 : 0 ) if and only if λ 𝜆\lambda italic_λ is such that η=p 𝜂 𝑝\eta=p italic_η = italic_p. It is tangent at N 𝑁 N italic_N if and only if c⁢λ−1=±λ 𝑐 𝜆 1 plus-or-minus 𝜆 c\lambda-1=\pm\lambda italic_c italic_λ - 1 = ± italic_λ. If p=p′𝑝 superscript 𝑝′p=p^{\prime}italic_p = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT then η=p 𝜂 𝑝\eta=p italic_η = italic_p implies c⁢λ−1=−λ 𝑐 𝜆 1 𝜆 c\lambda-1=-\lambda italic_c italic_λ - 1 = - italic_λ, and under such a specialization the cubic is reducible and the line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT is one of its components. If p≠p′𝑝 superscript 𝑝′p\neq p^{\prime}italic_p ≠ italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, for no specialization of λ 𝜆\lambda italic_λ can the curve contain the line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT (except if μ=0 𝜇 0\mu=0 italic_μ = 0, i.e. λ 2=−p/r superscript 𝜆 2 𝑝 𝑟\lambda^{2}=-p/r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = - italic_p / italic_r). Report issue for preceding element Proof. Report issue for preceding element The line joining the points (1:1:0):1 1:0(1:1:0)( 1 : 1 : 0 ) to the point (p:η:μ):𝑝 𝜂:𝜇(p:\eta:\mu)( italic_p : italic_η : italic_μ ) is parametrized as Report issue for preceding element (p+t:η+t:μ),t∈ℙ 1.(p+t:\eta+t:\mu),\qquad t\in\mathbb{P}_{1}.( italic_p + italic_t : italic_η + italic_t : italic_μ ) , italic_t ∈ blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT . Substituting into (5.9) we get a cubic equation with a factor μ⁢t 𝜇 𝑡\mu t italic_μ italic_t. Indeed the equation simplifies to Report issue for preceding element μ⁢t⋅(p⁢μ 2−p 2−2⁢p⁢t−p′⁢μ 2+η 2+2⁢η⁢t)=0.⋅𝜇 𝑡 𝑝 superscript 𝜇 2 superscript 𝑝 2 2 𝑝 𝑡 superscript 𝑝′superscript 𝜇 2 superscript 𝜂 2 2 𝜂 𝑡 0\mu t\cdot(p\mu^{2}-p^{2}-2pt-p^{\prime}\mu^{2}+\eta^{2}+2\eta t)=0.italic_μ italic_t ⋅ ( italic_p italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 italic_p italic_t - italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_η start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_η italic_t ) = 0 . If μ=0 𝜇 0\mu=0 italic_μ = 0 the cubic is degenerate and contains the whole line H=0 𝐻 0 H=0 italic_H = 0, a case which we are disregarding. The value t=0 𝑡 0 t=0 italic_t = 0 yields the point N 𝑁 N italic_N. The value t=∞𝑡 t=\infty italic_t = ∞ yields C 𝐶 C italic_C, whereas a third solution is obtained from the polynomial in brackets, which is 2⁢(η−p)⁢t+(p⁢μ 2−p 2)−(p′⁢μ 2−η 2).2 𝜂 𝑝 𝑡 𝑝 superscript 𝜇 2 superscript 𝑝 2 superscript 𝑝′superscript 𝜇 2 superscript 𝜂 2 2(\eta-p)t+(p\mu^{2}-p^{2})-(p^{\prime}\mu^{2}-\eta^{2}).2 ( italic_η - italic_p ) italic_t + ( italic_p italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) - ( italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_η start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) . By (5.10) this equals Report issue for preceding element 2⁢(η−p)⁢t+p⁢r⁢(λ 2−(c⁢λ−1)2).2 𝜂 𝑝 𝑡 𝑝 𝑟 superscript 𝜆 2 superscript 𝑐 𝜆 1 2 2(\eta-p)t+pr(\lambda^{2}-(c\lambda-1)^{2}).2 ( italic_η - italic_p ) italic_t + italic_p italic_r ( italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) . Then, if (η−p,λ 2−(c⁢λ−1)2)≠(0,0)𝜂 𝑝 superscript 𝜆 2 superscript 𝑐 𝜆 1 2 0 0(\eta-p,\lambda^{2}-(c\lambda-1)^{2})\neq(0,0)( italic_η - italic_p , italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ≠ ( 0 , 0 ) it gives Report issue for preceding element t=p⁢r⁢(c⁢λ−1)2−λ 2 2⁢(η−p).𝑡 𝑝 𝑟 superscript 𝑐 𝜆 1 2 superscript 𝜆 2 2 𝜂 𝑝 t=pr\frac{(c\lambda-1)^{2}-\lambda^{2}}{2(\eta-p)}.italic_t = italic_p italic_r divide start_ARG ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG start_ARG 2 ( italic_η - italic_p ) end_ARG . Then t 𝑡 t italic_t vanishes if and only if c⁢λ−1=±λ 𝑐 𝜆 1 plus-or-minus 𝜆 c\lambda-1=\pm\lambda italic_c italic_λ - 1 = ± italic_λ and is infinite if and only if η=p 𝜂 𝑝\eta=p italic_η = italic_p. In this second case the line is tangent to the point at infinity, proving the first assertion of the lemma. Report issue for preceding element Let us now analyze the cases when it happens that for some λ 𝜆\lambda italic_λ both η−p 𝜂 𝑝\eta-p italic_η - italic_p and λ 2−(c⁢λ−1)2 superscript 𝜆 2 superscript 𝑐 𝜆 1 2\lambda^{2}-(c\lambda-1)^{2}italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT vanish. Report issue for preceding element We can exclude that η−p=0=c⁢λ−1−λ 𝜂 𝑝 0 𝑐 𝜆 1 𝜆\eta-p=0=c\lambda-1-\lambda italic_η - italic_p = 0 = italic_c italic_λ - 1 - italic_λ because in this case c≠1 𝑐 1 c\neq 1 italic_c ≠ 1 and we obtain that p⁢(c−1)2+r=0 𝑝 superscript 𝑐 1 2 𝑟 0 p(c-1)^{2}+r=0 italic_p ( italic_c - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r = 0 which is impossible since p,r>0 𝑝 𝑟 0 p,r>0 italic_p , italic_r > 0. On the contrary, it can happen that η−p=0=(c+1)⁢λ−1 𝜂 𝑝 0 𝑐 1 𝜆 1\eta-p=0=(c+1)\lambda-1 italic_η - italic_p = 0 = ( italic_c + 1 ) italic_λ - 1. Indeed this amounts to (c+1)⁢λ=1 𝑐 1 𝜆 1(c+1)\lambda=1( italic_c + 1 ) italic_λ = 1 and to p⁢(c 2−1)+r=0 𝑝 superscript 𝑐 2 1 𝑟 0 p(c^{2}-1)+r=0 italic_p ( italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 1 ) + italic_r = 0, which in turn amounts to p=p′𝑝 superscript 𝑝′p=p^{\prime}italic_p = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT (and can never happen if c=−1 𝑐 1 c=-1 italic_c = - 1). Report issue for preceding element In this case the specialization λ↦1/(c+1)maps-to 𝜆 1 𝑐 1\lambda\mapsto 1/(c+1)italic_λ ↦ 1 / ( italic_c + 1 ) which makes η 𝜂\eta italic_η be equal to p 𝑝 p italic_p leads to a reducible cubic, containing the line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT. ∎ Report issue for preceding element In the generic case (that is, for generic values of λ 𝜆\lambda italic_λ) the line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT contains another point beyond C,N 𝐶 𝑁 C,N italic_C , italic_N; since d⁢i⁢v⁢(l C)∼A+B+C similar-to 𝑑 𝑖 𝑣 subscript 𝑙 𝐶 𝐴 𝐵 𝐶 div(l_{C})\sim A+B+C italic_d italic_i italic_v ( italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT ) ∼ italic_A + italic_B + italic_C, such point is equal to the sum A+B 𝐴 𝐵 A+B italic_A + italic_B (with respect to the group law with origin N 𝑁 N italic_N). The computations above exhibit this point: Report issue for preceding element A+B=(2 p(η−p)+p r(c λ−1)2−p r λ 2:2 η(η−p)+p r(c λ−1)2−p r λ 2:2 μ(η−p)).A+B=(2p(\eta-p)+pr(c\lambda-1)^{2}-pr\lambda^{2}:2\eta(\eta-p)+pr(c\lambda-1)^% {2}-pr\lambda^{2}:2\mu(\eta-p)).italic_A + italic_B = ( 2 italic_p ( italic_η - italic_p ) + italic_p italic_r ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT : 2 italic_η ( italic_η - italic_p ) + italic_p italic_r ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT : 2 italic_μ ( italic_η - italic_p ) ) .(5.11) Remark 5.6. Report issue for preceding element Letting λ 𝜆\lambda italic_λ vary correspondingly to rational points on our conic, this section already provides an infinite set of rational solutions for our problem. Indeed, we find that k=p/μ 𝑘 𝑝 𝜇 k=p/\mu italic_k = italic_p / italic_μ or k′=η/μ superscript 𝑘′𝜂 𝜇 k^{\prime}=\eta/\mu italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_η / italic_μ only if t=0 𝑡 0 t=0 italic_t = 0, which, as we have seen, happens only if (c±1)⁢λ=1 plus-or-minus 𝑐 1 𝜆 1(c\pm 1)\lambda=1( italic_c ± 1 ) italic_λ = 1. Report issue for preceding element Next, we want to prove that the sections found so far are not all torsion, with the purpose to use the group law on the elliptic curves of the family in order to produce a Zariski-dense set of rational points. Report issue for preceding element In principle there are several methods for this, for instance one could specialize at some rational points; however there are parameters which complicate the situation. A suitable method is to explore points of the conic S 𝑆 S italic_S where a given section becomes torsion of low order: if this occurs at a point of good reduction, then either the section is torsion of that order, or it is non-torsion. Report issue for preceding element With this in mind, we look for a specialization of λ 𝜆\lambda italic_λ for which the elliptic curve has good reduction, and an extra relation between A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C holds; namely, we want that A:=(1:0:0)A:=(1:0:0)italic_A := ( 1 : 0 : 0 ) and B:=(0:1:0)B:=(0:1:0)italic_B := ( 0 : 1 : 0 ) become torsion of order 2 2 2 2 with respect to the ‘moving’ point N:=(p:η:μ)N:=(p:\eta:\mu)italic_N := ( italic_p : italic_η : italic_μ ). This will occur when the point N 𝑁 N italic_N is a flex and all the tangents drawn from A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C contain N 𝑁 N italic_N. In view of the lemma just proved, this happens if and only if p=η 𝑝 𝜂 p=\eta italic_p = italic_η. However, we must check that this is a good specialization, namely that the specialized cubic has genus 1 1 1 1. Report issue for preceding element Lemma 5.7. Report issue for preceding element For no (complex) value of λ 𝜆\lambda italic_λ the corresponding cubic curve is singular at the point N=(p:η:μ)N=(p:\eta:\mu)italic_N = ( italic_p : italic_η : italic_μ ). The unique tangent to the cubic at N 𝑁 N italic_N meets the point A=(1:0:0)A=(1:0:0)italic_A = ( 1 : 0 : 0 ) if and only if λ=0 𝜆 0\lambda=0 italic_λ = 0 and meets B 𝐵 B italic_B if and only if c⁢λ=1 𝑐 𝜆 1 c\lambda=1 italic_c italic_λ = 1. In these two cases the cubic becomes reducible and the tangent at N 𝑁 N italic_N is one of its components. Report issue for preceding element Proof. Report issue for preceding element Working in affine coordinates, i.e. putting H=1 𝐻 1 H=1 italic_H = 1, the equation of the cubic curve reads Report issue for preceding element f⁢(k,k′):=(p−k 2)⁢(η−μ⁢k′)−(p′−k′⁣2)⁢(p−μ⁢k)=0,assign 𝑓 𝑘 superscript 𝑘′𝑝 superscript 𝑘 2 𝜂 𝜇 superscript 𝑘′superscript 𝑝′superscript 𝑘′2 𝑝 𝜇 𝑘 0 f(k,k^{\prime}):=(p-k^{2})(\eta-\mu k^{\prime})-(p^{\prime}-k^{\prime 2})(p-% \mu k)=0,italic_f ( italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) := ( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_η - italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) - ( italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) = 0 , and the point N 𝑁 N italic_N in question is the point (p/μ,η/μ)𝑝 𝜇 𝜂 𝜇(p/\mu,\eta/\mu)( italic_p / italic_μ , italic_η / italic_μ ). We have Report issue for preceding element ∂f∂k=−2⁢k⁢(η−μ⁢k′)+μ⁢(p′−k′⁣2),∂f∂k′=−μ⁢(p−k 2)+2⁢k′⁢(p−μ⁢k);formulae-sequence 𝑓 𝑘 2 𝑘 𝜂 𝜇 superscript 𝑘′𝜇 superscript 𝑝′superscript 𝑘′2 𝑓 superscript 𝑘′𝜇 𝑝 superscript 𝑘 2 2 superscript 𝑘′𝑝 𝜇 𝑘\dfrac{\partial f}{\partial k}=-2k(\eta-\mu k^{\prime})+\mu(p^{\prime}-k^{% \prime 2}),\qquad\dfrac{\partial f}{\partial k^{\prime}}=-\mu(p-k^{2})+2k^{% \prime}(p-\mu k);divide start_ARG ∂ italic_f end_ARG start_ARG ∂ italic_k end_ARG = - 2 italic_k ( italic_η - italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) + italic_μ ( italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) , divide start_ARG ∂ italic_f end_ARG start_ARG ∂ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_ARG = - italic_μ ( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + 2 italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ( italic_p - italic_μ italic_k ) ; specializing at the point (p/μ,η/μ)𝑝 𝜇 𝜂 𝜇(p/\mu,\eta/\mu)( italic_p / italic_μ , italic_η / italic_μ ) we obtain for the gradient the expression: Report issue for preceding element ∇f⁢(p μ,η μ)=1 μ⁢(μ 2⁢p′−η 2,−μ 2⁢p+p 2).∇𝑓 𝑝 𝜇 𝜂 𝜇 1 𝜇 superscript 𝜇 2 superscript 𝑝′superscript 𝜂 2 superscript 𝜇 2 𝑝 superscript 𝑝 2\nabla f\left(\frac{p}{\mu},\frac{\eta}{\mu}\right)=\frac{1}{\mu}(\mu^{2}p^{% \prime}-\eta^{2},-\mu^{2}p+p^{2}).∇ italic_f ( divide start_ARG italic_p end_ARG start_ARG italic_μ end_ARG , divide start_ARG italic_η end_ARG start_ARG italic_μ end_ARG ) = divide start_ARG 1 end_ARG start_ARG italic_μ end_ARG ( italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_η start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , - italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_p + italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) . As already observed we have Report issue for preceding element μ 2⁢p−p 2=p⁢r⁢λ 2,μ 2⁢p′−η 2=p⁢r⁢(c⁢λ−1)2,formulae-sequence superscript 𝜇 2 𝑝 superscript 𝑝 2 𝑝 𝑟 superscript 𝜆 2 superscript 𝜇 2 superscript 𝑝′superscript 𝜂 2 𝑝 𝑟 superscript 𝑐 𝜆 1 2\mu^{2}p-p^{2}=pr\lambda^{2},\qquad\mu^{2}p^{\prime}-\eta^{2}=pr(c\lambda-1)^{% 2},italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_p - italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_η start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p italic_r ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , arriving at Report issue for preceding element μ⁢∇f⁢(p μ,η μ)=p⁢r⁢((c⁢λ−1)2,−λ 2).𝜇∇𝑓 𝑝 𝜇 𝜂 𝜇 𝑝 𝑟 superscript 𝑐 𝜆 1 2 superscript 𝜆 2\mu\nabla f\left(\frac{p}{\mu},\frac{\eta}{\mu}\right)=pr((c\lambda-1)^{2},-% \lambda^{2}).italic_μ ∇ italic_f ( divide start_ARG italic_p end_ARG start_ARG italic_μ end_ARG , divide start_ARG italic_η end_ARG start_ARG italic_μ end_ARG ) = italic_p italic_r ( ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , - italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) . Note that the vector in the right hand side is never zero. This proves that N 𝑁 N italic_N is a smooth point. Report issue for preceding element This formula also proves that the (unique) tangent line has the direction of the vector ((c⁢λ−1)2,−λ 2)superscript 𝑐 𝜆 1 2 superscript 𝜆 2((c\lambda-1)^{2},-\lambda^{2})( ( italic_c italic_λ - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , - italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ), so its point at infinity is A=(1:0:0)A=(1:0:0)italic_A = ( 1 : 0 : 0 ) if and only if λ 𝜆\lambda italic_λ vanishes and is B 𝐵 B italic_B if and only if c⁢λ−1′𝑐 𝜆 superscript 1′c\lambda-1^{\prime}italic_c italic_λ - 1 start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT vanishes. Report issue for preceding element Finally, in these two cases such a tangent is a bitangent of the cubic by Lemma 5.5, so it must be a component. ∎ Report issue for preceding element 5.2.2. Good reduction Report issue for preceding element First of all, as we already observed, μ=0 𝜇 0\mu=0 italic_μ = 0 is a point of bad reduction, since for that value the cubic degenerates in the union of a conic and the line H=0 𝐻 0 H=0 italic_H = 0. This is the only case in which the curve contains the line H=0 𝐻 0 H=0 italic_H = 0. In this case, the second component is a smooth conic. (In fact, otherwise we have p⁢η=p⁢p′𝑝 𝜂 𝑝 superscript 𝑝′p\eta=pp^{\prime}italic_p italic_η = italic_p italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT and μ=0 𝜇 0\mu=0 italic_μ = 0, leading to r⁢λ 2=−p 𝑟 superscript 𝜆 2 𝑝 r\lambda^{2}=-p italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = - italic_p and r⁢p+(p⁢c⁢(c−1)+r)2=0 𝑟 𝑝 superscript 𝑝 𝑐 𝑐 1 𝑟 2 0 rp+(pc(c-1)+r)^{2}=0 italic_r italic_p + ( italic_p italic_c ( italic_c - 1 ) + italic_r ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0, which is impossible since p,r>0 𝑝 𝑟 0 p,r>0 italic_p , italic_r > 0.) Report issue for preceding element Let us classify the other cases of reducibility. Let l A subscript 𝑙 𝐴 l_{A}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT (resp. l B,l C subscript 𝑙 𝐵 subscript 𝑙 𝐶 l_{B},l_{C}italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT) denote the lines joining A 𝐴 A italic_A (resp. B,C 𝐵 𝐶 B,C italic_B , italic_C) to the point N 𝑁 N italic_N and l 𝑙 l italic_l denote the line containing A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C (i.e. the line H=0 𝐻 0 H=0 italic_H = 0). Note that l 𝑙 l italic_l intersects the cubic precisely at these three distinct points for every specialization of λ 𝜆\lambda italic_λ with μ≠0 𝜇 0\mu\neq 0 italic_μ ≠ 0. Hence these points remain smooth for every specialization (outside the case μ=0 𝜇 0\mu=0 italic_μ = 0 when l 𝑙 l italic_l is a component of 𝒞 𝒞\mathcal{C}caligraphic_C). By Lemma 5.7 the point N 𝑁 N italic_N too remains smooth for every specialization. Report issue for preceding element Suppose now that for a specialization of λ 𝜆\lambda italic_λ the cubic becomes reducible. Let us first exclude it is the union of three lines. We have already seen that l 𝑙 l italic_l is a component of 𝒞 𝒞\mathcal{C}caligraphic_C if and only if μ=0 𝜇 0\mu=0 italic_μ = 0, and in this case the second component is a smooth conic. So we can suppose that l 𝑙 l italic_l is not a component of 𝒞 𝒞\mathcal{C}caligraphic_C. Then A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C are smooth points, as we observed; then, since the tangents from A 𝐴 A italic_A and B 𝐵 B italic_B must be unique and contain N 𝑁 N italic_N, two of the three components would be l A,l B subscript 𝑙 𝐴 subscript 𝑙 𝐵 l_{A},l_{B}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT, making N 𝑁 N italic_N become a singular point, contrary to Lemma 5.7. Report issue for preceding element Suppose now that 𝒞 𝒞\mathcal{C}caligraphic_C specializes to the union of a smooth conic and a line. Since we are excluding μ=0 𝜇 0\mu=0 italic_μ = 0, such a line cannot be l 𝑙 l italic_l. Hence two among A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C belong to the smooth conic and the other one to the line. Report issue for preceding element If the line contains A 𝐴 A italic_A, then it is l A subscript 𝑙 𝐴 l_{A}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT and the smooth conic must contain B,C 𝐵 𝐶 B,C italic_B , italic_C, not N 𝑁 N italic_N (which already belongs to l A subscript 𝑙 𝐴 l_{A}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT and is smooth). This is the case λ=0 𝜆 0\lambda=0 italic_λ = 0. Report issue for preceding element Symmetrically, if the line contains B 𝐵 B italic_B, then it is l B subscript 𝑙 𝐵 l_{B}italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT and A,C 𝐴 𝐶 A,C italic_A , italic_C belong to the smooth conic. This occurs if and only if c⁢λ−1=0 𝑐 𝜆 1 0 c\lambda-1=0 italic_c italic_λ - 1 = 0. Report issue for preceding element If the line contains C 𝐶 C italic_C, let us show that it must be l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT, so by Lemma 5.5p=η 𝑝 𝜂 p=\eta italic_p = italic_η and λ=−c⁢λ+1 𝜆 𝑐 𝜆 1\lambda=-c\lambda+1 italic_λ = - italic_c italic_λ + 1 and p=p′𝑝 superscript 𝑝′p=p^{\prime}italic_p = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. Indeed, otherwise A 𝐴 A italic_A and N 𝑁 N italic_N would belong to the smooth conic. However l A subscript 𝑙 𝐴 l_{A}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT would be tangent to that conic at A 𝐴 A italic_A and intersect that conic also at N 𝑁 N italic_N. Report issue for preceding element Finally, the line cannot contain only the point N 𝑁 N italic_N among the four considered points, since the three aligned points A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C cannot lie on a smooth conic. Report issue for preceding element Finally, reducibility sometimes arises: for μ=0 𝜇 0\mu=0 italic_μ = 0 (the curve contains l 𝑙 l italic_l), for λ=0 𝜆 0\lambda=0 italic_λ = 0 (the curve contains l A subscript 𝑙 𝐴 l_{A}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT), for η=0 𝜂 0\eta=0 italic_η = 0 (the curve contains l B subscript 𝑙 𝐵 l_{B}italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT) and, in the case p=p′𝑝 superscript 𝑝′p=p^{\prime}italic_p = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, also for λ=−c⁢λ+1 𝜆 𝑐 𝜆 1\lambda=-c\lambda+1 italic_λ = - italic_c italic_λ + 1 (the curve contains l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT). Report issue for preceding element Recall that we are particularly interested in the specialization p=η 𝑝 𝜂 p=\eta italic_p = italic_η. We now prove: Report issue for preceding element Lemma 5.8. Report issue for preceding element If p,p′,s 𝑝 superscript 𝑝′𝑠 p,p^{\prime},s italic_p , italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_s are pairwise distinct, the specialization p=η 𝑝 𝜂 p=\eta italic_p = italic_η, i.e. λ=p⁢(1−c)/r 𝜆 𝑝 1 𝑐 𝑟\lambda=p(1-c)/r italic_λ = italic_p ( 1 - italic_c ) / italic_r, leaves the cubic curve irreducible. Report issue for preceding element Proof. Report issue for preceding element Suppose that O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT satisfy the hypotheses of the lemma. By what we said, when the curve 𝒞 𝒞\mathcal{C}caligraphic_C becomes reducible, it contains one of the lines l,l A,l B,l C 𝑙 subscript 𝑙 𝐴 subscript 𝑙 𝐵 subscript 𝑙 𝐶 l,l_{A},l_{B},l_{C}italic_l , italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT. Report issue for preceding element Since p≠p′𝑝 superscript 𝑝′p\neq p^{\prime}italic_p ≠ italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT by Lemma 5.5 the curve specialized at p=η 𝑝 𝜂 p=\eta italic_p = italic_η does not contain l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT. Report issue for preceding element Suppose that it contains l A subscript 𝑙 𝐴 l_{A}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT under the specialization p=η 𝑝 𝜂 p=\eta italic_p = italic_η; by Lemma 5.7 this means that λ=0 𝜆 0\lambda=0 italic_λ = 0, so c=1 𝑐 1 c=1 italic_c = 1, whence (in the above notation) P′=P+R superscript 𝑃′𝑃 𝑅 P^{\prime}=P+R italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_P + italic_R where (R.P)=0(R.P)=0( italic_R . italic_P ) = 0. This implies s=(P.P′)=\|P\|2=p s=(P.P^{\prime})=\|P\|^{2}=p italic_s = ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p, contrary to assumptions. Report issue for preceding element The argument for B 𝐵 B italic_B is symmetric: if the cubic contains l B subscript 𝑙 𝐵 l_{B}italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT (for p=η 𝑝 𝜂 p=\eta italic_p = italic_η) then we derive from Lemma 5.7 that c⁢λ=1 𝑐 𝜆 1 c\lambda=1 italic_c italic_λ = 1. This, together with p=η 𝑝 𝜂 p=\eta italic_p = italic_η yields p′=p⁢c 2+r=p⁢c=s superscript 𝑝′𝑝 superscript 𝑐 2 𝑟 𝑝 𝑐 𝑠 p^{\prime}=pc^{2}+r=pc=s italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_p italic_c start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r = italic_p italic_c = italic_s. Report issue for preceding element So far we have excluded that the curve contains l C,l A,l B subscript 𝑙 𝐶 subscript 𝑙 𝐴 subscript 𝑙 𝐵 l_{C},l_{A},l_{B}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT. Report issue for preceding element Let us exclude that it may contain l 𝑙 l italic_l; otherwise μ=0 𝜇 0\mu=0 italic_μ = 0, so λ 2=−p/r superscript 𝜆 2 𝑝 𝑟\lambda^{2}=-p/r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = - italic_p / italic_r. This is not compatible with p=η 𝑝 𝜂 p=\eta italic_p = italic_η, which corresponds to a real value of λ 𝜆\lambda italic_λ. ∎ Report issue for preceding element We now have a general geometrical lemma. Report issue for preceding element Lemma 5.9. Report issue for preceding element Let 𝒞 𝒞\mathcal{C}caligraphic_C be an irreducible cubic curve, Q∈𝒞 𝑄 𝒞 Q\in\mathcal{C}italic_Q ∈ caligraphic_C any point. Suppose that there exist three distinct smooth points A,B,C∈𝒞∖{Q}𝐴 𝐵 𝐶 𝒞 𝑄 A,B,C\in\mathcal{C}\setminus{Q}italic_A , italic_B , italic_C ∈ caligraphic_C ∖ { italic_Q } such that the tangents to the curve at these points all contain the point Q 𝑄 Q italic_Q. Then 𝒞 𝒞\mathcal{C}caligraphic_C is smooth. If moreover A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C are collinear, then Q 𝑄 Q italic_Q is a flex. Report issue for preceding element Proof. Report issue for preceding element The projection from Q 𝑄 Q italic_Q provides a rational map 𝒞→ℙ 1→𝒞 subscript ℙ 1\mathcal{C}\to{\mathbb{P}}_{1}caligraphic_C → blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT of degree 2 2 2 2 ramified (at least) at the points A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C. Then by Hurwitz genus formula the geometric genus of 𝒞 𝒞\mathcal{C}caligraphic_C is (at least) 1 1 1 1, so it coincides with the arithmetic genus and the curve is smooth. Report issue for preceding element This proves the first assertion of the lemma. Report issue for preceding element Suppose now A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C are collinear and consider the three lines l A,l B,l C subscript 𝑙 𝐴 subscript 𝑙 𝐵 subscript 𝑙 𝐶 l_{A},l_{B},l_{C}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT joining the three points A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C to Q 𝑄 Q italic_Q, which by assumption are tangent to 𝒞 𝒞\mathcal{C}caligraphic_C at A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C respectively. Denote also by l A,l B,l C subscript 𝑙 𝐴 subscript 𝑙 𝐵 subscript 𝑙 𝐶 l_{A},l_{B},l_{C}italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT , italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT three corresponding linear forms providing their equations. Let now l 𝑙 l italic_l be the line containing A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C and again use the same symbol for a corresponding linear form. Finally, let l∗superscript 𝑙 l^{}italic_l start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT denote both the tangent to the curve at Q 𝑄 Q italic_Q and a corresponding linear form. Report issue for preceding element The rational function l 2⋅l∗l A⋅l B⋅l C⋅superscript 𝑙 2 superscript 𝑙⋅subscript 𝑙 𝐴 subscript 𝑙 𝐵 subscript 𝑙 𝐶\displaystyle{\frac{l^{2}\cdot l^{}}{l_{A}\cdot l_{B}\cdot l_{C}}}divide start_ARG italic_l start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ⋅ italic_l start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT end_ARG start_ARG italic_l start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT ⋅ italic_l start_POSTSUBSCRIPT italic_B end_POSTSUBSCRIPT ⋅ italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT end_ARG, viewed as a function on the curve 𝒞 𝒞\mathcal{C}caligraphic_C, is regular and non vanishing at A,B,C 𝐴 𝐵 𝐶 A,B,C italic_A , italic_B , italic_C; if Q 𝑄 Q italic_Q were not a flex, it would have a simple pole at the point Q 𝑄 Q italic_Q and no other pole, resulting in a function of exact degree 1 1 1 1, which is impossible on any smooth cubic curve. ∎ Report issue for preceding element The sections are not torsion. As above, take for the origin the point N=(p:η:μ)N=(p:\eta:\mu)italic_N = ( italic_p : italic_η : italic_μ ) and consider the three sections provided by the (smooth) points A:=(1:0:0),B:=(0:1:0),C:=(1:1:0)A:=(1:0:0),\,B:=(0:1:0),\,C:=(1:1:0)italic_A := ( 1 : 0 : 0 ) , italic_B := ( 0 : 1 : 0 ) , italic_C := ( 1 : 1 : 0 ). Report issue for preceding element We want to specialize λ 𝜆\lambda italic_λ so that p=η 𝑝 𝜂 p=\eta italic_p = italic_η; in that case the line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT will be tangent at C 𝐶 C italic_C and the values of the three sections will satisfy A+B=C 𝐴 𝐵 𝐶 A+B=C italic_A + italic_B = italic_C for the group law, since the linear equivalence A+B+C∼2⁢C+N similar-to 𝐴 𝐵 𝐶 2 𝐶 𝑁 A+B+C\sim 2C+N italic_A + italic_B + italic_C ∼ 2 italic_C + italic_N then holds. Report issue for preceding element Proposition 5.10. Report issue for preceding element If p,p′,s 𝑝 superscript 𝑝′𝑠 p,p^{\prime},s italic_p , italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_s are pairwise distinct, the two sections A 𝐴 A italic_A and B 𝐵 B italic_B are not both torsion. Report issue for preceding element Note that we always have A−B=C 𝐴 𝐵 𝐶 A-B=C italic_A - italic_B = italic_C, which is torsion, hence that the two sections are not both torsion amounts that none of them is torsion. Report issue for preceding element Proof. Report issue for preceding element First note that they are not identically torsion of order 2 2 2 2: indeed if any of A,B 𝐴 𝐵 A,B italic_A , italic_B were identically torsion of order 2 2 2 2, then N 𝑁 N italic_N would be identically a flex, since the tangents at A,B 𝐴 𝐵 A,B italic_A , italic_B pass through N 𝑁 N italic_N, so we would have 2⁢A+N∼3⁢N similar-to 2 𝐴 𝑁 3 𝑁 2A+N\sim 3N 2 italic_A + italic_N ∼ 3 italic_N or 2⁢B+N∼3⁢N similar-to 2 𝐵 𝑁 3 𝑁 2B+N\sim 3N 2 italic_B + italic_N ∼ 3 italic_N, and the tangent at N 𝑁 N italic_N would not meet the curve elsewhere. That this is not the case can be proved by explicit computation, but we can also argue as follows. Report issue for preceding element Suppose by contradiction that both A 𝐴 A italic_A and B 𝐵 B italic_B are identically 2 2 2 2-torsion; since A+C=B 𝐴 𝐶 𝐵 A+C=B italic_A + italic_C = italic_B we would have also identically C=A+B 𝐶 𝐴 𝐵 C=A+B italic_C = italic_A + italic_B; however, the tangent drawn from C 𝐶 C italic_C does not always meet the origin N 𝑁 N italic_N, which (in view of calculations above) prevents C 𝐶 C italic_C to be equal to A+B 𝐴 𝐵 A+B italic_A + italic_B. Report issue for preceding element Coming back to our main issue, let us now specialize λ 𝜆\lambda italic_λ to the unique point such that p=η 𝑝 𝜂 p=\eta italic_p = italic_η, i.e. λ=p⁢(1−c)/r 𝜆 𝑝 1 𝑐 𝑟\lambda=p(1-c)/r italic_λ = italic_p ( 1 - italic_c ) / italic_r, so that by Lemma 5.5 the tangent at C 𝐶 C italic_C passes through N 𝑁 N italic_N. Report issue for preceding element Since we are assuming the hypotheses of the previous Lemma 5.8, the curve is irreducible. Then Lemma 5.9 asserts it is smooth, i.e. this value of λ 𝜆\lambda italic_λ is a specialization of good reduction. Report issue for preceding element On the other hand, under this good specialization the sections A,B 𝐴 𝐵 A,B italic_A , italic_B satisfy the extra relation A+B=C 𝐴 𝐵 𝐶 A+B=C italic_A + italic_B = italic_C (which provides also 2⁢A=2⁢B=O 2 𝐴 2 𝐵 𝑂 2A=2B=O 2 italic_A = 2 italic_B = italic_O). Report issue for preceding element By general (elementary) theory of reduction of elliptic schemes, this proves that A,B 𝐴 𝐵 A,B italic_A , italic_B are not identically torsion, for otherwise the torsion order would be preserved by reduction, contrary to the opening assertion. ∎ Report issue for preceding element In view of this result, in order to achieve our goal (i.e. to find a non-torsion rational section), it suffices to work on the assumption that two among p=\|P\|2 𝑝 superscript 𝑃 2 p=\|P\|^{2}italic_p = \| italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, p′=\|P′\|2 superscript 𝑝′superscript superscript 𝑃′2 p^{\prime}=\|P^{\prime}\|^{2}italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = \| italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, s=(P.P′)s=(P.P^{\prime})italic_s = ( italic_P . italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) are equal. Report issue for preceding element However note that our goal remains unchanged if we permute the three points O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, and the same holds for our assumptions. Indeed, changing the origin of ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT amounts to add a same vector to O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT and this leaves invariant the space ℚ⁢(P−O)+ℚ⁢(P′−O)ℚ 𝑃 𝑂 ℚ superscript 𝑃′𝑂{\mathbb{Q}}(P-O)+{\mathbb{Q}}(P^{\prime}-O)blackboard_Q ( italic_P - italic_O ) + blackboard_Q ( italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_O ), which can be defined as the ℚ ℚ{\mathbb{Q}}blackboard_Q-vector subspace of ℝ 2 superscript ℝ 2{\mathbb{R}}^{2}blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT made up of linear combinations of the three points, with zero sum of coordinates, which is independent of the origin. Report issue for preceding element Suppose then that the triangle O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT is not equilateral. If not all three sides have equal lengths, then we may choose the origin so that p≠p′𝑝 superscript 𝑝′p\neq p^{\prime}italic_p ≠ italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. Also, we may assume that p=s 𝑝 𝑠 p=s italic_p = italic_s. Then P′−P superscript 𝑃′𝑃 P^{\prime}-P italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_P is orthogonal to P 𝑃 P italic_P. and \|P′−P\|2=p′−p superscript superscript 𝑃′𝑃 2 superscript 𝑝′𝑝\|P^{\prime}-P\|^{2}=p^{\prime}-p\| italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_P \| start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_p. If this is different from p 𝑝 p italic_p, then we may move the origin in P 𝑃 P italic_P and obtain a triangle with two distinct orthogonal sides, falling into the ‘good’ case already discussed. Therefore we may assume that p′−p=p superscript 𝑝′𝑝 𝑝 p^{\prime}-p=p italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_p = italic_p, i.e. that p′=2⁢p superscript 𝑝′2 𝑝 p^{\prime}=2p italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = 2 italic_p. Report issue for preceding element This corresponds to the fact that the triangle O⁢P⁢P′𝑂 𝑃 superscript 𝑃′OPP^{\prime}italic_O italic_P italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT is isosceles and right-angled in P 𝑃 P italic_P. Let us first treat this case. Report issue for preceding element The case of the rectangular isosceles triangle. In this case the strategy will be the same as above, but we shall use another specialization. Note that in the present situation we have c=1 𝑐 1 c=1 italic_c = 1, P′=P+R superscript 𝑃′𝑃 𝑅 P^{\prime}=P+R italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_P + italic_R, r=p 𝑟 𝑝 r=p italic_r = italic_p, and p′=2⁢p superscript 𝑝′2 𝑝 p^{\prime}=2p italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = 2 italic_p. Report issue for preceding element We take λ=1−c⁢λ=1−λ 𝜆 1 𝑐 𝜆 1 𝜆\lambda=1-c\lambda=1-\lambda italic_λ = 1 - italic_c italic_λ = 1 - italic_λ, i.e. λ=1/2 𝜆 1 2\lambda=1/2 italic_λ = 1 / 2, so that, , by Lemma 5.5 the line l C subscript 𝑙 𝐶 l_{C}italic_l start_POSTSUBSCRIPT italic_C end_POSTSUBSCRIPT joining C 𝐶 C italic_C to N 𝑁 N italic_N becomes tangent to the cubic at N 𝑁 N italic_N. Since then A+B+C∼C+2⁢N similar-to 𝐴 𝐵 𝐶 𝐶 2 𝑁 A+B+C\sim C+2N italic_A + italic_B + italic_C ∼ italic_C + 2 italic_N, this implies A+B=0 𝐴 𝐵 0 A+B=0 italic_A + italic_B = 0, which is an extra relation between A,B 𝐴 𝐵 A,B italic_A , italic_B. Report issue for preceding element We have then to prove that this specialization leads to a smooth curve. Report issue for preceding element Note that in the present situation we have η=p⁢c+r⁢λ=3⁢p/2 𝜂 𝑝 𝑐 𝑟 𝜆 3 𝑝 2\eta=pc+r\lambda=3p/2 italic_η = italic_p italic_c + italic_r italic_λ = 3 italic_p / 2, μ 2=p⁢(1+λ 2)=5⁢p/4 superscript 𝜇 2 𝑝 1 superscript 𝜆 2 5 𝑝 4\mu^{2}=p(1+\lambda^{2})=5p/4 italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p ( 1 + italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) = 5 italic_p / 4. Report issue for preceding element Our cubic becomes (in affine coordinates) Report issue for preceding element (p−k 2)⁢(3⁢p 2−μ⁢k′)=(2⁢p−k′⁣2)⁢(p−μ⁢k).𝑝 superscript 𝑘 2 3 𝑝 2 𝜇 superscript 𝑘′2 𝑝 superscript 𝑘′2 𝑝 𝜇 𝑘(p-k^{2})(\frac{3p}{2}-\mu k^{\prime})=(2p-k^{\prime 2})(p-\mu k).( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( divide start_ARG 3 italic_p end_ARG start_ARG 2 end_ARG - italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = ( 2 italic_p - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) . Note that it suffices to exclude singularities at finite points since first of all, the curve is smooth at infinity, since the line H=0 𝐻 0 H=0 italic_H = 0 is not a component of the curve and it intersects the curve at three distinct points. Report issue for preceding element We write the equation as X⁢Y=Z⁢W 𝑋 𝑌 𝑍 𝑊 XY=ZW italic_X italic_Y = italic_Z italic_W. Differentiating with respect to k 𝑘 k italic_k and k′superscript 𝑘′k^{\prime}italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT we see that a point is singular if and only if the two conditions 2⁢k⁢Y=μ⁢Z 2 𝑘 𝑌 𝜇 𝑍 2kY=\mu Z 2 italic_k italic_Y = italic_μ italic_Z and μ⁢X=2⁢k′⁢W 𝜇 𝑋 2 superscript 𝑘′𝑊\mu X=2k^{\prime}W italic_μ italic_X = 2 italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_W hold. Report issue for preceding element Suppose first that W=0 𝑊 0 W=0 italic_W = 0, so k=p/μ 𝑘 𝑝 𝜇 k=p/\mu italic_k = italic_p / italic_μ; then k 2=p 2/μ 2=4⁢p/5≠p superscript 𝑘 2 superscript 𝑝 2 superscript 𝜇 2 4 𝑝 5 𝑝 k^{2}=p^{2}/\mu^{2}=4p/5\neq p italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4 italic_p / 5 ≠ italic_p. So X≠0 𝑋 0 X\neq 0 italic_X ≠ 0 and we the second condition does not hold. Hence W≠0 𝑊 0 W\neq 0 italic_W ≠ 0. Multiplying by W 𝑊 W italic_W the first condition we get 2⁢k⁢W⁢Y=μ⁢Z⁢W=μ⁢X⁢Y 2 𝑘 𝑊 𝑌 𝜇 𝑍 𝑊 𝜇 𝑋 𝑌 2kWY=\mu ZW=\mu XY 2 italic_k italic_W italic_Y = italic_μ italic_Z italic_W = italic_μ italic_X italic_Y. Report issue for preceding element If Y=0 𝑌 0 Y=0 italic_Y = 0, then Z=0 𝑍 0 Z=0 italic_Z = 0 as well and we find k′=3⁢p/(2⁢μ)superscript 𝑘′3 𝑝 2 𝜇 k^{\prime}=3p/(2\mu)italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = 3 italic_p / ( 2 italic_μ ) and k′⁣2=2⁢p superscript 𝑘′2 2 𝑝 k^{\prime 2}=2p italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT = 2 italic_p, hence 9⁢p 2/(4⁢μ 2)=2⁢p 9 superscript 𝑝 2 4 superscript 𝜇 2 2 𝑝 9p^{2}/(4\mu^{2})=2p 9 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / ( 4 italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) = 2 italic_p, which in turn yields 9⁢p/5=2⁢p 9 𝑝 5 2 𝑝 9p/5=2p 9 italic_p / 5 = 2 italic_p, which is impossible. Report issue for preceding element Hence Y≠0 𝑌 0 Y\neq 0 italic_Y ≠ 0 and 2⁢k⁢W=μ⁢X 2 𝑘 𝑊 𝜇 𝑋 2kW=\mu X 2 italic_k italic_W = italic_μ italic_X. Since W≠0 𝑊 0 W\neq 0 italic_W ≠ 0 this yields k=k′𝑘 superscript 𝑘′k=k^{\prime}italic_k = italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, and we remain with the conditions Report issue for preceding element (p−k 2)⁢(3⁢p 2−μ⁢k)=(2⁢p−k 2)⁢(p−μ⁢k),2⁢k⁢(p−μ⁢k)=μ⁢(p−k 2).formulae-sequence 𝑝 superscript 𝑘 2 3 𝑝 2 𝜇 𝑘 2 𝑝 superscript 𝑘 2 𝑝 𝜇 𝑘 2 𝑘 𝑝 𝜇 𝑘 𝜇 𝑝 superscript 𝑘 2(p-k^{2})(\frac{3p}{2}-\mu k)=(2p-k^{2})(p-\mu k),\qquad 2k(p-\mu k)=\mu(p-k^{% 2}).( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( divide start_ARG 3 italic_p end_ARG start_ARG 2 end_ARG - italic_μ italic_k ) = ( 2 italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) , 2 italic_k ( italic_p - italic_μ italic_k ) = italic_μ ( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) . They respectively yield p⁢k 2−2⁢p⁢μ⁢k+p 2=0 𝑝 superscript 𝑘 2 2 𝑝 𝜇 𝑘 superscript 𝑝 2 0 pk^{2}-2p\mu k+p^{2}=0 italic_p italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 italic_p italic_μ italic_k + italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 and μ⁢k 2−2⁢p⁢k+μ⁢p=0 𝜇 superscript 𝑘 2 2 𝑝 𝑘 𝜇 𝑝 0\mu k^{2}-2pk+\mu p=0 italic_μ italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 italic_p italic_k + italic_μ italic_p = 0. The discriminants up to a factor 4 4 4 4 are p 2⁢μ 2−p 3=p 3/4 superscript 𝑝 2 superscript 𝜇 2 superscript 𝑝 3 superscript 𝑝 3 4 p^{2}\mu^{2}-p^{3}=p^{3}/4 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT = italic_p start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT / 4 and p 2−p⁢μ 2=−p 2/4 superscript 𝑝 2 𝑝 superscript 𝜇 2 superscript 𝑝 2 4 p^{2}-p\mu^{2}=-p^{2}/4 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = - italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT / 4, hence one equation has two real roots, the other one two non-real conjugate roots. So there is no common solution whatever p 𝑝 p italic_p. Report issue for preceding element We are left with the case of an equilateral triangle. Now p=p′𝑝 superscript 𝑝′p=p^{\prime}italic_p = italic_p start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, s=p/2 𝑠 𝑝 2 s=p/2 italic_s = italic_p / 2, c=1/2 𝑐 1 2 c=1/2 italic_c = 1 / 2, r=5⁢p/4 𝑟 5 𝑝 4 r=5p/4 italic_r = 5 italic_p / 4. Report issue for preceding element The case of the equilateral triangle. In this case we choose the specialization λ=c⁢λ−1=(λ/2)−1 𝜆 𝑐 𝜆 1 𝜆 2 1\lambda=c\lambda-1=(\lambda/2)-1 italic_λ = italic_c italic_λ - 1 = ( italic_λ / 2 ) - 1, i.e. λ=−2 𝜆 2\lambda=-2 italic_λ = - 2, and prove it leads to a smooth cubic. We have μ 2=p+4⁢r=6⁢p superscript 𝜇 2 𝑝 4 𝑟 6 𝑝\mu^{2}=p+4r=6p italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p + 4 italic_r = 6 italic_p. Report issue for preceding element Again, we can work with the affine model, which takes the shape Report issue for preceding element (p−k 2)⁢(−2⁢p−μ⁢k′)=(3⁢p/2−k′⁣2)⁢(p−μ⁢k).𝑝 superscript 𝑘 2 2 𝑝 𝜇 superscript 𝑘′3 𝑝 2 superscript 𝑘′2 𝑝 𝜇 𝑘(p-k^{2})(-2p-\mu k^{\prime})=(3p/2-k^{\prime 2})(p-\mu k).( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( - 2 italic_p - italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = ( 3 italic_p / 2 - italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) . In order to compute possible singularities, we argue as before, writing this as X⁢Y=Z⁢W 𝑋 𝑌 𝑍 𝑊 XY=ZW italic_X italic_Y = italic_Z italic_W. The two derivatives yield the equations 2⁢k⁢Y=μ⁢Z 2 𝑘 𝑌 𝜇 𝑍 2kY=\mu Z 2 italic_k italic_Y = italic_μ italic_Z and μ⁢X=2⁢k′⁢W 𝜇 𝑋 2 superscript 𝑘′𝑊\mu X=2k^{\prime}W italic_μ italic_X = 2 italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_W. Report issue for preceding element Again, suppose that W=0 𝑊 0 W=0 italic_W = 0, then X=0 𝑋 0 X=0 italic_X = 0, hence p 2=μ 2⁢k 2=6⁢p⁢k 2=6⁢p 2 superscript 𝑝 2 superscript 𝜇 2 superscript 𝑘 2 6 𝑝 superscript 𝑘 2 6 superscript 𝑝 2 p^{2}=\mu^{2}k^{2}=6pk^{2}=6p^{2}italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 6 italic_p italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 6 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, which does not hold. Multiplying by W 𝑊 W italic_W the first condition and using the cubic equation we get 2⁢k⁢Y⁢W=μ⁢X⁢Y 2 𝑘 𝑌 𝑊 𝜇 𝑋 𝑌 2kYW=\mu XY 2 italic_k italic_Y italic_W = italic_μ italic_X italic_Y. If Y=0 𝑌 0 Y=0 italic_Y = 0 then Z=0 𝑍 0 Z=0 italic_Z = 0, hence (2⁢p)2=(μ⁢k′)2=6⁢p⁢k′⁣2=9⁢p 2 superscript 2 𝑝 2 superscript 𝜇 superscript 𝑘′2 6 𝑝 superscript 𝑘′2 9 superscript 𝑝 2(2p)^{2}=(\mu k^{\prime})^{2}=6pk^{\prime 2}=9p^{2}( 2 italic_p ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_μ italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 6 italic_p italic_k start_POSTSUPERSCRIPT ′ 2 end_POSTSUPERSCRIPT = 9 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, which once more does not hold. Hence we can divide out by Y 𝑌 Y italic_Y and find 2⁢k⁢W=μ⁢X 2 𝑘 𝑊 𝜇 𝑋 2kW=\mu X 2 italic_k italic_W = italic_μ italic_X, whereas also μ⁢X=2⁢k′⁢W 𝜇 𝑋 2 superscript 𝑘′𝑊\mu X=2k^{\prime}W italic_μ italic_X = 2 italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_W holds. Since W≠0 𝑊 0 W\neq 0 italic_W ≠ 0 we get again k=k′𝑘 superscript 𝑘′k=k^{\prime}italic_k = italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, leading to the system Report issue for preceding element (k 2−p)⁢(2⁢p+μ⁢k)=(3⁢p/2−k 2)⁢(p−μ⁢k),2⁢k⁢(p−μ⁢k)=μ⁢(p−k 2),formulae-sequence superscript 𝑘 2 𝑝 2 𝑝 𝜇 𝑘 3 𝑝 2 superscript 𝑘 2 𝑝 𝜇 𝑘 2 𝑘 𝑝 𝜇 𝑘 𝜇 𝑝 superscript 𝑘 2(k^{2}-p)(2p+\mu k)=(3p/2-k^{2})(p-\mu k),\qquad 2k(p-\mu k)=\mu(p-k^{2}),( italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_p ) ( 2 italic_p + italic_μ italic_k ) = ( 3 italic_p / 2 - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( italic_p - italic_μ italic_k ) , 2 italic_k ( italic_p - italic_μ italic_k ) = italic_μ ( italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) , which may be rewritten as Report issue for preceding element 6⁢p⁢k 2+p⁢μ⁢k−7⁢p 2=0,μ⁢k 2−2⁢p⁢k+p⁢μ=0.formulae-sequence 6 𝑝 superscript 𝑘 2 𝑝 𝜇 𝑘 7 superscript 𝑝 2 0 𝜇 superscript 𝑘 2 2 𝑝 𝑘 𝑝 𝜇 0 6pk^{2}+p\mu k-7p^{2}=0,\qquad\mu k^{2}-2pk+p\mu=0.6 italic_p italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_p italic_μ italic_k - 7 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 , italic_μ italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 2 italic_p italic_k + italic_p italic_μ = 0 . The discriminants are respectively p 2⁢μ 2+164⁢p 3>0 superscript 𝑝 2 superscript 𝜇 2 164 superscript 𝑝 3 0 p^{2}\mu^{2}+164p^{3}>0 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 164 italic_p start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT > 0 and 4⁢p 2−4⁢p⁢μ 2=−18⁢p 2<0 4 superscript 𝑝 2 4 𝑝 superscript 𝜇 2 18 superscript 𝑝 2 0 4p^{2}-4p\mu^{2}=-18p^{2}<0 4 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 4 italic_p italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = - 18 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT < 0. As before there cannot be common solutions, which concludes the verification. Report issue for preceding element End of the proof of Theorem 5.2. By means of the last Proposition and the special cases just treated, we have shown that the sections provided by one (hence both) of the points A,B 𝐴 𝐵 A,B italic_A , italic_B are not identically torsion on the elliptic scheme over the conic S:μ 2=p+r⁢λ 2:𝑆 superscript 𝜇 2 𝑝 𝑟 superscript 𝜆 2 S:\mu^{2}=p+r\lambda^{2}italic_S : italic_μ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_p + italic_r italic_λ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. Choose then the section A 𝐴 A italic_A for definiteness, By a well-known theorem of Silverman-Tate (see , Prop. 3.2 p. 69 or Appendix C by Masser) the algebraic points x 𝑥 x italic_x of the conic S 𝑆 S italic_S such that A x subscript 𝐴 𝑥 A_{x}italic_A start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT becomes torsion (on the elliptic curve ℰ x subscript ℰ 𝑥\mathcal{E}{x}caligraphic_E start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT corresponding to x 𝑥 x italic_x) have bounded height. Therefore if we let x 𝑥 x italic_x vary though the rational points S⁢(ℚ)𝑆 ℚ S({\mathbb{Q}})italic_S ( blackboard_Q ) of S 𝑆 S italic_S, for all but finitely many ones A x subscript 𝐴 𝑥 A{x}italic_A start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT will be non-torsion on ℰ x subscript ℰ 𝑥\mathcal{E}{x}caligraphic_E start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT. Therefore by taking multiples n⁢A x,n∈ℤ 𝑛 subscript 𝐴 𝑥 𝑛 ℤ nA{x},n\in{\mathbb{Z}}italic_n italic_A start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT , italic_n ∈ blackboard_Z we shall obtain infinitely many rational points on ℰ x subscript ℰ 𝑥\mathcal{E}{x}caligraphic_E start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT for such points x 𝑥 x italic_x. Since S⁢(ℚ)𝑆 ℚ S({\mathbb{Q}})italic_S ( blackboard_Q ) is non-empty by assumption, we have that S 𝑆 S italic_S is birationally equivalent to ℙ 1 subscript ℙ 1\mathbb{P}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT over ℚ ℚ{\mathbb{Q}}blackboard_Q, hence S⁢(ℚ)𝑆 ℚ S({\mathbb{Q}})italic_S ( blackboard_Q ) is Zariski dense in S 𝑆 S italic_S, whence the rational points are Zariski dense in the cubic family, concluding the argument. Report issue for preceding element 5.3. An alternative construction Report issue for preceding element We present an alternative construction of a section of the above cubic fibration. It also provides rational points. Report issue for preceding element The equations (5.5) and (5.6) and give the following linear relation between x 𝑥 x italic_x and y 𝑦 y italic_y with coefficients in ℚ⁢[k,k′]ℚ 𝑘 superscript 𝑘′{\mathbb{Q}}[k,k^{\prime}]blackboard_Q [ italic_k , italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ] Report issue for preceding element 2⁢p⁢φ 1⁢x+2⁢r⁢k⁢y=φ 0,2 𝑝 subscript 𝜑 1 𝑥 2 𝑟 𝑘 𝑦 subscript 𝜑 0 2p\varphi_{1}x+2rky=\varphi_{0},2 italic_p italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_x + 2 italic_r italic_k italic_y = italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT ,(5.12) where Report issue for preceding element φ 1=k′−k⁢c,φ 0=k′⁢δ−k⁢δ′.formulae-sequence subscript 𝜑 1 superscript 𝑘′𝑘 𝑐 subscript 𝜑 0 superscript 𝑘′𝛿 𝑘 superscript 𝛿′\varphi_{1}=k^{\prime}-kc,\qquad\varphi_{0}=k^{\prime}\delta-k\delta^{\prime}.italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT - italic_k italic_c , italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT = italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT italic_δ - italic_k italic_δ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT .(5.13) Then (5.12) allows to express y 𝑦 y italic_y as 2⁢r⁢k⁢y=φ 0−2⁢p⁢φ 1⁢x 2 𝑟 𝑘 𝑦 subscript 𝜑 0 2 𝑝 subscript 𝜑 1 𝑥 2rky=\varphi_{0}-2p\varphi_{1}x 2 italic_r italic_k italic_y = italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT - 2 italic_p italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_x. Multiplying the equation p⁢x 2+r⁢y 2=z 2 𝑝 superscript 𝑥 2 𝑟 superscript 𝑦 2 superscript 𝑧 2 px^{2}+ry^{2}=z^{2}italic_p italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT by 4⁢r⁢k 2 4 𝑟 superscript 𝑘 2 4rk^{2}4 italic_r italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and substituting for 2⁢r⁢k⁢y 2 𝑟 𝑘 𝑦 2rky 2 italic_r italic_k italic_y from the last equation one gets Report issue for preceding element 4⁢p⁢r⁢k 2⁢x 2+(2⁢p⁢φ 1⁢x−φ 0)2=4⁢r⁢k 2⁢z 2;4 𝑝 𝑟 superscript 𝑘 2 superscript 𝑥 2 superscript 2 𝑝 subscript 𝜑 1 𝑥 subscript 𝜑 0 2 4 𝑟 superscript 𝑘 2 superscript 𝑧 2 4prk^{2}x^{2}+(2p\varphi_{1}x-\varphi_{0})^{2}=4rk^{2}z^{2};4 italic_p italic_r italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( 2 italic_p italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_x - italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 4 italic_r italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ; using now the equation 2⁢k⁢z=2⁢p⁢x−δ 2 𝑘 𝑧 2 𝑝 𝑥 𝛿 2kz=2px-\delta 2 italic_k italic_z = 2 italic_p italic_x - italic_δ one obtains Report issue for preceding element 4⁢p⁢r⁢k 2⁢x 2+(2⁢p⁢φ 1⁢x−φ 0)2=r⁢(2⁢p⁢x−δ)2 4 𝑝 𝑟 superscript 𝑘 2 superscript 𝑥 2 superscript 2 𝑝 subscript 𝜑 1 𝑥 subscript 𝜑 0 2 𝑟 superscript 2 𝑝 𝑥 𝛿 2 4prk^{2}x^{2}+(2p\varphi_{1}x-\varphi_{0})^{2}=r(2px-\delta)^{2}4 italic_p italic_r italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( 2 italic_p italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_x - italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_r ( 2 italic_p italic_x - italic_δ ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(5.14) Hence we get the following quadratic equation for x 𝑥 x italic_x: Report issue for preceding element A⁢x 2+B⁢x+C=0,𝐴 superscript 𝑥 2 𝐵 𝑥 𝐶 0 Ax^{2}+Bx+C=0,italic_A italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_B italic_x + italic_C = 0 ,(5.15) where Report issue for preceding element {A=4⁢p⁢r⁢k 2+4⁢p 2⁢φ 1 2−4⁢p 2⁢r B=4⁢p⁢r⁢δ−4⁢p⁢φ 0⁢φ 1 C=φ 0 2−r⁢δ 2.\left{\begin{matrix}A&=&4prk^{2}+4p^{2}\varphi_{1}^{2}-4p^{2}r\ B&=&4pr\delta-4p\varphi_{0}\varphi_{1}\ C&=&\varphi_{0}^{2}-r\delta^{2}.\end{matrix}\right.{ start_ARG start_ROW start_CELL italic_A end_CELL start_CELL = end_CELL start_CELL 4 italic_p italic_r italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 4 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - 4 italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_r end_CELL end_ROW start_ROW start_CELL italic_B end_CELL start_CELL = end_CELL start_CELL 4 italic_p italic_r italic_δ - 4 italic_p italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_CELL end_ROW start_ROW start_CELL italic_C end_CELL start_CELL = end_CELL start_CELL italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_r italic_δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . end_CELL end_ROW end_ARG(5.16) We want to produce a rational solution to (5.15). One way to do this is to consider the case when A=0 𝐴 0 A=0 italic_A = 0, thus obtaining a linear equation in x 𝑥 x italic_x that has obviously a rational solution. We note that the condition A=0 𝐴 0 A=0 italic_A = 0 amounts to Report issue for preceding element p⁢(k/p)2+r⁢(φ 1/r)2=1.𝑝 superscript 𝑘 𝑝 2 𝑟 superscript subscript 𝜑 1 𝑟 2 1 p(k/p)^{2}+r(\varphi_{1}/r)^{2}=1.italic_p ( italic_k / italic_p ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r ( italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT / italic_r ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 1 .(5.17) This represents a point on the conic p⁢X 2+r⁢Y 2=1 𝑝 superscript 𝑋 2 𝑟 superscript 𝑌 2 1 pX^{2}+rY^{2}=1 italic_p italic_X start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r italic_Y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 1, which by assumption has a rational point, and thus can be parametrized rationally, say in terms of a parameter t 𝑡 t italic_t, as p⁢f⁢(t)2+r⁢g⁢(t)2=1 𝑝 𝑓 superscript 𝑡 2 𝑟 𝑔 superscript 𝑡 2 1 pf(t)^{2}+rg(t)^{2}=1 italic_p italic_f ( italic_t ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_r italic_g ( italic_t ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 1, where f,g 𝑓 𝑔 f,g italic_f , italic_g are certain (quadratic) non-constant rational functions in ℚ⁢(t)ℚ 𝑡{\mathbb{Q}}(t)blackboard_Q ( italic_t ). Putting k=p⁢f⁢(t)𝑘 𝑝 𝑓 𝑡 k=pf(t)italic_k = italic_p italic_f ( italic_t ), φ 1=r⁢g⁢(t)subscript 𝜑 1 𝑟 𝑔 𝑡\varphi_{1}=rg(t)italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = italic_r italic_g ( italic_t ), we can express both k 𝑘 k italic_k and k′superscript 𝑘′k^{\prime}italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT as rational functions of t 𝑡 t italic_t; in turn, we can express φ 0 subscript 𝜑 0\varphi_{0}italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT as well rationally in terms of t 𝑡 t italic_t. Note that we can also use the former coordinates on the conic S 𝑆 S italic_S, which would yield r⁢k⁢λ=p⁢φ 1 𝑟 𝑘 𝜆 𝑝 subscript 𝜑 1 rk\lambda=p\varphi_{1}italic_r italic_k italic_λ = italic_p italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, k=p/μ 𝑘 𝑝 𝜇 k=p/\mu italic_k = italic_p / italic_μ, whence k′=φ 1+k⁢c=(r⁢λ+c⁢p)/μ superscript 𝑘′subscript 𝜑 1 𝑘 𝑐 𝑟 𝜆 𝑐 𝑝 𝜇 k^{\prime}=\varphi_{1}+kc=(r\lambda+cp)/\mu italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT + italic_k italic_c = ( italic_r italic_λ + italic_c italic_p ) / italic_μ. Report issue for preceding element The above formula then yields an expression of x=−C/B 𝑥 𝐶 𝐵 x=-C/B italic_x = - italic_C / italic_B in terms of t 𝑡 t italic_t (or of λ,μ 𝜆 𝜇\lambda,\mu italic_λ , italic_μ), provided that B 𝐵 B italic_B is not identically zero as a function of t 𝑡 t italic_t, and therefore allows to express all the variables x,y 𝑥 𝑦 x,y italic_x , italic_y and z 𝑧 z italic_z as rational functions of t 𝑡 t italic_t, using the above linear equations for y 𝑦 y italic_y and for z 𝑧 z italic_z. Report issue for preceding element In order to verify that B 𝐵 B italic_B is not identically zero with this choice, we may argue directly, or else note that in view of equation (5.15) B=0 𝐵 0 B=0 italic_B = 0 also implies C=0 𝐶 0 C=0 italic_C = 0, since we are in the case A=0 𝐴 0 A=0 italic_A = 0. Hence, we have to exclude that B≡C≡0 𝐵 𝐶 0 B\equiv C\equiv 0 italic_B ≡ italic_C ≡ 0 under the present choice of k,φ 1 𝑘 subscript 𝜑 1 k,\varphi_{1}italic_k , italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. Note that C=0 𝐶 0 C=0 italic_C = 0 amounts to φ 0 2=r⁢δ 2 superscript subscript 𝜑 0 2 𝑟 superscript 𝛿 2\varphi_{0}^{2}=r\delta^{2}italic_φ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_r italic_δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and under this condition B=0 𝐵 0 B=0 italic_B = 0 implies (r⁢δ)2=r⁢δ 2⁢φ 1 2 superscript 𝑟 𝛿 2 𝑟 superscript 𝛿 2 superscript subscript 𝜑 1 2(r\delta)^{2}=r\delta^{2}\varphi_{1}^{2}( italic_r italic_δ ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_r italic_δ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT. But δ=p−k 2≠0 𝛿 𝑝 superscript 𝑘 2 0\delta=p-k^{2}\neq 0 italic_δ = italic_p - italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ≠ 0, whence φ 1=r subscript 𝜑 1 𝑟\varphi_{1}=r italic_φ start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = italic_r would be constant on the conic, which is not the case. Report issue for preceding element This completes the verification that x,y 𝑥 𝑦 x,y italic_x , italic_y, and so z 𝑧 z italic_z, are well-defined in terms of t 𝑡 t italic_t. Geometrically, this means that we have defined a rational curve on the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT, which can be viewed as a section of our elliptic fibration. If we verify that this is not identically torsion, this section would give rise to a Zariski-dense set of rational points, as in our former argument. We have not checked this property, since it appears a bit complicated, and since we have already done this for the former sections; however in principle one can argue as above, exploring for instance when the section becomes of order 2 2 2 2, or of any chosen order, and trying to prove that this happens outside the bad reduction. Report issue for preceding element 6. The Geometry of the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT Report issue for preceding element The goal of this section is to study the geometry of the surface Report issue for preceding element 𝒮 3=𝒮⁢(O,P,P′)={Q∈𝔸 2⁢(ℚ):d⁢(O,Q)∈ℚ,d⁢(P,Q)∈ℚ⁢and⁢d⁢(P′,Q)∈ℚ},subscript 𝒮 3 𝒮 𝑂 𝑃 superscript 𝑃′conditional-set 𝑄 superscript 𝔸 2 ℚ formulae-sequence 𝑑 𝑂 𝑄 ℚ 𝑑 𝑃 𝑄 ℚ and 𝑑 superscript 𝑃′𝑄 ℚ{\mathcal{S}}_{3}={\mathcal{S}}(O,P,P^{\prime})={Q\in{\mathbb{A}}^{2}({% \mathbb{Q}}):d(O,Q)\in{\mathbb{Q}},\ d(P,Q)\in{\mathbb{Q}}\text{ and }d(P^{% \prime},Q)\in{\mathbb{Q}}},caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT = caligraphic_S ( italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) = { italic_Q ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( blackboard_Q ) : italic_d ( italic_O , italic_Q ) ∈ blackboard_Q , italic_d ( italic_P , italic_Q ) ∈ blackboard_Q and italic_d ( italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_Q ) ∈ blackboard_Q } , i.e. the set of points Q 𝑄 Q italic_Q with rational distances from O,P 𝑂 𝑃 O,P italic_O , italic_P and P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. For this analysis we restrict to the case in which P,P′∈ℚ 2 𝑃 superscript 𝑃′superscript ℚ 2 P,P^{\prime}\in{\mathbb{Q}}^{2}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ∈ blackboard_Q start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT so that any point with rational distance from O,P 𝑂 𝑃 O,P italic_O , italic_P and P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT is automatically defined over ℚ ℚ{\mathbb{Q}}blackboard_Q (provided that the points are not aligned, which we have already discussed in a previous section). Report issue for preceding element A rational point Q=(x,y)∈𝒮 3⁢(ℚ)𝑄 𝑥 𝑦 subscript 𝒮 3 ℚ Q=(x,y)\in{\mathcal{S}}_{3}({\mathbb{Q}})italic_Q = ( italic_x , italic_y ) ∈ caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ( blackboard_Q ) corresponds to a rational solution to the system Report issue for preceding element x 2+y 2=z 2,superscript 𝑥 2 superscript 𝑦 2 superscript 𝑧 2\displaystyle x^{2}+y^{2}=z^{2},italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(6.1) (x−a)2+(y−b)2=u 2,superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 superscript 𝑢 2\displaystyle(x-a)^{2}+(y-b)^{2}=u^{2},( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(6.2) (x−a′)2+(y−b′)2=v 2,superscript 𝑥 superscript 𝑎′2 superscript 𝑦 superscript 𝑏′2 superscript 𝑣 2\displaystyle(x-a^{\prime})^{2}+(y-b^{\prime})^{2}=v^{2},( italic_x - italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(6.3) where z=d⁢(O,Q)𝑧 𝑑 𝑂 𝑄 z=d(O,Q)italic_z = italic_d ( italic_O , italic_Q ), u=d⁢(P,Q)𝑢 𝑑 𝑃 𝑄 u=d(P,Q)italic_u = italic_d ( italic_P , italic_Q ) and v=d⁢(P′,Q)𝑣 𝑑 superscript 𝑃′𝑄 v=d(P^{\prime},Q)italic_v = italic_d ( italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT , italic_Q ). In this presentation the set 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT lies in the intersection of 3 3 3 3 quadrics in 𝔸 5 superscript 𝔸 5{\mathbb{A}}^{5}blackboard_A start_POSTSUPERSCRIPT 5 end_POSTSUPERSCRIPT, defined by the three equations above. We shall show that (a smooth projective completion of) this variety is a K⁢3 𝐾 3 K3 italic_K 3 elliptic surface with a dense set of rational points over ℚ ℚ{\mathbb{Q}}blackboard_Q. Report issue for preceding element In the sequel we shall also prove that the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is endowed with (several) elliptic fibrations parametrized by a line. The presence of a second fibration is related to the density of rational points even in the sense of the Euclidean topology, as first noticed by Swinnerton-Dyer in . Report issue for preceding element 6.1. The surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT as a ramified cover of ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT Report issue for preceding element Let us show how to re-obtain the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT as an abstract ramified covering of the (x,y)𝑥 𝑦(x,y)( italic_x , italic_y )-plane. In the first part of this paragraph we argue birationally, describing the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT only up to birational transformations. In the last part of the paragraph, we shall work out a smooth complete model. Report issue for preceding element We first note that the three equations above correspond each to a quadratic cover of the plane ramified over the union of two lines: for instance, (6.1) amounts to taking a cover ramified over the singular conic of equation x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0, which consists of a pair of complex conjugate lines. Report issue for preceding element The two other pairs of lines, defined by the vanishing of the left-hand side in equations (6.2), (6.3), have the same slopes of the first pair; hence the six lines intersect in triples at the two points at infinity (1:±i:0):1 plus-or-minus 𝑖:0(1:\pm i:0)( 1 : ± italic_i : 0 ) (under the embedding 𝔸 2↪ℙ 2↪superscript 𝔸 2 subscript ℙ 2{\mathbb{A}}^{2}\hookrightarrow{\mathbb{P}}_{2}blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ↪ blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT given by (x,y)↦(x:y:1)(x,y)\mapsto(x:y:1)( italic_x , italic_y ) ↦ ( italic_x : italic_y : 1 )). Report issue for preceding element As we said, the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is obtained as a degree-eight covering of the plane ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, actually an abelian covering of type (2,2,2)2 2 2(2,2,2)( 2 , 2 , 2 ). This covering can then be decomposed as Report issue for preceding element 𝒮 3→X 2→X 1→ℙ 2→subscript 𝒮 3 subscript 𝑋 2→subscript 𝑋 1→subscript ℙ 2{\mathcal{S}}{3}\to X{2}\to X_{1}\to{\mathbb{P}}_{2}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → italic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT → italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT where each arrow denotes a degree-two covering. (Note that we can choose the surface X 1 subscript 𝑋 1 X_{1}italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT to be birationally equivalent to the surface 𝒮 1⁢(O)subscript 𝒮 1 𝑂{\mathcal{S}}{1}(O)caligraphic_S start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_O ) and X 2 subscript 𝑋 2 X{2}italic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT to be birationally equivalent to the surface 𝒮 2⁢(O,P)subscript 𝒮 2 𝑂 𝑃{\mathcal{S}}_{2}(O,P)caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_O , italic_P ).) Report issue for preceding element For instance, for the first covering X 1→ℙ 2→subscript 𝑋 1 subscript ℙ 2 X_{1}\to{\mathbb{P}}{2}italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT we can take the one ramified over the pair of lines x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0; the second covering is algebraically described by adjoining to the function field of X 1 subscript 𝑋 1 X{1}italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT the square root of the function (x−a)2+(y−b)2 superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2(x-a)^{2}+(y-b)^{2}( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT; it ramifies over the zero set of this function, i.e. the pre-image in X 1 subscript 𝑋 1 X_{1}italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT of the mentioned pair of lines intersecting on P 𝑃 P italic_P; the last one ramifies over the pre-image in X 2 subscript 𝑋 2 X_{2}italic_X start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT of the pair of lines intersecting on P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT. Report issue for preceding element Let us consider the pencil Λ 0 subscript Λ 0\Lambda_{0}roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT of lines in ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT passing through the origin 0 0 of 𝔸 2⊂ℙ 2 superscript 𝔸 2 subscript ℙ 2{\mathbb{A}}^{2}\subset{\mathbb{P}}{2}blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ⊂ blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. Each such line has a reducible pre-image in X 1 subscript 𝑋 1 X_{1}italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, consisting of two components. Each such component l 𝑙 l italic_l is a smooth curve of genus zero above which the cover 𝒮 3→X 1→subscript 𝒮 3 subscript 𝑋 1{\mathcal{S}}{3}\to X{1}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT induces a cover of type (2,2)2 2(2,2)( 2 , 2 ). Since the cover 𝒮 3→X 1→subscript 𝒮 3 subscript 𝑋 1{\mathcal{S}}{3}\to X{1}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ramifies above the four curves in X 1 subscript 𝑋 1 X_{1}italic_X start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT defined by the vanishing of the functions (x−a)2+(y−b)2=0 superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 0(x-a)^{2}+(y-b)^{2}=0( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0 and (x−a′)2+(y−b′)2=0 superscript 𝑥 superscript 𝑎′2 superscript 𝑦 superscript 𝑏′2 0(x-a^{\prime})^{2}+(y-b^{\prime})^{2}=0( italic_x - italic_a start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0, and l 𝑙 l italic_l intersects each such curve at a single point, the ramification on l 𝑙 l italic_l occurs above four points. Hence, by Riemann-Hurwitz, the pre-image in 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT of each such curve l 𝑙 l italic_l is then a genus-one curve. In this way, we obtain a family of genus-one curves on 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. Changing the role of the three equations (6.1), (6.2), (6.3) one can consider the pencils Λ P subscript Λ 𝑃\Lambda_{P}roman_Λ start_POSTSUBSCRIPT italic_P end_POSTSUBSCRIPT (resp. Λ P′subscript Λ superscript 𝑃′\Lambda_{P^{\prime}}roman_Λ start_POSTSUBSCRIPT italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT) of lines through P 𝑃 P italic_P (resp. P′superscript 𝑃′P^{\prime}italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT), obtaining a total of three families of genus-one curves. Report issue for preceding element Let us consider the projection 𝒮 3⇢Λ 0⇢subscript 𝒮 3 subscript Λ 0{\mathcal{S}}{3}\dashrightarrow\Lambda{0}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ⇢ roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT. It is defined by projecting a point s=(x,y,z,u,v)∈𝒮 3 𝑠 𝑥 𝑦 𝑧 𝑢 𝑣 subscript 𝒮 3 s=(x,y,z,u,v)\in{\mathcal{S}}{3}italic_s = ( italic_x , italic_y , italic_z , italic_u , italic_v ) ∈ caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT to the corresponding point (x,y)𝑥 𝑦(x,y)( italic_x , italic_y ) of the plane and then associating the line joining the origin to the point (x,y)𝑥 𝑦(x,y)( italic_x , italic_y ) (which can be identified with the projective point (0:0):0 0(0:0)( 0 : 0 ). As mentioned, the pre-image of any line in Λ 0 subscript Λ 0\Lambda{0}roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT is reducible, unless such line is a component of the singular conic x 2+y 2=0 superscript 𝑥 2 superscript 𝑦 2 0 x^{2}+y^{2}=0 italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 0; in that case, the pre-image is a double curve (a non-reduced divisor). Hence there cannot exist sections Λ 0→𝒮 3→subscript Λ 0 subscript 𝒮 3\Lambda_{0}\to{\mathcal{S}}{3}roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT → caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT of the above defined projection 𝒮 3⇢Λ 0⇢subscript 𝒮 3 subscript Λ 0{\mathcal{S}}{3}\dashrightarrow\Lambda_{0}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ⇢ roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT. However, considering the Stein factorization Report issue for preceding element 𝒮 3⇢ℙ 1→Λ 0⇢subscript 𝒮 3 subscript ℙ 1→subscript Λ 0{\mathcal{S}}{3}\dashrightarrow{\mathbb{P}}{1}\to\Lambda_{0}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ⇢ blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT → roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT(6.4) where ℙ 1→Λ 0→subscript ℙ 1 subscript Λ 0{\mathbb{P}}{1}\to\Lambda{0}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT → roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT is the quadratic cover ramified over the two points (1:±i):1 plus-or-minus 𝑖(1:\pm i)( 1 : ± italic_i ), the first projection Report issue for preceding element π 0:𝒮 3⇢ℙ 1:subscript 𝜋 0⇢subscript 𝒮 3 subscript ℙ 1\pi_{0}:{\mathcal{S}}{3}\dashrightarrow{\mathbb{P}}{1}italic_π start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT : caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ⇢ blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT(6.5) which is a fibration (i.e. its fibers are generically irreducible) does admit sections, as we shall show below. Report issue for preceding element Let us now describe a smooth projective model of the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT which regularizes the projection 𝒮⇢Λ 0⇢𝒮 subscript Λ 0{\mathcal{S}}\dashrightarrow\Lambda{0}caligraphic_S ⇢ roman_Λ start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT. By abuse of notation, we shall denote again by 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT such a model. To construct this model, we shall proceed as in section 2. We start from blowing-up the three points O,P,P′𝑂 𝑃 superscript 𝑃′O,P,P^{\prime}italic_O , italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT on ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, together with the two points at infinity (1:±i:0):1 plus-or-minus 𝑖:0(1:\pm i:0)( 1 : ± italic_i : 0 ). We obtain a smooth surface 𝒮 𝒮{\mathcal{S}}caligraphic_S. The surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is then defined as the (2,2,2)2 2 2(2,2,2)( 2 , 2 , 2 ) covering of 𝒮 𝒮{\mathcal{S}}caligraphic_S ramified over the strict transform of the singular conics defined by the quadratic forms appearing on the left-hand side of (6.1), (6.2), (6.3). More precisely, 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT will be the normalization of the fiber product of the three surfaces each of which obtained as the quadratic cover of 𝒮 𝒮{\mathcal{S}}caligraphic_S ramified over each of the mentioned degenerate conics. Report issue for preceding element A simple (but lengthy) application of the Riemann-Hurwitz formula for surfaces shows that the canonical class of 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT vanishes, so the surface is a K⁢3 𝐾 3 K3 italic_K 3 surface (or an abelian surface, but this possibility is ruled out by the fact that it contains non-constant families of elliptic curves). Alternatively, one can consider a generic line on ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT: its pre-image on 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is a curve 𝒞 𝒞\mathcal{C}caligraphic_C covering the line eight-to-one; more precisely, 𝒞→ℙ 1→𝒞 subscript ℙ 1\mathcal{C}\to{\mathbb{P}}{1}caligraphic_C → blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT is an abelian cover of type (2,2,2)2 2 2(2,2,2)( 2 , 2 , 2 ) ramified over eight points; then its Euler characteristic is 8 8 8 8; since 𝒞 2=8 superscript 𝒞 2 8\mathcal{C}^{2}=8 caligraphic_C start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = 8, by adjunction it follows that the product 𝒞⋅K 𝒮 3⋅𝒞 subscript 𝐾 subscript 𝒮 3\mathcal{C}\cdot K_{{\mathcal{S}}{3}}caligraphic_C ⋅ italic_K start_POSTSUBSCRIPT caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_POSTSUBSCRIPT vanishes, where K 𝒮 3 subscript 𝐾 subscript 𝒮 3 K{{\mathcal{S}}{3}}italic_K start_POSTSUBSCRIPT caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_POSTSUBSCRIPT denotes a canonical divisor of 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. Hence the canonical class is numerically trivial. Report issue for preceding element 6.2. Alternative models for 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT Report issue for preceding element Let us consider once again the diagram (6.4) leading to the projection π O:𝒮 3→ℙ 1:subscript 𝜋 𝑂→subscript 𝒮 3 subscript ℙ 1\pi_{O}:{\mathcal{S}}{3}\to{\mathbb{P}}{1}italic_π start_POSTSUBSCRIPT italic_O end_POSTSUBSCRIPT : caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT of equation (6.5). Changing O 𝑂 O italic_O with the two other points P,P′𝑃 superscript 𝑃′P,P^{\prime}italic_P , italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT leads to a rational map Report issue for preceding element 𝒮 3⇢ℙ 1×ℙ 1×ℙ 1.⇢subscript 𝒮 3 subscript ℙ 1 subscript ℙ 1 subscript ℙ 1{\mathcal{S}}{3}\dashrightarrow{\mathbb{P}}{1}\times{\mathbb{P}}{1}\times{% \mathbb{P}}{1}.caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ⇢ blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT . Regularizing this map, and abusing again of the notation by using again the symbol 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT for the new domain, one obtains a regular map of 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT into ℙ 1×ℙ 1×ℙ 1 subscript ℙ 1 subscript ℙ 1 subscript ℙ 1{\mathbb{P}}{1}\times{\mathbb{P}}{1}\times{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, actually an embedding. We now show that the image has multi-degree (2,2,2)2 2 2(2,2,2)( 2 , 2 , 2 ). This amounts to saying that each of the three projections 𝒮 3→(ℙ 1)2→subscript 𝒮 3 superscript subscript ℙ 1 2{\mathcal{S}}{3}\to({\mathbb{P}}{1})^{2}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → ( blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, obtained by omitting one factor in the projection to (ℙ 1)3 superscript subscript ℙ 1 3({\mathbb{P}}{1})^{3}( blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT, has degree two. Let us prove this fact. The projection 𝒮 3→ℙ 1×ℙ 1→subscript 𝒮 3 subscript ℙ 1 subscript ℙ 1{\mathcal{S}}{3}\to{\mathbb{P}}{1}\times{\mathbb{P}}_{1}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT can be composed on the left to obtain a projection Report issue for preceding element 𝒮 3→Λ P×Λ P′→subscript 𝒮 3 subscript Λ 𝑃 subscript Λ superscript 𝑃′{\mathcal{S}}{3}\to\Lambda{P}\times\Lambda_{P^{\prime}}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → roman_Λ start_POSTSUBSCRIPT italic_P end_POSTSUBSCRIPT × roman_Λ start_POSTSUBSCRIPT italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT(6.6) (see again diagram (6.4)), whose degree is four times the degree of the map 𝒮 3→(ℙ 1)2→subscript 𝒮 3 superscript subscript ℙ 1 2{\mathcal{S}}{3}\to({\mathbb{P}}{1})^{2}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → ( blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT we are interested in. Now, the fiber of a generic point in Λ P×Λ P′subscript Λ 𝑃 subscript Λ superscript 𝑃′\Lambda_{P}\times\Lambda_{P^{\prime}}roman_Λ start_POSTSUBSCRIPT italic_P end_POSTSUBSCRIPT × roman_Λ start_POSTSUBSCRIPT italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT is a single point in ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT (since a point in Λ P subscript Λ 𝑃\Lambda{P}roman_Λ start_POSTSUBSCRIPT italic_P end_POSTSUBSCRIPT (resp. in Λ P′subscript Λ superscript 𝑃′\Lambda_{P^{\prime}}roman_Λ start_POSTSUBSCRIPT italic_P start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT) represents a line in ℙ 2 subscript ℙ 2{\mathbb{P}}{2}blackboard_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT and two lines intersect at one point) and so consists in eight points in 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. Hence the projection (6.6) has degree eight and so the projection 𝒮 3→(ℙ 1)2→subscript 𝒮 3 superscript subscript ℙ 1 2{\mathcal{S}}{3}\to({\mathbb{P}}{1})^{2}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → ( blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT has degree two as wanted. Report issue for preceding element Now, smooth surfaces in ℙ 1×ℙ 1×ℙ 1 subscript ℙ 1 subscript ℙ 1 subscript ℙ 1{\mathbb{P}}{1}\times{\mathbb{P}}{1}\times{\mathbb{P}}_{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT of multi-degree (2,2,2)2 2 2(2,2,2)( 2 , 2 , 2 ) are known to be K 𝐾 K italic_K 3 surfaces. Actually, the surface studied in by the first and the third author in the frame of the Hilbert Property belongs to this family of K 𝐾 K italic_K 3 surfaces. Report issue for preceding element 6.3. The Kummer model for the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT Report issue for preceding element In this section we shall realize the surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT as a quartic in ℙ 3 subscript ℙ 3{\mathbb{P}}{3}blackboard_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. First note that working over the the complex field (and actually even over the reals) we can choose coordinates so that the second point P 𝑃 P italic_P is the point (1,0)1 0(1,0)( 1 , 0 ); of course, for our Diophantine problem we must then consider twists of the surface we are constructing. The three equations defining 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT become then Report issue for preceding element x 2+y 2=z 2 superscript 𝑥 2 superscript 𝑦 2 superscript 𝑧 2\displaystyle x^{2}+y^{2}=z^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT (x−1)2+y 2=u 2 superscript 𝑥 1 2 superscript 𝑦 2 superscript 𝑢 2\displaystyle(x-1)^{2}+y^{2}=u^{2}( italic_x - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(6.7) (x−a)2+(y−b)2=v 2,superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2 superscript 𝑣 2\displaystyle(x-a)^{2}+(y-b)^{2}=v^{2},( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , where a∈ℝ 𝑎 ℝ a\in{\mathbb{R}}italic_a ∈ blackboard_R and b≠0 𝑏 0 b\neq 0 italic_b ≠ 0. Using the first equation to eliminate x 2,y 2 superscript 𝑥 2 superscript 𝑦 2 x^{2},y^{2}italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT from the second and third equations, we obtain Report issue for preceding element 2⁢x=z 2−u 2+1,2⁢a⁢x+2⁢b⁢y=a 2+b 2+z 2−v 2,formulae-sequence 2 𝑥 superscript 𝑧 2 superscript 𝑢 2 1 2 𝑎 𝑥 2 𝑏 𝑦 superscript 𝑎 2 superscript 𝑏 2 superscript 𝑧 2 superscript 𝑣 2 2x=z^{2}-u^{2}+1,\qquad 2ax+2by=a^{2}+b^{2}+z^{2}-v^{2},2 italic_x = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 , 2 italic_a italic_x + 2 italic_b italic_y = italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , hence Report issue for preceding element x=z 2−u 2+1 2 𝑥 superscript 𝑧 2 superscript 𝑢 2 1 2\displaystyle x=\dfrac{z^{2}-u^{2}+1}{2}italic_x = divide start_ARG italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 1 end_ARG start_ARG 2 end_ARG y=a 2+b 2+(1−a)⁢z 2−v 2+a⁢u 2−a 2⁢b 𝑦 superscript 𝑎 2 superscript 𝑏 2 1 𝑎 superscript 𝑧 2 superscript 𝑣 2 𝑎 superscript 𝑢 2 𝑎 2 𝑏\displaystyle y=\dfrac{a^{2}+b^{2}+(1-a)z^{2}-v^{2}+au^{2}-a}{2b}italic_y = divide start_ARG italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( 1 - italic_a ) italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_a italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_a end_ARG start_ARG 2 italic_b end_ARG which together with the first equation (6.7) give rise to a single quartic equation in z,u,v 𝑧 𝑢 𝑣 z,u,v italic_z , italic_u , italic_v. For instance when a=0,b=1 formulae-sequence 𝑎 0 𝑏 1 a=0,b=1 italic_a = 0 , italic_b = 1 we obtain the following equation, that was already present in [19, Section D19]: Report issue for preceding element 2⁢z 4+v 4+u 4+2=2⁢z 2⁢(v 2+u 2)+2⁢u 2+2⁢v 2.2 superscript 𝑧 4 superscript 𝑣 4 superscript 𝑢 4 2 2 superscript 𝑧 2 superscript 𝑣 2 superscript 𝑢 2 2 superscript 𝑢 2 2 superscript 𝑣 2 2z^{4}+v^{4}+u^{4}+2=2z^{2}(v^{2}+u^{2})+2u^{2}+2v^{2}.2 italic_z start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_v start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + 2 = 2 italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) + 2 italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + 2 italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT .(6.8) The surface defined by this equation, and by those obtained for arbitrary choices of a,b 𝑎 𝑏 a,b italic_a , italic_b with b≠0 𝑏 0 b\neq 0 italic_b ≠ 0, is a singular quartic surface, having sixteen simple nodes . Report issue for preceding element Recall that a Kummer surface is a smooth projective minimal model of a quotient of the form A/{±1}𝐴 plus-or-minus 1 A/{\pm 1}italic_A / { ± 1 }, where A 𝐴 A italic_A is an abelian surface and {±1}plus-or-minus 1{\pm 1}{ ± 1 } denotes the group generated by the map A∋p↦−p∈A contains 𝐴 𝑝 maps-to 𝑝 𝐴 A\ni p\mapsto-p\in A italic_A ∋ italic_p ↦ - italic_p ∈ italic_A; sometimes one calls a Kummer surface also such a quotient, which has sixteen isolated singularities, corresponding to the 2 2 2 2-torsion of the abelian surface A 𝐴 A italic_A. The corresponding smooth model has sixteen −2 2-2- 2-curves, projecting to such singular points. Report issue for preceding element We shall denote by Kum⁢(A)Kum 𝐴\mathrm{Kum}(A)roman_Kum ( italic_A ) the Kummer surface associated to the abelian surface A 𝐴 A italic_A. Report issue for preceding element Theorem 6.1. Report issue for preceding element The surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is a Kummer surface. Report issue for preceding element It is a classical fact, see for example , that every complex quartic surface on ℙ 3 subscript ℙ 3{\mathbb{P}}_{3}blackboard_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT with sixteen nodes, and no other singularities, is a (singular model of a) Kummer surface; such a surface is actually isomorphic to a surface of the form A/{±1}𝐴 plus-or-minus 1 A/{\pm 1}italic_A / { ± 1 }, where the sixteen nodes correspond to the sixteen points of 2 2 2 2-torsion on A 𝐴 A italic_A, which are fixed by the involution on A 𝐴 A italic_A. Report issue for preceding element This would suffice to prove the result. However, we shall give an explicit proof of Theorem 6.1, constructing the abelian surface A 𝐴 A italic_A which covers 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT and describing the cover A→𝒮 3→𝐴 subscript 𝒮 3 A\to{\mathcal{S}}{3}italic_A → caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT. Report issue for preceding element Proof. Report issue for preceding element As we said, over the complex numbers we can suppose the equations of 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT are those given in (6.7). After setting Report issue for preceding element x+i⁢y=ξ,x−i⁢y=η formulae-sequence 𝑥 𝑖 𝑦 𝜉 𝑥 𝑖 𝑦 𝜂 x+iy=\xi,\qquad x-iy=\eta italic_x + italic_i italic_y = italic_ξ , italic_x - italic_i italic_y = italic_η the three equations (6.7) take the form Report issue for preceding element ξ⁢η=z 2 𝜉 𝜂 superscript 𝑧 2\displaystyle\xi\eta=z^{2}italic_ξ italic_η = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(6.9) (ξ−1)⁢(η−1)=u 2 𝜉 1 𝜂 1 superscript 𝑢 2\displaystyle(\xi-1)(\eta-1)=u^{2}( italic_ξ - 1 ) ( italic_η - 1 ) = italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT(6.10) (ξ−a−b⁢i)⁢(η−a+b⁢i)=v 2,𝜉 𝑎 𝑏 𝑖 𝜂 𝑎 𝑏 𝑖 superscript 𝑣 2\displaystyle(\xi-a-bi)(\eta-a+bi)=v^{2},( italic_ξ - italic_a - italic_b italic_i ) ( italic_η - italic_a + italic_b italic_i ) = italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(6.11) Consider now the two elliptic curves E 1 subscript 𝐸 1 E_{1}italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and E 2 subscript 𝐸 2 E_{2}italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT given in affine equations as Report issue for preceding element E 1:y 1 2=ξ⁢(ξ−1)⁢(ξ−a−b⁢i),E 2:y 2 2=η⁢(η−1)⁢(η−a+b⁢i).:subscript 𝐸 1 superscript subscript 𝑦 1 2 𝜉 𝜉 1 𝜉 𝑎 𝑏 𝑖 subscript 𝐸 2:superscript subscript 𝑦 2 2 𝜂 𝜂 1 𝜂 𝑎 𝑏 𝑖 E_{1}:\,y_{1}^{2}=\xi(\xi-1)(\xi-a-bi),\qquad E_{2}:\,y_{2}^{2}=\eta(\eta-1)(% \eta-a+bi).italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT : italic_y start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_a - italic_b italic_i ) , italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT : italic_y start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_η ( italic_η - 1 ) ( italic_η - italic_a + italic_b italic_i ) .(6.12) We note at once that E 1,E 2 subscript 𝐸 1 subscript 𝐸 2 E_{1},E_{2}italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT are not defined over ℚ ℚ{\mathbb{Q}}blackboard_Q and they contain only finitely many ℚ ℚ{\mathbb{Q}}blackboard_Q-rational points (in fact only the points of order 2 2 2 2): in particular the rational points on the associated Kummer surface Kum⁢(E 1×E 2)Kum subscript 𝐸 1 subscript 𝐸 2\mathrm{Kum}(E_{1}\times E_{2})roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ), do not come from the rational points on E 1×E 2 subscript 𝐸 1 subscript 𝐸 2 E_{1}\times E_{2}italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. Report issue for preceding element The function field of the surface 𝒮 3/ℚ subscript 𝒮 3 ℚ{\mathcal{S}}_{3}/{\mathbb{Q}}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT / blackboard_Q contains the square root of the product Report issue for preceding element ξ⁢(ξ−1)⁢(ξ−a−b⁢i)⁢η⁢(η−1)⁢(η−a+b⁢i)=(x 2+y 2)⁢((x−1)2+y 2)⁢((x−a)2+(y−b)2);𝜉 𝜉 1 𝜉 𝑎 𝑏 𝑖 𝜂 𝜂 1 𝜂 𝑎 𝑏 𝑖 superscript 𝑥 2 superscript 𝑦 2 superscript 𝑥 1 2 superscript 𝑦 2 superscript 𝑥 𝑎 2 superscript 𝑦 𝑏 2\xi(\xi-1)(\xi-a-bi)\eta(\eta-1)(\eta-a+bi)=(x^{2}+y^{2})((x-1)^{2}+y^{2})((x-% a)^{2}+(y-b)^{2});italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_a - italic_b italic_i ) italic_η ( italic_η - 1 ) ( italic_η - italic_a + italic_b italic_i ) = ( italic_x start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( ( italic_x - 1 ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_y start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ( ( italic_x - italic_a ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_b ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ; this square root together with the functions x 𝑥 x italic_x and y 𝑦 y italic_y generate the function field (even over ℚ ℚ{\mathbb{Q}}blackboard_Q) of the Kummer surface Kum⁢(E 1×E 2)Kum subscript 𝐸 1 subscript 𝐸 2\mathrm{Kum}(E_{1}\times E_{2})roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ). In particular, after putting α=a+b⁢i∈ℂ∖ℝ 𝛼 𝑎 𝑏 𝑖 ℂ ℝ\alpha=a+bi\in{\mathbb{C}}\setminus{\mathbb{R}}italic_α = italic_a + italic_b italic_i ∈ blackboard_C ∖ blackboard_R, we obtain Report issue for preceding element ℂ⁢(Kum⁢(E 1×E 2))=ℂ⁢(ξ,η)⁢(ξ⁢(ξ−1)⁢(ξ−α)⁢η⁢(η−1)⁢(η−α¯))ℂ Kum subscript 𝐸 1 subscript 𝐸 2 ℂ 𝜉 𝜂 𝜉 𝜉 1 𝜉 𝛼 𝜂 𝜂 1 𝜂¯𝛼{\mathbb{C}}(\mathrm{Kum}(E_{1}\times E_{2}))={\mathbb{C}}(\xi,\eta)(\sqrt{\xi% (\xi-1)(\xi-\alpha)\eta(\eta-1)(\eta-\bar{\alpha})})blackboard_C ( roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) ) = blackboard_C ( italic_ξ , italic_η ) ( square-root start_ARG italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_α ) italic_η ( italic_η - 1 ) ( italic_η - over¯ start_ARG italic_α end_ARG ) end_ARG ) and the inclusion of functions fields Report issue for preceding element ℂ⁢(ξ,η)ℂ 𝜉 𝜂\displaystyle{\mathbb{C}}(\xi,\eta)blackboard_C ( italic_ξ , italic_η )⊂ℂ⁢(ξ,η)⁢(ξ⁢(ξ−1)⁢(ξ−α)⁢η⁢(η−1)⁢(η−α¯))absent ℂ 𝜉 𝜂 𝜉 𝜉 1 𝜉 𝛼 𝜂 𝜂 1 𝜂¯𝛼\displaystyle\subset{\mathbb{C}}(\xi,\eta)(\sqrt{\xi(\xi-1)(\xi-\alpha)\eta(% \eta-1)(\eta-\bar{\alpha})})⊂ blackboard_C ( italic_ξ , italic_η ) ( square-root start_ARG italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_α ) italic_η ( italic_η - 1 ) ( italic_η - over¯ start_ARG italic_α end_ARG ) end_ARG ) ⊂ℂ⁢(ξ,η)⁢(ξ⁢(ξ−1)⁢(ξ−α),η⁢(η−1)⁢(η−α¯))=ℂ⁢(E 1×E 2).absent ℂ 𝜉 𝜂 𝜉 𝜉 1 𝜉 𝛼 𝜂 𝜂 1 𝜂¯𝛼 ℂ subscript 𝐸 1 subscript 𝐸 2\displaystyle\subset{\mathbb{C}}(\xi,\eta)(\sqrt{\xi(\xi-1)(\xi-\alpha)},\sqrt% {\eta(\eta-1)(\eta-\bar{\alpha})})={\mathbb{C}}(E_{1}\times E_{2}).⊂ blackboard_C ( italic_ξ , italic_η ) ( square-root start_ARG italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_α ) end_ARG , square-root start_ARG italic_η ( italic_η - 1 ) ( italic_η - over¯ start_ARG italic_α end_ARG ) end_ARG ) = blackboard_C ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) . Geometrically, this corresponds to the dominant maps Report issue for preceding element ℙ 1×ℙ 1←Kum⁢(E 1×E 2)←E 1×E 2.←subscript ℙ 1 subscript ℙ 1 Kum subscript 𝐸 1 subscript 𝐸 2←subscript 𝐸 1 subscript 𝐸 2{\mathbb{P}}{1}\times{\mathbb{P}}{1}\leftarrow\mathrm{Kum}(E_{1}\times E_{2}% )\leftarrow E_{1}\times E_{2}.blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ← roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) ← italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT .(6.13) We also have a rational dominant map of degree 4 4 4 4, actually an abelian cover of type (2,2)2 2(2,2)( 2 , 2 ), Report issue for preceding element 𝒮 3→Kum⁢(E 1×E 2).→subscript 𝒮 3 Kum subscript 𝐸 1 subscript 𝐸 2{\mathcal{S}}{3}\to\mathrm{Kum}(E{1}\times E_{2}).caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) . This map corresponds to the inclusion Report issue for preceding element ℂ⁢(ξ,η)⁢(ξ⁢(ξ−1)⁢(ξ−α)⁢η⁢(η−1)⁢(η−α¯))⊂ℂ⁢(ξ,η)⁢(ξ⁢η,(ξ−1)⁢(η−1),(ξ−α)⁢(η−α¯))ℂ 𝜉 𝜂 𝜉 𝜉 1 𝜉 𝛼 𝜂 𝜂 1 𝜂¯𝛼 ℂ 𝜉 𝜂 𝜉 𝜂 𝜉 1 𝜂 1 𝜉 𝛼 𝜂¯𝛼{\mathbb{C}}(\xi,\eta)(\sqrt{\xi(\xi-1)(\xi-\alpha)\eta(\eta-1)(\eta-\bar{% \alpha})})\subset{\mathbb{C}}(\xi,\eta)(\sqrt{\xi\eta},\sqrt{(\xi-1)(\eta-1)},% \sqrt{(\xi-\alpha)(\eta-\bar{\alpha})})blackboard_C ( italic_ξ , italic_η ) ( square-root start_ARG italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_α ) italic_η ( italic_η - 1 ) ( italic_η - over¯ start_ARG italic_α end_ARG ) end_ARG ) ⊂ blackboard_C ( italic_ξ , italic_η ) ( square-root start_ARG italic_ξ italic_η end_ARG , square-root start_ARG ( italic_ξ - 1 ) ( italic_η - 1 ) end_ARG , square-root start_ARG ( italic_ξ - italic_α ) ( italic_η - over¯ start_ARG italic_α end_ARG ) end_ARG ) Consider now the degree-2 2 2 2 map E 1×E 2→Kum⁢(E 1×E 2)→subscript 𝐸 1 subscript 𝐸 2 Kum subscript 𝐸 1 subscript 𝐸 2 E_{1}\times E_{2}\to\mathrm{Kum}(E_{1}\times E_{2})italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT → roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ); performing the fiber product of the two maps pointing to Kum⁢(E 1×E 2)Kum subscript 𝐸 1 subscript 𝐸 2\mathrm{Kum}(E_{1}\times E_{2})roman_Kum ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) we obtain a commutative diagram, where A→𝒮 3→𝐴 subscript 𝒮 3 A\to{\mathcal{S}}{3}italic_A → caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is a degree-2 2 2 2 map and the arrow ϕ:A→E 1×E 2:italic-ϕ→𝐴 subscript 𝐸 1 subscript 𝐸 2\phi:A\to E{1}\times E_{2}italic_ϕ : italic_A → italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is a degree-4 4 4 4 Galois covering of type (2,2)2 2(2,2)( 2 , 2 ). Report issue for preceding element Report issue for preceding element(6.14) We now prove that ϕ italic-ϕ\phi italic_ϕ is unramified, thus proving that A 𝐴 A italic_A is an abelian surface. Report issue for preceding element Actually, this cover corresponds to the function field extension Report issue for preceding element ℂ⁢(E 1×E 2)⊂ℂ⁢(ξ,η)⁢(ξ⁢(ξ−1)⁢(ξ−α),η⁢(η−1)⁢(η−α¯),ξ⁢η⁢(ξ−1)⁢(η−1)).ℂ subscript 𝐸 1 subscript 𝐸 2 ℂ 𝜉 𝜂 𝜉 𝜉 1 𝜉 𝛼 𝜂 𝜂 1 𝜂¯𝛼 𝜉 𝜂 𝜉 1 𝜂 1{\mathbb{C}}(E_{1}\times E_{2})\subset{\mathbb{C}}(\xi,\eta)(\sqrt{\xi(\xi-1)(% \xi-\alpha)},\sqrt{\eta(\eta-1)(\eta-\bar{\alpha})},\sqrt{\xi\eta}\sqrt{(\xi-1% )(\eta-1)}).blackboard_C ( italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) ⊂ blackboard_C ( italic_ξ , italic_η ) ( square-root start_ARG italic_ξ ( italic_ξ - 1 ) ( italic_ξ - italic_α ) end_ARG , square-root start_ARG italic_η ( italic_η - 1 ) ( italic_η - over¯ start_ARG italic_α end_ARG ) end_ARG , square-root start_ARG italic_ξ italic_η end_ARG square-root start_ARG ( italic_ξ - 1 ) ( italic_η - 1 ) end_ARG ) . Now, if we look at the possible ramification of this extension, this can occur only over the curves on E 1×E 2 subscript 𝐸 1 subscript 𝐸 2 E_{1}\times E_{2}italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT lying over the lines ξ=0,η=0,ξ−1=0,η−1=0 formulae-sequence 𝜉 0 formulae-sequence 𝜂 0 formulae-sequence 𝜉 1 0 𝜂 1 0\xi=0,\eta=0,\xi-1=0,\eta-1=0 italic_ξ = 0 , italic_η = 0 , italic_ξ - 1 = 0 , italic_η - 1 = 0 of ℙ 1×ℙ 1 subscript ℙ 1 subscript ℙ 1{\mathbb{P}}{1}\times{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT; moreover the ramification index can be at most 2 2 2 2. However, these lines are already ramified with index 2 2 2 2 on the cover 𝒮 3→ℙ 1×ℙ 1→subscript 𝒮 3 subscript ℙ 1 subscript ℙ 1{\mathcal{S}}{3}\to{\mathbb{P}}{1}\times{\mathbb{P}}{1}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT; by Abyankhar’s lemma, the cover A→E 1×E 2→𝐴 subscript 𝐸 1 subscript 𝐸 2 A\to E{1}\times E_{2}italic_A → italic_E start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × italic_E start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT turns out to be unramified. ∎ Report issue for preceding element Weak approximation. A natural question, also related to the distribution of rational points in 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT, is whether weak-approximation holds. This has been an extensively studied topic both for elliptic K3 surfaces and for Kummer surfaces. For example, there are known instances of elliptic K3 surfaces for which there is a ‘transcendental obstruction’ to Weak Approximation, in the sense of the paper . More generally Colliot-Thélène, Skorobogatov and Swinnerton-Dyer in , building on previous work of Swinnerton-Dyer , studied the Brauer-Manin obstruction to weak approximation for elliptic surfaces, and Skorobogatov and Swinnerton-Dyer analyzed, in particular, the case of a family of Kummer surfaces, whose associated abelian variety is the product of two elliptic curves. They proved in [24, Theorem 1] that, under certain hypotheses (for example that the Jacobians of the two elliptic curves have all their 2-division points defined over the ground field, plus additional more technical conditions), the equations defining these families of Kummer surfaces are soluble in the base field k 𝑘 k italic_k. Report issue for preceding element Field moduli and field of definition. As we remarked, the condition (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ) about the points P 1,P 2,P 3 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3 P_{1},P_{2},P_{3}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT in Theorem 1.1 is equivalent to saying that the relevant (Kummer K⁢3 𝐾 3 K3 italic_K 3) surface 𝒮 3=𝒮 P 1,P 2,P 3 subscript 𝒮 3 subscript 𝒮 subscript 𝑃 1 subscript 𝑃 2 subscript 𝑃 3{\mathcal{S}}{3}={\mathcal{S}}{P_{1},P_{2},P_{3}}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT = caligraphic_S start_POSTSUBSCRIPT italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_P start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT end_POSTSUBSCRIPT can be defined over ℚ ℚ{\mathbb{Q}}blackboard_Q. The following example shows that such surfaces can have the rational field ℚ ℚ{\mathbb{Q}}blackboard_Q as a field of moduli without being definable over ℚ ℚ{\mathbb{Q}}blackboard_Q. Consider the triangle 0,(1,1⁢2),(1,1−2)0 1 1 2 1 1 2 0,(1,1\sqrt{2}),(1,1-\sqrt{2})0 , ( 1 , 1 square-root start_ARG 2 end_ARG ) , ( 1 , 1 - square-root start_ARG 2 end_ARG ): it is defined over ℚ⁢(2)ℚ 2{\mathbb{Q}}(\sqrt{2})blackboard_Q ( square-root start_ARG 2 end_ARG ) and is invariant by Galois conjugation. Hence the corresponding surface is defined over ℚ⁢(2)ℚ 2{\mathbb{Q}}(\sqrt{2})blackboard_Q ( square-root start_ARG 2 end_ARG ) and is isomorphic to its Galois conjugate. However, the quadratic form appearing in (i⁢i)𝑖 𝑖(ii)( italic_i italic_i ) of Theorem 1.1 is not defined over ℚ ℚ{\mathbb{Q}}blackboard_Q, so the surface is not definable over ℚ ℚ{\mathbb{Q}}blackboard_Q. Report issue for preceding element The Hilbert property. The rational points in the surfaces of type 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT turn out also to satisfy another density condition, stronger then Zariski-density and not comparable with density in the real topology. Namely they satisfy the Hilbert Property; equivalently, they form a non-thin set. The Hilbert property for such surfaces can be derived, after proving the Zariski-density of rational points and non-torsion of the sections of the elliptic fibrations considered in this paper, by the methods introduced in and developed by J. Demeio in , . Report issue for preceding element 7. Distance from four or more points Report issue for preceding element Let us now consider the problem of the rationality of the distances from four (or more) points. This corresponds to adding a quadratic equation of the form Report issue for preceding element (x−α)2+(y−β)2=w 2 superscript 𝑥 𝛼 2 superscript 𝑦 𝛽 2 superscript 𝑤 2(x-\alpha)^{2}+(y-\beta)^{2}=w^{2}( italic_x - italic_α ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_y - italic_β ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_w start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT to the system of the three equations (5.2), (5.3), (5.4). Geometrically, adding the extra equation corresponds to operating a quadratic cover 𝒮 4→𝒮 3→subscript 𝒮 4 subscript 𝒮 3{\mathcal{S}}{4}\to{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT → caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT of the K⁢3 𝐾 3 K3 italic_K 3-surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ramified over the pull-back of the curve defined by the vanishing of the left-hand side of the above equation. This equation represents again a pair of lines on the plane, and a two-component curve on 𝒮 𝒮{\mathcal{S}}caligraphic_S of self-intersection 2 2 2 2. This is a big (but non-ample) divisor on 𝒮 𝒮{\mathcal{S}}caligraphic_S. Its pull-back to 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT is again a big divisor on 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT, say D 𝐷 D italic_D. Then any cover of the K⁢3 𝐾 3 K3 italic_K 3-surface 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ramified over D 𝐷 D italic_D is a surface of general type. Report issue for preceding element According to the celebrated Bombieri-Lang conjecture, its rational points (over any fixed number field) should be degenerate. In particular we have the following fact: Report issue for preceding element Proposition 7.1. Report issue for preceding element The Bombieri-Lang conjecture implies that the k 𝑘 k italic_k-rational points on the surface 𝒮 4 subscript 𝒮 4{\mathcal{S}}_{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT are not Zariski dense, for any number field k 𝑘 k italic_k. Report issue for preceding element However, in our situation, we lack techniques to study the degeneracy of rational points. In particular there is no known case of the conjecture for non-singular simply connected surfaces 4 4 4 see for a recent example over function fields. and 𝒮 4 subscript 𝒮 4{\mathcal{S}}_{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT is indeed simply connected. Report issue for preceding element Proposition 7.2. Report issue for preceding element The surface 𝒮 4 subscript 𝒮 4{\mathcal{S}}_{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT is simply connected. Report issue for preceding element Proof. Report issue for preceding element Using the same arguments as in Section 6 we can describe the surface 𝒮 4 subscript 𝒮 4{\mathcal{S}}{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT as a ramified cover of the blow up of ℙ 1×ℙ 1 subscript ℙ 1 subscript ℙ 1{\mathbb{P}}{1}\times{\mathbb{P}}_{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT over (the strict transform of) 5 horizontal lines and 5 vertical lines, of type (2,2,2,2)2 2 2 2(2,2,2,2)( 2 , 2 , 2 , 2 ). To conclude that the surface is simply-connected one can use the description of the fundamental group of a bidouble cover given in (see for example Proposition 2.7 or more generally Proposition 1.8). ∎ Report issue for preceding element An alternative approach is via the so-called product-quotient surfaces studied by Catanese et al., where the surface 𝒮 4 subscript 𝒮 4{\mathcal{S}}{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT has the same fundamental group as the quotient of a product of two curves of genus 5 (which one can construct explicitly from the map 𝒮 4→ℙ 1×ℙ 1→subscript 𝒮 4 subscript ℙ 1 subscript ℙ 1{\mathcal{S}}{4}\to{\mathbb{P}}{1}\times{\mathbb{P}}{1}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT → blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT). The latter quotient has trivial fundamental group using the main Theorem of Armstrong in . Indeed, using the description of 𝒮 4 subscript 𝒮 4{\mathcal{S}}{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT as a (2,2,2,2)2 2 2 2(2,2,2,2)( 2 , 2 , 2 , 2 ) Galois cover of the blow up ℙ 1×ℙ 1 subscript ℙ 1 subscript ℙ 1{\mathbb{P}}{1}\times{\mathbb{P}}{1}blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × blackboard_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, with group (ℤ/2⁢ℤ)4 superscript ℤ 2 ℤ 4({\mathbb{Z}}/2{\mathbb{Z}})^{4}( blackboard_Z / 2 blackboard_Z ) start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT, extending the similar construction for 𝒮 3 subscript 𝒮 3{\mathcal{S}}{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT given in Section 6, one obtains two genus 5 5 5 5 curves C 1,C 2 subscript 𝐶 1 subscript 𝐶 2 C_{1},C_{2}italic_C start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_C start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, both covers of ℙ 1 superscript ℙ 1{\mathbb{P}}^{1}blackboard_P start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT of type (2,2,2,2)2 2 2 2(2,2,2,2)( 2 , 2 , 2 , 2 ) (these are constructed using the 5 vertical, and 5 horizontal lines in the branch divisor of the cover 𝒮 4 subscript 𝒮 4{\mathcal{S}}{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT). Their product, 𝒞 1×𝒞 2 subscript 𝒞 1 subscript 𝒞 2{\mathcal{C}}{1}\times{\mathcal{C}}{2}caligraphic_C start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT × caligraphic_C start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, dominates 𝒮 4 subscript 𝒮 4{\mathcal{S}}{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT, and satisfy the hypotheses of , thus obtaining that 𝒮 4 subscript 𝒮 4{\mathcal{S}}_{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT, is simply connected. Report issue for preceding element We end this section with an easy observation: a proof of the fact that 𝒮 4⁢(ℚ)subscript 𝒮 4 ℚ{\mathcal{S}}{4}({\mathbb{Q}})caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT ( blackboard_Q ) is not Zariski dense, would answer in particular the famous Erdős-Ulam problem in the negative, i.e. there would be no Zariski-dense rational distance set in the plane. Recall that a rational distance set is a set of points whose mutual distances are all rational. Given such a set X⊂ℝ 2 𝑋 superscript ℝ 2 X\subset{\mathbb{R}}^{2}italic_X ⊂ blackboard_R start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, after choosing four points P 1,…,P 4 subscript 𝑃 1…subscript 𝑃 4 P{1},\ldots,P_{4}italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT in general position in X 𝑋 X italic_X, supposing that X 𝑋 X italic_X does contain at least four points in general position, the set X∖{P 1,…,P 4}𝑋 subscript 𝑃 1…subscript 𝑃 4 X\setminus{P_{1},\ldots,P_{4}}italic_X ∖ { italic_P start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_P start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT } can be lifted to a set of rational points on a surface 𝒮 4 subscript 𝒮 4{\mathcal{S}}{4}caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT. The conjectural degeneracy of the set 𝒮 4⁢(ℚ)subscript 𝒮 4 ℚ{\mathcal{S}}{4}({\mathbb{Q}})caligraphic_S start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT ( blackboard_Q ) immediately implies the degeneracy (in the Zariski topology) of the set X 𝑋 X italic_X. In turn, this last fact easily implies that all but finitely many points of X 𝑋 X italic_X must be contained in a line or a circle. Report issue for preceding element 8. Integral Distances from two points Report issue for preceding element As mentioned above, in it was shown that the integral (=lattice) points of the plane having integral distance from two lattice points O,P 𝑂 𝑃 O,P italic_O , italic_P are usually infinite in number, but are never Zariski dense (and in fact always contained in a finite union of hyperbolae and lines). This last fact may be read as a case of Runge’s theorem for surfaces. For larger rings of integers the situation is different (as it happens for Runge’s theorem for curves): in fact, as we now show, over certain other rings of algebraic integers, the integral points in question form a Zariski-dense set. Report issue for preceding element In this section we study solutions to the system given by equations (3.1) and (3.2) in rings of integers of certain quadratic number fields. We shall work out a couple of explicit cases, both real and non-real. In that last case then the meaning of ‘distance’ loses its usual significance. Report issue for preceding element 8.1. Integral points with coordinates in a real quadratic number field Report issue for preceding element Let δ>0 𝛿 0\delta>0 italic_δ > 0 be any positive integer, not a perfect square. Then the quadratic ring ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ] admits infinitely many units. This fact is crucial in the proof of the following result: Report issue for preceding element Theorem 8.1. Report issue for preceding element Let O,P∈ℤ 2 𝑂 𝑃 superscript ℤ 2 O,P\in{\mathbb{Z}}^{2}italic_O , italic_P ∈ blackboard_Z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT be integer points of the plane and let δ>0 𝛿 0\delta>0 italic_δ > 0 be a non-square positive integer. The set of points with coordinates in the ring ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ] whose distances from O 𝑂 O italic_O and from P 𝑃 P italic_P belong to the ring ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ] is Zariski-dense in the plane. Report issue for preceding element Proof. Report issue for preceding element Choosing coordinates, as before, such that O=(0,0)𝑂 0 0 O=(0,0)italic_O = ( 0 , 0 ) and P=(a,b)𝑃 𝑎 𝑏 P=(a,b)italic_P = ( italic_a , italic_b ), consider the line l 𝑙 l italic_l of equation x=a 𝑥 𝑎 x=a italic_x = italic_a. For every point Q=(a,t)∈l 𝑄 𝑎 𝑡 𝑙 Q=(a,t)\in l italic_Q = ( italic_a , italic_t ) ∈ italic_l with t∈ℤ⁢[δ]𝑡 ℤ delimited-[]𝛿 t\in{\mathbb{Z}}[\sqrt{\delta}]italic_t ∈ blackboard_Z [ square-root start_ARG italic_δ end_ARG ], its distance from P 𝑃 P italic_P equals \|t−b\|𝑡 𝑏\|t-b\|\| italic_t - italic_b \|, hence lies in the quadratic ring ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ]. The condition that d⁢(O,Q)𝑑 𝑂 𝑄 d(O,Q)italic_d ( italic_O , italic_Q ) also belongs to that ring amounts to saying that Report issue for preceding element a 2+(b−t)2=s 2,superscript 𝑎 2 superscript 𝑏 𝑡 2 superscript 𝑠 2 a^{2}+(b-t)^{2}=s^{2},italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + ( italic_b - italic_t ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_s start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , for some s∈ℤ⁢[δ]𝑠 ℤ delimited-[]𝛿 s\in{\mathbb{Z}}[\sqrt{\delta}]italic_s ∈ blackboard_Z [ square-root start_ARG italic_δ end_ARG ]. This is the equation of a conic admitting the integral points (t,s)=(b,a)𝑡 𝑠 𝑏 𝑎(t,s)=(b,a)( italic_t , italic_s ) = ( italic_b , italic_a ). Due to the presence of infinitely many units in the ring ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ], that conic admits infinitely many points with coordinates in that ring. This is a well-known fact, however we give a short proof for simplicity. Let us write the above equation as Report issue for preceding element (s−b+t)⁢(s+b−t)=a 2.𝑠 𝑏 𝑡 𝑠 𝑏 𝑡 superscript 𝑎 2(s-b+t)(s+b-t)=a^{2}.( italic_s - italic_b + italic_t ) ( italic_s + italic_b - italic_t ) = italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . For every unit ω∈(ℤ⁢[δ])∗𝜔 superscript ℤ delimited-[]𝛿\omega\in({\mathbb{Z}}[\sqrt{\delta}])^{}italic_ω ∈ ( blackboard_Z [ square-root start_ARG italic_δ end_ARG ] ) start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT, solving for s,t 𝑠 𝑡 s,t italic_s , italic_t the equations Report issue for preceding element s−b+t=a⁢ω,s+b−t=b−a⁢ω−1 formulae-sequence 𝑠 𝑏 𝑡 𝑎 𝜔 𝑠 𝑏 𝑡 𝑏 𝑎 superscript 𝜔 1 s-b+t=a\omega,\qquad s+b-t=b-a\omega^{-1}italic_s - italic_b + italic_t = italic_a italic_ω , italic_s + italic_b - italic_t = italic_b - italic_a italic_ω start_POSTSUPERSCRIPT - 1 end_POSTSUPERSCRIPT provides a rational solution. This solution is integral whenever ω≡1(mod 2)𝜔 annotated 1 pmod 2\omega\equiv 1\pmod{2}italic_ω ≡ 1 start_MODIFIER ( roman_mod start_ARG 2 end_ARG ) end_MODIFIER, which happens for an infinite subgroup of units. Report issue for preceding element Hence the line l 𝑙 l italic_l contains an infinite set X⊂l 𝑋 𝑙 X\subset l italic_X ⊂ italic_l of ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ]-integral points whose distances from O,P 𝑂 𝑃 O,P italic_O , italic_P belong to ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ]. The same of course holds for the horizontal line y=0 𝑦 0 y=0 italic_y = 0. To produce a Zariski-dense set we argue as in section 3.1. Report issue for preceding element Consider the family of confocal conics with foci O 𝑂 O italic_O and P 𝑃 P italic_P. Now, two such conics pass through every point x∈X 𝑥 𝑋 x\in X italic_x ∈ italic_X. Let us choose e.g. the hyperbola 𝒞 x subscript 𝒞 𝑥\mathcal{C}{x}caligraphic_C start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT passing through x 𝑥 x italic_x; as in the previous argument, such hyperbola 𝒞 x subscript 𝒞 𝑥\mathcal{C}{x}caligraphic_C start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT contains infinitely many integral points (with respect to the ring ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ]). For each such integral point y 𝑦 y italic_y we have that the three quantities d⁢(O,y)2,d⁢(P,y)2 𝑑 superscript 𝑂 𝑦 2 𝑑 superscript 𝑃 𝑦 2 d(O,y)^{2},d(P,y)^{2}italic_d ( italic_O , italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d ( italic_P , italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and d⁢(O,y)−d⁢(P,y)𝑑 𝑂 𝑦 𝑑 𝑃 𝑦 d(O,y)-d(P,y)italic_d ( italic_O , italic_y ) - italic_d ( italic_P , italic_y ) are all integral. Then d⁢(O,y)+d⁢(P,y)𝑑 𝑂 𝑦 𝑑 𝑃 𝑦 d(O,y)+d(P,y)italic_d ( italic_O , italic_y ) + italic_d ( italic_P , italic_y ) is rational; since it is an algebraic integer (in view of the fact that d⁢(O,y)2,d⁢(P,y)2 𝑑 superscript 𝑂 𝑦 2 𝑑 superscript 𝑃 𝑦 2 d(O,y)^{2},d(P,y)^{2}italic_d ( italic_O , italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d ( italic_P , italic_y ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT belong to ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ]), it is also an element of ℤ⁢[δ]ℤ delimited-[]𝛿{\mathbb{Z}}[\sqrt{\delta}]blackboard_Z [ square-root start_ARG italic_δ end_ARG ]. It follows that the integral points of 𝒞 y subscript 𝒞 𝑦\mathcal{C}{y}caligraphic_C start_POSTSUBSCRIPT italic_y end_POSTSUBSCRIPT are still solutions to our problem. Since the union of the 𝒞 x subscript 𝒞 𝑥\mathcal{C}{x}caligraphic_C start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT, x∈X 𝑥 𝑋 x\in X italic_x ∈ italic_X, is Zariski-dense, the theorem is proved. ∎ Report issue for preceding element It might be interesting to look at the points with coordinates in the ring ℤ[δ]{\mathbb{Z}}[\sqrt{\delta]}blackboard_Z [ square-root start_ARG italic_δ ] end_ARG having ℚ ℚ{\mathbb{Q}}blackboard_Q-rational distance from O 𝑂 O italic_O and P 𝑃 P italic_P. However, in this case the distances turn out to be rational integers (in ℤ ℤ{\mathbb{Z}}blackboard_Z), and by a Runge-type argument (as in the mentioned case of integral points with integral distances) the degeneracy follows. Report issue for preceding element A more compelling problem consists in studying the distribution of the points Q 𝑄 Q italic_Q having coordinates in the number field ℚ⁢(δ)ℚ 𝛿{\mathbb{Q}}(\sqrt{\delta})blackboard_Q ( square-root start_ARG italic_δ end_ARG ) and having ℚ ℚ{\mathbb{Q}}blackboard_Q-rational distances from both O 𝑂 O italic_O and P 𝑃 P italic_P. This amounts to intersecting circles centered in O 𝑂 O italic_O and P 𝑃 P italic_P with rational radii and asking that the intersection points, which generically are quadratic over ℚ ℚ{\mathbb{Q}}blackboard_Q, belong to a fixed number ring ℚ⁢(δ)ℚ 𝛿{\mathbb{Q}}(\sqrt{\delta})blackboard_Q ( square-root start_ARG italic_δ end_ARG ). Of course this set of points contains the set 𝒮 2⁢(ℚ)subscript 𝒮 2 ℚ{\mathcal{S}}_{2}({\mathbb{Q}})caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Q ) considered in the previous section, and consisting in rational points. We shall then concentrate on those points which are irrational but have rational distances from the given points. Report issue for preceding element Writing Report issue for preceding element x=x 1+δ⁢x 2,y=y 1+δ⁢y 2,formulae-sequence 𝑥 subscript 𝑥 1 𝛿 subscript 𝑥 2 𝑦 subscript 𝑦 1 𝛿 subscript 𝑦 2 x=x_{1}+\sqrt{\delta}x_{2},\qquad y=y_{1}+\sqrt{\delta}y_{2},italic_x = italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT + square-root start_ARG italic_δ end_ARG italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_y = italic_y start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT + square-root start_ARG italic_δ end_ARG italic_y start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , for rational numbers x 1,x 2,y 1,y 2 subscript 𝑥 1 subscript 𝑥 2 subscript 𝑦 1 subscript 𝑦 2 x_{1},x_{2},y_{1},y_{2}italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_y start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_y start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, the condition that d⁢(O,Q)2 𝑑 superscript 𝑂 𝑄 2 d(O,Q)^{2}italic_d ( italic_O , italic_Q ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, where Q=(x,y)𝑄 𝑥 𝑦 Q=(x,y)italic_Q = ( italic_x , italic_y ), is rational leads to the relation Report issue for preceding element x 1⁢x 2+y 1⁢y 2=0.subscript 𝑥 1 subscript 𝑥 2 subscript 𝑦 1 subscript 𝑦 2 0 x_{1}x_{2}+y_{1}y_{2}=0.italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT + italic_y start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_y start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = 0 . We shall neglect the already mentioned case x 2=y 2=0 subscript 𝑥 2 subscript 𝑦 2 0 x_{2}=y_{2}=0 italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = italic_y start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = 0 (i.e. Q 𝑄 Q italic_Q rational), as well as the case x i⁢y j=0 subscript 𝑥 𝑖 subscript 𝑦 𝑗 0 x_{i}y_{j}=0 italic_x start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT italic_y start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = 0 for any choice of i,j∈{1,2}𝑖 𝑗 1 2 i,j\in{1,2}italic_i , italic_j ∈ { 1 , 2 }. The rationality of the squared distance d⁢(P,Q)2 𝑑 superscript 𝑃 𝑄 2 d(P,Q)^{2}italic_d ( italic_P , italic_Q ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT gives an analogous relation. The two relations together lead to an expression of y 𝑦 y italic_y in terms of x 𝑥 x italic_x, namely: Report issue for preceding element y 1=b a⁢x 1,y 2=−a b⁢x 2.formulae-sequence subscript 𝑦 1 𝑏 𝑎 subscript 𝑥 1 subscript 𝑦 2 𝑎 𝑏 subscript 𝑥 2 y_{1}=\frac{b}{a}x_{1},\qquad y_{2}=-\frac{a}{b}x_{2}.italic_y start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = divide start_ARG italic_b end_ARG start_ARG italic_a end_ARG italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_y start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = - divide start_ARG italic_a end_ARG start_ARG italic_b end_ARG italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT . Then the rationality of the distance d⁢(O,Q)𝑑 𝑂 𝑄 d(O,Q)italic_d ( italic_O , italic_Q ) gives Report issue for preceding element x 1 2⁢(1+b 2 a 2)+δ⁢x 2 2⁢(1+a 2 b 2)=z 2,superscript subscript 𝑥 1 2 1 superscript 𝑏 2 superscript 𝑎 2 𝛿 superscript subscript 𝑥 2 2 1 superscript 𝑎 2 superscript 𝑏 2 superscript 𝑧 2 x_{1}^{2}\left(1+\frac{b^{2}}{a^{2}}\right)+\delta x_{2}^{2}\left(1+\frac{a^{2% }}{b^{2}}\right)=z^{2},italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( 1 + divide start_ARG italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG start_ARG italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG ) + italic_δ italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( 1 + divide start_ARG italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG start_ARG italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG ) = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , which can be re-written, after renaming the variables, as Report issue for preceding element u 2+δ⁢v 2=(a 2+b 2)⁢w 2.superscript 𝑢 2 𝛿 superscript 𝑣 2 superscript 𝑎 2 superscript 𝑏 2 superscript 𝑤 2 u^{2}+\delta v^{2}=(a^{2}+b^{2})w^{2}.italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_δ italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = ( italic_a start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_b start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) italic_w start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT . It is clear that for certain choices of δ,a,b 𝛿 𝑎 𝑏\delta,a,b italic_δ , italic_a , italic_b, the above equation has no non-trivial rational solution. The rationality of the distance d⁢(P,Q)𝑑 𝑃 𝑄 d(P,Q)italic_d ( italic_P , italic_Q ) provides another (non-homogeneous) quadratic equation. The two equations together define again a Del Pezzo surface of degree four, as in the previous section. Given the points P=(a,b)𝑃 𝑎 𝑏 P=(a,b)italic_P = ( italic_a , italic_b ), for some choice of δ 𝛿\delta italic_δ the set of points defined on ℚ⁢(δ)ℚ 𝛿{\mathbb{Q}}(\sqrt{\delta})blackboard_Q ( square-root start_ARG italic_δ end_ARG ), but not over ℚ ℚ{\mathbb{Q}}blackboard_Q, and having rational distance from both O 𝑂 O italic_O and P 𝑃 P italic_P will be dense. Report issue for preceding element 8.2. Points with coordinates in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] Report issue for preceding element Let us now consider an imaginary quadratic ring. The simplest such case occurs when both the coordinates of the points Q 𝑄 Q italic_Q and the ‘distances’ d⁢(O,Q),d⁢(P,Q)𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 d(O,Q),d(P,Q)italic_d ( italic_O , italic_Q ) , italic_d ( italic_P , italic_Q ) are Gaussian integers, i.e. elements of ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ]. In this setting, we keep the algebraic definition of distance, namely the distance between two points A,B∈(ℤ⁢[i])2 𝐴 𝐵 superscript ℤ delimited-[]𝑖 2 A,B\in({\mathbb{Z}}[i])^{2}italic_A , italic_B ∈ ( blackboard_Z [ italic_i ] ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT is a square-root of the sum of the squares of the entries of the vector A⁢B→→𝐴 𝐵\vec{AB}over→ start_ARG italic_A italic_B end_ARG; this is defined only up to sign, but that is immaterial for the field of definition. Basically, we are interested on the points of the surface 𝒮 2 subscript 𝒮 2{\mathcal{S}}_{2}caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT defined over the ring ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ]. We will show that such set is Zariski dense: Report issue for preceding element Proposition 8.2. Report issue for preceding element Given P=(a,b)∈ℤ 2 𝑃 𝑎 𝑏 superscript ℤ 2 P=(a,b)\in{\mathbb{Z}}^{2}italic_P = ( italic_a , italic_b ) ∈ blackboard_Z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, P≠O 𝑃 𝑂 P\neq O italic_P ≠ italic_O, the set 𝒮 2⁢(ℤ⁢[i])subscript 𝒮 2 ℤ delimited-[]𝑖{\mathcal{S}}_{2}({\mathbb{Z}}[i])caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Z [ italic_i ] ) is Zariski dense, where as usual Report issue for preceding element 𝒮 2⁢(ℤ⁢[i])={Q∈𝔸 2⁢(ℤ⁢[i]):d⁢(O,Q)∈ℤ⁢[i],d⁢(P,Q)∈ℤ⁢[i]}.subscript 𝒮 2 ℤ delimited-[]𝑖 conditional-set 𝑄 superscript 𝔸 2 ℤ delimited-[]𝑖 formulae-sequence 𝑑 𝑂 𝑄 ℤ delimited-[]𝑖 𝑑 𝑃 𝑄 ℤ delimited-[]𝑖{\mathcal{S}}_{2}({\mathbb{Z}}[i])={Q\in{\mathbb{A}}^{2}({\mathbb{Z}}[i]):d(O% ,Q)\in{\mathbb{Z}}[i],d(P,Q)\in{\mathbb{Z}}[i]}.caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Z [ italic_i ] ) = { italic_Q ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( blackboard_Z [ italic_i ] ) : italic_d ( italic_O , italic_Q ) ∈ blackboard_Z [ italic_i ] , italic_d ( italic_P , italic_Q ) ∈ blackboard_Z [ italic_i ] } . Proof. Report issue for preceding element We start by observing that, similarly to the case of rational distances, the set 𝒮 2⁢(ℤ⁢[i])subscript 𝒮 2 ℤ delimited-[]𝑖{\mathcal{S}}_{2}({\mathbb{Z}}[i])caligraphic_S start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( blackboard_Z [ italic_i ] ) corresponds to the system given by equations (3.1) and (3.2). Since now −1 1-1- 1 is a square in our ring, the system is equivalent to the following: Report issue for preceding element w 1⁢w 2=z 2,subscript 𝑤 1 subscript 𝑤 2 superscript 𝑧 2\displaystyle w_{1}w_{2}=z^{2},italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = italic_z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(8.1) (w 1−α)⁢(w 2−α¯)=(z−k)2,subscript 𝑤 1 𝛼 subscript 𝑤 2¯𝛼 superscript 𝑧 𝑘 2\displaystyle(w_{1}-\alpha)(w_{2}-\bar{\alpha})=(z-k)^{2},( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_α ) ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - over¯ start_ARG italic_α end_ARG ) = ( italic_z - italic_k ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ,(8.2) where we denote with α¯¯𝛼\bar{\alpha}over¯ start_ARG italic_α end_ARG the number a−i⁢b 𝑎 𝑖 𝑏 a-ib italic_a - italic_i italic_b (which is the complex conjugate of α 𝛼\alpha italic_α if a,b∈ℝ 𝑎 𝑏 ℝ a,b\in{\mathbb{R}}italic_a , italic_b ∈ blackboard_R, as we are assuming). By unique factorization in the ring ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ], we can rewrite the system as follows: we set w 1=d 1⁢r 2 subscript 𝑤 1 subscript 𝑑 1 superscript 𝑟 2 w_{1}=d_{1}r^{2}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_r start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and w 2=d 1⁢u 2 subscript 𝑤 2 subscript 𝑑 1 superscript 𝑢 2 w_{2}=d_{1}u^{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, and similarly w−α=d 2⁢s 2 𝑤 𝛼 subscript 𝑑 2 superscript 𝑠 2 w-\alpha=d_{2}s^{2}italic_w - italic_α = italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_s start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT and w 2−α¯=d 2⁢v 2 subscript 𝑤 2¯𝛼 subscript 𝑑 2 superscript 𝑣 2 w_{2}-\bar{\alpha}=d_{2}v^{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - over¯ start_ARG italic_α end_ARG = italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT for some d 1,d 2,r,s,u,v∈ℤ⁢[i]subscript 𝑑 1 subscript 𝑑 2 𝑟 𝑠 𝑢 𝑣 ℤ delimited-[]𝑖 d_{1},d_{2},r,s,u,v\in{\mathbb{Z}}[i]italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_r , italic_s , italic_u , italic_v ∈ blackboard_Z [ italic_i ] (where for instance d 1 subscript 𝑑 1 d_{1}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT is a gcd\gcd roman_gcd of w 1,w 2 subscript 𝑤 1 subscript 𝑤 2 w_{1},w_{2}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT). Every time we can write w 1,w 2,w 1−α,w 2−α¯subscript 𝑤 1 subscript 𝑤 2 subscript 𝑤 1 𝛼 subscript 𝑤 2¯𝛼 w_{1},w_{2},w_{1}-\alpha,w_{2}-\bar{\alpha}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_α , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - over¯ start_ARG italic_α end_ARG in such a way for suitable (d 1,d 2,r,s,u,v)subscript 𝑑 1 subscript 𝑑 2 𝑟 𝑠 𝑢 𝑣(d_{1},d_{2},r,s,u,v)( italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_r , italic_s , italic_u , italic_v ) we obtain a solution of the system of equations (8.1), (8.2). Eliminating w 1,w 2 subscript 𝑤 1 subscript 𝑤 2 w_{1},w_{2}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT this amounts to the following system: Report issue for preceding element d 1⁢r 2−d 2⁢s 2=α,subscript 𝑑 1 superscript 𝑟 2 subscript 𝑑 2 superscript 𝑠 2 𝛼\displaystyle d_{1}r^{2}-d_{2}s^{2}=\alpha,italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_r start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_s start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = italic_α ,(8.3) d 1⁢u 2−d 2⁢v 2=α¯.subscript 𝑑 1 superscript 𝑢 2 subscript 𝑑 2 superscript 𝑣 2¯𝛼\displaystyle d_{1}u^{2}-d_{2}v^{2}=\bar{\alpha}.italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT - italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_v start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT = over¯ start_ARG italic_α end_ARG .(8.4) We want to show that the above system has infinitely many solutions (d 1,d 2,r,s,u,v)∈(ℤ⁢[i])6 subscript 𝑑 1 subscript 𝑑 2 𝑟 𝑠 𝑢 𝑣 superscript ℤ delimited-[]𝑖 6(d_{1},d_{2},r,s,u,v)\in({\mathbb{Z}}[i])^{6}( italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_r , italic_s , italic_u , italic_v ) ∈ ( blackboard_Z [ italic_i ] ) start_POSTSUPERSCRIPT 6 end_POSTSUPERSCRIPT, and that these solutions give rise to a Zariski-dense set of points (w 1,w 2)=(d 1⁢r 2,d 1⁢u 2)∈𝔸 2 subscript 𝑤 1 subscript 𝑤 2 subscript 𝑑 1 superscript 𝑟 2 subscript 𝑑 1 superscript 𝑢 2 superscript 𝔸 2(w_{1},w_{2})=(d_{1}r^{2},d_{1}u^{2})\in{\mathbb{A}}^{2}( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) = ( italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_r start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) ∈ blackboard_A start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT; of course, the density of the pairs (w 1,w 2)subscript 𝑤 1 subscript 𝑤 2(w_{1},w_{2})( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) is equivalent to the density of the pairs (x,y)𝑥 𝑦(x,y)( italic_x , italic_y ), which are obtained from (w 1,w 2)subscript 𝑤 1 subscript 𝑤 2(w_{1},w_{2})( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) via a fixed linear transformation. Report issue for preceding element This amounts to show that for given α∈ℤ⁢[i],α≠0 formulae-sequence 𝛼 ℤ delimited-[]𝑖 𝛼 0\alpha\in{\mathbb{Z}}[i],\alpha\neq 0 italic_α ∈ blackboard_Z [ italic_i ] , italic_α ≠ 0 there exists choices of non-zero d 1,d 2 subscript 𝑑 1 subscript 𝑑 2 d_{1},d_{2}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT such that each of the equations (8.3), (8.4) has infinitely many solutions. Indeed, in view of the fact that the two equations have separated variables, if each of two equations (8.3), (8.4) has infinitely many solutions the pairs (r,u)𝑟 𝑢(r,u)( italic_r , italic_u ) produced from these solutions will be Zariski-dense on the plane; consequently also the pairs (w 1,w 2)=d 1⁢(r 2,u 2)subscript 𝑤 1 subscript 𝑤 2 subscript 𝑑 1 superscript 𝑟 2 superscript 𝑢 2(w_{1},w_{2})=d_{1}(r^{2},u^{2})( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) = italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_r start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , italic_u start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) will be Zariski-dense. Report issue for preceding element The form of the two equations above is amenable to be studied using the theory of Pell’s equations. Report issue for preceding element One can see from the above system that there always exist a choice of d 1,d 2 subscript 𝑑 1 subscript 𝑑 2 d_{1},d_{2}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT such that the two equations (8.3) and (8.4) have both a solution: it is enough to evaluate the left hand sides at a specific choice of r,s,u 𝑟 𝑠 𝑢 r,s,u italic_r , italic_s , italic_u and v 𝑣 v italic_v and solve for d 1 subscript 𝑑 1 d_{1}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and d 2 subscript 𝑑 2 d_{2}italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT (for example choosing r,s,u,v 𝑟 𝑠 𝑢 𝑣 r,s,u,v italic_r , italic_s , italic_u , italic_v that form a matrix with determinant invertible in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ]). Therefore we are in the situation in which we have two conics defined over ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] both with an integral point over ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ]. However, such conics might contain only finitely many points with coordinates in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ]. Report issue for preceding element To avoid this case, we can use (a case of) a theorem of Alvanos, Bilu and Poulakis that, in this special case, states the following: for an affine curve 𝒞 𝒞{\mathcal{C}}caligraphic_C of genus zero, defined over a number field K 𝐾 K italic_K, to have infinitely many integral points is sufficient to have two points at infinity which are conjugate over K 𝐾 K italic_K and whose field of definition is not a CM-extension of K 𝐾 K italic_K (see also for the analogue statement over the integers, due to Gauss). Recall that an extension of number fields L/K 𝐿 𝐾 L/K italic_L / italic_K is called a CM-extension if K 𝐾 K italic_K is totally real and L 𝐿 L italic_L is totally imaginary. In our case, there are two points at infinity of equations (8.3) and (8.4), namely [±D:1:0]delimited-[]:plus-or-minus 𝐷 1:0[\pm\sqrt{D}:1:0][ ± square-root start_ARG italic_D end_ARG : 1 : 0 ], where D=d 2/d 1 𝐷 subscript 𝑑 2 subscript 𝑑 1 D=d_{2}/d_{1}italic_D = italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT / italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. Hence our system will have infinitely many solutions provided that d 2/d 1 subscript 𝑑 2 subscript 𝑑 1 d_{2}/d_{1}italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT / italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT is not a square in ℚ⁢[i]ℚ delimited-[]𝑖{\mathbb{Q}}[i]blackboard_Q [ italic_i ]. (Note that this condition is relevant: in fact, Theorem 1.2 in shows that when d 2/d 1∈ℚ⁢(i)subscript 𝑑 2 subscript 𝑑 1 ℚ 𝑖\sqrt{d_{2}/d_{1}}\in{\mathbb{Q}}(i)square-root start_ARG italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT / italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG ∈ blackboard_Q ( italic_i ) the integral points will be finite in number. Note also that, when d 2/d 1∉ℚ⁢(i)subscript 𝑑 2 subscript 𝑑 1 ℚ 𝑖\sqrt{d_{2}/d_{1}}\notin{\mathbb{Q}}(i)square-root start_ARG italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT / italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG ∉ blackboard_Q ( italic_i ), since ℚ⁢(i)ℚ 𝑖{\mathbb{Q}}(i)blackboard_Q ( italic_i ) is totally imaginary, the field of definition of the points at infinity is never a CM extension of ℚ⁢(i)ℚ 𝑖{\mathbb{Q}}(i)blackboard_Q ( italic_i ).) Report issue for preceding element Finally, it it remains to prove the following Report issue for preceding element Lemma 8.3. Report issue for preceding element Given a non-zero number α∈ℤ⁢[i]𝛼 ℤ delimited-[]𝑖\alpha\in{\mathbb{Z}}[i]italic_α ∈ blackboard_Z [ italic_i ] there exist d 1,d 2,r,s,u,v∈ℤ⁢[i]subscript 𝑑 1 subscript 𝑑 2 𝑟 𝑠 𝑢 𝑣 ℤ delimited-[]𝑖 d_{1},d_{2},r,s,u,v\in{\mathbb{Z}}[i]italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_r , italic_s , italic_u , italic_v ∈ blackboard_Z [ italic_i ] such that d 1/d 2 subscript 𝑑 1 subscript 𝑑 2 d_{1}/d_{2}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT / italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is not a square and the relations (8.3), (8.4) hold. Report issue for preceding element Proof. Report issue for preceding element Let us separate the proof into two cases, according as α⁢α¯𝛼¯𝛼\alpha\bar{\alpha}italic_α over¯ start_ARG italic_α end_ARG being a perfect square or not. Report issue for preceding element If α⁢α¯𝛼¯𝛼\alpha\bar{\alpha}italic_α over¯ start_ARG italic_α end_ARG is not a square, which is equivalent to saying that α/α¯𝛼¯𝛼\alpha/\bar{\alpha}italic_α / over¯ start_ARG italic_α end_ARG is not a square in ℚ⁢(i)ℚ 𝑖{\mathbb{Q}}(i)blackboard_Q ( italic_i ), then take d 1=α,d 2=α¯formulae-sequence subscript 𝑑 1 𝛼 subscript 𝑑 2¯𝛼 d_{1}=\alpha,d_{2}=\bar{\alpha}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = italic_α , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = over¯ start_ARG italic_α end_ARG, r=1,s=o,u=0,v=i formulae-sequence 𝑟 1 formulae-sequence 𝑠 𝑜 formulae-sequence 𝑢 0 𝑣 𝑖 r=1,s=o,u=0,v=i italic_r = 1 , italic_s = italic_o , italic_u = 0 , italic_v = italic_i. Report issue for preceding element If, on the contrary α⁢α¯𝛼¯𝛼\alpha\bar{\alpha}italic_α over¯ start_ARG italic_α end_ARG is a square, by unique factorization in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] we can write Report issue for preceding element α=ρ⋅ξ 2,α~=ρ¯⋅ξ¯2,formulae-sequence 𝛼⋅𝜌 superscript 𝜉 2~𝛼⋅¯𝜌 superscript¯𝜉 2\alpha=\rho\cdot\xi^{2},\quad\tilde{\alpha}=\bar{\rho}\cdot{\bar{\xi}}^{2},italic_α = italic_ρ ⋅ italic_ξ start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , over~ start_ARG italic_α end_ARG = over¯ start_ARG italic_ρ end_ARG ⋅ over¯ start_ARG italic_ξ end_ARG start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT , where ρ 𝜌\rho italic_ρ is square-free. The decomposition is unique up to units. Also, the fact that α⁢α¯𝛼¯𝛼\alpha\bar{\alpha}italic_α over¯ start_ARG italic_α end_ARG is a perfect square forces ρ⁢ρ¯𝜌¯𝜌\rho\bar{\rho}italic_ρ over¯ start_ARG italic_ρ end_ARG to be a perfect square; then ρ 𝜌\rho italic_ρ, being square-free, cannot be divisible by non-rational primes; it follows that ρ 𝜌\rho italic_ρ is real (hence a rational integer) or purely imaginary. In the first case, choose d 1,d 2∈ℤ subscript 𝑑 1 subscript 𝑑 2 ℤ d_{1},d_{2}\in{\mathbb{Z}}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∈ blackboard_Z with d 1−d 2=ρ subscript 𝑑 1 subscript 𝑑 2 𝜌 d_{1}-d_{2}=\rho italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = italic_ρ and d 1/d 2 subscript 𝑑 1 subscript 𝑑 2 d_{1}/d_{2}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT / italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT not a square, which is certainly possible. Then put r=s=ξ 𝑟 𝑠 𝜉 r=s=\xi italic_r = italic_s = italic_ξ, u=v=ξ¯𝑢 𝑣¯𝜉 u=v=\bar{\xi}italic_u = italic_v = over¯ start_ARG italic_ξ end_ARG. In the second case, i.e. when ρ 𝜌\rho italic_ρ is purely imaginary, take two purely imaginary numbers d 1,d 2 subscript 𝑑 1 subscript 𝑑 2 d_{1},d_{2}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT with the above property and r=s=ξ 𝑟 𝑠 𝜉 r=s=\xi italic_r = italic_s = italic_ξ, u=v=i⁢ξ¯𝑢 𝑣 𝑖¯𝜉 u=v=i\bar{\xi}italic_u = italic_v = italic_i over¯ start_ARG italic_ξ end_ARG. This completes the proof of the lemma. ∎ Report issue for preceding element To conclude the proof of the proposition, note that given a point P⁢(a,b)≠(0,0)∈ℤ 2 𝑃 𝑎 𝑏 0 0 superscript ℤ 2 P(a,b)\neq(0,0)\in{\mathbb{Z}}^{2}italic_P ( italic_a , italic_b ) ≠ ( 0 , 0 ) ∈ blackboard_Z start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT, the complex number α=a+b⁢i∈ℤ⁢[i]𝛼 𝑎 𝑏 𝑖 ℤ delimited-[]𝑖\alpha=a+bi\in{\mathbb{Z}}[i]italic_α = italic_a + italic_b italic_i ∈ blackboard_Z [ italic_i ] is non-zero. Then by the above lemma and the previous considerations the system (8.3), (8.4) has infinitely many solutions in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] giving rise, as explained, to a dense set of points Q∈ℤ⁢[i]×ℤ⁢[i]𝑄 ℤ delimited-[]𝑖 ℤ delimited-[]𝑖 Q\in{\mathbb{Z}}[i]\times{\mathbb{Z}}[i]italic_Q ∈ blackboard_Z [ italic_i ] × blackboard_Z [ italic_i ] in the plane with d⁢(O,Q),d⁢(P,Q)𝑑 𝑂 𝑄 𝑑 𝑃 𝑄 d(O,Q),d(P,Q)italic_d ( italic_O , italic_Q ) , italic_d ( italic_P , italic_Q ) both in ℤ⁢[i]ℤ delimited-[]𝑖{\mathbb{Z}}[i]blackboard_Z [ italic_i ] as wanted. ∎ Report issue for preceding element As we mentioned in the introduction, the set of points having integral distances from three or more fixed points, over any ring of S 𝑆 S italic_S-integers in any number field, is expected to be degenerate. Indeed, it is (the projection to the plane of) the set of integral points on an affine K3 surface; since the canonical divisor of a K3 surface is zero, the sum of the divisor at infinity and the canonical divisor turns out to be ample and Vojta’s conjecture predicts degeneracy. However, no method is known to prove degeneracy in general; even in the special case of equation (6.8) we do not know how to prove that the S 𝑆 S italic_S-integral solutions are not Zariski-dense. Report issue for preceding element One might try to prove the weaker statement that among the rational points of 𝒮 3 subscript 𝒮 3{\mathcal{S}}_{3}caligraphic_S start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT constructed via elliptic fibrations in Section 5 only finitely many of them (or maybe only those in a degenerate set) can have integral distances from the three given points. Report issue for preceding element References Report issue for preceding element ↑ P. Alvanos, Y. Bilu, D. 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Catanese, On the moduli spaces of surfaces of general type. J. Differential Geom. 19 (1984), no.2, 483–515. ↑ S. Coccia, The Hilbert property for integral points of affine smooth cubic surfaces, J. Number Theory 200 (2019), 353-379. ↑ S. Coccia, A Hilbert irreducibility theorem for integral points on del Pezzo surfaces, Int. Math. Res. Not. IMRN (2024), no.6, 5005–5049. ↑ J.-L. Colliot-Thélène, A.N. Skorobogatov and H.P.F. Swinnerton-Dyer, Hasse principle for pencils of curves of genus one whose Jacobians have rational 2 2 2 2-division points, Invent. Math. 134 (1998), no.3, 579–650. ↑ P. Corvaja, U. Zannier, On the Hilbert Property and the fundamental group of algebraic varieties, Math. Z. 286 (2017), 579-602. ↑ P. Corvaja, U. Zannier, Finiteness theorems on elliptical billiards and a variant of the dynamical Mordell-Lang conjecture, Proc. Lond. Math. Soc. (3) 127 (2023), n.5, 1268–1337. ↑ M. Davis, One equation to rule them all, Trans. New York Acad. Sci. Ser. II 30 (1968), no. 6, 766–773. ↑ J.L. Demeio, Non-rational varieties with the Hilbert property, Int. J. Number Theory 16 (2020), no.4, 803–822. ↑ J.L. Demeio, Elliptic fibrations and the Hilbert property, Int. Math. Res. Not. IMRN (2021), no.13, 10260–10277. ↑ C. Gasbarri, E. Rousseau, A. Turchet, J. Wang, Simply Connectedness And Hyperbolicity, ArXiv preprint ↑ R.K. Guy, Unsolved problems in number theory, third edition, Problem Books in Mathematics, Springer, New York, 2004. ↑ B. Hassett, Yu. Tschinkel, Density of integral points on algebraic varieties, in: Rational Points on Algebraic Varieties, in: Progr. Math., vol. 199, Birkhäuser, Basel, 2001, pp. 169-197. ↑ B. Hassett, A. Varilly-Alvarado, P. Varilly, Transcendental obstructions to weak approximation on general K3 surfaces, Adv. Math. 228 (2011), 1377–1404. ↑ V. V. Nikulin, Kummer surfaces.(Russian), Izv. Akad. Nauk SSSR Ser. Mat.39 (1975), no.2, 278–293, 471. ↑ J. H. Silverman, On the distribution of integer points on curves of genus zero, Theoret. Comput. Sci. 235 (2000), no. 1, 163-170. ↑ A.N. Skorobogatov and H.P.F. Swinnerton-Dyer, 2-descent on elliptic curves and rational points on certain Kummer surfaces, Adv. Math. 198 (2) (2005) 448–483. ↑ H.P.F. Swinnerton-Dyer, A 4+B 4=C 4+D 4 superscript 𝐴 4 superscript 𝐵 4 superscript 𝐶 4 superscript 𝐷 4 A^{4}+B^{4}=C^{4}+D^{4}italic_A start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_B start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT = italic_C start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT + italic_D start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT revisited. J. Lond. Math. Soc. 43 (1968), 149-151. ↑ H.P.F. Swinnerton-Dyer, Rational points on certain intersections of two quadrics, in Abelian varieties, Barth, Hulek and Lange ed., Walter de Gruyter, Berlin, New York, 1995. ↑ T. Tao, The Erdős-Ulam Problem, varieties of general type, and the Bombieri-Lang conjecture, at ↑ U. Zannier, Integral distances from (two) given lattice points, Enseign. Math. 68 (2022), no 1/2, 201–216. ↑ U. Zannier, Some problems of unlikely intersections in arithmetic and geometry, Annals of Mathematics Studies, 181, Princeton Univ. Press, Princeton, NJ, 2012. Report Issue Report Github Issue Title: Content selection saved. Describe the issue below: Description: Submit without Github Submit in Github Report Issue for Selection Generated by L A T E xml Instructions for reporting errors We are continuing to improve HTML versions of papers, and your feedback helps enhance accessibility and mobile support. To report errors in the HTML that will help us improve conversion and rendering, choose any of the methods listed below: Click the "Report Issue" button. Open a report feedback form via keyboard, use "Ctrl + ?". Make a text selection and click the "Report Issue for Selection" button near your cursor. 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17511	https://www.youtube.com/watch?v=Tub_1FIhJ70	Tricks of the Trade IV: Row and Column Sums Lorenzo Sadun 18100 subscribers 37 likes Description 4008 views Posted: 25 Sep 2013 We explain how, if the sums of the entries of each row of a matrix are the same, or if the sums of the entries of each column are the same, then the common sum is an eigenvalue of the matrix. Transcript: This is the last in our series of videos on 'Tricks of the Trade', ways to diagonalize matrices. In this video, we're going to look at how to use transposes, row sums, and column sums to find eigenvalues. The idea about column sums is the following. Take a look at this matrix. You'll notice that each column adds up to 2. 1 + 3 + -2 makes 2, 4 + -3 + 1 is 2, -1 + 1 + 2 is 2, so what happens if you multiply this matrix by (1 1 1)? You get 1 + 3 + -2, 4 - 3 + 1, -2 + 1 + 2, and that is (2 2 2), so (1 1 1) was an eigenvector with eigenvalue 2. If the columns had all added up to 17, then (1 1 1) would have been an eigenvector, with eigenvalue 17. Whenever all the rows of a matrix have the same sum, then (1 1 1 1 1), however many ones you need, you know, n ones for an n by n matrix, that's always gonna be an eigenvector, and whatever that sum is is going to be an eigenvalue. So, for example, suppose we have the matrix ((0 1 1) (1 0 1) (1 1 0)), we've seen this matrix before, but the sum of the first row is two, the sum of the second row is two, the sum of the third row is two, so all the rows add up to two, and that means that two has to be an eigenvalue. And now we can use our other tricks to find the other eigenvalues. The trace is 0 + 0 + 0, so all the eigenvalues have to add up to 0, 2 plus the other two eigenvalues is 0, so the other two eigenvalues have to add up to -2. The determinant is 2, so 2 times lambda 2 times lambda 3, is 2, so lambda 2 times lambda 3 has to be 1, and the only numbers that multiply to 1 and add to -2 are -1 and -1, so we have 2 is an eigenvalue, with multiplicity 1, -1 is an eigenvalue with algebraic multiplicity 2. Turns out to also have geometric multiplicity 2, but that's - you have to do more work to figure that out. Okay, now, I talked about transposes. A property of determinants is that if you take the transpose of a matrix and then take it's determinant, that's the same thing as the determinant of the original matrix, because - well, then if you take the determinant of lambda times the identity minus the transpose matrix, well that's the same thing as the determinant of lambda identity minus A transpose, because the identity transpose is just the identity, and that's gonna be the determinant of lambda I minus A, because when you take the transpose of a matrix, you don't change the determinant. But, this is the characteristic polynomial of A transpose, and this is the characteristic polynomial of A, so A and A transpose have the same characteristic polynomial, well that means they have to have the same eigenvalues, so the eigenvalues of A transpose are always the same as the eigenvalues of A. Now, the eigenvectors can be very different, but the eigenvalues are the same. So now, let's take a look at this matrix. You look at the rows, and you say oh, this row adds up to 4, and this row adds up to 1, and this row adds up to 1, that doesn't look like there's anything happening with the rows. But the columns all add up to the same thing. This column adds up to 2, this column adds up to 2, this column adds up to 2. Since the columns all add up to 2, well the columns of M are the rows of M transpose, so all the rows of M transpose add up to 2, but that means that 2 has to be an eigenvalue of M transpose, well that means that 2 has to be an eigenvalue of M. The general principle is that if you ever have a situation where all the columns of the matrix are - have the same sum, then that sum is an eigenvalue. You can't write down the eigenvector, there's no simple formula for the eigenvector, but the common sum of the columns is an eigenvalue. Now, it often happens in probability that you're dealing with matrices that describe the probability of a situation tomorrow, as a function of the probability of what things are today, and you have that each column adds up to 1, that's called a probability matrix, a probability matrix is a matrix where each entry is non-negative, and the columns add up to 1, by this principle, such matrices always have 1 as an eigenvalue. Okay, so let's combine all the tricks we've learned so far to figure out the eigenvalues of this big 5 by 5 matrix. The first thing we want to do, is we want to partition it. It's block triangular, so we really only have to find the eigenvalues of A, and we have to find the eigenvalues of D, so let's do these one at a time, let's look at A. Oops, this is a MINUS 5. The trace of A is 0. The determinant of A is 5 times -5, minus 3 times -3, so that's -25 plus 9, so it's -16, so the eigenvalues of A have the property that they add up to 0, and their product is -16, so that means the eigenvalues of A have to be 4 and -4, and that means that the eigenvalues of M have to include 4 and -4. Now, let's take a look at D. This column adds up to 3, this column adds up to 3, this column adds up to 3, so one of the eigenvalues is 3. Sorry, these are eigenvalues 1 and 2, and this is gonna be eigenvalue 3. Now, the trace is 6, and that means that 3 plus the 4th eigenvalue plus the 5th eigenvalue has to be 6, so the fourth eigenvalue plus the fifth eigenvalue has to be 3. Now, if you work out the determinant, you get the determinant of D is also 6, and that has to be 3 times lambda 4 times lambda 5, so lambda 4 times lambda 5 have to be 2, so what are two numbers that add up to 3 and who's product is 2? Well, 1 and 2. So there we have it. The eigenvalues of this big 5 by 5 matrix are 4, -4, 3, 1, and 2. Now how do you find the eigenvectors, well, if you find the eigenvectors of this piece, you pad them with 0's and you get an eigenvector of the whole thing. If you have the eigenvectors of this piece, you can't pad them with 0's to get eigenvectors of the whole thing, you have to take M minus 3 times the identity and row reduce it to figure out this eigenvector, M minus 1 times the identity and row reduce it to get this eigenvector, and minus 2 times the identity and row reduce it to get this eigenvector. The tricks of the trade generally help a lot with finding eigenvalues, but once you've got the eigenvalues, you still have to work to get the eigenvectors.
17512	https://www.instagram.com/p/DPHb7cRjeUV/	Instagram Log In Sign Up noahjsad27 sunderlandafc • Follow sunderlandafc1d Approximately 71% of the Earth’s surface is covered by water. You know the rest... hendo19801d What a performance, every single player out there today was outstanding. Sadiki on a yellow early doors to then play like that for 90 mins, what a lad. Like Reply lukhany.o_101d What a player. We take pride knowing he's ours🇨🇩🇨🇩 Like Reply View all 1 replies nathan_kimbs1d Sadiki🇨🇩🇨🇩🇨🇩🇨🇩 Like Reply jrauch971d Ils commencent à le découvrir 🥹 Like Reply charliewilkinson43911d Our recruitment this summer has been worldclass Like Reply aan_andre_afriza Like Reply claire_penrice16h And the other 29% is covered by Noah Like Reply rob.barber2121d Given he was on a yellow for most of the game, he was immense Like Reply hawaylk1d What a player this kid is. He's everywhere like a rash. Immense. Like Reply andymile101d What a player Like Reply chelseafans_cymru Like Reply jubskiii1d Through this game and last week's game Noah probably covered the whole earth's distance Like Reply garrethfitzgram1d 27 more points to go! 👏 Like Reply View all 4 replies legarsquetuneconnaitpas1d 😂😂 do you like ? It's congolese ! 😍🇨🇩 Like Reply fredwardeze23h I cant wait for sunderland vs Newcastle derby match,it’s gonna be 🔥🔥🔥 Like Reply View all 7 replies 25,591 likes 1 day ago Log in to like or comment. More posts from sunderlandafc See more posts Meta About Blog Jobs Help API Privacy Consumer Health Privacy Terms Locations Instagram Lite Meta AI Meta AI Articles Threads Contact Uploading & Non-Users Meta Verified English © 2025 Instagram from Meta See more from sunderlandafc See photos, videos and more from Sunderland AFC. Sign up for Instagram Log in By continuing, you agree to Instagram's Terms of Use and Privacy Policy.
17513	https://math.stackexchange.com/questions/1098669/let-w-x-y-z-be-natural-numbers-find-the-correct-alternative	combinatorics - Let $w, x, y, z$ be natural numbers. Find the correct alternative. - 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 Let w,x,y,z w,x,y,z be natural numbers. Find the correct alternative. Ask Question Asked 10 years, 8 months ago Modified10 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Let w w, x x, y y, z z be four natural numbers such that their sum is 8⋅m+10 8⋅m+10, where m m is a natural number. Given m m which of the following is possible: The max. possible value of w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 is 6⋅m 2+40⋅m+26 6⋅m 2+40⋅m+26. The max. possible value of w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 is 16⋅m 2+40⋅m+28 16⋅m 2+40⋅m+28. The min. possible value of w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 is 16⋅m 2+40⋅m+28 16⋅m 2+40⋅m+28. The min. possible value of w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 is 16⋅m 2+40⋅m+26 16⋅m 2+40⋅m+26. The answer given is option 4. I tried to expand (w+x+y+z)2(w+x+y+z)2. This will be 0(mod 4)0(mod⁡4). Now the only way sum of four integers give even number is if 1. All even, 2. All odd, 3. 2 odd and 2 even. For the first 2 cases if we expand the expression everything except w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 will be a multiple of four. So, w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 should be divisible by 4 4. Only case 3 requires w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 to be divisible by 2 2. Now if we put m=1 m=1 and take w,x,y,z w,x,y,z as 1,3,5,9 1,3,5,9 we can rule out first 2 options of the question. From here, can we argue that option 4 is the possible choice? Since option 4 gives 2(mod 4)2(mod⁡4) case. Is there any other general way to prove this? This is a question from a chapter of number theory. But solution in any form is heartily welcome. combinatorics elementary-number-theory discrete-mathematics Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 18, 2015 at 21:24 user2097 2,455 1 1 gold badge 18 18 silver badges 19 19 bronze badges asked Jan 10, 2015 at 8:29 archangel89archangel89 903 10 10 silver badges 32 32 bronze badges 2 Can it is necessary to solve the Diophantine equation? w 2+x 2+y 2+z 2=16 m 2+40 m+a w 2+x 2+y 2+z 2=16 m 2+40 m+a individ –individ 2015-01-19 04:40:02 +00:00 Commented Jan 19, 2015 at 4:40 Someone voting down the answers....Tahir Imanov –Tahir Imanov 2015-01-20 20:02:50 +00:00 Commented Jan 20, 2015 at 20:02 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. You can use the AM-QM inequality; 2 m+5 2=w+x+y+z 4≤w 2+x 2+y 2+z 2 4−−−−−−−−−−−−−−−√2 m+5 2=w+x+y+z 4≤w 2+x 2+y 2+z 2 4 Hence you easily get w 2+x 2+y 2+z 2≥16 m 2+40 m+25 w 2+x 2+y 2+z 2≥16 m 2+40 m+25 Your book reports 26 26 instead of 25 25; this can be fixed by noting that to have equality you should have w=x=y=z w=x=y=z, but his would imply 8 m+10=4 w 8 m+10=4 w. The left hand side is not divisible by 4 4 though, so we cannot have equality and the minimun attainable is at lest one above the previous bound, that is to say 16 m 2+40 m+26 16 m 2+40 m+26. Of course one should now prove that you can actually reach this; you can do so by following the general line guide "the more the numbers are close to each other, the less their quadratic mean will be" (if the arithmetic mean is fixed, of course) So following this idea you can try to set x=w=2 m+2 x=w=2 m+2, y=z=2 m+3 y=z=2 m+3. You can easily see that with this choice {w+x+y+z=8 m+10 w 2+x 2+y 2+z 2=16 m 2+40 m+26{w+x+y+z=8 m+10 w 2+x 2+y 2+z 2=16 m 2+40 m+26 and it is indeed the minimum for what we have said before Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 10, 2015 at 9:39 answered Jan 10, 2015 at 8:45 AntAnt 21.5k 6 6 gold badges 50 50 silver badges 104 104 bronze badges 2 Nice. But as you said is there any way to prove it?archangel89 –archangel89 2015-01-10 09:32:20 +00:00 Commented Jan 10, 2015 at 9:32 @archangel89 I edited the answer.. it should be all good now :)Ant –Ant 2015-01-10 09:39:27 +00:00 Commented Jan 10, 2015 at 9:39 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. When w=8 m+7 w=8 m+7 and x=y=z=1 x=y=z=1 then w 2+x 2+y 2+z 2=64 m 2+112 m+52.w 2+x 2+y 2+z 2=64 m 2+112 m+52. This is clearly greater than both options 1. and 2.; whence these are falsified. When y−x≥2 y−x≥2 then replacing x x, y y by x′:=x+1 x′:=x+1, y′:=y−1 y′:=y−1 respectively leads to x′2+y′2=x 2+y 2−2(y−x)+2<x 2+y 2.x′2+y′2=x 2+y 2−2(y−x)+2<x 2+y 2. Therefore w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 can be strictly decreased without changing the sum w+x+y+z=8 m+10 w+x+y+z=8 m+10, until the variables differ by at most 1 1. If at this point they were all even or all odd they would have to be equal; whence their sum would have to be divisible by 4 4. And if 1 1 or 3 3 of them would be odd their sum would be odd. The only configuration that remains for the minimum of w 2+x 2+y 2+z 2 w 2+x 2+y 2+z 2 is that w=2 n,x=2 n,y=2 n±1,z=2 n±1 w=2 n,x=2 n,y=2 n±1,z=2 n±1 for some n≥1 n≥1. This leads to the condition 8 n±2=8 m+10 8 n±2=8 m+10 and enforces the ++-sign in the above alternative. Therefore we arrive at w=x=2 m+2,y=z=2 m+3,w=x=2 m+2,y=z=2 m+3, which indeed leads to w 2+x 2+y 2+z 2=16 m 2+40 m+26.w 2+x 2+y 2+z 2=16 m 2+40 m+26. It follows that option 4. is correct, and option 3. (with 28 28 instead of 26 26) is false as well. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 14, 2015 at 14:50 answered Jan 14, 2015 at 13:27 Christian BlatterChristian Blatter 233k 14 14 gold badges 204 204 silver badges 490 490 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The simplest aproach would be using Arithmetic Mean - Geometric Mean Inequality. Let, A=w+x+y+z=8 m+10 A=w+x+y+z=8 m+10 . Therefore, A 2=64 m 2+160 m+100 A 2=64 m 2+160 m+100 Also, A 2=w 2+2 w x+2 w y+2 w z+x 2+2 x y+2 x z+y 2+2 y z+z 2 A 2=w 2+2 w x+2 w y+2 w z+x 2+2 x y+2 x z+y 2+2 y z+z 2 A 2≤4 w 2+4 x 2+4 y 2+4 z 2∗A 2≤4 w 2+4 x 2+4 y 2+4 z 2∗ Therefore, w 2+x 2+y 2+z 2≥1 4 A 2=16 m 2+40 m+25 w 2+x 2+y 2+z 2≥1 4 A 2=16 m 2+40 m+25 The equality is possible, if and only if w=x=y=z=2 m+2.5 w=x=y=z=2 m+2.5 . Therefore, m i n(w 2+x 2+y 2+z 2)=16 m 2+40 m+26 m i n(w 2+x 2+y 2+z 2)=16 m 2+40 m+26 ∗2 w x≤w 2+x 2,2 w y≤w 2+y 2,a n d e t c.∗2 w x≤w 2+x 2,2 w y≤w 2+y 2,a n d e t c. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 18, 2015 at 16:26 Tahir ImanovTahir Imanov 313 1 1 silver badge 12 12 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics elementary-number-theory discrete-mathematics 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 10Find the highest power of two in the expression. 1Construction of Natural Numbers 1Let S S be the set of first 100 natural numbers,L L is the least common multiple of all the elements in S,P S,P is the product of all the primes in S S 2For any natural numbers a, b, c, d if ab = cd is it possible that a + b + c + d is prime number 3How many words of length n n over the alphabet {0,1,2}{0,1,2} contain an even number of zeros? 2Number of nine digits numbers whose sum of the digits is even 3Recursive formula to find the number of natural numbers in which there are no two adjacent even digits. 6Analyzing a Diophantine equation: A k+1=B!A k+1=B! 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17514	https://vocal.media/education/top-5-strategies-for-tackling-fill-in-the-blanks-in-the-duolingo-english-test-ih1u0j8k	Main navigation Main navigation Top 5 Strategies for Tackling Fill In The Blanks in the Duolingo English Test Duolingo Fill In The Blanks Duolingo English test is becoming popular among students because of its adaptive testing method. Students who are willing to study abroad must choose this exam as it has many advantages. Though the test format is accessible, students often find the fill-in-the-blanks section quite challenging. So, to help you score well, here are tips to crack the Duolingo English test fill in the blanks. 1 Understand the Context: The first step to crack in the blanks question type is to read the entire paragraph carefully. You have to understand the meaning of the sentences so that you will get a clue about which word fits best. Look at the surrounding words and try to infer the context. For example: The sky was bright and clear after the storm." Here, the word "bright" makes sense because it matches the context of a clear sky after a storm. 2. Use Grammar Clues: In Duolingo English test fill in the blanks question type, grammatical structure plays a big role. The missing word must agree with the rest of the sentence in terms of tense, subject-verb agreement, and word form (noun, verb, adjective, etc.). While filling in the blanks, keep this context in mind: What is to be filled out, like nouns, verbs, adjectives, or adverbs? Tense of the sentence. Example: "She enjoys running every morning. Here, the word "enjoys" fits because the subject "she" is singular, and the sentence is in the present tense. 3 Enhance your vocabulary Having a strong vocabulary is crucial for filling in the blanks questions. The more words you know, the easier it is to recognize the correct options. To build your vocabulary: The more familiar you are with different words, the more confident you'll be when choosing the right word during the test. 4. Look for Collocations Collocations are words that commonly go together. For example, we say "make a decision" and not "make a decision." Recognizing these word pairings will help you quickly identify the right words to fill in the blanks. Some common collocations include: The more you practice identifying these, the better your chances of selecting the right words in Fill in the Blanks questions. 5 Eliminate Wrong Answers This is the most important thing that you must not miss. If you are not sure about the correct answer, use the elimination method. It will help you narrow down your options and choose the correct word. Start by identifying the choices that don't fit grammatically or contextually. For instance: "He quickly runs to the store." Here, "run" is incorrect because the subject "he" requires the verb to be in its singular form: "runs." Once you've eliminated the wrong options, you'll be left with fewer choices, making it easier to choose the correct answer. Bonus Tip: Practice as much as you can. Take the Duolingo Practice test, and you will get an idea of how the test will work. You can practice on online platforms like Gurully. Here, you will get real-time exam simulation with AI scoring. Practice on the mock test, and you will be able to determine where you are lacking and how to strengthen those points. The Fill in the Blanks section of the Duolingo English Test can be tricky, but with the right strategies, you can excel. By understanding the context, using grammar clues, enhancing your vocabulary, recognizing collocations, and eliminating wrong answers, you'll be better prepared to tackle this challenging part of the exam. Don't forget to practice regularly on platforms like Gurully to simulate real test conditions. With consistent effort and focus, you'll be well on your way to achieving a great score in the Duolingo English Test! About the Creator Brijesh Dhanani I am Brijesh Dhanani, co-founder of Gurully, I have been instrumental in building a platform that empowers students to excel in global English proficiency tests like PTE, IELTS, Duolingo, and CELPIP. Reader insights Be the first to share your insights about this piece. How does it work? Comments There are no comments for this story Be the first to respond and start the conversation. Keep reading More stories from Brijesh Dhanani and writers in Education and other communities. Last Week Preparation Tips for PTE Core Exam: Secure Your Canadian PR Do you have only a week to take your PTE exam? You have already taken classes and English-speaking tutorials and practiced online with paid and free resources, but you still have doubts. I mean, this is your life and your maple-soaked future we are talking about. By Brijesh Dhananiabout a year ago in Education How do adults become emotionally intelligent and use it to build stronger relationships and better mental health? Emotional intelligence (EI or EQ) refers broadly to one's ability to recognize, understand, regulate, and effectively use emotions—both their own and others'. By Naveen Garg3 days ago in Education How to Write Efficient & Optimized MATLAB Code for Complex Projects? As you know, crafting efficient and optimized MATLAB code is crucial. Especially when you are working on a big project. Efficiency is not only about making our code faster; it is also about making it easy to read and fix. However, if you still face any issues in efficiency of your MATLAB assignment, get assistance from MATLAB assignment help. By henry scott2 days ago in Education Knotted Roots - Part 2 My sweat-soaked and mud-stained shirt draped over Spot’s saddle, as Grace and I walked with our arms linked. The burnt amber glow behind the mountains quickly faded, and the starry night and full moon illuminated our dirt trail. For a few minutes, we walked silently, both of us keeping a curious eye on the sky, as if hoping this moment would never end. “You know, I don’t think I mentioned this before, but…” Grace hesitated, a small smile shadowed by the brim of her hat, which was pulled upwards as her eyes watched our boots. “It’s really nice having you back.” She stated, a slight hint of relief in her voice, peering upwards towards me to meet my eyes, as I looked over at her. Unable to really respond, I smiled at her as the stubble from my beard prickled my cheeks. I wish I were here because I truly wanted to be here, not out of remorse from a dying father. But what kind of son would I be if I allowed his legacy to die, to let him die alone? Ever since mom passed, he was always focused on the ranch. I grew weary of it all and wanted to experience life outside of this country lifestyle. Would Grace understand that? Should I tell her? Deciding to shake off the uneasiness, I decided to push that conversation off for another time. “Yeah, Grace, it’s good to be home.” Her eyes sparkled under the night sky, lulling me into her warmth. I placed my hand on her arm, as if assuring her. “Remember when we would run through the wildflowers?” She began, “You mean, chasing you for stealing my lasso?” With a shared moment of laughter she continued, “Okay, I may have borrowed…” “Borrowed? Is that what we call it now?” I teased. “Okay, first off, I at least know how to use a lasso!” Her snarky recoil caused a shift in me, a warmth in my center. The truth is, I missed Grace, but I didn’t miss the painful memories from losing my mom and now my ill father. “Second,” She continued, and her voice faded as I became consumed by her witty playfulness, the way her dimples revealed her bluff. “You think you can lasso better than me?!” I asked her, my voice low, husky. We came to a stop, my eyes challenging her, my smile welcoming. “If I didn’t know any better, I’d say you were presenting a challenge.” Grace had moved in a way where we were facing each other, her voice low, alluring as her accent still thick. She leaned inwards and shifted upwards from her tiptoes. “You know I love a good competition.” The way her smile grew and the feeling of pure electricity between us intensified. Tucking a loose blonde curl behind Grace’s ear, I cupped her chin in my rough hand, her skin soft and radiant. “Jasper…” The breathlessness in her voice was being restrained by something, “Yes, my kitten…” My eyes darkened, my mouth dry. “I…” “Grace!” A male’s voice rang out, cutting the tension between us, a voice I didn’t recognize. My face, shadowed and hidden by my hat brim, hid my annoyance from Grace, a moment to adjust before she could see. “Who’s that little kitten?” Her head hung low and then upwards with a reassuring smile, one that was almost apologetic in a way, “That’s uh, well…” “There you are, Grace!” A lean, muscled man came running up, his jeans dark and clean, his boots without a scuff, his blonde wavy hair tucked behind his ears, a loose white button-down shirt tucked behind an oversized belt buckle. He swept by me, embracing her in his arms, spinning her around as her feet lifted from the ground. “I thought something had happened to you…” “Greg, I…I’m just fine, and I can stand on my own; you needn’t sweep me up like that.” Grace clamored with poise and elegance as she charmingly demanded to be put down. This Greg, whoever he was, doesn’t appear to be a cowboy; he’s too clean-cut, too polished. “Oh! I’m sorry, man, I was just so caught up seeing my little sunshine here, I lost my manners. The name’s Greg, Greg Combs.” He extended his hand, reluctantly, I shook his hand. It was like waving a limp noodle, as my firm hand squeezed his as if silently warning him. “Jasper Black.” My tone was sharp like a blade. “Well, Grace, it looks like you’ll find your way from here,” I said as I gave Spot a few pats. Tipping my hat to Grace respectfully, I walked off towards my porch, digging my hands in my pockets, not looking back. Of course she had a boyfriend; hell, he could be her husband for all I know! I don't want to ruin her any more than I already have. I realized as my boots thudded across the wooden porch, I left more than just my father that day. I left her. By Sibley Shamra2 days ago in Chapters Find us on social media Miscellaneous links © 2025 Creatd, Inc. All Rights Reserved.
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With the curriculum-aligned word problems games on SplashLearn, your kids will be prepared to tackle all the multiplication problems. Personalized Learning Fun Rewards Actionable Reports Parents, Sign Up for Free Teachers, Use for Free Join millions of learners learning with a smile Sign Up GRADE Pre-K K 1 2 3 4 5 CONTENT TYPE Resources Games Worksheets Glossary Lesson Plans SUBJECT Math (2,046) Number Sense (381) Number Recognition (30) Number Recognition Within 5 (12) Number Recognition Within 10 (12) Number Recognition Within 20 (6) Number Tracing (20) Number Tracing Within 5 (5) Number Tracing Within 10 (5) Number Tracing Within 20 (10) Number Sequence (56) Counting (141) Counting Objects Within 5 (47) Counting Objects Within 10 (43) Counting Objects Within 20 (6) Compare Numbers (47) Compare Objects (7) Compare Numbers Using Place Value (5) Compare 2-Digit Numbers (6) Compare 3-Digit Numbers (10) Order Numbers (15) Skip Counting (36) Skip Count By 2 (8) Skip Count By 5 (8) Skip Count By 10 (13) Skip Count By 100 (3) Even And Odd Numbers (3) Place Value (78) Teen Numbers (4) Word Form (5) Expanded And Standard 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Kids will solve word problems by making sense of "multiply by" situations and selecting the right answers from given options. This interactive experience helps kids apply decimal arithmetic in real-world scenarios, boosting their problem-solving skills. Perfect for curious young minds! 5 5.NBT.7 VIEW DETAILS Multiplication Word Problems Solving the Word Problems Related to Multiplication Game Tackle multiplication word problems with this engaging game! Kids will strengthen their problem-solving skills by working on multiplicative comparison scenarios. They'll solve for unknown quantities while having fun. This game makes understanding multiplication easy and enjoyable, helping kids build confidence in math. 4 5 4.OA.2 VIEW DETAILS < View all result 2 GRADE Pre-K K 1 2 3 4 5 Introduction Traditionally, learning multiplication was associated with rote learning of the multiplication tables. With the pandemic bringing the world on an online platform, the focus of education has shifted from rote learning to digital learning. Today, children are encouraged to apply academic concepts in everyday life. In such a situation, online learning helps your kids prepare for an uncertain future by ensuring that their conceptual roots are firm. A word problem presents real-world scenarios to your kids in a few sentences. The kid is expected to comprehend the sentences and offer a solution to the problem after performing the necessary mathematical calculations. In this article, we will dwell upon solving multiplication word problems in fun and games. Understanding Word Problem Games Word problems are an essential part of a kid’s mathematical journey as they teach the child to apply academic concepts to real-world problems. By fourth grade, the child is familiar with mathematical words such as fewer, difference, share, reduced, equal, etc. This makes it the ideal age to be introduced to the world of word problems. With games on word problems, the child will be familiar with the RUCSAC method of solving word problems. With word problems, children take a systematic approach where the problem is first read, understood, and the child then chooses the mathematical keywords. The little one solves the problem to find the answer and finally validates the same. By grade four, the word problem includes ones where the child is given a scenario where he or she has X apples and is asked to share the same among Y people. The child needs to compute how many apples each person gets. Taking a Fun Approach to Learn The greatest teachers agree that children are the most receptive when they are having fun. Some online educational websites introduce the concepts of multiplication in grade 3, where the child is given a basic understanding of arrays, grouping, and the size of the groupings. As the child graduates to higher classes, the multiplication concepts they picked up will provide a solid foundation for mathematics learning. The Hurdle of Word Problems The typical challenge of word problems is that middle schoolers struggle to understand the question. With the COVID pandemic bringing education to an online platform, students lack the guidance to show them what a problem expects. When such a fear is not dealt with, it can translate to a lifetime of mathematics phobia. Help Your Child Develop Math Skills As a parent, you need to understand the mental stress of learning and grasping concepts that affect your child. Understanding math concepts is difficult and can be really hard for your kid. With SplashLearn, you can help ensure that the online learning experience is fun for your little ones and that they do not miss out on the enjoyment of the best years of their life. The SplashLearn solutions are planned so that the kid gets ample practice on a topic before moving to the next. The fun element of the mathematical word games ensures that your child enjoys the activity. EndNote As parents, your focus is on providing a holistic learning environment to your kids so that they pick up on new skills every day. That way, your little ones will be off to a productive start in the journey of life. The multiplication word problems by SplashLearn presents the subject in a fun manner so that your kids get the necessary practice. Not only does this prepare them for the academic tests of the lower grades, but it also helps to develop your child’s analytical skills and make them receptive learners for life. We hope that you enjoyed this content. 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17516	https://runestone.academy/ns/books/published/dmoi-4/sec_comb-proofs.html	Skip to main content \usepackage Section 3.6 Combinatorial Proofs Objectives After completing this section, you should be able to do the following. 🔗 Write counting problems that have a given answer. 🔗 🔗 2. Write two different solutions to a counting problem. 🔗 🔗 3. Prove binomial identities using combinatorial proofs. 🔗 🔗 🔗 Subsection Section Preview Investigate! Look at any cell in the interior Pascal’s triangle and the six numbers that surround it. For example, you might look at this cell: 🔗 Of the six numbers surrounding our selected cell, we will divide them into two groups of three, alternating between the groups. So for this example, we have a group with 4, 10, and 15, and a second group with 5, 6, and 20. But notice: . Does this work no matter what center cell you pick? Why?? 🔗 🔗 One of the coolest things about combinatorics is that you can often answer the same counting question in dramatically different ways. When we recognize this about a particular problem, we can often generalize the question to reveal two different expressions that must represent the same quantity. The counting problem itself becomes a proof of the equality of the two expressions. This style of proof is called a combinatorial proof. 🔗 Worksheet Preview Activity  It is often possible to find the answer to a counting question in two different ways. Doing so results in two formulas that give the answer, which might look different, but must be equal (since they are both the correct answer to the same question). This provides a proof of the equality of the two formulas, called a combinatorial proof. 🔗 Let’s explore a couple of examples. 🔗 1. Consider the set of 7-bit strings of weight 3. That is, strings of 0s and 1s that are seven characters long and have exactly three 1s. How many such strings are there? 🔗 (a) Give the answer as a single binomial coefficient. 🔗 🔗 (b) Now count just those 7-bit strings of weight 3 that start with a 1. How many are there? 🔗 🔗 (c) Now count just those 7-bit strings of weight 3 that start with 01. How many are there? 🔗 🔗 (d) Now count just those 7-bit strings of weight 3 that start with 001. How many are there? 🔗 🔗 (e) Continue this process until you have counted all 7-bit strings of weight 3, as a sum of binomial coefficients. What is this sum? 🔗 🔗 🔗 2. Consider the counting question “How many ways can you permute the letters of the word STATISTICS?” Note that this is not just a permutation, since there are repeated letters. 🔗 (a) How many ways can you select three of the ten positions in the anagram to be occupied by the letter S? 🔗 🔗 (b) How many ways can you select three of the remaining seven positions in the anagram to be occupied by the letter T? 🔗 🔗 (c) Continue with this approach until you have found an expression for the number of ways to permute the letters of STATISTICS as a product of binomial coefficients. Write out this product. 🔗 🔗 (d) Now answer the counting question again, this time starting by asking how many ways you can select positions for the letter A first. Continue in any way you like until you have found a different expression for the number of ways to permute the letters of STATISTICS as a product of binomial coefficients. Write out this product. 🔗 🔗 (e) Try yet another approach. What is wrong with saying the answer is ? This is too large, but we can correct it by dividing to account for outcomes that are equivalent. What should you divide by? 🔗 Write your answer as a quotient of factorials. Do you get the same answer as before? 🔗 🔗 🔗 🔗 🔗 Subsection Patterns in Pascal’s Triangle Have a look again at Pascal’s triangle. Forget for a moment where it comes from. Just look at it as a mathematical object. What do you notice? 🔗 There are lots of patterns hidden away in the triangle, enough to fill a reasonably sized book. Here are just a few of the most obvious ones: The entries on the border of the triangle are all 1. 🔗 🔗 2. Any entry not on the border is the sum of the two entries above it. 🔗 🔗 3. The triangle is symmetric. In any row, entries on the left side are mirrored on the right side. 🔗 🔗 4. The sum of all entries on a given row is a power of 2. (You should check this!) 🔗 🔗 🔗 We would like to state these observations in a more precise way, and then prove that they are correct. Now each entry in Pascal’s triangle is in fact a binomial coefficient. The 1 on the very top of the triangle is . The next row (which we will call row 1, even though it is not the top-most row) consists of and . Row 4 (the row 1, 4, 6, 4, 1) consists of the binomial coefficients . 🔗 Given this description of the elements in Pascal’s triangle, we can rewrite the above observations as follows: and . 🔗 2. . 🔗 3. . 🔗 4. . 🔗 🔗 Each of these is an example of a binomial identity : an identity (i.e., equation) involving binomial coefficients. 🔗 Our goal is to establish these identities. We wish to prove that they hold for all values of and . These proofs can be done in many ways. One option would be to give algebraic proofs, using the formula for : . 🔗 Here’s how you might do that for the second identity above. 🔗 Example 3.6.1. Give an algebraic proof for the binomial identity . 🔗 Solution. Proof. By the definition of , we have and . 🔗 Thus, starting with the right-hand side of the equation: . 🔗 The second line (where the common denominator is found) works because and . 🔗 🔗 🔗 🔗 This is certainly a valid proof but also is entirely useless. Even if you understand the proof perfectly, it does not tell you why the identity is true. A better approach would be to explain what means and then say why that is also what means. Let’s see how this works for the four identities we observed above. 🔗 Example 3.6.2. Explain why and . 🔗 Solution. What do these binomial coefficients tell us? Well, gives the number of ways to select 0 objects from a collection of objects. There is only one way to do this, namely to not select any of the objects. Thus . Similarly, gives the number of ways to select objects from a collection of objects. There is only one way to do this: Select all objects. Thus . 🔗 Alternatively, we know that is the number of -bit strings with weight 0. There is only one such string, the string of all 0’s. So . Similarly is the number of -bit strings with weight . There is only one string with this property, the string of all 1’s. 🔗 Another way: gives the number of subsets of a set of size containing 0 elements. There is only one such subset, the empty set. gives the number of subsets containing elements. The only such subset is the original set (of all elements). 🔗 🔗 🔗 Example 3.6.3. Explain why . 🔗 Solution. The easiest way to see this is to consider bit strings. is the number of bit strings of length containing 1’s. Of all of these strings, some start with a 1 and the rest start with a 0. First consider all the bit strings which start with a 1. After the 1, there must be more bits (to get the total length up to ) and exactly of them must be 1’s (as we already have one, and we need total). How many strings are there like that? There are exactly such bit strings, so of all the length bit strings containing 1’s, of them start with a 1. Similarly, there are which start with a 0 (we still need bits and now of them must be 1’s). Since there are bit strings containing bits with 1’s, that is the number of length bit strings with 1’s which start with a 0. Therefore . 🔗 Another way: Consider the question, how many ways can you select pizza toppings from a menu containing choices? One way to do this is just . Another way to answer the same question is to first decide whether or not you want anchovies. If you do want anchovies, you still need to pick toppings, now from just choices. That can be done in ways. If you do not want anchovies, then you still need to select toppings from choices (the anchovies are out). You can do that in ways. Since the choices with anchovies are disjoint from the choices without anchovies, the total choices are . But wait. We answered the same question in two different ways, so the two answers must be the same. Thus . 🔗 You can also explain (prove) this identity by counting subsets, or even lattice paths. 🔗 🔗 🔗 Example 3.6.4. Prove the binomial identity . 🔗 Solution. Why is this true? counts the number of ways to select things from choices. On the other hand, counts the number of ways to select things from choices. Are these really the same? Well, what if instead of selecting the things you choose to exclude them. How many ways are there to choose things to exclude from choices? Clearly this is as well (it doesn’t matter whether you include or exclude the things once you have chosen them). And if you exclude things, then you are including the other things. So the set of outcomes should be the same. 🔗 Let’s try the pizza counting example like we did above. How many ways are there to pick toppings from a list of choices? On the one hand, the answer is simply . Alternatively, you could make a list of all the toppings you don’t want. To end up with a pizza containing exactly toppings, you need to pick toppings to not put on the pizza. You have choices for the toppings you don’t want. Both of these ways give you a pizza with toppings, in fact all the ways to get a pizza with toppings. Thus these two answers must be the same: . 🔗 You can also prove (explain) this identity using bit strings, subsets, or lattice paths. The bit string argument is nice: counts the number of bit strings of length with 1’s. This is also the number of bit strings of length with 0’s (just replace each 1 with a 0 and each 0 with a 1). But if a string of length has 0’s, it must have 1’s. And there are exactly strings of length with 1’s. 🔗 🔗 🔗 Example 3.6.5. Prove the binomial identity . 🔗 Solution. Let’s do a “pizza proof” again. We need to find a question about pizza toppings which has as the answer. How about this: If a pizza joint offers toppings, how many pizzas can you build using any number of toppings from no toppings to all toppings, using each topping at most once? 🔗 On one hand, the answer is . For each topping, you can say “yes” or “no,” so you have two choices for each topping. 🔗 On the other hand, divide the possible pizzas into disjoint groups: the pizzas with no toppings, the pizzas with one topping, the pizzas with two toppings, etc. If we want no toppings, there is only one pizza like that (the empty pizza, if you will), but it would be better to think of that number as since we choose 0 of the toppings. How many pizzas have 1 topping? We need to choose 1 of the toppings, so . We have: 🔗 Pizzas with 0 toppings: 🔗 Pizzas with 1 topping: 🔗 Pizzas with 2 toppings: 🔗 🔗 Pizzas with toppings: . 🔗 🔗 The total number of possible pizzas will be the sum of these, which is exactly the left-hand side of the identity we are trying to prove. 🔗 Again, we could have proved the identity using subsets, bit strings, or lattice paths (although the lattice path argument is a little tricky). 🔗 🔗 🔗 Hopefully this gives some idea of how explanatory proofs of binomial identities can go. It is worth pointing out that more traditional proofs can also be beautiful.4 Most every binomial identity can be proved using mathematical induction, using the recursive definition for . We will discuss induction in Section 4.5. For example, consider the following rather slick proof of the last identity. 🔗 Expand the binomial : . 🔗 Let and . We get: . 🔗 Of course this simplifies to: . 🔗 Something fun to try: Let and . Neat huh? 🔗 🔗 Subsection More Proofs The explanatory proofs given in the above examples are typically called combinatorial proofs. In general, to give a combinatorial proof for a binomial identity, say , you do the following: 🔗 Find a counting problem you will be able to answer in two ways. 🔗 🔗 2. Explain why one answer to the counting problem is . 🔗 🔗 3. Explain why the other answer to the counting problem is . 🔗 🔗 🔗 Since both and are the answers to the same question, we must have . 🔗 The tricky thing is coming up with the question. This is not always obvious, but it gets easier the more counting problems you solve. You will start to recognize types of answers as the answers to types of questions. More often what will happen is that you will be solving a counting problem and happen to think up two different ways of finding the answer. Now you have a binomial identity, and the proof is right there. The proof is the problem you just solved together with your two solutions. 🔗 For example, consider this counting question: 🔗 How many 10-letter words use exactly four A’s, three B’s, two C’s, and one D? 🔗 🔗 Let’s try to solve this problem. We have 10 spots for letters to go. Four of those need to be A’s. We can pick the four A-spots in ways. Now where can we put the B’s? Well there are only 6 spots left; we need to pick of them. This can be done in ways. The two C’s need to go in two of the 3 remaining spots, so we have ways of doing that. That leaves just one spot of the D, but we could write that 1 choice as . Thus the answer is: . 🔗 But why stop there? We can find the answer another way too. First let’s decide where to put the one D: we have 10 spots and we need to choose 1 of them, so this can be done in ways. Next, choose one of the ways to place the two C’s. We now have spots left, and three of them need to be filled with B’s. There are ways to do this. Finally the A’s can be placed in (that is, only one) ways. So another answer to the question is . 🔗 Interesting. This gives us the binomial identity: . 🔗 Here are a couple more binomial identities with combinatorial proofs. 🔗 Example 3.6.6. Prove the identity . 🔗 Solution. To give a combinatorial proof we need to think up a question we can answer in two ways: one way needs to give the left-hand side of the identity, and the other way needs to be the right-hand side of the identity. Our clue to what question to ask comes from the right-hand side: counts the number of ways to select 3 things from a group of things. Let’s name those things . In other words, we want to find 3-element subsets of those numbers (since order should not matter, subsets are exactly the right thing to think about). We will have to be a bit clever to explain why the left-hand side also gives the number of these subsets. Here’s the proof. 🔗 Proof. Consider the question “How many 3-element subsets are there of the set ?” We answer this in two ways: 🔗 Answer 1: We must select 3 elements from the collection of elements. This can be done in ways. 🔗 Answer 2: Break this problem up into cases by what the middle number in the subset is. Say each subset is written in increasing order. We count the number of subsets for each distinct value of . The smallest possible value of is , and the largest is . 🔗 When , there are subsets: 1 choice for and choices (3 through ) for . 🔗 When , there are subsets: 2 choices for and choices for . 🔗 When , there are subsets: 3 choices for and choices for . 🔗 And so on. When , there are choices for and only 1 choice for , so subsets. 🔗 Therefore the total number of subsets is . 🔗 Since Answer 1 and Answer 2 are answers to the same question, they must be equal. Therefore . 🔗 🔗 🔗 🔗 Example 3.6.7. Prove the binomial identity . 🔗 Solution. We will give two different proofs of this fact. The first will be very similar to the previous example (counting subsets). The second proof is a little slicker, using lattice paths. 🔗 Proof. Consider the question: “How many pizzas can you make using toppings when there are toppings to choose from?” 🔗 Answer 1: There are toppings, from which you must choose . This can be done in ways. 🔗 Answer 2: Divide the toppings into two groups of toppings (perhaps meats and veggies). Any choice of toppings must include some number from the first group and some number from the second group. Consider each possible number of meat toppings separately: 🔗 0 meats: , since you need to choose 0 of the meats and of the veggies. 🔗 1 meat: , since you need 1 of meats so of veggies. 🔗 2 meats: . Choose 2 meats and the remaining toppings from the veggies. 🔗 And so on. The last case is meats, which can be done in ways. 🔗 Thus the total number of pizzas possible is . 🔗 This is not quite the left-hand side … yet. Notice that and and so on, by the identity in Example 3.6.4. Thus we do indeed get . 🔗 Since these two answers are answers to the same question, they must be equal, and thus . 🔗 🔗 For an alternative proof, we use lattice paths. This is reasonable to consider because the right-hand side of the identity reminds us of the number of paths from to . 🔗 Proof. Consider the question: How many lattice paths are there from to ? 🔗 Answer 1: We must travel steps, and of them must be in the up direction. Thus there are paths. 🔗 Answer 2: Note that any path from to must cross the line . That is, any path must pass through exactly one of the points: , , , …, . For example, this is what happens in the case : 🔗 How many paths pass through ? To get to that point, you must travel units, and of them are to the right, so there are ways to get to . From to takes steps, and of them are up. So there are ways to get from to . Therefore there are paths from to through the point . 🔗 What about through ? There are paths to get there ( steps, 1 to the right) and paths to complete the journey to ( steps, up). So there are paths from to through . 🔗 In general, to get to through the point we have paths to the midpoint and then paths from the midpoint to . So there are paths from to through . 🔗 All together then, the total paths from to passing through exactly one of these midpoints is . 🔗 Since these two answers are answers to the same question, they must be equal, and thus . 🔗 🔗 🔗 🔗 🔗 Reading Questions Reading Questions 1. Which of the following describes the overall strategy for a combinatorial proof? 🔗 Multiple Choice (dmoi-4__rq-counting-proofs-format) 🔗 2. shortanswer (dmoi-4__rq-counting-proofs-question) 🔗 3. shortanswer (dmoi-4__rq-counting-proofs-q) 🔗 🔗 Exercises Practice Problems 1. Create a combinatorial proof of the identity . 🔗 Arrow keys to navigate. Space to select / deselect block to move. Drag from here This is because you have 2 choices for the first digit, and 10 choices for the second digit. 🔗 This is because you must either choose 2 identical toppings or two different toppings. 🔗 Consider the question, “How many 2-topping pizzas can you make choosing from 10 toppings?” 🔗 Consider the question, “How many two-digit numbers start with a 3 or 4?” 🔗 A second answer to the question is . 🔗 This is because there are 10 choices for the first topping, and 10 choices for the second topping. 🔗 This is because there are 10 numbers that start with 3, and another 10 that start with 4. 🔗 The first way to answer this is . 🔗 Since both expressions answer the same question, they must be equal. Therefore . 🔗 or{ or{ or{ Drop blocks here Parsons (dmoi-4__parsons-comb-proof-basic) 🔗 2. If you were asked to give a combinatorial proof of the identity , which of the following would be reasonable questions to use? 🔗 Multiple Choice (dmoi-4__rs-comb-proof-question) 🔗 🔗 Exercises Additional Exercises 1. Give a combinatorial proof of the identity . 🔗 🔗 2. Suppose you own fezzes and bow ties. Of course, and are both greater than 1. How many combinations of fez and bow tie can you make? You can wear only one fez and one bow tie at a time. Explain. 🔗 🔗 2. Explain why the answer is also . (If this is what you claimed the answer was in part (a), try it again.) 🔗 🔗 3. Use your answers to parts (a) and (b) to give a combinatorial proof of the identity . 🔗 🔗 🔗 🔗 3. How many triangles can you draw using the dots below as vertices? 🔗 Find an expression for the answer which is the sum of three terms involving binomial coefficients. 🔗 🔗 2. Find an expression for the answer which is the difference of two binomial coefficients. 🔗 🔗 3. Generalize the above to state and prove a binomial identity using a combinatorial proof. Say you have points on the horizontal axis and points in the semi-circle. 🔗 🔗 🔗 Hint. There will be 185 triangles. But to find them … How many vertices of the triangle can be on the horizontal axis? 🔗 🔗 2. Will any three dots work as the vertices? 🔗 🔗 🔗 🔗 🔗 4. Consider all the triangles you can create using the points shown below as vertices. Note that we are not allowing degenerate triangles (ones with all three vertices on the same line), but we do allow non-right triangles. 🔗 Find the number of triangles, and explain why your answer is correct. 🔗 🔗 2. Find the number of triangles again, using a different method. Explain why your new method works. 🔗 🔗 3. State a binomial identity that your two answers above establish (that is, give the binomial identity that your two answers are a proof for). Then generalize this using ’s and ’s. 🔗 🔗 🔗 Hint. The answer is 120. 🔗 🔗 🔗 5. A woman is getting married. She has 15 best friends but can only select 6 of them to be her bridesmaids, one of which needs to be her maid of honor. How many ways can she do this? 🔗 What if she first selects the 6 bridesmaids, and then selects one of them to be the maid of honor? 🔗 🔗 2. What if she first selects her maid of honor, and then 5 other bridesmaids? 🔗 🔗 3. Explain why . 🔗 🔗 🔗 6. Consider the identity: . Is this true? Try it for a few values of and . 🔗 🔗 2. Use the formula for to give an algebraic proof of the identity. 🔗 🔗 3. Give a combinatorial proof of the identity. 🔗 🔗 🔗 Hint. Try Exercise 5. 🔗 🔗 🔗 7. Give a combinatorial proof of the identity . 🔗 Hint. What if you wanted a pair of co-maids-of-honor? 🔗 🔗 🔗 8. Consider the binomial identity . Give a combinatorial proof of this identity. Hint: What if some number of a group of people wanted to go to an escape room, and among those going, one needed to be the team captain? 🔗 🔗 2. Give an alternate proof by multiplying out and taking derivatives of both sides. 🔗 🔗 🔗 Hint. For the combinatorial proof: What if you don’t yet know how many bridesmaids you will have? 🔗 🔗 🔗 9. Give a combinatorial proof for the identity . 🔗 Hint. Count handshakes. 🔗 🔗 🔗 10. Consider the bit strings in (bit strings of length 6 and weight 2). 🔗 How many of those bit strings start with 1? 🔗 🔗 2. How many of those bit strings start with 01? 🔗 🔗 3. How many of those bit strings start with 001? 🔗 🔗 4. Are there any other strings we have not counted yet? Which ones, and how many are there? 🔗 🔗 5. How many bit strings are there total in ? 🔗 🔗 6. What binomial identity have you just given a combinatorial proof for? 🔗 🔗 🔗 11. Let’s count ternary digit strings, that is, strings in which each digit can be 0, 1, or 2. How many ternary digit strings contain exactly digits? 🔗 🔗 2. How many ternary digit strings contain exactly digits and 2’s. 🔗 🔗 3. How many ternary digit strings contain exactly digits and 2’s. (Hint: Where can you put the non-2 digit, and then what could it be?) 🔗 🔗 4. How many ternary digit strings contain exactly digits and 2’s. (Hint: See previous hint.) 🔗 🔗 5. How many ternary digit strings contain exactly digits and 2’s. 🔗 🔗 6. How many ternary digit strings contain exactly digits and no 2’s. (Hint: What kind of a string is this?) 🔗 🔗 7. Use the above parts to give a combinatorial proof for the identity . 🔗 🔗 🔗 🔗 12. How many ways are there to rearrange the letters in the word “rearrange”? Answer this question in at least two different ways to establish a binomial identity. 🔗 🔗 13. Establish the identity below using a combinatorial proof. . 🔗 Hint. This one might remind you of Example 3.6.6 🔗 🔗 🔗 14. In Example 3.6.5 we established that the sum of any row in Pascal’s triangle is a power of two. Specifically, . The argument given there used the counting question, “How many pizzas can you build using any number of different toppings?” To practice, give new proofs of this identity using different questions. Use a question about counting subsets. 🔗 🔗 2. Use a question about counting bit strings. 🔗 🔗 3. Use a question about counting lattice paths. 🔗 🔗 🔗 Hint. For the lattice paths, think about what sort of paths would count. Not all the paths will end at the same point, but you could describe the set of end points as a line. 🔗 🔗 🔗 15. The Stanley Cup is decided in a best of 7 tournament between two teams. In how many ways can your team win? Let’s answer this question two ways: How many of the 7 games does your team need to win? How many ways can this happen? 🔗 🔗 2. What if the tournament goes all 7 games? So you win the last game. How many ways can the first 6 games go down? 🔗 🔗 3. What if the tournament goes just 6 games? How many ways can this happen? What about 5 games? 4 games? 🔗 🔗 4. What are the two different ways to compute the number of ways your team can win? Write down an equation involving binomial coefficients (that is, ’s). What pattern in Pascal’s triangle is this an example of? 🔗 🔗 🔗 🔗 2. Generalize. What if the rules changed, and you played a best of tournament (5 wins required)? What if you played an game tournament with wins required to be named champion? 🔗 🔗 🔗 🔗 16. Let be a list of positive integers that sum to (i.e., ). Use two graphs containing vertices to explain why . 🔗 Hint. How many edges does have? One of the two graphs will not be connected (unless ). 🔗 🔗 🔗 🔗 You have attempted 1 of 7 activities on this page. 🔗 PrevTopNext
17517	https://www.ncbi.nlm.nih.gov/books/NBK544316/	Histology, Schwann Cells - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Histology, Schwann Cells Matthew Fallon; Prasanna Tadi. Author Information and Affiliations Authors Matthew Fallon; Prasanna Tadi 1. Affiliations 1 Asram Medical College, Eluru, India Last Update: May 1, 2023. Go to: Introduction Schwann cells embryologically derive from the neural crest. They myelinate peripheral nerves and serve as the primary glial cells of the peripheral nervous system (PNS), insulating and providing nutrients to axons. Myelination increases conduction velocity along the axon, allowing for the saltatory conduction of impulses.Nonmyelinating Schwann cells do not wrap axons to improve conduction, but still, provide trophic support and cushioning to the unmyelinated axons. Go to: Structure Each Schwann cell makes up a single myelin sheath on a peripheral axon, with each ensuing myelin sheath made by a different Schwann cell, such that numerous Schwann cells are needed to myelinate the length of an axon. This arrangement is in contrast to oligodendrocytes, the myelinating cell of the central nervous system (CNS), which form myelin sheaths for multiple surrounding axons. Schwann cells are surrounded by a basal lamina, while oligodendrocytes are not. Between adjacent myelin sheaths, there are gaps of approximately 1 micrometer, called nodes of Ranvier. There is a concentration of voltage-gated sodium channels at the node, which is the site of saltatory conduction.Schmidt-Lanterman incisures are cytoplasmic outpouchings that interrupt compact myelin in heavily myelinated neurons. They contain a high density of gap junctions and other cell junctions, serving a role in communication and maintenance of the Schwann cell. Go to: Function Schwann cells serve as the myelinating cell of the PNS and support cells of peripheral neurons. A Schwann cell forms a myelin sheath by wrapping its plasma membrane concentrically around the inner axon. While the nucleus remains fixed, the inner turn of the glial cell membrane spirals around the axon to add membrane layers, or lamellae, to the myelin sheath. The plasma membrane of Schwann cells has an extremely high lipid content, and cholesterol is particularly important for assembling the myelin sheath. The compact myelin sheath insulates the axon segment, significantly reducing membrane capacitance and increasing conduction velocity. Neuregulin-type III expression on axons is essential for survival and maturation of Schwann cell precursors, and the degree of myelination depends on the amount of neuregulin on the surface of the axon.Schwann cells also provide energy metabolites to axons, shuttling them through monocarboxylate transports available along the surface of the axon and inner membrane of the Schwann cell. Schwann cells are critical in response to PNS axon damage and axon regeneration. Wallerian degeneration will occur distal to the injury site. The distal axon segment dies and Schwann cells, followed by macrophages, clear the dead cell contents, and promote axon regeneration. Schwann cells undergo several phenotypic changes at this time: they activate myelin breakdown, up-regulate the expression of cytokines (including TNF-a) to recruit macrophages to the injury site, up-regulate neurotrophic factors to stimulate axon regeneration and neuron survival, and organize a regeneration pathway along their basal lamina tube to guide axon growth. Axonotmesis and neurotmesis are the main types of PNS nerve injury. In axonotmesis, such as in a crush injury, the axon suffers disruption, but the basal lamina tube of the Schwann cells remain. The lumen of the tube provides guidance cues to the regenerating axon sprout as it grows, promoting highly effective axon regeneration and restoration of function in 3 to 4 weeks. In neurotmesis, such as in a cut injury, the axon, Schwann cell basal lamina, and surrounding connective tissue sheath are disrupted. The regenerating axon and its associated Schwann cells still grow from the proximal to the distal nerve stump. Because of targeting errors in the absence of the basal lamina tube, correct reinnervation and recovery of function are poor in neurotmesis. Go to: Tissue Preparation Aldehyde is the preferred routine fixative for nervous tissue. Electron microscopy requires an aldehyde fixative with high purity.After fixation, the sample gets embedded in either paraffin or epoxy. Paraffin allows for the study of the entire cross-sectional area of a nerve and is the preferred medium for light microscopy and larger nerve sections. Epoxy is preferable for smaller nerve branches and visualization with electron microscopy. Recent utilization of cryofixation, high pressure freezing and freeze substitution, is a beneficial supplement to aldehyde fixation in electron microscopy and may improve the preservation of structure detail and contrast. Go to: Histochemistry and Cytochemistry Immunohistochemical stains are valuable tools to differentiate Schwann cells from other cell types. S-100 is a protein unique to neural crest-derived cells, so anti-S-100 antibodies can be used to stain for healthy Schwann cells or nervous tissue neoplasms, such as schwannomas. Myelin basic protein (MBP) neutralizes phospholipid charges on the inner surface of the membrane and is present in Schwann cells but not satellite cells, the other major PNS glial cell.Anti-MBP can be used to differentiate Schwann cells or oligodendrocytes from other glial cells. P0, a peripheral nerve myelin protein, is a transmembrane adhesion protein that promotes the extracellular lamellar apposition that forms the intraperiod lines.Anti-P0 can be used to identify granular cell tumors, which derive from Schwann cells. Go to: Microscopy, Light Under light microscopy, Schwann cell nuclei and myelin sheath are visible, as well as their basal laminae and associated axons. Both light and electron microscopy demonstrate myelin sheaths of various thicknesses according to neuregulin expression by the axon. More heavily myelinated axons may have more than 40 lamellae, as visualized by alternating intraperiod and dense lines. Intraperiod lines demonstrate the apposition of extracellular surfaces of compact lamellae of the Schwann cell plasma membrane. Dense lines demonstrate the close apposition of the cytoplasmic surfaces of the membrane in compact myelin. Go to: Microscopy, Electron Under transmission electron microscopy, the lamellar structure and Schwann cell cytoplasmic content can be visualized clearly, including mitochondria, microtubules, and microfilaments. Osmium tetroxide is used to stain the myelin, allowing the dark (osmiophilic) myelin of myelinating Schwann cells to be clearly distinguished from the lighter membranes of the non-myelinating Schwann cells. Myelinating Schwann cells can be seen surrounding a single axon with a myelin sheath, though it may also have unmyelinated axons associated with its outer cytoplasm. Bundles of unmyelinated axons are visible, settled into the cytoplasm of a non-myelinating Schwann cell. Endoneurium, the loose connective tissue sheaths that surround individual nerve fibers, can also be visualized. Go to: Pathophysiology The major diseases involving Schwann cells are demyelinating or neoplastic processes. Disorders that cause damage to the myelin sheath in the PNS, affecting the function of Schwann cells and axons, are called peripheral demyelinating diseases. Various insults, such as genetic mutations, infections, trauma, and autoimmune processes can trigger this demyelination and eventual neurodegeneration. Guillain-Barre Syndrome is a rare autoimmune peripheral demyelinating disease characterized by acute ascending flaccid paralysis, which can be life-threatening if the disease affects the muscles of respiration. It is often associated with a preceding infection of the gastrointestinal or respiratory tract, particularly C. jejuni and associated anti-GM1 and anti-GD1a antibodies. The association with infection and accumulation of anti-ganglioside antibodies suggests that ganglioside-like antigens found on C. jejuni lead to the production of antibodies that are cross-reactive with myelinating cells of the PNS. Guillain-Barre can also be caused by other pathogens, trauma, surgery, monoclonal antibody treatment, and rarely by vaccination. The patients usually present with proximal muscle weakness of lower extremity. The most common variant of Guillain-Barre is acute inflammatory demyelinating polyradiculopathy (AIDP), which presents histologically with segmental demyelination with lymphocytic and monocytic infiltration. Charcot-Marie-Tooth disease (CMT) is a rare hereditary peripheral demyelinating disorder, most commonly with autosomal dominant inheritance. Several subtypes affect different proteins and can affect both sensory and motor nerves, but all disrupt Schwann cell structure and function. PMP22 is the most commonly affected protein and causes CMT1A, leading to growth arrest in Schwann cells and abnormal Schwann cell number between nodes of Ranvier. CMT1 is characterized by segmental demyelination and remyelination, causing onion skin appearance on biopsy, and greatly reduced conduction velocity of nerves. Diabetes mellitus is associated with hyperglycemia, hyperlipidemia, hypertension, and impaired insulin signaling, which can damage microvasculature, leading to the common complication of diabetic peripheral neuropathy. Diabetic neuropathy is due to damage to Schwann cells and axons in both sensory and motor nerves. Schwann cells appear to be more susceptible to direct damage caused by hyperglycemia. In contrast, neurons are highly metabolically active, and function better in a hyperglycemic environment but are at greater risk of degeneration caused by hypoxia and loss of trophic support from Schwann cells. Hyperglycemia causes Schwann cell dysfunction, production of reactive oxygen species, initiation of the inflammatory cascade, disruptions in axon conduction, and impaired regeneration after nerve damage. Schwannomas, neurofibromas, and malignant peripheral nerve sheath tumors (MPNSTs) are all neoplastic conditions that arise from Schwann cells. Schwannomas are typically solitary encapsulated lesions made exclusively of neoplastic Schwann cells. Schwannomas do not invade the associated nerve but may produce symptoms caused by mass effect. Neurofibromas and MPNSTs are made of multiple cell types, including Schwann cells, and commonly infiltrate the associated nerve. Neurofibromas commonly arise in patients with neurofibromatosis 1 (NF1), an autosomal dominant disorder caused by a mutation in the NF1 tumor suppressor gene, which may present with dermal and/or plexiform neurofibromas. Dermal neurofibromas are hormone-sensitive tumors that begin to appear as NF1 patients enter puberty, and these tumors have little to no malignant potential. Plexiform neurofibromas are often congenital, not hormone-responsive, and can undergo malignant transformation to MPNSTs. Go to: Clinical Significance Guillain-Barre manifests clinically with symmetric ascending paralysis and paresthesia, which may progress to dyspnea and choking over hours to days. Management is supportive and, with ventilatory support and monitoring for cardiac arrhythmias and other complications, the prognosis is good with patients typically recovering function within 12 months.The earlier the clinician identifies and treats the condition, the better the prognosis.In randomized controlled trials, there are two treatment options currently considered the standard of care in Guillain-Barre syndrome (GBS). These include either intravenous immunoglobulin (IVIG) or plasma exchange. IVIG is thought to act by its immune-modulating action; however, the exact mechanism remains unclear. IVIG dosing is 2 grams/kilogram divided over 5 days. [Level I] Plasma exchange is thought to act by removing pathogenic antibodies, humoral mediators, and complement proteins involved in the pathogenesis of GBS. Similar to IVIG,its exact mechanism of action in the treatment of GBS remains unproven. The patient generally receives plasma exchange as a volume of an exchange over five sessions. Patients with CMT may present with distal muscle weakness, foot drop, depressed or absent deep tendon reflexes, atrophy of muscles of below the knee, and atrophy of the muscles of the thenar eminence. CMT does not reduce the lifespan, and management is supportive. Diabetic neuropathy classically arises in patients with long-standing diabetes as a sensory neuropathy with loss of temperature, vibration, touch, and pain sensation. Patients may also have accompanying neuropathic pain. Nerve damage progresses from small sensory fibers to large sensory fibers, to large motor fibers, causing weakness, loss of function, and paralysis. Nerve damage also is more pronounced in distal extremities, a characteristic “stocking-glove” pattern. Treatment of diabetic neuropathy is limited to symptomatic treatment of neuropathic pain and maintenance of euglycemia to prevent the development of diabetic neuropathy or slow its progression. Schwann cell neoplasms can be identified by immunohistochemistry for Schwann cell markers such as S-100. Management varies from monitoring and supportive care for asymptomatic dermal neurofibromas to surgery, chemotherapy, and radiation for metastatic MPNSTs. Go to: Review Questions Access free multiple choice questions on this topic. Comment on this article. Figure Schwann cell Image courtesy Dr Chaigasame Go to: References 1. Salzer JL, Zalc B. Myelination. Curr Biol. 2016 Oct 24;26(20):R971-R975. [PubMed: 27780071] 2. Gonçalves NP, Vægter CB, Andersen H, Østergaard L, Calcutt NA, Jensen TS. Schwann cell interactions with axons and microvessels in diabetic neuropathy. Nat Rev Neurol. 2017 Mar;13(3):135-147. [PMC free article: PMC7391875] [PubMed: 28134254] 3. Salzer JL. Schwann cell myelination. Cold Spring Harb Perspect Biol. 2015 Jun 08;7(8):a020529. [PMC free article: PMC4526746] [PubMed: 26054742] 4. Feltri ML, Poitelon Y, Previtali SC. How Schwann Cells Sort Axons: New Concepts. Neuroscientist. 2016 Jun;22(3):252-65. [PMC free article: PMC5181106] [PubMed: 25686621] 5. Jessen KR, Mirsky R, Lloyd AC. Schwann Cells: Development and Role in Nerve Repair. Cold Spring Harb Perspect Biol. 2015 May 08;7(7):a020487. [PMC free article: PMC4484967] [PubMed: 25957303] 6. Fix AS, Garman RH. Practical aspects of neuropathology: a technical guide for working with the nervous system. Toxicol Pathol. 2000 Jan-Feb;28(1):122-31. [PubMed: 10668998] 7. Miko TL, Gschmeissner SE. Histological methods for assessing myelin sheaths and axons in human nerve trunks. Biotech Histochem. 1994 Mar;69(2):68-77. [PubMed: 8204769] 8. Möbius W, Nave KA, Werner HB. Electron microscopy of myelin: Structure preservation by high-pressure freezing. Brain Res. 2016 Jun 15;1641(Pt A):92-100. [PubMed: 26920467] 9. Nakazato Y, Ishizeki J, Takahashi K, Yamaguchi H. Immunohistochemical localization of S-100 protein in granular cell myoblastoma. Cancer. 1982 Apr 15;49(8):1624-8. [PubMed: 6175391] 10. Dupin E, Baroffio A, Dulac C, Cameron-Curry P, Le Douarin NM. Schwann-cell differentiation in clonal cultures of the neural crest, as evidenced by the anti-Schwann cell myelin protein monoclonal antibody. Proc Natl Acad Sci U S A. 1990 Feb;87(3):1119-23. [PMC free article: PMC53422] [PubMed: 1967835] 11. Mukai M. Immunohistochemical localization of S-100 protein and peripheral nerve myelin proteins (P2 protein, P0 protein) in granular cell tumors. Am J Pathol. 1983 Aug;112(2):139-46. [PMC free article: PMC1916267] [PubMed: 6192721] 12. Gao H, You Y, Zhang G, Zhao F, Sha Z, Shen Y. The use of fiber-reinforced scaffolds cocultured with Schwann cells and vascular endothelial cells to repair rabbit sciatic nerve defect with vascularization. Biomed Res Int. 2013;2013:362918. [PMC free article: PMC3893804] [PubMed: 24490158] 13. Kamil K, Yazid MD, Idrus RBH, Das S, Kumar J. Peripheral Demyelinating Diseases: From Biology to Translational Medicine. Front Neurol. 2019;10:87. [PMC free article: PMC6433847] [PubMed: 30941082] 14. Steck AJ, Czaplinski A, Renaud S. Inflammatory demyelinating neuropathies and neuropathies associated with monoclonal gammopathies: treatment update. Neurotherapeutics. 2008 Oct;5(4):528-34. [PMC free article: PMC4514701] [PubMed: 19019303] 15. Fontés M. Charcot Marie Tooth Disease. A Single Disorder? Int J Mol Sci. 2018 Nov 29;19(12) [PMC free article: PMC6321061] [PubMed: 30501086] 16. Carroll SL. Molecular mechanisms promoting the pathogenesis of Schwann cell neoplasms. Acta Neuropathol. 2012 Mar;123(3):321-48. [PMC free article: PMC3288530] [PubMed: 22160322] 17. Zhu Y, Ghosh P, Charnay P, Burns DK, Parada LF. Neurofibromas in NF1: Schwann cell origin and role of tumor environment. Science. 2002 May 03;296(5569):920-2. [PMC free article: PMC3024710] [PubMed: 11988578] 18. Arányi Z, Kovács T, Sipos I, Bereczki D. Miller Fisher syndrome: brief overview and update with a focus on electrophysiological findings. Eur J Neurol. 2012 Jan;19(1):15-20, e1-3. [PubMed: 21631649] 19. Staedtke V, Bai RY, Blakeley JO. Cancer of the Peripheral Nerve in Neurofibromatosis Type 1. Neurotherapeutics. 2017 Apr;14(2):298-306. [PMC free article: PMC5398990] [PubMed: 28349408] 20. Gutmann DH, Ferner RE, Listernick RH, Korf BR, Wolters PL, Johnson KJ. Neurofibromatosis type 1. Nat Rev Dis Primers. 2017 Feb 23;3:17004. [PubMed: 28230061] Disclosure:Matthew Fallon declares no relevant financial relationships with ineligible companies. Disclosure:Prasanna Tadi declares no relevant financial relationships with ineligible companies. Introduction Structure Function Tissue Preparation Histochemistry and Cytochemistry Microscopy, Light Microscopy, Electron Pathophysiology Clinical Significance Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK544316 PMID: 31335036 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Introduction Structure Function Tissue Preparation Histochemistry and Cytochemistry Microscopy, Light Microscopy, Electron Pathophysiology Clinical Significance Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed Histology, Axon.[StatPearls. 2025]Histology, Axon.Muzio MR, Fakoya AO, Cascella M. StatPearls. 2025 Jan Akt Regulates Axon Wrapping and Myelin Sheath Thickness in the PNS.[J Neurosci. 2016]Akt Regulates Axon Wrapping and Myelin Sheath Thickness in the PNS.Domènech-Estévez E, Baloui H, Meng X, Zhang Y, Deinhardt K, Dupree JL, Einheber S, Chrast R, Salzer JL. J Neurosci. 2016 Apr 20; 36(16):4506-21. Molecular Probes for PNS Neurotoxicity, Degeneration, and Regeneration.[Methods Mol Med. 1999]Molecular Probes for PNS Neurotoxicity, Degeneration, and Regeneration.Toews AD. Methods Mol Med. 1999; 22:67-87. Review Bioenergetics of Axon Integrity and Its Regulation by Oligodendrocytes and Schwann Cells.[Mol Neurobiol. 2024]Review Bioenergetics of Axon Integrity and Its Regulation by Oligodendrocytes and Schwann Cells.Mishra SK, Tiwari SP. Mol Neurobiol. 2024 Aug; 61(8):5928-5934. Epub 2024 Jan 22. Review Schwann Cell Development and Myelination.[Cold Spring Harb Perspect Biol...]Review Schwann Cell Development and Myelination.Salzer J, Feltri ML, Jacob C. Cold Spring Harb Perspect Biol. 2024 Sep 3; 16(9). Epub 2024 Sep 3. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Histology, Schwann Cells - StatPearlsHistology, Schwann Cells - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... 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17518	https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/81634d22f185742647f70f1d8969feee_MIT18_781S12_lec18.pdf	Lecture 18 Continued Fractions I Continued Fractions - different way to represent real numbers. 415 43 1 1 1 1 = 4 + = 4 + = 4 + = 4 + = 4 + 93 93 93 7 1 1 2 + 2 + 2 + 43 43 43 1 6 + 7 7 = [4, 2, 6, 7] In general: 1 a0 + = [a0, a1, a2, . . . an] 1 a1 + 1 a2 + .. 1 a3 + . an Simple continued fraction if ai ∈Z and ai > 0 for i > 0. Contains the same information as an application of Euclid’s Algorithm 415 43 415 = 4 · 93 + 43 ⇒ = 4 + 93 93 93 7 93 = 2 · 43 + 7 ⇒ = 2 + 43 43 43 1 43 = 6 · 7 + 1 ⇒ = 6 + 7 7 7 = 7 · 1 With this we see that the simple continued fraction of a rational number is always ﬁnite. Never terminates for an irrational number. Eg. π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, . . . ] Eg. e = [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, . . . ] 1 √ Eg. Golden Ratio φ = 1+ 5 2 ≈1.618 . . . satisﬁes φ2 = φ + 1 ⇒ 1 = φ φ−1 . 1 1 φ = 1 + (φ −1) = 1 + = = [1, 1, 1, 1, 1, 1, . . . ] 1 1 + φ φ −1 Finite simple continued fraction ⇐ ⇒rational number. Periodic simple continued fraction ⇐ ⇒quadratic irrational (like φ) Eg. What about √ 2? Look at 1 1 √ 1 1 + 2 = 2 + ( √ 2 −1) = 2 + = 2 + = 2 + 1 1 + √ 2 1 + √ 2 √ 2 −1 (1 + √ 2)(1 √ − 2) 1 + √ 2 = [2, 2, 2, 2, 2, 2, . . . ], so √ 2 = [1, 2, 2, 2, 2, 2, . . . ] What about other algebraic numbers such as √ 3 2? It’s a complete mystery. (Deﬁnition) Convergent: [a0, a1, . . . ak] is called a convergent to [a0, a1, . . . an] for 0 ≤k ≤n. An inﬁnite simple continued fraction [a0, a1, . . . ] equals limk [a →∞ 0, a1, . . . ak]. (We will prove this limit exists.) For 415 93 : 4 2 6 7 0 1 4 9 58 415 1 0 1 2 13 93 Determinants are −1, 1, −1, 1, −1 Recurrence: pk = akpk−1 + pk−2 qk = akqk−1 + qk−2 a0 a1 a2 . . . p 2 = 0 p 1 = 1 p0 p1 p2 . . . − − q 2 = 1 q 1 = 0 q0 q1 q2 . . . − − Theorem 59. pk [a0, a1 . . . ak] = qk 2 Proof. By induction, base case k = 0 p0 a0p a0 = −1 + p−2 1 = · = a0 q0 a0q 1 + q − −2 1 Now assume holds for all k: 1 [a0, a1, . . . ak+1] = a0, a1, . . . ak 1, a − k + ak+1 p′ = k qk ′ ak + 1 p ak+1 k ′ + −1 pk ′ −2 = ak + 1 ak+1 qk ′ −1 + qk ′ −2 (akak+1 + 1)p′ = k−1 + ak+1p′ k−2 (akak+1 + 1)qk ′ −1 + ak+1qk ′ −2 ak+1(akp′ = k−1 + p′ k−2) + p′ k−1 ak+1(akqk ′ −1 + qk ′ −2) + qk ′ −1 ak+1(akpk 1 + p = − k−2) + pk−1 ak+1(akqk−1 + qk ) −2 + qk−1 ak+1pk + pk = −1 ak+1qk + qk−1 pk+1 = qk+1 ■ Theorem 60. pk 1qk −qk 1pk = ( 1) − − − k Proof. By induction, base case is easy to check. Assume to hold for k pkqk+1 −qkpk+1 = pk(ak+1qk + qk 1) −qk(ak+1p + − k pk−1) = pkqk−1 −qkpk−1 = −(qkpk−1 −pkqk−1) = (−1)(−1)k = (−1)k+1 ■ 3 Proof 2. pk pk−1 ak+1 1 pk + pk−1 p = ak+1 k = qk qk−1 1 0 ak+1qk + qk−1 qk p p a pk+1 pk qk+1 qk n n−1 = 0 1 a1 1 a2 1 . . . qn qn−1 1 0 1 0 1 0 n = k Y =0 ak 1 1 0 n p a n pn−1 1 = k qn qn−1 Y k=0 1 0 = ( n −1) +1 ■ Note: Take the transpose pn qn an pn 1 qn 1 = an 1 −1 1 a . . . 0 1 1 0 0 − − 1 1 0 a k Y 0 = =n k 1 1 0 We get that pn = [an, an 1, . . . a0] pn−1 − qn = [an, an 1, . . . a1] qn−1 − Corollary 61. p k k−1 p qk 1 − k ( = −1) qk q q − k k−1 Corollary 62. (pk, qk) = 1 Corollary 63. p k 1 k 2qk −qk−2pk = (−1) −a − k 4 Proof. pk−1qk −qk−1pk = (−1)k akp k k−1qk −akqk 1pk = ( − −1) ak (pk −pk−2)qk −(qk −q k k ) −2 pk = (−1) ak p +1 k q −2 k −q −2p k k k = (−1) ak ■ pk 2 p (−1)k−1 k ak ( < 0 k even Corollary 3 ⇒ − qk−2 − = = qk qk−2qk > 0 k odd p0 p2 p ⇒ 4 < < . . . q0 q2 q4 p ⇒ 1 p3 p5 > > . . . q1 q3 q5 Even terms increasing, bounded above by odd terms, odd terms decreasing, bounded below by even terms, so they both converge. From Corollary 1 the even and odd convergents get arbitrarily close. So both even and odd sequences converge to the same real number x. pk pk p q − k+1 x k − ≤qk qk+1 very good approximations. 1 1 = qkqk−1 ≤q2 k ⇒ Theorem 64. One of every 2 consecutive convergents satisﬁes pk 1 qk −x ≤2q2 k Theorem 65. One of every 3 consecutive convergents satisﬁes pk −x 1 ≤√ 5q2 k (Pr qk oofs in next lecture) 5 MIT OpenCourseWare 18.781 Theory of Numbers Spring 2012 For information about citing these materials or our Terms of Use, visit:
17519	https://www.chegg.com/homework-help/questions-and-answers/aluminum-al-molar-mass-2698-g-mol-pure-block-substance-weighing-357-g-contains-many-atoms--q118649917	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Aluminum (Al) has a molar mass of 26.98 g/mol. A pure block of this substance weighing 3.57 g contains how many atoms? Units of single atoms are assumed. Question 8 3 pts How many moles of Silver (Ag) atoms are in a pure 5.5 gram silver ring? Hint: The class periodic table shows an atomic mass (which is also a molar mass) of silver as 107.87. The periodic Molar mass of aluminum is 26.98 g/mol. Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
17520	https://www.frontiersin.org/journals/microbiology/articles/10.3389/fmicb.2023.1121781/full	Your new experience awaits. Try the new design now and help us make it even better ORIGINAL RESEARCH article Front. Microbiol., 30 March 2023 Sec. Microbe and Virus Interactions with Plants Volume 14 - 2023 \| This article is part of the Research TopicInsights on Plant-Associated Microorganisms: Diversity, Systematics and GenomicsView all 23 articles Morphological characterization, pathogenicity screening, and molecular identification of Fusarium spp. isolates causing post-flowering stalk rot in maize J. Harish1Prashant P. Jambhulkar1Ruchira Bajpai1Meenakshi Arya1Piyoosh K. Babele2Sushil K. Chaturvedi2Anil Kumar2Dilip K. Lakshman3 1Department of Plant Pathology, College of Agriculture, Rani Lakshmi Bai Central Agricultural University, Jhansi, Uttar Pradesh, India 2College of Agriculture, Rani Lakshmi Bai Central Agricultural University, Jhansi, Uttar Pradesh, India 3USDA-ARS, Beltsville, MD, United States Post flowering stalk rot (PFSR) of maize caused by the Fusarium species complex is a serious threat to maize production worldwide. The identification of Fusarium species causing PFSR based on morphology traditionally relies on a small set of phenomic characteristics with only minor morphological variations among distinct Fusarium species. Seventy-one isolates were collected from 40 sites in five agro-climatic zones of India to assess the diversity of Fusarium spp. associated with maize crops showing symptoms of PFSR in the field. To investigate the pathogenicity of Fusarium spp. causing PFSR sixty isolates were toothpick inoculated between the first and second node at 55 days after sowing during the tassel formation stage of the crop in Kharif (Rainy season), and Rabi (Winter season) season field trials. Ten most virulent Fusarium isolates, based on the highest observed disease index, were identified by homology and phylogenetic analyses of partial sequences of the translation elongation factor 1 α (Tef-1α). Based on morphological traits such as mycelial growth patterns and mycelial pigmentation, Fusarium isolates were divided into nine clusters. The isolates were judged to be virulent based on their ability to decrease seedling vigour in in-vivo situations and high disease severity in field experiments. Pathogenicity test during the Kharif season showed 12 isolates with virulent disease symptoms with a mean severity ranging between 50 to 67 percent disease index (PDI) whereas in Rabi season, only five isolates were considered virulent, and the mean severity ranged between 52 to 67 PDI. Based on pathological characterization and molecular identification, 10 strains of Fusarium species namely, Fusarium acutatum (2/10), Fusarium verticillioides (Syn. Gibberella fujikuroi var. moniliformis) (7/10), Fusarium andiyazi (2/10) recorded the highest diseases index. All these species are part of the Fusarium fujikuroi species complex (FFSC). The distribution of virulent isolates is specific to a geographical location with a hot humid climate. Increased knowledge regarding the variability of Fusarium spp. responsible for PFSR of maize occurring across wide geographical locations of India will enable more informed decisions to be made to support the management of the disease, including screening for resistance in maize-inbred lines. Introduction Maize (Zea mays L.) is the third most cultivated cereal grain in the world and contributes greatly to global food security. Maize can be grown in temperate tropical and subtropical climates (Souki et al., 2011). With the development of agriculture, infectious plant diseases have become an increasingly significant factor affecting crop yield and economic efficiency. Maize crop is often infected by several plant pathogens (bacteria, viruses, fungi, nematodes, etc.) that are detrimental to the yield and quality of grains and eventually threaten food security around the globe (Tiru et al., 2021). Pathogenic diseases of maize are one of the major obstacles leading to hefty economic, nutritional, and livelihood impacts. Among several pathogenic diseases, fungal diseases caused by Fusarium spp. are most threatening to the yield and quality of grains. Moreover, fungal infection and consequent contamination with mycotoxins pose serious human and animal health hazards, following ingestion of contaminated food and feeds. Post flowering stalk rot (PFSR) of maize is a serious diseases affecting maize production worldwide. PFSR is mainly caused by the Fusarium fujikuroi species complex (FFSC) which includes more than 60 Fusarium species (both phytopathogenic and clinical importance) (Yilmaz et al., 2021). The pathogens of FFSC responsible for causing PFSR are Fusarium verticillioides (Sacc.) Nirenberg, Fusarium subglutinans (Wollenw. and Reinking) P.E. Nelson, Toussoun and Marasas, Fusarium graminearum Schwabe, Fusarium proliferatum (Matsush.) Nirenberg ex Gerlach and Nirenberg, and Fusarium oxysporum Schltdl (Gherbawy et al., 2002; Harleen et al., 2016). This complex is a collective group of fungi that lives in the soil and spreads through plant roots and crowns. PFSR-infected maize plants exhibit symptoms of drooping, wilting, and drying of leaves, empty cob development, and an increase in the angle between stalks and cobs in the field. Fusarium verticillioides has been reported as a major fungal species of PFSR (Venturini et al., 2013; Yu et al., 2017; Jambhulkar et al., 2022). The isolates of Fusarium spp. from various agroclimatic regions exhibit significant variability in their pathogenicity, genetics, reproductive, and toxicity characteristics (Danielsen et al., 1998; Chandra Nayaka et al., 2008; Venturini et al., 2013; Qiu et al., 2015). The pathogenic potential and cultural variation of these continuously evolving plant pathogens depend on geographical locations, season/weather, and cropping patterns (Kavas et al., 2022). The majority of maize-growing states in India, especially rainfed regions, such as Jammu and Kashmir, Punjab, Haryana, Delhi, Rajasthan, Madhya Pradesh, Uttar Pradesh, Bihar, West Bengal, Andhra Pradesh, Tamil Nadu, and Karnataka are prone to PFSR infection (Kaur and Mohan, 2012). During the Kharif season (Rainy season), Macrophomina phaseolina (Tassi) Goid is the principal pathogen responsible for PFSR in India, which develops when there is a protracted dry spell. In other parts of the country where maize is grown under guaranteed irrigation, F. verticillioides is primarily responsible for PFSR. However, several other Fusarium spp. are also associated with this disease, which affects approximately 27.0 to 77% of maize plants (Mamatha et al., 2020). All India Co-ordinated Research Project on Maize (AICRP-Maize) 2014, reported that this disease causes 22 to 64% yield losses. In different agroecological zones of India, multiple Fusarium spp. exist with varying pathogenic potential, making PFSR control strategies very challenging and often unsuccessful. The virulence pattern of Fusarium spp. isolates in not unclear. Moreover, being a soilborne pathogen, management of the disease through a fungicide application is difficult, costly, and destructive to ecology (Rahjoo et al., 2008; Harleen et al., 2016). Therefore, morphological characterization and pathogenicity screening of Fusarium spp. causing PFSR is critical for preliminary identification and classification and eventually for developing efficient disease management strategies. Several prior studies showed variability in cultural and microscopic characteristics of several Fusarium spp. isolated from Rajasthan (Khokhar et al., 2014), Punjab (Harleen et al., 2016), Karnataka (Ramesha and Naik, 2017), and Telangana (Mamatha et al., 2020) states of India. These reports indicate region-specific variability in several Fusarium spp. of these states. We argue that a more comprehensive and comparative investigation of these PFSR-causing pathogens across wide geographical locations of India is necessary to delineate the virulence behaviour of these isolates and to design their management strategies. Moreover, being a soilborne pathogen, management of the disease through conventional measures is difficult such as fungicide application which is cost-prohibitive and destructive to ecology. A substantial amount of work has been done to investigate the use of biological control to address the issue of PFSR (Chandra Nayaka et al., 2010; Wu et al., 2015; Lu et al., 2020; Jambhulkar et al., 2022), but breeding resistant germplasm is the ultimate solution to reduce crop loss due to PFSR. Therefore, stable resistant inbred lines against virulent isolates of Fusarium spp. will be useful as resistant sources against PFSR. This study aims to comprehensively characterize morphologies of Fusarium spp., isolated from five agro-climatically distinct states viz. Rajasthan, Gujarat, Madhya Pradesh, Telangana, Karnataka, and Andhra Pradesh of India. The respective Fusarium isolates were cultured from rotten maize stems exhibiting signs of PFSR, such as drooping cobs and desiccated plants. Moreover, we examined the pathogenicity of the collected isolates to identify the most virulent ones. Increased knowledge regarding the variability of Fusarium spp. responsible for PFSR of maize occurring across wide geographical locations of India will enable more informed decisions to be made to support the management of the disease. Materials and methods Sampling and fungal strain isolation During 2020 and 2021, maize plants infected by PFSR were randomly gathered from 40 sites in southern Rajasthan, eastern Gujarat, western Madhya Pradesh, Karnataka, and Telangana. A few locations of sample collection performed during Rabi 2020 are Banswara, Survaniya, Bodla, Karji, Hamirpura, Rakho, Badodiya Kali Pahari, Kalinjara, Sagdod, Devliya, Kushalgarh, Sajjangarh, Monadungar villages Ghatol, Senavasa, Badgaon, Chanduji ka Gada in Southern Rajasthan. Eastern Gujarat: Dahod, Godhra, Pavagarh, Vadodara. In Karnataka, samples were taken from Mandya, Haveri, Davanagere, Mysore, Raichur, Belagavi, and Bagalkot, among other locations. One isolate was recovered from Warangal in Telangana. Geographical locations depicting the latitude and longitude of Fusarium spp. sample collection is given in Supplementary Table 1. These locations are part of five agro-climatic zones viz. Eastern plateau and hills, Central Plateau and hills, Western plateau and hills, Southern plateau and hills, and Gujarat plain and hills. On the India map, each collection site’s latitude and longitudinal position (GPS data) were recorded and mapped (Figure 1). FIGURE 1 Figure 1. Geographical locations indicating sampling sites of Fusarium spp. Isolates. These sample collection locations are part of five agro-climatic zones in India viz. Eastern plateau and hills, Central Plateau and hills, Western plateau and hills, Southern plateau and hills, and Gujarat plain and hills. To isolate the pathogens, all adhering soil particles were removed from the vascular pith tissues of infected maize plants by washing them with tap water. The pith of the infected stalk was surface sterilized by soaking in 0.1% sodium hypochlorite for 30 s before being rinsed twice in sterile distilled water, dried on sterile blotting paper, and placed on potato dextrose agar plates supplemented with streptomycin sulphate (100 μg ml–1) to prevent bacterial contamination and incubated in the dark. The isolates were moved from potato dextrose agar (PDA) to Spezieller Nahrstof farmer agar (SNA) medium (1 g KH2PO4, 1 g KNO3, 0.5 g MgSO4.7H2O, 0.5 g KCl, 0.2 g glucose, 0.2 g sucrose, and 20 g agar in 1 L SDW), to enhance sporulation, according to Leslie and Summerell (2006). Seventy-one isolates of Fusarium pathogen cultures from infected stalks were purified by isolating a single spore. The isolates were sub-cultured on potato dextrose agar slants and allowed to grow at 27 ± 1°C in the dark for 15 days. These slants were stored in a refrigerator at 4°C and utilized throughout the study. Cultural and morphological characterization of isolates To investigate the morphology of various Fusarium isolates, isolates were individually cultured on PDA. A 5 mm culture disc was cut from the edge of the actively developing culture plates and positioned at the centre of fresh PDA plates and incubated in the dark at 27 ± 1°C. After 7–8 days, observations of culture phenotypes were recorded considering colony diameter, colour, margins, and general appearance from all the Petri dishes. Each isolate was maintained in triplicate. Thereafter, micro-morphological characterization was conducted by observing fungal cells under a microscope. Eventually, to study conidial morphology, culture explants were transferred and grown onto SNA Petri dishes. Slides were prepared from four-day-old fungal cultures on SNA medium to promote the sporulation and production of both microconidia and macroconidia. The length, width, and number of septations per conidia were recorded and all the microphotographs were captured using a CILIKA® digital microscope (Medprime, Thane, Mumbai, India) at 40X magnification. In-vitro pathogenicity test The germination paper/paper towel method (Flint-Garcia et al., 2005) was used to determine the pathogenicity of specific Fusarium isolates in vitro. A total of 71 Fusarium isolates were examined on PC-4 maize seeds. Conidial suspensions of 2x107cfu/ml were made by scraping mycelium into sterile distilled water in a test tube. Surface-sterilized maize seeds were soaked for 30 min in test tubes containing a suspension of each inoculum. After pathogen treatment, 10 maize seeds per inoculum are placed in moist germination paper, labelled, rolled, and placed in a seed germinator with controlled conditions (28°C temperature and 80% relative humidity). The initial count for seed germination, shoot, and root length, was recorded on the eighth day, followed by the tenth day, and the final count on the twelfth day, respectively. Pathogenicity of the Fusarium isolates and their influence on seed germination, viability, and vitality were recorded. The Dezfuli et al. (2008) formula was utilized to calculate seedling vigour. VigourindexVI=Germination(%)×Totalseedlinglength(cm) Pathogenicity studies during Kharif and Rabi seasons in field condition The field experiments for pathogenicity evaluation were conducted in the research farm of Rani Lakshmi Bai Central Agricultural University, Jhansi (N 25°30′55.75′′ E 78°32′47.31′′), India. Seeds of a cultivated variety with intermediate susceptibility “Pusa Composite 4” were seeded in the field adopting a Randomized Block Design with three replications of each isolate during Kharif (rainy season) 2020 and Rabi (winter season) 2020-21. Standard agronomic practices adhered to for the soil preparation and fertilizer application. The spacing between plants was 30 cm, while the spacing between rows was maintained at 60 cm. Toothpick inoculation method Round wooden 6 cm-long toothpicks were sterilized by boiling in distilled water for 1 h. Any impurities such as gum, resin, or any other particles that may limit fungus growth were discarded by decanting the water, and the process was repeated three times. Following sterilization, the toothpicks were dried on sterile blotting paper. Ten toothpicks were put into each test tube, and PDB (Potato dextrose broth; Hi-Media®) was added to each tube to thoroughly moisten the toothpicks. The pointed tips of the toothpicks were maintained in an upward position and the volume of broth in a test tube were such that the 1–1.5 cm tip of the toothpicks remained above the broth level. The test tubes were autoclaved, cooled, and then infected with Fusarium isolate plugs. The inoculated test tubes were incubated at 27 ± 1 °C with 12 h of alternating light and darkness for 10 days until the maximal mycelial growth took place on toothpick tips. Sixty isolates [FUR14, FUG16, FUR13, F38, F27, F3, F2, F43, FUG7, F19, FUG4, F9, F55, F58, F44, F16, F47, FUG1, F45, FUG2, F10, F7, F6, F31, F36, F28, FUG5, F34, F8, F22 (Dungarpur), F46, FUG6, F12, F33, F14, F26, F48, F20, F4, F11, F49, FUG8, F42, D2, FUG3, F52, F57, FUR10, F21, F18, FUR15, F13, F1, F25, FUR12, F59, Chokhla, F32, FUR11, and F35] were used in pathogenicity test during Kharif season 2020 and the nine isolates FUR14, FUG16, FUR13, F38, F27, F3, F2, F43, FUG7 which showed less virulence during Kharif season 2020, were excluded for the subsequent trial. For the pathogenicity trial in the Rabi season of 2020-21, the 51 Fusarium isolates from the 2020 Kharif season and nine new isolates collected from the southern region of India, viz-a-viz, Mandya, Mandya 2, Haveri, Mysore, W3-2, G1-3, B1-1, Davanagere, and Raichur were included in the study. At the tasselling stage, 45 to 50-day-old maize plants were artificially inoculated. Five plants per replication were inoculated with each Fusarium isolate. The lower internodes, particularly the second internode above the soil’s surface in each plant, were inoculated with mycelium. The fungus-colonized toothpicks were inserted diagonally after puncturing the stem and creating a 2-centimetre hole in the first internodes with a jabber. The onset of symptoms initiated 20 to 25 days following vaccination. At the time of harvest, disease intensity and severity (PDI) were computed using Payak and Sharma (1983) 1 to 9 scale. Lesion length was measured by split opening the inoculated stalk at harvest to determine the percentage disease index (Chester, 1950; Chaube and Singh, 1991; Vieira et al., 2012). Percentdiseaseindex= sumofnumericalratingTotalnumberofsampletaken×Maximumgrade×100 The disease rating scale to measure the disease severity of PFSR is given by Payak and Sharma (1983). (1) No discoloration or discoloration only at the point of inoculation; (2) less than 25% of the inoculated internode discoloured; (3) 25 to < 50 % of the inoculated internode discoloured; (4) 50 to < 100 % of the inoculated internode discoloured; (5) 25% of adjacent internode discoloured; (6) half discoloration of the adjacent internode; (7) Discolouration of three internodes; (8) Discolouration of four internodes, and (9) Discolouration of five internodes or plants prematurely killed. Isolation of genomic DNA The genomic DNAs of isolates were extracted using the CTAB method (Li et al., 2013). The 10 most virulent Fusarium isolates were incubated on PDA plates at 27 ± 2°C under the dark for 6–7 days. An approximately 1 cm2 of fungal mycelium was transferred from culture plates to a sterile 2 ml Eppendorf tube. Mycelium was macerated using a tissue homogenizer and 500 μl CTAB buffer, followed by incubation of the Eppendorf tube at 60°C for 1 h in a hot water bath (Cole-Palmer India Pvt. Ltd. Mumbai) with a gentle shaking at 10 min interval. The composition of the buffer was (2.5% CTAB, 4 M NaCl, 20 mM EDTA, 100 mM Tris-HCl, 0.2% β- Mercaptoethanol; pH 8.0). All components for the buffer preparation were procured from (Hi-media, India). The concentration and purity of DNA were determined using Nanodrop (NanoDrop™ ThermoFisher™ Scientific, Mumbai, India) and recording its absorbance at (260/280 nm) and the quality of DNA through agarose gel electrophoresis (0.8%) (w/v) (iGene Labserve, Delhi, India). The gel was observed under the Gel documentation system (Syngene®InGenius3, Frederick, USA). The final volume of DNA was made to 50 ng. Phylogenetic analysis of the TEF-1α gene sequence Sequences of the translation elongation factor 1α (TEF-1α) gene from each isolate were amplified using the Tef-1α EF-1 [5’-ATGGGTAAGGA (A/G) GACAEAGAC-3’] and EF-2 [5’- GGA (G/A) GTACCAGT (G/C) ATCATGTT-3’] primer pairs (O’Donnell et al., 1998). PCR amplification (50 μl reaction volume) was performed with an initial denaturation at 94 °C for 5 min followed by 35 cycles each at 94 °C for 1 min, annealing at 50 °C for 1 min, extension at 72 °C for 2 mins followed by a final extension at 72°C for 10 mins, in a Thermocycler (Veriti™, Applied Biosystem™, New Delhi, India). The Tef-1α gene sequences of the ten most virulent Fusarium spp. isolates were sequenced through the Sanger sequencing method at Medauxin™, Bangalore. The obtained sequences were subjected to nBLAST to molecular identify at species levels through homology with available nucleotide sequences from the NCBI fungal DNA database1. The homology-identified sequences were deposited in the GenBank database of the National Centre for Biotechnology Information (NCBI), and Accession numbers were obtained. The reference Tef-1α nucleotide sequences of the Fusarium spp. showing closest homologies in the nBLAST search were retrieved from the FUSARIOID-ID database (Crous et al., 2021) for molecular phylogenetic relationships (Fusarium verticillioides Reference species: MW401977 CBS 117.28 strain, MW402080 CBS 141.59 strain; F. andiyazi Reference species: MN533989 CBS 119856 strain, MN193854 NRRL 31727 strain; F. acutatum Reference species: MW402124 CBS 401.97 strain, MW402125 CBS 402.97 strain). Alternaria burnsii isolate Alt-MP6 (Sequence ID: ON993391) was used as an outgroup. The phylogenetic tree was constructed by using MEGA X software using 1,000 bootstrap values and beautified by using ITOL software (Letunic and Bork, 2007). Statistical analysis Arc GIS 10.1 platform (ESRI Inc., United States) was used for preparing the map for the study area and sample location sites. For hierarchical cluster analysis, all measurements of morphological and microscopic characters (including pigmentation, pattern, and type of mycelium, colony colour, macro and microconidia size, shape, and septation), were averaged and given values. A cluster analysis was conducted, and Unweighted Pair Group Method with Arithmetic mean (UPGMA) based dendrogram constructed (Jaccard’s coefficient) by using the software PAST 4.03. For each trait, descriptive statistics were calculated. Analysis of variance (ANOVA) was done by verifying the Shapiro-Wilk tests by SAS 9.1 (SAS, Inc., North Carolina, USA) software at 5% probability. Pair-wise mean comparison of disease severity between isolates during Kharif and Rabi seasons was calculated according to Tukey’s HSD test at p < 0.05. Data on pathogenic variability of virulent isolates under in-vitro, Kharif, and Rabi seasons were analysed by analysis of variance (ANOVA) and compared differences between percent reduction in seedling vigour and disease severity by virulent isolates using R (Gentleman, 2008). Arc sine transformation was done for percent data to make residuals normal and then back-transformed for graphical presentation. The significance of mean difference within percent reduction in vigour and disease severity by virulent strains was tested by Student’s t-test in combination with Bonferroni correction at P = 0.05 level of probability. Box plots were created to depict the distributions of seed germination parameters and shoot, root, and vigour index throughout growth days and trials. The analysis was performed at 0.05 or 0.01 significance level as indicated. The correlation study was performed using Jamovi version 1.2.27 at 5% level of significance. Results Identification of isolated cultures from maize stalk rot samples Morphological characterization Fusarium isolates display variability in several phenotypic characteristics such as colony colours, mycelium, pigmentation, sporulation, branching, conidial size, and shape. All the Fusarium isolates displayed a pronounced difference in their colony colours ranging in hue from violet, light violet, light pink, and light pink to light violet, dark violet, and filthy white. The colony colour of 47 of the Fusarium isolates was white to dirty white. The distribution of the pigments diffused in the culture agar of the 71 isolates were as follows: 42 purple, 7 dark purple, 6 pink, 8 yellow, and 1 light purple-2 isolates did not produce pigments. A significant variation in their colours and conidia among different isolates is depicted in Figure 2. The mycelial development of the Fusarium was considerably more variable. Some isolates exhibited white mycelial growth that was fluffy, white mycelial growth that was sparse, and white and purple mycelial growth that was mixed (Figure 2). All 71 isolates displayed a significant degree of variation in their mycelial size, colour, pigmentation, shape and size of macro and microconidia, and their septation. Among them, 51 isolates displayed fluffy mycelial development, 13 isolates were less fluffy, and 7 isolates were appressed. Among all 71 isolates, 30 had pointed macroconidia, 6 blunt-end conidia, and 35 sickle-shaped conidia. Fusarium isolates were divided into distinct clusters based on pigmentation, colony colour, mycelium pattern, mycelium type, conidial size, and shape. Based on morphological traits, 71 isolates were grouped into nine clusters (Figure 3). Cluster I consisted of six isolates with purple to dark purple pigmentation, purple to light purple, and a fluffy, condensed mycelium. Pointed to sickle-shaped macroconidia size was in the range of 13–15.3 μm with 2 septa. Microconidial size was in the range of 5.3–10.0 μm. Cluster II comprised of five isolates with purple, white, orange, and light-yellow pigmentation, white to dirty white colony colour, and fluffy to appressed mycelium. The size of sickle-shaped macroconidia ranged between 22–51 μm having 2–7 septa. Microconidial size was highly variable between 6–30 μm. Nine isolates from cluster III exhibited purple to dark purple pigmentation, purple to light purple colony colour, and fluffy to less fluffy mycelium. The size of macroconidia was between 18–40 μm having 2–7 septa. The virulent isolates Raichur and Davanagere belonged to this cluster having pointed and blunt conidia respectively. Seven isolates from cluster IV exhibited purple pigmentation, except F32 and G1-3 exhibited yellow and pink pigmentation, respectively, white to dirty white and light purple colony colour, fluffy to less fluffy, and appressed mycelium. Pointed shape macroconidial size was in the range of 26 to 32 μm with 2–3 septa. The FUG-7 isolates had 6 septa. All isolates in this cluster were pointed in shape. The size of microconidia was in the range of 11–22 μm. Eleven isolates form cluster V having purple pigment, white to dirty-white colony with fluffy growth. The size of sickle-shaped macroconidia was from 22 to as large as 53 μm with up to 2–7 septa. Microconidial size was in the range of 10–21 μm. The virulent isolate F13 belonged to this cluster. Nine isolates from the VI cluster contained variable pigments of pink, yellow, white, and orange. Mycelial colour was white to dirty white, and fluffy mycelium. Macroconidial size was in the range of 21 to 44 μm with 2–6 septa. Microconidial size was in the range of 10–26 μm. The FUG8 had large size conidia with 4 septa. Cluster VII has seven isolates having variable pigment range. Isolates belonging this cluster had white to dirty white and light purple colony colour, and fluffy and dense mycelium. Macroconidial size was in the range of 21–36 μm with 2–7 septa. The virulent isolate FUG15 had the largest conidia in this cluster with seven septa. In cluster VIII, eleven isolates exhibited purple pigmentation with white and dirty white colony coloration and fluffy mycelium. This cluster had larger sickle-shaped conidia with 20 to 44 μm with as high as 8 septa. Microconidial size was in the range of 17 to 20 μm. Six isolates in cluster IX exhibited purple to dark purple pigmentation, having purple to light purple colony colour, fluffy mycelium. The size of macroconidia was in the range of 22–45 μm and 2–6 septa. This cluster consisted of pointed and blunt-end conidia. This cluster had a maximum of 3 virulent isolates including F18, FUR 11, and F59 (Figure 3). FIGURE 2 Figure 2. Cultural and morphological characteristics of Fusarium spp. causing PFSR in maize. For each isolate in the top row is the upper surface of the culture plate, the middle row is the lower surface, and the bottom row is a microscopic photograph of spores. A- F6; B- F27; C- F-47; D- F52; E, F- F59; G-F41; H- Raichur; I- Mysore; J- F19; K-F2; L- F13; M-F26; N-F33; O- F13; P-F14; Q- F8; R-F9; S- F58; T-F49; U-F28. FIGURE 3 Figure 3. Dendrogram derived from morphological characters of 71 isolates of Fusarium spp. collected from 5 agroclimatic zones based on pigmentation, colony colour, mycelial pattern, shape and size of macro, and microconidia and number of septa in macroconidia grown on PDA medium. The production of microconidia occurred in chains and branches upon being grown on the SNA media. On a CILIKA™ digital microscope, the length and width of randomly selected microconidia and macroconidia were measured. Macroconidia produced three distinct types of spores, including those with sickle- or tapering-shaped, pointy, and blunted ends. There was a greater formation of sickle-shaped macroconidia in 32 Fusarium isolates, while 30 isolates produced pointed macroconidia, and five isolates produced blunt macroconidia. Sickle-shaped macroconidia ranged in size from 53.0 μm ± 8.0 × 1.5 μm ± 0.5 m (FUR 13) with a Quotient value (Qv) of 35 μm ± 7.4 to 13 μm ± 2.5 × 2 ± 1.0 m (F12) with a Qv of 7 ± 2.6. Comparatively, blunt-shaped macroconidia are smaller, ranging between 30.0 μm ± 3.0 × 2.5 μm ± 1.0 m (FUR 11) Qv 12.0 ± 4.4 to 18.0 ± 2.6 × 2.0 ± 1.3 μm (F 58). The size of pointed macroconidia ranged from 44 μm ± 3.5 × 1.5 μm ± 0.5 m (F52) Qv 29.0 ± 8.4 to 13.0 ± 2.0 × 2.0 ± 0.5 m (Qv 6.5 ± 1.3) (F48). Microconidial dimensions ranged from 23.0 μm ± 3.1 × 2.5 μm ± 0.5 m (F8) Qv 9.3 ± 0.7 to 5.0 ± 1.5 × 1.5 ± 0.5 m (F48) Qv 3.6 ± 0.3 (Supplementary Table 2). The number of septa detected in macroconidia varies between 2 and 8 septations. With 0-1 septa, microconidia were circular to oval in shape. The F52, F13, FUR 15, F16, F22, F14, F22 (D2), and B1-1are examples of isolates with 7 septa in their macroconidia, while only isolate F20 was reported to have 8 septa. The range of macroconidia is 10–53 μm × 1–4 μm. Microconidia size ranges from 4–31 μm × 1–3 μm. The F. acutatum isolate FUR 13 had macroconidia of size 53.0 μm ± 8.0 × 1.5 μm ± 0.5, Isolate F13 developed conidia with lengths of sizes 51.0 μm ± 15.4 × 2.0 μm ± 0.5. The F. oxysporum isolate F27 had large macroconidia 50.0 μm ± 17.5 × 1.7 μm ± 0.8 with Qv of 30.2 ± 14.9 (Figure 4). FIGURE 4 Figure 4. Conidial morphology of Fusarium spp. (A) Sickel-shaped Macroconidia of Fusarium oxysporum (50.3 μm ± 17.5 × 1.7 μm ± 0.8) 1000X, (B). Sickle-shaped F. verticillioides (40.0 μm ± 1.0 × 2.3 μm ± 1.0) 1000X, (C) False head and monophialide conidiogenesis of F. acutatum, 400 X, and (D) Blunt ended macroconidia of F. andiyazi (18.0 μm ± 2.6 × 2.0 μm ± 1.3) 1000 X observed in the present study. In-vitro pathogenicity evaluation of Fusarium isolates Effect of treatment with Fusarium spp. on maize seed germination Fusarium isolates significantly altered the seed germination of maize, the germination rate of different Fusarium spp. treated maize seeds varied between 0 and 70% on different days following fungal inoculation. Based on the extent of germination, these isolates were categorized as less virulent (F39, F7, FUG1, F49, F58, F32, F19, Davanagere), moderately virulent [FUR13, F28, F52, FUG16, F38, F45, Mandya, F3, F44, F13, Chokhla, F1, F57, FUG6, F47, F43, F22 (D2), F8, Mandya 2 and Haveri]. Virulent isolates F46, F12, F21, F9, F11, FUR50, F31, F35, FUG4, FUG3, F9, FUG8, F14, F10, F20, F42, FUG7, F48, F6, FUG5, F4, F55, F59, F26, Mysore, Raichur, and F36 inhibited germination of maize seeds on a paper towel by one hundred percent. Non-virulent isolates consistently promoted germination up to 12 days after inoculation (p ≤ 0.05), but virulent isolates inhibited germination. The maximum rise in germination was found 10 days after inoculation, as indicated by the Box Plot trend for change in germination percentage (p ≤ 0.05; Figure 5A). The non-germinating seeds were determined to be decaying or weakly germinated. FIGURE 5 Figure 5. Box plot showing the distribution of germination percent (A), root length (B), shoot length (C), seedling length (D), and vigour (E) on different days after challenging maize seeds with Fusarium isolates. Effect of treatment with Fusarium spp. on maize seedling root length Based on the pathogen’s virulence, there was a significant (p ≤ 0.05) difference in root length between Fusarium spp.-treated seeds and untreated control seeds. The treated seeds had a mean root length between 0–24.8 cm. Seeds inoculated with Fusarium spp. isolates showed a steady increase in root length for the first 10 days, and, thereafter, a decline (Figure 5B). The untreated seeds produced roots measuring 25 cm in length. The roots inoculated with isolates of reduced virulence [F52, F38, Mandya, F44, Chokhla, F1, FUG, FUR12, F16, F49, F43, F19, F22 (D2) Mandya 2, Davanagere] grew relatively vigorously. These isolates have the potential to reduce root length by up to 20% compared to the control. FUG1, F16, F44, FUG10, F43, F19, FUR12, F1, F38, Davanagere, FUG16, FUR13, F27, FUG9, F13, FUG2, F47, G1-3, F39, F34, FUR14, F8, F57, F18, F34, FUG6, F7, F3, F2, F45, F58, B1-1, F22, F28, and F25 were the isolates which reduced root length by 22 to 60 %. The remaining isolates belonged to a very virulent group that caused root lengths of fewer than 10 cm. W3-2, F36, Raichur, Mysore, F26, F59, F55, F4, FUG5, F6, F48, FUG 7, F42, F20, F10, F14, FUG 8, F9, FUG3, FUG4, F35, F31, FUR 15, F11, F21, F12, and F46 were the isolates that produced a substantial reduction in root length (Supplementary Table 3). Other than W3-2, all these isolates inhibited seed germination. Thus, these isolates reduced root length by 100%t and are considered highly virulent isolates. Data points in the Box plot depict the reduction in root length on the 12th day after treatment with all Fusarium spp. (Figure 5B). Effect of treatment with Fusarium spp. on maize seedling shoot length The shoot length of the seedlings was measured at 8, 10, and 12 days after treatment with Fusarium spp. isolates. The variation in the shoot length between three different periods of the evaluation was not significant. There was a correlation between the seedlings’ length and the isolates’ virulence. The greater the length of the seedling, the less virulent the inoculated Fusarium isolate. The shoot length of the seedlings varied significantly, ranging from 0 to 43 cm. The shoot length of the seedling treated with the F38 Fusarium isolate F38 was 42.1 cm which was at par with the untreated control having a shoot length of 45 cm. The shoot length inoculated with the studied isolates F52, FUR14, F19, F38, Mandya, F3, F1, F18, F22 (D2), F22 (2C), B1-1, F47, and F43 ranged between 30 to 43 cm which caused a 6.4 to 32.1% reduction in shoot length as compared to untreated control (Figure 5C). Thus, these isolates were considered less virulent. The average shoot length of seedlings treated with F8, F34, F22 (2C), FUG1, FUG 16, FUG 9, FUG10, FUG2, F45, FUR11, F57, G1-3, Mandya 2, Chokhla, F27, F34, F44, F16, F39, F7, F45, F58, F49, Haveri, F32, F2, W3-2, FUR 13, F32, F33, and Davanagere, ranged between 20–29.9 cm. Thereby recording a 33.8 to 55.4% reduction in shoot length and referred to as moderately virulent isolates (Figure 5C). The drastically shorter shoot length ranged between 0−19.6 cm, caused by the Fusarium spp. isolates FUG2, F28, F25, F12, F21, F9, F11, FUR15, F31, F35, FUG4, FUG3, FUG8, F14, F10, F20, F42, FUG7, F48, F6, FUG5, F4, F55, F59, F26, Mysore, Raichur, and F36 thereby causing 56.3–100 % reduction in shoot length (Supplementary Table 4). Thus, these isolates were deemed the most virulent isolates. Except for three isolates (FUG2, F28, and F25), the remaining 25 isolates did not permit the germination of seeds. Effect of treatment with Fusarium spp. on maize seedling vigour According to the measurements of seed germination, shoot length, and root length, the seedling vigour of each isolate of Fusarium spp. there were considerable differences in germination and shoot-root length, resulting in significant differences in seedling vigour. Using the formula given in the materials and methods, the vigour of seedlings was estimated. The untreated control seeds produced the most vigorous seedlings with a vigour index of 2347 ± 10.86. Isolates of Fusarium spp. with a high vigour index were rated less virulent. F44, F7, F3, FUG16, F39, Davanagere, F13, F49, FUG 1, F58, F32, and F19 which had a vigour index of more than 1,500, were responsible for up to 36% decrease in vigour compared to the control (Figures 5D, E). The 29 moderately virulent isolates had a vigour index between 822.6 ± 9.94 and 1463.9 ± 9.53. These isolates are F34, F25, F27, B1-1, G1-3, F18, F22 (D2), FUR14, FUG9, F47, Mandya 2, F52, F16, F22 (2C), FUG10, F43, F7, F8, F57, F45, F28, FUR13, FUR12, Chokhla, FUG6, Mandya, F1, FUR11 and F38 (Supplementary Table 5). These isolates reduced seedling vigour by 37.6 to 64.9 % in an in vitro paper towel test. The 30 isolates were recorded to cause 65–100 % reduction in seedling vigour index with a vigour index from 822.2 ± 9.12 to 0. These 30 isolates F46, F12, F21, F11, FUR15, F31, F35, FUG4, FUG3, F9, FUG8, F14, F10, F20, F42, FUG7, F48, F6, FUG5, F4, F59, F26, Mysuru, Raichur, F36, F55, W3-2, FUG2, F33, and F2 with least vigour index were considered as most virulent (Supplementary Table 5). Data points in the box plot depict that higher vigour was observed 10 days after seed treatment with Fusarium isolates. The median of boxes on 10th and 12th day after inoculation was near 1,000. Our observations showed that 38 isolates were below the median line with the vigour index of less than 1,000 while 33 isolates were above the median line with a vigour index more than 1000 (Figure 5E). Pathogenicity of Fusarium isolates during Kharif 2020 and Rabi 2020-21 seasons under field conditions The pathogenicity of sixty Fusarium isolates was evaluated with the development of visible symptoms such as drooping, wilting, and drying of leaves, empty cob development, and an increase in the angle between stalks and cobs in the field. All the isolates were able to cause infection. The intensity of disease symptoms varied among the isolates. At the harvest, all the inoculated plants were split open to measure lesion length (Figure 6). During the Kharif 2020 season, out of 60 isolates, nine were less virulent (11–25% PDI, average lesion length = 2.83–6 cm), 39 were moderately virulent (25–50% PDI, average lesion length = 5.5–12.03), and 12 were virulent (50–67% PDI, average lesion length = 6–14 cm; Figure 7A). The virulent isolates F21, FUR15, F18, F35, F1, F32, Chokhla, FUR12, F59, F25, FUR11, F13; exhibited virulent reactions with mean severity ranging from 4.7 to 6 cm lesion length (50 to 67% PDI). Isolates FUR14, FUG16, FUR13, F38, F27, F3, F43, FUG7, and F2 were less virulent or avirulent, with mean severity ranging from 1–2 (11–25% PDI). The F25, F35, and F59 were found to be virulent both in vitro and in the field during the Kharif season (Supplementary Table 5). There were 20% virulent isolates, 65% moderately virulent isolates, and 15% less virulent isolates. During Rabi season 2020-21, nearly 60 isolates, including nine distinct Fusarium isolates, were obtained from South India during this season and were evaluated for pathogenicity (Davanagere, Mysore, Mandya, Mandya 2, Raichur, Haveri, B1-1, W3-2, and G1-3). The Kharif season isolates with a lower PDI and severity of symptoms were eliminated during the Rabi season (FUR14, FUG16, FUR13, F38, F27, F3, F2, F43, and FUG7). Isolates with mean disease severity in the range of 19–22% and causing lesion length of 2–2.6 cm (FUG2, FUG1, F44, F52, FUR10, F39, FUR12, Mandya 2) were deemed to be less virulent. As many as 47 isolates were moderately virulent with the mean disease severity of these isolates ranging between 25.9 and 48.1 % and lesion lengths ranging between 2.3 and 4.3 cm (Supplementary Table 5). The Raichur, Chokhla, Davanagere, F1 and F13 isolates have an average symptom severity ranging between 51.9 and 66.7% and were considered virulent isolates (Figure 7B; Supplementary Table 5). Three isolates, F1, Chokhla, and F13, were virulent throughout the Kharif and Rabi seasons. In the Rabi season, virulent isolates like F35, F21, F18, FUR15, F25, F32, and FUR11 from the previous season were determined to be moderately virulent. During the Rabi season, five isolates F13, F1, Davanagere, Chokhla, and Raichur were found to be virulent. Of the total analysed isolates 8% were virulent, 77% were moderately virulent, and 15% were less virulent during Rabi season (Figure 7B). The inoculated stem showed vascular discoloration, from where the pathogen was reisolated to confirm Koch’s postulates and to confirm the similarity of cultural and microscopic morphology with those of the inoculated Fusarium isolates. FIGURE 6 Figure 6. Comparative lesion size of virulent (F1,F18, Raichur), Moderately virulent (D2, F10, FUG49), and less virulent (F3, FUG 16, F39). Virulent isolates caused lesions covering 2–3 nodes, moderately virulent isolates covered entire pith between nodes, and less virulent isolates a lesion restricted to the inoculated area. FIGURE 7 Figure 7. Comparative virulence among Fusarium isolates observed during the in-vitro, Kharif 2020 and Rabi 2020-21 season. Values for histograms sharing the same letter label are not significantly different (P > 0.05). Comparative severity in vitro, in Kharif and Rabi seasons Seventy-one Fusarium isolates were tested in vitro using the paper towel method. In both the Kharif and Rabi seasons, the pathogenicity of sixty Fusarium isolates was evaluated using the toothpick inoculation method. The in vitro investigation revealed that, of the 71 isolates examined for pathogenicity, 30 isolates inhibited germination and decreased seedling growth. Based on less vigour of the seedlings, these isolates were considered virulent isolates. Under in vitro conditions, as mentioned above F46, F12, F21, F11, FUR15, F31, F35, FUG4, FUG3, F9, FUG8, F14, F10, F20, F42, FUG7, F48, F6, FUG5, F4, F59, F26, Mysore, Raichur, and F36, exhibited a 100% drop in seedling vigour, whereas F55, W3-2, FUG2, F33, and F2 exhibited a reduction in vigour in the range of 65 to 99.5% (Supplementary Table 5). While the in vitro pathogenicity of the isolates was determined by their ability to diminish seedling vigour, the virulence of the field-level evaluation was determined by the severity of the disease. Pathogenicity evaluation of the 60 isolates during Kharif 2020 reported 12 virulent isolates with disease severity of more than 50% including F21 (52 ± 0.58), F18 (52 ± 0.58), FUR15 (52 ± 0.58), F13 (52 ± 0.58), F1 (55 ± 0.58), F25 (55 ± 0.58), FUR12 (55 ± 1.15), F59 (59 ± 1.15), Chokhla (59.3 ± 1.15), F32 (59.3 ± 0.58), FUR 11 (63.0 ± 0.58), and F35 (66.7 ± 1.15). During the Rabi 2020-21 season experiment of pathogenicity evaluation, only five isolates exhibited disease severity above 50%. Those include F13 (52 ± 0.58), F1 (52 ± 1.15), Davanagere (52 ± 0.58), Chokhla (55.6 ± 1.15), and Raichur (67 ± 0.02) (Supplementary Table 5). Based on their performance in in-vitro, Kharif, and Rabi field studies, the ten most virulent isolates were selected for resistance evaluation of inbred lines. These isolates were F10, FUR11, Davanagere, F59, Raichur, F21, F18, Chokhla, FUR15 and FUG9 (Figure 8). FIGURE 8 Figure 8. Box plot showing disease severity levels of Fusarium spp. isolates during (A) Kharif 2020 and (B) Rabi 2020-21; Within each box horizontal black line denote median values. Molecular characterization of Fusarium spp. isolates The Tef 1α partial gene sequences of the ten most virulent isolates were searched for closest homologies in the NCBI. Based on the reference sequences of the closest species a phylogenetic tree was constructed. The ten most virulent isolates were identified as strains of Fusarium acutatum (FUR11, F10), Fusarium verticillioides (Syn. Gibberella fujikuroi var. moniliformis) (Davanagere, Raichur, FUG9, F13, FUR15, F21, and F59), and Fusarium andiyazi (F18). The deposited GenBank accession numbers of the respective isolates (or strains) were OP725849, OP748381, ON385434, OP748380, ON385437, ON457741, OP748376, OP748376, OP748382, OP748383, and OP651068. The phylogenetic tree demonstrated that the virulent strains recovered from different locations in India with similar species identity clustered together, and confirmed the homology-based identification by forming three distinct clades with Fusarium verticillioides Fusarium verticillioides (Reference species: MW402103 CBS 181.31 strain, KF499582 CBS 218.76 strain), F. andiyazi (Reference species: MN533989 CBS119856 strain, OP486865 LLC1194 strain), and F. acutatum (Reference species: KR071754 CBS 402.97 strain, MW401971 CBS 113964 strain) (Figure 9). FIGURE 9 Figure 9. Phylogenetic analysis of Fusarium spp. virulent strains based on translation elongation factor (Tef-1α) sequences by MEGA X software using neighbour joining method with 1,000 bootstrap replication. Accessions in bold are strains characterized in the present experiment. Reference sequences were obtained from Crous et al. (2021). Two reference strains for each species are: Fusarium verticillioides (Reference species: MW401977 CBS 117.28 strain, MW402080 CBS 141.59 strain), F. andiyazi (Reference species: MN533989 CBS 119856 strain, MN193854 NRRL 31727 strain), and F. acutatum (Reference species: MW402124 CBS 401.97 strain, MW402125 CBS 402.97 strain). The Alternaria burnsii isolate Alt-MP6 (Sequence ID: ON993391) was used as an outgroup. Fusarium verticillioides; Fusarium andiyazi; Fusarium acutatum. Discussion Fusarium spp. are major pathogen of maize causing symptoms at different growth stages such as crown root rot, seedling rot, stalk rot, wilt, and ear rot of maize in many countries. Among various growth-stage diseases, PFSR which include Fusarium stalk rot, charcoal rot, and late wilt is a major threat to the cultivation of maize crop that affects crop just after tasselling and cob-filling stage in different parts of the world including India, leading to severe yield losses and reduction in grain quality (Payak and Sharma, 1983; Krishna et al., 2013). Frequent incidences of this disease have been reported in many Indian states such as Rajasthan, Gujarat, Madhya Pradesh, Karnataka, Telangana, and Andhra Pradesh. Moreover, recent incidences of PFSR are also observed in Punjab and Maharashtra. Due to the increasing incidences of this fungal pathogen and the extent of pathogenicity in maize crops, we designed a study to thoroughly investigate the morphological diversity and screen the pathogenic potential of Fusarium spp. causing PFSR in various Indian states having different agro-climatic conditions. Changes in agro-climatic conditions alter the attributes of the fungus and can drive the emergence of novel, and climate-resilient fungal species with negative consequences for food security (Shekhar and Singh, 2021). Our study began with a survey of different PFSR-infected maize fields to cover diverse locations with different agro-climatic conditions of Indian states where 71 Fusarium spp. were isolated from symptomatic maize stalks from five agro-climatic zones viz. Eastern plateau and hills, Central Plateau and hills, Western plateau and hills, Southern plateau and hills, and Gujarat plain and hills. Variations in colony appearance were observed between the isolates, such as mycelial variations such as fluffy and sparse white growth, and white and purple mycelial growth having a range of colour vacations such as violet, light violet, light pink, and light pink to light violet, dark violet, and filthy white. In a previous study, authors also observed pink, violet, purple, and brown mycelial colours at later stages on PDA plates (Pavlovic et al., 2016). Harleen et al. (2016) investigated the morphological characteristics and distribution of 56 Fusarium isolates from various maize-growing locations in Punjab. Ramesha and Naik (2017) examined the colony morphology and pathogenicity of Fusarium isolates from Karnataka. Mamatha et al. (2020) surveyed 13 Telangana villages and evaluated the incidence of post-flowering stalk rot and the morphological variety of Fusarium spp. Traditional studies of cultural characteristics and microscopic features of micro and macroconidia and chlamydospores as well as the presence or absence of other morphological structures assist in taxonomic classification and identification of Fusarium species. The morphological characteristics of complex Fusarium spp. are challenging to observe. Thus, from the phenotype dendrogram (Figure 3) the F. verticillioides strains were observed to spread throughout the various clades. Moreover, the F. acutatum and F. verticillioides strains appeared in the same clades. Considering these variations in morphological features among strains it is very difficult to identify strains. Molecular techniques help in a more reliable identification of species, but it is pertinent to complement classical methods. Sporulation of micro and macroconidia was observed on SNA media in the range of sickle- or tapering-shaped, pointy, and blunted ends. Nevertheless, several researchers obtained comparable results (Varela et al., 2013; Shin et al., 2014; Zhang et al., 2021). Ramesha and Naik (2017) found that macro-conidia was sickle-shaped, pedicellate, and distributed, with a high degree of morphological variation in terms of the shape of the micro and macroconidia, as well as variation in the number of septa in each strain and micro conidiophore branching. Fusarium verticillioides which was predominantly isolated from the stalk rot tissues formed long slender macroconidia having a curved apical cell to a tapered point with basal cell notched hill bearing 2–5 septa. Sempere and Santamarina (2009) also observed similar conidial features with false head formation. Based on multivariate cluster analysis of morphological characteristics, 71 isolates were grouped in 9 clusters. Some of the virulent isolates were molecularly characterized and confirmed the identity. In our investigations, out of the ten most virulent isolates, 7 were identified as F. verticillioides. This showed that F. verticillioides is the major pathogen causing PFSR. Our results support the previous finding that F. verticillioides is the dominant Fusarium spp causing PFSR in India (Swamy et al., 2019; Jambhulkar et al., 2022). Among the virulent strain F. verticillioides (Davanagere, Raichur, F13, FUG9, FUR15, F21, and F59), except the FUG9 and FUR 15, remaining strains had purple to dark purple pigmentation. FUG9 and FUR15 were in close proximity but differ in sub-clade due to change in pigmentation and shape of conidia. Davanagere, Raichur and F59 had light pink to light purple colony colour, remaining strains had white to dull white colony colour. Largest among seven F. verticillioides strain was F13 and smallest was F21. Maximum number of septa were observed in F13 and FUR15 while minimum number observed was in F21. Two strains of F. acutatum harboured variation in morphology and micrometry still molecular characterization identified them as one strain of F. acutatum. These variations in cultural morphology and micrometry did not lead to accurate identification of species. Such variation in the cultural morphology of F. verticillioides was reported by previous researchers (Prasad and Padwick, 1939; Patil et al., 2007; Harleen et al., 2016). The pathogenicity of 71 isolates was evaluated in an in-vitro test and subsequently, 60 isolates in Kharif and Rabi field experiments. The present study was the first nationwide evaluation of the pathogenic variability of Fusarium spp. isolated from six states and five agroclimatic zones of India. Pathogenic variabilities among isolates of Fusarium spp. were ascertained based on each isolate causing lesions at the inoculated node and temporal variations in appearance lesions. Fusarium isolates exhibit brown to black discoloration, whereas sterile toothpicks are less infected (Yu et al., 2017). Mesterhazy et al. (2020) evaluated resistance to F. graminearum, F. culmorum, and F. verticillioides through toothpick inoculation and toxin response. The toothpick inoculation technique yielded accurate and efficient pathogenicity results, and discovered more genotypic diversity. The results of the pathogenicity tests revealed that Fusarium spp. isolates were less virulent to highly virulent against cv. Pusa Composite 4. Kharif 2020 had the highest infection rate of the two experimental years, possibly due to meteorological conditions. Similar to our findings, pathogenic potentials of 56 F. verticillioides isolates during Kharif and spring seasons on maize were documented in Punjab (Harleen et al., 2016). Ramesha and Naik (2017) proved the pathogenicity of Fusarium isolates to produce disease symptoms with the development of blackening of vascular bundles in Karnataka. There are claims that Fusarium spp. inoculums enhance root growth without compromising germination (Roman et al., 2020), but we noticed a reduction in root growth using the paper towel approach. There are reports of a constant decline in maize root length following Fusarium spp. inoculation (Ye et al., 2013; Kuhnem et al., 2015). Hassani et al. (2019) discovered that natural infection with Fusarium spp. decreases shoot and root length. Similar outcomes were reported under in-vitro circumstances in the present study. In vitro pathogenicity evaluation reported 30 virulent isolates with seedling vigour index in the range of 0 to 822.2 which suggested that there was a 65 to 100% reduction in seedling vigour index. Only 8 percent of the isolates with a vigour index of more than 1,500 remained avirulent. Significant differences in stalk rot index were observed among 71 isolates. In vitro inoculation with F. moniliforme and Aspergillus niger showed not only a pathogenic effect on the germination of maize seeds but also retarded seedling growth (Hussain et al., 2013). The pathogenic effect of Fusarium spp. isolates from different agroclimatic zones showed variable pathogen virulence in field experiments which ultimately led to symptoms such as drooping, wilting, drying of leaves, empty cob development, and increased angle between stalks and cobs in the field. The virulence of Fusarium species isolates Raichur (66.6%), and F35 (66.7%) were the highest. Under in vitro conditions, FUR15, F35, F59, Raichur, and F21 were considered highly virulent isolates. In the Kharif season, FUR11 (62.9%), Chokhla (59.3%), F59 (59.2%), F1 (55.5%), F18 (51.8 %), F21 (51.8%), F13 (51.8%), and in Rabi season Raichur (66.6%) and Chokhla (55.6%) were reported as most virulent. In this regard, some reports suggest that virulency may differ in different agroclimatic zones (Olowe et al., 2018). The most virulent isolates were from the eastern side of the Central Plateau and hills, the northern part of the Western plateau, and hills. The eastern side of the Central Plateau and hills is a hot, humid zone of southern Rajasthan. Therefore, the greatest observed incidence of the pathogen virulence was predominant in southern Rajasthan, suggesting that PFSR is the most important disease in maize. High temperature and relative humidity of southern Rajasthan than in other Rabi maize-growing states make it a hot spot for PFSR in maize. Thus, such climatic conditions are responsible for causing significant economic loss by PFSR in the eastern side of the Central Plateau and hills, the northern part of the Western plateau and hills, and southern plains and hills as compared to other agroclimatic zones (Czembor et al., 2015; Jambhulkar et al., 2022). Taken together, in the present study, we compared the disease index upon inoculations with 60 isolates of Fusarium spp. collected from geographically diverse states of India, in vitro and in the Kharif and Rabi seasons. Ten isolates recorded the highest disease index during pathogenicity evaluations and were considered the most virulent isolates. Those isolates were molecularly identified as strains of Fusarium acutatum (FUR11, F10), Fusarium verticillioides (Davanagere, Raichur, FUG9, F13, FUR15, F21, and F59), and Fusarium andiyazi (F18). Due to the pronounced intraspecies morphological variabilities and the presence of multiple species within the F. fujikuroi Species Complex (Yilmaz et al., 2021), sequencing of Tef 1-α is recommended to reliably identify Fusarium pathogens, and future studies should include this information as a baseline before making recommendations in guiding pathologists for disease management. Moreover, the respective Fusarium isolates will be used in our program to evaluate the resistance of different maize inbred lines as a critical component of PFSR disease management. Data availability statement The data presented in the study are deposited in the NCBI GenBank repository, accession numbers: OP725849, OP748381, ON385434, OP748380, ON385437, ON457741, OP748376, OP748382, OP748383, and OP651068. Author contributions PJ conceived and obtained funding from DST-SERB. PJ and JH collected isolates from diseased plants from different agroclimatic regions of India and prepared the first draft of the manuscript. JH performed morphological characterization, microscopy and did the phylogenetic analysis. PJ, JH, and RB did pathogenicity. RB and JH did molecular characterization and carried out clustering. PB, PJ, and DL edited the manuscript. MA, SC, DL, and AK provided valuable insights into the manuscript. All authors approved the final version of the manuscript. Funding This work was funded by the Department of Science and Technology SERB, New Delhi, India, for their financial support under Project F. No. EEQ/00181/2019. Acknowledgments We thankful to Maharishi Tomar and Shailendra Kumar for the data analysis. We also grateful to Vice-Chancellor, RLBCAU, Jhansi, India, for providing facilities to conduct the experiments of the student’s thesis work as well as for the research project. Conflict of interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Publisher’s note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. 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Bilodeau, Canadian Food Inspection Agency (CFIA), Canada Reviewed by: Bishnu Maya Bashyal, Indian Agricultural Research Institute (ICAR), India Marcin Wit, Warsaw University of Life Sciences, Poland David Overy, Agriculture and Agri-Food Canada (AAFC), Canada Copyright © 2023 Harish, Jambhulkar, Bajpai, Arya, Babele, Chaturvedi, Kumar and Lakshman. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Prashant P. Jambhulkar, ppjambhulkar@gmail.com; Dilip K. Lakshman, dilip.lakshman@usda.gov Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
17521	https://www.wolframalpha.com/input/?i=minimize	minimize - Wolfram\|Alpha UPGRADE TO PRO APPS TOUR Sign in Use Wolfram\|Alpha Pro for reality-checked results Go Pro Now minimize Natural Language Math Input Extended KeyboardExamplesUpload Random Assuming "minimize" refers to a computation \| Use as a general topic or a word instead Computational Inputs: » curve function: Also include: search domain Compute Input interpretation Global minima Step-by-step solution Local minimum Decimal form Step-by-step solution Plot Download Page POWERED BY THE WOLFRAM LANGUAGE Related Queries: random yoga pose curve Mathematica optimization functions series of 5 - 3 x^4 + x^5wrt x 5 - 3 x^4 + x^5 vs d (5 - 3 x^4 + x^5)/dx Have a question about using Wolfram\|Alpha?Contact Pro Premium Expert Support »Give us your feedback » Pro Mobile Apps Products Business API & Developer Solutions LLM Solutions Resources & Tools About Contact Connect ©2025 Wolfram Terms Privacy wolfram.com Wolfram Language Mathematica Wolfram Demonstrations Wolfram for Education MathWorld
17522	https://tvst.arvojournals.org/article.aspx?articleid=2632039	Assessing the Accuracy of Foveal Avascular Zone Measurements Using Optical Coherence Tomography Angiography: Segmentation and Scaling \| TVST \| ARVO Journals tvst Issues Topics For Authors About Editorial Board Journals Home MANAGE ALERTS Forgot password? Create an Account To View More... Purchase this article with an account. or Subscribe Now Advanced Search All Journals All Journals IOVS JOV TVST Issues Topics For Authors About Editorial Board May 2017 Volume 6, Issue 3 ‹ Issue › Jump To... Introduction Methods Results Discussion Acknowledgments References Open Access Articles\| August 2017 Assessing the Accuracy of Foveal Avascular Zone Measurements Using Optical Coherence Tomography Angiography: Segmentation and Scaling Rachel Linderman; Alexander E. Salmon; Margaret Strampe; Madia Russillo; Jamil Khan; Joseph Carroll Author Affiliations & Notes Rachel Linderman Department of Ophthalmology & Visual Sciences, Medical College of Wisconsin, Milwaukee, WI, USA Alexander E. Salmon Department of Cell Biology, Neurobiology, & Anatomy, Medical College of Wisconsin, Milwaukee, WI, USA Margaret Strampe Department of Ophthalmology & Visual Sciences, Medical College of Wisconsin, Milwaukee, WI, USA University of Minnesota Medical School, Minneapolis, MN, USA Madia Russillo Medical College of Wisconsin, Milwaukee, WI, USA Jamil Khan Medical College of Wisconsin, Milwaukee, WI, USA Joseph Carroll Department of Ophthalmology & Visual Sciences, Medical College of Wisconsin, Milwaukee, WI, USA Department of Cell Biology, Neurobiology, & Anatomy, Medical College of Wisconsin, Milwaukee, WI, USA Correspondence: Joseph Carroll, Department of Ophthalmology & Visual Sciences, Medical College of Wisconsin, 925 N. 87 th Street, Milwaukee, WI 53226-0509, USA. e-mail: jcarroll@mcw.edu Connected Content Errata: Erratum Translational Vision Science & Technology August 2017, Vol.6, 16. doi: Views Full Article Figures Tables PDF Share email facebook twitter linkedin digg tumblr Tools AlertsUser Alerts You are adding an alert for: Assessing the Accuracy of Foveal Avascular Zone Measurements Using Optical Coherence Tomography Angiography: Segmentation and Scaling You will receive an email whenever this article is corrected, updated, or cited in the literature. You can manage this and all other alerts in My Account The alert will be sent to: Confirm × ###### This feature is available to authenticated users only. Sign In or Create an Account× Get Citation Citation Rachel Linderman, Alexander E. Salmon, Margaret Strampe, Madia Russillo, Jamil Khan, Joseph Carroll; Assessing the Accuracy of Foveal Avascular Zone Measurements Using Optical Coherence Tomography Angiography: Segmentation and Scaling. Trans. Vis. Sci. Tech. 2017;6(3):16. Download citation file: Ris (Zotero) EndNote BibTex Medlars ProCite RefWorks Reference Manager © ARVO (1962-2015); The Authors (2016-present) × Get Permissions Abstract Purpose: The foveal avascular zone (FAZ) is altered in numerous diseases. We assessed factors (axial length, segmentation method, age, sex) impacting FAZ measurements from optical coherence tomography (OCT) angiography images. Methods: We recruited 116 Caucasian subjects without ocular disease, and acquired two 3 × 3 mm AngioVue scans per each right eye (232 total scans). In images of the superficial plexus, the FAZ was segmented using the AngioVue semiautomatic nonflow measurement tool and ImageJ manual segmentation. In images from the full retinal thickness, the FAZ was segmented using the AngioAnalytics automatic FAZ tool. Repeatability, reliability, and reproducibility were calculated for FAZ measurements (acircularity, area). Results: FAZ area (mean ± SD) for manual segmentation was 0.257 ± 0.104 mm 2, greater than both semiautomatic (0.231 ± 0.0939 mm 2) and automatic (0.234 ± 0.0933 mm 2) segmentation (P< 0.05). Not correcting for axial length introduced errors up to 31% in FAZ area. Manual area segmentation had better repeatability (0.022 mm 2) than semiautomatic (0.046 mm 2) or automatic (0.060 mm 2). FAZ acircularity had better repeatability with automatic than manual segmentation (0.086 vs. 0.114). Reliability of all area measurements was excellent (intraclass correlation coefficient [ICC] = 0.994 manual, 0.969 semiautomatic, 0.948 automatic). Reliability of acircularity measurements was 0.879 for manual and 0.606 for automatic. Conclusion: We identified numerous factors affecting FAZ measurements. These errors confound comparisons across studies and studies examining factors that may correlate with FAZ measures. Translational Relevance: Using FAZ measurements as biomarkers for disease progression requires assessing and controlling for different sources of error. Not correcting for ocular magnification can result in significant inaccuracy in FAZ measurements, while choice of segmentation method affects both repeatability and accuracy. Introduction The foveal avascular zone (FAZ) is an area within the central macula that is devoid of retinal capillaries. The size of the FAZ varies dramatically among normal individuals and is correlated with the size of the foveal pit.1,2 The FAZ is known to be reduced in size or altogether absent in individuals born prematurely,3,4 along with patients with albinism,5 idiopathic foveal hypoplasia,6,7 and aniridia.8 The FAZ is enlarged in patients with sickle cell disease.9 In addition, retinal vascular changes around the FAZ also occur early in diabetes mellitus with capillary dropout occurring even before the onset of retinopathy,10,11 leading to a significant increase in the size of the FAZ. Furthermore, FAZ size is correlated with visual acuity in patients with diabetic retinopathy,12 suggesting that assessment of the FAZ could be a useful biomarker for studying diabetic patients. The variability of the FAZ in health and disease has made it an often-studied structure using a variety of techniques, including fluorescein angiography (FA), retinal function imager, and adaptive optics (AO)–based methods.13 While these approaches have facilitated a diverse array of studies of the FAZ, the recent advent of optical coherence tomography-angiography (OCT-A) has made noninvasive FAZ visualization mainstream. With this has come a plethora of studies examining FAZ size and shape in a myriad of diseases. Across a number of studies, mean area of the superficial FAZ using OCT-A ranges from 0.24 to 0.30 mm 2 in healthy eyes.11,14,15 However, as with other OCT-based measurements (such as retinal nerve fiber layer [RNFL] thickness, foveal morphology, and retinal thickness), it is critical to evaluate factors affecting the accuracy and reliability of FAZ measurements. One of the major factors affecting FAZ measurements is the segmentation or marking of the FAZ boundary. Commercial systems have algorithms that provide estimates of the “nonflow” area, defining the FAZ. Various studies have demonstrated excellent repeatability and reliability of such measurements.16–19 However, as new devices and different segmentation algorithms become available, it is important to continually reassess the reliability and repeatability of FAZ measurements. Furthermore, there are conflicting data on whether semiautomatic measurements agree with measurements obtained using manual segmentation. La Spina et al.20 reported that semiautomatic measurements were larger on average than manual measurements, while Magrath et al.21 reported no difference between semiautomatic and manual measurements when using the AngioVue OCT-A system. Beyond the specific segmentation method, correct scaling of the retinal image also could impact the accuracy of FAZ measurements. While scans typically are 3 × 3 or 6 × 6 mm in size, individual differences in axial length (and, thus, ocular magnification) affect the absolute dimensions of the scan, meaning that measurements of the FAZ obtained from uncorrected scans will be inaccurate. Recently, FAZ area was shown to be correlated negatively with axial length, yet the investigators did not correct their images for ocular magnification.15 As such, the relationship could be due, at least in part, to the differences in ocular magnification across eyes. Thus, the purpose of our study was to assess the effect of axial length on FAZ measurements as well as to compare our previously published manual FAZ segmentation method22 to semiautomatic and automatic methods. We also explored the effect of sex and age on FAZ measurements in this relatively large Caucasian cohort, working towards the long-term goal of establishing robust normative databases for open dissemination. Methods Subjects This study was approved by the Institutional Review Board (IRB) of the Medical College of Wisconsin and was conducted in accordance with the tenets of the Declaration of Helsinki. Informed consent was obtained for all subjects once the nature and risks of the study were explained. Exclusion criteria included any prior history or clinical evidence of retinal or systemic vascular disease. We imaged 116 Caucasian subjects (51 male, 65 female). The mean (± standard deviation) age was 30.5 ± 14.5 years (range, 8–77 years). Axial length measurements were obtained from all subjects using an IOL Master (Carl Zeiss, Meditec, Dublin, CA). Assessing the Foveal Avascular Zone Subjects were imaged with the AngioVue OCT-A system (Optovue, Inc., Fremont, CA). Two scans (one horizontal and one vertical, each consisting of 304 B-scans at 304 A-scans per B-scan) were acquired at the fovea of the right eye with a nominal scan size of 3 × 3 mm. These two scans then were coregistered by the device to minimize motion artifact and create a single volume from which an image of the superficial plexus and an image of the full retinal thickness angiogram was extracted.23,24 Two such volumes were obtained for each subject. The superficial plexus image was created by integrating motion contrast data from 3 μm below the internal limiting membrane (ILM) to 16 μm above the inner plexiform layer (IPL), while the full retinal thickness angiogram was created by integrating motion contrast data from the ILM to 75 μm above the retinal pigment epithelium (RPE). The 232 superficial plexus images were manually segmented by a single masked observer (R.L.) using ImageJ (National Institutes of Health [NIH], Bethesda, MD). The coordinates from the manual segmentation along with the image dimensions were entered into a previously described custom Matlab script (Mathworks, Natick, MA) using the function poly2mask to produce a mask defining the area of the FAZ.2,22 The nominal area of the FAZ (in mm 2) was calculated according to the formula: A n o m i n a l=(S I,0 S J,0 I J)×A p x 2 A n o m i n a l=(S I,0 S J,0 I J)×A p x 2 where Display FormulaA n o m i n a l A n o m i n a l is the nominal area of the FAZ in mm 2, Display FormulaS I,0 S I,0 and Display FormulaS J,0 S J,0 are the nominal scan dimensions in mm, Display FormulaI I and Display FormulaJ J are the number of samples in each dimension in pixels, and Display FormulaA p x 2 A p x 2 is the nominal area of the FAZ in in pixels 2, which was calculated using the Matlab function regionprops. The actual area of the FAZ (in mm 2) was then calculated as follows: A c o r r e c t e d=A n o m i n a l(A L S A L M)2 A c o r r e c t e d=A n o m i n a l(A L S A L M)2 where Display FormulaA L s A L s is the axial length of the subject in mm, and Display FormulaA L M A L M is the axial length assumed for the model eye by the manufacturer (23.95 mm). Nominal FAZ perimeter in pixels (Display FormulaP n o m i n a l P n o m i n a l) was also obtained using the Matlab function regionprops and adjusted for ocular magnification to obtain the FAZ perimeter in mm (Display FormulaP c o r r e c t e d P c o r r e c t e d): P c o r r e c t e d=P n o m i n a l(S I,0 I)(A L S A L M)P c o r r e c t e d=P n o m i n a l(S I,0 I)(A L S A L M) FAZ acircularity then was calculated as the ratio of the FAZ perimeter to the circumference of a circle with an area equivalent to that of the FAZ.25 Using the AngioVue review software (Optovue, Inc.; ver. 2016.2.0.16), the FAZ area for each superficial plexus image was found using the built-in nonflow measurement tool using a single seed point (i.e., semiautomatic). The FAZ acircularity was not calculated using this algorithm due to the software not reporting perimeter values. Finally, the FAZ area and FAZ acircularity for each full retinal thickness angiogram were found using the AngioVue review software (clinical ver. 2016.200.0.37) with “AngioAnalytics” enabled (i.e., automatic). Statistics Intrasession repeatability and reliability of the FAZ area and FAZ acircularity (where possible) for the three segmentation methods were assessed.26 Repeatability was calculated as 2.77σ w, where σ w represents the average within-subject standard deviation, and measurement error was calculated as 1.96σ w. The 95% confidence intervals were calculated using the formula Display FormulaC I 95%=1.96 σ w 2 n(m−1)√C I 95%=1.96 σ w 2 n(m−1) where Display Formulan n is the number of subjects (116) and Display Formulam m is the number of measurements (2). Reliability was assessed by finding the intraclass correlation coefficient (ICC) using the R statistical package (The R Foundation for Statistical Computing, Vienna, Austria). Sex differences were assessed using Mann-Whitney U tests, while linear regressions were used to assess for a relationship between FAZ area or FAZ acircularity and age using R statistical package. Reproducibility among the three methods was assessed using the Friedman test with post-test and further analyzed using Wilcoxon signed-rank tests and Bland-Altman plots.26 Results Effect of Axial Length on FAZ Measurements The average axial length was 24.04 ± 1.25 mm (mean ± SD; range, 21.45–27.45 mm). Assuming a nominal 3 × 3 mm scan area, we observed a slight negative trend between FAZ area and axial length computed using the semiautomatic method, though this did not reach statistical significance (R 2 = 0.023, P = 0.10, Fig. 1A). When correcting the scan size for differences in ocular magnification, the trend is abolished (R 2 = 0.0061, P = 0.87, Fig. 1B). The error in FAZ area estimates as a function of axial length is shown in Figure 1C; the average error was 8.29% with a maximum error of 31.36%. The absolute maximum error was 0.07 mm 2. Similar trends were seen for the manual and automatic segmentation methods (data not shown); thus, for subsequent FAZ area analyses, we used data that were corrected for axial length. Differences in ocular magnification would not affect FAZ acircularity measurements, owing to the method by which it is calculated. Figure 1 View OriginalDownload Slide The effect of axial length on semiautomatic FAZ area measurements. (A) FAZ area assuming a nominal 3 × 3 mm scan area. There is a slight negative trend as axial length increases (solid line is trendline, dashed lines are 95% confidence intervals), though not significant (P = 0.10). (B) FAZ area when accounting for axial length. The downward trend disappears when axial length is considered (P = 0.87). (C) The error in FAZ area measurement resulting from a failure to account for axial length. As the deviation in axial length increases, so does the error in the FAZ area. (Error model, E = [1 − (AL S/AL M)2] ∗ 100%; solid line.) Data plotted separately for males (open circles) and females (crosses). Figure 1 The effect of axial length on semiautomatic FAZ area measurements. (A) FAZ area assuming a nominal 3 × 3 mm scan area. There is a slight negative trend as axial length increases (solid line is trendline, dashed lines are 95% confidence intervals), though not significant (P = 0.10). (B) FAZ area when accounting for axial length. The downward trend disappears when axial length is considered (P = 0.87). (C) The error in FAZ area measurement resulting from a failure to account for axial length. As the deviation in axial length increases, so does the error in the FAZ area. (Error model, E = [1 − (AL S/AL M)2] ∗ 100%; solid line.) Data plotted separately for males (open circles) and females (crosses). View OriginalDownload Slide Effect of Different Segmentation Methods When comparing different segmentation methods, the manual segmentation performed better than the semiautomatic or the automatic segmentation, though all methods showed excellent repeatability and reliability (Fig. 2, Table 1) for FAZ area. Figure 2 View OriginalDownload Slide Effect of segmentation method on intrasession repeatability. Data are expressed in Bland-Altman plots to show the repeatability of FAZ area measurements using either (A) manual segmentation, (B) semiautomatic segmentation software (version 2016.2.0.16), or (C) automatic segmentation software (version 2016.200.0.37). Solid lines represent the average difference between the two trials, while dashed lines represent 95% confidence intervals. Figure 2 Effect of segmentation method on intrasession repeatability. Data are expressed in Bland-Altman plots to show the repeatability of FAZ area measurements using either (A) manual segmentation, (B) semiautomatic segmentation software (version 2016.2.0.16), or (C) automatic segmentation software (version 2016.200.0.37). Solid lines represent the average difference between the two trials, while dashed lines represent 95% confidence intervals. View OriginalDownload Slide Table 1 View Table A Comparison between Manual, Semi-Automatic and Automatic Segmentation Methods Table 1 A Comparison between Manual, Semi-Automatic and Automatic Segmentation Methods This difference is likely due to reduced image quality (from motion artifacts) and low vessel contrast affecting the algorithm-driven methods more than the human observer (Fig. 3). We used average values for each method to compare the FAZ area measurements and observed a significant difference between the manual segmentation and both automatic segmentations (Friedman test with post-test, P< 0.0001) but not between the two automatic segmentation methods (P> 0.05). When comparing the automatic to the manual segmentation methods, the mean difference (± SD) was 0.0240 ± 0.0259 mm 2 (Wilcoxon signed-rank test, P< 0.0001) with manual being, on average, larger than automatic segmentation for FAZ area. As a percentage of the automatic FAZ area, the difference ranged from 2.2% to 54.4% (the absolute maximum error was 0.145 mm 2). The difference between the two methods increased as a function of FAZ area (y = 0.874 x + 0.0086; P = 0.0001). A similar difference was observed between the manual and semiautomatic methods, while no difference was observed between the semiautomatic and automatic methods (data not shown). Figure 3 View OriginalDownload Slide The effect of image quality on manual, semiautomatic, and automatic measurements. Manual segmentation is shown on the OCT-A images of the superficial plexus marked with red dots. Subject JC_10567 had similar image quality between images and had a corresponding similarity between all methods. A large motion artifact is present in the second image from JC_10580, which has a greater effect on the semiautomatic segmentation than the manual or automatic segmentation. The second image from JC_10585 exhibited decreased contrast that affected all methods, though the automatic segmentation appeared most severely affected. Figure 3 The effect of image quality on manual, semiautomatic, and automatic measurements. Manual segmentation is shown on the OCT-A images of the superficial plexus marked with red dots. Subject JC_10567 had similar image quality between images and had a corresponding similarity between all methods. A large motion artifact is present in the second image from JC_10580, which has a greater effect on the semiautomatic segmentation than the manual or automatic segmentation. The second image from JC_10585 exhibited decreased contrast that affected all methods, though the automatic segmentation appeared most severely affected. View OriginalDownload Slide When comparing FAZ acircularity between automatic and manual segmentations, the mean difference (± SD) was 0.045 ± 0.0975 with the manual, on average, being larger than the automatic segmentation. As a percentage of the automatic FAZ acircularity, the difference ranged from 0.1% to 32.0% (the absolute maximum error was 0.36). The difference between the two methods increased as a function of FAZ acircularity (y = −0.230 x + 0.2574; P = 0.0001). Across all subjects, we observed good repeatability and reliability of the FAZ acircularity index (Table 1), though it was worse than that observed for FAZ area measurements, indicating it is more sensitive to small errors in segmentation. Other Biological Variables Affecting FAZ Measurements Females had larger FAZ area than males for all segmentation methods, though no difference in FAZ acircularity was observed between the groups (Table 2). No correlation between age and FAZ area was observed (manual, P = 0.920; semiautomatic, P = 0.996; automatic, P = 0.920), nor was there any correlation between age and FAZ acircularity (manual, P = 0.912; automatic, P = 0.337). Table 2 View Table The Effect of Gender on Measurements of the FAZ Table 2 The Effect of Gender on Measurements of the FAZ Discussion We examined various factors impacting FAZ measurements, including axial length, sex, age, and segmentation method. Not correcting for axial length represents a significant source of error for measuring FAZ area. While this may not be critical for longitudinal studies comparing multiple scans from one person, these errors limit the ability to compare or combine FAZ area measurements across studies. In addition, our analysis suggests that a previous report15 of a negative correlation between axial length and FAZ area is likely due to the differences in ocular magnification across eyes. These errors are likely even more significant when studying populations with high refractive error. Our estimates of FAZ area obtained with the manual segmentation method generally agreed with previous reports.13–15,17–19,22 As has been reported previously,15,27 we found increased FAZ area in females, though this is in contrast with data from Samara et al.14 Similarly, our observation of no relationship between FAZ size and age is in agreement with that of Tan et al.15 and Samara et al.14 but conflicts with other studies.27–29 These differences could be due to different sex, age, and/or racial distribution. Further, none of the aforementioned studies corrected for axial length.14,15,28 As such, prior conclusions drawn from studies are in many cases confounded by the lack of correction for ocular magnification, and revisiting these data may be worthwhile. Despite excellent repeatability across all three segmentation methods, we observed a significant difference in FAZ area when comparing methods, with the algorithm-driven methods tending to produce smaller FAZ areas on average. This is in contrast with Magrath et al.,21 who found no difference between semiautomatic and manual measurements when using the Optovue system, and La Spina et al.,20 who found an increase in FAZ area when using the semiautomatic algorithm.20 It is important to note that these studies used different versions of the semiautomatic algorithm (La Spina et al.20 used version 2014.4.0.68; Magrath et al.21 used version 2014.4.0.13; this study used versions 2016.2.0.16 [semiautomatic] and clinical version 2016.200.0.37 [automatic]). Reporting the version of analysis software used is critical to facilitate comparisons across studies. FAZ area obtained from the automatic or semiautomatic segmentation algorithms used here can result in errors of nearly 60%. Different algorithms may produce different results, thus similar validation studies would need to be performed on one's segmentation algorithm of interest. It is important to note that all subjects in this study are Caucasian. This could limit the observed range of FAZ areas due to African Americans having significantly larger foveal pits (and, thus, FAZs).30 With the difference between the segmentation methods increasing as the FAZ area becomes larger, it is important to further examine the performance of the automatic algorithm in subjects with larger pits. Similarly, normative databases comprising data from Caucasians should not be used to assess FAZ measurements from non-Caucasian subjects. Also, we had a significant number of subjects (52%) between 20 and 29 years old. More work must occur to increase the number of younger and older individuals to better understand the true variability across all age groups. In addition, all subjects had normal vision resulting in subjectively good image quality and minimal motion artifacts. While this led to excellent repeatability for FAZ area and FAZ acircularity, the introduction of unsteady fixation will lead to an increase in motion artifacts and a corresponding decreased image quality.31 Thus, the differences observed here between the different segmentation approaches may not hold for different patient populations for whom fixation is known to be unstable. In conclusion, the two different Optovue FAZ measurement algorithms assessed herein have similar repeatability when compared to manual segmentation. However, accurate measurements require correction for axial length and careful review of automatic segmentation results (possibly with manual correction). Other metrics, like vessel density, may be affected similarly, especially when reported over Early Treatment of Diabetic Retinopathy Study (ETDRS)-like retinal areas of some fixed distance. The fact that FAZ acircularity and axis ratio are not impacted by ocular magnification make them attractive metrics to explore further, considering ocular magnification adjustments currently are not available in clinical devices.32 Acknowledgments The authors thank Erin Curran, Mara Goldberg, Phyllis Summerfelt, and Vesper Williams for help recruiting and imaging subjects, and Tina Hsiao and Zhou Qienyuan for help using the Optovue machine and software, as well as helpful comments on the manuscript. Supported in part by the National Eye Institute of the NIH under award numbers R01EY024969 and P30EY001931, and by the National Institute of Aging of the NIH under award number T35AG029793. The content is solely the responsibility of the authors and does not necessarily represent the official views of the NIH. Rachel Linderman is the recipient of a Fight for Sight – Nick Cacciola Summer Student Fellowship Award. References Chui TYP, Zhong Z, Song H, Burns SA. 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Pakzad-Vaezi K, Keane PA, Cardoso JN, Egan C, Tufail A. Optical coherence tomography angiography of foveal hypoplasia [published online ahead of print November 29, 2016]. Br J Ophthalmol. Nelson LB, Spaeth GL, Nowinski TS, Margo CE, Jackson, L. Aniridia. A review. Surv Ophthalmol. 1984; 28: 621–642. Sanders RJ, Brown GC, Rosenstein RB, Margargal L. Foveal avascular zone diameter and sickle cell disease. Arch Ophthalmol. 1991; 109: 812–815. Tam J, Dhamdhere KP, Tiruveedhula P, et al. Subclinical capillary changes in non-proliferative diabetic retinopathy. Optom Vis Sci. 2012; 89: E692–E703. Takase N, Nozaki M, Kato A, Ozeki H, Yoshida M, Ogura Y. Enlargement of foveal avascular zone in diabetic eyes evaluated by en face optical coherence tomography angiography. Retina. 2015; 35: 2377–2383. Balaratnasingam C, Inoue M, Ahn S, et al. Visual acuity is correlated with the area of the foveal avascular zone in diabetic retinopathy and retinal vein occlusion. Ophthalmology. 2016; 123: 2352–2367. Shahlaee A, Pefkianaki M, Hsu J, Ho AC. Measurement of foveal avascular zone dimensions and its reliability in healthy eyes using optical coherence tomography angiography. Am J Ophthalmol. 2016; 161: 50–55. Samara WA, Say EA, Khoo CT, et al. Correlation of foveal avascular zone size with foveal morphology in normal eyes using optical coherence tomography angiography. Retina. 2015; 35: 2188–2195. Tan CS, Lim LW, Chow VS, et al. Optical coherence tomography angiography evaluation of the parafoveal vasculature and its relationship with ocular factors. Invest Ophthalmol Vis Sci. 2016; 57: OCT224–OCT234. Hwang TS, Gao SS, Liu L, et al. Automated quantification of capillary nonperfusion using optical coherence tomography angiography in diabetic retinopathy. JAMA Ophthalmol. 2016; 134: 367–373. Lupidi M, Coscas F, Cagini C, et al. Automated quantitative analysis of retinal microvasculature in normal eyes on optical coherence tomography angiography. Am J Ophthalmol. 2016; 169: 9–23. Coscas F, Sellam A, Glacet-Bernard A, et al. Normative data for vascular density in superficial and deep capillary plexuses of healthy adults assessed by optical coherence tomography angiography. Invest Ophthalmol Vis Sci. 2016; 57: OCT211–OCT223. Carpineto P, Mastropasqua R, Marchini G, Toto L, Di Nicola M, Di Antonio L. Reproducibility and repeatability of foveal avascular zone measurements in healthy subjects by optical coherence tomography angiography. Br J Ophthalmol. 2016; 100: 671–676. La Spina C, Carnevali A, Marchese A, Querques G, Bandello F. Reproducibility and reliability of optical coherence tomography angiography for foveal avascular zone evaluation and measurement in different settings [published online ahead of print December 20, 2016]. Retina. Magrath GN, Say EA, Sioufi K, Ferenczy S, Samara WA, Shields CL. Variability in foveal avascular zone and capillary density using optical coherence tomography machines in healthy eyes [published online ahead of print December 16, 2016]. Retina. Wilk MA, Dubis AM, Cooper RF, Summerfelt P, Dubra A, Carroll J. Assessing the spatial relationships between fixation and foveal specializations. Vision Res. 2017; 132: 53–61. Kraus MF, Potsaid B, Mayer MA, et al. Motion correction in optical coherence tomography volumes on a per A-scan basis using orthogonal scan patterns. Biomed Opt Express. 2012; 3: 1182–1199. Kraus MF, Liu JJ, Schottenhamml J, et al. Quantitative 3D-OCT motion correction with tilt and illumination correction, robust similarity measure and regularization. Biomed Opt Express. 2014; 5: 2591–2613. Tam J, Dhamdhere KP, Tiruveedhula P, et al. Disruption of the retinal parafoveal capillary network in type 2 diabetes before the onset of diabetic retinopathy. Invest Ophthalmol Vis Sci. 2011; 52: 9257–9266. Bland JM, Altman DG. Statistics notes: Measurement error. Br Med J (Clin Res Ed). 1996; 313: 744. Yu J, Jiang C, Wang X, et al. Macular perfusion in healthy Chinese: an optical coherence tomography angiogram study. Invest Ophthalmol Vis Sci. 2015; 56: 3212–3217. Iafe NA, Phasukkijwatana N, Chen X, Sarraf D. Retinal capillary density and foveal avascular zone area are age-dependent: Quantitative analysis using optical coherence tomography angiography. Invest Ophthalmol Vis Sci. 2016; 57: 5780–5787. Gong D, Zou X, Zhang X, Yu W, Qu Y, Dong F. The influence of age and central foveal thickness on foveal zone size in healthy people. Ophthal Surg Lasers Imag Retina. 2016; 47: 142–148. Wagner-Schuman M, Dubis AM, Nordgren RN, et al. Race- and sex-related differences in retinal thickness and foveal pit morphology. Invest Ophthalmol Vis Sci. 2011; 52: 625–634. Ghasemi Falavarjani K, Al-Sheikh M, Akil H, Sadda SR . Image artefacts in swept-sources optical coherence tomography angiography. Br J Ophthalmol. 2016; 101: 564–568. Krawitz BD, Mo S, Geyman LS, et al. Acircularity index and axis ratio of the foveal avascular zone in diabetic eyes and healthy controls measured by optical coherence tomography angiography [published online ahead of print February 26, 2017]. Vision Res. Figure 1 View OriginalDownload Slide The effect of axial length on semiautomatic FAZ area measurements. (A) FAZ area assuming a nominal 3 × 3 mm scan area. There is a slight negative trend as axial length increases (solid line is trendline, dashed lines are 95% confidence intervals), though not significant (P = 0.10). (B) FAZ area when accounting for axial length. The downward trend disappears when axial length is considered (P = 0.87). (C) The error in FAZ area measurement resulting from a failure to account for axial length. As the deviation in axial length increases, so does the error in the FAZ area. (Error model, E = [1 − (AL S/AL M)2] ∗ 100%; solid line.) Data plotted separately for males (open circles) and females (crosses). Figure 1 The effect of axial length on semiautomatic FAZ area measurements. (A) FAZ area assuming a nominal 3 × 3 mm scan area. There is a slight negative trend as axial length increases (solid line is trendline, dashed lines are 95% confidence intervals), though not significant (P = 0.10). (B) FAZ area when accounting for axial length. The downward trend disappears when axial length is considered (P = 0.87). (C) The error in FAZ area measurement resulting from a failure to account for axial length. As the deviation in axial length increases, so does the error in the FAZ area. (Error model, E = [1 − (AL S/AL M)2] ∗ 100%; solid line.) Data plotted separately for males (open circles) and females (crosses). View OriginalDownload Slide Figure 2 View OriginalDownload Slide Effect of segmentation method on intrasession repeatability. Data are expressed in Bland-Altman plots to show the repeatability of FAZ area measurements using either (A) manual segmentation, (B) semiautomatic segmentation software (version 2016.2.0.16), or (C) automatic segmentation software (version 2016.200.0.37). Solid lines represent the average difference between the two trials, while dashed lines represent 95% confidence intervals. Figure 2 Effect of segmentation method on intrasession repeatability. Data are expressed in Bland-Altman plots to show the repeatability of FAZ area measurements using either (A) manual segmentation, (B) semiautomatic segmentation software (version 2016.2.0.16), or (C) automatic segmentation software (version 2016.200.0.37). Solid lines represent the average difference between the two trials, while dashed lines represent 95% confidence intervals. View OriginalDownload Slide Figure 3 View OriginalDownload Slide The effect of image quality on manual, semiautomatic, and automatic measurements. Manual segmentation is shown on the OCT-A images of the superficial plexus marked with red dots. Subject JC_10567 had similar image quality between images and had a corresponding similarity between all methods. A large motion artifact is present in the second image from JC_10580, which has a greater effect on the semiautomatic segmentation than the manual or automatic segmentation. The second image from JC_10585 exhibited decreased contrast that affected all methods, though the automatic segmentation appeared most severely affected. Figure 3 The effect of image quality on manual, semiautomatic, and automatic measurements. Manual segmentation is shown on the OCT-A images of the superficial plexus marked with red dots. Subject JC_10567 had similar image quality between images and had a corresponding similarity between all methods. A large motion artifact is present in the second image from JC_10580, which has a greater effect on the semiautomatic segmentation than the manual or automatic segmentation. The second image from JC_10585 exhibited decreased contrast that affected all methods, though the automatic segmentation appeared most severely affected. View OriginalDownload Slide Table 1 View Table A Comparison between Manual, Semi-Automatic and Automatic Segmentation Methods Table 1 A Comparison between Manual, Semi-Automatic and Automatic Segmentation Methods Table 2 View Table The Effect of Gender on Measurements of the FAZ Table 2 The Effect of Gender on Measurements of the FAZ Copyright 2017 The Authors This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. 3,752 Views 104Web of Science View Metrics × Related Articles Variability of Foveal Avascular Zone Metrics Derived From Optical Coherence Tomography Angiography Images Prevalence of Subclinical CNV and Choriocapillaris Nonperfusion in Fellow Eyes of Unilateral Exudative AMD on OCT Angiography Comparative Study of Optical Coherence Tomography Angiography and Phase-Resolved Doppler Optical Coherence Tomography for Measurement of Retinal Blood Vessels Caliber Intrasession Repeatability and Interocular Symmetry of Foveal Avascular Zone and Retinal Vessel Density in OCT Angiography Polypoidal Choroidal Vasculopathy on Swept-Source Optical Coherence Tomography Angiography with Variable Interscan Time Analysis From Other Journals Optical Coherence Tomographic Correlates of Angiographic Subtypes of Occult Choroidal Neovascularization Quantitative Optical Coherence Tomography Findings in Various Subtypes of Neovascular Age-Related Macular Degeneration Correlation of High-Definition Optical Coherence Tomography and Fluorescein Angiography Imaging in Neovascular Macular Degeneration Peripapillary Microvascular and Neural Changes in Diabetes Mellitus: An OCT-Angiography Study Color Fundus Image Guided Artery-Vein Differentiation in Optical Coherence Tomography Angiography Related Topics Imaging Translational Research Advertisement TVST Home Issues Topics For Authors About Editorial Board Online ISSN: 2164-2591 Investigative Ophthalmology & Visual Science Journal of Vision Translational Vision Science & Technology JOURNALS HOME TOPICS ABOUT ARVO JOURNALS Rights & Permissions Privacy Statement Advertising Submit a Manuscript Disclaimer Contact Us ARVO.org Copyright © 2025 Association for Research in Vision and Ophthalmology. 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17523	https://www.houseofmath.com/encyclopedia/functions/theory-of-functions/linear-functions/what-are-linear-functions	What Are Linear Functions? Log inSign up for free Home League Search Menu Learning Tools Bootcamps Curriculum Show more Show all games Go to games All Math Algebra Geometry Numbers & Quantities Statistics & Probability Functions Proofs Show more Show all games Go to games Games Junior Math Multiplication Master Show more Show all games Go to games Curriculum California CCSSM NorgeУкраїнськаPhilippinesInternational Log Out Encyclopedia>Functions>Theory of Functions>Linear Functions>What Are Linear Functions? Ready to unlock your full learning experience? Start today - and unlock your math superpower! Finally master math - for real Full access to the entire curriculum Video tutorials that actually work The fastest way to get a better math grade Unlock now What Are Linear Functions? Video Crash Courses Want to watch animated videos and solve interactive exercises about linear functions? Click here to try Video Crash Courses called “Linear Functions”! Linear functions express straight lines. The straight lines can be gradual or steep, and they may go up (positive), go down (negative), or be horizontal. They are horizontal when the slope a=0. Formula The Formula for Linear Functions f(x)=a x+b, where a is the slope of the line and b is called the constant term, and is the y-coordinate where the graph intersects the y-axis. The slope a tells you how much the graph increases or decreases when the x-value increases by 1. Example 1 The function f(x)=−0.5 x+2 tells you that the graph intersects the y-axis in the point (0,2) and that the slope a=−0.5. That means that when x increases by 1, y decreases by 0.5. The graph is therefore sloping downwards from left to right. It looks like this: Rule Important Properties of the Linear Function The slope a tells you how much the graph increases/decreases when x increases by 1. If a>0, then the graph slopes upwards from left to right, and if a<0, it slopes downwards. The graph intersects the y-axis at the point b. The graph is a straight line with coordinates (x,y)=(x,f(x)). Example 2 Chelsea Clinton had a babysitter when she was a kid.Her father,Bill,was a bit stingy and decided to pay the babysitter$7.5 0 per hour plus$5 for travel to their home.Find a function that shows how much Bill had to pay the babysitter for x hours. You know that Bill has to pay $5 every time the babysitter comes to take care of Chelsea. You also know that he pays $7.5 0 per worked hour. The babysitter thus gets $7.5 0 for one hour, 7.5 0⋅$2=$1 5 for two hours, and so on. Bill pays thus 7.5 0 x dollars when the babysitter works x hours. The expression for how much Bill pays the babysitter looks like this: y=7.5 x+5 How much does Bill have to pay for the afternoon when he and Hillary have to go to a meeting at the White House and need a babysitter for 3 hours? Here x=3, so you get the following calculation: y=7.5⋅(3)+5=$2 7.5 0 The babysitter gets $2 7.5 0 for 3 hours of work. The babysitter planned to go to a U2 concert a few hours after she had to babysit.Tickets for the concert cost$3 5.She forgot her wallet at home,and hoped that she would earn enough money during the afternoon to cover the ticket.How late do Bill and Hillary have to be in order for the babysitter to have enough money for the concert? Here you know that the babysitter must earn at least $3 5 to be able to afford the ticket, meaning y=3 5. What you need to find is how long she must babysit— that is, a value for x. You put in the value for y and solve the equation for x: 3 5=7.5 x+5 3 5−5=7.5 x 3 0=7.5 x\|÷7.5 x=4 To be able to afford the concert, she had to babysit for four hours. So she hoped that Bill and Hillary were 4−3=1 hour late. If they were not, she would have to borrow money from a friend, or ask Bill for a pay rise. You need two points to find the function of a line. If you know two points that are on the graph, you can use the two formulas below to find the slope a and the constant term b. The symbol Δ (delta) is a Greek letter. In mathematics you use it to describe change or difference. This means that you read the formula below as “a is equal to the change in y divided by the change in x”. Formula How to Find a Function From Two Points The line through the points (x 1,y 1) and (x 2,y 2) has the slope a=Δ y Δ x=y 2−y 1 x 2−x 1, where y 1=f(x 1) and y 2=f(x 2). Note! Study the figure closely! To find the constant term b, you need to use the following formula: b=y 1−a x 1. Example 3 Find the slope of the line that goes through the points(5,2)and(3,6),and its intersection with the y-axis You let (x 1,y 1)=(3,6) and (x 2,y 2)=(5,2). You would get the same result if you swap (x 1,y 1) and (x 2,y 2). By inserting those values into the formula, you get a=y 2−y 1 x 2−x 1=2−6 5−3=−4 2=−2. You now know that your line decreases by 2 when you move one place to the right. Let’s see what the intersection with the y-axis is: b=y 1−a x 1=6−(−2)⋅3=6+6=1 2 b=y 1−a x 1=6−(−2)⋅3=6+6=1 2 The intersection with the y-axis is then (0,1 2). Next entry How Are Linear Functions Used in Real Life? 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17524	https://brainly.com/question/43052108	[FREE] An operation is defined on the set of real numbers by x y = x + y - 2xy. If the identity element is 0, - brainly.com 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 +21,6k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +12,2k Ace exams faster, with practice that adapts to you Practice Worksheets +7,3k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified An operation is defined on the set of real numbers by x∗y=x+y−2 x y. If the identity element is 0, find the inverse of the element p under . 1 See answer Explain with Learning Companion NEW Asked by Dou10 • 11/19/2023 0:02 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 16002187 people 16M 0.0 0 Upload your school material for a more relevant answer To find the inverse of element p under the operation , we set up the equation p p' = 0, solve for p' and find that the inverse p' equals -p / (1 - 2p), provided that p ≠ 1/2. Explanation The student is asking to find the inverse of the element __p_ under the operation where _x_ y is defined as _x+y-2xy_ and the identity element is 0. To find the inverse, we want to find a real number _p'_ such that _p_ p'_ results in the identity element 0. By the definition of the operation, we have: p p' = p + p' - 2pp' = 0. We can solve this equation for p' to find the inverse. Rearrange the terms to isolate p': p' - 2 pp' = -p p'(1 - 2 p) = -p Divide both sides by (1 - 2 p) assuming p ≠ 1/2 (since division by zero is not allowed): p' = -p / (1 - 2 p) Therefore, the inverse of the element p under the operation is _p'_ = -_p / (1 - 2 p_). Learn more about Inverse of an Element here: brainly.com/question/37431611 SPJ1 Answered by JoeEthan •49.4K answers•16M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 16002187 people 16M 0.0 0 Intermediate Microeconomics with Excel - Humberto Barreto Essential Graduate Physics - Quantum Mechanics - Konstantin K. Likharev Advanced Quantum Mechanics - Pieter Kok Upload your school material for a more relevant answer The inverse of the element p under the operation ∗ is given by the formula p′=1−2 p−p, provided that p=2 1. This means that for any valid p, we can find its corresponding inverse using this formula. The process involves setting the operation with its inverse equal to the identity element, which is 0. Explanation To find the inverse of the element p under the operation ∗ defined by x∗y=x+y−2 x y, we need to identify a number p′ such that p∗p′=0, where 0 is the identity element. This means we have: p∗p′=p+p′−2 p p′=0. Rearranging this equation gives us: p′−2 p p′=−p. We can factor out p′ on the left side: p′(1−2 p)=−p. Assuming p=2 1 (to avoid division by zero), we divide both sides by 1−2 p to isolate p′: p′=1−2 p−p. Thus, the inverse of the element p under the operation ∗ is: p′=1−2 p−p. This formula allows us to find the inverse for any real number p, as long as p is not equal to 2 1. Examples & Evidence For example, if p=1, then the inverse would be p′=1−2⋅1−1=−1−1=1. If p=0.5, we cannot use the formula, as it leads to division by zero, indicating that 0.5 does not have an inverse under this operation. This calculation is correct as it stems from the definition of the operation, and the logic follows standard mathematical procedures for finding inverses in operations defined on sets. Thanks 0 0.0 (0 votes) Advertisement Dou10 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 5.0 1 The operator V on the set R of real numbers is defined by: xVy = 7xy, for x, y ER. (a) Find under the operation V: (i.) the identity element; (ii.) the inverse of the element a E R. (b) Does every element a E R have an inverse? Community Answer Let be an operation on the set R of real numbers elementh defined by ab = a+b for all a,b £R Find the identity element. Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 New questions in Mathematics Evaluate tan(tan−1(−1)+3 π) Use the Leading Coefficient Test to determine the end behavior of the graph of the given polynomial function. f(x)=−3 x 4+2 x 3−7 x+5 A. The graph of f(x) rises to the left and rises to the right. B. The graph of f(x) rises to the left and falls to the right. C. The graph of f(x) falls to the left and falls to the right. D. The graph of f(x) falls to the left and rises to the right. Which of the following geometric objects occupy two dimensions? A. Ray B. Plane C. Cube D. Line E. Triangle F. Pyramid A group of students is given a 10 by 10 grid to cut into individual unit squares. The challenge is to create two squares using all of the unit squares. Their teacher states that after the two new squares are formed, one should have a side length two units greater than the other. Which equation represents x, the side length of the greater square? A. x 2+2 x 2=10 B. x 2+(x−2)2=100 C. x 2+2 x 2=100 D. $x^2+(x-2)^2=10 Find the area bounded by the lines y=2 3x−4 and y=−2 x+7 and the x axis. 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17526	https://www.uptodate.com/contents/society-guideline-links-adult-with-chest-pain-in-the-emergency-department	Society guideline links: Adult with chest pain in the emergency department - UpToDate Subscribe Sign in English Deutsch Español 日本語 Português Why UpToDate? Product Editorial Subscription Options How can UpToDate help you?Select the option that best describes you Medical Professional Resident, Fellow, or Student Hospital or Institution Group Practice Subscribe Society guideline links: Adult with chest pain in the emergency department View Topic Share Font Size Small Normal Large Bookmark Rate Feedback Tools Formulary drug information for this topic No drug references linked in this topic. Find in topic Formulary Print Share Feedback Font Size Small Normal Large Outline Introduction Canada United States Europe United Kingdom RELATED TOPICS Society guideline links: Aortic dissection and other acute aortic syndromes Society guideline links: Chronic coronary syndrome Society guideline links: Community-acquired pneumonia in adults Society guideline links: Deep sternal wound infection Society guideline links: Dyspnea Society guideline links: Non-ST-elevation acute coronary syndromes (non-ST-elevation myocardial infarction) Society guideline links: Pericardial disease Society guideline links: Pneumothorax Society guideline links: ST-elevation myocardial infarction (STEMI) Society guideline links: Venous thromboembolism (VTE) Society guideline links: Adult with chest pain in the emergency department Introduction This topic includes links to society and government-sponsored guidelines from selected countries and regions around the world. 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17527	https://aplusphysics.com/community/index.php?/blogs/entry/874-equation-dump-for-magnetism/	Equation Dump for Magnetism! - Blog bdavis - APlusPhysics Community Jump to content Blog bdavis This Entry Everywhere This Blog This Entry Status Updates Pages Topics Files Blog Entries Events Products Members Forums Blogs Books AP Physics 1 Essentials AP Physics 2 Essentials The AP Physics C Companion - Mechanics Regents Physics Essentials Regents Q&A Book Honors Physics Essentials iPad Honors Physics Downloads Activity All Activity My Activity Streams stream_title_9 stream_title_10 stream_title_11 stream_title_12 stream_title_13 Unread Content Content I Started Search More More Leaderboard Existing user? Sign In Sign In - [x] Remember me Not recommended on shared computers Sign In Forgot your password? Or sign in with one of these services Sign in with Google Sign in with Facebook Sign in with X Sign Up #2f4979 004d65 006547 3a751d 856716 853616 82191a 811f4d 5d116d 3c4859 All Activity Home Blog General Blog bdavis Equation Dump for Magnetism! Facebook Twitter Youtube Instagram Blog bdavis A blog by bdavis in General Followers 0 entries 42 comments 32 views 12,148 Equation Dump for Magnetism! Entry posted by bdavisApril 1, 2013 5,073 views Share More sharing options... Followers 0 F=qVXB F=ILXB Motion of Point Charges: A particle of mass m and charge q moving with speed v in a plane perpendicular to a uniform magnetic field moves in a circular orbit. The period and frequency of this circular motion are independent of the radius of the orbit and of the speed of the particle. Newton's 2nd law: qvB=m((v^2)/r) Cyclotron period: T= 2(pi)m/(qB) Cyclotron Frequency: f= 1/T = (qB)/(2(pi)m) Velocity Selector: consists of corssed electric and magnetic fields so that the electric and magnetic forces balance for a partivle moving with speed v. E=vB Mass Spectrometer: The mass-to-charge ratio of an ion of known speed can be determined by measuring the radius of the circular path taken by the ion in a known magnetic field. Magnetic dipole moment: u= NIAn Torque= t= u x B Potential energy of a magnetic dipole: U= -u . B Net force on a current loop in a uniform magnetic field is 0. Biot Savart Law: B= (u_0)Lxr/(4(pi)r^2) Magnetic field lines: the magnetic field is indicated by lines parallel to B at any point whose density is proportional to the magnitude of N. Magnetic lines do not begin or end at any point in space. Instead, they form continuous loops. Gauss's Law for magnetism: Net flux= integral over the closed surface( BdA)= 0 Ampere's Law: Integral over the closed surface (B . dl)= (u_0)I_perm These are the main and basic laws and concepts of magnetism. Other equations can be derived for different objects with current flowing through them and deriving them helps to gain a better understanding of the relationships between different objects and their magnetic fields when current is induced. Share More sharing options... Followers 0 Previous entry maximizing pitching speed Next entry The problem with u (mew) 0 Comments Recommended Comments There are no comments to display. Add a comment... × Pasted as rich text. Paste as plain text instead Only 75 emoji are allowed. × Your link has been automatically embedded. Display as a link instead × Your previous content has been restored. Clear editor × You cannot paste images directly. Upload or insert images from URL. 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17528	https://fred.stlouisfed.org/tags/series?t=discount%3Brate	Rate, Discount - Economic Data Series \| FRED \| St. Louis Fed Skip to main content Federal Reserve Economic Data Explore Our Apps Image 4 ### FRED Tools and resources to find and use economic data worldwide Image 5 ### FRASER U.S. financial, economic, and banking history Image 6 ### ALFRED Vintages of economic data from specific dates in history Image 7 ### CASSIDI View banking market concentrations and perform HHI analysis Release Calendar Tools FRED Add-in for Excel FRED API FRED Mobile Apps News Blog About What is FRED Tutorials Digital Badges Contact Us My Account Explore Our Apps Explore Our Apps Image 9 ### FRED Tools and resources to find and use economic data worldwide Image 10 ### FRASER U.S. financial, economic, and banking history Image 11 ### ALFRED Vintages of economic data from specific dates in history Image 12 ### CASSIDI View banking market concentrations and perform HHI analysis STL Fed Home Page Release Calendar Tools FRED Add-in for Excel FRED API FRED Mobile Apps News Blog About What is FRED Tutorials Digital Badges Contact Us Home>Tags 81 Series Skip to main content With Tags: Discount Rate Without Tag: None Selected Clear All Tags Filter by: View All Concepts View All Discontinued(70)Interest(41)Interest Rate(41)Average(40)Commercial(32)Commercial Paper(32)Fees(24)Mortgage(24)Origination(24)Mortgage Discount Points(24) Geography Types View All Nation(61)Freddie Mac Region(20) Geographies View All United States of America(77)North Central Freddie Mac Region(4)Northeast Freddie Mac Region(4)Southeast Freddie Mac Region(4)Southwest Freddie Mac Region(4)United Kingdom(4)West Freddie Mac Region(4) Frequencies View All Weekly(29)Daily(25)Annual(14)Monthly(11)Quarterly(2) Sources View All Board of Governors(52)Freddie Mac(24)Bank of England(4)International Monetary Fund(1) Releases View All H.15 Selected Interest Rates(40)International Financial Statistics(1) Citation & Copyright View 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Offering Rate, Quoted on a Discount Basis (DISCONTINUED) Percent, Not Seasonally Adjusted [x] Monthly Apr 1971 to Aug 1997 (2022-06-03) [x] Annual 1971 to 1997 (2022-06-03) [x] Daily 1971-04-08 to 1997-08-29 (2022-06-03) Treasury Bill Discount Rate in the United Kingdom Percent per Annum, Not Seasonally Adjusted [x] Monthly Jan 1923 to Jan 2017 (2017-06-09) [x] Quarterly Q1 1923 to Q4 2016 (2017-06-09) Average Dealer Offering Rate on 6-Month Domestic Private Bankers' Acceptances, Quoted on a Discount Basis (DISCONTINUED) Percent, Not Seasonally Adjusted [x] Annual 1976 to 2000 (2022-06-03) [x] Daily 1976-01-02 to 2000-06-30 (2022-06-03) [x] Monthly Jan 1976 to Jun 2000 (2022-06-03) Average Dealer Offering Rate on 3-Month Domestic Private Bankers' Acceptances, Quoted on a Discount Basis (DISCONTINUED) Percent, Not Seasonally Adjusted [x] Annual 1965 to 2000 (2022-06-03) [x] Weekly 1946-01-04 to 2000-06-30 (2022-06-03) [x] Daily 1965-02-04 to 2000-06-30 (2022-06-03) [x] Monthly Feb 1965 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17529	https://www.chegg.com/homework-help/questions-and-answers/overall-formation-constant-b2-ag-nh3-2-complex-16x10-000040-mol-agnos-00010-mol-nh3-dissol-q31319726	Solved The overall formation constant, B2, for the Ag(NH3)2+ \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Chemistry Chemistry questions and answers The overall formation constant, B2, for the Ag(NH3)2+ complex is 1.6x10 0.00040 mol of AgNOs and 0.0010 mol of NH3 are dissolved in water to giv a final volume of 2.00 L, determine the concentration of Ag ion at equilibrium. Include effects of ionic strength and solve by the method of successive approximations. Assume that NHs(aq) is undissociated. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: The overall formation constant, B2, for the Ag(NH3)2+ complex is 1.6x10 0.00040 mol of AgNOs and 0.0010 mol of NH3 are dissolved in water to giv a final volume of 2.00 L, determine the concentration of Ag ion at equilibrium. Include effects of ionic strength and solve by the method of successive approximations. Assume that NHs(aq) is undissociated. Show transcribed image text Here’s the best way to solve it.Solution Share Share Share done loading Copy link View the full answer Previous questionNext question Transcribed image text: The overall formation constant, B2, for the Ag(NH3)2+ complex is 1.6x10 0.00040 mol of AgNOs and 0.0010 mol of NH3 are dissolved in water to giv a final volume of 2.00 L, determine the concentration of Ag ion at equilibrium. Include effects of ionic strength and solve by the method of successive approximations. Assume that NHs(aq) is undissociated. Not the question you’re looking for? Post any question and get expert help quickly. 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17530	https://teksguide.org/resource/92-mechanical-energy-and-conservation-energy	Skip to main content 9.2 Mechanical Energy and Conservation of Energy Learning Objectives Mechanical Energy and Conservation of Energy Calculations involving Mechanical Energy and Conservation of Energy Practice Problems Check Your Understanding Learning Objectives Learning Objectives By the end of this section, you will be able to do the following: Explain the law of conservation of energy in terms of kinetic and potential energy Perform calculations related to kinetic and potential energy. Apply the law of conservation of energy Section Key Term \| law of conservation of energy \| Mechanical Energy and Conservation of Energy Mechanical Energy and Conservation of Energy We saw earlier that mechanical energy can be either potential or kinetic. In this section we will see how energy is transformed from one of these forms to the other. We will also see that, in a closed system, the sum of these forms of energy remains constant. Quite a bit of potential energy is gained by a roller coaster car and its passengers when they are raised to the top of the first hill. Remember that the potential part of the term means that energy has been stored and can be used at another time. You will see that this stored energy can either be used to do work or can be transformed into kinetic energy. For example, when an object that has gravitational potential energy falls, its energy is converted to kinetic energy. Remember that both work and energy are expressed in joules. Refer back to Figure 9.3. The amount of work required to raise the TV from point A to point B is equal to the amount of gravitational potential energy the TV gains from its height above the ground. This is generally true for any object raised above the ground. If all the work done on an object is used to raise the object above the ground, the amount work equals the object’s gain in gravitational potential energy. However, note that because of the work done by friction, these energy–work transformations are never perfect. Friction causes the loss of some useful energy. In the discussions to follow, we will use the approximation that transformations are frictionless. Now, let’s look at the roller coaster in Figure 9.6. Work was done on the roller coaster to get it to the top of the first rise; at this point, the roller coaster has gravitational potential energy. It is moving slowly, so it also has a small amount of kinetic energy. As the car descends the first slope, its PE is converted to KE. At the low point much of the original PE has been transformed to KE, and speed is at a maximum. As the car moves up the next slope, some of the KE is transformed back into PE and the car slows down. Figure 9.6 During this roller coaster ride, there are conversions between potential and kinetic energy. Virtual Physics Energy Skate Park Basics This simulation shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KE and PE by clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. Figure 9.7 Click here for the simulation Grasp Check The bar graphs show how KE and PE are transformed back and forth. Which statement best explains what happens to the mechanical energy of the system as speed is increasing? The mechanical energy of the system increases, provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. The mechanical energy of the system increases provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. On an actual roller coaster, there are many ups and downs, and each of these is accompanied by transitions between kinetic and potential energy. Assume that no energy is lost to friction. At any point in the ride, the total mechanical energy is the same, and it is equal to the energy the car had at the top of the first rise. This is a result of the law of conservation of energy, which says that, in a closed system, total energy is conserved—that is, it is constant. Using subscripts 1 and 2 to represent initial and final energy, this law is expressed as KE1+PE1=KE2+PE2. Either side equals the total mechanical energy. The phrase in a closed system means we are assuming no energy is lost to the surroundings due to friction and air resistance. If we are making calculations on dense falling objects, this is a good assumption. For the roller coaster, this assumption introduces some inaccuracy to the calculation. Calculations involving Mechanical Energy and Conservation of Energy Calculations involving Mechanical Energy and Conservation of Energy Tips For Success When calculating work or energy, use units of meters for distance, newtons for force, kilograms for mass, and seconds for time. This will assure that the result is expressed in joules. Watch Physics Conservation of Energy This video discusses conversion of PE to KE and conservation of energy. The scenario is very similar to the roller coaster and the skate park. It is also a good explanation of the energy changes studied in the snap lab. Click to view content Grasp Check Did you expect the speed at the bottom of the slope to be the same as when the object fell straight down? Which statement best explains why this is not exactly the case in real-life situations? The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. Worked Example Applying the Law of Conservation of Energy A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m? Strategy Choose the equation. 9.6 KE1+PE1=KE2+PE2 9.7 KE=12mv2; PE=mgh 9.8 12mv21+mgh1=12mv22+mgh2 List the knowns. m = 10 kg, v1 = 0, g = 9.80 9.9 ms2, h1 = 20 m, h2 = 10 m Identify the unknowns. KE2 and PE2 Substitute the known values into the equation and solve for the unknown variables. Solution 9.10 PE2=mgh2=10(9.80)10=980 J 9.11 KE2=PE2−(KE1+PE1)=980−{[0−[10(9.80)20]]}=980 J Discussion Alternatively, conservation of energy equation could be solved for v2 and KE2 could be calculated. Note that m could also be eliminated. Tips For Success Note that we can solve many problems involving conversion between KE and PE without knowing the mass of the object in question. This is because kinetic and potential energy are both proportional to the mass of the object. In a situation where KE = PE, we know that mgh = (1/2)mv2. Dividing both sides by m and rearranging, we have the relationship 2gh = v2. Practice Problems Practice Problems A child slides down a playground slide. If the slide is 3 m high and the child weighs 300 N, how much potential energy does the child have at the top of the slide? (Round g to 10 m/s2. ) 0 J 100 J 300 J 900 J A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to the ground, converting all of its PE to kinetic energy. What is the velocity of the apple just before it hits the ground? 0 m/s 2 m/s 10 m/s 50 m/s Snap Lab Converting Potential Energy to Kinetic Energy In this activity, you will calculate the potential energy of an object and predict the object’s speed when all that potential energy has been converted to kinetic energy. You will then check your prediction. You will be dropping objects from a height. Be sure to stay a safe distance from the edge. Don’t lean over the railing too far. Make sure that you do not drop objects into an area where people or vehicles pass by. Make sure that dropping objects will not cause damage. You will need the following: Materials for each pair of students: Four marbles (or similar small, dense objects) Stopwatch Materials for class: Metric measuring tape long enough to measure the chosen height A scale Instructions Procedure Work with a partner. Find and record the mass of four small, dense objects per group. Choose a location where the objects can be safely dropped from a height of at least 15 meters. A bridge over water with a safe pedestrian walkway will work well. Measure the distance the object will fall. Calculate the potential energy of the object before you drop it using PE = mgh = (9.80)mh. Predict the kinetic energy and velocity of the object when it lands using PE = KE and so, mgh=mv22; v=2(9.80)h−−−−−−−√=4.43h−−√. One partner drops the object while the other measures the time it takes to fall. Take turns being the dropper and the timer until you have made four measurements. Average your drop multiplied by and calculate the velocity of the object when it landed using v = at = gt = (9.80)t. Compare your results to your prediction. Grasp Check Galileo’s experiments proved that, contrary to popular belief, heavy objects do not fall faster than light objects. How do the equations you used support this fact? Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. Check Your Understanding Check Your Understanding Exercise 3 Describe the transformation between forms of mechanical energy that is happening to a falling skydiver before his parachute opens. Kinetic energy is being transformed into potential energy. Potential energy is being transformed into kinetic energy. Work is being transformed into kinetic energy. Kinetic energy is being transformed into work. Exercise 4 True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. True False Exercise 5 Identify equivalent terms for stored energy and energy of motion. Stored energy is potential energy, and energy of motion is kinetic energy. Energy of motion is potential energy, and stored energy is kinetic energy. Stored energy is the potential as well as the kinetic energy of the system. Energy of motion is the potential as well as the kinetic energy of the system. Print Share Copy and paste the link code above.
17531	https://www.merckmanuals.com/professional/pediatrics/perinatal-problems/small-for-gestational-age-sga-infant	honeypot link IN THIS TOPIC Etiology Symptoms and Signs Treatment Prognosis Key Points OTHER TOPICS IN THIS CHAPTER Gestational Age Growth Parameters in Neonates Neonatal Resuscitation Birth Injuries Hypothermia in Neonates Large-for-Gestational-Age (LGA) Infant Small-for-Gestational-Age (SGA) Infant Postterm Infants Preterm Infants Small-for-Gestational-Age (SGA) Infant (Intrauterine Growth Restriction) By Arcangela Lattari Balest, MD, University of Pittsburgh, School of Medicine Reviewed By Alicia R. Pekarsky, MD, State University of New York Upstate Medical University, Upstate Golisano Children's Hospital Reviewed/Revised Modified Feb 2025 v1087269 View Patient Education Infants whose weight is < the 10th percentile for gestational age are classified as small for gestational age. Complications include perinatal asphyxia, meconium aspiration, polycythemia, and hypoglycemia. Etiology \| Symptoms and Signs \| Treatment \| Prognosis \| Key Points \| Topic Resources Fenton Growth Chart for Preterm... Fenton Growth Chart for Preterm... The standard definition of neonatal gestational age is the number of weeks from the first day of a pregnant patient's last normal menstrual period to the date of delivery. Defining gestational age by the last menstrual period is the universal standard used by obstetricians and neonatologists for discussing fetal maturation, but it is not an accurate measure of weeks of fetal development. This is because ovulation and conception occur in the middle of the menstrual cycle, thus the gestational age is counted beginning approximately 2 weeks before conception. Also, determining gestational age based on the last menstrual period may be even more inaccurate if a pregnant patient has irregular menses. The Fenton growth charts provide a more precise assessment of growth vs gestational age; there are separate charts for boys and girls. Fenton Growth Chart for Preterm Boys \| \| \| Fenton T, Kim J : A systematic review and meta-analysis to revise the Fenton growth chart for preterm infants. BMC Pediatrics 13:59, 2013. doi: 10.1186/1471-2431-13-59; used with permission. \| Fenton Growth Chart for Preterm Girls \| \| \| Fenton T, Kim J : A systematic review and meta-analysis to revise the Fenton growth chart for preterm infants. BMC Pediatrics 13:59, 2013. doi: 10.1186/1471-2431-13-59; used with permission. \| Etiology of SGA Infant Causes may be divided into those in which the growth restriction is Symmetric: Height, weight, and head circumference are about equally affected. Asymmetric: Weight is most affected, with a relative sparing of growth of the brain, cranium, and long bones. Symmetric growth restriction usually results from a fetal problem that begins early in gestation, often during the first trimester. When the cause begins relatively early in gestation, the entire body is affected, resulting in fewer cells of all types. Common causes include Genetic disorders, which may include central nervous system , cardiac , or renal anomalies First-trimester congenital infections (eg, with cytomegalovirus , rubella virus , or Toxoplasma gondii ) Asymmetric growth restriction usually results from placental or maternal problems that typically manifest in the late second or the third trimester. When the cause begins relatively late in gestation, organs and tissues are not equally affected, resulting in asymmetric growth restriction. Common causes include Placental insufficiency resulting from maternal disease involving the small blood vessels (eg, preeclampsia , hypertension , renal disease , antiphospholipid antibody syndrome , long-standing diabetes ) Relative placental insufficiency caused by multiple gestation Placental involution accompanying postmaturity Chronic maternal hypoxemia caused by pulmonary disease or cardiac disease Maternal malnutrition Conception using assisted reproductive technology Age (adolescent or over age 35) Previous SGA delivery An infant may also have asymmetric growth restriction and be small for gestational age (SGA) if the mother is a heavy user of opioids, cocaine , alcohol, and/or tobacco during pregnancy (see An infant may also have asymmetric growth restriction and be small for gestational age (SGA) if the mother is a heavy user of opioids, cocaine, alcohol, and/or tobacco during pregnancy (see Social and Illicit Drugs During Pregnancy ). Many SGA infants are healthy but just constitutionally small, and not all infants whose growth was restricted in utero are SGA (ie, weight is < the 10th percentile for gestational age). Symptoms and Signs of SGA Infant Despite their size, SGA infants have physical characteristics (eg, skin appearance, ear cartilage, sole creases) and behavior (eg, alertness, spontaneous activity, zest for feeding) similar to those of infants of like gestational age who are typical in size. However, they may appear thin with decreased muscle mass and subcutaneous fat tissue. Facial features may appear sunken, resembling those of an elderly person ("wizened facies"). The umbilical cord can appear thin and small. Complications Full-term SGA infants do not have the complications related to organ system immaturity that preterm infants of similar size have. They are, however, at risk of Perinatal asphyxia Meconium aspiration Hypoglycemia Polycythemia Hypothermia Increased risk of infection Perinatal asphyxia during labor is the most serious potential complication. It is a risk if intrauterine growth restriction is caused by placental insufficiency (with marginally adequate placental perfusion) because each uterine contraction slows or stops maternal placental perfusion by compressing the spiral arteries. Therefore, when placental insufficiency is suspected, the fetus should be assessed before labor and the fetal heart rate should be monitored during labor. If fetal compromise is detected, rapid delivery, often by cesarean delivery , is indicated. Meconium aspiration may occur during perinatal asphyxia. SGA infants, especially those who are postterm, may pass meconium into the amniotic sac and begin deep gasping movements. The consequent aspiration may result in meconium aspiration syndrome . Meconium aspiration syndrome is often most severe in growth-restricted or postterm infants because the meconium is contained in a smaller volume of amniotic fluid and thus more concentrated. Hypoglycemia often occurs in the early hours and days of life because of a lack of adequate glycogen synthesis and thus decreased glycogen stores and must be treated quickly with IV glucose. Polycythemia may occur when SGA fetuses have chronic mild hypoxia caused by placental insufficiency. Erythropoietin release is increased, leading to an increased rate of erythrocyte production. The neonate with polycythemia at birth appears ruddy and may be tachypneic or lethargic. This ruddy appearance can still be visible in neonates with dark skin, but it may be less obvious than in neonates with light skin. Hypothermia may occur because of impaired thermoregulation, which involves multiple factors including increased heat loss due to the decrease in subcutaneous fat, decreased heat production due to intrauterine stress and depletion of nutrient stores, and increased surface to volume ratio because of small size. SGA infants should be in a thermoneutral environment to minimize oxygen consumption. Infection risk is increased in SGA neonates because they may have an impaired immune system, which increases their risk of developing infections while in the hospital. Treatment of SGA Infant Supportive care Underlying conditions and complications are treated. There is no specific intervention for the SGA state, but prevention is aided by prenatal advice on the importance of avoiding alcohol, tobacco, and illicit drugs. Growth hormone is sometimes given to certain SGA infants who remain quite small at 2 to 4 years of age. Hormone therapy must be given for several years and must be considered on a case-by-case basis ( 1 ). Treatment reference Ferrigno R, Savanelli MC, Cioffi D, Pellino V, Klain A . Auxological and metabolic effects of long-term treatment with recombinant growth hormone in children born small for gestational age: a retrospective study. Endocrine . 2024;84(1):213-222. doi:10.1007/s12020-023-03665-4 Prognosis for SGA Infant If asphyxia can be avoided, neurologic prognosis for term SGA infants is quite good. However, later in life there is probably increased risk of ischemic heart disease, hypertension, and stroke, which are thought to be caused by abnormal vascular development. If intrauterine growth restriction is caused by chronic placental insufficiency, adequate nutrition may allow SGA infants to demonstrate remarkable “catch-up” growth after delivery. Infants who are SGA because of genetic factors, congenital infection, or maternal substance use often have a worse prognosis, depending on the specific diagnosis. Key Points Infants whose weight is < the 10th percentile for gestational age are small for gestational age (SGA). Disorders early in gestation cause symmetric growth restriction, in which height, weight, and head circumference are about equally affected. Disorders late in gestation cause asymmetric growth restriction, in which weight is most affected, with relatively normal growth of the brain, cranium, and long bones. Although small, SGA infants do not have the complications related to organ system immaturity that preterm infants of similar size have. Complications are mainly those of the underlying cause but generally also include perinatal asphyxia, meconium aspiration, hypoglycemia, polycythemia, and hypothermia. Drugs Mentioned In This Article Test your Knowledge Take a Quiz! 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17532	https://www.jacksonsd.org/cms/lib/NJ01912744/Centricity/Domain/560/SE%206.7A.pdf	Section 6.7 Recursively Defi ned Sequences 339 6.7 Essential Question Essential Question How can you defi ne a sequence recursively? A recursive rule gives the beginning term(s) of a sequence and a recursive equation that tells how an is related to one or more preceding terms. Describing a Pattern Work with a partner. Consider a hypothetical population of rabbits. Start with one breeding pair. After each month, each breeding pair produces another breeding pair. The total number of rabbits each month follows the exponential pattern 2, 4, 8, 16, 32, . . .. Now suppose that in the fi rst month after each pair is born, the pair is too young to reproduce. Each pair produces another pair after it is 2 months old. Find the total number of pairs in months 6, 7, and 8. 1 2 3 4 5 1 Month Number of pairs 1 2 3 5 Red pair produces green pair. Red pair produces orange pair. Red pair produces blue pair. Red pair is too young to produce. Blue pair produces purple pair. Using a Recursive Equation Work with a partner. Consider the following recursive equation. an = an − 1 + an − 2 Each term in the sequence is the sum of the two preceding terms. Copy and complete the table. Compare the results with the sequence of the number of pairs in Exploration 1. a1 a2 a3 a4 a5 a6 a7 a8 1 1 Communicate Your Answer Communicate Your Answer 3. How can you defi ne a sequence recursively? 4. Use the Internet or some other reference to determine the mathematician who fi rst described the sequences in Explorations 1 and 2. RECOGNIZING PATTERNS To be profi cient in math, you need to look closely to discern a pattern or structure. Recursively Defi ned Sequences 340 Chapter 6 Exponential Functions and Sequences 6.7 Lesson What You Will Learn What You Will Learn Write terms of recursively defi ned sequences. Write recursive rules for sequences. Translate between recursive rules and explicit rules. Write recursive rules for special sequences. Writing Terms of Recursively Defi ned Sequences So far in this book, you have defi ned arithmetic and geometric sequences explicitly. An explicit rule gives an as a function of the term’s position number n in the sequence. For example, an explicit rule for the arithmetic sequence 3, 5, 7, 9, . . . is an = 3 + 2(n − 1), or an = 2n + 1. Now, you will defi ne arithmetic and geometric sequences recursively. A recursive rule gives the beginning term(s) of a sequence and a recursive equation that tells how an is related to one or more preceding terms. Writing Terms of Recursively Defi ned Sequences Write the fi rst six terms of each sequence. Then graph each sequence. a. a1 = 2, an = an − 1 + 3 b. a1 = 1, an = 3an − 1 SOLUTION You are given the fi rst term. Use the recursive equation to fi nd the next fi ve terms. a. a1 = 2 b. a1 = 1 a2 = a1 + 3 = 2 + 3 = 5 a2 = 3a1 = 3(1) = 3 a3 = a2 + 3 = 5 + 3 = 8 a3 = 3a2 = 3(3) = 9 a4 = a3 + 3 = 8 + 3 = 11 a4 = 3a3 = 3(9) = 27 a5 = a4 + 3 = 11 + 3 = 14 a5 = 3a4 = 3(27) = 81 a6 = a5 + 3 = 14 + 3 = 17 a6 = 3a5 = 3(81) = 243 n an 18 12 6 00 2 4 6 n an 240 160 80 00 2 4 6 STUDY TIP A sequence is a discrete function. So, the points on the graph are not connected. explicit rule, p. 340 recursive rule, p. 340 Previous arithmetic sequence geometric sequence Core Vocabulary Core Vocabulary Core Core Concept Concept Recursive Equation for an Arithmetic Sequence an = an − 1 + d, where d is the common difference Recursive Equation for a Geometric Sequence an = r ⋅ an − 1, where r is the common ratio Section 6.7 Recursively Defi ned Sequences 341 Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Write the fi rst six terms of the sequence. Then graph the sequence. 1. a1 = 0, an = an − 1 − 8 2. a1 = −7.5, an = an − 1 + 2.5 3. a1 = −36, an = 1 — 2 an − 1 4. a1 = 0.7, an = 10an − 1 Writing Recursive Rules Writing Recursive Rules Write a recursive rule for each sequence. a. −30, −18, −6, 6, 18, . . . b. 500, 100, 20, 4, 0.8, . . . SOLUTION Use a table to organize the terms and fi nd the pattern. a. Position, n 1 2 3 4 5 Term, an −30 −18 −6 6 18 + 12 + 12 + 12 + 12 The sequence is arithmetic, with fi rst term a1 = −30 and common difference d = 12. an = an − 1 + d Recursive equation for an arithmetic sequence an = an − 1 + 12 Substitute 12 for d. So, a recursive rule for the sequence is a1 = −30, an = an − 1 + 12. b. × 1 — 5 × 1 — 5 × 1 — 5 × 1 — 5 Position, n 1 2 3 4 5 Term, an 500 100 20 4 0.8 The sequence is geometric, with fi rst term a1 = 500 and common ratio r = 1 — 5 . an = r ⋅ an − 1 Recursive equation for a geometric sequence an = 1 — 5 an − 1 Substitute 1 — 5 for r. So, a recursive rule for the sequence is a1 = 500, an = 1 — 5 an − 1. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Write a recursive rule for the sequence. 5. 8, 3, −2, −7, −12, . . . 6. 1.3, 2.6, 3.9, 5.2, 6.5, . . . 7. 4, 20, 100, 500, 2500, . . . 8. 128, −32, 8, −2, 0.5, . . . 9. Write a recursive rule for the height of the sunfl ower over time. 4 months: 6.5 feet 3 months: 5 feet 2 months: 3.5 feet 1 month: 2 feet COMMON ERROR When writing a recursive rule for a sequence, you need to write both the beginning term(s) and the recursive equation. 342 Chapter 6 Exponential Functions and Sequences Translating between Recursive and Explicit Rules Translating from Recursive Rules to Explicit Rules Write an explicit rule for each recursive rule. a. a1 = 25, an = an − 1 − 10 b. a1 = 19.6, an = −0.5an − 1 SOLUTION a. The recursive rule represents an arithmetic sequence, with fi rst term a1 = 25 and common difference d = −10. an = a1 + (n − 1)d Explicit rule for an arithmetic sequence an = 25 + (n − 1)(−10) Substitute 25 for a1 and −10 for d. an = −10n + 35 Simplify. An explicit rule for the sequence is an = −10n + 35. b. The recursive rule represents a geometric sequence, with fi rst term a1 = 19.6 and common ratio r = −0.5. an = a1r n − 1 Explicit rule for a geometric sequence an = 19.6(−0.5)n − 1 Substitute 19.6 for a1 and −0.5 for r. An explicit rule for the sequence is an = 19.6(−0.5)n − 1. Translating from Explicit Rules to Recursive Rules Write a recursive rule for each explicit rule. a. an = −2n + 3 b. an = −3(2)n − 1 SOLUTION a. The explicit rule represents an arithmetic sequence, with fi rst term a1 = −2(1) + 3 = 1 and common difference d = −2. an = an − 1 + d Recursive equation for an arithmetic sequence an = an − 1 + (−2) Substitute −2 for d. So, a recursive rule for the sequence is a1 = 1, an = an − 1 − 2. b. The explicit rule represents a geometric sequence, with fi rst term a1 = −3 and common ratio r = 2. an = r ⋅ an − 1 Recursive equation for a geometric sequence an = 2an − 1 Substitute 2 for r. So, a recursive rule for the sequence is a1 = −3, an = 2an − 1. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Write an explicit rule for the recursive rule. 10. a1 = −45, an = an − 1 + 20 11. a1 = 13, an = −3an − 1 Write a recursive rule for the explicit rule. 12. an = −n + 1 13. an = −2.5(4)n − 1 Section 6.7 Recursively Defi ned Sequences 343 Writing Recursive Rules for Special Sequences You can write recursive rules for sequences that are neither arithmetic nor geometric. One way is to look for patterns in the sums of consecutive terms. Writing Recursive Rules for Other Sequences Use the sequence shown. 1, 1, 2, 3, 5, 8, . . . a. Write a recursive rule for the sequence. b. Write the next three terms of the sequence. SOLUTION a. Find the difference and ratio between each pair of consecutive terms. 1 1 2 3 1 − 1 = 0 2 − 1 = 1 3 − 2 = 1 There is no common difference, so the sequence is not arithmetic. 1 1 2 3 1 — 1 = 1 2 — 1 = 2 3 — 2 = 1 1 — 2 There is no common ratio, so the sequence is not geometric. Find the sum of each pair of consecutive terms. a1 + a2 = 1 + 1 = 2 2 is the third term. a2 + a3 = 1 + 2 = 3 3 is the fourth term. a3 + a4 = 2 + 3 = 5 5 is the fi fth term. a4 + a5 = 3 + 5 = 8 8 is the sixth term. Beginning with the third term, each term is the sum of the two previous terms. A recursive equation for the sequence is an = an − 2 + an − 1. So, a recursive rule for the sequence is a1 = 1, a2 = 1, an = an − 2 + an − 1. b. Use the recursive equation an = an − 2 + an − 1 to fi nd the next three terms. a7 = a5 + a6 a8 = a6 + a7 a9 = a7 + a8 = 5 + 8 = 8 + 13 = 13 + 21 = 13 = 21 = 34 The next three terms are 13, 21, and 34. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Write a recursive rule for the sequence. Then write the next three terms of the sequence. 14. 5, 6, 11, 17, 28, . . . 15. −3, −4, −7, −11, −18, . . . 16. 1, 1, 0, −1, −1, 0, 1, 1, . . . 17. 4, 3, 1, 2, −1, 3, −4, . . . The sequence in Example 5 is called the Fibonacci sequence. This pattern is naturally occurring in many objects, such as fl owers. U a b S a Th i E l 5 i 344 Chapter 6 Exponential Functions and Sequences Exercises 6.7 Dynamic Solutions available at BigIdeasMath.com 1. COMPLETE THE SENTENCE A recursive rule gives the beginning term(s) of a sequence and a(n) _______ that tells how an is related to one or more preceding terms. 2. WHICH ONE DOESN’T BELONG? Which rule does not belong with the other three? Explain your reasoning. a1 = 9, an = 4an − 1 a1 = −1, an = 5an − 1 a1 = −3, an = an − 1 + 1 an = 6n − 2 Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–6, determine whether the recursive rule represents an arithmetic sequence or a geometric sequence. 3. a1 = 2, an = 7an − 1 4. a1 = 18, an = an − 1 + 1 5. a1 = 5, an = an − 1 − 4 6. a1 = 3, an = −6an − 1 In Exercises 7–12, write the fi rst six terms of the sequence. Then graph the sequence. (See Example 1.) 7. a1 = 0, an = an − 1 + 2 8. a1 = 10, an = an − 1 − 5 9. a1 = 2, an = 3an − 1 10. a1 = 8, an = 1.5an − 1 11. a1 = 80, an = − 1 — 2 an − 1 12. a1 = −7, an = −4an − 1 In Exercises 13–20, write a recursive rule for the sequence. (See Example 2.) 13. n 1 2 3 4 an 7 16 25 34 14. n 1 2 3 4 an 8 24 72 216 15. 243, 81, 27, 9, 3, . . . 16. 3, 11, 19, 27, 35, . . . 17. 0, −3, −6, −9, −12, . . . 18. 5, −20, 80, −320, 1280, . . . 19. n an 0 −24 −48 (4, −64) (3, −16) 3 5 (1, −1) (2, −4) 20. n an 36 24 12 00 2 4 (1, 35) (2, 24) (3, 13) (4, 2) 21. MODELING WITH MATHEMATICS Write a recursive rule for the number of bacterial cells over time. 1 hour 2 hours 3 hours 4 hours 22. MODELING WITH MATHEMATICS Write a recursive rule for the length of the deer antler over time. 1 day: 4 in. 1 2 2 days: 4 in. 3 4 3 days: 5 in. 4 days: 5 in. 1 4 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Section 6.7 Recursively Defi ned Sequences 345 In Exercises 23–28, write an explicit rule for the recursive rule. (See Example 3.) 23. a1 = −3, an = an − 1 + 3 24. a1 = 8, an = an − 1 − 12 25. a1 = 16, an = 0.5an − 1 26. a1 = −2, an = 9an − 1 27. a1 = 4, an = an − 1 + 17 28. a1 = 5, an = −5an − 1 In Exercises 29–34, write a recursive rule for the explicit rule. (See Example 4.) 29. an = 7(3)n − 1 30. an = −4n + 2 31. an = 1.5n + 3 32. an = 6n − 20 33. an = (−5)n − 1 34. an = −81 ( 2 — 3 ) n − 1 In Exercises 35–38, graph the fi rst four terms of the sequence with the given description. Write a recursive rule and an explicit rule for the sequence. 35. The fi rst term of a sequence is 5. Each term of the sequence is 15 more than the preceding term. 36. The fi rst term of a sequence is 16. Each term of the sequence is half the preceding term. 37. The fi rst term of a sequence is −1. Each term of the sequence is −3 times the preceding term. 38. The fi rst term of a sequence is 19. Each term of the sequence is 13 less than the preceding term. In Exercises 39–44, write a recursive rule for the sequence. Then write the next two terms of the sequence. (See Example 5.) 39. 1, 3, 4, 7, 11, . . . 40. 10, 9, 1, 8, −7, 15, . . . 41. 2, 4, 2, −2, −4, −2, . . . 42. 6, 1, 7, 8, 15, 23, . . . 43. n an 30 20 10 00 2 4 (1, 1) (2, 3) (3, 3) (4, 9) (5, 27) 44. n an 60 40 20 00 2 4 (1, 64) (4, 4) (5, 1) (3, 4) (2, 16) 45. ERROR ANALYSIS Describe and correct the error in writing an explicit rule for the recursive rule a1 = 6, an = an − 1 − 12. an = a1 + (n − 1)d an = 6 + (n − 1)(12) an = 6 + 12n − 12 an = −6 + 12n ✗ 46. ERROR ANALYSIS Describe and correct the error in writing a recursive rule for the sequence 2, 4, 6, 10, 16, . . .. 2, 4, 6, . . . + 2 + 2 The sequence is arithmetic, with fi rst term a1 = 2 and common diff erence d = 2. an = an − 1 + d a1 = 2, an = an − 1 + 2 ✗ In Exercises 47–51, the function f represents a sequence. Find the 2nd, 5th, and 10th terms of the sequence. 47. f (1) = 3, f (n) = f (n − 1) + 7 48. f (1) = −1, f (n) = 6f (n − 1) 49. f (1) = 8, f (n) = −f (n − 1) 50. f (1) = 4, f (2) = 5, f (n) = f (n − 2) + f (n − 1) 51. f (1) = 10, f (2) = 15, f (n) = f (n − 1) − f (n − 2) 52. MODELING WITH MATHEMATICS The X-ray shows the lengths (in centimeters) of bones in a human hand. 2.5 3.5 6 9.5 a. Write a recursive rule for the lengths of the bones. b. Measure the lengths of different sections of your hand. Can the lengths be represented by a recursively defi ned sequence? Explain. 346 Chapter 6 Exponential Functions and Sequences 53. USING TOOLS You can use a spreadsheet to generate the terms of a sequence. A = A2 B 3 2 1 4 3 =A1+2 5 C a. To generate the terms of the sequence a1 = 3, an = an − 1 + 2, enter the value of a1, 3, into cell A1. Then enter “=A1+2” into cell A2, as shown. Use the fi ll down feature to generate the fi rst 10 terms of the sequence. b. Use a spreadsheet to generate the fi rst 10 terms of the sequence a1 = 3, an = 4an − 1. (Hint: Enter “=4A1” into cell A2.) c. Use a spreadsheet to generate the fi rst 10 terms of the sequence a1 = 4, a2 = 7, an = an − 1 − an − 2. (Hint: Enter “=A2-A1” into cell A3.) 54. HOW DO YOU SEE IT? Consider Squares 1–6 in the diagram. 1 2 4 5 6 3 a. Write a sequence in which each term an is the side length of square n. b. What is the name of this sequence? What is the next term of this sequence? c. Use the term in part (b) to add another square to the diagram and extend the spiral. 55. REASONING Write the fi rst 5 terms of the sequence a1 = 5, an = 3an − 1 + 4. Determine whether the sequence is arithmetic, geometric, or neither. Explain your reasoning. 56. THOUGHT PROVOKING Describe the pattern for the numbers in Pascal’s Triangle, shown below. Write a recursive rule that gives the mth number in the nth row. 1 2 6 1 3 10 1 4 1 5 1 1 1 3 10 1 4 1 5 1 1 57. REASONING The explicit rule an = a1 + (n − 1)d defi nes an arithmetic sequence. a. Explain why an − 1 = a1 + [(n − 1) − 1]d. b. Justify each step in showing that a recursive equation for the sequence is an = an − 1 + d. an = a1 + (n − 1)d = a1 + [(n − 1) + 0]d = a1 + [(n − 1) − 1 + 1]d = a1 + [((n − 1) − 1) + 1]d = a1 + [(n − 1) − 1]d + d = an − 1 + d 58. MAKING AN ARGUMENT Your friend claims that the sequence −5, 5, −5, 5, −5, . . . cannot be represented by a recursive rule. Is your friend correct? Explain. 59. PROBLEM SOLVING Write a recursive rule for the sequence. 3, 7, 15, 31, 63, . . . Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Simplify the expression. (Skills Review Handbook) 60. 5x + 12x 61. 9 − 6y −14 62. 2d − 7 − 8d 63. 3 − 3m + 11m Write a linear function f with the given values. (Section 4.2) 64. f (2) = 6, f (−1) = −3 65. f (−2) = 0, f (6) = −4 66. f (−3) = 5, f (−1) = 5 67. f (3) = −1, f (−4) = −15 Reviewing what you learned in previous grades and lessons
17533	https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/21%3A_The_p-Block_Elements/21.04%3A_The_Elements_of_Group_16_(The_Chalcogens)	Skip to main content 21.4: The Elements of Group 16 (The Chalcogens) Last updated : Jul 4, 2022 Save as PDF 21.3: The Elements of Group 15 (The Pnicogens) 21.5: The Elements of Group 17 (The Halogens) Page ID : 6511 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To understand the trends in properties and reactivity of the group 16 elements: the chalcogens. The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element. Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S8 in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO2 fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility. Group 16 is the first group in the p block with no stable metallic elements. Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO3, Ag2CO3, and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. Another important industrial use of oxygen is in the production of TiO2, which is commonly used as a white pigment in paints, paper, and plastics. Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices. Jöns Jakob Berzelius (1779–1848) Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium). The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium. Preparation and General Properties of the Group 16 Elements Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO2: 2KClO3(s)⇌ΔMnO2(s)2KCl(s)+3O2(g)(21.4.1) Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (Figure 21.4.1). Sulfur is also recovered from H2S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS2). Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH3)2Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.” With their ns2np4 electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in Table Figure 21.4.1. As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po2+ or Po4+ ion in aqueous solution, depending on conditions. Table 21.4.1: Selected Properties of the Group 16 Elements \| Property \| Oxygen \| Sulfur \| Selenium \| Tellurium \| Polonium \| \| The configuration shown does not include filled d and f subshells. \| \| \| \| \| \| \| †The values cited for the hexacations are for six-coordinate ions and are only estimated values. \| \| \| \| \| \| \| atomic mass (amu) \| 16.00 \| 32.07 \| 78.96 \| 127.60 \| 209 \| \| atomic number \| 8 \| 16 \| 34 \| 52 \| 84 \| \| atomic radius (pm) \| 48 \| 88 \| 103 \| 123 \| 135 \| \| atomic symbol \| O \| S \| Se \| Te \| Po \| \| density (g/cm3) at 25°C \| 1.31 (g/L) \| 2.07 \| 4.81 \| 6.24 \| 9.20 \| \| electron affinity (kJ/mol) \| −141 \| −200 \| −195 \| −190 \| −180 \| \| electronegativity \| 3.4 \| 2.6 \| 2.6 \| 2.1 \| 2.0 \| \| first ionization energy (kJ/mol) \| 1314 \| 1000 \| 941 \| 869 \| 812 \| \| ionic radius (pm)† \| 140 (−2) \| 184 (−2), 29 (+6) \| 198 (−2), 42 (+6) \| 221 (−2), 56 (+6) \| 230 (−2), 97 (+4) \| \| melting point/boiling point (°C) \| −219/−183 \| 115/445 \| 221/685 \| 450/988 \| 254/962 \| \| normal oxidation state(s) \| −2 \| +6, +4, −2 \| +6, +4, −2 \| +6, +4, −2 \| +2 (+4) \| \| product of reaction with H2 \| H2O \| H2S \| H2Se \| none \| none \| \| product of reaction with N2 \| NO, NO2 \| none \| none \| none \| none \| \| product of reaction with O2 \| — \| SO2 \| SeO2 \| TeO2 \| PoO2 \| \| product of reaction with X2 \| O2F2 \| SF6, S2Cl2, S2Br2 \| SeF6, SeX4 \| TeF6, TeX4 \| PoF4, PoCl2, PoBr2 \| \| standard reduction potential (E°, V) (E0 → H2E in acidic solution) \| +1.23 \| +0.14 \| −0.40 \| −0.79 \| −1.00 \| \| type of oxide \| — \| acidic \| acidic \| amphoteric \| basic \| \| valence electron configuration \| 2s22p4 \| 3s23p4 \| 4s24p4 \| 5s25p4 \| 6s26p4 \| Reactions and Compounds of Oxygen As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O2, all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O3), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H2O2) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure: 2H2O2(l)→2H2O(l)+O2(g) ΔGo=−119kJ/mol(21.4.2) As in groups 14 and 15, the lightest element in group 16 has the greatest tendency to form multiple bonds. Despite the strength of the O=O bond (DO2 = 494 kJ/mol), O2 is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O2 and related species, such as the peroxide and superoxide ions, are in Table 21.4.2. With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO4 is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens. Table 21.4.2: Some Properties of O2 and Related Diatomic Species \| Species \| Bond Order \| Number of Unpaired e− \| O–O Distance (pm) \| \| Source of data: Lauri Vaska, “Dioxygen-Metal Complexes: Toward a Unified View,” Accounts of Chemical Research 9 (1976): 175. \| \| \| \| \| O2+ \| 2.5 \| 1 \| 112 \| \| O2 \| 2 \| 2 \| 121 \| \| O2− \| 1.5 \| 1 \| 133 \| \| O22− \| 1 \| 0 \| 149 \| Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO2, are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (Eδ+–Oδ−). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid: H2O(l)+SO3(g)→H2SO4(aq)(21.4.3) The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect: Al2O3(s)+6H+(aq)→2Al3+(aq)+3H2O(l)(21.4.4) Al2O3(s)+2OH−(aq)+3H2O(l)→2Al(OH)−4(aq)(21.4.5) Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric. Example 21.4.1 For each reaction, explain why the given products form. Ga2O3(s) + 2OH−(aq) + 3H2O(l) → 2Ga(OH)4−(aq) 3H2O2(aq) + 2MnO4−(aq) + 2H+(aq) → 3O2(g) + 2MnO2(s) + 4H2O(l) KNO3(s) −→Δ KNO(s) + O2(g) Given: balanced chemical equations Asked for: why the given products form Strategy: Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form. Solution Gallium is a metal. We expect the oxides of metallic elements to be basic and therefore not to react with aqueous base. A close look at the periodic table, however, shows that gallium is close to the diagonal line of semimetals. Moreover, aluminum, the element immediately above gallium in group 13, is amphoteric. Consequently, we predict that gallium will behave like aluminum (Equation 21.4.5). Hydrogen peroxide is an oxidant that can accept two electrons per molecule to give two molecules of water. With a strong oxidant, however, H2O2 can also act as a reductant, losing two electrons (and two protons) to produce O2. Because the other reactant is permanganate, which is a potent oxidant, the only possible reaction is a redox reaction in which permanganate is the oxidant and hydrogen peroxide is the reductant. Recall that reducing permanganate often gives MnO2, an insoluble brown solid. Reducing MnO4− to MnO2 is a three-electron reduction, whereas the oxidation of H2O2 to O2 is a two-electron oxidation. This is a thermal decomposition reaction. Because KNO3 contains nitrogen in its highest oxidation state (+5) and oxygen in its lowest oxidation state (−2), a redox reaction is likely. Oxidation of the oxygen in nitrate to atomic oxygen is a two-electron process per oxygen atom. Nitrogen is likely to accept two electrons because oxoanions of nitrogen are known only in the +5 (NO3−) and +3 (NO2−) oxidation states. Exercise 21.4.2 Predict the product(s) of each reaction and write a balanced chemical equation for each reaction. SiO2(s) + H+(aq) → NO(g) + O2(g) → SO3(g) + H2O(l) → H2O2(aq) + I–(aq) → Answer SiO2(s) + H+(aq) → no reaction 2NO(g) + O2(g) → 2NO2(g) SO3(g) + H2O(l) → H2SO4(aq) H2O2(aq) + 2I−(aq) → I2(aq) + 2OH−(aq) Reactions and Compounds of the Heavier Chalcogens Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column. Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S8 rings packed together in a complex “crankshaft” arrangement (Figure 21.4.2), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (Sn2−) and polyselenides (Sen2−), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group. As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO3 to form oxoacids such as H2SO4. In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po2+ ion. Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group. Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF6), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (Figure 21.4.2). In contrast, only four fluorides of selenium (SeF6, SeF4, FSeSeF, and SeSeF2) and only three of tellurium (TeF4, TeF6, and Te2F10) are known. Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO2 is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO2 is a white solid with an infinite chain structure (each Se is three coordinate), TeO2 is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO2 is a yellow ionic solid in which each Po4+ ion is eight coordinate. The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H2YO3—sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H2SeO4) are strong acids, but telluric acid [Te(OH)6] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH)6, with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid. The stability of the highest oxidation state of the chalcogens decreases down the column. Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS2 and CSe2 are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO2. Because these double bonds are significantly weaker than the C=O bond, however, CS2, CSe2, and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y2−), as in Na2S and FeS, or polychalcogenide ions (Yn2−), as in FeS2 and Na2S5. The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group. Ionic chalcogenides like Na2S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H2S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous. Example 21.4.2 For each reaction, explain why the given product forms or no reaction occurs. SO2(g) + Cl2(g) → SO2Cl2(l) SF6(g) + H2O(l) → no reaction 2Se(s) + Cl2(g) → Se2Cl2(l) Given: balanced chemical equations Asked for: why the given products (or no products) form Strategy: Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs. Solution One of the reactants (Cl2) is an oxidant. If the other reactant can be oxidized, then a redox reaction is likely. Sulfur dioxide contains sulfur in the +4 oxidation state, which is 2 less than its maximum oxidation state. Sulfur dioxide is also known to be a mild reducing agent in aqueous solution, producing sulfuric acid as the oxidation product. Hence a redox reaction is probable. The simplest reaction is the formation of SO2Cl2 (sulfuryl chloride), which is a tetrahedral species with two S–Cl and two S=O bonds. Sulfur hexafluoride is a nonmetallic halide. Such compounds normally react vigorously with water to produce an oxoacid of the nonmetal and the corresponding hydrohalic acid. In this case, however, we have a highly stable species, presumably because all of sulfur’s available orbitals are bonding orbitals. Thus SF6 is not likely to react with water. Here we have the reaction of a chalcogen with a halogen. The halogen is a good oxidant, so we can anticipate that a redox reaction will occur. Only fluorine is capable of oxidizing the chalcogens to a +6 oxidation state, so we must decide between SeCl4 and Se2Cl2 as the product. The stoichiometry of the reaction determines which of the two is obtained: SeCl4 or Se2Cl2. Exercise 21.4.2 Predict the products of each reaction and write a balanced chemical equation for each reaction. Te(s) + Na(s) −→Δ SF4(g) + H2O(l) → CH3SeSeCH3(soln) + K(s) → Li2Se(s) + H+(aq) → Answer Te(s) + 2Na(s) → Na2Te(s) SF4(g) + 3H2O(l) → H2SO3(aq) + 4HF(aq) CH3SeSeCH3(soln) + 2K(s) → 2KCH3Se(soln) Li2Se(s) + 2H+(aq) → H2Se(g) + 2Li+(aq) Summary The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group. 21.3: The Elements of Group 15 (The Pnicogens) 21.5: The Elements of Group 17 (The Halogens)
17534	https://artofproblemsolving.com/wiki/index.php/2023_IOQM/Problem_12?srsltid=AfmBOor6P0n3jiq2EcIjUlXwvOpGDBU-xQTFj7TbNWrzCsDqMmYqg4-6	Art of Problem Solving 2023 IOQM/Problem 12 - 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 2023 IOQM/Problem 12 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2023 IOQM/Problem 12 P(x) = x³ + ax² + bx + c. a,b,c ∈ ℤ And, c is odd. p₁ = P(1) = (1)³ + a(1)² + b(1) + c p₁ = 1 + a + b + c p₂ = P(2) = (2)³ + a(2)² + b(2) + c p₂ = 8 + 4a + 2b + c p₃ = P(3) = (3)³ + a(3)² + b(3) + c p₃ = 27 + 9a + 3b + c And, p₀ = P(0) = (0)³ + a(0)² + b(0) + c p₀ = c Now, it is given that, p₁³ + p₂³ + p₃³ = 3p₁p₂p₃ this is a well known identity. If the SUM OF CUBES of 3 numbers is equal to THREE TIMES their product, then either their sum is 0 or the given numbers are EQUAL to each other. –> p₁ = p₂ = p₃ --> p₁ = p₂ --> 1 + a + b + c = 8 + 4a + 2b + c ==> 7 + 3a + b = 0 --------------> (1) and, p₂ = p₃ --> 8 + 4a + 2b + c = 27 + 9a + 3b + c ==> 19 + 5a + b = 0 --------------> (2) Now, subtract Equation (1) from (2) 12 + 2a = 0 ==> a = (-6) Using [a = (-6)] in equation (1) 7 + 3(-6) + b = 0 => b = 11 Finally, calculating the value of -> [p₂ + 2p₁ - 3p₀] p₂ + 2p₁ - 3p₀ = 8 + 4a + 2b + c + 2[1 + a + b + c] - 3[c] = 8 + 4a + 2b + c + 2 + 2a + 2b + 2c - 3c Rearranging the terms, we get p₂ + 2p₁ - 3p₀ = 8 + 2 + 4a + 2a + 2b + 2b + c + 2c - 3c = 10 + 6a + 4b Using values of (a) and (b) which were calculated above p₂ + 2p₁ - 3p₀ = 10 + 6(-6) + 4(11) Solving further, we finally get p₂ + 2p₁ - 3p₀ = 18 ~ Arpit 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.
17535	https://www.scribd.com/document/411570006/327745817-Log-and-Antilog-Tables	Log and Antilog Tables \| PDF \| Elementary Mathematics \| Discrete Mathematics 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 Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 512 views 67 pages Log and Antilog Tables This document contains a logarithm table listing values from 0 to 99 in the left column and corresponding logarithm values in the right column. It includes 100 rows of data organized into a … Full description Uploaded by Usha Devi AI-enhanced title and description Go to previous items Go to next items Download Save Save 327745817 Log and Antilog Tables For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save 327745817 Log and Antilog Tables For Later You are on page 1/ 67 Search Fullscreen LOGARITHM TABLES 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 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 0 0000 0414 0792 1139 1461 1761 2041 2304 2553 2788 adDownload to read ad-free 3010 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 6021 6128 6232 6335 6435 6532 6628 6721 6812 6902 6990 7076 7160 7243 7324 7404 1 0043 0453 0828 1173 1492 1790 2068 2330 2577 2810 3032 3243 3444 3636 3820 3997 4166 4330 4487 4639 4786 adDownload to read ad-free 4928 5065 5198 5328 5453 5575 5694 5809 5922 6031 6138 6243 6345 6444 6542 6637 6730 6821 6911 6998 7084 7168 7251 7332 7412 2 0086 0492 0864 1206 1523 1818 2095 2355 2601 2833 3054 3263 3464 3655 3838 4014 4183 4346 4502 4654 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 6042 6149 adDownload to read ad-free 6253 6355 6454 6551 6646 6739 6830 6920 7007 7093 7177 7259 7340 7419 3 0128 0531 0899 1239 1553 1847 2122 2380 2625 2856 3075 3284 3483 3674 3856 4031 4200 4362 4518 4669 4814 4955 5092 5224 5353 5478 5599 5717 5832 5944 6053 6160 6263 6365 6464 6561 6656 6749 6839 6928 7016 7101 7185 adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Akash Pyq Maths No ratings yet Akash Pyq Maths 11 pages Geometry & Trigonometry AA HL W1 No ratings yet Geometry & Trigonometry AA HL W1 3 pages Explaining Logarithms 100% (11) Explaining Logarithms 112 pages Code Exploiter No ratings yet Code Exploiter 664 pages Log Tables and Their Operations No ratings yet Log Tables and Their Operations 6 pages Log Tables and Their Operations No ratings yet Log Tables and Their Operations 6 pages Flame and The Rebel Riders 100% (1) Flame and The Rebel Riders 35 pages Business Statistics No ratings yet Business Statistics 9 pages Show Me The Father 100% (1) Show Me The Father 35 pages Multiplication Table 1 To 25 No ratings yet Multiplication Table 1 To 25 2 pages 1 To 10 Tables No ratings yet 1 To 10 Tables 4 pages 1 To 10 Tables No ratings yet 1 To 10 Tables 4 pages Tables, Square, Cubes Chart No ratings yet Tables, Square, Cubes Chart 1 page Table of Decimal Logarithms No ratings yet Table of Decimal Logarithms 9 pages DOC-20250519-WA0003-merge (1) - 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17536	https://www.masterorganicchemistry.com/2016/11/03/alkene-nomenclature-cis-and-trans-and-e-and-z/	Skip to content Tutoring Login Home / E and Z Notation For Alkenes (+ Cis/Trans) Alkene Reactions By James Ashenhurst E and Z Notation For Alkenes (+ Cis/Trans) Last updated: February 28th, 2025 \| E and Z Notation For Alkenes Unlike C–C single bonds, C–C double bonds can’t undergo rotation without breaking the pi bond One consequence of this is geometric isomerism –the existence of alkene stereoisomers that differ solely in how their substituents are arranged in space about the double bond In simple cases where there are two identical substituents on each carbon of the alkene, we can use cis– and trans– to designate the isomers where those substituents are on the sameand opposite sides of the double bond, respectively. For geometric isomers that lack two identical substituents, we rank the two substituents on each end of the double bond according to the Cahn-Ingold-Prelog (CIP) rules. The Z isomer (“zusammen“, same) is the geometric isomer where the #1 ranked substituents are on the same side of the double bond. Mnemonic: “zee zame zide“ E isomer (“entgegen“) is the geometric isomer where the #1 ranked substituents are on the opposite side of the double bond, Table of Contents When do we use cis– and trans– Notation In Rings? cis– and trans– Isomerism In Alkenes Watch out for ambiguous names when geometrical isomerism is possible! cis– and trans– isomerism in cyclic alkenes When “cis“- and “trans‘” fails: E and Z Notation E and Z Notation For Alkenes Breaking Ties: The Method of Dots Conclusion: E and Z Notation For Alkenes Notes Quiz Yourself! This post was co-authored with Matt Pierce of Organic Chemistry Solutions. Ask Matt about scheduling an online tutoring session here. Quick Review: cis– And trans- Isomerism (“Geometrical Isomerism”) In Rings Earlier on our MOC series on cycloalkanes, we saw that a key feature of small rings is that they can’t be turned “inside out” without breaking bonds.(See post: Cycloalkanes – Dashes and Wedges) One of the most important consequence of this is that it can lead to the existence of stereoisomers –molecules which share the same molecular formula and the same connectivity but have a different arrangement of atoms in space. These two versions of 1,2 dichlorocyclopentane (below) are an example. They have the same connectivity – both are 1,2-dichlorocyclopentane – but have different arrangements of their atoms in space. The chlorines are on the same side of the ring in the left-hand isomer (both “wedges”, coming out of the page) and on the opposite sides (one wedged, one dashed) on the right-hand isomer. These two molecules cannot be interconverted through rotation of the C-C bond without rupturing the ring (use a model kit and try, if you like). They are therefore isomers. Molecules which have the same connectivity but different arrangement in space are known as stereoisomers. Specifically, the relationship between the two molecules above is that of diastereomers: stereoisomers which are not mirror images of each other. (See post: Types of Isomers) These two molecules have different physical properties – different boiling points, melting points, reactivities, spectral characteristics and so on. [Just to note, the other subclass of stereoisomer is “enantiomers”. We apply this to two stereoisomers which are (non-superimposable) mirror images of each other. Also: keep in mind that the terms “diastereomer” and “enantiomer” denote comparative relationships, like the terms “brother” or “cousin”. ] 1. When Do We Use cis-And trans-Notation In Rings? We use the terms cis- and trans– to denote therelativeconfiguration of two groups to each other in situations where there is restricted rotation. [Side note: the “restricted rotation” is how cis- and trans- subtly differs from syn and anti, which we use in cases where there is free rotation, such as the orientation of methyl groups in “eclipsed” and “staggered” butane. Bottom line: syn and anti forms can generally be interconverted through bond rotation: cis and trans forms cannot. ] In nomenclature, “cis” is used to distinguish the isomer where two identical groups (e.g. the two chlorines in 1,2-dichlorocyclopentane) are pointing in the same direction from the plane of the ring, and trans to distinguish the isomer where they point in opposite directions. [You might also hear organic chemists say, “the chlorines are cis to each other” or “the hydrogens are trans to one another”.] A common name for these so-called “cis-trans” isomers is “geometric isomers”. Those scolds at IUPAC actually discourage the term “geometric isomers”, and for once, I agree: the term is somewhat redundant and can cause confusion. In the rest of this post I’ll just use the term “cis-trans” isomers. In order for cis- trans-isomerism to exist in rings, we need two conditions: two (and only two) carbons each bearing non-identical substituents above and below the ring the two carbons have at least one of those substituents in common In 1,2-dichlorocyclopentane we saw that C-1 and C-2 each had non-identical substituents (H and Cl) above and below the ring, and they each had at least one substituent in common (in fact they have two substituents in common: H and Cl ). Here’s another example: cis-and trans– 1-ethyl-2-methylcyclobutane. Note that they each have two carbons which each bear non-identical substituents above and below the ring (H and CH3; H and CH2CH3). They also have at least one substituent in common (H). So we can refer to cis-1-ethyl-2-methylcyclohexane as the isomer where the two hydrogens are pointing in the same direction, and trans where they point in opposite directions. If you’ve covered chirality, you might also note an interesting fact: there are two ways to draw each of the cis-and trans– isomers, and they can’t be superimposed on each other. These are enantiomers, by the way. (See post: Enantiomers, Diastereomers or the Same) So cis-and trans- doesn’t specify which enantiomer (it can be applied to either). It’s just describing the relative configuration of the two groups (Hin this case). If we want to specify a particular enantiomer, we need to use the Cahn-Ingold-Prelog (CIP) system of assigning R and S configurations, which provides us with the “absolute” configuration. In that case, cis– and trans-is redundant. (See post: Cahn-Ingold-Prelog System) Because cis– and trans– is relative, it doesn’t work if the two carbons don’t share a common substituent. In that case you also have to use (R)/(S). We’re taking too long to go through rings here, so let’s just illustrate 2 examples where “cis” and trans” doesn’t work in rings and leave it there. 2. cis– and trans- Isomerism (Geometric Isomerism) In Alkenes cis-trans isomerism is also possible for alkenes. As in small rings, rotation about pi bonds is also constrained: due to the “side-on” overlap of pi bonds, one can’t rotate a pi bond without breaking it. This stands in contrast to conventional sigma bonds (single bonds) in acyclic molecules, where free rotation is possible: witness 1,2-dichloroethane (below left). Hence we can have molecules such as cis-1,2-dichloroethene [boiling point 60°C] and trans-1,2-dichloroethene [boiling point: 48°C] which can be separated from each other due to their differing physical properties. We can also use the cis–transnomenclature to distinguish isomers such as 2-methyl-3-hexene (above right). In the cis isomer, the two hydrogens are on the same side of the pi bond, and in the trans isomer, the two hydrogens are on the opposite side of the bond. [Note: this risks a “tsk-tsk” with accompanying finger-wag from IUPAC , but it nevertheless gets the right structure: see the Note 1 below for a digression as to why] As with rings, the minimum requirement for cis-transisomerism in alkenes is that each carbon is bonded to two different groups,and that the two carbons have at least one substituent in common. As with rings, cis-trans isomerism isn’t possible if one of the carbons of the double bond is attached to two identical groups, as with 1,1-dibromo-1-propene, below. Try it for yourself if you’re not convinced. 3. Watch Out For Ambiguous Names Where Cis/Trans Isomerism Is Possible A quick digression: one consequence of our newfound appreciation of geometrical isomerism is that many simple-sounding molecule names are actually ambiguous. For instance, the descriptor “3-hexene” does not unambiguously describe a specific molecule. [The same is true for 2-butene: try it! ]. To nail down the specific molecule, we need to specify cis– or trans– 3-hexene. Note that 1-hexene is still OK, since the 1-position of 1-hexene is attached to two identical groups (hydrogens) and thus no cis–trans isomers are possible. 4. Cis– Trans- Isomerism For Cyclic Alkenes cis- and trans can also be applied to alkenes in rings. For example, on paper it’s possible to draw cis– and trans– cyclohexene, since the pi bond fulfills the requirements for cis- trans-isomerism. In reality, trans-cyclohexene is impossibly strained. Try kissing yourself on the tailbone. That will give you some idea of the strain involved in trying to accommodate a trans– double bond in a six membered ring . [Note 2] For this reason, for ring sizes 7 and below, it’s safe to ignore writing “cis” : the configuration is assumed. At ring sizes of 8 and above, we do need to put a cis– or trans-in the name, because the trans– isomer becomes feasible. (Imagine trying to kiss yourself on the tailbone if you had the neck of a giraffe: suddenly not impossible!) 5. A Solution For When “Cis” and “Trans” Fails: The E/Z System We saw that cis and trans fails in rings when the two carbons lacked a common substituent. It also fails for alkenes under these circumstances. Case in point: try to apply cis and trans to the alkene below: See the problem? In the absence of two identical groups, we have no reference point! On the left, the chlorine is cis to Br and transto F. But does that really justify calling the isomer “cis” ? How do we decide? What we need is some way to determine priorities in these situations. [note: some textbooks may still refer to this alkene as exhibiting “cis-trans isomerism” even though we must use E and Z] 6. The E and ZNotation For Alkenes Thankfully, we can apply the ranking system developed by Cahn, Ingold, and Prelog for chiral centers (as touched on in this earlier post on (R)/(S) nomenclature) for this purpose. The protocol is as follows: Each carbon in the pi bond is attached to two substituents. For each carbon, these two substituents are ranked (1 or 2) according to the atomic numbers of the atom directly attached to the carbon. (e.g. Cl > F ) If both substituents ranked 1 are on the same side of the pi bond, the bond is given the descriptor Z (short for German Zusammen, which means “together”). If both substituents ranked 1 are on the opposite sideof the pi bond, the bond is given the descriptor E(short for German Entgegen, which means “opposite”). So Z resembles “cis” and E resembles “trans” . (Note: they are not necessarily the same and do not always correlate: see Note 2 for an example of a cis alkene which is E . The E/Z system is comprehensive for all alkenes capable of geometric isomerism, including the cis/trans alkene examples above. We often use cis/trans for convenience, but E/Z is the “official”, IUPAC approved way to name alkene stereoisomers]. One easy way to remember Z is to say “Zee Zame Zide” in a German accent. My way of doing it was pretending that the Z stands for “zis”.Whatever works for you. Here’s a practical example: As with chiral centers, ranking according to atomic number can result in ties if we restrict ourselves merely to the atoms directly attached to the pi bonds. 7. Breaking Ties: The Method of Dots For instance, the alkene below presents us with a dilemma: one of the carbons of the alkene is attached to two carbon atoms. So how do we determine priorities in this case. How do we break ties? In the case of ties, we must apply the method of dots.Dots are handy placeholders which is why I like to use this method. Place a dot on each of the two atoms you are comparing. List the 3 atoms each atom is attached to, in order of atomic number. Compare the lists, much like you would compare a set of three playing cards. Just as a hand of (8, 8, 7) would beat (8, 7, 7), so would (C, C, H) beat (C, H, H). If the lists are identical, move the dots outward to the highest priority atom on the list. At the first point of difference,assign (E or Z). If there is no difference… then the groups are identical, and E / Zdoes not apply. Here’s a practical example of the “method of dots”. Here’s a more complex example with multiple alkenes. In this case each pi bond is designated by a number with its own separate E or Z configuration. OK, this was long. But hopefully useful. Watch out for a future post in which we go into more detail on the “method of dots”. 8. Conclusion: E and Z Notation For Alkenes cis-trans- is OK for describing simple alkene stereoisomers, but only works in certain cases. Furthermore, it only gives relative configurations. The E/Zsystem is comprehensive and describes the absoluteconfiguration of the molecule. See below for an example of an Ealkene which is “cis” and a Z alkene which is “trans”. Just a reminder: this post was co-authored by Matt Pierce of Organic Chemistry Solutions. Ask Matt about scheduling an online tutoring session here. Notes Related Articles Introduction to Assigning (R) and (S): The Cahn-Ingold-Prelog Rules Assigning Cahn-Ingold-Prelog (CIP) Priorities (2) – The Method of Dots Alkene Stability Types of Isomers: Constitutional Isomers, Stereoisomers, Enantiomers, and Diastereomers Enantiomers vs Diastereomers vs The Same? Two Methods For Solving Problems Geometric Isomers In Small Rings: Cis And Trans Cycloalkanes Stereoselective and Stereospecific Reactions Alkene Addition Reactions: “Regioselectivity” and “Stereoselectivity” (Syn/Anti) Note 1: It’s possible to have an alkene we’d describe as ‘cis’be E and vice versa. E/Zis the preferred, more comprehensive nomenclature since it describes absolute configuration, whereas cis- trans-merely describes relativeconfiguration. Note 2: trans-cyclopropene, trans-cyclobutene, and trans-cyclopentene have never been synthesized or observed. trans-cyclohexene is a laboratory curiosity, stable at a few degrees above absolute zero. trans-cycloheptene has an extremely short half-life at room temperature. trans-cyclooctene is a stable molecule [it also exhibits axial chirality, which is interesting! ]. Quiz Yourself! Become a MOC member to see the clickable quiz with answers on the back. Become a MOC member to see the clickable quiz with answers on the back. Become a MOC member to see the clickable quiz with answers on the back. Become a MOC member to see the clickable quiz with answers on the back. Become a MOC member to see the clickable quiz with answers on the back. Become a MOC member to see the clickable quiz with answers on the back. 00 General Chemistry Review Lewis Structures Ionic and Covalent Bonding Chemical Kinetics Chemical Equilibria Valence Electrons of the First Row Elements How Concepts Build Up In Org 1 ("The Pyramid") 01 Bonding, Structure, and Resonance How Do We Know Methane (CH4) Is Tetrahedral? Hybrid Orbitals and Hybridization How To Determine Hybridization: A Shortcut Orbital Hybridization And Bond Strengths Sigma bonds come in six varieties: Pi bonds come in one A Key Skill: How to Calculate Formal Charge The Four Intermolecular Forces and How They Affect Boiling Points 3 Trends That Affect Boiling Points How To Use Electronegativity To Determine Electron Density (and why NOT to trust formal charge) Introduction to Resonance How To Use Curved Arrows To Interchange Resonance Forms Evaluating Resonance Forms (1) - The Rule of Least Charges How To Find The Best Resonance Structure By Applying Electronegativity Evaluating Resonance Structures With Negative Charges Evaluating Resonance Structures With Positive Charge Exploring Resonance: Pi-Donation Exploring Resonance: Pi-acceptors In Summary: Evaluating Resonance Structures Drawing Resonance Structures: 3 Common Mistakes To Avoid How to apply electronegativity and resonance to understand reactivity Bond Hybridization Practice Structure and Bonding Practice Quizzes Resonance Structures Practice 02 Acid Base Reactions Introduction to Acid-Base Reactions Acid Base Reactions In Organic Chemistry The Stronger The Acid, The Weaker The Conjugate Base Walkthrough of Acid-Base Reactions (3) - Acidity Trends Five Key Factors That Influence Acidity Acid-Base Reactions: Introducing Ka and pKa How to Use a pKa Table The pKa Table Is Your Friend A Handy Rule of Thumb for Acid-Base Reactions Acid Base Reactions Are Fast pKa Values Span 60 Orders Of Magnitude How Protonation and Deprotonation Affect Reactivity Acid Base Practice Problems 03 Alkanes and Nomenclature Meet the (Most Important) Functional Groups Condensed Formulas: Deciphering What the Brackets Mean Hidden Hydrogens, Hidden Lone Pairs, Hidden Counterions Don't Be Futyl, Learn The Butyls Primary, Secondary, Tertiary, Quaternary In Organic Chemistry Branching, and Its Affect On Melting and Boiling Points The Many, Many Ways of Drawing Butane Wedge And Dash Convention For Tetrahedral Carbon Common Mistakes in Organic Chemistry: Pentavalent Carbon Table of Functional Group Priorities for Nomenclature Summary Sheet - Alkane Nomenclature Organic Chemistry IUPAC Nomenclature Demystified With A Simple Puzzle Piece Approach Boiling Point Quizzes Organic Chemistry Nomenclature Quizzes 04 Conformations and Cycloalkanes Staggered vs Eclipsed Conformations of Ethane Conformational Isomers of Propane Newman Projection of Butane (and Gauche Conformation) Introduction to Cycloalkanes Geometric Isomers In Small Rings: Cis And Trans Cycloalkanes Calculation of Ring Strain In Cycloalkanes Cycloalkanes - Ring Strain In Cyclopropane And Cyclobutane Cyclohexane Conformations Cyclohexane Chair Conformation: An Aerial Tour How To Draw The Cyclohexane Chair Conformation The Cyclohexane Chair Flip The Cyclohexane Chair Flip - Energy Diagram Substituted Cyclohexanes - Axial vs Equatorial Ranking The Bulkiness Of Substituents On Cyclohexanes: "A-Values" Cyclohexane Chair Conformation Stability: Which One Is Lower Energy? Fused Rings - Cis-Decalin and Trans-Decalin Naming Bicyclic Compounds - Fused, Bridged, and Spiro Bredt's Rule (And Summary of Cycloalkanes) Newman Projection Practice Cycloalkanes Practice Problems 05 A Primer On Organic Reactions The Most Important Question To Ask When Learning a New Reaction Curved Arrows (for reactions) Nucleophiles and Electrophiles The Three Classes of Nucleophiles Nucleophilicity vs. Basicity What Makes A Good Nucleophile? What Makes A Good Leaving Group? 3 Factors That Stabilize Carbocations Equilibrium and Energy Relationships 7 Factors that stabilize negative charge in organic chemistry 7 Factors That Stabilize Positive Charge in Organic Chemistry What's a Transition State? Hammond's Postulate Learning Organic Chemistry Reactions: A Checklist (PDF) Introduction to Oxidative Cleavage Reactions 06 Free Radical Reactions Free Radical Reactions 3 Factors That Stabilize Free Radicals Bond Strengths And Radical Stability Free Radical Initiation: Why Is "Light" Or "Heat" Required? Initiation, Propagation, Termination Monochlorination Products Of Propane, Pentane, And Other Alkanes Selectivity In Free Radical Reactions Selectivity in Free Radical Reactions: Bromination vs. Chlorination Halogenation At Tiffany's Allylic Bromination Bonus Topic: Allylic Rearrangements In Summary: Free Radicals Synthesis (2) - Reactions of Alkanes Free Radicals Practice Quizzes 07 Stereochemistry and Chirality Types of Isomers: Constitutional Isomers, Stereoisomers, Enantiomers, and Diastereomers How To Draw The Enantiomer Of A Chiral Molecule How To Draw A Bond Rotation Introduction to Assigning (R) and (S): The Cahn-Ingold-Prelog Rules Assigning Cahn-Ingold-Prelog (CIP) Priorities (2) - The Method of Dots Enantiomers vs Diastereomers vs The Same? Two Methods For Solving Problems Assigning R/S To Newman Projections (And Converting Newman To Line Diagrams) How To Determine R and S Configurations On A Fischer Projection The Meso Trap Optical Rotation, Optical Activity, and Specific Rotation Optical Purity and Enantiomeric Excess What's a Racemic Mixture? Chiral Allenes And Chiral Axes Stereochemistry Practice Problems and Quizzes 08 Substitution Reactions Nucleophilic Substitution Reactions - Introduction Two Types of Nucleophilic Substitution Reactions The SN2 Mechanism Why the SN2 Reaction Is Powerful The SN1 Mechanism The Conjugate Acid Is A Better Leaving Group Comparing the SN1 and SN2 Reactions Polar Protic? Polar Aprotic? Nonpolar? All About Solvents Steric Hindrance is Like a Fat Goalie Common Blind Spot: Intramolecular Reactions Substitution Practice - SN1 Substitution Practice - SN2 09 Elimination Reactions Elimination Reactions (1): Introduction And The Key Pattern Elimination Reactions (2): The Zaitsev Rule Elimination Reactions Are Favored By Heat Two Elimination Reaction Patterns The E1 Reaction The E2 Mechanism E1 vs E2: Comparing the E1 and E2 Reactions Antiperiplanar Relationships: The E2 Reaction and Cyclohexane Rings Bulky Bases in Elimination Reactions Comparing the E1 vs SN1 Reactions Elimination (E1) Reactions With Rearrangements E1cB - Elimination (Unimolecular) Conjugate Base Elimination (E1) Practice Problems And Solutions Elimination (E2) Practice Problems and Solutions 10 Rearrangements Introduction to Rearrangement Reactions Rearrangement Reactions (1) - Hydride Shifts Carbocation Rearrangement Reactions (2) - Alkyl Shifts Pinacol Rearrangement The SN1, E1, and Alkene Addition Reactions All Pass Through A Carbocation Intermediate 11 SN1/SN2/E1/E2 Decision Identifying Where Substitution and Elimination Reactions Happen Deciding SN1/SN2/E1/E2 (1) - The Substrate Deciding SN1/SN2/E1/E2 (2) - The Nucleophile/Base SN1 vs E1 and SN2 vs E2 : The Temperature Deciding SN1/SN2/E1/E2 - The Solvent Wrapup: The Key Factors For Determining SN1/SN2/E1/E2 Alkyl Halide Reaction Map And Summary SN1 SN2 E1 E2 Practice Problems 12 Alkene Reactions E and Z Notation For Alkenes (+ Cis/Trans) Alkene Stability Alkene Addition Reactions: "Regioselectivity" and "Stereoselectivity" (Syn/Anti) Stereoselective and Stereospecific Reactions Hydrohalogenation of Alkenes and Markovnikov's Rule Hydration of Alkenes With Aqueous Acid Rearrangements in Alkene Addition Reactions Halogenation of Alkenes and Halohydrin Formation Oxymercuration Demercuration of Alkenes Hydroboration Oxidation of Alkenes m-CPBA (meta-chloroperoxybenzoic acid) OsO4 (Osmium Tetroxide) for Dihydroxylation of Alkenes Palladium on Carbon (Pd/C) for Catalytic Hydrogenation of Alkenes Cyclopropanation of Alkenes A Fourth Alkene Addition Pattern - Free Radical Addition Alkene Reactions: Ozonolysis Summary: Three Key Families Of Alkene Reaction Mechanisms Synthesis (4) - Alkene Reaction Map, Including Alkyl Halide Reactions Alkene Reactions Practice Problems 13 Alkyne Reactions Acetylides from Alkynes, And Substitution Reactions of Acetylides Partial Reduction of Alkynes With Lindlar's Catalyst Partial Reduction of Alkynes With Na/NH3 To Obtain Trans Alkenes Alkyne Hydroboration With "R2BH" Hydration and Oxymercuration of Alkynes Hydrohalogenation of Alkynes Alkyne Halogenation: Bromination and Chlorination of Alkynes Oxidation of Alkynes With O3 and KMnO4 Alkenes To Alkynes Via Halogenation And Elimination Reactions Alkynes Are A Blank Canvas Synthesis (5) - Reactions of Alkynes Alkyne Reactions Practice Problems With Answers 14 Alcohols, Epoxides and Ethers Alcohols - Nomenclature and Properties Alcohols Can Act As Acids Or Bases (And Why It Matters) Alcohols - Acidity and Basicity The Williamson Ether Synthesis Ethers From Alkenes, Tertiary Alkyl Halides and Alkoxymercuration Alcohols To Ethers via Acid Catalysis Cleavage Of Ethers With Acid Epoxides - The Outlier Of The Ether Family Opening of Epoxides With Acid Epoxide Ring Opening With Base Making Alkyl Halides From Alcohols Tosylates And Mesylates PBr3 and SOCl2 Elimination Reactions of Alcohols Elimination of Alcohols To Alkenes With POCl3 Alcohol Oxidation: "Strong" and "Weak" Oxidants Demystifying The Mechanisms of Alcohol Oxidations Protecting Groups For Alcohols Thiols And Thioethers Calculating the oxidation state of a carbon Oxidation and Reduction in Organic Chemistry Oxidation Ladders SOCl2 Mechanism For Alcohols To Alkyl Halides: SN2 versus SNi Alcohol Reactions Roadmap (PDF) Alcohol Reaction Practice Problems Epoxide Reaction Quizzes Oxidation and Reduction Practice Quizzes 15 Organometallics What's An Organometallic? Formation of Grignard and Organolithium Reagents Organometallics Are Strong Bases Reactions of Grignard Reagents Protecting Groups In Grignard Reactions Synthesis Problems Involving Grignard Reagents Grignard Reactions And Synthesis (2) Organocuprates (Gilman Reagents): How They're Made Gilman Reagents (Organocuprates): What They're Used For The Heck, Suzuki, and Olefin Metathesis Reactions (And Why They Don't Belong In Most Introductory Organic Chemistry Courses) Reaction Map: Reactions of Organometallics Grignard Practice Problems 16 Spectroscopy Degrees of Unsaturation (or IHD, Index of Hydrogen Deficiency) Conjugation And Color (+ How Bleach Works) Introduction To UV-Vis Spectroscopy UV-Vis Spectroscopy: Absorbance of Carbonyls UV-Vis Spectroscopy: Practice Questions Bond Vibrations, Infrared Spectroscopy, and the "Ball and Spring" Model Infrared Spectroscopy: A Quick Primer On Interpreting Spectra IR Spectroscopy: 4 Practice Problems 1H NMR: How Many Signals? Homotopic, Enantiotopic, Diastereotopic Diastereotopic Protons in 1H NMR Spectroscopy: Examples 13-C NMR - How Many Signals Liquid Gold: Pheromones In Doe Urine Natural Product Isolation (1) - Extraction Natural Product Isolation (2) - Purification Techniques, An Overview Structure Determination Case Study: Deer Tarsal Gland Pheromone 17 Dienes and MO Theory What To Expect In Organic Chemistry 2 Are these molecules conjugated? Conjugation And Resonance In Organic Chemistry Bonding And Antibonding Pi Orbitals Molecular Orbitals of The Allyl Cation, Allyl Radical, and Allyl Anion Pi Molecular Orbitals of Butadiene Reactions of Dienes: 1,2 and 1,4 Addition Thermodynamic and Kinetic Products More On 1,2 and 1,4 Additions To Dienes s-cis and s-trans The Diels-Alder Reaction Cyclic Dienes and Dienophiles in the Diels-Alder Reaction Stereochemistry of the Diels-Alder Reaction Exo vs Endo Products In The Diels Alder: How To Tell Them Apart HOMO and LUMO In the Diels Alder Reaction Why Are Endo vs Exo Products Favored in the Diels-Alder Reaction? Diels-Alder Reaction: Kinetic and Thermodynamic Control The Retro Diels-Alder Reaction The Intramolecular Diels Alder Reaction Regiochemistry In The Diels-Alder Reaction The Cope and Claisen Rearrangements Electrocyclic Reactions Electrocyclic Ring Opening And Closure (2) - Six (or Eight) Pi Electrons Diels Alder Practice Problems Molecular Orbital Theory Practice 18 Aromaticity Introduction To Aromaticity Rules For Aromaticity Huckel's Rule: What Does 4n+2 Mean? Aromatic, Non-Aromatic, or Antiaromatic? Some Practice Problems Antiaromatic Compounds and Antiaromaticity The Pi Molecular Orbitals of Benzene The Pi Molecular Orbitals of Cyclobutadiene Frost Circles Aromaticity Practice Quizzes 19 Reactions of Aromatic Molecules Electrophilic Aromatic Substitution: Introduction Activating and Deactivating Groups In Electrophilic Aromatic Substitution Electrophilic Aromatic Substitution - The Mechanism Ortho-, Para- and Meta- Directors in Electrophilic Aromatic Substitution Understanding Ortho, Para, and Meta Directors Why are halogens ortho- para- directors? Disubstituted Benzenes: The Strongest Electron-Donor "Wins" Electrophilic Aromatic Substitutions (1) - Halogenation of Benzene Electrophilic Aromatic Substitutions (2) - Nitration and Sulfonation EAS Reactions (3) - Friedel-Crafts Acylation and Friedel-Crafts Alkylation Intramolecular Friedel-Crafts Reactions Nucleophilic Aromatic Substitution (NAS) Nucleophilic Aromatic Substitution (2) - The Benzyne Mechanism Reactions on the "Benzylic" Carbon: Bromination And Oxidation The Wolff-Kishner, Clemmensen, And Other Carbonyl Reductions More Reactions on the Aromatic Sidechain: Reduction of Nitro Groups and the Baeyer Villiger Aromatic Synthesis (1) - "Order Of Operations" Synthesis of Benzene Derivatives (2) - Polarity Reversal Aromatic Synthesis (3) - Sulfonyl Blocking Groups Birch Reduction Synthesis (7): Reaction Map of Benzene and Related Aromatic Compounds Aromatic Reactions and Synthesis Practice Electrophilic Aromatic Substitution Practice Problems 20 Aldehydes and Ketones What's The Alpha Carbon In Carbonyl Compounds? Nucleophilic Addition To Carbonyls Aldehydes and Ketones: 14 Reactions With The Same Mechanism Sodium Borohydride (NaBH4) Reduction of Aldehydes and Ketones Grignard Reagents For Addition To Aldehydes and Ketones Wittig Reaction Hydrates, Hemiacetals, and Acetals Imines - Properties, Formation, Reactions, and Mechanisms All About Enamines Breaking Down Carbonyl Reaction Mechanisms: Reactions of Anionic Nucleophiles (Part 2) Aldehydes Ketones Reaction Practice 21 Carboxylic Acid Derivatives Nucleophilic Acyl Substitution (With Negatively Charged Nucleophiles) Addition-Elimination Mechanisms With Neutral Nucleophiles (Including Acid Catalysis) Basic Hydrolysis of Esters - Saponification Transesterification Proton Transfer Fischer Esterification - Carboxylic Acid to Ester Under Acidic Conditions Lithium Aluminum Hydride (LiAlH4) For Reduction of Carboxylic Acid Derivatives LiAlH[Ot-Bu]3 For The Reduction of Acid Halides To Aldehydes Di-isobutyl Aluminum Hydride (DIBAL) For The Partial Reduction of Esters and Nitriles Amide Hydrolysis Thionyl Chloride (SOCl2) And Conversion of Carboxylic Acids to Acid Halides Diazomethane (CH2N2) Carbonyl Chemistry: Learn Six Mechanisms For the Price Of One Making Music With Mechanisms (PADPED) Carboxylic Acid Derivatives Practice Questions 22 Enols and Enolates Keto-Enol Tautomerism Enolates - Formation, Stability, and Simple Reactions Kinetic Versus Thermodynamic Enolates Aldol Addition and Condensation Reactions Reactions of Enols - Acid-Catalyzed Aldol, Halogenation, and Mannich Reactions Claisen Condensation and Dieckmann Condensation Decarboxylation The Malonic Ester and Acetoacetic Ester Synthesis The Michael Addition Reaction and Conjugate Addition The Robinson Annulation Haloform Reaction The Hell–Volhard–Zelinsky Reaction Enols and Enolates Practice Quizzes 23 Amines The Amide Functional Group: Properties, Synthesis, and Nomenclature Basicity of Amines And pKaH 5 Key Basicity Trends of Amines The Mesomeric Effect And Aromatic Amines Nucleophilicity of Amines Alkylation of Amines (Sucks!) Reductive Amination The Gabriel Synthesis Some Reactions of Azides The Hofmann Elimination The Hofmann and Curtius Rearrangements The Cope Elimination Protecting Groups for Amines - Carbamates The Strecker Synthesis of Amino Acids Introduction to Peptide Synthesis Reactions of Diazonium Salts: Sandmeyer and Related Reactions Amine Practice Questions 24 Carbohydrates D and L Notation For Sugars Pyranoses and Furanoses: Ring-Chain Tautomerism In Sugars What is Mutarotation? Reducing Sugars The Big Damn Post Of Carbohydrate-Related Chemistry Definitions The Haworth Projection Converting a Fischer Projection To A Haworth (And Vice Versa) Reactions of Sugars: Glycosylation and Protection The Ruff Degradation and Kiliani-Fischer Synthesis Isoelectric Points of Amino Acids (and How To Calculate Them) Carbohydrates Practice Amino Acid Quizzes 25 Fun and Miscellaneous A Gallery of Some Interesting Molecules From Nature Screw Organic Chemistry, I'm Just Going To Write About Cats On Cats, Part 1: Conformations and Configurations On Cats, Part 2: Cat Line Diagrams On Cats, Part 4: Enantiocats On Cats, Part 6: Stereocenters Organic Chemistry Is Shit The Organic Chemistry Behind "The Pill" Maybe they should call them, "Formal Wins" ? Why Do Organic Chemists Use Kilocalories? The Principle of Least Effort Organic Chemistry GIFS - Resonance Forms Reproducibility In Organic Chemistry What Holds The Nucleus Together? How Reactions Are Like Music Organic Chemistry and the New MCAT 26 Organic Chemistry Tips and Tricks Common Mistakes: Formal Charges Can Mislead Partial Charges Give Clues About Electron Flow Draw The Ugly Version First Organic Chemistry Study Tips: Learn the Trends The 8 Types of Arrows In Organic Chemistry, Explained Top 10 Skills To Master Before An Organic Chemistry 2 Final Common Mistakes with Carbonyls: Carboxylic Acids... Are Acids! Planning Organic Synthesis With "Reaction Maps" Alkene Addition Pattern #1: The "Carbocation Pathway" Alkene Addition Pattern #2: The "Three-Membered Ring" Pathway Alkene Addition Pattern #3: The "Concerted" Pathway Number Your Carbons! The 4 Major Classes of Reactions in Org 1 How (and why) electrons flow Grossman's Rule Three Exam Tips A 3-Step Method For Thinking Through Synthesis Problems Putting It Together Putting Diels-Alder Products in Perspective The Ups and Downs of Cyclohexanes The Most Annoying Exceptions in Org 1 (Part 1) The Most Annoying Exceptions in Org 1 (Part 2) The Marriage May Be Bad, But the Divorce Still Costs Money 9 Nomenclature Conventions To Know Nucleophile attacks Electrophile 27 Case Studies of Successful O-Chem Students Success Stories: How Corina Got The The "Hard" Professor - And Got An A+ Anyway How Helena Aced Organic Chemistry From a "Drop" To B+ in Org 2 – How A Hard Working Student Turned It Around How Serge Aced Organic Chemistry Success Stories: How Zach Aced Organic Chemistry 1 Success Stories: How Kari Went From C– to B+ How Esther Bounced Back From a "C" To Get A's In Organic Chemistry 1 And 2 How Tyrell Got The Highest Grade In Her Organic Chemistry Course This Is Why Students Use Flashcards Success Stories: How Stu Aced Organic Chemistry How John Pulled Up His Organic Chemistry Exam Grades Success Stories: How Nathan Aced Organic Chemistry (Without It Taking Over His Life) How Chris Aced Org 1 and Org 2 Interview: How Jay Got an A+ In Organic Chemistry How to Do Well in Organic Chemistry: One Student's Advice "America's Top TA" Shares His Secrets For Teaching O-Chem "Organic Chemistry Is Like..." - A Few Metaphors How To Do Well In Organic Chemistry: Advice From A Tutor Guest post: "I went from being afraid of tests to actually looking forward to them". Comments Comment section 41 thoughts on “E and Z Notation For Alkenes (+ Cis/Trans)” HiCould you please explain why can’t we find the Z/E or cis/trans in cyclic alkenes ? Is there any cyclic alkene for which I do precise Z/E or cis/trans ? Reply In small, cyclic alkenes, the two ring carbons must be on the same side of the alkene (cis) due to ring strain. If you try and make trans-cyclopentene for example you will quickly have a very bent molecular model kit. The first cyclic alkene with a stable trans- geometric isomer is trans cyclooctene. Once the ring size increases past eight, then geometrical isomerism is possible. So in larger rings we should be specific about whether the double bond is cis or trans (or E/Z). Reply 2. Hi, Here why is this a trans compound if the two identical methyl groups are on the same side? Reply It is (E) and also cis. I wonder if this is a glitch with the nomenclature software at pubchem. No human being would name this “trans” Reply 3. Pingback: Classify The Following Alkenes.In Terms Of E/Z Stereoisomer Configurations. - [100% Official Pages] 4. Greetings…58 years old here, applying to a graduate program, am reviewing my organic chemistry and found your website helpful. Thank you Reply Glad you find it helpful, Tom. Best wishes on your graduate program. James Reply 5. Hi James, I have a question want to ask you, does cycloalkene have a geometrical isomer, for cyclohexene or pentene? I think is no, I can’t change the substituent at the double bond to get another isomer, can you tell me am I wrong? I really want to know. Reply It’s not energetically possible for 5- or 6- membered rings. Seven membered rings can be trans but have a short lifetime. Only at 8-membered rings do they start to be reasonably stable. Reply 6. Can you please say which is (cis-trans or E-Z) is accepted in IUPAC nomenclature Reply E-Z is preferred and is non-ambiguous, but in simple cases cis-trans- will do. Reply 7. but some molculeswe take the longest chain of carbon to determine if it is cis or transrather than take the similar group as a point referencefor examplecis-3,4-Dimethyl-2-pentene Reply 8. does a cis and E configuration of a molecule make it less polar? Reply If we’re just talking about alkenes, then yes, very slightly. Carbon is more electronegative than H. So there is some polarization along the C-H (but not that much). The boiling point of a molecule is often used as an indicator for how polar it is. Given this, for which molecule would you expect the boiling point to be higher: cis-2-butene or trans-2-butene? Hint: think about the dipole moments Reply 9. If a molecule only has one double bond that has EZ stereochemistry do we signify that with either an E or Z without putting a locant in front? Does this idea apply with molecules that have multiple double bonds where only one has EZ stereochemistry?I am curious because I have seen both ways.I have put chemical names into molecule generators online that give the same molecule with or without the locant, so I assume it does not matter. Reply I do not always set the best example for nomenclature. From a practical standpoint the important thing is that the answer refer to a unique compound. If there is only one double bond capable of E/Z then it is not absolutely necessary to put it in, IMO. Sometimes E or Z does not need to be mentioned at all. IUPAC gives the example of cyclohexene, where Z is assumed since the E form is extremely unstable. Reply 10. very well explained, I loved the quiz. Thanks Reply 11. Can you please explain that part where you said that if we use R/S absolute configuration cis/trans is redundant. My understanding was you need a chiral centre for the RS one and the double bond would be a part of one of the four groups so wouldn’t you still have to say cis this R and cis this S and so on ? And thank you for the efforts you put into this site . Reply That was in a section on cis trans in rings. In rings cis/trans is redundant if there are two chiral centers and you determine R/S on them both. Reply 12. You said geometric isomers (cis/trans) are a subset of diastereoisomers but aren’t they a subset of stereoisomers instead? Reply They are a subset of diastereomers, and diastereomers are a subset of stereoisomers. Reply 13. Can you tell me whether the following compounds have E or Z configuration?1) CH3 and H on one side of C-C double bond and CHO and Br on the other2)CH3 and H on one side of C-N double bond and OH on the other3)CH3 and H on one side of the C-C double bond and Br and OH on the other Reply Do any of them have identical substituents on one side of the double bond? If not, then they are capable of E/Z configurations. Reply 14. which is more stable? like which will be the natural product of a reaction? will cis ever form and be more stable than trans? Reply Which is more stable, cis or trans? – trans is more stable (less steric hindrance between alkyl groups). I don’t know how to answer the question “which will be the natural product of a reaction” because it depends on multiple factors. In the E2 for example there can be cases where only the cis can form and the trans is not possible due to the requirement that the C-H bond and C-leaving group bond are “anti”. Cis can form in the reduction of alkynes with Lindlar and H2, for example. When both cis and trans are possible trans is generally preferred. For example in the elimination (E2) reaction of say, 3-chloropentane, the trans will be the dominant product. I can’t give you an exact proportion at this moment but a good estimate would be at least 4:1. Reply 15. Can we apply the Z, E nomenclature to orotic acid? What is the Z isomer? How do the rules apply?Thank you Reply I had to look up the structure. E and Z descriptors are not required for cycloalkenes with ring sizes less than 8, since the trans- double bond isomers are extremely unstable. Reply 16. Top notch explanation!!!! Reply 17. Thank you for the helpful description, but what do we do when we have a RACEMIC substituted-ring that does not have 2 identical substituents. Example a cyclopentane substituted with an amino acid at C1 (2 substituents are amine and carboxcylic acid) and a alkyl sidechanie at C2 (2 substituents are hydrogen and ethyl)? Thanks,Michael Reply From your description the way to do it would be to employ the CIP system and assign R and S configurations to the chiral centers. That will unambiguously assign the configuration of each molecule. Reply 18. Amazing explanations as always man, you’re a life saver Reply Glad you find it helpful, DL! Reply 19. I love the way you explain the concepts. It’s quite hard to find someone teaching organic chemistry in such a simple way. Thank you for creating the materials, these are helpful indeed. Reply Thank you Kripi! Reply 20. When a molecule has both a chiral center (R/S) and an alkene functional group (E/Z), which comes first in its name? The R/S before E/Z, other way around, or do we just go by the locant? Thanks! Reply You just go by the locant. The attached example is (2Z, 7R, 11E)-trideca-2,11-dien-7-ol from the IUPAC Blue book, Page 1190 (P-92.4.2.1). Reply 21. Is the E/Z protocol appilcable to allene systems? Reply No it is not. Allenes can have a chiral axis. For chiral allenes, enantiomers are designated by the terms M and P . Reply 22. So if you have a molecule that is Z konfigurated and another molecule that is E konfigurated, they are diastereomers? What if you have a molecule that is Z konfigurated and the other is Z konfigurated as well? Would they still be diastereomers? Reply -If the IUPAC descriptor for two molecules is identical except for the fact that they ONLY differ in E and Z then they are diastereomers. e.g. (E)-4-Methyl-3-octene and (Z)-4-Methyl-3-octene.-If they have different IUPAC names then they are not stereoisomers. (E)-4-Methyl-3-octene and (Z)-4-Methyl-3-nonene-If the IUPAC name is the same and the E/Z descriptor are the same, they are the same molecule. Reply 23. Thank you! Reply Leave a Reply This site uses Akismet to reduce spam. Learn how your comment data is processed.
17537	https://www.youtube.com/watch?v=J2YhCuCI7wc	Common Misconceptions with Exponents GRETCHEN ROBERTS 179 subscribers 1 likes Description 79 views Posted: 15 Sep 2020 Transcript: okay so let's have some discussion around the difference between the two expressions that i have written here the first one says three to the fourth times three squared the second one says the quantity three to the fourth squared why don't you pause the video and take a few minutes to look at the similarities and differences between the two expressions that i have written here so what you should notice about the first one is that this is the property of exponents the product property so i have two exponential terms with the same base and different exponents so we learned that you can solve this by expanding each of those terms so three to the fourth is three times three times three times three and then times another three times three so the first four were the first that's the first term and then the second two belong to the second term since i'm multiplying them um i could just expand them all the way out and i have a total of six threes now that i'm multiplying so we use exponents to represent repeated multiplication so this is now three to the sixth the product property of integer exponents that you learned was that if you have two exponential terms with the same base and different exponents you keep the base and you add the exponents four plus two is six but let's look at what's different about the second expression so this one actually says three to the fourth times three to the fourth so it's actually taking that quantity 3 to the 4th and we're squaring that so this is the power property of integer exponents so i'm raising an exponential term to a power which means that i can expand it out this way or if you want to take it down and expand it out this way and actually write three to the fourth all the way out you can count back and say how many threes do you now have well you have eight of them so this is three to the eighth or you could use the product property that you know and say three to the four plus four is three to the eighth so you should also know the power property right so this quantity three to the fourth is being squared so we can multiply four times two which is eight to get three to the eighth so hopefully you notice the differences between the two expressions that i've just written down and you've taken some time to think about their similarities and differences let's try another one so let's talk about let's talk about 5 squared negative 5 squared and negative 5 squared okay so the first one is just 5 times 5 which you know because you can expand 5 times 5 equals 25 you've seen that one a lot before but then let's look at the second one so this one is saying that we're multiplying that 5 squared by a negative 1. so this can be expanded as negative 1 times 5 times 5 right and the parentheses are here so this is saying 25 times negative 1 which is negative 25. but now let's look at this third one so this is saying negative 5 times negative 5. well you know that two negatives make it positive negative 5 times negative 5 is 25. hopefully you can see the difference around where the parentheses give us our groupings so just a normal 5 squared like you're used to seeing and then negative 5 squared and then negative 5 squared hopefully you can see the difference in those three
17538	https://www.khanacademy.org/science/shs-biology-1/xca2b82fdfba136b8:2nd-quarter/xca2b82fdfba136b8:atp-adp-cycle/v/atp-synthase	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
17539	https://es.slideshare.net/slideshow/ejes-teoria-fatiga/246791867	Ejes teoria fatiga \| PDF 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: Cookie Policy Download free for 30 days Sign in UploadLanguage (EN)Support BusinessMobileSocial MediaMarketingTechnologyArt & PhotosCareerDesignEducationPresentations & Public SpeakingGovernment & NonprofitHealthcareInternetLawLeadership & ManagementAutomotiveEngineeringSoftwareRecruiting & HRRetailSalesServicesScienceSmall Business & EntrepreneurshipFoodEnvironmentEconomy & FinanceData & AnalyticsInvestor RelationsSportsSpiritualNews & PoliticsTravelSelf ImprovementReal EstateEntertainment & HumorHealth & MedicineDevices & HardwareLifestyle Change Language Language English Español Português Français Deutsche Cancel Save Submit search EN Upload Download free for 30 days Sign in FF Uploaded byFernando Fuente 414 views Ejes teoria fatiga AI-enhanced description Este documento trata sobre la teoría de fatiga y el mecanismo de falla por fatiga. Explica que muchos elementos mecánicos están sujetos a cargas variables que pueden causar grietas y falla con el tiempo. También describe la historia del descubrimiento de la fatiga, incluyendo los primeros casos de falla de ejes de ferrocarril, y los experimentos pioneros de Wohler que establecieron la relación entre esfuerzo y vida útil. Además, explica el proceso de crecimiento de grietas bajo c Automotive◦ Read more 0 Save Share Embed Download Download to read offline 1 / 58 2 / 58 Most read 3 / 58 4 / 58 5 / 58 6 / 58 7 / 58 8 / 58 9 / 58 10 / 58 11 / 58 12 / 58 13 / 58 14 / 58 15 / 58 16 / 58 17 / 58 18 / 58 19 / 58 20 / 58 Most read 21 / 58 22 / 58 23 / 58 24 / 58 Most read 25 / 58 26 / 58 27 / 58 28 / 58 29 / 58 30 / 58 31 / 58 32 / 58 33 / 58 34 / 58 35 / 58 36 / 58 37 / 58 38 / 58 39 / 58 40 / 58 41 / 58 42 / 58 43 / 58 44 / 58 45 / 58 46 / 58 47 / 58 48 / 58 49 / 58 50 / 58 51 / 58 52 / 58 53 / 58 54 / 58 55 / 58 56 / 58 57 / 58 58 / 58 ![Image 9: CAPÍTULO 5 CARGAS VARIABLES - TEORÍA DE FATIGA 9 fatiga es la mitad del esfuerzo último. La otra línea es horizontal y parte desde el punto (1380, 690) MPa; se asume, entonces, que para los aceros con Su > 1380 MPa, el límite de fatiga es aproximadamente 690 MPa. Figura 5.7 Relación entre el límite de fatiga y el esfuerzo último de los aceros Podemos concluir que para la mayoría de los aceros (de bajo carbono, aleados, inoxidables): (5.2) (5.3) Una gráfica similar a la de la figura 5.7, se obtiene para la relación entre la dureza de un acero y el límite de fatiga. La diferencia principal e importante es que la curva no permanece horizontal para valores grandes de dureza, sino que tiende a caer. Al aumentar la dureza de un acero, no necesariamente aumenta el límite de fatiga; para valores altos de dureza, un aumento de ésta puede implicar una reducción del límite de fatiga. La siguiente expresión es una relación aproximada entre el límite de fatiga, en MPa, y la dureza, en grados Brinell (HB), para los aceros : (5.4) Los datos experimentales muestran que la relación entre el límite de fatiga y el esfuerzo último de los aceros depende de la microestructura : Aceros al carbono Se’  0.6Su (ferrita) (5.5.a) Se’  0.4Su (perlita) (5.5.b) Se’  0.25Su (martensita) (5.5.c) Aceros aleados Se’  0.35Su (martensita) (5.6) ksi). (200 MPa 1380 si , 5 . 0   u u e S S ' S ksi). (200 MPa 1380 si , ksi 100 MPa 690    u e S ' S HB. 400 Dureza si [HB]) Dureza )( 72 . 1 ( ] MPa   ' Se Su (MPa) Se’ (MPa) 900 750 600 450 300 150 300 600 900 1200 1500 1800 Muy pocos casos Mayoría de datos 690 MPa 1380 MPa Pendiente: 0.5 ![Image 28: 28 CONCEPTOS BÁSICOS SOBRE DISEÑO DE MÁQUINAS 5.9 RESISTENCIA A LA FATIGA CORREGIDA PARA VIDA FINITA E INFINITA El límite de fatiga, Se’, y la resistencia a la fatiga para vida finita, Sf’, estudiadas en las secciones 5.5.1 y 5.5.2, son aplicables a probetas normalizadas y pulidas girando sometidas a flexión. Como se vio en las secciones anteriores, los elementos de máquinas distan mucho de tener las características de estas probetas de ensayo, ya que pueden tener superficies rugosas, estar sometidas a otros tipos de carga, trabajar en condiciones ambientales severas, etc.. Vamos a denominar resistencia a la fatiga corregida, Sn, a aquella que tiene en cuenta el efecto del estado superficial, del tamaño, de la confiabilidad, de la temperatura, del tipo de carga y de los efectos varios (pero no de los concentradores de esfuerzos). Parecería lógico pensar que Sn es el producto de los coeficientes Ka, Kb, Kc, Kd, Ke y Kcar por Se’ o Sf’. Esto es cierto para vida infinita, pero los datos experimentales han mostrado que estos factores inciden de manera diferente para vida finita. En esta sección se estudian las ecuaciones generales para determinar la resistencia a la fatiga corregida en función del número de ciclos. La figura 5.18 muestra un diagrama típico Sn-nc para los aceros, en escala logarítmica (se grafica la resistencia corregida). La curva está dividida en tres rectas AB, BC y CD. La línea AB está en el rango de bajo ciclaje (LCF), es decir, en el rango 100  nc  103 . Cuando se diseña un elemento para un vida menor o igual a 103 ciclos, puede tomarse, conservadoramente, como resistencia a la fatiga corregida el menor valor de ésta en ese rango; es decir, si 100  nc  103 , Sn = Sn103, donde          1] en citado [9, 1] en citado 9, 10 Mises von de es equivalent esfuerzos los calculan se si torsión para 9 . 0 torsión para 577 . 0 9 . 0 axial carga para 75 . 0 flexión para 9 . 0 3 u u u u n S S S S S (5.36) Nótese que a pesar de que con la ecuación anterior se calcula la resistencia a la fatiga corregida (que tiene en cuenta los efectos de ciertos factores), los factores Ka, Kb, Kc, Kd y Ke no están incluidos en ella. Esto se debe a que estos factores parecen no tener efecto en la resistencia para vidas menores a 103 ciclos. Figura 5.18 Diagrama Sn-nc típico de muchos aceros 10 0 10 3 nc 10 6 nc (log) Sn (log) Su KSe’ A B C Sn103 Sn D LCF HCF ![Image 29: CAPÍTULO 5 CARGAS VARIABLES - TEORÍA DE FATIGA 29 Para el rango 103 < nc < 106 , debe encontrase la ecuación de Sn en función de nc (ecuación de la línea BC). La ecuación de una línea recta en el diagrama logarítmico Sn - nc está dada por: (5.37) donde a y b son constantes. De acuerdo con las propiedades de los logaritmos, esta ecuación puede expresarse como: (5.38) donde b = logc, entonces (5.39.a y b) Las constantes a y c se determinan mediante los puntos conocidos B y C: en B, Sn = Sn103 y nc = 103 , y en C, Sn = KSe’ y nc = 106 , donde, como se vio . car e d c b a K K K K K K K  (5.20R ) Para 106 ciclos, los factores Ka, Kb, Kc, Kd, Ke y Kcar sí están involucrados en la ecuación para Sn, indicando que sí tienen efecto sobre ésta. Reemplazando estos dos pares de valores en la ecuación 5.39.b obtenemos: (5.40.a y b) De este sistema de ecuaciones se obtienen las constantes a y c: de la ecuación 5.40.a: (5.41) Reemplazando en la ecuación 5.40.b: (5.42) entonces: (5.43) De ésta se obtiene que: (5.44) Reemplazando la ecuación 5.43 en la 5.41 se obtiene: (5.45) , ) log( ) log( b n a S c n   , log ) log( ) log( c n S a c n   . ó ), log( ) log( a c n a c n n c S n c S     . ) 10 ( y , ) 10 ( 6 3 103 a e a n c ' KS c S   , 10 ó , 10 / 10 ) 3 6 ( 10 3 6 10 3 3 a n e a a n e S ' KS S ' KS     . / log 3 ó , / 10 3 3 10 10 3 n e n e a S ' KS a S ' KS    . / log 3 1 3 10 n e S ' KS a  . ] ' 2 10 10 10 3 3 3 ' KS S S KS S c e n n e n   . 103 103 a n S c  ![Image 30: 30 CONCEPTOS BÁSICOS SOBRE DISEÑO DE MÁQUINAS Reemplazando las ecuaciones 5.44 y 5.45 en la ecuación 5.39.b, tenemos: . ] [ 3 10 3 log 3 1 2 10                  n e S ' KS c e n n n ' KS S S (5.46) Ésta es la ecuación de la resistencia a la fatiga corregida, Sn, en función del número de ciclos, nc, para el rango 103 < nc < 106 . En el último rango, nc  106 , Sn es constante (figura 5.18) e igual a: (5.47) De acuerdo con esto, si se quiere diseñar para una duración mayor o igual a un millón de ciclos, debe usarse el límite de fatiga corregido; por lo tanto, la pieza duraría indefinidamente. En las ecuaciones 5.36 a 5.47 y el diagrama de la figura 5.18 no se ha incluido el efecto de los concentradores de esfuerzos. En las ecuaciones de diseño estudiadas más adelante se incluirá dicho efecto. En resumen, para los aceros o materiales que exhiben el codo C en nc  106 (figura 5.18), Sn está dado por: , 10 Si 3  c n . 3 10 n n S S  (5.48) , 10 10 Si 6 3   c n . ] [ 3 10 3 log 3 1 2 10                  n e S ' KS c e n n n ' KS S S (5.49) , 10 Si 6  c n . ' KS S e n  (5.50) Puede seguirse un procedimiento similar para materiales que no posean límite de fatiga. Puede demostrarse que para materiales que no exhiben el codo C (véase la figura 5.6): , 10 Si 3  c n . 3 10 n n S S  (5.51) , 10 Si 3 cref c n n   . ] [ ] 3 10 3 log 1 / 3 / 3 1 10                    ' KS S z c z f z n n f n n ' KS S S (5.52) , Si cref c n n  . ' KS S f n  (5.53) donde z = log(103 ) – log(ncref) = 3 – log(ncref) y Sf’ = Sf’@ncref es la resistencia a la fatiga para una vida ncref. Por ejemplo, para una aleación de aluminio se podría tomar ncref = 5108 , ya que se podría estimar Sf’@5108 mediante la ecuación 5.8. Nótese que no se presenta ecuación para nc > ncref. En este caso podría usarse la ecuación 5.52 (lo cual constituiría una extrapolación), pero el resultado tiene una confiabilidad cuestionable y, probablemente, es conservador. El estudiante puede comprobar que para ncref = 106 , la ecuación 5.52 es equivalente a la 5.49, excepto que en ésta se usa Se’ en lugar de Sf’. . ' KS S e n        ![Image 31: CAPÍTULO 5 CARGAS VARIABLES - TEORÍA DE FATIGA 31 Si ncref = 5108 , z = -5.699 y la ecuación 5.52 es equivalente a: , 10 5 10 Si 8 3    c n . ] [ ] 8 10 5 @ 3 10 8 3 log 699 . 5 1 5264 . 0 10 5 @ 5264 . 1 10                         ' KS S c f n n f n n ' KS S S (5.54) donde, Sn103 en las ecuaciones 5.48 a 5.54 está dado por la ecuación 5.36. 5.10 LÍNEAS DE FALLA 5.10.1 Introducción El límite de fatiga y la resistencia a la fatiga para vida finita constituyen propiedades base para el diseño de elementos sometidos a cargas variables. Sin embargo, Se’ o Sf’ no puede ser utilizado directamente en el diseño, ya que es obtenido bajo condiciones especiales de esfuerzo: probeta normalizada y pulida, girando sometida a flexión bajo condiciones ambientales favorables. Particularmente, nos interesa referirnos aquí a las condiciones de flexión giratoria. Bajo este tipo de carga, la probeta sufre un variación sinusoidal de esfuerzo repetido invertido para la cual el esfuerzo medio es igual a cero (véase la figura 5.10.a ó la 5.5.b). Se necesitan ecuaciones de diseño que sirvan no sólo para un esfuerzo repetido invertido, sino también para cualquier tipo de variación sinusoidal, donde Sm pueda ser diferente de cero. Para encontrar dichas ecuaciones fueron necesarias más pruebas experimentales, de las cuales se concluyó que, en general, si se agrega una componente media del esfuerzo, el elemento falla con una componente alternativa menor. Las figuras 5.19.a y b muestran las tendencias típicas que siguen los puntos de ensayo en diagramas de esfuerzo medio - esfuerzo alternativo, Sm-Sa y Sms-Sas, respectivamente, cuando se someten probetas normalizadas y pulidas a diferentes combinaciones de estos esfuerzos. Las cruces en los diagramas indican las combinaciones (Sm, Sa) o (Sms, Sas) para las cuales un pequeño aumento en el esfuerzo medio o en el alternativo produciría la falla de la probeta. Las combinaciones de esfuerzos que estén entre la zona de las cruces y el origen del diagrama no producirían falla, mientras que combinaciones de esfuerzos que estén por fuera de la zona de las cruces producirían la falla de la probeta de ensayo. Figura 5.19 Diagramas esfuerzo medio contra esfuerzo alternativo Analicemos algunos puntos de los diagramas. Al someter una probeta a un esfuerzo estático, ésta fallará cuando el esfuerzo (máximo) sea igual a la resistencia. Si el esfuerzo es estático, Smax = Smin = Sm y Sa = 0 (no hay amplitud de onda) o Ssmax = Ssmin = Sms y Sas = 0 (para esfuerzos cortantes); entonces, la falla en  Sms Sas 0.577Se’ Sus Falla por fatiga (a) Diagrama Sm-Sa (esfuerzos normales) (b) Diagrama Sms-Sas (esfuerzos cortantes) Sm Sa Se’ Su Falla inmediata por compresión Falla por fatiga ![Image 44: 44 CONCEPTOS BÁSICOS SOBRE DISEÑO DE MÁQUINAS El índice de sensibilidad a la entalla del material se obtiene con la ecuación 5.31. La constante de Neuber se toma de la tabla 5.3; para acero con Su = 550 MPa y carga de torsión, a = 0.31 mm0.5 . Reemplazando este valor en la ecuación 5.31 se obtiene: . 836 . 0 mm 5 . 2 mm 31 . 0 1 1 0.5    q De la ecuación 5.30 se obtiene el factor de concentración de esfuerzos por fatiga: . 69 . 1 ) 1 83 . 1 ( 836 . 0 1 ) 1 ( 1        t f K q K El factor de concentración de esfuerzos por fatiga para vida finita, Kff, es igual a Kf ya que se está diseñando para vida infinita (ecuación 5.34). Resistencia a la fatiga corregida: La resistencia a la fatiga corregida está dada por la ecuación 5.50: MPa. 4 . 73 MPa 275 267 . 0     ' KS S e n Cálculo de la potencia: El factor de seguridad se toma del rango adecuado de la tabla 3.1 (capítulo 3). Tomamos N = 1.5. Reemplazando los valores correspondientes en la ecuación 5.80, se obtiene: , ) m 07 . 0 ( 16 Pa 10 4 . 73 69 . 1 0 5 . 1 1 3 6  max us T S    de donde m. N 1950 16 69 . 1 5 . 1 ) m 07 . 0 ( Pa) 10 4 . 73 ( 3 6        max T Nótese que no es necesario verificar la condición dada por la ecuación 5.80, porque al ser Sms = 0, el punto (Sms, Sas) en el diagrama Sms-Sas estará por debajo de la línea de seguridad correspondiente a la falla inmediata por fluencia en tracción. La potencia la obtenemos mediante la ecuación 3.17 (capítulo 3): donde de , [r/min] 2 ] W [ 60 ] m N [ n P T    kW. 17 . 8 W 60 ) 1950 )( 40 ( 2 W 60 ] m N [ ] r/min 2       max max T n P La potencia máxima que se puede transmitir con el escalón del árbol (alternando el sentido de giro) sin que se produzca falla después de un gran número de ciclos es de 8.17 kW. ![Image 46: 46 CONCEPTOS BÁSICOS SOBRE DISEÑO DE MÁQUINAS (el momento calculado de esta manera es un poco mayor al real) y produce un resultado casi idéntico al que se obtendría descomponiendo la fuerza en dos, ya que la distancia entre las ruedas del carro es pequeña comparada con la luz de la viga. Ecuaciones de equilibrio y cálculo de las reacciones en los apoyos: x R R x M B B A ) kN/m 20 ( donde de ; 0 m) 10 ( kN) (200 ; 0      . ) kN/m 20 ( kN 200 donde de ; 0 kN 200 ; 0 x R R R F A B A y        Diagramas de fuerza cortante y momento flector: Figura 5.29 Diagrama de fuerza cortante de la viga Figura 5.30 Diagrama de momento flector de la viga Momento flector máximo: Los diagramas de fuerza cortante y momento flector se han construido con la fuerza máxima sobre la viga en una posición arbitraria. El momento máximo ocurre en la sección B y está dado por: , [m]) ( 20 [m] 200 m] kN 2 x x M    Aunque por la simplicidad de la configuración de la viga, se puede deducir directamente con qué distancia x se maximiza el momento flector, para encontrar esta distancia efectuamos lo siguiente: , 0  dx dM entonces m. 5 entonces , 0 40 200 entonces ; 0 ) 20 200 ( 2      x x dx x x d + V (kN) x (m) 200 – 20x – 20x A B C x 10 – x M (kNm) x (m) (200 – 20x)x A B C ![Image 53: CAPÍTULO 5 CARGAS VARIABLES - TEORÍA DE FATIGA 53   1 1    t f K q K , donde Kt se obtiene del apéndice 5. Para chaveteros y roscas, Kf se obtiene directamente de las tablas 5.4 y 5.5 respectivamente     0 entonces 2 ó 2 si entonces y si entonces y si entonces ó si                 fm ys smin smax f y min max f ms as f ys fm ys smin f ys smax f m a f y fm y min f y max f f fm s y max s f y max f K S S S K S S S K S S K S K S S K S S K S S K S K S S K S S K K K S S K S S K RESISTENCIA A LA FATIGA CORREGIDA (VIDA FINITA E INFINITA) 3 10 10 si , 3  c n n S  n S 6 3 log 3 1 2 10 10 10 si , ] [ 3 10 3                   c S ' KS c e n n n ' KS S n e 6 10 si ,  c e n ' KS 3 10 10 si , 3  c n n S cref c ' KS S z c z f z n n n n ' KS S f n                     3 log 1 / 3 / 3 1 10 10 si , ] [ ] [ 3 10 3 , donde z = 3 – log(ncref) 8 3 log 699 . 5 1 5264 . 0 10 5 @ 5264 . 1 10 10 5 10 si , ] [ ] 8 10 5 @ 3 10 8 3                           c ' KS S c f n n n ' KS S f n cref c n f n n ' KS cref  si , @ donde          Mises von de es equivalent esfuerzos los secalculan si torsión para 9 . 0 torsión para 577 . 0 9 . 0 axial carga para 75 . 0 flexión para 9 . 0 3 10 u u u u n S S S S S ciclos 10 si , 1 3  c n 6 3 log 3 1 10 10 si ,   c f K c n K n f ciclos 10 si , 6  c f n K c f ff n K K todo para ,   ff K Aceros y materiales dúctiles de baja resistencia Materiales frágiles o de alta resistencia Aceros o materiales con codo en nc = 106 Materiales que no poseen límite de fatiga Se puede asumir conservadoramente: material cualquier y todo para , c f ff n K K   n S Ad Recommended PDF Polea y Correas byAdrian Perez PPTX Diseño de flechas o ejes (calculo del factor de seguridad empleado para flechas) byAngel Villalpando PDF Mecanica_de_Fluidos_Fundamentos_y_Aplica.pdf byAlvaroGarcia947446 PPTX Diseño de flechas o ejes (selección de materiales) byAngel Villalpando PPT Diseño 13 factores que modifican el límite de resistencia a la fatiga-utp byMarc Llanos DOCX Promesa unilateral-mejor 2 bycqam DOCX Paralelos y meridianos, coordenadas geográficas byMartin Alberto Belaustegui PPTX Técnicas e Instrumentos de recolección de datos o información para la investi... by-_Oriana C. 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Odoznabwbbdizianqbvysud ajos9zuhswkao bynazaflores2023 Introducción. La historia de al Ándalus. Entre el rechazo y la mitificació... byAlvaroGonzalezGarcia10 PLAN DE CONTINGENCIA, SALUD OCUPACIONAL.pptx byfreddycardenas23 Sesión 4. Diapositivas 202501 (Parte 3).pptx byalexabel12 2do_Encuentro_Familiar_de_Aprendizaje.doc byNancyPanduroGarca slidesaver.app_hjdeih.GDFGDFGDFGDFGDFGpptx byDavidLeonardoGalindo1 Ejes teoria fatiga CAPÍTULO 5 CARGAS VARIABLES- TEORÍA DE FATIGA 5.1 INTRODUCCIÓN Muchos de los elementos de máquinas, tales como cigüeñales, árboles, ejes, bielas y resortes, son sometidos a cargas variables. El comportamiento de los materiales bajo este tipo de carga es diferente a aquel bajo cargas estáticas1 ; mientras que una pieza soporta una gran carga estática, la misma puede fallar con una carga mucho menor si ésta se repite un gran número de veces. Los esfuerzos variables en un elemento tienden a producir grietas que crecen a medida que éstos se repiten, hasta que se produce la falla total; este fenómeno se denomina fatiga. Por lo tanto, el diseño de elementos sometidos a cargas variables debe hacerse mediante una teoría que tenga en cuenta los factores que influyen en la aparición y desarrollo de las grietas, las cuales pueden producir la falla después de cierto número de repeticiones (ciclos) de esfuerzo. La teoría que estudia el comportamiento de los materiales sometidos a cargas variables se conoce como teoría de fatiga. En este libro se estudia el procedimiento de esfuerzo-vida, el cual es uno de los tres modelos de falla por fatiga que existen actualmente y es el más utilizado para elementos de maquinaria rotativos. Aunque en el capítulo 3 se resolvieron problemas de elementos sometidos a cargas variables, utilizando el procedimiento de diseño para cargas estáticas y factores de seguridad grandes, se afirmó que los resultados de este procedimiento son válidos como un diseño preliminar o previo, y que no sustituye el análisis por fatiga. En este capítulo, el estudio de la teoría de fatiga (modelo esfuerzo-vida) comienza por una breve historia y la descripción del mecanismo de falla por fatiga. Después se introducen los modelos de falla por fatiga y los conceptos de límite de fatiga y resistencia a la fatiga. Se estudia el modelado de los esfuerzos variables y los factores que modifican la resistencia a la fatiga. Luego se presentan las ecuaciones para calcular la resistencia a la fatiga corregida en función del número de ciclos. Se estudian diferentes modelos y líneas de falla, hasta llegar a las ecuaciones de diseño por fatiga para esfuerzos simples. Al final, se describe un procedimiento para diseñar elementos sometidos a cargas variables combinadas. 5.2 HISTORIA DE LA FATIGA El término fatiga se le denomina a la falla de un material sometido a cargas variables, después de cierto número de repeticiones (ciclos) de carga. Podría decirse que este tipo de falla fue observado por primera vez en el siglo XIX, cuando los ejes de los carros de ferrocarril comenzaron a fallar después de un corto tiempo de servicio. A pesar de haber sido construidos con acero dúctil, se observó una falla súbita de tipo frágil . 1 Cargas estáticas son aquellas que se aplican gradualmente, una sola vez y permanecen constantes en magnitud y dirección. 2 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Los ejes están sometidos a cargas transversales que generan flexión, tal como se ilustra en la figura 5.1.a. Debido al giro del eje, cualquier punto de la periferia pasará por el punto t1 (figura 5.1.b) (tiempo t1 en la figura 5.1.c), soportando un esfuerzo de tracción máximo. Luego pasará por el eje neutro (en t2) soportando cero esfuerzo. Cuando haya girado un cuarto de vuelta más soportará un esfuerzo máximo de compresión (en t3) (ya que estará al otro lado del eje neutro). Un cuarto de vuelta después, el punto pasará nuevamente por el eje neutro (en t4). Finalmente, el punto regresará a su posición inicial completando un ciclo de esfuerzo, donde comenzará el siguiente ciclo. Por lo tanto, este tipo de ejes está sometido a esfuerzos normales cíclicos. Figura 5.1 Esfuerzos variables en un eje girando sometido a un momento flector constante Suponiendo que las ruedas giraban con una frecuencia de cinco vueltas por segundo, cualquier punto de la periferia (punto crítico) de la sección crítica del eje sufrirá 5 ciclos de esfuerzo en un segundo. En un minuto sufrirá 560 = 300 ciclos; en una hora 30060 = 18000, en un día 1800024 = 4.32105 . En tres días de trabajo continuo, cada eje soportaría más de un millón de ciclos de esfuerzo. Entonces, a pesar del poco tiempo de funcionamiento de los ejes de las ruedas de los carros de ferrocarril, dichos ejes estuvieron sometidos a un gran número de ciclos de esfuerzo. Los ejes habían sido diseñados con los procedimientos de diseño de la fecha, los cuales estaban basados en la experiencia adquirida con estructuras cargadas estáticamente; sin embargo, éstos no eran correctos para las cargas variables que tenían que soportar los ejes y por eso fallaron. El término “fatiga” fue usado por primera vez por Poncelet en 1839, para describir la situación de falla de los materiales sometidos a cargas variables. Debido a que la falla por fatiga tiene apariencia frágil, se pensaba que el material se había “cansado” y hecho frágil después de soportar un cierto número de fluctuaciones de esfuerzo. Similarmente, en 1843, Rankine publicó un estudio sobre las causas de la ruptura inesperada de los muñones de los ejes de ferrocarril, en el cual decía que el material dúctil se había cristalizado y hecho frágil debido a la fluctuación de los esfuerzos. Luego, en 1870, después de 12 años de investigación sobre las fallas por fatiga, el ingeniero alemán August Wohler publicó los resultados de su estudio, en el cual se mostraba como “culpable” de la falla al número de ciclos de esfuerzo. Al realizar pruebas sobre las mitades rotas de los ejes que habían fallado por fatiga, observó que el material tenía la misma resistencia y ductilidad bajo carga de tensión que el material original; es decir, el material no se había “cansado” ni fragilizado como se creía. Sin embargo, el término fatiga se sigue utilizando para referirse a las fallas de elementos sometidos a cargas variables. Este ingeniero alemán encontró, además, la existencia de un límite de resistencia a la fatiga (o límite de fatiga) para los aceros. Wohler realizó pruebas sobre probetas de acero sometidas a “flexión giratoria”, denominada así al tipo de carga que se genera en un elemento que gira sometido a un momento flector (a) Eje giratorio sometido a flexión (b) Cualquier punto en la periferia soporta un esfuerzo que varía desde el máximo (en tracción) hasta el mínimo (en compresión) y viceversa (c) Variación sinusoidal del esfuerzo en cualquier punto de la sección; en los puntos de la periferia se da la mayor amplitud M n (r/min) t1 t2 t3 t4 E.N. S t1 t2 t3 t4 t5 t Smax Smin CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 3 constante. En dichas pruebas se pretendía relacionar los niveles de esfuerzo a los cuales se sometían las probetas, con el número de ciclos de carga que soportaban hasta la falla. Wohler obtuvo un diagrama como el de la figura 5.2, el cual es conocido como diagrama S-nc (esfuerzo - número de ciclos) o diagrama de vida-resistencia de Wohler. El esfuerzo (o resistencia) S corresponde al valor del esfuerzo máximo (véase la figura 5.1.c) al cual se somete la probeta, y nc es el número de ciclos de esfuerzo. Las líneas del diagrama representan aproximaciones a los puntos reales de falla obtenidos en los ensayos. Figura 5.2 Diagrama S-nc o diagrama de Wohler En la figura 5.2 se observa que para el rango donde la pendiente de la curva continua es negativa, entre menor sea el esfuerzo al cual se somete la probeta, mayor es su duración. Si se somete una probeta al esfuerzo último, Su, la probeta sólo soporta la primera aplicación de la carga máxima. Para esfuerzos menores se tendrán mayores duraciones. El diagrama para muchos aceros es como el dado por la curva ABC. La curva tiene un codo en S = Se’ y nc  106 ciclos, a partir del cual el esfuerzo que produce la falla permanece constante. Esto indica que si la probeta se somete a un esfuerzo menor que Se’, ésta no fallará; es decir, la probeta tendrá una vida infinita. A niveles superiores de esfuerzo, la probeta fallará después de un número de ciclos de carga y, por lo tanto, tendrá vida finita. Como Se’ es el límite por debajo del cual no se produce falla, se le conoce como límite de fatiga; éste se definirá en la sección 5.5.1. Muchos aceros al carbono y aleados, algunos aceros inoxidables, hierros, aleaciones de molibdeno, aleaciones de titanio y algunos polímeros[2, citado en 1] poseen un codo a partir del cual la pendiente de la curva es nula. Otros materiales como el aluminio, el magnesio, el cobre, las aleaciones de níquel, algunos aceros inoxidables y el acero de alto contenido de carbono y de aleación[2, citado en 1] no poseen límite de fatiga, teniendo comportamientos similares al dado por la curva ABD de la figura 5.2. A pesar de que la pendiente de la curva puede ser menor para nc mayor de aproximadamente 107 ciclos, teóricamente no existe un nivel de esfuerzo, por pequeño que éste sea, que nunca produzca la falla en la probeta. La prueba de flexión giratoria, que arroja los datos de la curva S-nc, se convirtió en estándar. Para muchos materiales de ingeniería se han desarrollado estas pruebas con el fin de determinar sus comportamientos cuando se someten a cargas variables y, específicamente, para encontrar los límites de fatiga o la resistencia a la fatiga para un número de ciclos determinado. 10 0 10 3 10 6 10 9 nc (log) S (log) Su Se’ Materiales que poseen límite de fatiga (por ejemplo, el acero) Materiales que no poseen límite de fatiga (por ejemplo, el aluminio) A B C D 4 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Para terminar con la historia de la fatiga, consideremos los siguientes casos: en 1954, muchos años después del descubrimiento del fenómeno de la fatiga, el Comet Británico, la primera aeronave a reacción comercial para pasajeros, tuvo dos accidentes graves debidos a fallas por fatiga causadas por ciclos de presurización y despresurización de la cabina. Más recientemente, en 1988, un Boeing 737 de Hawaiian Airlines perdió un tercio de la parte superior de la cabina en pleno vuelo; aterrizó con pérdidas mínimas de vidas. Éstos, sin embargo, no son los únicos ejemplos recientes de falla por fatiga de tipo catastrófico. 5.3 MECANISMO DE FALLA POR FATIGA Como se dijo en la sección 5.2, la fatiga es la falla de un material sometido a cargas variables, después de cierto número de ciclos de carga. Dos casos típicos en los cuales podría ocurrir falla por fatiga son los ejes, como los de los carros de ferrocarril, y los árboles2 . Normalmente, estos elementos giran sometidos a flexión, que es el mismo tipo de carga al cual se someten las probetas en la técnica de ensayo de fatiga por flexión giratoria: las fibras pasan de tracción a compresión y de compresión a tracción en cada revolución del elemento (figura 5.1), y muchas veces bastan menos de unos pocos días para que el número de ciclos de carga alcance un millón. La figura 5.3 muestra un árbol de transmisión de potencia en el cual las fuerzas involucradas generan torsión, cortante directo y flexión; las dos últimas generan esfuerzos variables. Figura 5.3 Árbol de transmisión de potencia. Las fuerzas sobre la rueda dentada y el piñón, y las reacciones en los apoyos, generan torsión, flexión y fuerza cortante. Particularmente, se producen esfuerzos variables por flexión, debido al giro del árbol Si las cargas variables sobre un elemento, como el árbol de la figura 5.3, son de magnitud suficiente como para producir fluencia en ciertos puntos, es posible que después de cierto tiempo aparezca una grieta microscópica. Normalmente, ésta se genera en la vecindad de una discontinuidad (véase la figura 5.4) o en un punto que soporta un gran esfuerzo o una gran fluctuación de éste. La grieta que se inicia es un concentrador de esfuerzos altamente nocivo; por lo tanto, tiende a expandirse con cada fluctuación de los esfuerzos. Efectivamente, la grieta crece gradualmente (con cada ciclo de esfuerzo), a lo largo de planos normales al esfuerzo máximo a tracción . A pesar de que en materiales dúctiles los esfuerzos cortantes son los encargados de iniciar las grietas, los esfuerzos normales de tracción son los que actúan en la grieta tratando de abrirla y haciendo que crezca. La sección del material se reduce con el crecimiento gradual de la grieta, hasta que finalmente se rompe cuando la combinación del tamaño de la grieta y de la 2 Los árboles y ejes son elementos que soportan piezas que giran solidariamente o entorno a ellos. La diferencia entre árbol y eje radica en que el primero transmite potencia mientras que el último no. Rodamiento o cojinete de contacto rodante Árbol Piñón (estrella) de una transmisión por cadena Rueda dentada Fuerzas CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 5 magnitud del esfuerzo nominal (que depende del tamaño remanente de la sección) produce una fractura súbita de tipo frágil.3 La falla por fatiga se puede dividir en tres etapas. La etapa de iniciación de grietas, en la cual el esfuerzo variable sobre algún punto genera una grieta después de cierto tiempo. La etapa de propagación de grietas, que consiste en el crecimiento gradual de la grieta. Finalmente, la etapa de fractura súbita, que ocurre por el crecimiento inestable de la grieta. La figura 5.4 ilustra la apariencia típica de la sección de un elemento que ha fallado por fatiga. La falla comienza alrededor de un punto de gran esfuerzo, en el chavetero (o cuñero), desde donde se extiende paulatinamente formando ralladuras denominadas marcas de playa. Durante la fractura progresiva del material, ocurre rozamiento entre las caras de la sección, produciéndose una superficie lisa y brillante. Finalmente, el elemento falla súbitamente dejando una superficie áspera como si fuera un material frágil. Figura 5.4 Ilustración de una sección de un árbol que ha fallado por fatiga El tipo de fractura producido por fatiga se le denomina comúnmente progresiva, debido a la forma paulatina en que ocurre, frágil, debido a que la fractura ocurre sin deformación plástica apreciable, y súbita, porque la falla final ocurre muy rápidamente. El mecanismo de falla por fatiga siempre empieza con una grieta (preexistente o que se forma) y ocurre cuando el esfuerzo repetido en algún punto excede algún valor crítico relacionado con la resistencia a la fatiga del material. Para los materiales que poseen límite de fatiga, teóricamente es posible que nunca se generen grietas y, por lo tanto, que no ocurra la falla, si los esfuerzos son tales que las deformaciones en el material sean siempre elásticas. Esto es lo deseable cuando se diseña para que un elemento soporte las cargas indefinidamente. Finalmente, es conveniente tener presente que los materiales poco dúctiles, los cuales tienen poca capacidad de deformación plástica, tienden a generar grietas con mayor rapidez que los materiales más dúctiles. Además, los materiales frágiles pueden llegar directamente a la propagación de grietas, a partir de microgrietas preexistentes . Los materiales frágiles no son adecuados para aplicaciones con carga variable. 3 La mecánica de fractura estudia el fenómeno de falla súbita en elementos con grietas. Este fenómeno de falla es el mismo, ya sea que el elemento esté sometido a carga estática o a carga variable, y depende, entre otros factores, del tamaño de la grieta y del valor del esfuerzo nominal en la sección. Fractura repentina (tipo frágil) Fractura progresiva Grieta inicial Chavetero (discontinuidad) 6 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS 5.4 MODELOS DE FALLA POR FATIGA Antes de presentar los modelos de falla por fatiga que se utilizan actualmente, es conveniente hablar sobre los regímenes de fatiga. Regímenes de fatiga Cuando se tienen elementos sometidos a esfuerzos cíclicos se habla de los regímenes de fatiga: fatiga de bajo ciclaje (LCF) y fatiga de alto ciclaje (HCF), los cuales tienen relación con el número de veces que se repiten los esfuerzos en un elemento. Un régimen de bajo ciclaje es aquel en el cual se somete un elemento a un número de ciclos de esfuerzo menor que aproximadamente 102 a 104 , según el material. Aunque es lógico pensar en que no existe una línea divisoria exacta entre los dos regímenes, es usual hablar de 103 ciclos como línea divisoria; es decir, si una pieza soporta menos de 103 ciclos, está en régimen de bajo ciclaje, mientras que si soporta más de 103 ciclos, está en régimen de alto ciclaje. Esta clasificación es conveniente desde el punto de vista de la aplicación de los modelos de falla por fatiga. Modelos de falla por fatiga Actualmente existen tres modelos de falla por fatiga: el procedimiento de vida-esfuerzo, el de vida- deformación y el de mecánica de fractura elástica lineal (LEFM); cada uno de ellos tiene sus ventajas y desventajas y tiene cabida en cierta aplicación. La teoría de fatiga que se estudia en este capítulo corresponde al modelo de vida-esfuerzo, que es el más antiguo. Este modelo es adecuado para el diseño de piezas en el régimen de alto ciclaje (HCF) en las cuales la variación de los esfuerzos sea conocida y consistente, como ocurre generalmente en las máquinas rotativas. Como se verá, este modelo consiste en limitar los esfuerzos a valores menores que los críticos y es fácil de aplicar; además, hay muchos datos empíricos disponibles. El método de vida-deformación se basa en las deformaciones del elemento. Es más aplicable a situaciones de bajo ciclaje (LCF) para predecir la iniciación de grietas y es bastante complejo, por lo que requiere del uso de computador. Finalmente, el modelo de mecánica de fractura elástica lineal (LEFM) es mejor para el estudio de la etapa de propagación de grietas; por lo tanto, es útil para predecir la vida de bajo ciclaje (LCF) de piezas ya agrietadas. Estas dos últimas teorías se utilizan, por ejemplo, en las máquinas de transporte en las cuales ocurren sobrecargas grandes, muy pocas veces durante la vida esperada. 5.5 LÍMITE DE FATIGA Y RESISTENCIA A LA FATIGA Cuando se efectúa el diseño de elementos sometidos a cargas estáticas, las propiedades que interesan son el esfuerzo último a tracción, compresión o torsión (Su, Suc o Sus) y la resistencia de fluencia en tracción, compresión o torsión (Sy, Syc o Sys). Estas propiedades se obtienen con ensayos con carga estática. Para carga variable se debe utilizar, además, una propiedad que tenga en cuenta la resistencia a las cargas variables. Esta propiedad podría ser el límite de fatiga. 5.5.1 Límite de fatiga El límite de fatiga es el esfuerzo máximo invertido que puede ser repetido un número indefinido de veces sobre una probeta normalizada y pulimentada girando sometida a flexión, sin que se produzca falla o rotura. CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 7 Para aclarar este concepto considere la figura 5.5. La figura 5.5.a muestra la forma típica de una probeta para el ensayo de fatiga, la cual es de sección circular con diámetro d en la parte más delgada. Esta probeta se somete a un momento flector constante M y se pone a girar a n revoluciones por minuto. En el instante mostrado, el punto más crítico (punto A) soporta un esfuerzo normal dado por: (5.1) Figura 5.5 Ensayo de fatiga por flexión giratoria y diagrama de Wohler Debido al giro de la probeta, el punto A (y cualquier punto en la periferia de la sección más delgada) soporta un esfuerzo que varía en la forma sinusoidal mostrada en la figura 5.5.b. A este tipo de variación de esfuerzo se le denomina repetido invertido. De acuerdo con la definición del límite de fatiga, al someter una probeta normalizada y pulimentada a flexión giratoria, el máximo esfuerzo S (dado por la ecuación 5.1) al cual se puede someter dicha probeta sin que falle, aún después de un gran número de ciclos de carga, es el límite de fatiga, denominado Se’. El límite de fatiga se obtiene realizando un gran número de veces la prueba de fatiga con valores diferentes de S (variando el momento flector aplicado). Para cada probeta se ubica una equis en el diagrama, con el esfuerzo aplicado y el número de vueltas que giró hasta romperse. La figura 5.5.c muestra un diagrama de Wohler típico de un acero. Los puntos de ensayo siguen una tendencia como la mostrada. A partir del punto A (1, Su) la línea desciende hasta al codo en B, y partir de allí los puntos tienden a ajustarse a la línea horizontal BC; el valor de esfuerzo correspondiente a esta línea es el límite de fatiga, Se’; al aplicar un esfuerzo menor a Se’ no ocurriría la falla, mientras que un esfuerzo mayor produciría la falla después de cierto número de vueltas. . 32 3 d M I Mc S    (a) Probeta típica para el ensayo de fatiga. La sección crítica es circular (b) Variación sinusoidal del esfuerzo en el punto A. Smax = –Smin = (Mc)/(I) = (32M)/(d 3 ) M M n (r/min) A d (c) Diagrama S-nc o de Wohler 10 0 10 3 10 6 nc (log) S (log) Su Se’ A B C Límite de fatiga S 0 ¼ ½ ¾ 1 No. de vueltas Smax Smin 8 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS 5.5.2 Resistencia a la fatiga para vida finita Las piezas sometidas a cargas variables pueden diseñarse para un número de ciclos determinado, dependiendo de la vida requerida. Particularmente, los materiales que no poseen límite de fatiga no se pueden diseñar para vida infinita, sino que deben diseñarse para una duración determinada. Entonces, podemos hablar de una “resistencia al fatiga” para vida finita. La figura 5.6 muestra un diagrama S-nc o de Wohler, típico de algunos materiales que no poseen límite de fatiga (aleaciones de aluminio, cobre, etc.). Si, por ejemplo, se quiere diseñar para una duración finita de 108 ciclos, la resistencia a usar es Sf’@1108, que corresponde a un nivel de esfuerzo que idealmente produciría una vida de nc = 108 . En general, denotamos Sf’ a la resistencia a la fatiga para vida finita. Figura 5.6 Diagrama S-nc o de Wohler para un material sin límite de fatiga La resistencia a la fatiga para vida finita, al igual que el límite de fatiga, es una propiedad que se basa en pruebas de flexión giratoria sobre probetas normalizadas y pulidas. Sin embargo, es necesario aclarar que pueden desarrollarse diagramas esfuerzo-deformación y obtenerse resistencias a la fatiga para carga axial, torsión y otros tipos de flexión. En este texto se trabajará sólo con límites o resistencias a la fatiga para flexión giratoria. 5.6 LÍMITES Y RESISTENCIAS A LA FATIGA En esta sección se presentan algunas ecuaciones y datos generales sobre los límites y resistencias a la fatiga de algunos materiales comunes. 5.6.1 Aceros La figura 5.7 ilustra la relación entre el límite de fatiga y el esfuerzo último para diferentes aceros. La zona de sombreado oscuro corresponde a la tendencia que sigue la mayoría de los aceros, de acuerdo con los datos experimentales; se observa que para valores de esfuerzo último menores de aproximadamente 1380 MPa, entre mayor es el Su del acero, mayor es su límite de fatiga. Sin embargo, para valores de esfuerzo último por encima de 1380 MPa, el límite de fatiga parece ser independiente de Su. Incluso, para aceros con Su muy por encima de este valor, la resistencia a la fatiga puede ser inferior a la de un acero con Su = 1380 MPa. Esto nos indica que, si de resistencia a la fatiga se trata, no parece ser conveniente utilizar un acero con Su > 1380 MPa. La tendencia de los datos se aproxima a las dos líneas rectas mostradas en la figura 5.7. Una línea tiene una pendiente de 0.5 y, al extrapolar, partiría desde el origen del diagrama; esto indica que el límite de 10 0 10 3 10 6 10 8 nc (log) S (log) Su Sf’@110 8 Resistencia a la fatiga para 108 ciclos CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 9 fatiga es la mitad del esfuerzo último. La otra línea es horizontal y parte desde el punto (1380, 690) MPa; se asume, entonces, que para los aceros con Su > 1380 MPa, el límite de fatiga es aproximadamente 690 MPa. Figura 5.7 Relación entre el límite de fatiga y el esfuerzo último de los aceros Podemos concluir que para la mayoría de los aceros (de bajo carbono, aleados, inoxidables): (5.2) (5.3) Una gráfica similar a la de la figura 5.7, se obtiene para la relación entre la dureza de un acero y el límite de fatiga. La diferencia principal e importante es que la curva no permanece horizontal para valores grandes de dureza, sino que tiende a caer. Al aumentar la dureza de un acero, no necesariamente aumenta el límite de fatiga; para valores altos de dureza, un aumento de ésta puede implicar una reducción del límite de fatiga. La siguiente expresión es una relación aproximada entre el límite de fatiga, en MPa, y la dureza, en grados Brinell (HB), para los aceros : (5.4) Los datos experimentales muestran que la relación entre el límite de fatiga y el esfuerzo último de los aceros depende de la microestructura : Aceros al carbono Se’  0.6Su (ferrita) (5.5.a) Se’  0.4Su (perlita) (5.5.b) Se’  0.25Su (martensita) (5.5.c) Aceros aleados Se’  0.35Su (martensita) (5.6) ksi). (200 MPa 1380 si , 5 . 0   u u e S S ' S ksi). (200 MPa 1380 si , ksi 100 MPa 690    u e S ' S HB. 400 Dureza si [HB]) Dureza )( 72 . 1 ( ] MPa [   ' Se Su (MPa) Se’ (MPa) 900 750 600 450 300 150 300 600 900 1200 1500 1800 Muy pocos casos Mayoría de datos 690 MPa 1380 MPa Pendiente: 0.5 10 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Una ecuación más elaborada es recomendada por Lessells[citado en 3] 4 , para aceros aleados de alta resistencia: (5.7) donde a y b son valores que se obtienen mediante interpolación rectilínea de los datos de la tabla 5.1. Tabla 5.1 Coeficientes a y b para la ecuación 5.7. Sy (MPa) a Su (MPa) b 588 0.2 588 0.45 1310 0.4 1372 0 Las ecuaciones 5.2 a 5.7 se deben usar cuando no se posea información más exacta sobre el límite de fatiga o cuando no sea necesario obtenerla, ya que como se dijo, los valores obtenidos de Se’ son aproximados. Cuando se conozca el límite de fatiga de un determinado acero (de libros, catálogos, etc.) debe utilizarse dicho valor, ya que seguramente fue encontrado con datos experimentales para dicho acero. Finalmente, ciertas aplicaciones podrían demandar ensayos experimentales para obtener un valor más exacto del límite de fatiga para un acero o material en particular. Para los ejemplos y ejercicios de este capítulo, se utilizará preferiblemente la ecuación 5.2 ó 5.3, ya que es sencilla de recordar y usar. En caso de que se disponga del límite de fatiga, se utilizará dicho valor. 5.6.2 Otros materiales A continuación se dan las ecuaciones para determinar los límites de fatiga aproximados de algunos materiales ferrosos fundidos y las resistencias a la fatiga de las aleaciones de aluminio y de cobre, para una duración de 5108 ciclos. Al igual que para el acero, estas ecuaciones son aproximaciones de datos experimentales. Aleaciones de aluminio Sf’@5108  0.4 Su, para Su < 330 MPa (48 ksi). (5.8.a) Sf’@5108  132 MPa = 19 ksi, para Su  330 MPa (48 ksi). (5.8.b) Aleaciones de cobre Sf’@5108  0.4 Su, para Su < 276 MPa (40 ksi). (5.9.a) Sf’@5108  96 MPa = 14 ksi, para Su  276 MPa (40 ksi). (5.9.b) Algunos materiales ferrosos fundidos De acuerdo con Faires : Acero fundido Se’  0.4 Su. (5.10) Hierro fundido gris Se’  0.35 Su. (5.11) Hierro fundido nodular Se’  0.4 Su. (5.12) Hierro fundido nodular normalizado Se’  0.33 Su. (5.13) 4 ASTM, Metals, Pub. esp. núm. 196. , u y e bS aS ' S   CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 11 Adicionalmente, Norton plantea lo siguiente para los hierros: Se’  0.4 Su, para Su < 400 MPa (60 ksi). (5.14.a) Se’  160 MPa = 24 ksi, para Su  400 MPa (60 ksi). (5.14.b) Una conclusión general que puede extraerse de los datos de esta sección es que los límites de fatiga o las resistencias a la fatiga son bastante menores que el esfuerzo último. La relación entre el límite de fatiga o la resistencia a la fatiga para 5108 ciclos y el esfuerzo último para los materiales mencionados varía entre 0.25 (ó menos) y 0.6 aproximadamente. Además, si se tiene en cuenta que existen varios factores que reducen la resistencia a las cargas variables (sección 5.8), las resistencias a la fatiga de las piezas de máquinas serán aún mucho menores. 5.7 VARIACIÓN DE LOS ESFUERZOS La variación de los esfuerzos de los elementos de ingeniería no necesariamente es igual a la que ocurre en flexión giratoria. Por lo tanto, es necesario considerar distintos casos de variación de esfuerzos. En un elemento sometido a cargas variables, los esfuerzos pueden variar con respecto al tiempo, t, de una forma muy irregular como se observa en la figura 5.8, y generalmente es difícil predecir con exactitud cómo es tal variación. Por ejemplo, el esfuerzo normal máximo en una viga de un ala de una aeronave puede variar de manera muy irregular, al ser sometida a la fuerza del viento y a las vibraciones que debe soportar. Figura 5.8 Curva esfuerzo – tiempo (S-t) del punto crítico de un elemento El modelo de falla por fatiga estudiado en este libro se basa en el diagrama de Wohler (figura 5.2, por ejemplo), el cual se obtiene con pruebas sobre probetas sometidas a flexión giratoria. Este tipo de carga produce una variación sinusoidal de los esfuerzos (figura 5.5.b); por lo tanto, en esta teoría se modela cualquier tipo de variación de esfuerzos de una manera sinusoidal5 . La figura 5.9 muestra un modelo de variación sinusoidal para la variación real de esfuerzos de la figura 5.8. Los valores de los esfuerzos máximo y mínimo se han mantenido iguales a los valores reales, y el periodo (o la frecuencia) fue escogido de tal manera que parezca similar al de la curva real. 5 En ausencia de corrosión, la forma exacta en que varían los esfuerzos no es tan importante; lo que incide en la falla por fatiga es el número de ciclos y los valores máximos y mínimos de los esfuerzos. S t Smax Smin 12 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Figura 5.9 Modelo sinusoidal de la curva S-t del punto crítico de un elemento. En este caso particular, el esfuerzo es fluctuante (con compresión más pequeña) Las variables o parámetros de un modelo como el de la figura 5.9 son: Smax : esfuerzo máximo. Smin : esfuerzo mínimo. Sm : esfuerzo medio (o promedio). Sa : esfuerzo alternativo (amplitud de la onda). R : relación de esfuerzos; es la relación entre el esfuerzo mínimo y el esfuerzo máximo. De la figura 5.9 pueden obtenerse las dos primeras de las siguientes relaciones: , 2 min max m S S S   , 2 min max a S S S   . max min S S R  (5.15) Para esfuerzos cortantes, las ecuaciones son: , 2 smin smax ms S S S   , 2 smin smax as S S S   , smax smin S S R  (5.16) donde Ssmax, Ssmin, Sms y Sas son los esfuerzos cortantes máximo, mínimo, medio y alternativo, respectivamente. Otras expresiones que se derivan de las anteriores son: , a m max S S S   . a m min S S S   (5.17) , as ms smax S S S   . as ms smin S S S   (5.18) Para el manejo de estas variables, debe tenerse en cuenta que un esfuerzo normal de tracción es positivo y un esfuerzo normal de compresión es negativo. Para el caso de esfuerzos cortantes, el esfuerzo es positivo en la dirección en la que éste alcanza su mayor magnitud, y es negativo en dirección contraria. Las variables en la cuales nos debemos concentrar para la utilización de las ecuaciones de diseño son el esfuerzo medio y el esfuerzo alternativo. De lo dicho hasta aquí y de las ecuaciones se puede inferir que (i) el esfuerzo alternativo nunca es negativo, ya que representa la amplitud de una onda; (ii) para el caso de esfuerzos normales, el esfuerzo S t Smax Smin Sm Sa Sa 0 CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 13 medio puede ser negativo o positivo, mientras que para esfuerzos cortantes el esfuerzo medio es siempre positivo (ya que la dirección positiva del esfuerzo es en la dirección del esfuerzo máximo). Algunas variaciones típicas de esfuerzo se muestran en la figura 5.10, con sus nombres y los rangos correspondientes para R. Figura 5.10 Variaciones sinusoidales de esfuerzo con respecto al tiempo De acuerdo con el procedimiento que sigue Norton para el caso de torsión o cortante, deben calcularse los esfuerzos equivalentes de von Mises. Para cortante simple, el estado de esfuerzo es como el mostrado en la figura 4.40.a (capítulo 4), y el esfuerzo equivalente de von Mises se obtiene de la ecuación 4.42: ; 3 2 2 2 s YY XX YY XX e S S S S S      (4.42R ) entonces ; 577 . 0 / 3 3 2 s s s e S S S     por lo tanto . 577 . 0 / y 577 . 0 / as ae ms me S S     (5.19) No se incluyó el valor absoluto en la ecuación 5.19, ya que tanto Sms como Sas son siempre positivos. S t Smax Smin Sm = 0 Sa Sa (a) Esfuerzo repetido invertido (totalmente alternante); por ejemplo, en una viga sometida a flexión giratoria R = –1 S Smax Smin Sm Sa Sa (c) Esfuerzo fluctuante; sólo tracción (línea continua) o sólo compresión (línea a trazos) 0 < R < 1 t R > 1 Sólo tracción Sólo compresión 0 S t Smax Smin = 0 Sm Sa Sa (b) Esfuerzo repetido en una dirección; Smin = 0 (ó Smax = 0, si el esfuerzo siempre es de compresión (R = )) R = 0 S t Smax Smin Sm Sa Sa (d) Esfuerzo fluctuante con tracción más pequeña. El caso de esfuerzo fluctuante con compresión más pequeña se muestra en la figura 5.9, para el que –1 < R < 0) R < –1 0 14 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS 5.8 FACTORES QUE AFECTAN LA RESISTENCIA A LA FATIGA En las secciones precedentes se habló del diagrama de Wohler, a partir del cual se obtienen el límite de fatiga y las resistencias a la fatiga. Como se dijo, estas propiedades están basadas en probetas de ensayo normalizadas (usualmente de 0.3 in de diámetro en la sección más delgada), de superficies altamente pulidas (pulido a espejo), trabajando bajo condiciones ambientales favorables, como por ejemplo, en ausencia de corrosión y a temperaturas “normales”. En general, las características de los elementos de máquinas y de su entorno difieren de aquellas de las probetas de ensayo. Las piezas suelen tener mayores rugosidades, ya que obtener una superficie pulida a espejo es un procedimiento costoso. Las temperaturas de trabajo pueden ser “bajas”, tendiendo a fragilizar los materiales, o “muy altas”, reduciendo la resistencia del material o produciendo el fenómeno de termofluencia (“creep”). Los elementos pueden tener concentradores de esfuerzos o pueden estar en presencia de agentes corrosivos. Factores como éstos se estudiarán en esta sección, y se presentarán las variables que involucran sus efectos sobre la falla de los materiales. Los factores que modifican la resistencia a la fatiga son: (a) Estado superficial (b) Tamaño de la pieza (c) Variabilidad en los procesos de manufactura (d) Temperatura (e) Efectos varios (corrosión, esfuerzos residuales y otros) (f) Tipo de carga (g) Concentradores de esfuerzos Los efectos de los aspectos (a) hasta (f) sobre la resistencia a la fatiga se cuantifican mediante los factores Ka, Kb, Kc, Kd, Ke y Kcar, cuyo producto se designa con el término K: . car e d c b a K K K K K K K  (5.20) 5.8.1 Factor de superficie (Ka) El estado superficial tiene efecto sobre la resistencia a la fatiga de los elementos; a mayor rugosidad de la superficie, menor será la resistencia, ya que las irregularidades de la superficie actúan como pequeñísimos concentradores de esfuerzos que pueden iniciar una grieta de manera más temprana. La manera de introducir el efecto del acabado superficial, así como el de otros aspectos que se estudian en las secciones 5.8.2 a 5.8.6, es definiendo factores multiplicadores de la resistencia a la fatiga. El factor de superficie, Ka, es el coeficiente que tiene en cuenta el efecto del acabado superficial sobre la resistencia del material a las cargas variables y está en el intervalo [0,1]. Para el caso de elementos pulidos a espejo Ka = 1, ya que este tipo de superficie es el que tienen las probetas para determinar el límite de fatiga; por lo tanto, no habría necesidad de hacer corrección por estado superficial. Un valor menor que uno implica que el estado superficial reduce en cierto grado la resistencia. Si, por ejemplo, Ka = 0.35, la resistencia a la fatiga corregida para vida infinita sería el 35% del límite de fatiga, si se considerara sólo el efecto de la rugosidad. La figura 5.11 presenta los resultados de ensayos experimentales efectuados sobre probetas de acero con diferentes acabados superficiales. Las curvas de los acabados más pulidos están por encima (los valores de Ka son mayores) de curvas de procesos que producen mayores rugosidades o que generan CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 15 descarburación de la superficie, como ocurre con el laminado en caliente y el forjado; una superficie descarburada contiene menor porcentaje de carbono, lo cual reduce la resistencia del acero. Figura 5.11 Factores de superficie para el acero. Fuente: Juvinall . Pueden aplicarse a otros metales dúctiles, pero como valores aproximados De la figura se puede observar también que Ka no sólo depende del acabado superficial sino también del esfuerzo último del acero. A mayor esfuerzo último, menor tiende a ser el factor de superficie, ya que los aceros más resistentes tienden a ser más sensibles a los efectos de concentración de esfuerzos producidos por las imperfecciones de la superficie. Esto se debe a que dichos materiales tienden a ser menos dúctiles, es decir, a tener menos capacidad de deformarse plásticamente. Como se dijo, las curvas de la figura 5.11 son válidas para el acero, con el cual se hicieron las pruebas experimentales, aunque también pueden aplicarse a otros metales dúctiles, pero como valores aproximados . Para el hierro fundido se puede tomar Ka = 1, porque las discontinuidades internas debidas a las inclusiones de grafito, hacen que la rugosidad de la superficie no reduzca de manera adicional la resistencia a la fatiga. 5.8.2 Factor de tamaño (Kb) El tamaño de la pieza en las secciones críticas también tiene efecto sobre su resistencia. En general, a mayor tamaño de la pieza menor es su resistencia, aunque para carga axial no existe este efecto. La pérdida de resistencia al aumentar los tamaños de las piezas se debe a que hay una mayor probabilidad de que exista un defecto en el volumen que soporta los mayores esfuerzos. Considere la figura 5.12 en la cual se muestran las secciones transversales de dos probetas; la segunda con el doble de diámetro que la 60 80 100 120 140 160 180 200 220 240 260 ksi 400 600 800 1000 1200 1400 1600 1800 MPa Esfuerzo último, Su Dureza Brinell (HB) 120 160 200 240 280 320 360 400 440 480 520 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Factor de superficie, Ka Pulido a espejo Rectificado fino o pulido comercial Mecanizado Laminado en caliente Forjado Corrosión con agua de la llave Corrosión en agua marina 16 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS primera. Si las probetas están sometidas a flexión giratoria, los puntos que soportan mayores esfuerzos son los que están entre la circunferencia a trazos y el contorno de la sección; si los puntos mostrados fueran defectos en la sección, la de mayor diámetro tendrá muchos más defectos en la zona crítica, con lo que tendría mayor probabilidad de que se iniciara una grieta por alguno de ellos. La curva continua (BC) de la figura 5.13 muestra la tendencia de datos experimentales, al someter probetas de diferentes tamaños (8 mm a 250 mm) a flexión giratoria. Por ejemplo, al someter una probeta de 250 mm de diámetro a flexión giratoria, se encuentra que el esfuerzo máximo que se puede aplicar sin que ocurra falla es aproximadamente 0.7Se’; es decir, el factor de tamaño es aproximadamente 0.7. Figura 5.13 Variación del factor de tamaño con respecto al diámetro equivalente 6 Para diámetros menores que 8 mm (que es aproximadamente el tamaño usual de las probetas de ensayo para fatiga), la resistencia a la fatiga de la pieza se toma igual al límite de fatiga, es decir, se toma Kb = 1 (línea AB). Los datos experimentales para diámetros mayores de 250 mm son muy escasos (por su elevado costo); se sugiere tomar Kb = 0.6 para diámetros mayores a 250 mm . Las siguientes ecuaciones[6, citado en 1] pueden usarse para determinar el factor de tamaño de piezas de acero sometidas a flexión o torsión: 6 Considere, por ahora, que el diámetro equivalente es el diámetro de la probeta sometida a flexión giratoria. Kb 1 0.8 0.6 0.4 0.2 0 0 50 100 150 200 250 de (mm) A B C Figura 5.12 El número de defectos en la zona de mayor esfuerzo de una probeta sometida a flexión giratoria tiende a ser mayor en una probeta de mayor diámetro CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 17 in), (10 mm 250 si , 6 . 0 in) 10 in 0.3 si , 869 . 0 ( mm 250 mm 8 si , 189 . 1 in) (0.3 mm 8 si , 1 097 . 0 097 . 0             e b e e b e e b e b d K d d K d d K d K Flexión o torsión (5.21) donde de es el diámetro de la probeta sometida a flexión giratoria (para otras secciones y otras solicitaciones diferentes de flexión giratoria, debe hallarse un diámetro equivalente como se discute más adelante). Las ecuaciones anteriores son dudosas para metales no ferrosos . Los datos experimentales sugieren que no existe efecto de tamaño para carga axial; por lo tanto, para piezas sometidas a carga axial: . 1  b K Carga axial (5.22) Lo cual significa que la resistencia a la fatiga no se afecta por el tamaño de la pieza. Debe tenerse en cuenta que las piezas de gran tamaño tienden a ser menos resistentes a todas las solicitaciones y tipos de carga; se recomienda ser precavidos al trabajar con piezas de gran tamaño. Para aplicar la ecuación 5.21 a secciones no circulares y solicitaciones diferentes a flexión giratoria, se debe determinar un diámetro equivalente, de. La manera de determinar los diámetros equivalentes no se explica aquí, pero se presentan los resultados para algunas secciones típicas. El estudiante puede consultar algunos textos de diseño[1,4] si quiere ahondar en el tema. Para secciones circulares de diámetro d, sometidos a: Flexión giratoria: de = d. (5.23.a) Flexión no giratoria7 : de = 0.37d. (5.23.b) Torsión: de = d. (5.23.c) Para secciones rectangulares o en “I” sometidas a flexión no giratoria: Sección rectangular de área hb: de = 0.808 hb . (5.24) Sección en “I” (figura 5.14): de = 0.808 hb , si t > 0.025h. (5.25) Figura 5.14 Sección en “I” 7 Existe “flexión no giratoria’ cuando se somete un elemento, que no gira, a un momento flector variable en magnitud pero que no cambia de dirección. E.N. t h b 18 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS 5.8.3 Factor de confiabilidad (Kc) Como se ha visto hasta ahora, la teoría de fatiga se basa, en gran medida, en datos experimentales. Como la mayoría de las curvas y datos obtenidos corresponden a la tendencia “media” que siguen los puntos de ensayo, la confiabilidad de estos datos es del 50%. Considere, por ejemplo, que alguien quiere determinar el límite de fatiga de un determinado material; lo que puede hacer es tomar varias muestras del material, elaborar probetas de ensayo de fatiga y anotar los diferentes límites de fatiga obtenidos. Con los datos se construye un histograma o distribución de frecuencias, que consiste en dividir el rango de valores obtenidos en un número pequeño de intervalos (seis, para el ejemplo de la figura 5.15). Se cuenta el número de datos que pertenecen a cada intervalo y se construye un rectángulo cuya base es igual a la amplitud del intervalo, y cuya altura es proporcional al número de datos en dicho intervalo. Esto se ilustra en la figura 5.15. En la práctica, los datos de las propiedades de un material, como el límite de fatiga, siguen una distribución “normal”, la cual tiene forma de campana. Figura 5.15 Dispersión de los valores experimentales del límite de fatiga de un material Si se toma como límite de fatiga del material el valor promedio obtenido, éste tendrá una confiabilidad del 50%, lo que significa que la mitad de las piezas construidas con este material tendrán un límite de fatiga igual o superior al valor promedio; la mitad, ya que a partir del valor promedio hacia la derecha, el área que queda (probabilidad) es la mitad del área total. Esto sucede también con otros datos, curvas y ecuaciones obtenidas experimentalmente; los valores calculados con las ecuaciones o determinados mediante curvas normalmente son promedios. El factor de confiabilidad, Kc, corrige la resistencia a la fatiga de tal manera que se tenga una mayor probabilidad (y confiabilidad) de que la resistencia real de una pieza sea mayor o igual que el valor corregido. Para la determinación de este factor se supone que la desviación estándar de la resistencia a la fatiga es de 8%. Utilizando ecuaciones estadísticas correspondientes a la campana de Gauss (distribución normal) se obtiene la siguiente tabla: Tabla 5.2 Factor de confiabilidad (Kc). Confiabilidad (%) 50 90 99 99.9 99.99 99.999 Kc 1 0.897 0.814 0.753 0.702 0.659 Al igual que Ka y Kb, el factor de confiabilidad varía entre 0 y 1. Teóricamente, una confiabilidad del 100% no se podría lograr ya que la campana de Gauss se extiende hasta menos infinito; sin embargo, como se observa en el diagrama de barras de la figura 5.15, los datos reales se extienden en un rango finito de valores. Para propósitos prácticos, una confiabilidad teórica del 99.9% sería suficiente en muchos casos. El diseñador decide con que confiabilidad trabaja, aunque es bueno recordar que el factor de seguridad, N, tiene en cuenta también las incertidumbres en las propiedades del material y en los datos Frecuencia Se’ Se’(promedio) CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 19 experimentales. La decisión de tomar cierta confiabilidad debe estar ligada a la selección de N; para valores bajos de N se podrán tomar confiabilidades altas, mientras que para valores muy conservadores, no sería necesario un valor grande de confiabilidad. 5.8.4 Factor de temperatura (Kd) Las propiedades de un material dependen de su temperatura, Temp. Por ejemplo, un acero puede fragilizarse al ser sometido a “bajas” temperaturas, y la resistencia a la fatiga puede reducirse notoriamente por encima de unos 500 °C. Para tener en cuenta el efecto de reducción de resistencia a la fatiga, se utiliza el factor de temperatura, Kd, que varía entre 0 y 1, dependiendo de la temperatura: cero cuando la resistencia es nula y uno cuando la resistencia para vida infinita es igual al límite de fatiga, es decir, cuando la temperatura no modifica la resistencia. De acuerdo con datos experimentales en los aceros, el límite de fatiga permanece más o menos constante entre la temperatura “ambiente” y 450 °C, y comienza a reducirse rápidamente por encima de este valor. La siguiente ecuación empírica[6, citado en 1] puede utilizarse para determinar el factor de temperatura de un acero: (5.26) El estudiante puede inferir de la ecuación 5.26, que mientras a 450 °C la resistencia a la fatiga de los aceros no se ha alterado con la temperatura, a 550 °C ésta se ha reducido en un 58%. La ecuación 5.26 no debe utilizarse para otros materiales diferentes al acero. 5.8.5 Factor de efectos varios (Ke) Existen otros factores que modifican la resistencia a la fatiga de los materiales; todos los efectos no considerados por los otros factores son cuantificados por el factor Ke. Sin embargo, es muy escasa la información cuantitativa sobre dichos efectos. En general, 0  Ke  1; en ausencia de corrosión, esfuerzos residuales, etc., se puede tomar Ke = 1. Algunos de los fenómenos a tener en cuenta en un diseño por fatiga incluyen: Corrosión Aunque hay información limitada sobre la resistencia de los materiales en entornos severos , la corrosión por agua o aire, por ejemplo, tiene un efecto altamente perjudicial sobre la resistencia a la fatiga. Una grieta esforzada en presencia de corrosión crecerá mucho más rápido; además, aún en ausencia de variación de esfuerzo las grietas tienden a crecer. Es por esto que el fenómeno de corrosión es tan perjudicial para los elementos de máquinas. El fenómeno de pérdida de resistencia a la fatiga por corrosión no ha sido suficientemente estudiado, pero hay algunos datos que pueden servir de ayuda al diseñador. Sin embargo, un estudio experimental bajo las condiciones particulares sería conveniente en los casos requeridos. No existe límite de fatiga en presencia de corrosión[1,4] (ya que las grietas aparecen y crecen con el tiempo). Contradictoriamente, las dos curvas a trazos de la figura 5.11 sugieren que sí existe límite de fatiga para el acero en presencia de corrosión (el efecto de corrosión se tendría en cuenta con el factor Ka). Por otro lado, Norton propone la siguiente ecuación para elementos de acero al carbono en agua limpia. (5.27) C 450 si , 1    emp d T K C 550 C 450 si ), C 450 ( C) / 0058 . 0 ( 1          emp emp d T T K MPa. 100  ' Se 20 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Es decir, el límite de fatiga es aproximadamente constante (100 MPa) independientemente de la cantidad de carbono del acero. Entonces, con respecto al límite de fatiga en un ambiente corrosivo, sería lo mismo un acero de bajo carbono que uno de alto carbono. Para el caso representado por la ecuación 5.27, no se utilizaría el factor de efectos varios, sino que se tomaría directamente 100 MPa como límite de fatiga. La resistencia a la fatiga tiende a reducirse si el agua que rodea el elemento es salada, si la temperatura es alta o si la frecuencia de aplicación de la carga es baja. También existe la corrosión por apriete (en ajustes forzados). Por ejemplo, en uniones atornilladas, cojinetes, cubos de ruedas dentadas y árboles o ejes. El proceso involucra cambio de color en la superficie, picadura y finalmente fatiga. El factor Ke varía entre 0.24 y 0.90, dependiendo del material de las piezas del ajuste. Proceso de manufactura Los procesos de manufactura tienen un efecto significativo en las propiedades de los materiales, incluyendo la resistencia a la fatiga. Las propiedades de un material dependerán de si éste es fundido, laminado, forjado, tratado térmicamente, etc., ya que estos procesos modifican la microestructura y las características de los granos (si los hay). Como la propagación de grietas se facilita a lo largo de los límites de grano, cualquier proceso de manufactura que modifique el tamaño, orientación y forma de los granos afectará la resistencia a la fatiga. Por ejemplo, procesos como el laminado, estirado y extrusión tienden a producir granos alargados en la dirección del estirado. Por lo tanto, la resistencia a la fatiga variará en direcciones diferentes. Procesos como el recocido tienden a liberar esfuerzos residuales, a hacer crecer los granos y a que éstos sean esféricos. La liberación de esfuerzos es generalmente benéfica, pero los otros dos aspectos tienden a producir un efecto negativo sobre la resistencia a la fatiga. Para terminar, hay que tener en cuenta que al procesar un material cuya resistencia a la fatiga inicial se conozca, la resistencia final será en general diferente. Esfuerzos residuales De acuerdo con lo dicho en el párrafo anterior, cuando se reprocesa un material, por ejemplo mediante forja o mecanizado, sus propiedades pueden cambiar, debido en parte a que los procesos de manufactura tienden a dejar esfuerzos residuales. Estos esfuerzos se deben a la recuperación elástica después de una deformación plástica no uniforme. Los esfuerzos residuales son perjudiciales si son de tracción, pero son benéficos si son de compresión, ya que éstos inhiben la generación de grietas y, por lo tanto, la falla por fatiga. Algunos procesos que producen esfuerzos residuales de compresión benéficos son el recocido y algunas veces el laminado, la extrusión y el martillado o forjado[4,7] ; esfuerzos residuales de tracción pueden aparecer después del forjado, extrusión, laminado, mecanizado y rectificado . Existen varias técnicas para introducir esfuerzos residuales de compresión que aumentan significativamente la resistencia a la fatiga. Algunas de estas técnicas son el graneado con perdigones, el preesforzado mecánico y los tratamientos térmicos y superficiales. La mayoría producen esfuerzos de compresión biaxiales en la superficie, esfuerzos de compresión triaxiales debajo de la superficie y esfuerzos de tracción triaxiales en la parte interna. Estos métodos son utilizados en muchos elementos de máquinas. Norton (páginas 394 a 399) describe algunas de estas técnicas con cierto detalle. Recubrimientos Los recubrimientos afectan significativamente la resistencia a la fatiga. La carburización, por ejemplo, produce un alto contenido de carbono en la superficie de un acero, aumentando la resistencia mecánica e impartiendo un esfuerzo residual compresivo en la superficie. Las superficies electrochapadas son muy porosas y tienden a reducir la resistencia, incluso en 50% . Similarmente, los recubrimientos metálicos CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 21 como el cromado, niquelado o cadmizado reducen la resistencia a la fatiga hasta en 50% . Por otro lado, el galvanizado (enchapado en zinc) no afecta significativamente la resistencia a la fatiga[4,7] . 5.8.6 Factor de carga (Kcar) El comportamiento a la fatiga de un elemento depende también del tipo de carga al cual se somete. Las resistencias a la rotura y a la fluencia de un material son diferentes para esfuerzos cortantes y normales; sucede lo mismo con la resistencia a la fatiga. Además, también hay diferencia entre carga axial y flexión, a pesar de que ambos tipos de carga generan esfuerzos normales. Tal como se dijo en la sección 5.5.1, el límite de fatiga es una propiedad determinada para flexión giratoria, y debemos calcular una resistencia a la fatiga para los tipos de carga restantes. De acuerdo con datos experimentales sobre aceros podemos afirmar lo siguiente: (i) para cualquier tipo de flexión, giratoria y no giratoria, la resistencia a la fatiga “ideal” (es decir, para una probeta normalizada y pulida) es igual al límite de fatiga. (ii) La resistencia a la fatiga “ideal” en torsión es aproximadamente 0.577Se’. (iii) Para carga axial existen varias versiones debido a la variabilidad de datos experimentales. Norton sugiere que la resistencia a la fatiga ideal bajo carga axial se tome como 0.7Se’; esto es más conservador que lo dicho por Faires y Shigley y Mischke que sugieren relaciones entre la resistencia a la fatiga en carga axial y el límite de fatiga de 0.8 y de 0.923 a 1, respectivamente. Mientras que en flexión y torsión los máximos esfuerzos ocurren en un pequeño volumen del elemento, todo el volumen de una pieza a carga axial está sometido al máximo esfuerzo. Se cree que la menor resistencia a la fatiga para carga axial se debe a la mayor probabilidad de que se presente una microgrieta en el mayor volumen esforzado. La relación entre la resistencia a la fatiga para cada tipo de carga y el límite de fatiga se denomina factor de carga, Kcar. Entonces, de acuerdo con lo dicho en el párrafo anterior, tenemos que: flexión. para , 1  car K (5.28.a) axial. carga para , 7 . 0  car K (5.28.b) cortante), (y torsión para , 577 . 0  car K (5.28.c) El caso de torsión se puede tratar calculando los esfuerzos de von Mises equivalentes a los esfuerzos aplicados . Entonces, se obtienen esfuerzos medios y alternativos normales y no es necesario usar el factor 0.577, indicado arriba para torsión. Entonces: Mises von de es equivalent esfuerzos los calculando cortante), (y torsión para , 1  car K (5.28.d) 5.8.7 Concentradores de esfuerzos - factores Kf , Kfm y Kff En el capítulo 3 (sección 3.6) se estudió el concepto de concentrador de esfuerzos. Los concentradores de esfuerzos son discontinuidades de las piezas, tales como chaveteros, agujeros, cambios de sección y ranuras, que producen un aumento localizado de los esfuerzos. Como se estudió en la sección 3.6.2, para cargas estáticas, el coeficiente teórico de concentración de esfuerzos, Kt, se tiene en cuenta en el diseño de materiales frágiles (con algunas excepciones), pero no en dúctiles. 22 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Por otro lado, como se vio en la sección 3.6.3, los concentradores de esfuerzos tienden a afectar a los elementos dúctiles y frágiles sometidos a cargas variables. Sin embargo, los diversos materiales tienen diferentes “sensibilidades” a los concentradores. Entonces, además de Kt, se usan unos factores de concentración de esfuerzos por fatiga (Kff, Kf, Kfm), los cuales se estudian a continuación. Factor de concentración de esfuerzos por fatiga, Kf (vida infinita) El factor de concentración de esfuerzos por fatiga, Kf, es un valor que multiplica al esfuerzo nominal, con el fin de obtener un valor “corregido” del esfuerzo, que tenga en cuenta el efecto de la discontinuidad. Este factor se aplica al esfuerzo alternativo . De acuerdo con estudios experimentales, Kf depende de (i) el coeficiente teórico de concentración de esfuerzos, Kt, (ii) el material y (iii) el radio, r, de la discontinuidad. Para tener en cuenta estos dos últimos, se define el índice de sensibilidad a la entalla, q, el cual es un coeficiente cuyo valor representa qué tan sensible es el material a la discontinuidad de radio r (qué tanto se afecta su resistencia a la fatiga). El coeficiente q varía desde 0, cuando el material no tiene sensibilidad a la discontinuidad, hasta 1, cuando el material es totalmente sensible a ésta. El índice de sensibilidad a la entalla se ha definido matemáticamente como: (5.29) de donde: (5.30) Como q varía entre 0 y 1, de acuerdo con la ecuación 5.30, el valor de Kf varía entre 1 (cuando el material no tiene sensibilidad a la entalla) y Kt (cuando el material es totalmente sensible a la entalla). Al encontrar los coeficientes Kt y q se obtiene, entonces, el valor de Kf. El coeficiente Kt se obtiene de las curvas del apéndice 5 o de algunas similares. El valor de q se obtiene a partir de: , 1 1 r a q   (5.31) donde r es el radio de la discontinuidad y a es una constante que depende del material y que se denomina constante de Neuber. La tabla 5.3 suministra valores de la constante de Neuber para aceros; Norton presenta también valores de esta constante para aluminios. El índice de sensibilidad a la entalla, q, de los hierros fundidos es muy pequeño (varía entre 0 y 0.2 aproximadamente); se recomienda tomar el valor conservador de q = 0.2 . La ecuación 5.31 puede utilizarse para construir curvas de sensibilidad a la entalla. Estas curvas se muestran en la figura 5.16. Por lo tanto, el índice de sensibilidad a la entalla puede calcularse mediante la ecuación 5.31 o puede obtenerse directamente de la figura 5.16. , 1 1    t f K K q ). 1 ( 1    t f K q K CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 23 (a) Aceros (b) Aluminio tratado térmicamente (-T) Figura 5.16 Curvas para la determinación del índice de sensibilidad a la entalla, q (parte 1) Radio de la discontinuidad, r (in) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 q 0.02 0.04 0.06 0.08 0.10 0.12 0 0.14 0.16 0.18 0.20 Su (MPa) (ksi) 621 90 414 60 276 40 207 30 138 20 Radio de la discontinuidad, r (in) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 q 0.02 0.04 0.06 0.08 0.10 0.12 0 0.14 0.16 0.18 0.20 Su (MPa) (ksi) (MPa) (ksi) 1379 200 1241 180 1103 160 965 140 965 140 827 120 827 120 689 100 689 100 552 80 552 80 414 60 483 70 345 50 414 60 345 50 Esfuerzo Torsión normal 24 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS (c) Aluminio recocido y endurecido por deformación (-O & -H) Figura 5.16 Curvas para la determinación del índice de sensibilidad a la entalla, q (parte 2) Tabla 5.3 Constante de Neuber para aceros (fuente: Norton ). Su a Carga axial y flexión Torsión (ksi) (MPa) (in0.5 ) (mm0.5 ) (in0.5 ) (mm0.5 ) 50 345 0.130 0.66 0.093 0.47 55 380 0.118 0.59 0.087 0.44 60 415 0.108 0.54 0.080 0.40 70 485 0.093 0.47 0.070 0.35 80 550 0.080 0.40 0.062 0.31 90 620 0.070 0.35 0.055 0.28 100 690 0.062 0.31 0.049 0.25 110 760 0.055 0.28 0.044 0.22 120 825 0.049 0.25 0.039 0.20 130 895 0.044 0.22 0.035 0.18 140 965 0.039 0.20 0.031 0.16 160 1100 0.031 0.16 0.024 0.12 180 1240 0.024 0.12 0.018 0.09 200 1380 0.018 0.09 0.013 0.07 220 1515 0.013 0.07 0.009 0.05 240 1655 0.009 0.05 Nota: los valores de la constante de Neuber para torsión equivalen a los de esfuerzos normales, cuando se toma un valor de Su que sea 20 ksi (138 MPa) mayor que el del material. Radio de la discontinuidad, r (in) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 q 0.02 0.04 0.06 0.08 0.10 0.12 0 0.14 0.16 0.18 0.20 Su (MPa) (ksi) 310 45 241 35 172 25 138 20 103 15 CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 25 En la figura 5.16, se observa que a mayor resistencia del material, mayor es el índice de sensibilidad a la entalla. Esto se debe a que a mayor resistencia del acero o aleación de aluminio, menor tiende a ser su ductilidad (menor capacidad de fluir plásticamente) y, por lo tanto, mayor su sensibilidad al concentrador de esfuerzos. Cuando se tienen chaveteros o roscas, muchas veces no se conoce el radio de la herramienta de corte, por lo cual se hace difícil la obtención de Kt, q y Kf. Las tablas 5.4 y 5.5 presentan valores aproximados de Kf para chaveteros y roscas en acero. Para clasificar un acero como endurecido o recocido, puede tomarse como criterio la dureza; endurecido si la dureza es mayor de 200 HB y recocido si es menor de 200 HB. Con estos valores, no se necesita hallar Kt, q, ni aplicar la ecuación 5.30. Tabla 5.4 Factores de concentración de esfuerzos por fatiga para chaveteros, válidos para aceros (fuente: Faires ). Clase de chavetero Kf Recocido, dureza menor que 200 HB Endurecido, dureza mayor que 200 HB Flexión Torsión Flexión Torsión Perfil 1.6 1.3 2.0 1.6 Patín 1.3 1.3 1.6 1.6 Tabla 5.5 Factores de concentración de esfuerzos por fatiga para roscas, válidos para elementos de acero sometidos a tracción o flexión (fuentes: Faires y Norton ). Clase de rosca Kf (tracción o flexión) Recocida Dureza < 200 HB Grado SAE  2 Clase SAE (ISO)  5.8 Endurecida Dureza > 200 HB Grado SAE  4 Clase SAE (ISO)  6.6 Laminada Tallada Laminada Tallada Unified National Standard (UNS) con raíces planas Cuadrada 2.2 2.8 3.0 3.8 Unified Nacional Standard (UNS) con raíces redondeadas 1.4 1.8 2.6 3.3 Los grados y clases de los pernos se dan en el capítulo 8 (tablas 8.3 y 8.4). Chavetero de perfil Chavetero de patín 26 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS EJEMPLO 5.1 Una pieza de acero, con Su = 550 MPa, tiene un cambio de sección con radio r = 2.5 mm y un coeficiente teórico de concentración de esfuerzos Kt = 1.6. La pieza está sometida a un momento flector en la sección de la discontinuidad. Hallar el índice de sensibilidad a la entalla y el factor de concentración de esfuerzos por fatiga. Figura 5.17 Elemento de sección circular con cambio de sección Solución: El índice de sensibilidad a la entalla del material se obtiene usando la ecuación 5.31. La constante de Neuber se toma de la tabla 5.3; para Su = 550 MPa, a = 0.40 mm0.5 . Reemplazando este valor en la ecuación 5.31 se obtiene: . 80 . 0 mm 5 . 2 mm 40 . 0 1 1 0.5    q Esto indica que esta pieza de acero es 80% sensible al concentrador de esfuerzos. De la ecuación 5.30 se obtiene el factor de concentración de esfuerzos por fatiga: . 48 . 1 ) 1 6 . 1 ( 8 . 0 1 ) 1 ( 1        t f K q K Factor de concentración de fatiga al esfuerzo medio, Kfm El factor de concentración de fatiga al esfuerzo medio, al igual que Kf, es un valor que multiplica al esfuerzo nominal, con el fin de obtener un valor “corregido” de esfuerzo. Este factor se aplica al esfuerzo medio para materiales dúctiles. El valor de Kfm depende de la fluencia localizada que pudiera ocurrir alrededor de la discontinuidad. Aquí se dan las ecuaciones para el cálculo de Kfm [2, citado en 1] ; la explicación de éstas se puede consultar en Norton : 0 entonces 2 si entonces y si entonces si          fm y min max f m a f y fm y min f y max f f fm y max f K S S S K S S K S K S S K S S K K K S S K (5.32) d D r = 2.5 mm CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 27 Adaptando esta ecuación al caso de esfuerzos cortantes: 0 entonces 2 si entonces y si entonces si          fm ys smin smax f ms as f ys fm ys smin f ys smax f f fm s y max s f K S S S K S S K S K S S K S S K K K S S K (5.33) Note que podría ser recomendable usar el valor conservador de Kfm = 1 cuando, de la ecuación 5.32 ó 5.33, Kfm < 1. Factor de concentración de esfuerzos por fatiga para vida finita, Kff El efecto del concentrador de esfuerzos sobre la resistencia a la fatiga varía con el número de ciclos. Recuérdese que una discontinuidad en un material dúctil sometido a una carga estática, prácticamente no afecta la resistencia de la pieza. Pues bien, al pasar de un ciclo de carga a un número indefinido de ciclos de carga sobre materiales dúctiles, el efecto de la discontinuidad sobre la resistencia pasa de ser nulo a ser máximo. El efecto de los concentradores sobre la resistencia a la fatiga en vida finita se tiene en cuenta mediante el factor de concentración de esfuerzos por fatiga para vida finita, Kff. Este valor, al igual que Kf, es mayor o igual a 1 y se aplica al esfuerzo alternativo. De acuerdo con datos experimentales sobre aceros de baja resistencia, Kff = 1 para vidas menores de 1000 ciclos, lo que significa que la resistencia a la fatiga no se ve afectada por la concentración de esfuerzos. El coeficiente Kff aumenta hasta alcanzar el valor de Kf cuando el número de ciclos es de 106 . Las ecuaciones para el cálculo de Kff, dadas a continuación, son válidas para aceros y materiales dúctiles de baja resistencia : (5.34) Para materiales frágiles o de alta resistencia : (5.35) Sin embargo, Norton usa siempre la ecuación 5.35, la cual sería conservadora para los aceros y materiales dúctiles de baja resistencia. En este texto se adopta este método conservador y, por lo tanto, se usa la ecuación 5.35 para todos los materiales. ciclos. 10 si , 1 3   c ff n K . 10 10 si , 6 3 log 3 1    c f K c ff n K n K f ciclos. 10 si , 6   c f ff n K K . todo para , c f ff n K K  28 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS 5.9 RESISTENCIA A LA FATIGA CORREGIDA PARA VIDA FINITA E INFINITA El límite de fatiga, Se’, y la resistencia a la fatiga para vida finita, Sf’, estudiadas en las secciones 5.5.1 y 5.5.2, son aplicables a probetas normalizadas y pulidas girando sometidas a flexión. Como se vio en las secciones anteriores, los elementos de máquinas distan mucho de tener las características de estas probetas de ensayo, ya que pueden tener superficies rugosas, estar sometidas a otros tipos de carga, trabajar en condiciones ambientales severas, etc.. Vamos a denominar resistencia a la fatiga corregida, Sn, a aquella que tiene en cuenta el efecto del estado superficial, del tamaño, de la confiabilidad, de la temperatura, del tipo de carga y de los efectos varios (pero no de los concentradores de esfuerzos). Parecería lógico pensar que Sn es el producto de los coeficientes Ka, Kb, Kc, Kd, Ke y Kcar por Se’ o Sf’. Esto es cierto para vida infinita, pero los datos experimentales han mostrado que estos factores inciden de manera diferente para vida finita. En esta sección se estudian las ecuaciones generales para determinar la resistencia a la fatiga corregida en función del número de ciclos. La figura 5.18 muestra un diagrama típico Sn-nc para los aceros, en escala logarítmica (se grafica la resistencia corregida). La curva está dividida en tres rectas AB, BC y CD. La línea AB está en el rango de bajo ciclaje (LCF), es decir, en el rango 100  nc  103 . Cuando se diseña un elemento para un vida menor o igual a 103 ciclos, puede tomarse, conservadoramente, como resistencia a la fatiga corregida el menor valor de ésta en ese rango; es decir, si 100  nc  103 , Sn = Sn103, donde          1] en citado [9, 1] en citado [9, 10 Mises von de es equivalent esfuerzos los calculan se si torsión para 9 . 0 torsión para 577 . 0 9 . 0 axial carga para 75 . 0 flexión para 9 . 0 3 u u u u n S S S S S (5.36) Nótese que a pesar de que con la ecuación anterior se calcula la resistencia a la fatiga corregida (que tiene en cuenta los efectos de ciertos factores), los factores Ka, Kb, Kc, Kd y Ke no están incluidos en ella. Esto se debe a que estos factores parecen no tener efecto en la resistencia para vidas menores a 103 ciclos. Figura 5.18 Diagrama Sn-nc típico de muchos aceros 10 0 10 3 nc 10 6 nc (log) Sn (log) Su KSe’ A B C Sn103 Sn D LCF HCF CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 29 Para el rango 103 < nc < 106 , debe encontrase la ecuación de Sn en función de nc (ecuación de la línea BC). La ecuación de una línea recta en el diagrama logarítmico Sn - nc está dada por: (5.37) donde a y b son constantes. De acuerdo con las propiedades de los logaritmos, esta ecuación puede expresarse como: (5.38) donde b = logc, entonces (5.39.a y b) Las constantes a y c se determinan mediante los puntos conocidos B y C: en B, Sn = Sn103 y nc = 103 , y en C, Sn = KSe’ y nc = 106 , donde, como se vio . car e d c b a K K K K K K K  (5.20R ) Para 106 ciclos, los factores Ka, Kb, Kc, Kd, Ke y Kcar sí están involucrados en la ecuación para Sn, indicando que sí tienen efecto sobre ésta. Reemplazando estos dos pares de valores en la ecuación 5.39.b obtenemos: (5.40.a y b) De este sistema de ecuaciones se obtienen las constantes a y c: de la ecuación 5.40.a: (5.41) Reemplazando en la ecuación 5.40.b: (5.42) entonces: (5.43) De ésta se obtiene que: (5.44) Reemplazando la ecuación 5.43 en la 5.41 se obtiene: (5.45) , ) log( ) log( b n a S c n   , log ) log( ) log( c n S a c n   . ó ), log( ) log( a c n a c n n c S n c S     . ) 10 ( y , ) 10 ( 6 3 103 a e a n c ' KS c S   , 10 ó , 10 / 10 ) 3 6 ( 10 3 6 10 3 3 a n e a a n e S ' KS S ' KS     . / log 3 ó , / 10 3 3 10 10 3 n e n e a S ' KS a S ' KS    . / log 3 1 3 10 n e S ' KS a  . ] [ ' 2 10 10 10 3 3 3 ' KS S S KS S c e n n e n   . 103 103 a n S c  30 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Reemplazando las ecuaciones 5.44 y 5.45 en la ecuación 5.39.b, tenemos: . ] [ 3 10 3 log 3 1 2 10                  n e S ' KS c e n n n ' KS S S (5.46) Ésta es la ecuación de la resistencia a la fatiga corregida, Sn, en función del número de ciclos, nc, para el rango 103 < nc < 106 . En el último rango, nc  106 , Sn es constante (figura 5.18) e igual a: (5.47) De acuerdo con esto, si se quiere diseñar para una duración mayor o igual a un millón de ciclos, debe usarse el límite de fatiga corregido; por lo tanto, la pieza duraría indefinidamente. En las ecuaciones 5.36 a 5.47 y el diagrama de la figura 5.18 no se ha incluido el efecto de los concentradores de esfuerzos. En las ecuaciones de diseño estudiadas más adelante se incluirá dicho efecto. En resumen, para los aceros o materiales que exhiben el codo C en nc  106 (figura 5.18), Sn está dado por: , 10 Si 3  c n . 3 10 n n S S  (5.48) , 10 10 Si 6 3   c n . ] [ 3 10 3 log 3 1 2 10                  n e S ' KS c e n n n ' KS S S (5.49) , 10 Si 6  c n . ' KS S e n  (5.50) Puede seguirse un procedimiento similar para materiales que no posean límite de fatiga. Puede demostrarse que para materiales que no exhiben el codo C (véase la figura 5.6): , 10 Si 3  c n . 3 10 n n S S  (5.51) , 10 Si 3 cref c n n   . ] [ ] [ 3 10 3 log 1 / 3 / 3 1 10                    ' KS S z c z f z n n f n n ' KS S S (5.52) , Si cref c n n  . ' KS S f n  (5.53) donde z = log(103 ) – log(ncref) = 3 – log(ncref) y Sf’ = Sf’@ncref es la resistencia a la fatiga para una vida ncref. Por ejemplo, para una aleación de aluminio se podría tomar ncref = 5108 , ya que se podría estimar Sf’@5108 mediante la ecuación 5.8. Nótese que no se presenta ecuación para nc > ncref. En este caso podría usarse la ecuación 5.52 (lo cual constituiría una extrapolación), pero el resultado tiene una confiabilidad cuestionable y, probablemente, es conservador. El estudiante puede comprobar que para ncref = 106 , la ecuación 5.52 es equivalente a la 5.49, excepto que en ésta se usa Se’ en lugar de Sf’. . ' KS S e n        CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 31 Si ncref = 5108 , z = -5.699 y la ecuación 5.52 es equivalente a: , 10 5 10 Si 8 3    c n . ] [ ] [ 8 10 5 @ 3 10 8 3 log 699 . 5 1 5264 . 0 10 5 @ 5264 . 1 10                         ' KS S c f n n f n n ' KS S S (5.54) donde, Sn103 en las ecuaciones 5.48 a 5.54 está dado por la ecuación 5.36. 5.10 LÍNEAS DE FALLA 5.10.1 Introducción El límite de fatiga y la resistencia a la fatiga para vida finita constituyen propiedades base para el diseño de elementos sometidos a cargas variables. Sin embargo, Se’ o Sf’ no puede ser utilizado directamente en el diseño, ya que es obtenido bajo condiciones especiales de esfuerzo: probeta normalizada y pulida, girando sometida a flexión bajo condiciones ambientales favorables. Particularmente, nos interesa referirnos aquí a las condiciones de flexión giratoria. Bajo este tipo de carga, la probeta sufre un variación sinusoidal de esfuerzo repetido invertido para la cual el esfuerzo medio es igual a cero (véase la figura 5.10.a ó la 5.5.b). Se necesitan ecuaciones de diseño que sirvan no sólo para un esfuerzo repetido invertido, sino también para cualquier tipo de variación sinusoidal, donde Sm pueda ser diferente de cero. Para encontrar dichas ecuaciones fueron necesarias más pruebas experimentales, de las cuales se concluyó que, en general, si se agrega una componente media del esfuerzo, el elemento falla con una componente alternativa menor. Las figuras 5.19.a y b muestran las tendencias típicas que siguen los puntos de ensayo en diagramas de esfuerzo medio - esfuerzo alternativo, Sm-Sa y Sms-Sas, respectivamente, cuando se someten probetas normalizadas y pulidas a diferentes combinaciones de estos esfuerzos. Las cruces en los diagramas indican las combinaciones (Sm, Sa) o (Sms, Sas) para las cuales un pequeño aumento en el esfuerzo medio o en el alternativo produciría la falla de la probeta. Las combinaciones de esfuerzos que estén entre la zona de las cruces y el origen del diagrama no producirían falla, mientras que combinaciones de esfuerzos que estén por fuera de la zona de las cruces producirían la falla de la probeta de ensayo. Figura 5.19 Diagramas esfuerzo medio contra esfuerzo alternativo Analicemos algunos puntos de los diagramas. Al someter una probeta a un esfuerzo estático, ésta fallará cuando el esfuerzo (máximo) sea igual a la resistencia. Si el esfuerzo es estático, Smax = Smin = Sm y Sa = 0 (no hay amplitud de onda) o Ssmax = Ssmin = Sms y Sas = 0 (para esfuerzos cortantes); entonces, la falla en  Sms Sas 0.577Se’ Sus Falla por fatiga (a) Diagrama Sm-Sa (esfuerzos normales) (b) Diagrama Sms-Sas (esfuerzos cortantes) Sm Sa Se’ Su Falla inmediata por compresión Falla por fatiga 32 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS estas condiciones estará representada por el punto (Su, 0) o (Sus, 0). Al observar los diagramas vemos que las cruces tienden hacia dichos puntos. La condición de falla de una probeta a flexión giratoria está representada por la combinación (0, Se’). Es decir, el esfuerzo medio es cero y el esfuerzo alternativo (que será igual al máximo, de acuerdo con la ecuación 5.17.a) que produce la falla es Se’. Efectivamente, en la figura 5.19.a, las cruces tienden a dicho punto. En la figura 5.19.b aparece 0.577 Se’, ya que, como se vio, la resistencia a la fatiga en torsión es aproximadamente 0.577Se’ (sección 5.8.6). Como se dijo en la sección 5.7, el esfuerzo medio cortante se toma siempre positivo, por lo tanto, en la figura 5.19.b sólo aparece el lado positivo de Sms; para esfuerzos normales, el esfuerzo medio puede ser positivo o negativo, tal como se muestra en la figura 5.19.a. Podemos agregar que para ambos diagramas de la figura 5.19, la tendencia de los puntos en el lado positivo del esfuerzo medio es similar. Para el lado negativo de Sm, las cruces tienden a subir un poco a medida que el esfuerzo medio aumenta en compresión, hasta llegar a unos valores de esfuerzos tales que no se produce falla por fatiga, sino que se produce una falla inmediata (fluencia o rotura) por compresión. Esto es debido a que los esfuerzos normales de compresión no expanden las grietas. Un punto sometido a esfuerzos cortantes variables puede desarrollar grietas, al hacerlo, los esfuerzos normales de tracción, no los de compresión, son los encargados de expandir la grieta. Como se dijo anteriormente, el caso de torsión se puede trabajar calculando los esfuerzos de von Mises equivalentes a los esfuerzos aplicados. Además, teniendo en cuenta que el caso de esfuerzos cortantes es más sencillo, debido a que Sms es siempre positivo, en este texto el estudio de las ecuaciones de fatiga se enfoca en esfuerzos normales. Para modelar los datos experimentales para Sm  0, se han propuesto, entre otros, tres modelos o aproximaciones diferentes, las cuales se muestran en la figura 5.20: - Línea o parábola de Gerber - Línea de Goodman modificada - Línea de Soderberg Figura 5.20 Líneas (modelos) de falla en el diagrama esfuerzo medio contra esfuerzo alternativo En las secciones 5.10.2 a 5.10.4, se estudian estas tres aproximaciones (válidas para Sm  0). En la sección 5.10.5 se trata el caso de esfuerzo medio negativo (Sm < 0). Después, en la sección 5.10.6, se estudian líneas de falla adicionales. Sm Sa Se’ Su Sy Parábola de Gerber Línea de Goodman modificada Línea de Soderberg CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 33 5.10.2 Línea o parábola de Gerber La línea de Gerber es una parábola que pasa por los puntos (0, Se’) y (Su, 0), en el diagrama Sm-Sa (figura 5.20), y está dada por: . 1 2 ' S S S S e a u m           (5.55) Esta aproximación es razonablemente fiel a los datos de ensayo y corresponde aproximadamente al 50% de confiabilidad; es decir, la mitad (50%) de los puntos de ensayo (cruces) está por encima de la parábola y la otra mitad está por debajo. Si se diseñara directamente con estas ecuaciones se tendría un 50% de probabilidad de no falla. Esta línea de falla es particularmente útil para el análisis de piezas que han fallado por fatiga, pero, Según Norton , no se usa mucho para diseñar. 5.10.3 Línea de Goodman modificada La línea de Goodman modificada es una recta que corta los ejes por los mismos puntos que la parábola de Gerber. Se deja al estudiante la deducción de la ecuación siguiente, partiendo de la ecuación general de una línea recta, Sa = aSm + b, utilizando los dos puntos conocidos en la recta: . 1 ' S S S S e a u m   (5.56) Esta aproximación es menos exacta que la de Gerber, pero tiene dos ventajas: (i) es más conservadora, ya que la mayoría de cruces quedan por encima de la línea y, por lo tanto, es más confiable; (ii) la ecuación de una línea recta es un poco más sencilla que la ecuación de una parábola. La línea de Goodman modificada es la más utilizada en el diseño por fatiga . 5.10.4 Línea de Soderberg La línea de Soderberg es una recta, que a diferencia de la línea de Goodman modificada, pasa por el punto (Sy, 0) en el diagrama Sm-Sa; es decir, utiliza como criterio de falla la resistencia de fluencia y no la de rotura. Es por esto que sólo se puede aplicar a materiales que posean límite de fluencia. Esta línea de falla es muy conservadora, pero como se verá más adelante, evita la verificación de que no ocurra fluencia inmediata. La ecuación para la línea de Soderberg es similar a la de Goodman modificada: . 1 ' S S S S e a y m   (5.57) 5.10.5 Líneas de falla para Sm < 0 Cuando el esfuerzo medio es negativo, los puntos de ensayo tienden a “subir” hacia la izquierda, tal como se muestra en la figura 5.19.a y 5.21. Para facilitar los cálculos, se asume que para valores negativos de Sm, la línea de falla es horizontal (pendiente cero) y se extiende hasta una zona para la cual tiende a ocurrir falla inmediata por deformación plástica o rotura en compresión. Esta es una asunción conservadora. La figura 5.21 muestra la línea correspondiente a esta falla, cuya ecuación es: (5.58) . ' S S e a  34 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Figura 5.21 Línea de falla por fatiga en el diagrama Sm-Sa para esfuerzos medios negativos Nótese que la componente media del esfuerzo, que es de compresión, no se tiene en cuenta (no está en la ecuación), cuando en realidad tiene cierto efecto benéfico sobre la resistencia a la fatiga. 5.10.6 Líneas de falla adicionales Las líneas de falla estudiadas en las secciones 5.10.2 a 5.10.5 son líneas de falla por fatiga. Es necesario considerar ahora unas líneas de falla adicionales, que tengan en cuenta la posibilidad de falla inmediata por deformación plástica o rotura (ya sea en tracción o compresión), cuando se intenta aplicar la carga variable. Falla inmediata – esfuerzo de tracción La falla de un elemento sometido a un esfuerzo de tracción está dada por: (5.59) Sabemos que Smax = Sm + Sa (ecuación 5.17.a), entonces, la ecuación 5.59 puede expresarse como: (5.60) Estas dos ecuaciones son líneas rectas en el diagrama Sm-Sa y se representan en la figura 5.22.a. Cualquier punto sobre las líneas o arriba de ellas representa un punto de falla, mientras que un punto debajo de ellas (hacia el origen) representa un punto de no falla. Figura 5.22 Líneas que representan la falla por un esfuerzo de tracción o compresión . ó u max y max S S S S   . ó u a m y a m S S S S S S     Sm Sa Se’ Línea de falla por fatiga cuando Sm < 0 (a) Falla por tracción Sm Sa Su Sy Rotura en tracción Fluencia en tracción Su Sy (b) Falla por compresión Sm Sa –Syc –Suc Rotura en compresión Fluencia en compresión Suc Syc CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 35 Falla inmediata – esfuerzo de compresión La falla de un elemento sometido a un esfuerzo de compresión está dada por: (5.61) El signo “–” debe utilizarse ya que el esfuerzo mínimo es negativo mientras que las resistencias a la compresión son valores positivos. Sabemos que Smin = Sm – Sa (ecuación 5.17.b), entonces, la ecuación 5.61 puede expresarse como: (5.62) Las líneas rectas correspondientes a estas dos ecuaciones se muestran en la figura 5.22.b. Cualquier punto sobre las líneas o arriba de ellas representa un punto de falla, mientras que un punto debajo de ellas (hacia el origen) representa un punto de no falla. 5.11 ECUACIONES DE DISEÑO 5.11.1 Introducción De las tres aproximaciones estudiadas en las secciones 5.10.2 a 5.10.4, la de Gerber es recomendada para el diseño por algunos autores, y como se dijo es útil para el análisis de piezas que han fallado por fatiga . La línea de Soderberg podría utilizarse para materiales dúctiles, los cuales poseen límite de fluencia. Aunque es demasiado conservadora, es recomendada por algunos autores. Según Norton , la línea de Goodman modificada es la preferida y puede aplicarse tanto para materiales dúctiles como frágiles. En este libro se utilizará principalmente la línea de Goodman modificada. En esta sección se estudia el panorama completo. Se integran todas las líneas de falla estudiadas y se plantean las ecuaciones de diseño, las cuales tienen en cuenta los factores que inciden en la resistencia a la fatiga de los elementos y los efectos de concentración de esfuerzos. La figura 5.23 muestra el diagrama Sm-Sa con las líneas de falla estudiadas anteriormente. Al efectuar un diseño, se espera que el elemento no falle, por lo tanto, las combinaciones (Sm, Sa) que no producen falla son las que caen dentro del polígono sombreado; cualquier punto de dicha zona está por debajo (o hacia el origen) de todas las líneas de falla, mientras que cualquier punto por fuera del polígono estaría por encima (o por fuera) de alguna línea de falla. El polígono de la figura 5.23 supone que el material posee límites de fluencia. Si no los posee, el polígono resultante está limitado por la línea de Goodman modificada y por la línea de rotura inmediata en compresión, desapareciendo las dos líneas de fluencia inmediata. Además, si se usa la aproximación de Soderberg (línea a trazos), el polígono no incluye la línea de fluencia en tracción. El diseñador debe trabajar con cautela ya que en cualquier diseño existen incertidumbres, debiéndose usar un factor de seguridad. El diagrama de la figura 5.23 puede adaptarse para el diseño, dividiendo cada propiedad del material en el diagrama por el factor de seguridad y trazando las líneas entre los valores que resulten. . ó uc min yc min S S S S     . ó uc a m yc a m S S S S S S       36 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Figura 5.23 Diagrama Sm-Sa con todos los tipos de falla por esfuerzo normal Además, debe tenerse en cuenta que las figuras 5.19 a 5.23 son aplicables a probetas de ensayo. Por lo tanto, para el diseño de elementos usados en la práctica, es necesario involucrar los factores que tienen en cuenta la reducción de la resistencia a la fatiga; es decir, debe usarse la resistencia a la fatiga corregida, Sn (sección 5.9). Finalmente, con el fin de incluir el efecto de los concentradores de esfuerzos, deben usarse los factores de concentración de esfuerzos (Kt, Kff y Kfm); estos se tendrán en cuenta en la sección 5.11.4. La figura 5.24 muestra las líneas de seguridad para materiales frágiles y dúctiles sometidos a esfuerzos normales variables. Para materiales frágiles se utilizan las líneas de seguridad correspondientes a la línea de Goodman modificada, la de falla por fatiga para Sm < 0 y la de rotura inmediata en compresión, ya que estos materiales no poseen resistencia de fluencia. Para materiales dúctiles se utilizan cuatro líneas de seguridad, si se usa Goodman modificada (o tres, si se usa Soderberg). Se debe incluir la línea de seguridad para la fluencia inmediata en tracción, y se trabaja con la línea de seguridad correspondiente a la fluencia inmediata en compresión, en lugar de aquella para rotura. Nótese que las líneas de seguridad han sido trazadas a partir de “resistencias” divididas por el factor de seguridad. Sm Sa Su Rotura inmediata en compresión Rotura inmediata en tracción Suc Su –Suc Syc Sy Se’ –Syc Sy Fluencia inmediata en tracción Fluencia inmediata en compresión Falla por fatiga (Sm < 0) Falla por fatiga según Goodman modificada Falla por fatiga según Soderberg Puntos de no falla CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 37 Figura 5.24 Líneas de seguridad para elementos (sin discontinuidades) sometidos a esfuerzos normales Un diseño seguro es aquel en el que la combinación esfuerzo medio - esfuerzo alternativo cae dentro de la zona sombreada correspondiente, la cual está limitada por tres o cuatro líneas. 5.11.2 Materiales dúctiles Para materiales dúctiles (figura 5.24.b), se puede seguir el siguiente procedimiento: Para Sm  0, la línea de seguridad de Goodman modificada, es decir, aquella en la cual se tiene en cuenta el factor de seguridad, es similar a la de la línea de Goodman modificada (ecuación 5.56), excepto que las propiedades del material se dividen por N y que se usa Sn en vez de Se’: . 1 o , / / 1 n a u m n a u m S S S S N N S S N S S     (5.63) Como para Sm > 0, parte del polígono de seguridad está limitado por la línea correspondiente a la fluencia inmediata en tracción, después de usar la ecuación anterior, debe verificarse la seguridad del elemento con respecto a esta falla (ya que parte de la línea de seguridad de fluencia está por debajo de la línea de seguridad de Goodman modificada). Esto se hace verificando que Sm + Sa  Sy/N (es decir, que Smax  (a) Líneas de seguridad para materiales frágiles Sm Suc/N Su –Suc Sn Sn/N –Suc/N Su/N Puntos seguros Líneas de seguridad Líneas de falla Sa Syc/N (b) Líneas de seguridad para materiales dúctiles Sm Sy/N Su/N –Syc A Sn/N –Syc/N Sy/N Puntos seguros Líneas de seguridad Líneas de falla Sa 38 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Sy/N). Si esta inecuación no se cumple, el factor de seguridad del resorte sería N = Sy/(Sm + Sa) = Sy/Smax. Entonces, usando el criterio de Goodman mofificada: . 1 usar si pero , 1 , 0 Para y a m y a m n a u m m S S S N N S S S S S S S N S        (5.64.a) Si se usa la línea de Soderberg, no es necesario verificar la fluencia en tracción: . 1 , 0 Para n a y m m S S S S N S    (5.64.b) Para Sm < 0, se puede optar por encontrar los rangos de cada línea, para lo cual debe determinarse el valor de Sm del punto en el cual se interceptan las líneas (punto A en la figura 5.24.b). La ecuación de la línea horizontal es: . 1 ó n a n a S S N N S S   (5.65) La ecuación de la línea inclinada es similar a la 5.62, pero también Syc va dividido por N: . 1 ó yc m a yc a m S S S N N S S S      (5.66) Las dos ecuaciones anteriores se cumplen simultáneamente sólo en el punto A. Por lo tanto, pueden utilizarse para encontrar el valor de Sm en A, SmA. Reemplazando la primera forma de la ecuación 5.65 en la primera forma de la 5.66 obtenemos: entonces , N S N S S yc n mA    (5.67) . N S S S yc n mA   (5.68) Podemos expresar, entonces, las ecuaciones 5.65 y 5.66 con sus respectivos rangos: . 1 0, Para n a m yc n S S N S N S S     (5.69) . 1 , Para yc m a yc n m S S S N N S S S     (5.70) Entonces, las ecuaciones 5.64, 5.69 y 5.70 son las que representan las cuatro líneas de seguridad para materiales dúctiles sometidos a esfuerzos normales (figura 5.24.b). CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 39 5.11.3 Materiales frágiles Al comparar las figuras 5.24.a y 5.24.b, puede concluirse que las ecuaciones de seguridad para materiales frágiles son similares a aquellas para materiales dúctiles, excepto que en vez de Syc aparecerá Suc y que no es necesario verificar la resistencia a la fluencia en tracción. Las ecuaciones 5.64, 5.69 y 5.70 se adaptan para materiales frágiles: . 1 , 0 Para n a u m m S S S S N S    (5.71) . 1 0, Para n a m uc n S S N S N S S     (5.72) . 1 , Para uc m a uc n m S S S N N S S S     (5.73) Según Hamrock et al. , hasta hace poco el único material frágil usado en un entorno a la fatiga fue el acero gris fundido a la compresión. En la actualidad, los materiales frágiles sometidos a cargas variables incluyen también las fibras de carbono y los cerámicos . Es de anotar que la teoría de fatiga ha sido desarrollada principalmente para materiales metálicos dúctiles. Por lo tanto, hay que ser precavidos cuando se diseñen materiales frágiles. Mott incluso se abstiene de presentar recomendaciones para el diseño por fatiga de materiales frágiles, ya que no es recomendable su uso para cargas variables. Mott recomienda realizar pruebas para verificar la seguridad de elementos frágiles bajo condiciones reales de servicio, si es necesario usarlos. 5.11.4 Ecuaciones de diseño generales Las ecuaciones estudiadas en las secciones 5.11.2 y 5.11.3 son aplicables a elementos sin concentradores de esfuerzos. Como se dijo en la sección 5.11.1, se debe incluir el efecto de los concentradores de esfuerzos. De acuerdo con la literatura, para las líneas de falla por fatiga se adopta lo siguiente: (i) la componente alternativa del esfuerzo nominal, Sa, se multiplica por Kff (tanto para materiales dúctiles como frágiles) y (ii) la componente media del esfuerzo nominal, Sm, se multiplica por Kfm, para materiales dúctiles, o por Kt, para materiales frágiles8 . Para las líneas de falla inmediata a compresión (rotura o fluencia): (i) en general, Sm y Sa se multiplican por Kt para materiales frágiles o (ii) Sm y Sa no se multiplican por ningún factor si el material es dúctil. A continuación se presentan las ecuaciones generales de diseño por fatiga, en las cuales se han aplicado los criterios descritos aquí. Esfuerzos normales en materiales dúctiles m.) G. (usando 1 usar si ; 1 , 0 Para y a m y a m n a ff u m fm m S S S N N S S S S S K S S K N S        (5.74.a) . Soderberg) usa se si , 0 (para 1 ó    m n a ff y m fm S S S K S S K N (5.74.b) 8 Es de anotar que no hay un consenso en la literatura con respecto al uso de los factores de concentración de esfuerzos para las componentes media y alternativa del esfuerzo nominal. Por ejemplo, Hamrock et al. propone, para materiales frágiles, multiplicar a Sa por Kt (en vez de Kff) y, para materiales dúctiles, multiplicar a Sm por 1 (en vez de Kfm). El procedimiento que se adopta en este texto es similar al propuesto por Norton . 40 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS . 1 0, Para n a ff m yc ff n S S K N S N S K S     (5.75) . 1 , Para yc m a yc ff n m S S S N N S K S S     (5.76) Esfuerzos normales en materiales frágiles . 1 , 0 Para n a ff u m t m S S K S S K N S    (5.77) . 1 0, Para n a ff m t uc ff n S S K N S N K S K S     (5.78) . ) ( 1 , Para uc m a t t uc ff n m S S S K N N K S K S S     (5.79) Esfuerzos cortantes en materiales dúctiles Para este caso puede usarse la ecuación 5.74, si se calculan los esfuerzos equivalentes medio y alternativo (ecuación 5.19), reemplazando Sm por me y Sa por ae. Alternativamente, puede usarse: m.) G. (usando 1 usar si ; 1 ys as ms ys as ms n as ff us ms fm S S S N N S S S S S K S S K N       (5.80.a) . Soderberg) (usando 1 ó n as ff ys ms fm S S K S S K N   (5.80.b) El esfuerzo cortante medio, Sms, es siempre mayor o igual a cero. Esfuerzos cortantes en materiales frágiles Según Norton , puede usarse la ecuación 5.77, si se calculan los esfuerzos equivalentes medio y alternativo (ecuación 5.19), reemplazando Sm por me y Sa por ae. Alternativamente, usar: . 1 n as ff us ms t S S K S S K N   (Siempre Sms  0) (5.81) Según el capítulo anterior, la teoría de la energía de distorsión o de von Mises-Hencky es adecuada para materiales dúctiles, pero no para frágiles. Esto se debe a que el mecanismo de falla estática es diferente para los dos tipos de materiales: falla por cortante, para materiales dúctiles, y por esfuerzo normal, para frágiles. Por otro lado, las fallas por fatiga son “fallas de tracción”, independientemente de si el material es dúctil o frágil . Debido a esto, para el caso de fatiga, Norton sugiere utilizar los esfuerzos equivalentes de von Mises tanto para materiales dúctiles como para frágiles. CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 41 Resistencia a la fatiga corregida Recuérdese que la resistencia a la fatiga corregida se calcula con las ecuaciones 5.48 a 5.54. Los valores de Se’ o Sf’, si son necesarios, se pueden obtener con las ecuaciones de la sección 5.6, y los coeficientes K, Kfm y Kff se calculan de la manera estudiada en la sección 5.8. 5.11.5 Comentarios finales Acerca del factor de seguridad Tal como se han planteado las ecuaciones de diseño anteriores, el factor de seguridad que se puede calcular con ellas es válido si, en una “sobrecarga”, las componentes media y alternativa de los esfuerzos aumentan proporcionalmente. En ciertos casos podría ocurrir que con una sobrecarga sólo aumente el esfuerzo medio, que sólo aumente el esfuerzo alternativo, o que aumenten ambos de una manera no proporcional; en estos casos el factor de seguridad se calcularía de una manera diferente y se recomienda que se haga de esta manera para diseños “reales”. Si el estudiante está interesado en este tema puede consultar a Norton y Shigley y Mischke . Selección de puntos críticos La selección de los puntos críticos de elementos sometidos a cargas variables se basa en las ecuaciones de diseño estudiadas en la sección anterior. Con base en éstas podemos decir que las secciones de menores dimensiones, con discontinuidades, con mayores cargas, con acabados superficiales más rugosos, sometidos a temperaturas críticas, etc., tienden a ser más críticas. También vale la pena recordar que los esfuerzos de tracción son más perjudiciales que los de compresión, ya que los primeros hacen crecer las grietas; sin embargo, los esfuerzos de compresión también pueden producir falla. Finalmente, se le recuerda al estudiante que las ecuaciones de diseño son aplicables a puntos de elementos y que, por lo tanto, todas las variables que se calculen para ser reemplazadas simultáneamente en una ecuación deben estar asociadas al punto que se esté considerando. Esfuerzos combinados variables Hasta este punto, se ha estudiado la teoría de fatiga para cargas variables simples. La sección 5.12 estudia el caso de esfuerzos combinados variables. EJEMPLO 5.2 El árbol escalonado mostrado en la figura gira a 40 r/min, transmitiendo potencia a una máquina a través de un acople flexible montado en el escalón derecho. Se sabe que el acero con el que se construirá el árbol tiene resistencias de Su = 550 MPa y Sy = 460 MPa. Hallar la potencia que se puede transmitir a través del acople flexible para que el escalón mostrado no falle, si el árbol cambia su sentido de giro, al igual que el sentido del par de torsión, repetidamente. Figura 5.25 Elemento de sección circular con cambio de sección 70 mm 90 mm r = 2.5 mm Puesto de acople 42 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Solución: Un acople flexible es un dispositivo que transmite potencia, sin transmitir cargas transversales, axiales ni momentos flectores significativos. Por lo tanto, el tramo de árbol mostrado en la figura soporta sólo un par de torsión, el cual induce esfuerzos cortantes simples. Como el árbol transmite potencia en ambos sentidos de giro, el par de torsión es variable, pasando desde un máximo en un sentido hasta un máximo (igual al primero) en sentido contrario. Punto crítico: La sección más crítica del tramo de árbol mostrado está en el escalón derecho (de menor diámetro), cerca al cambio de sección, ya que allí ocurre concentración de esfuerzos. Los puntos más críticos son los puntos de la periferia cercanos al redondeo, ya que éstos son los más alejados del eje neutro y están en las vecindades del concentrador de esfuerzos. Ecuación de diseño: La ecuación de diseño para este caso, material dúctil sometido a esfuerzos cortantes, es la 5.80.a, usando el criterio de Goodman modificada: . 1 usar si pero , 1 ys as ms ys as ms n as ff us ms fm S S S N N S S S S S K S S K N       Variación del par de torsión y del esfuerzo: Suponiendo que la potencia máxima que se transmite en cualquier sentido de giro es la misma, tendremos que el par de torsión, T (desconocido), varía entre un valor máximo, Tmax, y uno mínimo Tmin = –Tmax. Podemos aplicar las ecuaciones 5.16.a y b, al par de torsión, en vez de al esfuerzo: , 0 2 2      max max min max m T T T T T . 2 ) ( 2 max max max min max a T T T T T T       Para el cálculo del esfuerzo cortante máximo en los puntos críticos se utiliza la ecuación 2.12 (capítulo 2). El esfuerzo medio se calcula con Tm, y el esfuerzo alternativo con Ta: , 0 16 3   d T S m ms  . ) mm 70 ( 16 16 3 3   max a as T d T S   Nótese que el diámetro utilizado para calcular el esfuerzo alternativo (nominal) es el menor del cambio de sección. En la figura 5.26 se muestran modelos sinusoidales de las variaciones del par de torsión y del esfuerzo cortante. CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 43 Figura 5.26 Modelos sinusoidales del par de torsión, T, y del esfuerzo cortante, Ss, del escalón del árbol Propiedades del material: Las propiedades del acero son Su = 550 MPa y Sy = 460 MPa. El límite de fatiga se puede estimar a partir del esfuerzo último, con la ecuación 5.2 (ya que Su < 1380 MPa): MPa. 275 ) MPa 550 )( 5 . 0 ( 5 . 0    u e S ' S La resistencia máxima en torsión, Sus, no se requiere ya que Sms = 0. Determinación de factores: El factor de superficie, Ka, se determina con la figura 5.11, con Su = 550 MPa = 80 ksi. Se supone que el escalón se obtiene mediante mecanizado. Entonces Ka = 0.78. El factor de tamaño, Kb, se determina utilizando las ecuaciones 5.21 y 5.23.c: entonces , donde mm, 250 mm 8 si , 189 . 1 097 . 0 d d d d K e e e b      . 787 . 0 ) 70 ( 189 . 1 097 . 0    b K Asumiendo una confiabilidad del 99.9%, de la tabla 5.2 se obtiene que Kc = 0.753. El factor de temperatura, Kd, y el de efectos varios, Ke, se toman igual a la unidad, asumiendo que las condiciones de operación son “normales”. El factor de carga, Kcar, es igual a 0.577 para torsión (ecuación 5.28.c). Entonces: . 267 . 0 577 . 0 1 1 753 . 0 787 . 0 78 . 0         car e d c b a K K K K K K K Para hallar el factor de concentración de esfuerzos por fatiga para vida finita, Kff, se deben calcular el coeficiente teórico de concentración de esfuerzos, Kt, y el índice de sensibilidad a la entalla, q. De la figura A-5.10 (apéndice 5) se obtiene Kt, sabiendo que la relación de diámetros D/d = (90 mm)/(70 mm) = 1.29 y que la relación entre el radio del cambio de sección y el diámetro menor r/d = (2.5 mm)/(70 mm) = 0.036. Entonces Kt  1.83. T t Tmax Tmin Tm = 0 Ta Ta Ss t Ssmax Ssmin Sms = 0 Sas Sas 44 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS El índice de sensibilidad a la entalla del material se obtiene con la ecuación 5.31. La constante de Neuber se toma de la tabla 5.3; para acero con Su = 550 MPa y carga de torsión, a = 0.31 mm0.5 . Reemplazando este valor en la ecuación 5.31 se obtiene: . 836 . 0 mm 5 . 2 mm 31 . 0 1 1 0.5    q De la ecuación 5.30 se obtiene el factor de concentración de esfuerzos por fatiga: . 69 . 1 ) 1 83 . 1 ( 836 . 0 1 ) 1 ( 1        t f K q K El factor de concentración de esfuerzos por fatiga para vida finita, Kff, es igual a Kf ya que se está diseñando para vida infinita (ecuación 5.34). Resistencia a la fatiga corregida: La resistencia a la fatiga corregida está dada por la ecuación 5.50: MPa. 4 . 73 MPa 275 267 . 0     ' KS S e n Cálculo de la potencia: El factor de seguridad se toma del rango adecuado de la tabla 3.1 (capítulo 3). Tomamos N = 1.5. Reemplazando los valores correspondientes en la ecuación 5.80, se obtiene: , ) m 07 . 0 ( 16 Pa 10 4 . 73 69 . 1 0 5 . 1 1 3 6  max us T S    de donde m. N 1950 16 69 . 1 5 . 1 ) m 07 . 0 ( Pa) 10 4 . 73 ( 3 6        max T Nótese que no es necesario verificar la condición dada por la ecuación 5.80, porque al ser Sms = 0, el punto (Sms, Sas) en el diagrama Sms-Sas estará por debajo de la línea de seguridad correspondiente a la falla inmediata por fluencia en tracción. La potencia la obtenemos mediante la ecuación 3.17 (capítulo 3): donde de , [r/min] 2 ] W [ 60 ] m N [ n P T    kW. 17 . 8 W 60 ) 1950 )( 40 ( 2 W 60 ] m N [ ] r/min [ 2       max max T n P La potencia máxima que se puede transmitir con el escalón del árbol (alternando el sentido de giro) sin que se produzca falla después de un gran número de ciclos es de 8.17 kW. CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 45 EJEMPLO 5.3 Se quiere construir un puente grúa con una viga en I, formada por placas de espesor a = 50.8 mm de acero SAE 1020 laminado en caliente. La luz de la viga es de 10 m. La carga máxima a manipular en la grúa es de 200 kN (se incluye el peso del puente y de la grúa). Hallar el factor de seguridad de la viga de acuerdo con el criterio de Soderberg. Figura 5.27 Viga de un puente grúa Solución: Para el diseño de la viga deben tenerse en cuenta, además de las cargas normales de trabajo, las cargas pico producidas por sismos u otras causas. Como se prevé que las cargas pico sobre la viga del puente grúa tendrán lugar unas pocas veces durante su vida útil, se sigue el procedimiento de diseño para cargas estáticas. Por otro lado, las cargas de trabajo son repetitivas y pueden producir falla por fatiga; debe verificarse la resistencia a estas cargas mediante un criterio de falla por fatiga. Se deben encontrar los esfuerzos máximos y mínimos en la sección crítica. Como los esfuerzos en la viga dependen de la posición del carro sobre el puente, es necesario determinar la posición más crítica del carro, es decir, la que produce el momento máximo. Diagrama de cuerpo libre: La figura 5.28 muestra el diagrama de cuerpo libre de la viga, para una posición del carro dada por una distancia arbitraria x. Se toma el valor máximo de la fuerza. Figura 5.28 Diagrama de cuerpo libre de la viga del puente grúa Aunque la fuerza de 200 kN se divide en dos componentes, que actúan en los puntos de contacto de las ruedas con la viga, se ha asumido concentrada en un solo punto. Esta asunción es conservadora 6a a a 7a a Carga máxima de 200 kN Guía Columna Ruedas Carro A B C x 10 m – x F = 200 kN RA RB x y 46 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS (el momento calculado de esta manera es un poco mayor al real) y produce un resultado casi idéntico al que se obtendría descomponiendo la fuerza en dos, ya que la distancia entre las ruedas del carro es pequeña comparada con la luz de la viga. Ecuaciones de equilibrio y cálculo de las reacciones en los apoyos: x R R x M B B A ) kN/m 20 ( donde de ; 0 m) 10 ( kN) (200 ; 0      . ) kN/m 20 ( kN 200 donde de ; 0 kN 200 ; 0 x R R R F A B A y        Diagramas de fuerza cortante y momento flector: Figura 5.29 Diagrama de fuerza cortante de la viga Figura 5.30 Diagrama de momento flector de la viga Momento flector máximo: Los diagramas de fuerza cortante y momento flector se han construido con la fuerza máxima sobre la viga en una posición arbitraria. El momento máximo ocurre en la sección B y está dado por: , [m]) ( 20 [m] 200 m] [kN 2 x x M    Aunque por la simplicidad de la configuración de la viga, se puede deducir directamente con qué distancia x se maximiza el momento flector, para encontrar esta distancia efectuamos lo siguiente: , 0  dx dM entonces m. 5 entonces , 0 40 200 entonces ; 0 ) 20 200 ( 2      x x dx x x d + V (kN) x (m) 200 – 20x – 20x A B C x 10 – x M (kNm) x (m) (200 – 20x)x A B C CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 47 Esto quiere decir que el momento es máximo en la viga simplemente apoyada cuando la fuerza está ubicada en el centro de la viga. El momento flector máximo se obtiene reemplazando el valor de x en la ecuación para el momento: m. kN 500 ) 5 )( 20 ( ) 5 )( 200 ( 2     max M Puntos críticos: Los esfuerzos máximos en la viga actúan en los punto más alejados del eje neutro de la sección B, puntos 1 y 2 (figura 5.31), cuando x = 5 m. El punto 2 está sometido a compresión y el punto 1 está a tracción. Figura 5.31 Distribución de esfuerzos en la sección crítica de la viga Cálculo de los esfuerzos: El esfuerzo en el punto 1 varía entre S1max y S1min, los cuales están dados por: . 0 y 1 1 1   min max max S I c M S El esfuerzo mínimo ocurrirá cuando no se tenga carga en el puente y puede tomarse igual a cero, si se desprecia el peso de la viga y el de la grúa. Algo similar ocurre con el punto 2, a compresión. El esfuerzo máximo puede tomarse igual a cero (puente sin carga), y el esfuerzo mínimo cuando la carga máxima está en la posición más crítica: . 0 y 2 2 2    max max min S I c M S Como las distancias desde el eje neutro hasta los puntos críticos son iguales, c1 = c2 = 4a (figura 5.31), de las ecuaciones anteriores se concluye que S1max = –S2min = S. Los esfuerzos medios y alternativos están dados por las ecuaciones 5.15.a y b, entonces: . 2 2 y 2 2 1 1 1 1 1 1 S S S S S S S S min max a min max m       . 2 2 y 2 2 2 2 2 2 2 2 S S S S S S S S min max a min max m        c1 = 4a c2 = 4a Eje neutro Sección B S2 S1 (2) (1) 48 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Se anticipa que, aparte del factor de seguridad, la única variable de la ecuación de diseño en que difieren los puntos 1 y 2 es el esfuerzo medio (S1m = –S2m), ya que las propiedades y factores Sy, Sn y Kff son iguales para ambos puntos, al igual que los esfuerzos alternativos. Cuando dos puntos críticos 1 y 2 de un material uniforme sometido a esfuerzos normales sólo difieran en que S1m = –S2m, el punto más crítico es aquel cuyo esfuerzo medio sea positivo. Esto se debe a que los esfuerzos de tracción tienden a expandir las grietas que producen la falla por fatiga, mientras que los esfuerzos de compresión no. Para aclarar esto, los puntos críticos se ubican en el diagrama Sm - Sa. La figura 5.32 muestra las líneas de seguridad para este material dúctil uniforme (Sy = Syc) y la ubicación de los dos puntos críticos. Se puede concluir que el punto 1 es más crítico que el 2, ya que si el punto 1 está en un polígono dado por un factor de seguridad N, el 2 quedará adentro de dicho polígono o estaría en un polígono de seguridad dado por un mayor valor de N (polígono a trazos). Figura 5.32 Líneas de seguridad para la viga de material dúctil uniforme, usando el criterio de Soderberg Consecuentemente, sólo se requiere analizar el punto crítico 1, cuyo esfuerzo medio es de tracción. Para calcular los esfuerzos en el punto 1, se debe determinar el momento rectangular de inercia de la sección. El ejemplo 3.2 (capítulo 3) ilustró un procedimiento para calcular el momento rectangular de inercia de una sección compuesta por varias secciones simples, utilizando el teorema de los ejes paralelos. Debido a esto, aquí sólo se presenta el resultado, dejándose al estudiante el cálculo de I, si lo considera necesario: . m 10 27 . 1 ) mm 8 . 50 )( 67 . 190 ( 67 . 190 4 3 4 4      a I Los esfuerzos medio y alternativo en el punto 1 están dados por: MPa. 40 ) m 10 27 . 1 2( m) .0508 0 m)(4 kN 500 ( 2 2 4 3 1          I c M S S S max a m Propiedades del material: Las propiedades del material, acero SAE 1020 laminado en caliente, se obtienen de la tabla A-3.2 (apéndice 3) o se estiman a partir de otras propiedades de la misma tabla: Sy = 207 MPa Su = 379 MPa Se’  0.5Su = 189.5 MPa (de la ecuación 5.2) Sm Syc/N A Sn/N –Syc/N Sy/N Líneas de seguridad Sa (S/2, S/2) (-S/2, S/2) (1) (2) CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 49 Determinación de factores: Los factores que tienen en cuenta la reducción de la resistencia a la fatiga son: Ka  0.74 (figura 5.11, acabado: laminado en caliente, acero con Su = 379 MPa = 55 ksi) Kb = 0.6 (de = 0.808 bh , si t > 0.025h (ecuación 5.25); t = a = 50.8 mm > 0.0258a = 10.16 mm, entonces de = 0.808 mm 307 56 mm 8 . 50 808 . 0 8 7     a a . De la ecuación 5.21 se obtiene que Kb = 0.6) Kc = 0.753 (tabla 5.2, trabajando con una confiabilidad de 99.9%) Kd = 1 (ecuación 5.26, la temperatura en la viga es menor de 450 °C) Ke = 1 (no se prevé corrosión ni otros factores que reduzcan o aumenten la resistencia) Kcar = 1 (ecuación 5.28, el elemento está sometido a flexión) Kff = 1 (en el punto crítico no hay concentrador de esfuerzos. Nótese que Kt = 1 ya que no hay concentración de esfuerzos; de la ecuación 5.30 se obtiene que Kf = 1, y finalmente de la ecuación 5.34 se obtiene que Kff = 1) Kfm = 1 (ecuación 5.32, KfSmax= 1(80 MPa) = 80 MPa < Sy) K = 0.740.60.753111 = 0.334. Determinación del factor de seguridad usando el criterio de Soderberg: La resistencia a la fatiga corregida está dada por la ecuación 5.50, ya que se está diseñando para vida infinita; entonces Sn = K Se’ = 0.334189.5 MPa = 63.3 MPa. Reemplazando estos datos en la ecuación 5.74 (excepto que se usa Sy en lugar de Su, ya que se está utilizando la aproximación de Soderberg), tenemos: . 2 . 1 donde de , MPa 3 . 63 MPa 40 1 MPa 207 MPa 40 1 1      N N Este factor de seguridad parece un poco pequeño. Si las incertidumbres en el diseño son significativas, se corre el riesgo de falla por fatiga. Cuando se usa el criterio de Soderberg, no es necesario verificar la condición de la ecuación 5.74 (ni de la ecuación 5.80), ya que la línea de Soderberg está por debajo de la línea de falla inmediata por fluencia en tracción en todo el rango Sm > 0. La resistencia del puente a las cargas pico (combinación de cargas de trabajo, cargas sísmicas, etc.) debe verificarse usando el procedimiento de diseño estático; además, debe verificarse la resistencia de la soldadura y la estabilidad de las alas y el alma, es decir, verificar que no ocurra pandeo en éstas. 5.12 ESFUERZOS COMBINADOS VARIABLES En secciones precedentes se estudiaron las ecuaciones de diseño para cargas variables simples. En esta sección se termina el estudio de la teoría de fatiga, tratando el caso de esfuerzos multiaxiales, el cual es común en la práctica. Ejemplos típicos de elementos sometidos a esfuerzos combinados variables son los árboles para transmisión de potencia y las tuberías o sistemas sometidos a presión variable. 50 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS Los diferentes esfuerzos que actúan en un punto crítico de un elemento pueden ser: (a) Mutuamente sincrónicos en fase, es decir, actuando con la misma frecuencia y alcanzando sus valores máximos (y mínimos) simultáneamente. (b) Mutuamente sincrónicos fuera de fase, es decir, con igual frecuencia, pero los máximos (y mínimos) no se alcanzan simultáneamente. (c) Asincrónicos (con diferente frecuencia). (d) Aleatorios. (e) Alguna combinación de los anteriores. En el caso (a), los esfuerzos varían de una manera tal que los planos principales no cambian con el tiempo; es decir, los esfuerzos principales cambian en magnitud pero no en dirección. Este caso se denomina esfuerzo multiaxial simple. Los casos en los cuales los esfuerzos principales cambian de dirección con el tiempo, son asincrónicos o están fuera de fase se denominan esfuerzo multiaxial complejo. Este tópico está siendo investigado; se han analizado algunos casos específicos, pero no existe un procedimiento de diseño que sea aplicable a todas las situaciones de esfuerzo multiaxial complejo. Son muchas las posibles combinaciones y sólo unos pocos casos han sido estudiados. La literatura sugiere que asumir que las cargas son sincrónicas en fase (esfuerzo multiaxial simple) usualmente arroja resultados satisfactorios para el diseño de máquinas y usualmente, aunque no siempre, conservativos. Para el caso de esfuerzo multiaxial simple, Norton presenta dos métodos: el método Sines[10, citado en 1] y el método von Mises. El método Sines ha sido validado con datos experimentales para probetas de sección circular, pulidas y sin concentradores de esfuerzos sometidas a flexión y torsión combinadas. Sin embargo, este método podría ser no conservativo para probetas con concentradores de esfuerzos. Algunos autores recomiendan usar el método Von Mises. Este método, descrito a continuación, es más conservativo que el método Sines y podría ser aplicado conservadoramente en muchas aplicaciones de diseño de máquinas. Método von Mises En este método, se determina un esfuerzo equivalente de von Mises con los diferentes esfuerzos alternativos en el punto de análisis y el correspondiente esfuerzo equivalente para los esfuerzos medios: , 2 ) ( 6 ) ( ) ( ) ( 2 2 2 2 2 2 ZXm YZm XYm XXm ZZm ZZm YYm YYm XXm me                    (5.82) , 2 ) ( 6 ) ( ) ( ) ( 2 2 2 2 2 2 ZXa YZa a XY XXa ZZa ZZa YYa a YY a XX ae                    (5.83) o para un estado de esfuerzo plano: , 3 2 2 2 XYm YYm XXm YYm XXm me           (5.84) , 3 2 2 2 a XY YYa XXa a YY a XX ae           (5.85) donde me y ae son los esfuerzos equivalentes medio y alternativo, respectivamente, y las componentes medias y alternativas de los esfuerzos normales (XX, YY y ZZ) y cortantes (XY, YZ y ZX) se obtienen con CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 51 los diferentes esfuerzos que actúan en el punto de análisis (SXXm, SYYm, SZZm, SsXYm, SsYZm, SsZXm, SXXa, SYYa, SZZa, SsXYa, SsYZa y SsZXa), multiplicando cada uno de éstos por el correspondiente factor de concentración de esfuerzos por fatiga Kfm o Kff. Los esfuerzos equivalentes se reemplazan en una de las tres ecuaciones siguientes, las cuales son similares a la 5.74 ó 5.77, pero aquí no aparecen los factores Kfm y Kff, ya que éstos ya han sido involucrados en los cálculos de me y ae. Para materiales dúctiles: ó Soderberg) (usando 1 n ae y me S S N     (5.86) ), modificada Goodman (usando 1 usar si ; 1 y maxe y maxe n ae u me S σ N N S σ S S N       (5.87) donde maxe es el esfuerzo máximo equivalente de von Mises, calculado sin tener en cuenta los efectos de concentración de esfuerzos. Para materiales frágiles: . 1 n ae u me S S N     (5.88) Pueden existir conflictos en el cálculo de Sn, ya que Sn103 (ecuación 5.36) y los factores Kb (ecuaciones 5.21 a 5.25) y Kcar (ecuaciones 5.28.a, b y d) (no debe usarse la ecuación 5.28.c ya que se están calculando los esfuerzos equivalentes de von Mises) dependen del tipo de carga. Se recomienda usar los menores valores de Kb, Kcar y Sn103. Por ejemplo, usar Kcar = 0.7 y Sn103 = 0.75Su si se tiene una combinación de carga axial con flexión, torsión o ambas. 5.13 RESUMEN DEL CAPÍTULO El comportamiento de elementos sometidos a cargas variables es diferente al de elementos sometidos a cargas estáticas. Las cargas variables, aún siendo menores que las estáticas que producen la falla, podrían hacer fallar los materiales después de cierto tiempo, debido a la aparición y crecimiento de grietas. Actualmente existen tres modelos de falla por fatiga: el procedimiento de vida-esfuerzo, el de vida- deformación y el de mecánica de fractura elástica lineal (LEFM). El primero de ellos es el que se describe en este capítulo, siendo particularmente adecuado para el diseño en el régimen de alto ciclaje (HCF), cuando los esfuerzos varían de una manera consistente y conocida, como por ejemplo en las máquinas rotativas. Los otros dos métodos son más adecuados para el régimen de bajo ciclaje (LCF). 52 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS RESISTENCIAS A LA FATIGA (VALORES ESTIMADOS) 0.5 Su, si Su < 1380 MPa (200 ksi) 690 MPa = 100 ksi, si Su  1380 MPa (200 ksi) 0.4 Su, si Su < 330 MPa (48 ksi) 132 MPa = 19 ksi, si Su  330 MPa (48 ksi) ESFUERZOS VARIABLES FACTORES QUE REDUCEN LA RESISTENCIA A LA FATIGA K = Ka Kb Kc Kd Ke Kcar  Ka se obtiene de la figura 5.11 para los aceros. Esta figura puede usarse para otros metales dúctiles como valores aproximados  Ka = 1, para el hierro fundido Kb = Kc: según confiabilidad Kd = Kcar = donde la constante de Neuber, a , se obtiene de la tabla 5.3 El índice q puede obtenerse directamente de la figura 5.16 Se’  Aceros Carga axial Aceros en flexión y torsión 1 Sección circular: de = d , torsión o flexión giratoria de = 0.37 d , flexión no giratoria Sección rectangular o en I (si t > 0.025h): de = 0.808(hb)0.5 , flexión no giratoria 1, Temp  450 C 1 – 0.0058/°C (Temp – 450 °C), 450 °C < Temp  550 C SÓLO PARA ACEROS Confiabilidad (%) 50 90 99 99.9 99.99 99.999 Kc 1 0.897 0.814 0.753 0.702 0.659 1, flexión (o torsión si se calculan los esfuerzos equivalentes de von Mises) 0.577, torsión 0.7, carga axial in) (10 mm 250 si , 6 . 0 in) 10 in 0.3 si , 869 . 0 ( mm 250 mm 8 si , 189 . 1 in) (0.3 mm 8 si , 1 097 . 0 097 . 0             e b e e b e e b e b d K d d K d d K d K , 1 1 r a q   2 smin smax ms S S S   2 smin smax as S S S   2 min max m S S S   2 min max a S S S   Sf’@5108  Aleaciones de aluminio Se’  0.4 Su, Acero fundido Se’  0.35 Su, Hierro fundido gris Se’  0.4 Su, Hierro fundido nodular Se’  0.33 Su, Hierro fundido nodular normalizado CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 53   1 1    t f K q K , donde Kt se obtiene del apéndice 5. Para chaveteros y roscas, Kf se obtiene directamente de las tablas 5.4 y 5.5 respectivamente     0 entonces 2 ó 2 si entonces y si entonces y si entonces ó si                 fm ys smin smax f y min max f ms as f ys fm ys smin f ys smax f m a f y fm y min f y max f f fm s y max s f y max f K S S S K S S S K S S K S K S S K S S K S S K S K S S K S S K K K S S K S S K RESISTENCIA A LA FATIGA CORREGIDA (VIDA FINITA E INFINITA) 3 10 10 si , 3  c n n S  n S 6 3 log 3 1 2 10 10 10 si , ] [ 3 10 3                   c S ' KS c e n n n ' KS S n e 6 10 si ,  c e n ' KS 3 10 10 si , 3  c n n S cref c ' KS S z c z f z n n n n ' KS S f n                     3 log 1 / 3 / 3 1 10 10 si , ] [ ] [ 3 10 3 , donde z = 3 – log(ncref) 8 3 log 699 . 5 1 5264 . 0 10 5 @ 5264 . 1 10 10 5 10 si , ] [ ] [ 8 10 5 @ 3 10 8 3                           c ' KS S c f n n n ' KS S f n cref c n f n n ' KS cref  si , @ donde          Mises von de es equivalent esfuerzos los secalculan si torsión para 9 . 0 torsión para 577 . 0 9 . 0 axial carga para 75 . 0 flexión para 9 . 0 3 10 u u u u n S S S S S ciclos 10 si , 1 3  c n 6 3 log 3 1 10 10 si ,   c f K c n K n f ciclos 10 si , 6  c f n K c f ff n K K todo para ,   ff K Aceros y materiales dúctiles de baja resistencia Materiales frágiles o de alta resistencia Aceros o materiales con codo en nc = 106 Materiales que no poseen límite de fatiga Se puede asumir conservadoramente: material cualquier y todo para , c f ff n K K   n S 54 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS ECUACIONES DE DISEÑO MATERIALES DÚCTILES: Goodman modificada o Soderberg Esfuerzos normales 0 Para  m S : modificada Goodman . 1 usar si ; 1 y a m y a m n a ff u m fm S S S N N S S S S S K S S K N       : Soderberg . 1 n a ff y m fm S S K S S K N   . 1 0, Para n a ff m yc ff n S S K N S N S K S     . 1 , Para yc m a yc ff n m S S S N N S K S S     Esfuerzos cortantes : modificada Goodman . 1 usar si ; 1 ys as ms ys as ms n as ff us ms fm S S S N N S S S S S K S S K N       : Soderberg . 1 n as ff ys ms fm S S K S S K N   Esfuerzos combinados (método von Mises) : modificada Goodman . 1 usar si ; 1 y maxe y maxe n ae u me S σ N N S σ S S N       : Soderberg . 1 n ae y me S S N     MATERIALES FRÁGILES: Goodman modificada Esfuerzos normales Esfuerzos cortantes . 1 , 0 Para n a ff u m t m S S K S S K N S    . 1 n as ff us ms t S S K S S K N   . 1 0, Para n a ff m t uc ff n S S K N S N K S K S     . ) ( 1 , Para uc m a t t uc ff n m S S S K N N K S K S S     . 1 n ae u me S S N     Sms es siempre mayor o igual a cero Esfuerzos combinados (método von Mises) CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 55 donde me y ae están dados por: . 2 ) ( 6 ) ( ) ( ) ( 2 2 2 2 2 2 ZXm YZm XYm XXm ZZm ZZm YYm YYm XXm me                    . 2 ) ( 6 ) ( ) ( ) ( 2 2 2 2 2 2 ZXa YZa a XY XXa ZZa ZZa YYa a YY a XX ae                    o para estado biaxial: , 3 2 2 2 XYm YYm XXm YYm XXm me           . 3 2 2 2 a XY YYa XXa a YY a XX ae           maxe podría calcularse como . 2 ) ( 6 ) ( ) ( ) ( 2 2 2 2 2 2 ax m sZX ax m sYZ ax XYm s ax m XX ax m ZZ ax m ZZ ax m YY ax YYm ax XXm maxe S S S S S S S S S           5.14 REFERENCIAS Y BIBLIOGRAFÍA NORTON, Robert L.. Diseño de Máquinas. México: Ed. Prentice-Hall (Pearson), 1999. DOWLING, N. E.. Mechanical Behavior of Materials. N.J.: Prentice-Hall: Englewood Cliffs, 1993, págs. 416 y 418. FAIRES, V. M.. Diseño de Elementos de Máquinas. México: Editorial Limusa, 1995. 4ª Reimpresión. SHIGLEY, Joseph y MISCHKE, Charles. Diseño en Ingeniería Mecánica. México: McGRAW- HILL, 1991. JUVINALL, R. C.. Stress, Strain and Strength. Nueva York: McGraw-HILL, 1967, pág. 234. SHIGLEY, J. E. y MITCHELL, L. D.. Mechanical Engineering Design. Nueva York: McGraw- HILL, 1983, págs. 293 y 300. 4ª edición. HAMROCK, B. J., JACOBSON, B. y SCHMID, S. R.. Elementos de Máquinas. México: McGRAW-HILL, 2000. MOTT, R. L. Diseño de Elementos de Máquinas. 2ª ed. México: Prentice-Hall Hispanoamericana, S.A., 1992. JUVINALL, R. C. y MARSHEK, K. M.. Fundamentals of Machine Component Design. 2nd ed. Nueva York: John Wiley & Sons, 1967, pág. 270. SINES, G.. Failure of Materials under Combined Repeated Stresses Superimposed with Static Stresses, Technical Note 3495, NACA, 1955. 5.15 EJERCICIOS PROPUESTOS E-5.1 Suponga que el árbol de un motor debe soportar un par de torsión máximo de 2500 lbfin totalmente alternante. ¿Qué diámetro mínimo, d, debería tener el árbol mecanizado si tiene un chavetero de perfil y es de acero SAE 1050 laminado en frío? Tomar N = 2 (Soderberg) con una confiabilidad del 99%. Figura E-5.1 56 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS E-5.2 El árbol mostrado, de acero SAE 1045 laminado en frío, está sometido a un par de torsión que varía desde 5000 lbfin en un sentido hasta 3000 lbfin en sentido contrario. Hallar las dimensiones d2 y d3 del árbol mecanizado si N = 1.6 (Soderberg), Kc = 1, q = 0.9 y se prevé una vida útil de 5×105 ciclos. E-5.3 El eje escalonado de sección circular, de acero SAE 1050 laminado en frío, es mecanizado y soporta una fuerza F producida por una polea loca montada en el eje mediante un rodamiento. La fuerza F varía entre 500 lbf y 1500 lbf, y el eje se apoya en sus extremos, generándose las reacciones indicadas. Determinar el diámetro d para una duración indefinida, una confiabilidad del 50% y un factor de seguridad de 1.5. E-5.4 Resuelva el problema anterior, asumiendo que la polea loca se monta a presión sobre el eje, girando solidariamente con éste, y que el eje está apoyado mediante rodamientos en los extremos. Suponga además que F es constante e igual a 1500 lbf y que el radio de los redondeos es r = 0.25 in. E-5.5 El eslabón mostrado está sometido a una fuerza axial F que varía desde 70 kN en tracción hasta 90 kN en compresión y que se aplican mediante pasadores en los agujeros. Escoger una aleación de aluminio forjada de las de la tabla A-3.5, para que el eslabón soporte con seguridad la carga durante 5108 ciclos. Asuma N = 2, K = 0.4 y q = 1. Figura E-5.2 5 cm 5 cm 15 cm F F r = 4 mm 20 cm 10 cm F F Figura E-5.5 d F 0.8 in 0.8 in 3.2 in 3.2 in D = 1.5d r = 1/16 in d 0.5F 0.5F Figura E-5.3 d2 d3 = 1.2d2 d1 = d3/10 r = d2/10 d2 r CAPÍTULO 5 CARGASVARIABLES - TEORÍA DE FATIGA 57 E-5.6 La viga de la figura se construirá mediante dos placas de acero laminado en frío SAE 1040. Las dos entallas y el agujero son mecanizados. La viga está sometida a una fuerza constante F1 = 2 kN y una fuerza F2 que varía desde 2 kN hasta 4 kN. Calcular el factor de seguridad de la viga, asumiendo q = 1 y que no hay concentración de esfuerzos en el empotramiento. Trabajar con una confiabilidad del 99.9%. E-5.7 El resorte plano de la figura se utiliza para mantener en contacto un sistema leva plana - seguidor de rodillo. El resorte es de acero con Su = 180 ksi y Sy = 150 ksi, con una longitud L = 10 in y un peralte (altura de la sección rectangular) de 1/8 in. Debido a la precarga del resorte y a la forma de la leva, la deflexión del resorte varía entre ymin = 1.2 in y ymax = 2.0 in. Determinar el factor de seguridad del resorte asumiendo K = 0.65 y que no hay efecto de concentración de esfuerzos en el extremo empotrado del resorte. Usar la aproximación de: (a) Soderberg. (b) Goodman modificada. E-5.8 Resolver el problema anterior con ymin = 2.3 in y ymax = 2.8 in. Respuestas: E-5.1 d = 1.30 in (usando Goodman modificada o Soderberg). Figura E-5.6 F1 F2 1 m 1 m 1 m  4 cm r = 2 cm 20 cm 15 cm 7 cm 4 cm 15 cm 0.8 in Leva Seguidor h = 1/8 in Resorte de 10 in de longitud Figura E-5.7 58 CONCEPTOS BÁSICOSSOBRE DISEÑO DE MÁQUINAS E-5.2 d2 = 1.47 in y d3 = 1.76 in según Goodman modificada (d2 = 1.49 in y d3 = 1.79 in, usando Soderberg). E-5.3 d = 0.601 in usando Goodman modificada (d = 0.619 in, usando Soderberg). E-5.4 d = 0.746 in (usando Soderberg o Goodman modificada). E-5.5 Su  260 MPa. Puede seleccionarse la aleación de aluminio H36, la H38 u otra de mayor resistencia a la tracción (usando Soderberg o Goodman modificada). E-5.6 N = 11.8 usando Goodman modificada (N = 10.8, usando Soderberg). E-5.7 (a) N = 1.02; (b) N = 1.13. E-5.8 (a) N = 0.836 (según Soderberg, el resorte fallaría después de cierto número de ciclos); (b) N = 0.952 (el resorte fallaría inmediatamente por fluencia). Download AboutSupportTermsPrivacyCopyrightCookie PreferencesDo not sell or share my personal information English © 2025 Slideshare from Scribd
17540	https://www.youtube.com/watch?v=ZWrGNxoD1zw	Solving Inequalities Graphically (using two graphs) Professor Monte 1690 subscribers 6 likes Description 257 views Posted: 18 Mar 2024 In this short video, Professor Monte explains how to use the graph of two functions to solve inequalities. Basically, the idea is that whenever we have two functions in the form f(x) is less than g(x), then the graph of f(x) is below the graph of g(x). Similarly, if f(x) is greater than g(x), then the graph of f(x) is above g(x). Examples are given to make the idea as simple and easy as possible. Transcript: Hi, it's Professor Monte. Let's talk about solving inequalities graphically. Now I've got two examples up here. The graphs are already drawn. Obviously if you are presented the problem, they say solve the inequality graphically, but they don't give you the graphs. The first thing you're gonna do is to do is draw the graphs, but they're just graphs of lines. In this case, we had parabola. You'll have to do that from what we did previously, but once we have the graphs, let's talk about how to answer these questions. So the first one says, solve F of X is greater than G of X. Well, the blue graph is F of X, the red graph is G of X. What does this mean though? Well, remember the function value, whether it's F of X or G of X, it's the function value that's always the Y value. So the Y value of this first line is greater than the Y value of the second line. Here's the first line, here's the second line. So what are they talking about? Well, if the Y value's greater, it just means it's above. So in this case, this is saying since F of X is greater than G of X, it means the F graph is above the G graph. And so we just look where that happens. Well, the F graph here at whatever this X value is, say negative four. The red graph of the G graph is above the blue graph, the F graph. So it doesn't satisfy this. I wanna know where the blue graph or the F graph is above the red graph or the G graph. Well, it's over here for any of these X's above. So if I go here, starting at one and then anything to the right, the F graph is above the G graph and I just need to write that number notation. So since there's no equal sign here, I don't include the one. So I'll go parentheses one and then forever to the right, which is infinity. And that's my answer. So we're saying, okay, anytime I have an X value anywhere between one all the way to the right, the F graph is above the G graph or it's greater. All right, same idea. This time we have a less than. So now going with what we said there, this Y value's below that Y value. So now I'm just looking where the F graph is below the G graph. Well here the G graph is above the F graph. I wanna know when the F graph's below, oh, the F graph's below, where is blue below red, oh, it is here. So when it's here, this time I have the equal. So I use a bracket, anywhere over there the blue graph is below the red graph, the G graph. In between here, in between negative five and zero. Those x values, it's not below. The blue graph's above it, but once I get over here to zero, again, since I have the inequality, I'll have a bracket. And then anything to the right, the F graph or blue graph is below the red graph or the G graph. Okay, so that's the idea. I'm looking where this blue graph is below the red graph. And so interval notation is gonna say negative infinity to negative five. Remember, these are all related to X values for what X values is, is true. And then we'll say union or, it's also zero to infinity. That's my answer. Sometimes they'll say, or usually with set notation, they use the union symbol, which means or, okay, but that's, there's two different areas of the graph where the blue graph's below the red graph. So the basic idea, remember this, if it says above or if it says greater than it means above. If it says less than, it just means below. So don't make it more difficult than it is. Work with it, practice a little bit and I think you'll get it pretty quickly.
17541	https://bmcgenomics.biomedcentral.com/articles/10.1186/s12864-020-06962-8	Skip to main content BMC Genomics Download PDF Selected Topics in “Systems Biology and Bioinformatics” - 2019: genomics Review Open access Published: Differences between human and chimpanzee genomes and their implications in gene expression, protein functions and biochemical properties of the two species Maria V. Suntsova1 & Anton A. Buzdin1,2,3,4 BMC Genomics volume 21, Article number: 535 (2020) Cite this article 182k Accesses 79 Citations 269 Altmetric Metrics details Abstract Chimpanzees are the closest living relatives of humans. The divergence between human and chimpanzee ancestors dates to approximately 6,5–7,5 million years ago. Genetic features distinguishing us from chimpanzees and making us humans are still of a great interest. After divergence of their ancestor lineages, human and chimpanzee genomes underwent multiple changes including single nucleotide substitutions, deletions and duplications of DNA fragments of different size, insertion of transposable elements and chromosomal rearrangements. Human-specific single nucleotide alterations constituted 1.23% of human DNA, whereas more extended deletions and insertions cover ~ 3% of our genome. Moreover, much higher proportion is made by differential chromosomal inversions and translocations comprising several megabase-long regions or even whole chromosomes. However, despite of extensive knowledge of structural genomic changes accompanying human evolution we still cannot identify with certainty the causative genes of human identity. Most structural gene-influential changes happened at the level of expression regulation, which in turn provoked larger alterations of interactome gene regulation networks. In this review, we summarized the available information about genetic differences between humans and chimpanzees and their potential functional impacts on differential molecular, anatomical, physiological and cognitive peculiarities of these species. Background The divergence of human and chimpanzee ancestors dates back to approximately 6,5–7,5 million years ago or even earlier . It is still of a great interest to identify genetic elements that distinguish humans from chimpanzees and encode features of human physiological and mental identities [3,4,5]. It’s a difficult task to quantitate the exact percentage of differences between human and chimpanzee genomes. In early works, divergence of human and chimpanzee genomes was estimated as roughly 1% . This estimate was based on the comparison of protein-coding sequences and didn’t consider non-coding (major) part of DNA. However, the idea of ~ 99% similarity of genomes persisted for a long time, until 2005 when nearly complete initial sequencing results of both human and chimpanzee (Pan troglodytes) genomes became available. It was found that genome differences represented by single nucleotide alterations formed 1.23% of human DNA, whereas larger deletions and insertions constituted ~ 3% of our genome . Moreover, even higher proportion was shaped by chromosomal inversions and translocations comprising several megabase-long chromosomal regions or even entire chromosomes, as for the chromosomal fusion that took place when the human chromosome 2 was formed . Here we tried to review the major known structural and regulatory genetic alterations that had or might have a functional impact on the human and chimpanzee speciation (Table 1). Karyotype Human karyotype is represented by 46 chromosomes, whereas chimpanzees have 48 chromosomes . In general, both karyotypes are very similar. However, there is a major difference corresponding to the human chromosome 2. It has originated due to a fusion of two ancestral acrocentric chromosomes corresponding to chromosomes 2a and 2b in chimpanzee. Also, significant pericentric inversions were found in nine other chromosomes . Two out of nine are thought to occur in human chromosomes 1 and 18, and the other seven – in chimpanzee chromosomes 4, 5, 9, 12, 15, 16 and 17 [10,11,12]. In addition, there are numerous differences in the chromosomal organization of pericentric, paracentric, intercalary and Y type heterochromatin; for example, the chimpanzees have large additional telomeric heterochromatin region on chromosome 18 . Additionally, the majority of chimpanzee’s chromosomes contain subterminal constitutive heterochromatin (C-band) blocks (SCBs) that are absent in human chromosomes. SCBs predominantly consist of the subterminal satellite (StSat) repeats, they are found in African great apes but not in humans . The presence of such SCBs affects chimpanzees’ chromosomes behavior during meiosis causing persistent subtelomeric associations between homologous and non-homologous chromosomes. As a result of homologous and ectopic recombinations chimpanzees demonstrate greater chromatin variability in their subtelomeric regions . Studying sex chromosomes also revealed several peculiar traits. There are several regions of homology between X and Y chromosomes, so-called pseudoautosomal regions (PARs) most probably arisen due to translocation of DNA from X to Y chromosome . The term “pseudoautosomal” means that they can act as autosomes being involved in recombination between X and Y chromosomes. PAR1 is a 2,6 Mb long region located at the end of Y chromosome short arm. It is homologous to the terminal region of the short arm on X chromosome. PAR2 is a 330 kb-long sequence located on the termini of long arms of X and Y chromosomes. In contrast to PAR1 presenting in many mammalian genomes, PAR2 is human-specific . It includes four genes: SPRY3, SYBL1, IL9R and CXYorf1. The first two genes (SPRY3, SYBL1) are silent on the Y chromosome (SPRY3, SYBL1) and are subjects of X-inactivation-like mechanism. On the other hand, the genes IL9R and CXYorf1 are active in both sex chromosomes [55, 56]. Moreover, the short arm of Y contains a 4 Mb-long translocated region from the long arm of X chromosome, called X-translocated region (XTR) [14, 57]. A part of the XTR has undergone inversion due to recombination between the two mobile elements of LINE-1 family. Both translocation and inversion took place already after separation of human and chimpanzee ancestors [14, 58]. Finally, this region also includes genes PCDH11Y and TGIF2LY which correspond to X chromosome genes PCDH11X and TGIF2LX . Around 2% of human population have signs of recombination between X and Y chromosomes at the XTR. It should be considered, therefore, as an additional human-specific pseudoautosomal region PAR3 . Insertions, deletions and copy number variations Enzymatic machinery of LINE1 retrotransposons not only reverse transcribes its own RNA molecules, but also frequently produces cDNA copies of other cellular transcripts, e.g. host genes or non-coding RNAs [59, 60] Sometimes a template switch can occur due to reverse transcription, thus resulting in double or even triple chimeric retrotranscripts . Reverse-transcribed copies of the host genes are called processed pseudogenes . Immediately after primary assembly of the human and chimpanzee genomes, nearly 200 human- and 300 – chimpanzee-specific processed pseudogenes were identified. Most of them were copies of ribosomal protein genes which accounted for ~ 20% of species-specific pseudogenes . However, these numbers were significantly underestimated. For example, another study revealed already ~ 1800 and 1500 processed pseudogenes of ribosomal protein in the human and chimpanzee genomes, respectively, of which ~ 1300 were common . In addition to genome sequencing, DNA hybridization arrays were widely used for copy number variation studies [63, 64]. In human, microarray assay revealed a relatively increased copy number of 134 and decreased - of six genes compared to the genomes of other great apes such as chimpanzee (Pan troglodytes), bonobo (Pan paniscus), gorilla (Gorilla gorilla) and orangutan (Pongo pygmaeus) . However, the figure of six genes with decreased copy numbers was certainly an underestimation because hybridization was performed using the probes for human genes. This assay also couldn’t distinguish functional genes and pseudogenes. Anyway, the human-amplified group was found to be enriched in genes involved in central nervous system (CNS) functioning. These were NAIP (neuronal apoptosis inhibitory protein), SLC6A13 (gamma-aminobutyric acid (GABA) transporter), KIAAA0738 (zinc-finger transcription factor, expressed in brain), CHRFAM7A (fusion of acetylcholine receptor gene and FAM7), ARHGEF5 (guanine exchange factor), ROCK1 (Rho-dependent protein kinase), and also members of the gene families: ARHGEF, PAK, RhoGAP and USP10 (ubiquitin-specific protease) associated with various forms of mental retardation. Relatively to humans, chimpanzees had increased copy numbers of 37 and decreased copy numbers of 15 genes . The same study also revealed increased copy number of Rho GTPase-activating protein SRGAP2 gene in human genome . There were also two truncated human-specific homologs of this gene: SRGAP2B and SRGAP2C. The experiments with mouse embryos showed that SRGAP2 could facilitate maturation and limit density of dendrite spines in the developing neurons in neocortex. Truncated protein SRGAP2C forms a dimeric complex with the normal SRGAP2 and inhibits it. Apparently, physiological expression of SRGAP2C and SRGAP2B could impact human brain development by causing specific increase of spine density and extension of maturation of pyramidal neurons in human neocortex . Another study was focused on sequences conserved in chimpanzees and other primates but underrepresented in humans (termed hCONDELs) . Comparison of human, chimpanzee and macaque genomes revealed 510 conserved regions deleted in humans, all of them representing non-coding sequences except CMAHP gene, see below. The hCONDELs identified were enriched near genes involved in steroid hormone signaling and neuronal functioning. One hCONDEL was a sensory vibrissae and penile spines-specific enhancer of androgen receptor (AR) gene. Its absence caused the loss of vibrissae and spines in humans. Another deletion involved enhancer of a tumor suppressor gene GADD45G, which activated expression of this gene in the subventricular zone of the forebrain. It could relate to the specific pattern of expansion of brain regions in humans. In turn, the chimpanzee genome also lacks several conserved sequences. Among 344 such regions identified, significant enrichment was found for the localizations near genes related to synapse formation and functioning of glutamate receptors . Finally, substantial differences in copy numbers were reported for transposable elements (TEs). According to various estimates, the number of human-specific TE insertions varied from eight to 15,000 copies . It was estimated that humans have approximately twice as many unique TE copies as the chimpanzees [8, 26]. Since human-chimpanzee ancestral divergence, the most active TE groups were Alu, LINE1 and SVA which accounted for nearly 95% of all species-specific insertions . The most numerous group was Alu, which made over 5 thousand human-specific insertions and proliferated approx. Three times more intensely in humans than in chimpanzees [26, 27]. Most of chimpanzee-specific Alu copies are represented by subfamilies Alu Y and AluYc1, while human-specific insertions are predominantly the members of AluYa5 and AluYb8 subfamilies [8, 26]. However, both species also have specific inserts of AluS and AluYg6 family members. Besides insertional polymorphism, Alu also impacted divergence of the two genomes through homologous recombination. At least 492 human-specific deletions emerged because of recombinations between the different Alu copies that made ~ 400 kb of excised DNA. Of them, 295 deletions covered known or predicted genes . For example, the aforementioned CMAHP gene lost its 6th exon due to recombination event between the two Alu elements . Another example is tropoelastin gene. In most vertebrates, it has 36 exons. During the evolution, primate ancestors have lost the 35th exon, and then human ancestors additionally lost the exon 34, also most probably due to recombination between the Alu elements . On the other hand, Alu-Alu recombinations had significant impact also for the chimpanzee genome: at least 663 such chimpanzee-specific deletions lead to 771 kb DNA loss, and roughly a half of them took place inside gene regions . The activities of LINE-1 transposable elements were comparable in humans and chimpanzees and resulted in over 2000 species-specific integrations . LINE-1 is ~ 6 kb-long TE harboring two open reading frames. The majority of LINE-1 inserts are 5′-truncated, most probably due to apparently abortive reverse transcription . Interestingly, among the human-specific TEs there were several times more full-length LINE-1 elements with intact open reading frames. The species-specific insertions were made by the members of the LINE-1 subfamilies L1-Hs and L1-PA2 [26, 28, 67]. In addition, LINE-1 elements were responsible for at least 73 human-specific deletions collectively resulting in a loss of nearly 450 kb of genomic DNA [22, 23]. Another family termed SVA (SINE-VNTR-Alu) elements is represented in the human genome by about 1000 species-specific genomic copies, which is approximately twice higher than in the chimpanzee [26, 27]. Noteworthy, the human genome contains at least 84 insertions of a new, exclusively human-specific type of transposable elements called CpG-SVA or SVAF1, formed by CpG-island of human gene MAST2 fused with 5′-truncated fragment of SVA. This group most likely emerged through insertion of an SVA element into the first exon of MAST2 gene containing a CpG-island. Because of MAST2 promoter activity, a chimeric transcript was formed, processed and then reverse transcribed by LINE-1 enzymatic machinery followed by insertions into a plethora of new genomic positions. For these new copies of a hybrid element, MAST2 CpG island enabled male germ line-specific expression, thus facilitating fixation in the genome [29, 30]. Finally, like the other major groups of TEs SVA elements also mediated loss of human genomic DNA. At least 26 cases of SVA-associated human-specific deletions were mentioned in the literature, which totally resulted in ~ 46 kb of deleted DNA . After split of human and chimpanzee ancestors, there was also a HERV-K (HML-2) family of endogenous retroviruses that was proliferating in both genomes [31, 32, 68, 69]. Its insertional activity resulted in ~ 140 human-specific copies that formed ~ 330 kb of human DNA [31,32,33,34], some of them being polymorphic in human populations [69,70,71,72,73,74]. In turn, the chimpanzee genome has at least 45 species-specific insertions of these elements [37, 38]. In addition, two new specific retroviral families – PtERV1 and PtERV2 with 250 totally chimpanzee-specific copies, arose already in the chimpanzee genome [8, 39]. The new copies of transposable elements can appear in the genome not only through insertions but also due to duplications of genomic DNA. For example, several hundred copies of recently integrated HERV-K (HML-2) family provirus К111 were found in centromeres of 15 different human chromosomes. They amplified and spread due to recombinations of the enclosing progenitor locus. In contrast, there is only one copy of К111 in the chimpanzee genome and no copies in the other primates [35, 36]. Similarly, several dozen copies of a more ancient provirus K222 of the same family arose due to chromosomal recombination in pericentromeric regions of nine human chromosomes, versus only one copy in the chimpanzees and other higher primates . Furthermore, a human-specific endogenous retroviral (ERV) insert was demonstrated to serve as the tissue-specific enhancer driving hippocampal expression of PRODH gene responsible for proline degradation and metabolism of neuromediators in CNS . Finally, the ERVs can provide their promoters for expression of non-coding RNAs from the downstream genomic loci . Almost all ERV inserts in introns of human genes were fixed in the antisense orientation relative to gene transcriptional direction , most probably because of the interference of gene expression with their polyadenylation signals. However, it has a functional consequence of ERV-driven antisense transcripts overlapping with human genes. For two genes, SLC4A8 (for sodium bicarbonate cotransporter) and IFT172 (for intraflagellar transport protein 172), these human-specific antisense transcripts overlap with the exons and regulate their expression by specifically decreasing their mRNA levels . TE inserts also could play an important role in the speciation. TEs contain various regulatory elements such as promoters, enhancers, splice-sites and signals of transcriptional termination, which they use for their own expression and spread. Approximately 34% of all species-specific TEs in humans and chimpanzees are located close to known genes . Species-specific TE inserts, therefore, can strongly influence regulatory landscape of the host genome [79, 80]. In addition, TEs can disrupt gene structures by inserting themselves or through recombinations between their copies [21, 23]. These events could influence gene functioning and might cause the respective phenotypic differences [81, 82]. It is worth to note that the main complication of the earlier studies was connected with the quality of non-human genomes assembly. First of all, there were persisting several thousand gaps in the chimpanzee genome, which made a substantial fraction of DNA inaccessible for comparisons. Second, the final stages of apes genomes assemblies and annotations were performed using the human genome as a template . This obviously bias results by “humanizing” great ape genomes thereby concealing some human-specific structural variations. The combination of long-read sequence assembly and full-length cDNA sequencing for de novo chimpanzee genome assembly without guidance from the human genome allowed to overcome this problem . Comparison of de novo sequenced and independently assembled human and great ape genomes revealed 17,789 fixed human-specific structural variants (fhSVs), including 11,897 fixed human-specific insertions and 5892 fixed human-specific deletions. Among fhSVs, a loss of 13 start codons, 16 stop codons, and 61 exonic deletions in the human lineage were detected. Also, fhSVs affected 643 regulatory regions near 479 genes. Totally, 46 fhSVs deletions were detected that were expected to disrupt human genes, 41 of them were new. The affected genes included for example caspase recruitment domain family member 8 (CARD8), genes FADS1 and FADS2 involved in fatty acids biosynthesis, and two cell cycle genes WEE1 and CDC25C . Single nucleotide alterations Human specific single nucleotide alterations constitute ~ 1.23% of our genome. This value was found by directly comparing human with chimpanzee genomes. It was very close to the previous theoretical estimate of 1.2% calculated using average divergence rate for autosomes, for the time of human and chimpanzee ancestor’s divergence . In the human populations, ~ 86% of all human specific single nucleotide alterations is fixed and the rest 14% is polymorphic . Remarkably, the lowest and the highest human-chimpanzee nucleotide sequence divergences, 1.0 and 1.9%, respectively, were detected in the chromosomes X and Y. Outstandingly, as much as 15% of all ancestral CG-dinucleotides underwent mutations either in the human or in the chimpanzee lineage . Protein-coding sequences Protein coding sequences are 99.1% identical between the two species , and in two-thirds of the proteins amino acid sequences are absolutely the same . Generally, in comparison with the model of the latest common ancestor genome, the chimpanzee has more genes that underwent positive selection than human. This can be explained by the different effective sizes of ancestral populations of the two species . However, after divergence, transcription factors (TFs) were the fastest evolving group of genes, and human TFs had ~ 1,5 times higher amino acids substitution rate . Second, genes linked with neuronal functioning also evolved faster in the human lineage . There is a connection identified between mutations in the transcription factor FOXP2 gene and speech disorders, and an assumption was made that FOXP2 is responsible for speech and language development in humans. Indeed, the sequence analysis revealed that FOXP2 has signs of positive selection during human evolution having two human-specific amino acid substitutions: Thr303Asn and Asn325Ser, where the latter led to a new potential phosphorylation site . In vivo experiments showed that these substitutions may have important functional significance. Transgenic mice with humanized version of their FoxP2 gene demonstrated faster learning when both declarative and procedural mechanisms were involved. Also, they had peculiar dopamine levels and higher neuronal plasticity in the striatum . The microcephalin gene MCPH1 is involved in the regulation of brain development. Its mutations are linked with severe genetic disorders like microcephaly. During human speciation, this gene evolved under strong positive selection, which is still going on in the modern human population . Another gene connected with the brain size regulation, ASPM (abnormal spindle-like microcephaly associated, MCPH5), also evolved faster in hominids than in the other primates, having the highest rate of non-synonymous to synonymous substitutions in the human lineage . Several sexual reproduction genes were also among the most rapidly evolving and positively selected hits [44, 89], such as protamine genes PRM1 and PRM2 encoding histone analogs in sperm cells. Remarkably, human protamines evolve oppositely to histones, whose structures are highly conservative . Another group of highly diverged genes relates to immunity and cell recognition . A point mutation in the variable domain of T-cell gamma-receptor TCRGV10 destroyed a donor splice-site, which prevented splicing of the leader intron. Chimpanzees don’t have this mutation and their gene remains functional . Both species have many specific mutations in the genes involved in sialic acids metabolism - ST6GAL1, ST6GALNAC3, ST6GALNAC4, ST8SIA2 and HF1 . Sialic acids, or N-acetyl neuraminic (Neu5Ac) and N-glycolyl neuraminic acid (Neu5Gc), are common components of the carbohydrate cell surface complexes in mammals. Humans are exceptional because they completely lack Neu5Gc on their cell surfaces because their gene CMAHP coding an enzyme – cytidine monophosphate-N-acetylneuraminic acid hydroxylase – responsible for the conversion of CMP-Neu5Ac into CMP-Neu5Gc, has lost its activity. It happened because of the loss of a 92-nucleotide exon corresponding to the sixth ancestral exon, caused by insertion of an AluY element followed by recombination [20, 91]. Moreover, the mechanism of sialic acids recognition was also affected in the human lineage. Human gene SIGLEC11 for sialic acid receptor underwent a conversion with the pseudogene SIGLEC16 that significantly compromised its ability to bind sialic acids. However, it still can bind oligosialic acids (Neu5Acα2–8)2–3, that are highly abundant in the brain. Moreover, SIGLEC11 demonstrates human-specific expression in microglia . Similarly, the protein SIGLEC12 lost its sialic acid-binding activity due to human-specific substitution R122C. Nevertheless, SIGLEC12 gene is still expressed in macrophages and in several epithelial cell types . Another major affected group of genes is for the olfactory receptors. Humans and chimpanzees have a comparable number of olfactory receptor genes, around 800, and 689 of them are orthologous in the two species . However, in both species about half of them have lost their activities and became pseudogenes. Even though the final numbers of active genes are equal in human and chimpanzee, their repertoire is strikingly different – as much as 25% of the active olfactory receptor genes are species-specific. This has led to an assumption that the most recent common ancestor had more active olfactory receptor genes than modern humans and chimpanzees . Other examples include caspase 12, mannose-binding lectin gene MBL1P and keratin isoform KRTHAP1 that lost their activities due to human-specific mutations [8, 42, 94]. Non-coding sequences Non-coding sequences play crucial roles in gene regulation [95, 96]. Analysis of species-specific polymorphisms revealed that 96% of regions with the highest density of alterations (HAR, human accelerated region) map on non-coding DNA. The genes located near HARs are predominantly related to interaction with DNA, transcriptional regulation and neuronal development [48, 97]. The biggest number of HARs was observed for the NPAS3 (neuronal PAS domain-containing protein) gene. It codes for a transcription factor involved in brain development. The 14 HARsNPAS3 are located in non-coding regions and most of them may have regulatory functions, as confirmed by enhancer activities demonstrated in cell culture assay . Rapidly evolving human genome region HAR1 was found in the overlap of two non-coding RNA genes: HAR1F and HAR1R. The former is expressed at 7–19 weeks of embryonic development in the Cajal-Retzius cells of the emerging neocortex. At the later gestation period and in adulthood HAR1F is expressed also in the other parts of the brain. This expression pattern is conserved in all higher primates, but human-specific nucleotide alterations affected the secondary structure of this RNA [48, 99]. Another accelerated region HARE5 (HAR enhancer 5) is ~ 1,2 kb long enhancer of FZD8 gene. After human and chimpanzee ancestral divergence, their orthologous loci accumulated 10 and 6 nucleotide substitutions, respectively. FZD8 encodes a receptor protein in the WNT signaling pathway, which is involved in the regulation of brain development and size. In mouse, endogenous HARE5 homolog physically interacts with Fzd8 core promoter in the neocortex. In transgenic mice with Fzd8 under control of either human or chimpanzee enhancer, both demonstrated their activities in the developing neocortex, but the human enhancer became active at the earlier stages of development and its effect was more pronounced. Embryos with the human HARE5, therefore, showed a marked acceleration of neural progenitor cell cycle and increased brain size . There is also a particular fraction of non-coding sequences that was accelerated in humans but relatively conserved in the other species called HACNs (human accelerated conserved noncoding sequences) . They can overlap with the abovementioned HARs . HACNs are enriched near genes related to neuronal functioning, such as neuronal cell adhesion and brain development . Based on structural analyses of HACNs, HARs and their genomic contexts, around one third of them was predicted to be developmental enhancers . By functional role, they contribute in approximately equal proportions to brain and limb development and to a lesser extent - to heart development. Among 29 pairs of HARs and their chimpanzee orthologous regions tested in mouse embryos, 24 showed enhancer activity in vivo. Moreover, five of them demonstrated differential enhancer activities between human and chimpanzee sequences . In another study, all human enhancers predicted by the FANTOM project were aligned with the primate genomes in order to obtain human-specific fraction . Notably, the fastest evolving human enhancers predominantly regulated genes activated in neurons and neuronal stem cells. Totally, about 100 human-specific neuronal enhancers were identified, and one of them located on the 8q23.1 region was presumably related to Alzheimer’s disease development. It was assumed by the authors that recent human-specific enhancers, adaptive, on the one hand, may also impact age-related diseases . Transcriptional regulation It has been postulated few decades ago that differences between humans and chimpanzees are mostly caused by gene regulation changes rather than by alterations in their protein-coding sequences, and that these changes must affect embryo development . For example, evolutional acquisitions such as enlarged brain or modified arm emerged as a result of developmental changes during embryogenesis [102, 103]. Such changes include when, where and how genes are expressed. A plethora of genes involved in embryogenesis have pleiotropic effects and mutations within their coding sequence may cause complex, mostly negative, consequences for an organism. On the other hand, changes in gene regulation could be limited to a certain tissue or time frame that can enable fine tuning of a gene activity . Indeed, the fast-evolving sequences (HARs or HACNs) are often found close to the genes active during embryo- and neurogenesis [48,49,50, 100]. For example, HACNS1 (HAR2) demonstrates greater enhancer activity in limb buds of transgenic mice compared to orthologous sequences from chimpanzee or rhesus macaque . A similar pattern was observed for the aforementioned HARs related to genes NPAS3 and FZD8 that are active during CNS development in embryogenesis [51, 98]. Many studies were focused on finding differences between humans, chimpanzees and other mammals at the level of gene transcription [107,108,109]. Importantly, tissue-specific differences within the same species significantly exceeded in amplitude all species-specific differences in any tissue. The most transcriptionally divergent organs between humans and chimpanzees were liver and testis, and to a lesser extent – kidney and heart [107, 108]. A transcriptional distinction of liver may be a consequence of different nutritional adaptations in the two species. The major differences in testes are largely unexplained but may be related to predominantly monogamous behavior in humans. Surprisingly, the brain was the least divergent organ between humans and chimpanzees at the transcriptional level. In this regard, it is suggested that tighter regulation of signaling pathways in the brain underlies behavioral and cognitive differences [109, 110]. However, it was found that during evolution in the human cerebral cortex there were more transcriptional changes than in the chimpanzee . Among them, the prevailed difference was increased transcriptional activity [110, 111]. In addition, many differences were identified in the alternative splicing patterns including 6–8% of gene exons, thus supporting a concept that the differentially spliced transcripts have pronounced functional consequences for the speciation . Another study of transcriptional activity in the forebrain evidenced the higher difference between human and chimpanzee in the frontal lobe . The functions of frontal lobe-specific groups of co-expressed genes dealt mostly with neurogenesis and cell adhesion . Furthermore, the analysis of 230 genes associated with communication showed that about a quarter of them was differentially expressed in the brains of humans and other primates . KRAB-zinc finger (KRAB-ZNF) genes were overrepresented among the genes differentially expressed in the brain . Remarkably, the KRAB-ZNF gene family is known for its rapid evolution in primates, especially for its human- or chimpanzee-specific members . The studies of transcriptional timing in the postnatal brain development also revealed a number of human-specific features. A specific set of genes was found whose expression was delayed in humans compared to the other primates. For example, the maximum expression of synaptic genes in the human prefrontal cortex was shifted from 1 year of age as for the chimpanzees and macaques, to 5 years. It is congruent with the prolonged brain development period in humans relative to other primates [116, 117]. The results recently published by Pollen and colleagues allowed to look deeper into the developing human and chimpanzee brains by applying the organoid model . Cerebral organoids were generated from induced pluripotent stem cells (iPSCs) of humans and chimpanzees. Transcriptome analyses revealed 261 genes deferentially expressed in human versus chimpanzee cerebral organoids and macaque cortex. The PI3K/AKT/mTOR signaling axis appeared to be stronger activated in human, especially in radial glia . Epigenetic regulation is another factor that should be considered when looking at interspecies differences in gene expression. High throughput analysis of differentially methylated DNA in human and chimpanzee brains showed that human promoters had lower degree of methylation. A fraction of genes related to neurologic/psychiatric disorders and cancer was enriched among the differentially methylated entries . The analysis of H3K4me3 (trimethylated histone H3 is a marker of transcriptionally active chromatin) distribution in the neurons of prefrontal lobe revealed 471 human-specific regions, 33 of them were neuron-specific. Some of these regions were proximate to genes associated with neurologic and mental disorders, such as ADCYAP1, CACNA1C, CHL1, CNTN4, DGCR6, DPP10, FOXP2, LMX1B, NOTCH4, PDE4DIP, SLC2A3, SORCS1, TRIB3, TUBB2B and ZNF423 [119, 120]. Another active chromatin biomarker is the distribution of DNase I hypersensitivity sites (DHSs), that often indicate gene regulatory elements. It was found that 542 DHSs overlapped with HARs, thus being so-called human accelerated DHSs, haDHSs . Using chromatin immunoprecipitation assay, a number of haDHSs interacting genes were identified, many of which were connected with early development and neurogenesis [3, 121]. In a later study , about 3,5 thousand haDHSs were found, that were enriched near the genes related to neuronal functioning . Conclusions It is now generally accepted that both changes in gene regulation and alterations of protein coding sequences might have played a major role in shaping the phenotypic differences between humans and chimpanzees. In this context, complex bioinformatic approaches combining various OMICS data analyses, are becoming the key for finding genetic elements that contributed to human evolution. It is also extremely important to have relevant experimental models to validate the candidate species-specific genomic alterations. The currently developing experimental methods such as obtaining pluripotent stem cells and target genome modifications, like CRISPR-CAS , open exciting perspectives for finding a “needle in haystack” that was truly important for human functional evolution, or probably many such needles. However, at least for now using these experimental approaches for millions of species specific potentially impactful features reviewed here is impossible due to high costs and labor intensity. In turn, an alternative approach could be combining the refined data in a realistic model of human-specific development using a new generation systems biology approach trained on a functional genomic Big Data of humans and other primates. Such an approach could integrate knowledge of protein-protein interactions, biochemical pathways, spatio-temporal epigenetic, transcriptomic and proteomic patterns as well as high throughput simulation of functional changes caused by altered protein structures. The differences revealed could be also analyzed in the context of mammalian and primate-specific evolutionary trends, e.g. by using dN/dS approach to measure evolutionary rates of structural changes in proteins and enrichment by transposable elements in functional genomic loci to estimate regulatory evolution of genes . Apart from the single-gene level of data analysis, this information could be aggregated to look at the whole organismic, developmental or intracellular processes e.g. by using Gene Ontology terms enrichment analysis and quantitative analysis of molecular pathways . And finally, most of the results described here were obtained for the human and chimpanzee reference genomes, which were built each using DNAs of several individuals. Nowadays the greater availability of whole-genome sequencing highlighted the next challenge in human and chimpanzee comparison – populational genome diversity. For example, the recent study of 910 native African genomes was focused on the fraction of sequences absent from the reference Hg38 genome assembly. As many as 125,715 insertions missing in the Hg38 was identified with the average number of 859 insertions per individual, making up a total of 296,5 Mb. These findings clearly suggest that the current version of the human genome assembly can lack nearly 10% of the genome information. Furthermore, it also reflects the high degree of genome heterogeneity of the African population . Similar studies were performed for other populations as well. For example, in the Chinese population a total of 29,5 Mb new DNA and 167 predicted novel genes missing in the reference genome assembly was discovered . The chimpanzees also demonstrate substantial genome diversity with many population-specific traits: the central chimpanzees retain the highest diversity in the chimpanzee lineage, whereas the other subspecies show multiple signs of population bottlenecks . So far there were not so many studies published on the topic of non-reference human and chimpanzee genome comparison. However, some estimates can be made. In the recent study of 1000 genomes from the Swedish population there were identified totally 61,044 clusters totally making ~ 46 Mb of human DNA that were absent from the reference Hg38 human genome assembly. These clusters were called by the authors “new sequences” (NSs). As expected, NSs were enriched in simple repeats and satellites and varied greatly among the individuals. The most part of NSs (32,794) aligned confidently to the non-reference sequences from the aforementioned study of 910 African genomes . Finally, as many as 18,773 NSs were present also in the chimpanzee PT4 genome assembly. In terms of protein coding sequences, 143 orthologous chimpanzee genes contained a total of 2807 NSs, where four genes were strongly enriched: EPPK1, OR8U1, NINL, and METTL21C. Positioning of NS insertions in the human genome revealed that 2195 of them located within 2384 genes, where 85 NS insertion events were found within the exons of 82 genes . Another research consortium studied non-repetitive non-reference sequences (NRNR) in the genomes of 15,219 Icelanders . A total of 326,596 bp of NRNR DNA was found, where ~ 84% was formed by only 244 insertions longer than 200 bp. Notably, comparison with the chimpanzee genome revealed that over 95% of the NRNRs longer than 200 bp were present also in the chimpanzee genome assembly, thus indicating that they were ancestral . Thus, the lack of information on genome populational diversity could impact the total extent of human and chimpanzee interspecies divergence by misinterpretation of polymorphic sequences. However, it doesn’t abrogate most of the hypotheses and facts mentioned in this review. Still, these findings inevitably lead to the idea of the need, firstly, to create, and secondly, to compare human and chimpanzee pan-genomes. Availability of data and materials Not applicable. Abbreviations Mya: : Million years ago Mb: : Megabase (million base pairs) kb: : Kilobase (thousand base pairs) HAR: : Human accelerated region HERV: : Human endogenous retrovirus LINE: : Long interspersed nuclear element PAR: : Pseudoautosomal region TE: : Transposable element References Amster G, Sella G. 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About this supplement This article has been published as part of BMC Genomics Volume 21 Supplement 7, 2020: Selected Topics in “Systems Biology and Bioinformatics” - 2019: genomics. The full contents of the supplement are available online at Funding This study was supported by the Russian Foundation for Basic Research Grant 19–29-01108. Publication costs were funded by Moscow Institute of Physics and Technology (National Research University). The funding bodies played no role in the design of this study and collection, analysis, and interpretation of data and in writing of the manuscript. Author information Authors and Affiliations Institute for personalized medicine, I.M. Sechenov First Moscow State Medical University, Trubetskaya 8, Moscow, Russia Maria V. Suntsova & Anton A. Buzdin 2. Shemyakin-Ovchinnikov Institute of Bioorganic Chemistry of the Russian Academy of Sciences, Miklukho-Maklaya, 16/10, Moscow, Russia Anton A. Buzdin 3. Omicsway Corp, Walnut, CA, USA Anton A. Buzdin 4. Moscow Institute of Physics and Technology (National Research University), 141700, Moscow, Russia Anton A. Buzdin Authors Maria V. Suntsova View author publications 2. Anton A. Buzdin View author publications Search author on:PubMed Google Scholar Contributions AB and MS systematically analyzed the literature, interpreted the data, read and edited the manuscript. All authors read and approved the final manuscript. Corresponding author Correspondence to Anton A. Buzdin. Ethics declarations Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare that they have no competing interests. Additional information Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. 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Reprints and permissions About this article Cite this article Suntsova, M.V., Buzdin, A.A. Differences between human and chimpanzee genomes and their implications in gene expression, protein functions and biochemical properties of the two species. BMC Genomics 21 (Suppl 7), 535 (2020). Received: Accepted: Published: DOI: Share this article Anyone you share the following link with will be able to read this content: Provided by the Springer Nature SharedIt content-sharing initiative Keywords Human-specific Chimpanzee Genome alterations Genetic differences Molecular evolution
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17543	https://learn.valur.com/new-york-capital-gains-tax-explained/	TALK TO OUR TEAMLog In -> Estate Tax Capital Gains Ordinary income QSBS Advisors SOLUTIONS Guide to IDGTs Guide to Non-Grantor Trusts Guide to SLATs Guide to GRATs READ UP #### All estate tax articles Start from the ground up. Review our primers, guides to specific structures, and detailed case studies #### Estate tax video Check out our video series on planning to reduce your estate tax Estate Tax Planning Trusts: A Comprehensive Guide The purpose of estate tax planning is to maximize the assets you pass on to future generations by minimizing gift and estate taxes. Estate-tax strategies revolve around the use of... Read More SOLUTIONS Guide to CRUTs Guide to Exchange Funds Guide to Deferred Sales Trusts READ UP #### All Capital Gains articles Start from the ground up. Review our primers, guides to specific structures, and detailed case studies #### Capital Gains video Check out our video series on planning to reduce your Capital Gains Tax Deferral Strategy: Comparing the Big Three You can defer capital gain taxes with a Charitable Remainder Trust, Opportunity Zone, or Exchange Fund. CRTs get the best returns. Which is right for you? Read More SOLUTIONS Guide to Solar Investment Tax Credits Guide to Oil & Gas Funds READ UP #### All Ordinary Income articles Start from the ground up. Review our primers, guides to specific structures, and detailed case studies #### Ordinary Income video Check out our video series on planning to reduce your Ordinary Income Solar Tax Incentives vs. Oil and Gas Well Investments: A Comprehensive Comparison Taking advantage of solar tax incentives and investing in oil and gas wells are two popular strategies for offsetting ordinary income tax. How do you know which one is right... 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Read More FEATURED ARTICLE Tax Planning for Realized Gains and Ordinary Income Tax planning strategies for realized gains and ordinary income Read More FEATURED ARTICLE Charitable Remainder Trust Guide Read More FEATURED ARTICLE Guide To GRATs by Valur Read More Valur Library > Capital Gains > Employees & Investors > New York State Capital Gains Tax in 2025 Explained Mani Mahadevan Last updated: 22.02.25 New York State Capital Gains Tax in 2025 Explained If you have cashed out capital gains in New York State, you know you’ll lose something to taxes. But how much? It’s important to understand your capital gains tax liability and how it will impact your financial future, not least because that knowledge will empower you to take action to reduce your tax bill today. In this article, we’ll explain what capital gains are and how they are taxed in New York State. We’ll also show you different tax-planning strategies that can significantly reduce your state capital gains tax: Sell appreciated assets in a Charitable Remainder Trust in order to defer capital gains. Buy renewable energy projects that make you eligible for significant government tax incentives that lower your capital gains tax bill. Explore Oil and gas well investments that offer substantial tax benefits. Reduce your taxable income with charitable deduction tax strategies such as Charitable Lead Annuity Trusts. So let’s dive in! What are Capital Gains? A capital gain is a capital asset’s increase in value from the value at which it was purchased. Capital assets include stocks, real estate, crypto, and private businesses – in short, any significant property that could gain or lose value over time. Capital gains can be realized or unrealized. Realized capital gains are gains that you have captured by selling the asset. Unrealized gains, by contrast, refer to the increase in the value of an investment that you have not yet sold. For instance, if you hold stock that increases in value, but you haven’t sold it yet, that is considered an unrealized capital gain. In general, you will not pay taxes until you cash out or “realize” the gains. Long-Term and Short-Term Capital Gains There are two types of realized capital gains: Short-term capital gains: These are gains from selling assets that you’ve held for one year or less. At the federal level, short-term capital gains are typically taxed at the same (high) rate as ordinary income. Long-term capital gains: These are gains from selling assets that you’ve held for more than one year. At the federal level, long-term capital gains receive more favorable tax treatment than short-term gains. How are Capital Gains Taxed? Capital gains are not taxed until they are realized, meaning that even if your Apple stock has increased 50x from the day you bought the stock, you won’t owe any capital gains taxes until you sell the stock. Of course, once you do sell the stock, you will face federal and (depending on the state) state taxes. The tax rate will vary depending on your income and the type of asset you sold, but the rates are generally progressive, so individuals with higher incomes tend to face higher capital gains tax rates. Let’s look at how federal and state governments tax capital gains in more depth. Need some help to understand the most convenient tax planning structure to reduce your capital gains taxes? Our team of tax-planning experts can help! Talk to us What Is The Federal Capital Gains Tax (2025)? Short- and long-term capital gains are taxed differently; assets held for one year or less are taxed at ordinary income rates, while longer-held assets are taxed at lower rates. The short-term capital gains schedule matches the schedule for ordinary income, and your marginal and effective rates depend on your income and marital status, as shown below: \| Taxable income (Single Filers) \| Taxable income (Married Filing Jointly) \| Tax Rate \| --- \| $0 to $11,925 \| $0 to $23,850 \| 10% \| \| $11,925 to $48,475 \| $23,850 to $96,950 \| 12% \| \| $48,475 to $103,350 \| $96,950 to $206,700 \| 22% \| \| $103,350 to $197,300 \| $206,700 to $394,600 \| 24% \| \| $197,300 to $250,525 \| $394,600 to $501,050 \| 32% \| \| $250,525 to $626,350 \| $501,050 to $751,600 \| 35% \| \| $626,350 or more \| $751,600 or more \| 37% \| Short-Term Federal Capital Gains Tax Rates for 2025 Long-term capital gains, meanwhile, are taxed at a lower rate than ordinary income. Here, too, the precise rate depends on the individual’s income and marital status: \| Taxable income (Single Filers) \| Taxable income (Married Filing Jointly) \| Tax Rate \| --- \| $0 to $48,350 \| $0 to $96,700 \| 0% \| \| $48,350 – $533,400 \| $96,700 – $600,050 \| 15% \| \| $533,400 or more \| $600,050 or more \| 20% \| Long-Term Federal Capital Gains Tax Rates for 2025 In addition, some categories of capital assets fall entirely outside of this rubric: gains on collectibles such as art, jewelry, antiques, and stamp collections are taxed up to a maximum 28% rate. That’s not all: There’s an additional federal tax that was introduced in 2010, known as the Net Investment Income Tax (NIIT), that applies to most capital gains that exceed the exemption amount. The first $200,000 for a single filer, or $250,000 for married filers, are exempt from the NIIT. But everything in excess of those thresholds is taxed at 3.8%. What Is The New York State Capital Gains Tax? New York State taxes capital gains as income, and the capital gains tax rate reaches 10.9%(excludes New York City, which has its own separate tax brackets): \| Taxable Income (Single Filers) \| Taxable Income (Married Filing Jointly) \| Tax Rate on This Income \| --- \| $0 to $8,500 \| $0 to $17,150 \| 4% \| \| $8,500 to $11,700 \| $17,150 to $23,600 \| 4.5% \| \| $11,700 to $13,900 \| $23,600 to $27,900 \| 5.25% \| \| $13,900 to $80,650 \| $27,900 to $161,550 \| 5.50% \| \| $80,650 to $215,400 \| $161,550 to $323,200 \| 6.00% \| \| $215,400 to $1,077,550 \| $323,200 to $2,155,350 \| 6.85% \| \| $1,077,550 to $5,000,000 \| $2,155,350 to $5,000,000 \| 9.65% \| \| $5,000,000 to $25,000,000 \| $5,000,000 to $25,000,000 \| 10.3% \| \| $25,000,000 or more \| $25,000,000 or more \| 10.9% \| New York State Capital Gains Tax Rates Case Study So, what would these numbers look like in the real world? Let’s consider Jenna, a New York State investor who purchased 7,000 shares of Apple stock in April 2019 at $50 per share. She decides to sell her shares in January 2025 at a price of $100 each. Jenna held the stock for more than one year, so her realized income is considered long-term capital gain. Jenna realized a capital gain of $350,000. (She paid for 7,000 shares at $50 each, for a total of $350,000, and then sold them for $100 each, for a total of $700,000. That’s a net gain of $350,000). Federal taxes To simplify this example, let’s assume further that she doesn’t earn any other income. (If she did, it would be more complicated to figure out which bracket she falls into.) Given her $360,000 of gains, she would fall into the income bracket between $48,350 and $533,400, resulting in a long-term federal capital gains tax rate of 15%. As a result of the progressive tax system, however, not every dollar will be taxed at that rate. The amount below $48,350 won’t be taxed, so she would pay $46,748 in federal capital gains tax on this transaction (15% of every dollar over $48,350). In addition, Jenna would owe Net Investment Income Tax on the gains in excess of $200,000, resulting in another $6,080 of tax, bringing her total federal tax liability to $52,828. State taxes Jenna would also pay New York State taxes on her capital gains. Given her $350,000 capital gains, she falls into the 6.85% tax bracket. Like the federal government, New York State uses a progressive tax system, which means that different portions of the individual’s capital gains are taxed at different rates. Here’s how it breaks down for Jenna: The first $8,500 is taxed at 4% ($340) The portion from $8,500 to $11,700 is taxed at 4.5% ($144) The portion from $11,700 to $13,900 is taxed at 5.25% ($115.50) The portion from $13,900 to $80,650 is taxed at 5.50% ($3,663) The portion from $80,650 to $215,400 is taxed at 6.00% ($8,073) The portion from $215,400 to $350,000 is taxed at 6.85% ($9,229.90) Adding these amounts together, the individual would pay a total of $21,565.40 in New York State capital gains taxes for 2025. Short-term gains A quick counterfactual: If Jenna had sold her stock after holding for less than a year, her earnings would have been considered short-term capital gains, and she would have been subject to ordinary income taxes at both the federal and New York State levels. Don’t let taxes eat into your gains like Jenna’s did. Our Guided Planner will identify which strategies could save you thousands based on your specific situation. Understand your Tax Savings Opportunities What is Tax Planning? Capital gain taxes can significantly reduce your net earnings from the sale of an asset. Accordingly, it’s critical to identify strategies that can reduce these taxes. Tax planning is a strategic approach to reducing a person’s tax liability by leveraging various tax benefits and allowances. It’s about understanding the tax implications of your financial decisions so you can minimize your taxes and, ultimately, keep more of your hard-earned money. This might involve making investments that offer tax benefits, choosing the right type of retirement account, taking advantage of generally available deductions and credits, or creating a tax-advantaged trust or other vehicle. Tax-Planning Ideas to Reduce Capital Gains Tax There are many tax planning strategies that can help you reduce your federal and New York State capital gains tax liability. 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Read the detailed breakdown or get personalized guidance for your situation. Read the Full CRT Guide Get Personal Guidance Buy a Renewable Energy Project: Purchasing renewable energy projects can make you eligible for significant government tax incentives — tax credits and depreciation — to lower your income taxes. Taking into account tax savings and cash flow, this strategy can return 5.85x on your purchase price compared to choosing to pay your taxes directly. Learn more about renewable energy tax savings here or set up a call with us here to learn more. See how renewable energy can transform your tax burden into wealth-building opportunities. Browse projects tailored to different investment sizes and tax situations. Browse Solar Project Opportunities Invest in Oil & Gas Wells:These structures allow investors to put their money into the production and development of oil and gas wells. 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But unlike a donor advised fund or a typical charity, with a CLAT you can actually get a portion of your charitable gift back in the future in the form of a “remainder interest.” You can also compare the quantitative returns and tax savings of these different strategies using our capital gains tax savings comparison calculator and customize it to your own situation. Conclusion Capital gain taxes can significantly reduce the wealth you keep. Fortunately, there are several strategies available to minimize these taxes. Read more here and check out our Guided Planner tool, where we’ll point you toward the strategies that might apply to you. About Valur We’ve built a platform that makes advanced tax planning – once reserved for ultra-high-net-worth individuals – accessible to everyone. With Valur, you can reduce your taxes by six figures or more, at less than half the cost of traditional providers. 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17544	https://brainly.com/question/16673114	[FREE] The formula to calculate the area of a circle from its circumference is: A = \frac{c^2}{4\pi} - brainly.com 6 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 +21k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +36,2k Ace exams faster, with practice that adapts to you Practice Worksheets +6,8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified The formula to calculate the area of a circle from its circumference is: A=4 π c 2 2 See answers Explain with Learning Companion NEW Asked by hannasavan1241 • 05/29/2020 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4429817 people 4M 0.0 0 Upload your school material for a more relevant answer Using the formula A =c2/4π, you can calculate the area of a circle from its circumference. This formula is derived from the standard area formula, πr2, substituting the relationship of the circle's circumference and diameter. For instance, a circle with a circumference of 10 would have an area of 25/π square units. Explanation In this case, the formula given (A =c2/4π) is used to calculate the area of a circle given its circumference. It's derived from the relationship between a circle's circumference (c) and its diameter (d) where c = πd. So, the formula for the area (A), which is normally πr2, can be rewritten using circumference as A = π(c/2π)2 simplifies to A = c2/4π. Here's an example with a circle of circumference 10. Taking c = 10, you can calculate the area as A = (102)/4π = 25/π square units. Learn more about Circle Area Formula here: brainly.com/question/37848070 SPJ3 Answered by SkylerBliss •13.6K answers•4.4M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 4429817 people 4M 0.0 0 University Physics Volume 1 - William Moebs, Samuel J. Ling, Jeff Sanny Physics - From Stargazers to Starships Mathematics for Biomedical Physics - Jogindra M Wadehra Upload your school material for a more relevant answer The area of a circle can be calculated from its circumference using the formula A=4 π c 2. This relationship is derived from the expressions for circumference and area involving the radius of the circle. For instance, if the circumference is 10 units, the area would be approximately 7.96 square units. Explanation To calculate the area of a circle using its circumference, we can use the formula: A=4 π c 2 This formula comes from the relationship between the circumference c and the radius r of the circle. The circumference of a circle is calculated by the formula: c=2 π r From this, we know that the radius can be expressed as: r=2 π c The standard formula for the area of a circle is: A=π r 2 Substituting the expression for the radius into the area formula, we get: A=π(2 π c)2 This simplifies to: A=π(4 π 2 c 2)=4 π c 2 Thus, when we are given the circumference of a circle, we can directly apply this formula to find its area. Example: If the circumference of a circle is 10 units, then we can calculate the area as follows: Substitute c=10 into the formula: A=4 π 1 0 2=4 π 100=π 25≈7.96 square units This shows that the area of a circle with a circumference of 10 units is approximately 7.96 square units. Examples & Evidence For instance, if a circle's circumference is given as 10 units, substituting this value into the formula yields an area of approximately 7.96 square units. This mathematical relationship is based on well-established geometric formulas involving circles, and the derivation is commonly covered in high school mathematics. Thanks 0 0.0 (0 votes) Advertisement Community Answer This answer helped 1520 people 1K 5.0 1 the answer is b Explanation .... Answered by Anamaria1212 •6 answers•1.5K people helped Thanks 1 5.0 (2 votes) Advertisement ### 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 Explain the association and provide an explanation for the relationship between: • The number of letters in a person's name and the last digit of their phone number. • The number of ice cubes in a drink and the temperature of the drink. The system has no solution. The system has a unique solution: System A $\begin{array}{r} -x-4 y=4 \ x+4 y=4 \end{array}$ (x,y)=(1,□) The system has infinitely many solutions. They must satisfy the following equation: y=□ Identify the expression that uses adding the additive inverse to rewrite the expression 16−7+22−12. A. −16+(−7)+(−22)+(−12) B. −16+7+(−22)+12 C. 16+(−7)+22+(−12) D. 16+(−7)+22+12 (5 3⋅9542)3 A line passes through the point (4,2) and has a slope of 4 5. Write an equation in slope-intercept form for this line. Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
17545	http://www.seas.ucla.edu/~vandenbe/ee236a/homework/problems.pdf	Linear Programming Exercises Lieven Vandenberghe Electrical Engineering Department University of California, Los Angeles Fall Quarter 2013-2014 1 Hyperplanes and halfspaces Exercise 1. When does one halfspace contain another? Give conditions under which {x \| aT x ≤b} ⊆{x \| ˜ aT x ≤˜ b} (a ̸= 0, ˜ a ̸= 0). Also ﬁnd the conditions under which the two halfspaces are equal. Exercise 2. What is the distance between the two parallel hyperplanes {x ∈Rn \| aT x = b1} and {x ∈Rn \| aT x = b2}? Exercise 3. Consider a waveform s(x, t) = f(t −aT x) where t denotes time, x denotes position in R3, f : R →R is a given function, and a ∈R3 is a given nonzero vector. The surfaces deﬁned by t −aT x = constant are called wavefronts. What is the velocity (expressed as a function of a) with which wave-fronts propagate? As an example, consider a sinusoidal plane wave s(x, t) = sin(ωt −kT x). Exercise 4. Linear and piecewise-linear classiﬁcation. The ﬁgure shows a block diagram of a linear classiﬁcation algorithm. x1 x2 xn a1 a2 an b y The classiﬁer has n inputs xi. These inputs are ﬁrst multiplied with coeﬃcients ai and added. The result aT x = Pn i=1 aixi is then compared with a threshold b. If aT x ≥b, the output of the classiﬁer is y = 1; if aT x < b, the output is y = −1. The algorithm can be interpreted geometrically as follows. The set deﬁned by aT x = b is a hyperplane with normal vector a. This hyperplane divides Rn in two open halfspaces: one halfspace where aT x > b, and another halfspace where aT x < b. The output of the classiﬁer is y = 1 or y = −1 depending on the halfspace in which x lies. If aT x = b, we arbitrarily assign +1 to the output. This is illustrated below. a aT x = b aT x > b aT x < b 2 By combining linear classiﬁers, we can build classiﬁers that divide Rn in more complicated regions than halfspaces. In the block diagram below we combine four linear classiﬁers. The ﬁrst three take the same input x ∈R2. Their outputs y1, y2, and y3 are the inputs to the fourth classiﬁer. x1 x2 −1 1 1 −1 −1 −1 −1 2 −1 1 1 2 1 y y1 y2 y3 Make a sketch of the region of input vectors in R2 for which the output y is equal to 1. Simple linear programs Exercise 5. Consider the linear program minimize c1x1 + c2x2 + c3x3 subject to x1 + x2 ≥1 x1 + 2x2 ≤3 x1 ≥0, x2 ≥0, x3 ≥0. Give the optimal value and the optimal set for the following values of c : c = (−1, 0, 1), c = (0, 1, 0), c = (0, 0, −1). Exercise 6. For each of the following LPs, express the optimal value and the optimal solution in terms of the problem parameters (c, k, d, α, d1, d2, . . . ). If the optimal solution is not unique, it is suﬃcient to give one optimal solution. (a) minimize cT x subject to 0 ≤x ≤1. The variable is x ∈Rn. (b) minimize cT x subject to −1 ≤x ≤1. The variable is x ∈Rn. 3 (c) minimize cT x subject to −1 ≤1T x ≤1. The variable is x ∈Rn. (d) minimize cT x subject to 1T x = 1 x ≥0. The variable is x ∈Rn. (e) maximize cT x subject to 1T x = k 0 ≤x ≤1. The variable is x ∈Rn. k is an integer with 1 ≤k ≤n. (f) maximize cT x subject to 1T x ≤k 0 ≤x ≤1 The variable is x ∈Rn. k is an integer with 1 ≤k ≤n. (g) maximize cT x subject to dT x = α 0 ≤x ≤1. The variable is x ∈Rn. α and the components of d are positive. (h) minimize cT x subject to 0 ≤x1 ≤x2 ≤· · · ≤xn ≤1. The variable x ∈Rn. (i) maximize cT x subject to −y ≤x ≤y 1T y = k y ≤1. The variables are x ∈Rn and y ∈Rn. k is an integer with 1 ≤k ≤n. 4 (j) minimize 1T u + 1T v subject to u −v = c u ≥0, v ≥0 The variables are u ∈Rn and v ∈Rn. (k) minimize dT 1 u −dT 2 v subject to u −v = c u ≥0, v ≥0. The variables are u ∈Rn and v ∈Rn. We assume that d1 ≥d2. Exercise 7. An optimal control problem with an analytical solution. We consider the problem of maximizing a linear function of the ﬁnal state of a linear system, subject to bounds on the inputs: maximize dT x(N) subject to \|u(t)\| ≤U, t = 0, . . . , N −1 N−1 P t=0 \|u(t)\| ≤α, (1) where x and u are related via a recursion x(t + 1) = Ax(t) + Bu(t), x(0) = 0. The problem data are d ∈Rn, U, α ∈R, A ∈Rn×n and B ∈Rn. The variables are the input sequence u(0), . . . , u(N −1). (a) Express (1) as an LP. (b) Formulate a simple algorithm for solving this LP. (It can be solved very easily, without using a general LP solver, as a variation of the simple LPs of exercise 6, parts (d)–(g).) (c) Apply your method to the matrices A =      9.9007 10−1 9.9340 10−3 −9.4523 10−3 9.4523 10−3 9.9340 10−2 9.0066 10−1 9.4523 10−2 −9.4523 10−2 9.9502 10−2 4.9793 10−4 9.9952 10−1 4.8172 10−4 4.9793 10−3 9.5021 10−2 4.8172 10−3 9.9518 10−1     , (2) B =      9.9502 10−2 4.9793 10−3 4.9834 10−3 1.6617 10−4     . (3) (You can create these matrices by executing the MATLAB ﬁle ex7data.m on the class webpage.) Use d = (0, 0, 1, −1), N = 100, U = 2, α = 161. 5 Plot the optimal input and the resulting sequences x3(t) and x4(t). Remark. This model was derived as follows. We consider a system described by two second-order equations m1¨ v1(t) = −K(v1(t) −v2(t)) −D(˙ v1(t) −˙ v2(t)) + u(t) m2¨ v2(t) = K(v1(t) −v2(t)) + D(˙ v1(t) −˙ v2(t)). These equations describe the motion of two masses m1 and m2 with positions v1 ∈R and v2 ∈R, respectively, and connected by a spring with spring constant K and a damper with constant D. An external force u is applied to the ﬁrst mass. We use the values m1 = m2 = 1, K = 1, D = 0.1, so the state equations are      ¨ v1(t) ¨ v2(t) ˙ v1(t) ˙ v2(t)     =      −0.1 0.1 −1.0 1.0 0.1 −0.1 1.0 −1.0 1.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0           ˙ v1(t) ˙ v2(t) v1(t) v2(t)     +      1 0 0 0     u(t). We discretize the system by considering inputs u that are piecewise constant with sampling interval T = 0.1, i.e., we assume u is constant in the intervals [0.1k, 0.1(k+1)), for k = 0, 1, 2, . . .. It can be shown that the discretized state equations are z((k + 1)T) = Az(kT) + Bu(kT), k ∈Z, (4) where z(t) = (˙ v1(t), ˙ v2(t), v1(t), v2(t)), and A and B given by (2) and (3). Using the cost function dT x(N) with d = (0, 0, 1, −1) means that we maximize the distance between the two masses after N time steps. Exercise 8. Power allocation problem with analytical solution. Consider a system of n transmitters and n receivers. The ith transmitter transmits with power xi, i = 1, . . . , n. The vector x is the variable in this problem. The path gain from each transmitter j to each receiver i is denoted Aij and is assumed to be known. (Obviously, Aij ≥0, so the matrix A is elementwise nonnegative. We also assume that Aii > 0.) The signal received by each receiver i consists of three parts: the desired signal, arriving from transmitter i with power Aiixi, the interfering signal, arriving from the other transmitters with power P j̸=i Aijxj, and noise vi (vi is positive and known). We are interested in allocating the powers xi in such a way that the signal to noise plus interference ratio (SNIR) at each of the receivers exceeds a level α. (Thus α is the minimum acceptable SNIR for the receivers; a typical value might be around α = 3.) In other words, we want to ﬁnd x ≥0 such that for i = 1, . . . , n Aiixi ≥α  X j̸=i Aijxj + vi  . Equivalently, the vector x has to satisfy the set of linear inequalities x ≥0, Bx ≥αv (5) where B ∈Rn×n is deﬁned as Bii = Aii, Bij = −αAij, j ̸= i. 6 (a) Suppose you are given a desired level of α, so the right-hand side αv in (5) is a known positive vector. Show that (5) is feasible if and only if B is invertible and z = B−11 ≥0. Show how to construct a feasible power allocation x from z. (b) Show how to ﬁnd the largest possible SNIR, i.e., how to maximize α subject to the existence of a feasible power allocation. Hint. You can refer to the following result from linear algebra. Let T ∈Rn×n be a matrix with nonnegative elements, and s ∈R. Then the following statements are equivalent: (a) There exists an x ≥0 with (sI −T)x > 0. (b) sI −T is nonsingular and the matrix (sI −T)−1 has nonnegative elements. (c) s > maxi \|λi(T)\| where λi(T) (i = 1, . . . , n) are the eigenvalues of T. The quantity ρ(T) = maxi \|λi(T)\| is called the spectral radius of T. (For such s, the matrix sI −T is called a nonsingular M-matrix.) Remark. This problem gives an analytic solution to a very special form of transmitter power allocation problem. Speciﬁcally, there are exactly as many transmitters as receivers, and no power limits on the transmitters. One consequence is that the receiver noises vi play no role at all in the solution — just crank up all the transmitters to overpower the noises! Piecewise-linear optimization Exercise 9. Formulate the following problems as LPs: (a) minimize ∥Ax −b∥1 subject to ∥x∥∞≤1. (b) minimize ∥x∥1 subject to ∥Ax −b∥∞≤1. (c) minimize ∥Ax −b∥1 + ∥x∥∞. In each problem, A ∈Rm×n and b ∈Rm are given, and x ∈Rn is the optimization variable. Exercise 10. Formulate the following problems as LPs. (a) Given A ∈Rm×n, b ∈Rm, minimize m X i=1 max{0, aT i x + bi}. The variable is x ∈Rn. (b) Given p + 1 matrices A0, A1, . . . , Ap ∈Rm×n, ﬁnd the vector x ∈Rp that minimizes max ∥y∥1=1 ∥(A0 + x1A1 + · · · + xpAp)y∥1 . Exercise 11. Approximating a matrix in inﬁnity norm. The inﬁnity (induced) norm of a matrix A ∈Rm×n, denoted ∥A∥∞,i, is deﬁned as ∥A∥∞= max i=1,...,m n X j=1 \|aij\|. 7 The inﬁnity norm gives the maximum ratio of the inﬁnity norm of Ax to the inﬁnity norm of x: ∥A∥∞= max x̸=0 ∥Ax∥∞ ∥x∥∞ . This norm is sometimes called the max-row-sum norm, for obvious reasons. Consider the problem of approximating a matrix, in the max-row-sum norm, by a linear combination of other matrices. That is, we are given k + 1 matrices A0, . . . , Ak ∈Rm×n, and need to ﬁnd x ∈Rk that minimizes ∥A0 + x1A1 + · · · + xkAk∥∞. Express this problem as a linear program. Explain the signiﬁcance of any extra variables in your LP. Carefully explain why your LP formulation solves this problem, e.g., what is the relation between the feasible set for your LP and this problem? Exercise 12. We are given p matrices Ai ∈Rn×n, and we would like to ﬁnd a single matrix X ∈Rn×n that we can use as an approximate right-inverse for each matrix Ai, i.e., we would like to have AiX ≈I, i = 1, . . . , p. We can do this by solving the following optimization problem with X as variable: minimize maxi=1,...,p ∥I −AiX∥∞. (6) Here ∥H∥∞is the ‘inﬁnity-norm’ or ‘max-row-sum norm’ of a matrix H, deﬁned as ∥H∥∞= max i=1,...,m n X j=1 \|Hij\|, if H ∈Rm×n. Express problem (6) as an LP. You don’t have to reduce the LP to a canonical form, as long as you are clear about what the variables are, what the meaning is of any auxiliary variables that you introduce, and why the LP is equivalent to the problem (6). Exercise 13. Download the ﬁle ex13data.m from the class website and execute it in MATLAB. This will generate two vectors t, y ∈R42. We are interested in ﬁtting a linear function f(t) = α + βt through the points (ti, yi), i.e., we want to select α and β such that f(ti) ≈yi, i = 1, . . . , 42. We can calculate α and β by optimizing the following three criteria. (a) Least-squares: select α and β by minimizing 42 X i=1 (yi −α −βti)2. (b) ℓ1-norm approximation: select α and β by minimizing 42 X i=1 \|yi −α −βti\|. 8 (c) ℓ∞-norm approximation: select α and β by minimizing max i=1,...,42 \|yi −α −βti\|. Find the optimal values of α and β for each of the three optimization criteria. This yields three linear functions fls(t), fℓ1(t), fℓ∞(t). Plot the 42 data points, and the three functions f. What do you observe? Exercise 14. An illumination problem. We consider an illumination system of m lamps, at posi-tions l1, . . . , lm ∈R2, illuminating n ﬂat patches. lamp j lj vi vi+1 patch i rij θij The patches are line segments; the ith patch is given by [vi, vi+1] where v1, . . . , vn+1 ∈R2. The variables in the problem are the lamp powers p1, . . . , pm, which can vary between 0 and 1. The illumination at (the midpoint of) patch i is denoted Ii. We will use a simple model for the illumination: Ii = m X j=1 aijpj, aij = r−2 ij max{cos θij, 0}, (7) where rij denotes the distance between lamp j and the midpoint of patch i, and θij denotes the angle between the upward normal of patch i and the vector from the midpoint of patch i to lamp j, as shown in the ﬁgure. This model takes into account “self-shading” (i.e., the fact that a patch is illuminated only by lamps in the halfspace it faces) but not shading of one patch caused by another. Of course we could use a more complex illumination model, including shading and even reﬂections. This just changes the matrix relating the lamp powers to the patch illumination levels. The problem is to determine lamp powers that make the illumination levels close to a given desired illumination level Ides, subject to the power limits 0 ≤pi ≤1. (a) Suppose we use the maximum deviation φ(p) = max k=1,...,n \|Ik −Ides\| as a measure for the deviation from the desired illumination level. Formulate the illu-mination problem using this criterion as a linear programming problem. (b) There are several suboptimal approaches based on weighted least-squares. We consider two examples. 9 i. Saturated least-squares. We can solve the least-squares problem minimize n P k=1 (Ik −Ides)2 ignoring the constraints. If the solution is not feasible, we saturate it, i.e., set pj := 0 if pj ≤0 and pj := 1 if pj ≥1. Download the MATLAB ﬁle ex14data.m from the class webpage and generate problem data by [A,Ides] = ex14data. (The elements of A are the coeﬃcients aij in (7).) Compute a feasible p using this ﬁrst method, and calculate φ(p). ii. Weighted least-squares. We consider another least-squares problem: minimize n P k=1 (Ik −Ides)2 + µ m P i=1 (pi −0.5)2, where µ ≥0 is used to attach a cost to a deviation of the powers from the value 0.5, which lies in the middle of the power limits. For large enough µ, the solution of this problem will satisfy 0 ≤pi ≤1, i.e., be feasible for the original problem. Explain how you solve this problem in MATLAB. For the problem data generated by ex14data.m, ﬁnd the smallest µ such that p becomes feasible, and evaluate φ(p). (c) Using the same data as in part (b), solve the LP you derived in part (a). Compare the solution with the solutions you obtained using the (weighted) least-squares methods of part (b). Exercise 15. We consider the problem of ﬁnding optimal positions of n cells or modules on an integrated circuit. The variables are the coordinates xi, yi, i = 1, . . . , n, of the n cells. The cells must be placed in a square C = {(x, y) \| −1 ≤x ≤1, −1 ≤y ≤1}. Each cell has several terminals, which are connected to terminals on other cells, or to input/output (I/O) terminals on the perimeter of C. The positions of the I/O terminals are known and ﬁxed. The connections between the cells are speciﬁed as follows. We are given a matrix A ∈RN×n and two vectors bx ∈RN, by ∈RN. Each row of A and each component of bx and by describe a connection between two terminals. For each i = 1, . . . , N, we can distinguish two possibilities, depending on whether row i of A describes a connection between two cells, or between a cell and an I/O terminal. • If row i describes a connection between two cells j and k (with j < k), then ail =      1 if l = j −1 if l = k 0 otherwise , bx,i = 0, by,i = 0. In other words, we have aT i x −bx,i = xj −xk and aT i y −by,i = yj −yk for all x and y. • If row i describes a connection between a cell j and an I/O terminal with coordinates (¯ x, ¯ y), then ail = ( 1 if l = j 0 otherwise , bi,x = ¯ x, bi,y = ¯ y. Therefore aT i x −bx,i = xj −¯ x and aT i y −by,i = yj −¯ y for all x and y. The ﬁgure illustrates this notation for an example with n = 3, N = 6. 10 (−1, −1) (−1, 1) (1, 1) (1, −1) (−1, 0) (0.5, 1) (1, 0.5) (0, −1) (x1, y1) (x2, y2) (x3, y3) For this example, A, bx and by given by A =          1 −1 0 1 0 −1 1 0 0 0 1 0 0 0 1 0 0 1          , bx =          0.0 0.0 −1.0 0.5 0.0 1.0          , by =          0.0 0.0 0.0 1.0 −1.0 0.5          . The problem we consider is to determine the coordinates (xi, yi) that minimize some measure of the total wirelength of the connections. We can formulate diﬀerent variations. (a) Suppose we use the Euclidean distance between terminals to measure the length of a connection, and that we minimize the sum of the squares of the connection lengths. In other words we determine x and y by solving minimize N P i=1 (aT i x −bx,i)2 + (aT i y −by,i)2 or, in matrix notation, minimize ∥Ax −bx∥2 + ∥Ay −by∥2. (8) The variables are x ∈Rn and y ∈Rn. (Note that we don’t have to add the constraints −1 ≤xi ≤1 and −1 ≤yi ≤1 explicitly, since a solution with a cell outside C can never be optimal.) Since the two terms in (8) are independent, the solution can be obtained by solving two least-squares problems, one to determine x, and one to determine y. Equivalently, we can solve two sets of linear equations (AT A)x = AT bx, (AT A)y = AT by. (b) A second and more realistic choice is to use the Manhattan distance between two con-nected terminals as a measure for the length of the connection, i.e., to consider the optimization problem minimize N P i=1 \|aT i x −bx,i\| + \|aT i y −by,i\| . 11 In matrix notation, this can be written as minimize ∥Ax −bx∥1 + ∥Ay −by∥1. (c) As a third variation, suppose we measure the length of a connection between two ter-minals by the Manhattan distance between the two points, as in (b) , but instead of minimizing the sum of the lengths, we minimize the maximum length, i.e., we solve minimize maxi=1,...,N \|aT i x −bx,i\| + \|aT i y −by,i\| . (d) Finally, we can consider the problem minimize N P i=1 h(aT i x −bx,i) + h(aT i y −by,i) where h is a piecewise-linear function deﬁned as h(z) = max{z, −z, γ} and γ is a given positive constant. The function h is plotted below. z h(z) +γ −γ Give LP formulations for problems (b), (c) and (d). You may introduce new variables, but you must explain clearly why your formulation and the original problem are equivalent. Numerical example. We compare the solutions obtained from the four variations for a small example. For simplicity, we consider a one-dimensional version of the problem, i.e., the variables are x ∈Rn, and the goal is to place the cells on the interval [−1, 1]. We also drop the subscript in bx. The four formulations of the one-dimensional placement problem are the following. (a) ℓ2-placement: minimize ∥Ax −b∥2 = P i(aT i x −bi)2. (b) ℓ1-placement: minimize ∥Ax −b∥1 = P i \|aT i x −bi\|. (c) ℓ∞-placement: minimize ∥Ax −b∥∞= maxi \|aT i x −bi\|. (d) ℓ1-placement with ‘dead zone’: minimize P i h(aT i x −bi). We use a value γ = 0.02. To generate the data, download the ﬁle ex15data.m from the class webpage. The command [A,b] = ex15data(’large’) generates a problem with 100 cells and 300 connections; [A,b] = ex15data(’small’) generates a problem with with 50 cells and 150 connections. You can choose either problem. Compare the solutions obtained by the four methods. • Plot a histogram of the n positions xi for each solution (using the hist command). 12 • Also plot a histogram of the connnection lengths \|aT i x −bi\|. • Compute the total wire length P i \|aT i x −bi\| for each of the four solutions. • Compute the length of the longest connection maxi \|aT i x −bi\| for each of the four solutions. • So far we have assumed that the cells have zero width. In practice we have to take overlap between cells into account. Assume that two cells i and j overlap when \|xi−xj\| ≤ 0.01. For each of the four solutions, calculate how many pairs of cells overlap. You can express the overlap as a percentage of the total number n(n −1)/2 of pairs of cells. Are the results what you expect? Which of the four solutions would you prefer if the most important criteria are total wirelength P i \|aT i x −bi\| and overlap? Exercise 16. Formulate the following problem as an LP: maximize n P j=1 rj(xj) subject to n P j=1 Aijxj ≤cmax i , i = 1, . . . , m xj ≥0, j = 1, . . . , n. (9) The functions rj are deﬁned as rj(u) = ( pju 0 ≤u ≤qj pjqj + pdisc j (u −qj) u ≥qj, (10) where pj > 0, qj > 0 and 0 < pdisc j < pj. The variables in the problem are xj, j = 1, . . . , n. The parameters Aij, cmax i , pj, qj and pdisc j are given. The variables xj in the problem represent activity levels (for example, production levels for diﬀerent products manufactured by a company). These activities consume m resources, which are limited. Activity j consumes Aijxj of resource i. (Ordinarily we have Aij ≥0, i.e., activity j consumes resource i. But we allow the possibility that Aij < 0, which means that activity j actually generates resource i as a by-product.) The total resource consumption is additive, so the total of resource i consumed is ci = Pn j=1 Aijxj. Each resource consumption is limited: we must have ci ≤cmax i , where cmax i are given. Activity j generates revenue rj(xj), given by the expression (10). In this deﬁnition pj > 0 is the basic price, qj > 0 is the quantity discount level, and pdisc j is the quantity discount price, for (the product of) activity j. We have 0 < pdisc j < pj. The total revenue is the sum of the revenues associated with each activity, i.e., Pn j=1 rj(xj). The goal in (9) is to choose activity levels that maximize the total revenue while respecting the resource limits. Exercise 17. We consider a linear dynamical system with state x(t) ∈Rn, t = 0, . . . , N, and actuator or input signal u(t) ∈R, for t = 0, . . . , N −1. The dynamics of the system is given by the linear recurrence x(t + 1) = Ax(t) + bu(t), t = 0, . . . , N −1, where A ∈Rn×n and b ∈Rn are given. We assume that the initial state is zero, i.e., x(0) = 0. 13 The minimum fuel optimal control problem is to choose the inputs u(0), . . . , u(N −1) so as to minimize the total fuel consumed, which is given by F = N−1 X t=0 f(u(t)), subject to the constraint that x(N) = xdes, where N is the (given) time horizon, and xdes ∈ Rn is the (given) ﬁnal or target state. The function f : R →R is the fuel use map for the actuator, which gives the amount of fuel used as a function of the actuator signal amplitude. In this problem we use f(a) = ( \|a\| \|a\| ≤1 2\|a\| −1 \|a\| > 1. This means that fuel use is proportional to the absolute value of the actuator signal, for actuator signals between −1 and 1; for larger actuator signals the marginal fuel eﬃciency is half. (a) Formulate the minimum fuel optimal control problem as an LP. (b) Solve the following instance of the problem: A = " 1 1 0 0.95 # , b = " 0 0.1 # , x(0) = (0, 0), xdes = (10, 0), N = 20. We can interpret the system as a simple model of a vehicle moving in one dimension. The state dimension is n = 2, with x1(t) denoting the position of the vehicle at time t and x2(t) giving its velocity. The initial state is (0, 0), which corresponds to the vehicle at rest at position 0; the ﬁnal state is xdes = (10, 0), which corresponds to the vehicle being at rest at position 10. Roughly speaking, this means that the actuator input aﬀects the velocity, which in turn aﬀects the position. The coeﬃcient A22 = 0.95 means that velocity decays by 5% in one sample period, if no actuator signal is applied. Plot the input signal u(t) for t = 0, . . . , 19, and the position and velocity (i.e., x1(t) and x2(t)) for t = 0, . . . , 20. Exercise 18. Robot grasp problem with static friction. We consider a rigid object held by N robot ﬁngers. For simplicity we assume that the object and all forces acting on it lie in a plane. (0, 0) F ext x F ext y T ext Fi Gi θi (xi, yi) F1 G1 θ1 (x1, y1) F2 G2 θ2 (x2, y2) FN GN θN (xN, yN) 14 The ﬁngers make contact with the object at points (xi, yi), i = 1, . . . , N. (Although it does not matter, you can assume that the origin (0, 0) is at the center of gravity of the object.) Each ﬁnger applies a force with magnitude Fi on the object, in a direction normal to the surface at that contact point, and pointing towards the object. The horizontal component of the ith contact force is equal to Fi cos θi, and the vertical component is Fi sin θi, where θi is the angle between the inward pointing normal to the surface and a horizontal line. At each contact point there is a friction force Gi which is tangential to the surface. The horizontal component is Gi sin θi and the vertical component is −Gi cos θi. The orientation of the friction force is arbitrary (i.e., Gi can be positive or negative), but its magnitude \|Gi\| cannot exceed µFi, where µ ≥0 is a given constant (the friction coeﬃcient). Finally, there are several external forces and torques that act on the object. We can replace those external forces by equivalent horizontal and vertical forces F ext x and F ext y at the origin, and an equivalent torque T ext. These two external forces and the external torque are given. The static equilibrium of the object is characterized by the following three equations: N X i=1 (Fi cos θi + Gi sin θi) + F ext x = 0 (11) (the horizontal forces add up to zero), N X i=1 (Fi sin θi −Gi cos θi) + F ext y = 0 (12) (the vertical forces add up to zero), N X i=1 ((Fi cos θi + Gi sin θi)yi −(Fi sin θi −Gi cos θi)xi) + T ext = 0 (13) (the total torque is zero). As mentioned above, we assume the friction model can be expressed as a set of inequalities \|Gi\| ≤µFi, i = 1, . . . , N. (14) If we had no friction, then N = 3 ﬁngers would in general be suﬃcient to hold the object, and we could ﬁnd the forces Fi by solving the three linear equations (11)-(13) for the variables Fi. If there is friction, or N > 3, we have more unkown forces than equilibrium equations, so the system of equations is underdetermined. We can then take advantage of the additional degrees of freedom to ﬁnd a set of forces Fi that are ‘small’. Express the following two problems as LPs. (a) Find the set of forces Fi that minimizes PN i=1 Fi subject to the constraint that the object is in equilibrium. More precisely, the constraint is that there exist friction forces Gi that, together with Fi, satisfy (11)-(14). (b) Find a set of forces Fi that minimizes maxi=1,...,N Fi subject to the constraint that the object is in equilibrium. Which of these two problems do you expect will have a solution with a larger number of Fi’s equal to zero? 15 Exercise 19. Suppose you are given two sets of points {v1, v2, . . . , vK} and {w1, w2, . . . , wL} in Rn. Can you formulate the following two problems as LP feasibility problems? (a) Determine a hyperplane that separates the two sets, i.e., ﬁnd a ∈Rn and b ∈R with a ̸= 0 such that aT vi ≤b, i = 1, . . . , K, aT wi ≥b, i = 1, . . . , L. Note that we require a ̸= 0, so you have to make sure your method does not return the trivial solution a = 0, b = 0. You can assume that the matrices " v1 v2 · · · vK 1 1 · · · 1 # , " w1 w2 · · · wL 1 1 · · · 1 # have rank n + 1. (b) Determine a sphere separating the two sets of points, i.e., ﬁnd xc ∈Rn, R ≥0 such that (vi −xc)T (vi −xc) ≤R2, i = 1, . . . , K, (wi −xc)T (wi −xc) ≥R2, i = 1, . . . , L. (xc is the center of the sphere; R is its radius.) Exercise 20. Download the ﬁle ex20data.m from the class website and run it in MATLAB using the command [X,Y] = ex20data(id), where id is your student ID number (a nine-digit integer). This will create two matrices X ∈R4×100 and Y ∈R4×100. Let xi and yi be the ith columns of X and Y , respectively. (a) Verify (prove) that it is impossible to strictly separate the points xi from the points yi by a hyperplane. In other words, show that there exist no a ∈R4 and b ∈R such that aT xi + b ≤−1, i = 1, . . . , 100, aT yi + b ≥1, i = 1, . . . , 100. (b) Find a quadratic function that strictly separates the two sets, i.e., ﬁnd A = AT ∈R4×4, b ∈R4, c ∈R, such that xT i Axi + bT xi + c ≤−1, i = 1, . . . , 100, yT i Ayi + bT yi + c ≥1, i = 1, . . . , 100. (c) It may be impossible to ﬁnd a hyperplane that strictly separates the two sets, but we can try to ﬁnd a hyperplane that separates as many of the points as possible. Formulate a heuristic (i.e., suboptimal method), based on solving a single LP, for ﬁnding a ∈R4 and b ∈R that minimize the number of misclassiﬁed points. We consider xi as misclassiﬁed if aT xi + b > −1, and yi as misclassiﬁed if aT yi + b < 1. Describe and justify your method, and test it on the problem data. Exercise 21. Linear programming in decision theory. Suppose we have a choice of p available actions a ∈{1, . . . , p}, and each action has a certain cost (which can be positive, negative or zero). The costs depend on the value of an unknown parameter θ ∈{1, . . . , m} and are speciﬁed in the form of a loss matrix L ∈Rm×p, with Lij equal to the cost of action a = j when θ = i. 16 We do not know θ, but we can observe a random variable x with a distribution that depends on θ. We will assume that x is a discrete random variable with values in {1, 2, . . . , n}, so we can represent its distribution, for the m possible values of θ, by a matrix P ∈Rn×m with Pki = prob(x = k \|θ = i). A strategy is a rule for selecting an action a based on the observed value of x. A pure or deterministic strategy assigns to each of the possible observations a unique action a. A pure strategy can be represented by a matrix T ∈Rp×n, with Tjk = ( 1 action j is selected when x = k is observed 0 otherwise. Note that each column of a pure strategy matrix T contains exactly one entry equal to one, and the other entries are zero. We can therefore enumerate all possible pure strategies by enumerating the 0-1 matrices with this property. As a generalization, we can consider mixed or randomized strategies. In a mixed strategy we select an action randomly, using a distribution that depends on the observed x. A mixed strategy is represented by a matrix T ∈Rp×n, with Tjk = prob(a = j \| x = k). The entries of a mixed strategy matrix T are nonnegative and have column sums equal to one: Tjk ≥0, j = 1, . . . , p, k = 1, . . . , n, 1T T = 1T . A pure strategy is a special case of a mixed strategy with all the entries Tjk equal to zero or one. Now suppose the value of θ is i and we apply the strategy T. Then the expected loss is given by n X k=1 p X j=1 LijTjkPki = (LTP)ii. The diagonal elements of the matrix LTP are the expected losses for the diﬀerent values of θ = 1, . . . , m. We consider two popular deﬁnitions of an optimal mixed strategy, based on minimizing a function of the expected losses. (a) Minimax strategies. A minimax strategy minimizes the maximum of the expected losses: the matrix T is computed by solving minimize maxi=1,...,m(LTP)ii subject to Tjk ≥0, j = 1, . . . , p, k = 1, . . . , n 1T T = 1T . The variables are the pn entries of T. Express this problem as a linear program. (b) Bayes strategies. Assume that the parameter θ itself is random with a known distribu-tion qi = prob(θ = i). The Bayes strategy minimizes the average expected loss, where 17 the average is taken over θ. The matrix T of a Bayes strategy is the optimal solution of the problem minimize m P i=1 qi(LTP)ii subject to Tjk ≥0, j = 1, . . . , p, k = 1, . . . , n 1T T = 1T . This is a linear program in the pn variables Tjk. Formulate a simple algorithm for solving this LP. Show that it is always possible to ﬁnd an optimal Bayes strategy that is a pure strategy. Hint. First note that each column of the optimal T can be determined independently of the other columns. Then reduce the optimization problem over column k of T to one of the simple LPs in exercise 6. (c) As a simple numerical example, we consider a quality control system in a factory. The products that are examined can be in one of two conditions (m = 2): θ = 1 means the product is defective; θ = 2 means the product works properly. To examine the quality of a product we use an automated measurement system that rates the product on a scale of 1 to 4. This rating is the observed variable x: n = 4 and x ∈{1, 2, 3, 4}. We have calibrated the system to ﬁnd the probabilities Pij = prob(x = i \|θ = j) of producing a rating x = i when the state of the product is θ = j. The matrix P is P =      0.7 0.0 0.2 0.1 0.05 0.1 0.05 0.8     . We have a choice of three possible actions (p = 3): a = 1 means we accept the product and forward it to be sold; a = 2 means we subject it to a manual inspection to determine whether it is defective or not; a = 3 means we discard the product. The loss matrix is L = " 10 3 1 0 2 6 # . Thus, for example, selling a defective product costs us $10; discarding a good product costs $6, et cetera. i. Compute the minimax strategy for this L and P (using an LP solver). Is the minimax strategy a pure strategy? ii. Compute the Bayes strategy for q = (0.2, 0.8) (using an LP solver or the simple algorithm formulated in part 2). iii. Enumerate all (34 = 81) possible pure strategies T (in MATLAB), and plot the expected losses ((LTP)11, (LTP)22) of each of these strategies in a plane. iv. On the same graph, show the losses for the minimax strategy and the Bayes strategy computed in parts (a) and (b). v. Suppose we let q vary over all possible prior distributions (all vectors with q1+q2 = 1, q1 ≥0, q2 ≥0). Indicate on the graph the expected losses ((LTP)11, (LTP)22) of the corresponding Bayes strategies. 18 Exercise 22. Robust linear programming. (a) Let x ∈Rn be a given vector. Prove that xT y ≤∥x∥1 for all y with ∥y∥∞≤1. Is the inequality tight, i.e., does there exist a y that satisﬁes ∥y∥∞≤1 and xT y = ∥x∥1? (b) Consider the set of linear inequalities aT i x ≤bi, i = 1, . . . , m. (15) Suppose you don’t know the coeﬃcients ai exactly. Instead you are given nominal values ¯ ai, and you know that the actual coeﬃcient vectors satisfy ∥ai −¯ ai∥∞≤ρ for a given ρ > 0. In other words the actual coeﬃcients aij can be anywhere in the intervals [¯ aij −ρ, ¯ aij + ρ], or equivalently, each vector ai can lie anywhere in a rectangle with corners ¯ ai + v where v ∈{−ρ, ρ}n (i.e., v has components ρ or −ρ). The set of inequalities (15) must be satisﬁed for all possible values of ai, i.e., we re-place (15) with the constraints aT i x ≤bi for all ai ∈{¯ ai + v \| ∥v∥∞≤ρ} and for i = 1, . . . , m. (16) A straightforward but very ineﬃcient way to express this constraint is to enumerate the 2n corners of the rectangle of possible values ai and to require that ¯ aT i x + vT x ≤bi for all v ∈{−ρ, ρ}n and for i = 1, . . . , m. This is a system of m2n inequalities. Use the result in (a) to show that (16) is in fact equivalent to the much more compact set of nonlinear inequalities ¯ aT i x + ρ∥x∥1 ≤bi, i = 1, . . . , m. (17) (c) Consider the LP minimize cT x subject to aT i x ≤bi, i = 1, . . . , m. Again we are interested in situations where the coeﬃcient vectors ai are uncertain, but satisfy bounds ∥ai −¯ ai∥∞≤ρ for given ¯ ai and ρ. We want to minimize cT x subject to the constraint that the inequalities aT i x ≤bi are satisﬁed for all possible values of ai. We call this a robust LP: minimize cT x subject to aT i x ≤bi for all ai ∈{¯ ai + v \| ∥v∥∞≤ρ} and for i = 1, . . . , m. (18) It follows from (b) that we can express this problem as a nonlinear optimization problem minimize cT x subject to ¯ aT i x + ρ∥x∥1 ≤bi, i = 1, . . . , m. (19) Express (19) as an LP. Solving (19) is a worst-case approach to dealing with uncertainty in the data. If x⋆is the optimal solution of (19), then for any speciﬁc value of ai, it may be possible to ﬁnd feasible x with a lower objective value than x⋆. However such an x would be infeasible for some other value of ai. 19 Exercise 23. Robust Chebyshev approximation. In a similar way as in the previous problem, we can consider Chebyshev approximation problems minimize ∥Ax −b∥∞ in which A ∈Rm×n is uncertain. Suppose we can characterize the uncertainty as follows. The values of A depend on parameters u ∈Rp, which are unknown but satisfy ∥u∥∞≤ρ. Each row vector ai can be written as ai = ¯ ai + Biu where ¯ ai ∈Rn and Bi ∈Rn×p are given. In the robust Chebyshev approximation we minimize the worst-case value of ∥Ax−b∥∞. This problem can be written as minimize max∥u∥∞≤ρ maxi=1....,m (¯ ai + Biu)T x −bi . (20) Show that (20) is equivalent to minimize maxi=1....,m(\|¯ aT i x −bi\| + ρ∥BT i x∥1). (21) To prove this you can use the results from exercise 22. There is also a fairly straightforward direct proof. Express (21) as an LP. Exercise 24. Describe how you would use linear programming to solve the following problem. You are given an LP minimize cT x subject to Ax ≤b (22) in which the coeﬃcients of A ∈Rm×n are uncertain. Each coeﬃcient Aij can take arbitrary values in the interval [ ¯ Aij −∆Aij, ¯ Aij + ∆Aij], where ¯ Aij and ∆Aij are given with ∆Aij ≥0. The optimization variable x in (22) must be feasible for all possible values of A. In other words, we want to solve minimize cT x subject to Ax ≤b for all A ∈A where A ⊆Rm×n is the set A = {A ∈Rm×n \| ¯ Aij −∆Aij ≤Aij ≤¯ Aij + ∆Aij, i = 1, . . . , m, j = 1, . . . , n}. If you know more than one solution method, you should give the most eﬃcient one. Exercise 25. In the lecture we discussed the problem of ﬁnding a strictly separating hyperplane for a set of points with binary labels: si(aT vi + b) > 0, i = 1, . . . , N. (23) The variables are a ∈Rn and b ∈R. The n-vectors vi and the labels si ∈{−1, 1} are given. We can deﬁne the margin of separation of a strictly separating hyperplane as the maximum value of t such that si(aT (vi + w) + b) ≥0 for all w with ∥w∥∞≤t, i = 1, . . . , N. The idea is that if we replace each point vi with a hypercube (a ball in ∥· ∥∞-norm) centered at vi and with radius t, then the hyperplane separates the N hypercubes. 20 (a) Suppose a and b deﬁne a strictly separating hyperplane (i.e., satisfy (23)), and that the coeﬃcients are normalized so that min i=1,...,N si(aT vi + b) = 1. What is the margin of separation of the hyperplane? (b) Formulate the problem of ﬁnding a strictly separating hyperplane with maximum mar-gin of separation as a linear program. Exercise 26. Optimization problems with uncertain data sometimes involve two sets of variables that can be selected in two stages. When the ﬁrst set of variables is chosen, the problem data are uncertain. The second set of variables, however, can be selected after the actual values of the parameters have become known. As an example, we consider two-stage robust formulations of the Chebyshev approximation problem minimize ∥Ax + By + b∥∞, with variables x ∈Rn and y ∈Rp. The problem parameters A, B, b are uncertain, and we model the uncertainty by assuming that there are m possible scenarios (or instances of the problem). In scenario k, the values of A, B, b are Ak, Bk, bk. In the two-stage setting we ﬁrst select x before the scenario is known; then we choose y after learning the actual value of k. The optimal choice of y in the second stage is the value that minimizes ∥Akx+Bky +bk∥∞, for given x, Ak, Bk, bk. We denote by fk(x) the optimal value of this second-stage optimization problem for scenario k: fk(x) = min y ∥Akx + Bky + bk∥∞, k = 1, . . . , m. (a) We can minimize the worst-case objective by solving the optimization problem minimize max k=1,...,m fk(x) with x as variable. Formulate this problem as an LP. (b) If we know the probability distribution of the scenarios we can also minimize the ex-pected cost, by solving minimize m X k=1 πkfk(x) with x as variable. The coeﬃcient πk ≥0 is the probability that (A, B, b) is equal to (Ak, Bk, bk). Formulate this problem as an LP. Exercise 27. Feedback design for a static linear system. In this problem we use linear program-ming to design a linear feedback controller for a static linear system. (The method extends to dynamical systems but we will not consider the extension here.) The ﬁgure shows the system and the controller. 21 K P w u z y The elements of the vector w ∈Rnw are called the exogeneous inputs, z ∈Rnz are the critical outputs, y ∈Rny are the sensed outputs, and u ∈Rnu are the actuator inputs. These vectors are related as z = Pzww + Pzuu, y = Pyww + Pyuu (24) where the matrices Pzu, Pzw, Pyu, Pyw are given. The controller feeds back the sensed outputs y to the actuator inputs u. The relation is u = Ky (25) where K ∈Rnu×ny. The matrix K will be the design variable. Assuming I −Pyu is invertible, we can eliminate y from the second equation in (24). We have y = (I −PyuK)−1Pyww and substituting in the ﬁrst equation we can write z = Hw with H = Pzw + PzuK(I −PyuK)−1Pyw. (26) The matrix H is a complicated nonlinear function of K. Suppose that the signals w are disturbances or noises acting on the system, and that they can take any values with ∥w∥∞≤ρ for some given ρ. We would like to choose K so that the eﬀect of the disturbances w on the output z is minimized, i.e., we would like z to be as close as possible to zero, regardless of the values of w. Speciﬁcally, if we use the inﬁnity norm ∥z∥∞to measure the size of z, we are interested in determining K by solving the optimization problem minimize max ∥w∥∞≤ρ ∥Hw∥∞, (27) where H depends on the variable K through the formula (26). (a) We ﬁrst derive an explicit expression for the objective function in (27). Show that max ∥w∥∞≤ρ ∥Hw∥∞= ρ max i=1,...,nz X j=1,...,nw \|hij\| where hij are the elements of H. Up to the constant ρ, this is the maximum row sum of H: for each row of H we calculate the sum of the absolute values of its elements; we then select the largest of these row sums. 22 (b) Using this expression, we can reformulate problem (27) as minimize ρ maxi=1,...,nz P j=1,...,nw \|hij\|, (28) where hij depends on the variable K through the formula (26). Formulate (27) as an LP. Hint. Use a change of variables Q = K(I −PyuK)−1, and optimize over Q ∈Rnu×ny instead of K. You may assume that I+QPyu is invertible, so the transformation is invertible: we can ﬁnd K from Q as K = (I + QPyu)−1Q. Exercise 28. Formulate the following problem as an LP. Find the largest ball B(xc, R) = {x \| ∥x −xc∥≤R} enclosed in a given polyhedron P = {x \| aT i x ≤bi, i = 1, . . . , m}. In other words, express the problem maximize R subject to B(xc, R) ⊆P as an LP. The problem variables are the center xc ∈Rn and the radius R of the ball. Exercise 29. Let P1 and P2 be two polyhedra described as P1 = {x \| Ax ≤b} , P2 = {x \| −1 ≤Cx ≤1} , where A ∈Rm×n, C ∈Rp×n, and b ∈Rm. The polyhedron P2 is symmetric about the origin, i.e., if x ∈P2, then −x ∈P2. We say the origin is the center of P2. For t > 0 and xc ∈Rn, we use the notation tP2 + xc to denote the polyhedron tP2 + xc = {tx + xc \| x ∈P2}, which is obtained by ﬁrst scaling P2 by a factor t about the origin, and then translating its center to xc. Explain how you would solve the following two problems using linear programming. If you know diﬀerent formulations, you should choose the most eﬃcient method. (a) Find the largest polyhedron tP2 + xc enclosed in P1, i.e., maximize t subject to tP2 + xc ⊆P1. (b) Find the smallest polyhedron tP2 + xc containing P1, i.e., minimize t subject to P1 ⊆tP2 + xc. 23 In both problems the variables are t ∈R and xc ∈Rn. Exercise 30. Suppose you are given an infeasible set of linear inequalities aT i x ≤bi, i = 1, . . . , m, and you are asked to ﬁnd an x that satisﬁes many of the inequalities (ideally, as many as possible). Of course, the exact solution of this problem is diﬃcult and requires combinatorial or integer optimization techniques, so you should concentrate on heuristic or sub-optimal methods. More speciﬁcally, you are asked to formulate a heuristic method based on solving a single LP. Test the method on the example problem in the ﬁle ex30data.m available on the class web-page. (The command ex30data creates generates a sparse matrix A ∈R100×50 and a vector b ∈R100, that deﬁne an infeasible set of linear inequalities.) To count the number of inequal-ities satisﬁed by x, you can use the MATLAB command length(find(b-Ax > -1e-5)). Exercise 31. Explain how you would solve the following problem using linear programming. You are given two sets of points in Rn: S1 = {x1, . . . , xN}, S2 = {y1, . . . , yM}. You are asked to ﬁnd a polyhedron P = {x \| aT i x ≤bi, i = 1, . . . , m} that contains the points in S1 in its interior, and does not contain any of the points in S2: S1 ⊆{x \| aT i x < bi, i = 1, . . . , m}, S2 ⊆{x \| aT i x > bi for at least one i} = Rn \ P. An example is shown in the ﬁgure, with the points in S1 shown as open circles and the points in S2 as ﬁlled circles. You can assume that the two sets are separable in the way described. Your solution method should return ai and bi, i = 1, . . . , m, given the sets S1 and S2. The number of inequalities m is not speciﬁed, but it should not exceed M + N. You are allowed to solve one or more LPs or LP feasibility problems. The method should be eﬃcient, i.e., the dimensions of the LPs you solve should not be exponential as a function of N and M. 24 Exercise 32. Explain how you would solve the following problem using linear programming. Given two polyhedra P1 = {x \| Ax ≤b}, P2 = {x \| Cx ≤d}, prove that P1 ⊆P2, or ﬁnd a point in P1 that is not in P2. The matrices A ∈Rm×n and C ∈Rp×n, and the vectors b ∈Rm and d ∈Rp are given. If you know several solution methods, give the most eﬃcient one. Polyhedra Exercise 33. Which of the following sets S are polyhedra? If possible, express S in inequality form, i.e., give matrices A and b such that S = {x \| Ax ≤b}. (a) S = {y1a1 + y2a2 \| −1 ≤y1 ≤1, −1 ≤y2 ≤1} for given a1, a2 ∈Rn. (b) S = {x ∈Rn \| x ≥0, 1T x = 1, Pn i=1 xiai = b1, Pn i=1 xia2 i = b2}, where ai ∈R (i = 1, . . . , n), b1 ∈R, and b2 ∈R are given. (c) S = {x ∈Rn \| x ≥0, xT y ≤1 for all y with ∥y∥= 1}. (d) S = {x ∈Rn \| x ≥0, xT y ≤1 for all y with P i \|yi\| = 1}. (e) S = {x ∈Rn \| ∥x −x0∥≤∥x −x1∥} where x0, x1 ∈Rn are given. S is the the set of points that are closer to x0 than to x1. (f) S = {x ∈Rn \| ∥x −x0∥≤∥x −xi∥, i = 1, . . . , K} where x0, . . . , xK ∈Rn are given. S is the set of points that are closer to x0 than to the other xi. Exercise 34. Measurement with bounded errors. A series of K measurements y1, . . . , yK ∈R, are taken in order to estimate an unknown vector x ∈Rq. The measurements are related to the unknown vector x by yi = aT i x + vi, where vi is a measurement noise that satisﬁes \|vi\| ≤α but is otherwise unknown. The vectors ai and the measurement noise bound α are known. Let X denote the set of vectors x that are consistent with the observations y1, . . . , yK, i.e., the set of x that could have resulted in the measured values of yi. Show that X is a polyhedron. Now we examine what happens when the measurements are occasionally in error, i.e., for a few i we have no relation between x and yi. More precisely suppose that Ifault is a subset of {1, . . . , K}, and that yi = aT i x+vi with \|vi\| ≤α (as above) for i ̸∈Ifault, but for i ∈Ifault, there is no relation between x and yi. The set Ifault is the set of times of the faulty measurements. Suppose you know that Ifault has at most J elements, i.e., out of K measurements, at most J are faulty. You do not know Ifault; you know only a bound on its cardinality (size). Is X (the set of x consistent with the measurements) a polyhedron for J > 0? Exercise 35. (a) Is e x = (1, 1, 1, 1) an extreme point of the polyhedron P deﬁned by the linear inequalities        −1 −6 1 3 −1 −2 7 1 0 3 −10 −1 −6 −11 −2 12 1 6 −1 −3             x1 x2 x3 x4     ≤        −3 5 −8 −7 4        ? If it is, ﬁnd a vector c such that e x is the unique minimizer of cT x over P. 25 (b) Same question for the polyhedron deﬁned by the inequalities        0 −5 −2 −5 −7 −7 −2 −2 −4 −4 −7 −7 −8 −3 −3 −4 −4 −4 2 −2             x1 x2 x3 x4     ≤        −12 −17 −22 −18 −8        and the equality 8x1 −7x2 −10x3 −11x4 = −20. Feel free to use MATLAB (in particular the rank command). Exercise 36. We deﬁne a polyhedron P = {x ∈R5 \| Ax = b, −1 ≤x ≤1}, with A =    0 1 1 1 −2 0 −1 1 −1 0 2 0 1 0 1   , b =    1 1 1   . The following three vectors x are in P: (a) ˆ x = (1, −1/2, 0, −1/2, −1) (b) ˆ x = (0, 0, 1, 0, 0) (c) ˆ x = (0, 1, 1, −1, 0). Are these vectors extreme points of P? For each ˆ x, if it is an extreme point, give a vector c for which ˆ x is the unique solution of the optimization problem minimize cT x subject to Ax = b −1 ≤x ≤1. Exercise 37. Birkhoﬀ’s theorem. An n × n matrix X is called doubly stochastic if Xij ≥0, i, j = 1, . . . , n, n X i=1 Xij = 1, j = 1, . . . , n, n X j=1 Xij = 1, i = 1, . . . , n, In words, the entries of X are nonnegative, and its row and column sums are equal to one. The set of doubly stochastic matrices can be described as a polyhedron in Rn2, deﬁned as P = {x ∈Rn2 \| Cx = d, x ≥0} with x the matrix X stored as a vector in column-major order, x = vec(X) = (X11, X21, . . . , Xn1, X12, X22, . . . , Xn2, . . . , X1n, X2n, . . . , Xnn), 26 and C, d deﬁned as C =         I I · · · I 1T 0 · · · 0 0 1T · · · 0 . . . . . . ... . . . 0 0 · · · 1T         , d =         1 1 1 . . . 1         . (The identity matrices I have order n and the vectors 1 have length n.) The matrix C has size 2n × n2 and the vector d has length 2n. In this exercise we show that the extreme points of the set of doubly stochastic matrices are the permutation matrices. (a) A permutation matrix is a 0-1 matrix with exactly one element equal to one in each column and exactly one element equal to one in each row. Use the rank criterion for extreme points to show that all permutation matrices are extreme points of the polyhedron of doubly stochastic matrices. (More precisely, if X is a permutation matrix, then vec(X) is an extreme point of the polyhedron P of vectorized doubly stochastic matrices deﬁned above.) (b) Show that an extreme point X of the polyhedron of n × n doubly stochastic matrices has at most 2n −1 nonzero entries. Therefore, if X is an extreme point, it must has a row with exactly one nonzero element (with value 1) and a column with exactly one nonzero element (equal to 1). Use this observation to show that all extreme points are permutation matrices. Exercise 38. What is the optimal value of the LP maximize aT Xb subject to X doubly stochastic, with X as variable, for a general vector b ∈Rn and each of the following choices of a? • a = (1, 0, 0, . . . , 0). • a = (1, 1, 0, . . . , 0). • a = (1, −1, 0, . . . , 0). Exercise 39. Carath´ eodory’s theorem. A point of the form θ1v1+· · ·+θmvm, where θ1+· · ·+θm = 1 and θi ≥0, i = 1, . . . , m, is called a convex combination of v1, . . . , vm. Suppose x is a convex combination of points v1, . . . , vm in Rn. Show that x is a convex combination of a subset of r ≤n + 1 of the points v1, . . . , vm. In other words, show that x can be expressed as x = ˆ θ1v1 + · · · + ˆ θmvm, where ˆ θi ≥0, Pm i=1 ˆ θi = 1, and at most n + 1 of the coeﬃcients ˆ θi are nonzero. 27 Alternatives Exercise 40. Prove the following result. If a set of m linear inequalities in n variables is infeasible, then there exists an infeasible subset of no more than n + 1 of the m inequalities. Exercise 41. Let P ∈Rn×n be a matrix with the following two properties: • all elements of P are nonnegative: pij ≥0 for i = 1, . . . , n and j = 1, . . . , n • the columns of P sum to one: Pn i=1 pij = 1 for j = 1, . . . , n. Show that there exists a y ∈Rn such that Py = y, y ≥0, n X i=1 yi = 1. Remark. This result has the following application. We can interpret P as the transition probability matrix of a Markov chain with n states: if s(t) is the state at time t (i.e., s(t) is a random variable taking values in {1, . . . , n}), then pij is deﬁned as pij = prob(s(t + 1) = i \| s(t) = j). Let y(t) ∈Rn be the probability distribution of the state at time t, i.e., yi(t) = prob(s(t) = i). Then the distribution at time t + 1 is given by y(t + 1) = Py(t). The result in this problem states that a ﬁnite state Markov chain always has an equilibrium distribution y. Exercise 42. Arbitrage and theorems of alternatives. Consider an event (for example, a sports game, political elections, the evolution of the stockmarket over a certain period) with m possible outcomes. Suppose that n wagers on the outcome are possible. If we bet an amount xj on wager j, and the outcome of the event is i, then our return is equal to rijxj (this amount does not include the stake, i.e., we pay xj initially, and receive (1 + rij)xj if the outcome of the event is i, so rijxj is the net gain). We allow the bets xj to be positive, negative, or zero. The interpretation of a negative bet is as follows. If xj < 0, then initially we receive an amount of money \|xj\|, with an obligation to pay (1 + rij)\|xj\| if outcome i occurs. In that case, we lose rij\|xj\|, i.e., our net gain is rijxj (a negative number). We call the matrix R ∈Rm×n with elements rij the return matrix. A betting strategy is a vector x ∈Rn, with as components xj the amounts we bet on each wager. If we use a betting strategy x, our total return in the event of outcome i is equal to Pn j=1 rijxj, i.e., the ith component of the vector Rx. (a) The arbitrage theorem. Suppose you are given a return matrix R. Prove the following theorem: there is a betting strategy x ∈Rn for which Rx > 0 (29) if and only if there exists no vector p ∈Rm that satisﬁes RT p = 0, p ≥0, p ̸= 0. (30) 28 We can interpret this theorem as follows. If Rx > 0, then the betting strategy x guarantees a positive return for all possible outcomes, i.e., it is a sure-win betting scheme. In economics, we say there is an arbitrage opportunity. If we normalize the vector p in (30) so that 1T p = 1, we can interpret it as a probability vector on the outcomes. The condition RT p = 0 means that the expected return E Rx = pT Rx = 0 for all betting strategies. We can therefore rephrase the arbitrage theorem as follows. There is no sure-win betting strategy (or arbitrage opportunity) if and only if there is a probability vector on the outcomes that makes all bets fair (i.e., the expected gain is zero). (b) Options pricing. The arbitrage theorem is used in mathematical ﬁnance to determine prices of contracts. As a simple example, suppose we can invest in two assets: a stock and an option. The current unit price of the stock is S. The price ¯ S of the stock at the end of the investment period is unknown, but it will be either ¯ S = Su or ¯ S = Sd, where u > 1 and d < 1 are given numbers. In other words the price either goes up by a factor u, or down by a factor d. If the current interest rate over the investment period is r, then the present value of the stock price ¯ S at the end of the period is equal to ¯ S/(1 + r), and our unit return is Su 1 + r −S = S u −1 −r 1 + r if the stock goes up, and Sd 1 + r −S = S d −1 −r 1 + r if the stock goes down. We can also buy options, at a unit price of C. An option gives us the right to purchase one stock at a ﬁxed price K at the end of the period. Whether we exercise the option or not, depends on the price of the stock at the end of the period. If the stock price ¯ S at the end of the period is greater than K, we exercise the option, buy the stock and sell it immediately, so we receive an amount ¯ S −K. If the stock price ¯ S is less than K, we do not exercise the option and receive nothing. Combining both cases, we can say that the value of the option at the end of the period is max{0, ¯ S −K}, and the present value is max{0, ¯ S −K}/(1 + r). If we pay a price C per option, then our return is 1 1 + r max{0, ¯ S −K} −C per option. We can summarize the situation with the return matrix R = " (u −1 −r)/(1 + r) (max{0, Su −K})/((1 + r)C) −1 (d −1 −r)/(1 + r) (max{0, Sd −K})/((1 + r)C) −1 # . The elements of the ﬁrst row are the (present values of the) returns in the event that the stock price goes up. The second row are the returns in the event that the stock price goes down. The ﬁrst column gives the returns per unit investment in the stock. The second column gives the returns per unit investment in the option. 29 In this simple example the arbitrage theorem allows us to determine the price of the option, given the other information S, K, u, d, and r. Show that if there is no arbitrage, then the price of the option C must be equal to 1 1 + r (p max{0, Su −K} + (1 −p) max{0, Sd −K}) where p = (1 + r −d)/(u −d). Exercise 43. We consider a network with m nodes and n directed arcs. Suppose we can apply labels yr ∈R, r = 1, . . . , m, to the nodes in such a way that yr ≥ys if there is an arc from node r to node s. (31) It is clear that this implies that if yi < yj, then there exists no directed path from node i to node j. (If we follow a directed path from node i to j, we encounter only nodes with labels less than or equal to yi. Therefore yj ≤yi.) Prove the converse: if there is no directed path from node i to j, then there exists a labeling of the nodes that satisﬁes (31) and yi < yj. Duality Exercise 44. The main result of linear programming duality is that the optimal value of the LP minimize cT x subject to Ax ≤b is equal to the optimal value of the LP maximize −bT z subject to AT z + c = 0 z ≥0, except when they are both infeasible. Give an example in which both problems are infeasible. Exercise 45. Prove the following result. If the feasible set of a linear program minimize cT x subject to Ax ≤b is nonempty and bounded, then the feasible set of the corresponding dual problem maximize −bT z subject to AT z + c = 0 z ≥0 is nonempty and unbounded. 30 Exercise 46. Consider the LP minimize 47x1 + 93x2 + 17x3 −93x4 subject to        −1 −6 1 3 −1 −2 7 1 0 3 −10 −1 −6 −11 −2 12 1 6 −1 −3             x1 x2 x3 x4     ≤        −3 5 −8 −7 4        . Prove, without using any LP code, that x = (1, 1, 1, 1) is optimal. Exercise 47. Consider the polyhedron P = {x ∈R4 \| Ax ≤b, Cx = d} where A =      −1 −1 −3 −4 −4 −2 −2 −9 −8 −2 0 −5 0 −6 −7 −4     , b =      −8 −17 −15 −17      and C = h 13 11 12 22 i , d = 58. (a) Prove that ˆ x = (1, 1, 1, 1) is an extreme point of P. (b) Prove that ˆ x is optimal for the LP minimize cT x subject to Ax ≤b Cx = d with c = (59, 39, 38, 85). (c) Is ˆ x the only optimal point? If not, describe the entire optimal set. You can use any software, but you have to justify your answers analytically. Exercise 48. Consider the following optimization problem in x: minimize cT x subject to ∥Ax + b∥1 ≤1 (32) where A ∈Rm×n, b ∈Rm, c ∈Rn. (a) Formulate this problem as a an LP in inequality form and explain why your LP formu-lation is equivalent to problem (32). (b) Derive the dual LP, and show that it is equivalent to the problem maximize bT z −∥z∥∞ subject to AT z + c = 0. What is the relation between the optimal z and the optimal variables in the dual LP? 31 (c) Give a direct argument (i.e., not quoting any results from LP duality) that whenever x is primal feasible (i.e., ∥Ax + b∥1 ≤1) and z is dual feasible (i.e., AT z + c = 0), we have cT x ≥bT z −∥z∥∞. Exercise 49. Lower bounds in Chebyshev approximation from least-squares. Consider the Cheby-shev approximation problem minimize ∥Ax −b∥∞ (33) where A ∈Rm×n (m ≥n) and rank A = n. Let xcheb denote an optimal point for the Chebyshev approximation problem (there may be multiple optimal points; xcheb denotes one of them). The Chebyshev problem has no closed-form solution, but the corresponding least-squares problem does. We denote the least-squares solution xls as xls = argmin ∥Ax −b∥= (AT A)−1AT b. The question we address is the following. Suppose that for a particular A and b you have computed the least-squares solution xls (but not xcheb). How suboptimal is xls for the Cheby-shev problem? In other words, how much larger is ∥Axls −b∥∞than ∥Axcheb −b∥∞? To answer this question, we need a lower bound on ∥Axcheb −b∥∞. (a) Prove the lower bound ∥Axcheb −b∥∞≥ 1 √m∥Axls −b∥∞, using the fact that for all y ∈Rm, 1 √m∥y∥≤∥y∥∞≤∥y∥. (b) In the duality lecture we derived the following dual for (33): maximize bT z subject to AT z = 0 ∥z∥1 ≤1. (34) We can use this dual problem to improve the lower bound obtained in (a). • Denote the least-squares residual as rls = b −Axls. Assuming rls ̸= 0, show that ˆ z = −rls/∥rls∥1, ˜ z = rls/∥rls∥1, are both feasible in (34). • By duality bT ˆ z and bT ˜ z are lower bounds for ∥Axcheb −b∥∞. Which is the better bound? How does it compare with the bound obtained in part (a) above? One application is as follows. You need to solve the Chebyshev approximation problem, but only within, say, 10%. You ﬁrst solve the least-squares problem (which can be done faster), and then use the bound from part (b) to see if it can guarantee a maximum 10% error. If it can, great; otherwise solve the Chebyshev problem (by slower methods). 32 Exercise 50. A matrix A ∈R(mp)×n and a vector b ∈Rmp are partitioned in m blocks of p rows: A =       A1 A2 . . . Am       , b =       b1 b2 . . . bm       , with Ak ∈Rp×n, bk ∈Rp. (a) Express the optimization problem minimize m P k=1 ∥Akx −bk∥∞ (35) as an LP. (b) Suppose rank(A) = n and Axls −b ̸= 0, where xls is the solution of the least-squares problem minimize ∥Ax −b∥2. Show that the optimal value of (35) is bounded below by Pm k=1 ∥rk∥2 maxk=1,...,m ∥rk∥1 , where rk = Akxls −bk for k = 1, . . . , m. Exercise 51. Let x be a real-valued random variable which takes values in {a1, a2, . . . , an} where 0 < a1 < a2 < · · · < an, and prob(x = ai) = pi. Obviously p satisﬁes Pn i=1 pi = 1 and pi ≥0 for i = 1, . . . , n. (a) Consider the problem of determining the probability distribution that maximizes prob(x ≥ α) subject to the constraint E x = b, i.e., maximize prob(x ≥α) subject to E x = b, (36) where α and b are given (a1 < α < an, and a1 ≤b ≤an). The variable in problem (36) is the probability distribution, i.e., the vector p ∈Rn. Write (36) as an LP. (b) Take the dual of the LP in (a), and show that it is can be reformulated as minimize λb + ν subject to λai + ν ≥0 for all ai < α λai + ν ≥1 for all ai ≥α. The variables are λ and ν. Give a graphical interpretation of this problem, by interpret-ing λ and ν as coeﬃcients of an aﬃne function f(x) = λx + ν. Show that the optimal value is equal to ( (b −a1)/(¯ a −a1) b ≤¯ a 1 b ≥¯ a, where ¯ a = min{ai \| ai ≥α}. Also give the optimal values of λ and ν. 33 (c) From the dual solution, determine the distribution p that solves the problem in (a). Exercise 52. The max-ﬂow min-cut theorem. Consider the maximum ﬂow problem with nonneg-ative arc ﬂows: maximize t subject to Ax = te 0 ≤x ≤c. (37) Here e = (1, 0, . . . , 0, −1) ∈Rm, A ∈Rm×n is the node-arc incidence matrix of a directed graph with m nodes and n arcs, and c ∈Rn is a vector of positive arc capacities. The variables are t ∈R and x ∈Rn. In this problem we have an external supply of t at node 1 (the ‘source’ node) and −t at node m (the ‘target’ node), and we maximize t subject to the balance equations and the arc capacity constraints. A cut separating nodes 1 and m is a set of nodes that contains node 1 and does not contain node m, i.e., S ⊂{1, . . . , m} with 1 ∈S and m ̸∈S. The capacity of the cut is deﬁned as C(S) = X k∈A(S) ck, where A(S) is the set of arcs that start at a node in S and end at a node outside S. The problem of ﬁnding the cut with the minimum capacity is called the minimum cut problem. In this exercise we show that the solution of the minimum cut problem (with positive weights c) is provided by the dual of the maximum ﬂow problem (37). (a) Let p⋆be the optimal value of the maximum ﬂow problem (37). Show that p⋆≤C(S) (38) for all cuts S that separate nodes 1 and m. (b) Derive the dual problem of (37), and show that it can be expressed as minimize cT v subject to AT y ≤v y1 −ym = 1 v ≥0. (39) The variables are v ∈Rn and y ∈Rm. Suppose x and t are optimal in (37), and y and v are optimal in (39). Deﬁne the cut ˜ S = {i \| yi ≥y1}. Use the complementary slackness conditions for (37) and (39) to show that xk = ck if arc k starts at a node in ˜ S and ends at a node outside ˜ S, and that xk = 0 if arc k starts at a node outside ˜ S and ends at a node in ˜ S. Conclude that p⋆= C( ˜ S). Combined with the result of part 1, this proves that ˜ S is a minimum-capacity cut. 34 Exercise 53. A project consisting of n diﬀerent tasks can be represented as a directed graph with n arcs and m nodes. The arcs represent the tasks. The nodes represent precedence relations: If arc k starts at node i and arc j ends at node i, then task k cannot start before task j is completed. Node 1 only has outgoing arcs. These arcs represent tasks that can start immediately and in parallel. Node m only has incoming arcs. When the tasks represented by these arcs are completed, the entire project is completed. We are interested in computing an optimal schedule, i.e., in assigning an optimal start time and a duration to each task. The variables in the problem are deﬁned as follows. • yk is the duration of task k, for k = 1, . . . , n. The variables yk must satisfy the con-straints αk ≤yk ≤βk. We also assume that the cost of completing task k in time yk is given by ck(βk −yk). This means there is no cost if we we use the maximum allowable time βk to complete the task, but we have to pay if we want the task ﬁnished more quickly. • vj is an upper bound on the completion times of all tasks associated with arcs that end at node j. These variables must satisfy the relations vj ≥vi + yk if arc k starts at node i and ends at node j. Our goal is to minimize the sum of the completion time of the entire project, which is given by vm −v1, and the total cost P k ck(βk −yk). The problem can be formulated as an LP minimize −eT v + cT (β −y) subject to AT v + y ≤0 α ≤y ≤β, where e = (1, 0, . . . , 0, −1) and A is the node-arc incidence matrix of the graph. The variables are v ∈Rm, y ∈Rn. (a) Derive the dual of this LP. (b) Interpret the dual problem as a minimum cost network ﬂow problem with nonlinear cost, i.e., a problem of the form minimize n P k=1 fk(xk) subject to Ax = e x ≥0, where fk is a nonlinear function. Exercise 54. This problem is a variation on the illumination problem of exercise 14. In part (a) of exercise 14 we formulated the problem minimize maxk=1,...,n \|aT k p −Ides\| subject to 0 ≤p ≤1 as the following LP in p ∈Rm and an auxiliary variable w: minimize w subject to −w ≤aT k p −Ides ≤w, k = 1, . . . , n 0 ≤p ≤1. (40) 35 Now suppose we add the following constraint on the lamp powers p: no more than half the total power Pm i=1 pi is in any subset of r lamps (where r is a given integer with 0 < r < m). The idea is to avoid solutions where all the power is concentrated in very few lamps. Mathematically, the constraint can be expressed as r X i=1 p[i] ≤0.5 m X i=1 pi (41) where p[i] is the ith largest component of p. We would like to add this constraint to the LP (40). However the left-hand side of (41) is a complicated nonlinear function of p. We can write the constraint (41) as a set of linear inequalities by enumerating all subsets {i1, . . . , ir} ⊆{1, . . . , m} with r diﬀerent elements, and adding an inequality r X k=1 pik ≤0.5 m X i=1 pi for each subset. Equivalently, we express (41) as sT p ≤0.5 m X i=1 pi for all s ∈{0, 1}m with Pm i=1 si = r. This yields a set of m r ! linear inequalities in p. We can use LP duality to derive a much more compact representation. We will prove that (41) can be expressed as the set of 1 + 2m linear inequalities rt + m X i=1 xi ≤0.5 m X i=1 pi, pi ≤t + xi, i = 1, . . . , m, x ≥0 (42) in p ∈Rm, and auxiliary variables x ∈Rm and t ∈R. (a) Given a vector p ∈Rm, show that the sum of its r largest elements (i.e., p +· · ·+p[r]) is equal to the optimal value of the LP (in the variables y ∈Rm) maximize pT y subject to 0 ≤y ≤1 1T y = r. (43) (b) Derive the dual of the LP (43). Show that it can be written as minimize rt + 1T x subject to t1 + x ≥p x ≥0, (44) where the variables are t ∈R and x ∈Rm. By duality the LP (44) has the same optimal value as (43), i.e., p + · · · + p[r]. 36 It is now clear that the optimal value of (44) is less than 0.5 Pm i pi if and only if there is a feasible solution t, x in (44) with rt + 1T x ≤0.5 Pm i pi. In other words, p satisﬁes the constraint (41) if and only if the set of linear inequalities (42) in x and t are feasible. To include the nonlinear constraint (41) in (40), we can add the inequalities (42), which yields minimize w subject to −w ≤aT k p −Ides ≤w, k = 1, . . . , n 0 ≤p ≤1 rt + 1T x ≤0.5 1T p p ≤t1 + x x ≥0. This is an LP with 2m + 2 variables p, x, w, t, and 2n + 4m + 1 constraints. Exercise 55. In this problem we derive a linear programming formulation for the following vari-ation on ℓ∞- and ℓ1-approximation: given A ∈Rm×n, b ∈Rm, and an integer k with 1 ≤k ≤m, minimize k P i=1 \|Ax −b\|[i]. (45) The notation z[i] denotes the ith largest component of z ∈Rm, and \|z\|[i] denotes the ith largest component of the vector \|z\| = (\|z1\|, \|z2\|, . . . , \|zm\|) ∈Rm. In other words in (45) we minimize the sum of the k largest residuals \|aT i x −bi\|. For k = 1, this is the ℓ∞-problem; for k = m, it is the ℓ1-problem. Problem (45) can be written as minimize max 1≤i1<i2<···<ik≤m k X j=1 \|aT ijx −bij\|, or as the following LP in x and t: minimize t subject to sT (Ax −b) ≤t for all s ∈{−1, 0, 1}m, ∥s∥1 = k. Here we enumerate all vectors s with components −1, 0 or +1, and with exactly k nonzero elements. This yields an LP with 2k m k ! linear inequalities. We now use LP duality to derive a more compact formulation. (a) We have seen that for c ∈Rm and 1 ≤k ≤n, the the optimal value of the LP maximize cT v subject to −y ≤v ≤y 1T y = k y ≤1 (46) is equal to \|c\| + · · · + \|c\|[k]. Take the dual of the LP (46) and show that it can be simpliﬁed as minimize kt + 1T z subject to −t1 −z ≤c ≤t1 + z z ≥0 (47) 37 with variables t ∈R and z ∈Rm. By duality the optimal values of (47) and (46) are equal. (b) Now apply this result to c = Ax −b. From part (a), we know that the optimal value of the LP minimize kt + 1T z subject to −t1 −z ≤Ax −b ≤t1 + z z ≥0, (48) with variables t ∈R, z ∈Rm is equal to Pk i=1 \|Ax−b\|[i]. Note that the constraints (48) are linear in x, so we can simultaneously optimize over x, i.e., solve it as an LP with variables x, t and z. This way we can solve problem (45) by solving an LP with m+n+1 variables and 3m inequalities. Exercise 56. A portfolio optimization problem. We consider a portfolio optimization problem with n assets or stocks held over one period. The variable xi will denote the amount of asset i held at the beginning of (and throughout) the period, and pi will denote the price change of asset i over the period, so the return is r = pT x. The optimization variable is the portfolio vector x ∈Rn, which has to satisfy xi ≥0 and Pn i=1 xi ≤1 (unit total budget). If p is exactly known, the optimal allocation is to invest the entire budget in the asset with the highest return, i.e., if pj = maxi pi, we choose xj = 1, and xi = 0 for i ̸= j. However, this choice is obviously very sensitive to uncertainty in p. We can add various constraints to make the investment more robust against variations in p. We can impose a diversity constraint that prevents us from allocating the entire budget in a very small number of assets. For example, we can require that no more than, say, 90% of the total budget is invested in any 5% of the assets. We can express this constraint as ⌊n/20⌋ X i=1 x[i] ≤0.9 where x[i], i = 1, . . . , n, are the values xi sorted in decreasing order, and ⌊n/20⌋is the largest integer smaller than or equal to n/20. In addition, we can model the uncertainty in p by specifying a set P of possible values, and require that the investment maximizes the return in the worst-case scenario. The resulting problem is: maximize minp∈P pT x subject to 1T x ≤1, x ≥0, ⌊n/20⌋ P i=1 x[i] ≤0.9. (49) For each of the following sets P, can you express problem (49) as an LP? (a) P = {p(1), . . . , p(K)}, where p(i) ∈Rn are given. This means we consider a ﬁnite number of possible scenarios. (b) P = {¯ p + By \| ∥y∥∞≤1} where ¯ p ∈Rn and B ∈Rn×m are given. We can interpret ¯ p as the expected value of p, and y ∈Rm as uncertain parameters that determine the actual values of p. (c) P = {¯ p + y \| By ≤d} where ¯ p ∈Rn, B ∈Rr×m, and d ∈Rr are given. Here we consider a polyhedron of possible value of p. (We assume that P is nonempty.) 38 You may introduce new variables and constraints, but you must clearly explain why your formulation is equivalent to (49). If you know more than one solution, you should choose the most compact formulation, i.e., involving the smallest number of variables and constraints. Exercise 57. Let v be a discrete random variable with possible values c1, . . . , cn, and distribution pk = prob(v = ck), k = 1, . . . , n. The β-quantile of v, where 0 < β < 1, is deﬁned as qβ = min{α \| prob(v ≤α) ≥β}. For example, the 0.9-quantile of the distribution shown in the ﬁgure is q0.9 = 6.0. ck pk 1.0 1.5 2.5 4.0 4.5 5.0 6.0 7.0 9.0 10.0 0.08 0.14 0.18 0.14 0.16 0.14 0.08 0.04 0.02 0.02 A related quantity is fβ = 1 1 −β X ck>qβ pkck +  1 − 1 1 −β X ci>qβ pi  qβ. If P ci>qβ pi = 1 −β (and the second term vanishes), this is the conditional expected value of v, given that v is greater than qβ. Roughly speaking, fβ is the mean of the tail of the distribution above the β-quantile. In the example of the ﬁgure, f0.9 = 0.02 · 6.0 + 0.04 · 7.0 + 0.02 · 9.0 + 0.02 · 10.0 0.1 = 7.8. We consider optimization problems in which the values of ck depend linearly on some op-timization variable x. We will formulate the problem of minimizing fβ, subject to linear constraints on x, as a linear program. (a) Show that the optimal value of the LP maximize cT y subject to 0 ≤y ≤(1 −β)−1p 1T y = 1, (50) with variable y ∈Rn, is equal to fβ. The parameters c, p and β are given, with p > 0, 1T p = 1, and 0 < β < 1. 39 (b) Write the LP (50) in inequality form, derive its dual, and show that the dual is equivalent to the piecewise-linear minimization problem minimize t + 1 1 −β n X k=1 pk max{0, ck −t}, (51) with a single scalar variable t. It follows from duality theory and the result in part 1 that the optimal value of (51) is equal to fβ. (c) Now suppose ck = aT k x, where x ∈Rm is an optimization variable and ak is given, so qβ(x) and fβ(x) both depend on x. Use the result in part 2 to express the problem minimize fβ(x) subject to Fx ≤g, with variable x, as an LP. As an application, we consider a portfolio optimization problem with m assets or stocks held over a period of time. We represent the portfolio by a vector x = (x1, x2, . . . , xm), with xk the amount invested in asset k during the investment period. We denote by r the vector of returns for the m assets over the period, so the total return on the portfolio is rT x. The loss (negative return) is denoted v = −rT x. We model r as a discrete random variable, with possible values −a1, . . . , −an, and distribution pk = prob(r = −ak), k = 1, . . . , n. The loss of the portfolio v = −rT x is therefore a random variable with possible values ck = aT k x, k = 1, . . . , n, and distribution p. In this context, the β-quantile qβ(x) is called the value-at-risk of the portfolio, and fβ(x) is called the conditional value-at-risk. If we take β close to one, both functions are meaningful measures of the risk of the portfolio x. The result of part 3 implies that we can minimize fβ(x), subject to linear constraints in x, via linear programming. For example, we can minimize the risk (expressed as fβ(x)), subject to an upper bound on the expected loss (i.e., a lower bound on the expected return), by solving minimize fβ(x) subject to P k pkaT k x ≤R 1T x = 1 x ≥0. Exercise 58. Consider the problem minimize m P i=1 h(aT i x −bi) (52) where h is the function h(z) = ( 0 \|z\| ≤1 \|z\| −1 \|z\| > 1 40 and (as usual) x ∈Rn is the variable, and a1, . . . , am ∈Rn and b ∈Rm are given. Note that this problem can be thought of as a sort of hybrid between ℓ1- and ℓ∞-approximation, since there is no cost for residuals smaller than one, and a linearly growing cost for residuals larger than one. Express (52) as an LP, derive its dual, and simplify it as much as you can. Let xls denote the solution of the least-squares problem minimize m X i=1 (aT i x −bi)2, and let rls denote the residual rls = Axls −b. We assume A has rank n, so the least-squares solution is unique and given by xls = (AT A)−1AT b. The least-squares residual rls satisﬁes AT rls = 0. Show how to construct from xls and rls a feasible solution for the dual of (52), and hence a lower bound for its optimal value p⋆. Compare your lower bound with the trivial lower bound p⋆≥0. Is it always better, or only in certain cases? Exercise 59. The projection of a point x0 ∈Rn on a polyhedron P = {x \| Ax ≤b}, in the ℓ∞-norm, is deﬁned as the solution of the optimization problem minimize ∥x −x0∥∞ subject to Ax ≤b. The variable is x ∈Rn. We assume that P is nonempty. (a) Formulate this problem as an LP. (b) Derive the dual problem, and simplify it as much as you can. (c) Show that if x0 ̸∈P, then a hyperplane that separates x0 from P can be constructed from the optimal solution of the dual problem. Exercise 60. Describe a method for constructing a hyperplane that separates two given polyhedra P1 = {x ∈Rn\|Ax ≤b}, P2 = {x ∈Rn\|Cx ≤d}. Your method must return a vector a ∈Rn and a scalar γ such that aT x > γ for all x ∈P1, aT x < γ for all x ∈P2. a aT x = γ P1 P2 41 You can assume that P1 and P2 do not intersect. If you know several methods, you should give the most eﬃcient one. Exercise 61. Suppose the feasible set of the LP maximize bT z subject to AT z ≤c (53) is nonempty and bounded, with ∥z∥∞< µ for all feasible z. Show that any optimal solution of the problem minimize cT x + µ∥Ax −b∥1 subject to x ≥0 is also an optimal solution of the LP minimize cT x subject to Ax = b x ≥0, (54) which is the dual of problem (53). Exercise 62. An alternative to the phase-I/phase-II method for solving the LP minimize cT x subject to Ax ≤b, (55) is the “big-M”-method, in which we solve the auxiliary problem minimize cT x + Mt subject to Ax ≤b + t1 t ≥0. (56) M > 0 is a parameter and t is an auxiliary variable. Note that this auxiliary problem has obvious feasible points, for example, x = 0, t ≥max{0, −mini bi}. (a) Derive the dual LP of (56). (b) Prove the following property. If M > 1T z⋆, where z⋆is an optimal solution of the dual of (55), then the optimal t in (56) is zero, and therefore the optimal x in (56) is also an optimal solution of (55). Exercise 63. Robust linear programming with polyhedral uncertainty. Consider the robust LP minimize cT x subject to maxa∈Pi aT x ≤bi, i = 1, . . . , m, with variable x ∈Rn, where Pi = {a \| Cia ≤di}. The problem data are c ∈Rn, Ci ∈Rmi×n, di ∈Rmi, and b ∈Rm. We assume the polyhedra Pi are nonempty. Show that this problem is equivalent to the LP minimize cT x subject to dT i zi ≤bi, i = 1, . . . , m CT i zi = x, i = 1, . . . , m zi ≥0, i = 1, . . . , m 42 with variables x ∈Rn and zi ∈Rmi, i = 1, . . . , m. Hint. Find the dual of the problem of maximizing aT i x over ai ∈Pi (with variable ai). Exercise 64. We are given M + N polyhedra described by sets of linear inequalities Pi = {x ∈Rn \| Aix ≤bi}, i = 1, . . . , M + N. We deﬁne two sets S = P1 ∪P2 ∪· · · ∪PM and T = PM+1 ∪PM+2 ∪· · · ∪PM+N. (a) Explain how you can use linear programming to solve the following problem. Find a vector c and a scalar d such that cT x + d ≤−1 for x ∈S, cT x + d ≥1 for x ∈T (57) or show that no such c and d exist. Geometrically, the problem is to construct a hyperplane that strictly separates the polyhedra P1, . . . , PM from the polyhedra PM+1, . . . , PM+N. If you know several methods, give the most eﬃcient one. In particular, you should avoid methods based on enumerating extreme points, and methods that involve linear programs with dimensions that grow quadratically (or faster) with M or N. (b) The convex hull of a set S, denoted conv S, is deﬁned as the set of all convex combi-nations of points in S: conv S = {θ1v1 + · · · + θmvm \| θ1 + · · · + θm = 1, vi ∈S, θi ≥0, i = 1, . . . , m} . The convex hull of the shaded set S the ﬁgure is the polyhedron enclosed by the dashed lines. Show that if no separating hyperplane exists between S and T (i.e., there exists no c and d that satisfy (57)), then the convex hulls conv S and conv T intersect. Exercise 65. The polar of a polyhedron S is deﬁned as the set S∗= {y \| xT y ≤1 ∀x ∈S}. 43 (a) Show that the polar of the polyhedron P = {x \| Ax ≤1, Bx ≤0}, with A ∈Rm×n and B ∈Rp×n, is given by P ∗= conv{0, a1, . . . , am} + cone{b1, · · · , bp}. Here aT k and bT k denote the rows of the matrices A and B, and we use the notation conv{. . .} and cone{. . .} for the convex hull and the conic hull of a set of points (see lecture 4, page 4 and page 8). (b) Show that P ∗∗= P, where P ∗∗denotes the polar of the polyhedron P ∗in part 1. Exercise 66. Consider the LP minimize 47x1 + 93x2 + 17x3 −93x4 subject to          −1 −6 1 3 −1 −2 7 1 0 3 −10 −1 −6 −11 −2 12 1 6 −1 −3 11 1 −1 −8               x1 x2 x3 x4     ≤          −3 5 −8 −7 4 5          + ǫ          1 −3 13 46 −2 −75          . (58) where ǫ ∈R is a parameter. For ǫ = 0, this is the LP of exercise 46, with one extra inequality (the sixth inequality). This inequality is inactive at ˆ x = (1, 1, 1, 1), so ˆ x is also the optimal solution for (58) when ǫ = 0. (a) Determine the range of values of ǫ for which the ﬁrst four constraints are active at the optimum. (b) Give an explicit expression for the optimal primal solution, the optimal dual solution, and the optimal value, within the range of ǫ you determined in part (a). (If for some value of ǫ the optimal points are not unique, it is suﬃcient to give one optimal point.) Exercise 67. Consider the parametrized primal and dual LPs minimize (c + ǫd)T x subject to Ax ≤b, maximize −bT z subject to AT z + c + ǫd = 0 z ≥0 where A =        2 3 5 −4 2 −1 −3 4 −2 −1 3 1 −4 2 4 −2 2 −3 −9 1        , b =        6 2 1 0 −8        , c = (8, −32, −66, 14), d = (−16, −6, −2, 3). (a) Prove that x⋆= (1, 1, 1, 1) and z⋆= (9, 9, 4, 9, 0) are optimal when ǫ = 0. 44 (b) How does p⋆(ǫ) vary as a function of ǫ around ǫ = 0? Give an explicit expression for p⋆(ǫ), and specify the interval in which it is valid. (c) Also give an explicit expression for the primal and dual optimal solutions for values of ǫ around ǫ = 0. Remark: The problem is similar to the sensitivity problem discussed in the lecture notes. Here we consider the case where c is subject to a perturbation, while b is ﬁxed, so you have to develop the ‘dual’ of the derivation in the lecture notes. Exercise 68. Consider the pair of primal and dual LPs minimize (c + ǫd)T x subject to Ax ≤b + ǫf, maximize −(b + ǫf)T z subject to AT z + c + ǫd = 0 z ≥0 where A =        −4 12 −2 1 −17 12 7 11 1 0 −6 1 3 3 22 −1 −11 2 −1 −8        , b =        8 13 −4 27 −18        , c =      49 −34 −50 −5     , d =      3 8 21 25     , f =        6 15 −13 48 8        and ǫ is a parameter. (a) Prove that x⋆= (1, 1, 1, 1) is optimal when ǫ = 0, by constructing a dual optimal point z⋆that has the same objective value as x⋆. Are there any other primal or dual optimal solutions? (b) Express the optimal value p⋆(ǫ) as a continuous function of ǫ on an interval that con-tains ǫ = 0. Specify the interval in which your expression is valid. Also give explicit expressions for the primal and dual solutions as a function of ǫ over the same interval. Exercise 69. In some applications we are interested in minimizing two cost functions, cT x and dT x, over a polyhedron P = {x \| Ax ≤b}. For general c and d, the two objectives are competing, i.e., it is not possible to minimize them simultaneously, and there exists a trade-oﬀbetween them. The problem can be visualized as in the ﬁgure below. cT x dT x 45 The shaded region is the set of pairs (cT x, dT x) for all possible x ∈P. The circles are the values (cT x, dT x) at the extreme points of P. The lower part of the boundary, shown as a heavy line, is called the trade-oﬀcurve. Points (cT ˆ x, dT ˆ x) on this curve are eﬃcient in the following sense: it is not possible to improve both objectives by choosing a diﬀerent feasible x. Suppose (cT ˆ x, dT ˆ x) is a breakpoint of the trade-oﬀcurve, where ˆ x is a nondegenerate extreme point of P. Explain how the left and right derivatives of the trade-oﬀcurve at this breakpoint can be computed. Hint. Compute the largest and smallest values of γ such that ˆ x is optimal for the LP minimize dT x + γcT x subject to Ax ≤b. Exercise 70. Consider the ℓ1-norm minimization problem minimize ∥Ax + b + ǫd∥1 with A =          −2 7 1 −5 −1 3 −7 3 −5 −1 4 −4 1 5 5 2 −5 −1          , b =          −4 3 9 0 −11 5          , d =          −10 −13 −27 −10 −7 14          . (a) Suppose ǫ = 0. Prove, without using any LP code, that x⋆= 1 is optimal. Are there any other optimal points? (b) Give an explicit formula for the optimal value as a function of ǫ for small positive and negative values of ǫ. What are the values of ǫ for which your expression is valid? Exercise 71. We consider a network ﬂow problem on the simple network shown below. 1 3 4 5 2 u1 u2 u3 u4 u5 u6 u7 V1 V2 V3 V4 V5 Here u1, . . . , u7 ∈R denote the ﬂows or traﬃc along links 1, . . . , 7 in the direction indicated by the arrow. (Thus, u1 = 1 means a traﬃc ﬂow of one unit in the direction of the arrow on link 1, i.e., from node 1 to node 2.) V1, . . . , V5 ∈R denote the external inputs (or 46 outputs if Vi < 0) to the network. We assume that the net ﬂow into the network is zero, i.e., P5 i=1 Vi = 0. Conservation of traﬃc ﬂow states that at each node, the total ﬂow entering the node is zero. For example, for node 1, this means that V1 −u1 + u4 −u5 = 0. This gives one equation per node, so we have 5 traﬃc conservation equations, for the nodes 1, . . . , 5, respectively. (In fact, the equations are redundant since they sum to zero, so you could leave one, e.g., for node 5, out. However, to answer the questions below, it is easier to keep all ﬁve equations.) The cost of a ﬂow pattern u is given by P i ci\|ui\|, where ci > 0 is the tariﬀon link i. In addition to the tariﬀ, each link also has a maximum possible traﬃc level or link capacity: \|ui\| ≤Ui. (a) Express the problem of ﬁnding the minimum cost ﬂow as an LP in inequality form, for the network shown above. (b) Solve the LP from part (a) for the speciﬁc costs, capacities, and inputs c = (2, 2, 2, 1, 1, 1, 1), V = (1, 1, 0.5, 0.5, −3), U = (0.5, 0.5, 0.1, 0.5, 1, 1, 1). Find the optimal dual variables as well. (c) Suppose we can increase the capacity of one link by a small ﬁxed amount, say, 0.1. Which one should we choose, and why? (You’re not allowed to solve new LPs to answer this!) For the link you pick, increase its capacity by 0.1, and then solve the resulting LP exactly. Compare the resulting cost with the cost predicted from the optimal dual variables of the original problem. Can you explain the answer? (d) Now suppose we have the possibility to increase or reduce two of the ﬁve external inputs by a small amount, say, 0.1. To keep P i Vi = 0, the changes in the two inputs must be equal in absolute value and opposite in sign. For example, we can increae V1 by 0.1, and decrease V4 by 0.1. Which two inputs should we modify, and why? (Again, you’re not allowed to solve new LPs!) For the inputs you pick, change the value (increase or decrease, depending on which will result in a smaller cost) by 0.1, and then solve the resulting LP exactly. Compare the result with the one predicted from the optimal dual variables of the original problem. Exercise 72. Strict complementarity. We consider an LP minimize cT x subject to Ax ≤b, with A ∈Rm×n, and its dual maximize −bT z subject to AT z + c = 0, z ≥0. We assume the optimal value is ﬁnite. From duality theory we know that any primal optimal x⋆and any dual optimal z⋆satisfy the complementary slackness conditions z⋆ i (bi −aT i x⋆) = 0, i = 1, . . . , m. In other words, for each i, we have z⋆ i = 0, or aT i x⋆= bi, or both. 47 In this problem you are asked to show that there exists at least one primal-dual optimal pair x⋆, z⋆that satisﬁes z⋆ i (bi −aT i x⋆) = 0, z⋆ i + (bi −aT i x⋆) > 0, for all i. This is called a strictly complementary pair. In a strictly complementary pair, we have for each i, either z⋆ i = 0, or aT i x⋆= bi, but not both. To prove the result, suppose x⋆, z⋆are optimal but not strictly complementary, and aT i x⋆= bi, z⋆ i = 0, i = 1, . . . , M aT i x⋆= bi, z⋆ i > 0, i = M + 1, . . . , N aT i x⋆< bi, z⋆ i = 0, i = N + 1, . . . , m with M ≥1. In other words, m −M entries of b −Ax⋆and z⋆are strictly complementary; for the other entries we have zero in both vectors. (a) Use Farkas’ lemma to show that the following two sets of inequalities/equalities are strong alternatives: • There exists a v ∈Rn such that aT 1 v < 0 aT i v ≤ 0, i = 2, . . . , M aT i v = 0, i = M + 1, . . . , N. (59) • There exists a w ∈RN−1 such that a1 + N−1 X i=1 wiai+1 = 0, wi ≥0, i = 1, . . . , M −1. (60) (b) Assume the ﬁrst alternative holds, and v satisﬁes (59). Show that there exists a primal optimal solution ˜ x with aT 1 ˜ x < b1 aT i ˜ x ≤ bi, i = 2, . . . , M aT i ˜ x = bi, i = M + 1, . . . , N aT i ˜ x < bi, i = N + 1, . . . , m. (c) Assume the second alternative holds, and w satisﬁes (60). Show that there exists a dual optimal ˜ z with ˜ z1 > 0 ˜ zi ≥ 0, i = 2, . . . , M ˜ zi > 0, i = M + 1, . . . , N ˜ zi = 0, i = N + 1, . . . , m. (d) Combine (b) and (c) to show that there exists a primal-dual optimal pair x, z, for which b−Ax and z have at most ˜ M common zeros, where ˜ M < M. If ˜ M = 0, x, z are strictly complementary and optimal, and we are done. Otherwise, we apply the argument given above, with x⋆, z⋆replaced by x, z, to show the existence of a strictly complementary pair of optimal solutions with less than ˜ M common zeros in b −Ax and z. Repeating the argument eventually gives a strictly complementary pair. 48 Exercise 73. Self-dual homogeneous LP formulation. (a) Consider the LP minimize fT 1 u + fT 2 v subject to M11u + M12v ≤f1 −MT 12u + M22v = f2 u ≥0 (61) in the variables u ∈Rp and v ∈Rq. The problem data are the vectors f1 ∈Rp, f2 ∈Rq, and the matrices M11 ∈Rp×p, M12 ∈Rp×q, and M22 ∈Rq×q. Show that if M11 and M22 are skew-symmetric, i.e., MT 11 = −M11, MT 22 = −M22, then the dual of the LP (61) can be expressed as maximize −fT 1 w −fT 2 y subject to M11w + M12y ≤f1 −MT 12w + M22y = f2 w ≥0, (62) with variables w ∈Rp and y ∈Rq. Note that the dual problem is essentially the same as the primal problem. Therefore if u, v are primal optimal, then w = u, y = v are optimal in the dual problem. We say that the LP (61) with skew-symmetric M11 and M22 is self-dual. (b) Write down the optimality conditions for problem (61). Use the observation we made in part (a) to show that the optimality conditions can be simpliﬁed as follows: u, v are optimal for (61) if and only if M11u + M12v ≤f1 −MT 12u + M22v = f2 u ≥0 uT (f1 −M11u −M12v) = 0. In other words, u, v must be feasible in (61), and the nonnegative vectors u and s = f1 −M11u −M12v must satisfy the complementarity condition uT s = 0. It can be shown that if (61) is feasible, then it has an optimal solution that is strictly complementary, i.e., u + s > 0. (In other words, for each k either sk = 0 or uk = 0, but not both.) (c) Consider the LP minimize 0 subject to bT e z + cT e x ≤0 −be t + Ae x ≤0 AT e z + ce t = 0 e z ≥0, e t ≥0 (63) 49 with variables e x ∈Rn, e z ∈Rm, and e t ∈R. Show that this problem is self-dual. Use the result in part (b) to prove that (63) has an optimal solution that satisﬁes e t(cT e x + bT e z) = 0 and e t −(cT e x + bT e z) > 0. Suppose we have computed an optimal solution with these properties. We can distin-guish the following cases. • e t > 0. Show that x = e x/e t, z = e z/e t are optimal for the pair of primal and dual LPs minimize cT x subject to Ax ≤b (64) and maximize −bT z subject to AT z + c = 0 z ≥0. (65) • e t = 0 and cT e x < 0. Show that the dual LP (65) is infeasible. • e t = 0 and bT e z < 0. Show that the primal LP (65) is infeasible. This result has an important practical ramiﬁcation. It implies that we do not have to use a two-phase approach to solve the LP (64) (i.e., a phase-I to ﬁnd a feasible point, followed by a phase-II to minimize cT x starting at the feasible point). We can solve the LP (64) and its dual, or detect primal or dual infeasibility, by solving one single, feasible LP (63). The LP (63) is much larger than (64), but it can be shown that the cost of solving it is not much higher if one takes advantage of the symmetry in the constraints. Exercise 74. In the lecture we used linear programming duality to prove the minimax theorem for matrix games. The minimax theorem states that for any m × n matrix C (the payoﬀ matrix of the game), the optimal values of the optimization problems (Player 1) minimize max v∈Pn xT Cv subject to 1T x = 1 x ≥0 (Player 2) maximize min u∈Pm uT Cy subject to 1T y = 1 y ≥0 are equal. The optimal value of the two problems is called the value of the matrix game. Optimal solutions x and y are called optimal (randomized) strategies for the two players. Throughout this problem we assume that C is skew-symmetric (m = n and C = −CT ). (a) What is the value of the game (for skew-symmetric C)? What is the relation between the sets of optimal strategies for the two players? (b) Show that the Player 1 problem can be formulated as a self-dual LP minimize t subject to " 0 s # = " 0 −1T 1 C # " t x # + " 1 0 # x ≥0, s ≥0 with variables x, s, and t. 50 (c) Show that the Player 1 problem can also be formulated as a smaller self-dual LP minimize 0 subject to s = Cx x ≥0, s ≥0, (66) with variables x, s. Speciﬁcally, if x is a nonzero solution of this LP, then x/(1T x) is an optimal strategy for Player 1. Explain why nonzero solutions for the LP (66) are guaranteed to exist. Linear-fractional optimization Exercise 75. What is the optimal value of the linear-fractional optimization problem minimize a1x1 + a2x2 + · · · + anxn + b c1x1 + c2x2 + · · · + cnxn + d subject to x ≥0 with optimization variables x1, . . . , xn? The coeﬃcients ai, b, ci, d are given with ci > 0 and d > 0. Explain your answer. Exercise 76. Explain how you would use linear programming to solve the following optimization problems. (a) Given A ∈Rm×n, b ∈Rm, minimize maxi=1,...,m max{aT i x + bi, 1/(aT i x + bi)} subject to Ax + b > 0. The variable is x ∈Rn. (b) Given m numbers a1, a2, . . . , am ∈R, and two vectors l, u ∈Rm, ﬁnd the polynomial f(t) = c0 + c1t + · · · + cntn of lowest degree that satisﬁes the bounds li ≤f(ai) ≤ui, i = 1, . . . , m. The variables in the problem are the coeﬃcients ci of the polynomial. Exercise 77. In exercise 14, we encountered the problem minimize maxk=1,...,n \|aT k p −Ides\| subject to 0 ≤p ≤1 (67) (with variables p). We have seen that this is readily cast as an LP. In (67) we use the maximum of the absolute deviations \|Ik −Ides\| to measure the diﬀerence from the desired intensity. Suppose we prefer to use the relative deviations instead, where the relative deviation is deﬁned as max{Ik/Ides, Ides/Ik} −1 = ( (Ik −Ides)/Ides if Ik ≥Ides (Ides −Ik)/Ik if Ik ≤Ides. 51 This leads us to the following formulation: minimize maxk=1,...,n max{ aT k p/Ides, Ides/(aT k p) } subject to 0 ≤p ≤1 aT k p > 0, k = 1, . . . , n. (68) Explain how you can solve this using linear programming (i.e., by solving one or more LPs). Exercise 78. Consider the linear system of exercise 7, equation (4). We study two optimal control problems. In both problems we assume the system is initially at rest at the origin, i.e., z(0) = 0. (a) In the ﬁrst problem we want to determine the most eﬃcient input sequence u(kT), k = 0, . . . , 79, that brings the system to state (0, 0, 10, 10) in 80 time periods (i.e., at t = 8 the two masses should be at rest at position v1 = v2 = 10). We assume the cost (e.g., fuel consumption) of the input signal u is proportional to P k \|u(kT)\|. We also impose the constraint that the amplitude of the input must not exceed 2. This leads us to the following problem: minimize 79 P k=0 \|u(kT)\| subject to z(80T) = (0, 0, 10, 10) \|u(kT)\| ≤2, k = 0, . . . , 79. (69) The state z and the input u are related by (4) with z(0) = 0. The variables in (69) are u(0), u(T), . . . , u(79T). (b) In the second problem we want to bring the system to the state (0, 0, 10, 10) as quickly as possible, subject to the limit on the magnitude of u: minimize N subject to z(NT) = (0, 0, 10, 10) \|u(kT)\| ≤2, k = 0, . . . , N −1. The variables are N ∈Z, and u(0), u(T), . . . , u(N −1)T. Solve these two problems numerically. Plot the input u and the positions v1, v2 as functions of time. Exercise 79. Consider the linear-fractional program minimize (cT x + γ)/(dT x + δ) subject to Ax ≤b, (70) where A ∈Rm×n, b ∈Rm, c, d ∈Rn, and γ, δ ∈R. We assume that the polyhedron P = {x ∈Rn \| Ax ≤b} is bounded and that dT x + δ > 0 for all x ∈P. 52 Show that you can solve (70) by solving the LP minimize cT y + γz subject to Ay −zb ≤0 dT y + δz = 1 z ≥0 (71) in the variables y ∈Rn and z ∈R. More precisely, suppose ˆ y and ˆ z are a solution of (71). Show that ˆ z > 0 and that ˆ y/ˆ z solves (70). Exercise 80. Consider the problem minimize ∥Ax −b∥1/(cT x + d) subject to ∥x∥∞≤1, where A ∈Rm×n, b ∈Rm, c ∈Rn, and d ∈R. We assume that d > ∥c∥1. (a) Formulate this problem as a linear-fractional program. (b) Show that d > ∥c∥1 implies that cT x + d > 0 for all feasible x. (c) Show that the problem is equivalent to the convex optimization problem minimize ∥Ay −bt∥1 subject to ∥y∥∞≤t cT y + dt = 1, (72) with variables y ∈Rn, t ∈R. (d) Formulate problem (72) as an LP. Exercise 81. A generalized linear-fractional problem. Consider the problem minimize ∥Ax −b∥1/(cT x + d) subject to ∥x∥∞≤1 (73) where A ∈Rm×n, b ∈Rm, c ∈Rn and d ∈R are given. We assume that d > ∥c∥1. As a consequence, cT x + d > 0 for all feasible x. (a) Explain how you would solve this problem using linear programming. If you know more than one method, you should give the simplest one. (b) Prove that the following problem provides lower bounds on the optimal value of (73): maximize λ subject to ∥AT z + λc∥1 ≤bT z −λd ∥z∥∞≤1. (74) The variables are z ∈Rm and λ ∈R. (c) Use linear programming duality to show that the optimal values of (74) and (73) are in fact equal. 53 Exercise 82. We consider the problem minimize max i=1,...,m(aT i x + bi) min i=1,...,p(cT i x + di) subject to Fx ≤g with variable x ∈Rn. We assume that cT i x + di > 0 and maxi=1,...,m(aT i x + bi) ≥0 for all x satisfying Fx ≤g, and that the feasible set is nonempty and bounded. Show how the problem can be solved by solving one LP, using a trick similar to one described for linear-fractional problems in the lecture. The simplex method Exercise 83. Solve the following linear program using the simplex algorithm with Bland’s pivoting rule. Start the algorithm at the extreme point x = (2, 2, 0), with active set I = {3, 4, 5}. mininimize x1 + x2 −x3 subject to             −1 0 0 0 −1 0 0 0 −1 1 0 0 0 1 0 0 0 1 1 1 1                x1 x2 x3   ≤             0 0 0 2 2 2 4             . Exercise 84. Use the simplex method to solve the following LP: minimize −24x1 + 396x2 −8x3 −28x4 −10x5 subject to    12 4 1 −19 7 6 −7 18 −1 −13 1 17 3 18 −2           x1 x2 x3 x4 x5        =    12 6 1    x ≥0. Start with the initial basis {1, 2, 3}, and use Bland’s rule to make pivot selections. Also compute the dual optimal point from the results of the algorithm. Interior-point methods Exercise 85. The ﬁgure shows the feasible set of an LP minimize cT x subject to aT i x ≤bi, i = 1, . . . , 6 54 with two variables and six constraints. Also shown are the cost vector c, the analytic center, and a few contour lines of the logarithmic barrier function φ(x) = − m X i=1 log(bi −aT i x). c Sketch the central path as accurately as possible. Explain your answer. Exercise 86. Let x⋆(t0) be a point on the central path of the LP minimize cT x subject to Ax ≤b, with t0 > 0. We assume that A is m× n with rank(A) = n. Deﬁne ∆xnt as the Newton step at x⋆(t0) for the function t1cT x − m X i=1 log(bi −aT i x), where aT i denotes the ith row of A, and t1 > t0. Show that ∆xnt is tangent to the central path at x⋆(t0). x⋆(t0) x⋆(t1) ∆xnt c 55 Hint. Find an expression for the tangent direction ∆xtg = dx⋆(t0)/dt, and show that ∆xnt is a positive multiple of ∆xtg. Exercise 87. In the lecture on barrier methods, we noted that a point x∗(t) on the central path yields a dual feasible point z∗ i (t) = 1 t(bi −aT i x∗(t)), i = 1, . . . , m. (75) In this problem we examine what happens when x∗(t) is calculated only approximately. Suppose x is strictly feasible and v is the Newton step at x for the function tcT x + φ(x) = tcT x − m X i=1 log(b −aT i x). Let d ∈Rm be deﬁned as di = 1/(bi −aT i x), i = 1, . . . , m. Show that if λ(x) = ∥diag(d)Av∥≤1, then the vector z = (d + diag(d)2Av)/t is dual feasible. Note that z reduces to (75) if x = x∗(t) (and hence v = 0). This observation is useful in a practical implementation of the barrier method. In practice, Newton’s method provides an approximation of the central point x∗(t), which means that the point (75) is not quite dual feasible, and a stopping criterion based on the corresponding dual bound is not quite accurate. The results derived above imply that even though x∗(t) is not exactly centered, we can still obtain a dual feasible point, and use a completely rigorous stopping criterion. Exercise 88. Let P be a polyhedron described by a set of linear inequalities: P = {x ∈Rn \| Ax ≤b} , where A ∈Rm×n and b ∈Rm. Let φ denote the logarithmic barrier function φ(x) = − m X i=1 log(bi −aT i x). (a) Suppose b x is strictly feasible. Show that (x −b x)T ∇2φ(b x)(x −b x) ≤1 = ⇒ Ax ≤b, where ∇2φ(b x) is the Hessian of φ at b x. Geometrically, this means that the set Einner = {x \| (x −b x)T ∇2φ(b x)(x −b x) ≤1}, which is an ellipsoid centered at b x, is enclosed in the polyhedron P. 56 (b) Suppose b x is the analytic center of the inequalities Ax < b. Show that Ax ≤b = ⇒ (x −b x)T ∇2φ(b x)(x −b x) ≤m(m −1). In other words, the ellipsoid Eouter = {x \| (x −b x)T ∇2φ(b x)(x −b x) ≤m(m −1)} contains the polyhedron P. Exercise 89. Let ˆ x be the analytic center of a set of linear inequalities aT k x ≤bk, k = 1, . . . , m. Show that the kth inequality is redundant (i.e., it can be deleted without changing the feasible set) if bk −aT k ˆ x ≥m q aT k H−1ak where H is deﬁned as H = m X k=1 1 (bk −aT k ˆ x)2 akaT k . Exercise 90. The analytic center of a set of linear inequalities Ax ≤b depends not only on the geometry of the feasible set, but also on the representation (i.e., A and b). For example, adding redundant inequalities does not change the polyhedron, but it moves the analytic center. In fact, by adding redundant inequalities you can make any strictly feasible point the analytic center, as you will show in this problem. Suppose that A ∈Rm×n and b ∈Rm deﬁne a bounded polyhedron P = {x \| Ax ≤b} and that x⋆satisﬁes Ax⋆< b. Show that there exist c ∈Rn, γ ∈R, and a positive integer q, such that (a) P is the solution set of the m + q inequalities Ax ≤b cT x ≤γ cT x ≤γ . . . cT x ≤γ            q copies. (76) (b) x⋆is the analytic center of the set of linear inequalities given in (76). Exercise 91. We consider the problem of ﬁnding a solution of a set of linear inequalities Ax ≤b with A ∈Rm×n and b ∈Rm. 57 Suppose you know that there exists at least one solution with ∥x∥≤R for some given positive R. Show that the parameters α and d in the ‘phase-1’ linear program minimize y subject to Ax ≤b + y(α1 −b) dT x ≤1 y ≥−1, with variables x ∈Rn, y ∈R, can be chosen in such a way that: • the point (x, y) = (0, 1) is on the central path for the phase-1 LP; • the norm of d is less than or equal to 1/R. (The Cauchy-Schwarz inequality then implies that dT x ≤1 for all x with ∥x∥≤R. Therefore the inequalities Ax ≤b, dT x ≤1 are still feasible.) Exercise 92. Maximum-likelihood estimation with parabolic noise density. We consider the linear measurement model yi = aT i x + vi, i = 1, . . . , m. The vector x ∈Rn is a vector of parameters to be estimated, yi ∈R are the measured or observed quantities, and vi are the measurement errors or noise. The vectors ai ∈Rn are given. We assume that the measurement errors vi are independent and identically distributed with a parabolic density function p(v) = ( (3/4)(1 −v2) \|v\| ≤1 0 otherwise (shown below). −2 −1 0 1 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 v p(v) Let ¯ x be the maximum-likelihood (ML) estimate based on the observed values y, i.e., ¯ x = argmax x m X i=1 log 1 −(yi −aT i x)2 + m log(3/4) ! . 58 Show that the true value of x satisﬁes (x −¯ x)T H(x −¯ x) ≤4m2 where H = 2 m X i=1 1 + (yi −aT i ¯ x)2 (1 −(yi −aT i ¯ x)2)2 aiaT i . Exercise 93. Potential reduction algorithm. Consider the LP minimize cT x subject to Ax ≤b with A ∈Rm×n. We assume that rank A = n, that the problem is strictly feasible, and that the optimal value p⋆is ﬁnite. For l < p⋆and q > m, we deﬁne the potential function ϕpot(x) = q log(cT x −l) − m X i=1 log(bi −aT i x). The function ϕpot is deﬁned for all strictly feasible x, and although it is not a convex function, it can be shown that it has a unique minimizer. We denote the minimizer as x⋆ pot(l): x⋆ pot(l) = argmin Ax<b q log(cT x −l) − m X i=1 log(bi −aT i x) ! . (a) Show that x⋆ pot(l) lies on the central path, i.e., it is the minimizer of the function tcT x − m X i=1 log(bi −aT i x) for some value of t. (b) Prove that the following algorithm converges and that it returns a suboptimal x with cT x −p⋆< ǫ. given l < p⋆, tolerance ǫ > 0, q > m repeat 1. x := x⋆ pot(l) 2. if m(cT x −l) q < ǫ, return x 3. l := q −m q cT x + m q l Exercise 94. Consider the following variation on the barrier method for solving the LP minimize cT x subject to aT i x ≤bi, i = 1, . . . , m. We assume we are given a strictly feasible ˆ x (i.e., aT i ˆ x < bi for i = 1, . . . , m), a strictly dual feasible ˆ z (AT ˆ z + c = 0, ˆ z > 0), and a positive scalar ρ with 0 < ρ < 1. 59 initialize: x = ˆ x, wi = (bi −aT i ˆ x)ˆ zi, i = 1, . . . , m repeat: 1. x := argminy cT y − m P i=1 wi log(bi −aT i y) 2. w := ρw Give an estimate or a bound on the number of (outer) iterations required to reach an accuracy cT x −p⋆≤ǫ. Exercise 95. The inverse barrier. The inverse barrier of a set of linear inequalities aT i x ≤bi, i = 1, . . . , m, is the function ψ, deﬁned as ψ(x) = m X i=1 1 bi −aT i x for strictly feasible x. It can be shown that ψ is convex and diﬀerentiable on the set of strictly feasible points, and that ψ(x) tends to inﬁnity as x approaches the boundary of the feasible set. Suppose ˆ x is strictly feasible and minimizes cT x + ψ(x). Show that you construct from ˆ x a dual feasible point for the LP minimize cT x subject to aT i x ≤bi, i = 1, . . . , m. Exercise 96. Assume the primal and dual LPs (P) minimize cT x subject to Ax ≤b (D) maximize −bT z subject to AT z + c = 0 z ≥0 are strictly feasible. Let {x(t) \| t > 0} be the central path and deﬁne s(t) = b −Ax(t), z(t) = 1 t       1/s1(t) 1/s2(t) . . . 1/sm(t)       . (a) Suppose x∗, z∗are optimal for the primal and dual LPs, and deﬁne s∗= b −Ax∗. (If there are multiple optimal points, x∗, z∗denote an arbitrary pair of optimal points.) Show that z(t)T s∗+ s(t)T z∗= m t for all t > 0. From the deﬁnition of z(t), this implies that m X k=1 s∗ k sk(t) + m X k=1 z∗ k zk(t) = m. (77) 60 (b) As t goes to inﬁnity, the central path converges to the optimal points x∗ c = lim t→∞x(t), s∗ c = b −Ax∗ c = lim t→∞s(t), z∗ c = lim t→∞z(t). Deﬁne I = {k \| s∗ c,k = 0}, the set of active constraints at x∗ c. Apply (77) to s∗= s∗ c, z∗= z∗ c to get X k̸∈I s∗ c,k sk(t) + X k∈I z∗ c,k zk(t) = m. Use this to show that z∗ c,k > 0 for k ∈I. This proves that the central path converges to a strictly complementary solution, i.e., s∗ c + z∗ c > 0. (c) The primal optimal set is the set of all x that are feasible and satisfy complementary slackness with z∗ c: Xopt = {x \| aT k x = bk, k ∈I, aT k x ≤bk, k ̸∈I}. Let x∗be an arbitrary primal optimal point. Show that Y k̸∈I (bk −aT k x∗) ≤ Y k̸∈I (bk −aT k x∗ c). Hint. Use the arithmetic-geometric mean inequality m Y k=1 yk !1/m ≤1 m m X k=1 yk for nonnegative vectors y ∈Rm. Exercise 97. The most expensive step in one iteration of an interior-point method for an LP minimize cT x subject to Ax ≤b is the solution of a set of linear equations of the form AT DA ∆x = y, (78) where D is a positive diagonal matrix, the right-hand side y is a given vector, and ∆x is the unknown. The values of D and y depend on the method used and on the current iterate, and are not important for our purposes here. For example, the Newton equation in the barrier methodm ∇2φ(x)v = −tc −∇φ(x), is of the form (78). In the primal-dual method, we have to solve two sets of linear equations of the form (78) with D = X−1Z. It is often possible to speed up the algorithm signiﬁcantly by taking advantage of special structure of the matrix A when solving the equations (78). Consider the following three optimization problems that we encountered before in this course. 61 • ℓ1-minimization: minimize ∥Pu + q∥1 (P ∈Rr×s and q ∈Rr are given; u ∈Rs is the variable). • Constrained ℓ1-minimization: minimize ∥Pu + q∥1 subject to −1 ≤u ≤1 (P ∈Rr×s and q ∈Rr are given; u ∈Rs is the variable). • Robust linear programming (see exercise 22): minimize wT u subject to Pu + ∥u∥11 ≤q (P ∈Rr×s, q ∈Rr, and w ∈Rs are given; u ∈Rs is the variable). For each of these three problems, answer the following questions. (a) Express the problem as an LP in inequality form. Give the matrix A, and the number of variables and constraints. (b) What is the cost of solving (78) for the matrix A you obtained in part (a), if you do not use any special structure in A (knowing that the cost of solving a dense symmetric positive deﬁnite set of n linear equations in n variables is (1/3)n3 operations, and the cost of a matrix-matrix multiplication AT A, with A ∈Rm×n, is mn2 operations)? (c) Work out the product AT DA (assuming D is a given positive diagonal matrix). Can you give an eﬃcient method for solving (78) that uses the structure in the equations? What is the cost of your method (i.e., the approximate number of operations when r and s are large) as a function of the dimensions r and s ? Hint. Try to reduce the problem to solving a set of s linear equations in s variables, followed by a number of simple operations. For the third problem, you can use the following formula for the inverse of a matrix H + yyT , where y is a vector: (H + yyT )−1 = H−1 − 1 1 + yT H−1yH−1yyT H−1. Exercise 98. In this problem you are asked to write a MATLAB code for the ℓ1-approximation problem minimize ∥Pu + q∥1, (79) where P = Rr×s and q ∈Rr. The calling sequence for the code is u = l1(P,q). On exit, it must guarantee a relative accuracy of 10−6 or an absolute accuracy of 10−8, i.e., the code can terminate if ∥Pu + q∥1 −p⋆≤10−6 · p⋆ or ∥Pu + q∥1 −p⋆≤10−8, where p⋆is the optimal value of (79). You may assume that P has full rank (rank P = s). 62 We will solve the problem using Mehrotra’s method as described in applied to the LP minimize 1T v subject to " P −I −P −I # " u v # ≤ " −q q # . (80) We will take advantage of the structure in the problem to improve the eﬃciency. (a) Initialization. Mehrotra’s method can be started at infeasible primal and dual points. However good feasible starting points for the LP (80) are readily available from the solution uls of the least-squares problem minimize ∥Pu + q∥ (in MATLAB: u = -P\q). As primal starting point we can use u = uls, and choose v so that we have strict feasibility in (80). To ﬁnd a strictly feasible point for the dual of (80), we note that P T Puls = −P T q and therefore the least-squares residual rls = Puls + q satisﬁes P T rls = 0. This property can be used to construct a strictly feasible point for the dual of (80). You should try to ﬁnd a dual starting point that provides a positive lower bound on p⋆, i.e., a lower bound that is better than the trivial lower bound p⋆≥0. Since the starting points are strictly feasible, all iterates in the algorithm will remain strictly feasible, and we don’t have to worry about testing the deviation from feasibility in the convergence criteria. (b) As we have seen, the most expensive part of an iteration in Mehrotra’s method is the solution of two sets of equations of the form AT X−1ZA∆x = r1 (81) where X and Z are positive diagonal matrices that change at each iteration. One of the two equations is needed to determine the aﬃne-scaling direction; the other equation (with a diﬀerent right-hand side) is used to compute the combined centering-corrector step. In our application, (81) has r + s equations in r + s variables, since A = " P −I −P −I # , ∆x = " ∆u ∆v # . By exploiting the special structure of A, show that you can solve systems of the form (81) by solving a smaller system of the form P T DP∆u = r2, (82) followed by a number of inexpensive operations. In (82) D is an appropriately chosen positive diagonal matrix. This observation is important, since it means that the cost of one iteration reduces to the cost of solving two systems of size s × s (as opposed to (r + s) × (r + s)). In other words, although we have introduced r new variables to express (79) as an LP, the extra cost of introducing these variables is marginal. 63 (c) Test your code on randomly generated P and q. Plot the duality gap (on a logarithmic scale) versus the iteration number for a few examples and include a typical plot with your solutions. Exercise 99. This problem is similar to the previous problem, but instead we consider the con-strained ℓ1-approximation problem minimize ∥Pu + q∥1 subject to −1 ≤u ≤1 (83) where P = Rr×s and q ∈Rr. The calling sequence for the code is u = cl1(P,q). On exit, it must guarantee a relative accuracy of 10−5 or an absolute accuracy of 10−8, i.e., the code can terminate if ∥Pu + q∥1 −p⋆≤10−5 · p⋆ or ∥Pu + q∥1 −p⋆≤10−8, where p⋆is the optimal value of (83). You may assume that P has full rank (rank P = s). We will solve the problem using Mehrotra’s method as described in applied to the LP minimize 1T v subject to      P −I −P −I I 0 −I 0      " u v # ≤      −q q 1 1     . (84) We will take advantage of the structure in the problem to improve the eﬃciency. (a) Initialization. For this problem it is easy to determine strictly feasible primal and dual points at which the algorithm can be started. This has the advantage that all iterates in the algorithm will remain strictly feasible, and we don’t have to worry about testing the deviation from feasibility in the convergence criteria. As primal starting point, we can simply take u = 0, and a vector v that satisﬁes vi > \|(Pu + q)i\|, i = 1, . . . , r. What would you choose as dual starting point? (b) As we have seen, the most expensive part of an iteration in Mehrotra’s method is the solution of two sets of equations of the form AT X−1ZA∆x = r1 (85) where X and Z are positive diagonal matrices that change at each iteration. One of the two equations is needed to determine the aﬃne-scaling direction; the other equation (with a diﬀerent right-hand side) is used to compute the combined centering-corrector step. In our application, (85) has r + s equations in r + s variables, since A =      P −I −P −I I 0 −I 0     , ∆x = " ∆u ∆v # . 64 By exploiting the special structure of A, show that you can solve systems of the form (85) by solving a smaller system of the form (P T ˜ DP + ˆ D)∆u = r2, (86) followed by a number of inexpensive operations. The matrices ˜ D and ˆ D in (86) are appropriately chosen positive diagonal matrices. This observation is important, since the cost of solving (86) is roughly equal to the cost of solving the least-squares problem minimize ∥Pu + q∥. Since the interior-point method converges in very few iterations (typically less than 10), this allows us to conclude that the cost of solving (83) is roughly equal to the cost of 10 least-squares problems of the same dimension, in spite of the fact that we introduced r new variables to cast the problem as an LP. (c) Test your code on randomly generated P and q. Plot the duality gap (on a logarithmic scale) versus the iteration number for a few examples and include a typical plot with your solutions. Exercise 100. Consider the optimization problem minimize m P i=1 f(aT i x −bi) where f(u) =      0 \|u\| ≤1 \|u\| −1 1 ≤\|u\| ≤2 2\|u\| −3 \|u\| ≥2. The function f is shown below. u f(u) 1 2 −1 −2 1 The problem data are ai ∈Rn and bi ∈R. (a) Formulate this problem as an LP in inequality form minimize ¯ cT ¯ x subject to ¯ A¯ x ≤¯ b. (87) Carefully explain why the two problems are equivalent, and what the meaning is of any auxiliary variables you introduce. 65 (b) Describe an eﬃcient method for solving the equations ¯ AT D ¯ A∆¯ x = r that arise in each iteration of Mehrotra’s method applied to the LP (87). Here D is a given diagonal matrix with positive diagonal elements, and r is a given vector. Compare the cost of your method with the cost of solving the least-squares problem minimize m P i=1 (aT i x −bi)2. Exercise 101. The most time consuming step in a primal-dual interior-point method for solving an LP minimize cT x subject to Ax ≤b is the solution of linear equations of the form    0 A I AT 0 0 X 0 Z       ∆z ∆x ∆s   =    r1 r2 r3   , where X and Z are positive diagonal matrices. After eliminating ∆s from the last equation we obtain " −D A AT 0 # " ∆z ∆x # = " d f # where D = XZ−1, d = r1 −Z−1r3, f = r2. Describe an eﬃcient method for solving this equation for an LP of the form minimize cT x subject to Px ≤q −1 ≤x ≤1, where P ∈Rm×n is a dense matrix. Distinguish two cases: m ≫n and m ≪n. Exercise 102. A network is described as a directed graph with m arcs or links. The network supports n ﬂows, with nonnegative rates x1, . . . , xn. Each ﬂow moves along a ﬁxed, or pre-determined, path or route in the network, from a source node to a destination node. Each link can support multiple ﬂows, and the total traﬃc on a link is the sum of the rates of the ﬂows that travel over it. The total traﬃc on link i can be expressed as (Ax)i, where A ∈Rm×n is the ﬂow-link incidence matrix deﬁned as Aij = ( 1 ﬂow j passes through link i 0 otherwise. Usually each path passes through only a small fraction of the total number of links, so the matrix A is sparse. 66 Each link has a positive capacity, which is the maximum total traﬃc it can handle. These link capacity constraints can be expressed as Ax ≤b, where bi is the capacity of link i. We consider the network rate optimization problem maximize f1(x1) + · · · + fn(xn) subject to Ax ≤b x ≥0, where fk(xk) = ( xk xk ≤ck (xk + ck)/2 xk ≥ck, and ck > 0 is given. In this problem we choose feasible ﬂow rates xk that maximize a utility function P k fk(xk). (a) Express the network rate optimization problem as a linear program in inequality form. (b) Derive the dual problem and show that it is equivalent to minimize bT z + g1(aT 1 z) + · · · + gn(aT nz) subject to AT z ≥(1/2)1 z ≥0 with variables z ∈Rm, where ak is the kth column of A and gk(y) = ( (1 −y)ck y ≤1 0 y ≥1. (c) Suppose you are asked to write a custom implementation of the primal-dual interior-point method for the linear program in part 1. Give an eﬃcient method for solving the linear equations that arise in each iteration of the algorithm. Justify your method, assuming that m and n are very large, and that the matrix AT A is sparse. 67
17546	https://fiveable.me/key-terms/ap-hug/post-world-war-ii-baby-boom-phenomenon	Post-World War II baby boom phenomenon - (AP Human Geography) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms AP Human Geography Post-World War II baby boom phenomenon 🚜ap human geography review key term - Post-World War II baby boom phenomenon Citation: MLA Definition The Post-World War II baby boom phenomenon refers to the significant increase in birth rates that occurred in many countries, particularly in the United States, following the end of World War II around 1946. This surge in births was influenced by various factors, including returning soldiers, economic prosperity, and changing social norms that encouraged family growth. The baby boom has had lasting impacts on population composition, leading to shifts in demographics, workforce trends, and social policies as this generation aged. 5 Must Know Facts For Your Next Test The baby boom period lasted roughly from 1946 to 1964, during which millions of babies were born, significantly increasing the population. This demographic shift resulted in a large cohort entering schools, leading to increased demand for educational resources and facilities. The economic prosperity of the post-war years provided families with greater financial stability, encouraging higher birth rates during this time. As baby boomers aged, they have influenced various aspects of society, including healthcare needs, housing markets, and retirement systems. The baby boom has led to concerns about an aging population as this generation reaches retirement age, impacting social security and healthcare systems. Review Questions How did the Post-World War II baby boom phenomenon influence educational systems in the following decades? The Post-World War II baby boom phenomenon resulted in a significant increase in the number of children entering schools during the late 1940s through the 1960s. This surge put pressure on educational systems to expand capacity, requiring more teachers, classrooms, and resources. School districts had to adapt to accommodate this large cohort of students, leading to the construction of new schools and modifications in curriculum to meet their needs. Evaluate the economic factors that contributed to the baby boom following World War II and their long-term effects on society. The economic factors that contributed to the baby boom included rising post-war prosperity, increased job opportunities for returning soldiers, and access to affordable housing. These conditions created a sense of stability and optimism, encouraging families to have more children. The long-term effects have been profound, as the baby boom generation has significantly shaped consumer markets, influenced job markets, and created pressures on social services as they transitioned through different life stages. Assess how the aging of the baby boomer generation is reshaping societal structures and policies in contemporary times. The aging of the baby boomer generation is having significant implications for societal structures and policies. As this large demographic cohort moves into retirement age, there is increasing strain on healthcare systems and social security programs due to the growing number of retirees. Additionally, this shift is leading to changes in workforce dynamics as older workers retire, creating labor shortages in certain sectors while also prompting discussions around pension reforms and elder care services. Overall, these transformations challenge policymakers to adapt existing frameworks to meet the needs of an aging population. Related terms Demographic Transition Model: A theoretical model that describes the stages of population growth and decline in relation to economic development and social changes. Fertility Rate: The average number of children born to a woman over her lifetime, often used to measure population growth. Age Structure: The distribution of a population by age groups, which helps understand the potential for future population growth and social support systems. 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17547	https://brighterly.com/math/endpoint/	Math tutors / Knowledge Base / Endpoint in Math – Definition, Formula, Examples, Facts Endpoint in Math – Definition, Formula, Examples, Facts Jo-ann Caballes Updated on January 14, 2024 Table of Contents At Brighterly, we understand how the world of mathematics can seem complex and daunting. It is a language of its own, filled with various concepts and terminologies. But don’t fret! Our mission at Brighterly is to make learning these mathematical concepts not only easier, but fun as well. Today, we’re going to take a deep dive into a foundational geometry concept, the “endpoint”. This simple yet crucial concept forms the backbone of various other concepts in geometry. From line segments and rays to finding lengths and bisecting lines, we’re going to unravel the magic of endpoints in the realm of mathematics, illuminating your learning journey with Brighterly. So, buckle up and get ready to explore this fascinating world of endpoints, equipped with definitions, formulas, examples, and interesting facts! Endpoint in Math The word “endpoint” is derived from the words “end” and “point.” In mathematics, an endpoint often refers to a point that marks the end of a line segment or the starting point of a ray. Imagine you draw a line on a piece of paper, where you start and stop that line, those two points are called endpoints. Simple, isn’t it? Endpoints play a key role in different mathematical concepts, including line segments, rays, and intervals on a number line. Understanding this term will make solving geometric problems a breeze! What Is a Line Segment? A line segment is a part of a line that is bounded by two distinct end points, and contains every point on the line between its endpoints. Think of a stick – it has a starting point and an ending point. This is different from a “line,” which continues infinitely in both directions. What Is a Ray? A ray is a line that starts at a certain point and goes off in a certain direction to infinity. So, a ray has one endpoint where it starts, and then it goes on forever in one direction. Imagine a flashlight; it starts at one point (where the light is produced) and then the light spreads out infinitely. What Is an Endpoint in Geometry? In geometry, an endpoint can have two definitions depending on what it’s referring to. When talking about a line segment, the endpoint refers to either of the two points that mark the end of the line segment. In the case of a ray, the endpoint is where the ray starts before it goes off infinitely in one direction. Finding the Length of a Line Segment Using the Endpoints Calculating the length of a line segment is pretty straightforward when you know the coordinates of the endpoints. You can use the distance formula, derived from the Pythagorean theorem, to find the length of a line segment in a coordinate plane. Bisecting a Line Segment Using Endpoints Bisecting a line segment means dividing it into two equal parts. The midpoint formula is used to find the point that bisects a line segment, and it is the average of the x-coordinates and the y-coordinates of the endpoints. How to Find the Endpoint of a Line Segment? If you know the midpoint and one endpoint of a line segment, you can find the other endpoint. How? It’s as simple as using the midpoint formula in reverse! By treating the midpoint as the average and knowing one endpoint, we can find the other. The Formula to Find the Endpoint: Endpoint Formula The endpoint formula is derived from the midpoint formula. If M(x1, y1) is the midpoint and A(x, y) is the known endpoint, then the coordinates B(x2, y2) of the other endpoint are given by x2 = 2×1 – x and y2 = 2y1 – y. How Do We Name Objects Using the Endpoints? In geometry, we often name line segments and rays by their endpoints. For example, a line segment with endpoints A and B is named segment AB, while a ray with endpoint C going through point D is named ray CD. What is Endpoint Formula? In mathematics, especially geometry, you may come across situations where you know the midpoint of a line segment and one endpoint, and you are required to find the coordinates of the other endpoint. This is where the Endpoint Formula comes into play. The formula is derived from the Midpoint Formula which states that the midpoint M(x1, y1) of a line segment with endpoints A(x, y) and B(x2, y2) is given by the average of the x-coordinates and the y-coordinates of the endpoints, or: x1 = (x + x2) / 2, and y1 = (y + y2) / 2. By rearranging these equations, we can find the coordinates of the unknown endpoint B(x2, y2): x2 = 2×1 – x, and y2 = 2y1 – y. So, if you know the coordinates of the midpoint and one endpoint, you can calculate the coordinates of the other endpoint using the Endpoint Formula! Practice Problems on Endpoint Practicing is the best way to consolidate your understanding of a concept, and endpoints are no different. Let’s dive into a couple of practice problems: Problem: If the midpoint of a line segment is M(3, 4) and one endpoint is A(1, 2), find the coordinates of the other endpoint B. Solution: Using the endpoint formula: x2 = 2×1 – x = 23 – 1 = 5, and y2 = 2y1 – y = 24 – 2 = 6. Therefore, the coordinates of endpoint B are (5, 6). 2. Problem: Given that the midpoint of a line segment is M(-1, 2) and one endpoint is A(-3, 5), find the other endpoint B. Solution: Using the endpoint formula: x2 = 2×1 – x = 2(-1) – (-3) = -1, and y2 = 2y1 – y = 22 – 5 = -1. Therefore, the coordinates of endpoint B are (-1, -1). Conclusion Endpoints are one of those essential tools in the toolkit of mathematics. By defining the boundaries of line segments or directing the path of rays, they play a pivotal role in our understanding of geometry. Whether you’re calculating distances or bisecting lines, endpoints provide the much-needed foothold. At Brighterly, we believe in transforming mathematical complexities into simplified learning experiences. We hope this exploration of endpoints has turned this seemingly abstract concept into something tangible and easy to understand. As you venture further into the world of geometry, you’ll discover that mastering the concept of endpoints has given you the stepping stones to navigate through more advanced concepts. Remember, every expert was once a beginner. Keep practicing and let the world of math unfold its wonders to you! Frequently Asked Questions on Endpoint What is an endpoint in math? An endpoint in mathematics refers to the point that either terminates a line segment or initiates a ray. It is basically a “boundary marker” for line segments and a “starting marker” for rays. What is a line segment? A line segment is a part of a line that has two endpoints. It includes every point on the line that lies between its two endpoints. Think of a line segment as a closed-door corridor – it has a definite beginning and a definite end. What is a ray? A ray, in contrast to a line segment, is a part of a line that has one endpoint and extends indefinitely in one direction. If a line segment is a closed-door corridor, a ray is an open-door corridor – it has a starting point, but no ending point; it goes on forever. What is the endpoint formula? The endpoint formula is a practical tool in geometry that allows you to calculate the coordinates of an unknown endpoint if you know the coordinates of the midpoint and one endpoint of a line segment. The formula is as follows: If M(x1, y1) is the midpoint and A(x, y) is the known endpoint, then the coordinates B(x2, y2) of the other endpoint are given by x2 = 2×1 – x and y2 = 2y1 – y. This formula is derived from the midpoint formula and can be utilized in various geometry problems. Information Sources: Wolfram MathWorld National Council of Teachers of Mathematics (NCTM) The University of Arizona – Math Department Jo-ann Caballes 13 articles As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable. Table of Contents Endpoint in Math What Is a Line Segment? What Is a Ray? What Is an Endpoint in Geometry? Finding the Length of a Line Segment Using the Endpoints Bisecting a Line Segment Using Endpoints How to Find the Endpoint of a Line Segment? The Formula to Find the Endpoint: Endpoint Formula How Do We Name Objects Using the Endpoints? What is Endpoint Formula? Practice Problems on Endpoint Conclusion Frequently Asked Questions on Endpoint What is an endpoint in math? What is a line segment? What is the endpoint formula? Math & reading from 1st to 9th grade Looking for homework support for your child? Choose kid's grade Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Math & reading from 1st to 9th grade Looking for homework support for your child? Book free lesson Related math 26500 in Words The number 26500 is written in words as “twenty-six thousand five hundred”. It’s five hundred more than twenty-six thousand. For instance, if you have twenty-six thousand five hundred stamps, you start with twenty-six thousand stamps and then add five hundred more. Thousands Hundreds Tens Ones 26 5 0 0 How to Write 26500 in Words? […] Read more 110000 in Words We write 110000 in words as “one hundred ten thousand”. It’s ten thousand more than one hundred thousand. If a town has one hundred ten thousand residents, it means it has one hundred ten groups of one thousand residents each. Thousands Hundreds Tens Ones 110 0 0 0 How to Write 110000 in Words? The […] Read more Even Numbers – Definition with Examples Welcome to another enlightening brought to you by Brighterly, the beacon of fun and engaging education for children! Today, we dive into the fascinating world of even numbers. These numbers permeate every aspect of our daily lives, often without us realizing their significance. From the two wheels on your bike to the four legs of […] Read more Close a child’s math gaps with a tutor! Book a free demo lesson with our math tutor and see your kid fill math gaps with interactive lessons Book demo lesson Get full test results See Your Child’s Test Results Enter your name and email to view your child’s test results now! 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17548	https://books.google.com/books/about/Advanced_Strength_and_Applied_Elasticity.html?id=BqZRAAAAMAAJ	Advanced Strength and Applied Elasticity - A. C. Ugural, Saul K. Fenster - Google Books Sign in Hidden fields Try the new Google Books Books Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» ### Get Textbooks on Google Play Rent and save from the world's largest eBookstore. Read, highlight, and take notes, across web, tablet, and phone. Go to Google Play Now » My library My History Advanced Strength and Applied Elasticity A. C. Ugural, Saul K. Fenster Prentice Hall PTR, 2003 - Science - 544 pages This systematic exploration of real-world stress analysis has been completely revised and updated to reflect state-of-the-art methods and applications now in use throughout the fields of aeronautical, civil, and mechanical engineering and engineering mechanics. Distinguished by its exceptional visual interpretations of the solutions, it offers an in-depth coverage of the subjects for students and practicing engineers. The authors carefully balance comprehensive treatments of solid mechanics, elasticity, and computer-oriented numerical methods. In addition, a wide range of fully worked illustrative examples and an extensive problem sets-many taken directly from engineering practice-have been incorporated. Key additions to the Fourth Edition of this highly acclaimed textbook are materials dealing with failure theories, fracture mechanics, compound cylinders, numerical approaches, energy and variational methods, buckling of stepped columns, common shell types, and more. Contents include stress, strain and stress-strain relations, problems in elasticity, static and dynamic failure criteria, bending of beams and torsion of bars, finite difference and finite element methods, axisymmetrically loaded members, beams on elastic foundations, energy methods, elastic stability, plastic behavior of materials, stresses in plates and shells, and selected references to expose readers to the latest information in the field. More » From inside the book Contents Analysis of Stress 1 Strain and StressStrain Relations 51 TwoDimensional Problems in Elasticity 95 Copyright 15 other sections not shown Other editions - View all Advanced Strength and Applied Elasticity Ansel C. Ugural,Saul K. Fenster Limited preview - 2003 Common terms and phrases angle of twistappliedaxesaxialaxisbeambendingbending momentbody forcesboundary conditionsc₁cantilevercentroidcolumncompressionconstantcoordinatescross sectioncurvecylinderdeflectiondeformationDeterminediameterdiskdisplacementelasticequationsequilibriumExampleexpressionfactor of safetyFIGUREgiven by EqHooke's lawinertialoadM₁materialmaximum shearing stressmaximum stressMohr's circlemoment of inertianodalnormal stressobtainedplane strainplane stressplasticplateprincipal stressesradialradiusrectangularshear centershown in Figsolutionsteelstrain energystress componentsstress distributionsubjectedsubstitutingsurfacetangential stresstensiletensile stressthicknesstionTmaxtorquetorsionyieldzeroΕΙσ₁σασγσθσισοστσχдудх References to this book Stresses in Plates and Shells A. C. Ugural Snippet view - 1981 Mécanique des milieux continus: Thermoélasticité Jean Salençon Limited preview - 2007 All Book Search results » Bibliographic information Title Advanced Strength and Applied Elasticity Mechanical engineering AuthorsA. C. Ugural, Saul K. Fenster Edition 4, illustrated Publisher Prentice Hall PTR, 2003 Original from the University of Michigan Digitized Nov 29, 2007 ISBN 0130473928, 9780130473929 Length 544 pages SubjectsScience › Mechanics › Statics Science / Mechanics / Solids Science / Mechanics / Statics Technology & Engineering / Materials Science / General Technology & Engineering / Mechanical Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
17549	https://www.mathworks.com/help/stats/kruskalwallis.html	kruskalwallis - Kruskal-Wallis test - MATLAB Skip to content MATLAB Help Center MATLAB Help Center Community Learning Get MATLABMATLAB Sign In My Account My Community Profile Link License Sign Out Contact MathWorks Support Visit mathworks.com Search Help Center Help Center Help Center MathWorks MATLAB Help Center MathWorks MATLAB Answers File Exchange Videos Online Training Blogs Cody MATLAB Drive ThingSpeak Bug Reports Community Off-Canvas Navigation Menu Toggle Contents Documentation Home AI and Statistics Statistics and Machine Learning Toolbox Probability Distributions and Hypothesis Tests Hypothesis Tests Statistics and Machine Learning Toolbox ANOVA Analysis of Variance and Covariance kruskalwallis On this page Syntax Description Examples Test Data Samples for the Same Distribution Conduct Follow-up Tests for Unequal Medians Test for the Same Distribution Across Groups Input Arguments x group displayopt Output Arguments p tbl stats More About Kruskal-Wallis Test Version History See Also Documentation Examples Functions Blocks Apps Videos Answers Main Content kruskalwallis Kruskal-Wallis test collapse all in page Syntax p = kruskalwallis(x) p = kruskalwallis(x,group) p = kruskalwallis(x,group,displayopt) [p,tbl,stats] = kruskalwallis(___) Description p = kruskalwallis(x) returns the p-value for the null hypothesis that the data in each column of the matrix x comes from the same distribution, using the Kruskal-Wallis test. The alternative hypothesis is that not all samples come from the same distribution. The Kruskal-Wallis test provides a nonparametric alternative to a one-way ANOVA. For more information, see Kruskal-Wallis Test. example p = kruskalwallis(x,group) returns the p-value for a test of the null hypothesis that the data in each categorical group, as specified by the grouping variable group comes from the same distribution. The alternative hypothesis is that not all groups come from the same distribution. example p = kruskalwallis(x,group,displayopt) returns the p-value of the test and lets you display or suppress the ANOVA table and box plot. example [p,tbl,stats] = kruskalwallis(___) also returns the ANOVA table as the cell array tbl and the structure stats containing information about the test statistics. example Examples collapse all Test Data Samples for the Same Distribution Open in MATLAB Online Copy Code Copy Command Create two different normal probability distribution objects. The first distribution has mu = 0 and sigma = 1, and the second distribution has mu = 2 and sigma = 1. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online pd1 = makedist('Normal'); pd2 = makedist('Normal','mu',2,'sigma',1); Create a matrix of sample data by generating random numbers from these two distributions. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online rng('default'); % for reproducibility x = [random(pd1,20,2),random(pd2,20,1)]; The first two columns of x contain data generated from the first distribution, while the third column contains data generated from the second distribution. Test the null hypothesis that the sample data from each column in x comes from the same distribution. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online p = kruskalwallis(x) p = 3.6896e-06 The returned value of p indicates that kruskalwallis rejects the null hypothesis that all three data samples come from the same distribution at a 1% significance level. The ANOVA table provides additional test results, and the box plot visually presents the summary statistics for each column in x. Conduct Follow-up Tests for Unequal Medians Open in MATLAB Online Copy Code Copy Command Create two different normal probability distribution objects. The first distribution has mu = 0 and sigma = 1. The second distribution has mu = 2 and sigma = 1. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online pd1 = makedist('Normal'); pd2 = makedist('Normal','mu',2,'sigma',1); Create a matrix of sample data by generating random numbers from these two distributions. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online rng('default'); % for reproducibility x = [random(pd1,20,2),random(pd2,20,1)]; The first two columns of x contain data generated from the first distribution, while the third column contains data generated from the second distribution. Test the null hypothesis that the sample data from each column in x comes from the same distribution. Suppress the output displays, and generate the structure stats to use in further testing. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online [p,tbl,stats] = kruskalwallis(x,[],'off') p = 3.6896e-06 tbl=4×6 cell array {'Source' } {'SS' } {'df'} {'MS' } {'Chi-sq' } {'Prob>Chi-sq'} {'Columns'} {[7.6311e+03]} {[ 2]} {[3.8155e+03]} {[ 25.0200]} {[ 3.6896e-06]} {'Error' } {[1.0364e+04]} {} {[ 181.8228]} {0×0 double} {0×0 double } {'Total' } {[ 17995]} {} {0×0 double } {0×0 double} {0×0 double } stats = struct with fields: gnames: [3×1 char] n: [20 20 20] source: 'kruskalwallis' meanranks: [26.7500 18.9500 45.8000] sumt: 0 The returned value of p indicates that the test rejects the null hypothesis at the 1% significance level. You can use the structure stats to perform additional follow-up testing. The cell array tbl contains the same data as the graphical ANOVA table, including column and row labels. Conduct a follow-up test to identify which data sample comes from a different distribution. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online c = multcompare(stats); Note: Intervals can be used for testing but are not simultaneous confidence intervals. Display the multiple comparison results in a table. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online tbl = array2table(c,"VariableNames", ... ["Group A","Group B","Lower Limit","A-B","Upper Limit","P-value"]) tbl=3×6 table Group A Group B Lower Limit A-B Upper Limit P-value _ _ __ __ ____ 1 2 -5.1435 7.8 20.744 0.33446 1 3 -31.994 -19.05 -6.1065 0.0016282 2 3 -39.794 -26.85 -13.906 3.4768e-06 The results indicate that there is a significant difference between groups 1 and 3, so the test rejects the null hypothesis that the data in these two groups comes from the same distribution. The same is true for groups 2 and 3. However, there is not a significant difference between groups 1 and 2, so the test does not reject the null hypothesis that these two groups come from the same distribution. Therefore, these results suggest that the data in groups 1 and 2 come from the same distribution, and the data in group 3 comes from a different distribution. Test for the Same Distribution Across Groups Open in MATLAB Online Copy Code Copy Command Create a vector, strength, containing measurements of the strength of metal beams. Create a second vector, alloy, indicating the type of metal alloy from which the corresponding beam is made. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online strength = [82 86 79 83 84 85 86 87 74 82 ... 78 75 76 77 79 79 77 78 82 79]; alloy = {'st','st','st','st','st','st','st','st',... 'al1','al1','al1','al1','al1','al1',... 'al2','al2','al2','al2','al2','al2'}; Test the null hypothesis that the beam strength measurements have the same distribution across all three alloys. Get Copy Code Block Copy openExample Command Paste command in MATLAB to download and open example files Copy Open in MATLAB Online p = kruskalwallis(strength,alloy,'off') p = 0.0018 The returned value of p indicates that the test rejects the null hypothesis at the 1% significance level. Input Arguments collapse all x — Sample data vector \| matrix Sample data for the hypothesis test, specified as a vector or an m-by-n matrix. If x is an m-by-n matrix, each of the n columns represents an independent sample containing m mutually independent observations. Data Types:single \| double group — Grouping variable numeric vector \| logical vector \| character array \| string array \| cell array of character vectors Grouping variable, specified as a numeric or logical vector, a character or string array, or a cell array of character vectors. If x is a vector, then each element in group identifies the group to which the corresponding element in x belongs, and group must be a vector of the same length as x. If a row of group contains an empty value, that row and the corresponding observation in x are disregarded. NaN values in either x or group are similarly ignored. If x is a matrix, then each column in x represents a different group, and you can use group to specify labels for these columns. The number of elements in group and the number of columns in x must be equal. The labels contained in group also annotate the box plot. Example:{'red','blue','green','blue','red','blue','green','green','red'} Data Types:single \| double \| logical \| char \| string \| cell displayopt — Display option 'on' (default) \| 'off' Display option, specified as 'on' or 'off'. If displayopt is 'on', kruskalwallis displays the following figures: An ANOVA table containing the sums of squares, degrees of freedom, and other quantities calculated based on the ranks of the data in x. A box plot of the data in each column of the data matrix x. The box plots are based on the actual data values, rather than on the ranks. If displayopt is 'off', kruskalwallis does not display these figures. If you specify a value for displayopt, you must also specify a value for group. If you do not have a grouping variable, specify group as []. Example:'off' Output Arguments collapse all p — p-value scalar value in the range [0,1] p-value of the test, returned as a scalar value in the range [0,1]. p is the probability of observing a test statistic that is as extreme as, or more extreme than, the observed value under the null hypothesis. A small value of p indicates that the null hypothesis might not be valid. tbl — ANOVA table cell array ANOVA table of test results, returned as a cell array. tbl includes the sums of squares, degrees of freedom, and other quantities calculated based on the ranks of the data in x, as well as column and row labels. stats — Test data structure Test data, returned as a structure. You can perform follow-up multiple comparison tests on pairs of sample medians by using multcompare, with stats as the input value. More About collapse all Kruskal-Wallis Test The Kruskal-Wallis test is a nonparametric version of classical one-way ANOVA, and an extension of the Wilcoxon rank sum test to more than two groups. The Kruskal-Wallis test is valid for data that has two or more groups. It compares the medians of the groups of data in x to determine if the samples come from the same population (or, equivalently, from different populations with the same distribution). The Kruskal-Wallis test uses ranks of the data, rather than numeric values, to compute the test statistics. It finds ranks by ordering the data from smallest to largest across all groups, and taking the numeric index of this ordering. The rank for a tied observation is equal to the average rank of all observations tied with it. The F-statistic used in classical one-way ANOVA is replaced by a chi-square statistic, and the p-value measures the significance of the chi-square statistic. The Kruskal-Wallis test assumes that all samples come from populations having the same continuous distribution, apart from possibly different locations due to group effects, and that all observations are mutually independent. By contrast, classical one-way ANOVA replaces the first assumption with the stronger assumption that the populations have normal distributions. Version History Introduced before R2006a See Also anova1 \| boxplot \| friedman \| multcompare \| ranksum Topics Grouping Variables Thank you for your feedback! Why did you choose this rating? Submit How useful was this information? Unrated 1 star 2 stars 3 stars 4 stars 5 stars MATLAB Command You clicked a link that corresponds to this MATLAB command: Run the command by entering it in the MATLAB Command Window. Web browsers do not support MATLAB commands. Close Select a Web Site Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: United States. United States United States (English) United States (Deutsch) United States (Français) United States（简体中文） United States (English) You can also select a web site from the following list How to Get Best Site Performance Select the China site (in Chinese or English) for best site performance. 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17550	https://www.maine.gov/doe/sites/maine.gov.doe/files/bulk/prek4me/unit5pdfs/Week%203/16_%20LFOAI%20Opaque_Transparent_Translucent.pdf	Unit 5 Week 3 Let’s Find Out About It: Opaque/Transparent/Translucent Standards: ATL.RPS.PS.2-5 ELA.LS.VAU.PS.1 Materials: ● Moonbear’s Shadow ● a variety of opaque, transparent, translucent objects, including ziplock bag, paper bag, small containers, paint brush ● flashlight ● words written on the board or on a piece of paper, to reference during lesson: opaque, transparent, translucent Vocabulary: ● opaque ● transparent ● translucent ● shadow ● predict ● surface Preparation: Gather materials. Let’s Find Out About It: “In Moonbear’s Shadow, sunlight shone on Bear and created a shadow.” “Here is a (opaque object), a (transparent object) and a (translucent object). What do you predict will happen if we shine light on them? “The (object) is opaque--we see light on its surface, but we can’t see through it. The (object) is transparent-- we can see light shining through it and what’s on the other side of it. The (object) is translucent--we can see light shining through it, but we can’t see clearly what’s on the other side of it.” “Here are other materials. Do you think this container is opaque, transparent or translucent? “Let’s check.” “Yes, this container is opaque, its blocking the light so we can’t see the paintbrush inside.” Show materials. Children respond. Model. Point to words as they come up in discussion. Show opaque container. Children respond. Model sticking the paintbrush into the container (half in and half out). “Here is a clear plastic bag. Do you think the bag is opaque, transparent, or translucent? “Yes, the bag is transparent, we can see the light shining through it so we can see the paintbrush.” “Here is a paper bag. Do you predict this bag is transparent, translucent, or opaque?” “The bag is translucent, so we can see the paintbrush inside but not as clearly as through the clear plastic bag.” “Can you see what is inside this container? Why not?” “If I put this toy in this container, will you be able to see what it is? How do you know this container is translucent?” Show clear plastic bag. Children respond. Stick the paintbrush in the bag. Show a paper sandwich bag. Children respond. Stick the paintbrush in the bag. Show more transparent, opaque, and translucent objects. Children predict. Show an opaque box with a small toy inside. Show a translucent container. Children respond.
17551	https://www.amazon.com/Electronic-Devices-Circuit-Boylestad-Nashelsky/dp/9332542600	Electronic Devices and Circuit Theory: Boylestad Nashelsky: 9789332542600: Amazon.com: Books Skip to Main content About this item About this item About this item Buying options Compare with similar items Videos Reviews Keyboard shortcuts Search opt+/ Cart shift+opt+C Home shift+opt+H Orders shift+opt+O Add to cart shift+opt+K Show/Hide shortcuts shift+opt+Z To move between items, use your keyboard's up or down arrows. .us Delivering to North Cha... 29415 Update location Books Select the department you want to search in Search Amazon EN Hello, sign in Account & Lists Returns& Orders0 Cart Sign in New customer? Start here. 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17552	https://byjus.com/chemistry/chromate/	What is Chromate? Chromates are the salts of chromic acid which contains the chromate anion with the chemical formula CrO42– and usually have an intense yellow color. Chromate is the oxoanion which results from the removal of protons from chromic acid. It is also called chromium oxoanion or divalent inorganic anions. The most commonly used chromate salts are sodium and potassium salts. Table of Contents Chromate Structure Physical Properties of Chromate Chemical Properties of Chromate Uses of Chromate Frequently Asked Questions–FAQs IUPAC name – dioxido(dioxo)chromium \| \| \| --- \| \| CrO42- \| Chromate \| \| Density \| 2.73 g/cm³ \| \| Molecular Weight/ Molar Mass \| 194.1896 g/mol \| \| Conjugate acid‎ \| ‎Chromic acid (H2CrO4) \| \| Oxidation State \| +6 \| \| Chemical Formula \| CrO42- \| Chromate Structure – CrO42- Physical Properties of Chromate – CrO42- \| \| \| --- \| \| Odour \| Odourless \| \| Appearance \| Yellow powder \| \| Valency \| 2 \| \| Solubility \| Generally insoluble in water. \| Chemical Properties of Chromate – CrO42- Chromate ion acts as a strong oxidizing agent in acid solution. In an alkaline solution chromate ion combines with water to form chromium(III) hydroxide. CrO42− + 4 H2O + 3 e− → Cr(OH)3 + 5 OH− Chromate as its potassium salt reacts with barium nitrate to form barium chromate and potassium nitrate. K2CrO4 + Ba(NO3)2 ↔ BaCrO4 + 2KNO3 Uses of Chromate – CrO42- Sodium and potassium salts of chromate are highly corrosive and used in enamels, finishing leather and for rust-proofing metals. Yellow crystals of potassium chromate are used as a pigment in dye and in inks. It is toxic by ingestion. Children crayons are made by wax is usually paraffin and is non-toxic. Industrial crayons are made with lead chromate which is not harmless. Used in chrome plating to prevent corrosion. Frequently Asked Questions What is chromate used for? It is mainly used as an inhibitor of corrosion, as a primer, as a decorative finish, or to maintain electrical conductance. The process takes its name from the chromate present in chromic acid, also known as hexavalent chromium, the most commonly used chemical in the immersion bath phase by which the coating is applied. Q2 Is CrO4 an acid or base? The chromate ion is the predominant species in alkaline solutions but in acidic solutions, dichromate can become the predominant ion. Q3 How do you make a chromate solution? Formulated in yellow chromate. Dissolve the dichromate and the chromic acid first with a little water, then add the nitric and sulphuric and carry the volume with water up to 50 litres to dissolve both. Q4 Are chromate and chromium the same thing? Chrome is an item that is short for chromium, which is applied as an electroplated finish. On the other hand, Chromate is used to characterize a process resistant to corrosion that is known as a conversion finish, rather than an additive finish. Q5 What is potassium chromate used for? It is used as a fungicide in chemical processing, for making pigments for paints and inks, and for producing other chromium compounds. Potassium chromate is a potassium salt in a ratio of 2:1 consisting of potassium and chromate ions. It has a role as both a carcinogenic and an oxidizing agent. Test your Knowledge on Chromate! Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
17553	https://oxfordmedicaleducation.com/clinical-examinations/cardiovascular-examination-questions/	Examinations Fundoscopy Myasthenia Gravis (MG) – Neurological Examination Median nerve lesion – exam presentation Questions & Cases Questions about DVT (Deep Vein Thrombosis) Anaemia – Exam questions and answers Questions about myeloma and MGUS Skills & Procedures Tracheostomy Endotracheal tube (ETT) insertion (intubation) Supraglottic airway (e.g. laryngeal mask airway [LMA], i-Gel) Cardiovascular examination – Questions Common cardiovascular examination exam questions for medical finals, OSCEs and MRCP PACES Click on the the example questions below to reveal the answers Question 1. See below for descriptions of the waveform Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Note that in modern medicine, splitting of heart sounds is rarely clinically relevant. Sadly it still appears in exams. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. \| \| \| \| --- \| Degree of aortic stenosis \| Mean gradient – from Gorlin formula (mmHg) \| Aortic valve area (normal 3-4 cm2) \| \| Mild \| <25 \| >1.5 \| \| Moderate \| 25 – 50 \| 1.0 – 1.5 \| \| Severe \| >50 \| < 1.0 \| \| Critical \| >70 \| < 0.6 \| Question 24. Question 25. Question 26. Question 27. Example: A pacemaker in VVI mode denotes that it paces and senses the ventricle and is inhibited by a sensed ventricular event. The DDD mode denotes that both chambers are capable of being sensed and paced. Question 28. Question 29. Question 30. Note that none of the other treatments (e.g. diuretics and digoxin) have been shown to improve mortality. Further questions about the cardiovascular exam Click here for all the clinical examinations: how to examine, what to look for and how to present your findings ExaminationOSCEsPACESPLABQuestions Related Posts Anaemia – Exam questions and answers What are the abnormalities of the waveform of the JVP? Cranial nerve examination questions – facial nerve (VII) Status Epilepticus – Questions Cord compression – exam presentation Respiratory Examination Checklist Knee Examination – Orthopaedics Abdominal Examination One Response I like this information so much. Thank you for all this. This helps me in making questions .i was always facing the problm of question in firt time in exame now i can see the type of ques like these thank you again Leave a Reply Cancel Reply Your email address will not be published. Comment Name Email Website Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Sign Up! Follow Us! Follow us! Search Popular topics
17554	https://journalofoms.com/assets/paper_upload/paper/JOMS20192R.pdf	Journal of Orissa Mathematical Society ISSN: 0975-2323 Vol. 38, No. 01-02, 2019, 33-51 On Exponentially Convex Functions Muhammad Aslam Noor∗and Khalida Inayat Noor† Abstract In this paper, we deﬁne and introduce some new concepts of the exponentially convex functions. We investigate several properties of the exponentially convex functions and dis-cuss their relations with convex functions. Optimality conditions are characterized by a class of variational inequalities. Several interesting results characterizing the exponentially convex functions are obtained. Results obtained in this paper can be viewed as signiﬁcant improvement of previously known results 1 Introduction Convex functions and convex sets have played an important and fundamental part in the development of various ﬁelds of pure and applied sciences. Convexity theory describes a broad spectrum of very interesting developments involving a link among various ﬁelds of mathematics, physics, economics and engineering sciences. Some of these developments have made mutually enriching contacts with other ﬁelds. Ideas explaining these concepts led to the developments of new and powerful techniques to solve a wide class of linear and nonlinear problems. The development of convexity theory can be viewed as the simultaneous pursuit of two different lines of research. On the one hand, it reveals the fundamental facts ∗Mathematics Department, COMSATS University Islamabad, Islamabad, Pakistan, Email: noormaslam@gmail.com †Mathematics Department, COMSATS University Islamabad, Islamabad, Pakistan, Email: khalidan@gmail.com 34 M. A. Noor and K. I. Noor on the qualitative behaviour of solutions (regarding its existence, uniqueness and regularity) to important classes of problems; on the other hand, it also enables us to develop highly efﬁcient and powerful new numerical methods to solve nonlinear problems, see [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]. In recent years, various extensions and generalizations of convex functions and convex sets have been considered and studied using innovative ideas and techniques. It is known that more accurate and in-equalities can be obtained using the logarithmically convex functions than the convex functions. Closely related to the log-convex functions, we have the concept of exponentially convex(concave) functions, the origin of exponentially convex functions can be traced back to Bernstein . Avriel introduced and studied the concept of r-convex functions. For further properties of the r-convex functions, see Zhao et al and the references therein. which have important applications in information theory, big data analysis, machine learning and statistic. See, for example, [2, 3, 18, 19, 21, 22, 23, 24] and the references therein. Motivated and inspired by the ongoing research in this interesting, applicable and dynamic ﬁeld, we again consider the concept of exponentially convex functions. We discuss the basic properties of the ex-ponentially convex functions. It is has been shown that the exponentially convex(concave) have nice nice properties which convex functions enjoy. Several new concepts have been introduced and investigated. We show that the local minimum of the exponentially convex functions is the global minimum. The optimal conditions of the differentiable exponentially convex functions can be characterized by a class of variational inequalities, which is itself an interesting outcome of our main results. The difference (sum) of the exponentially convex function and exponentially afﬁne convex function is again a exponentially convex function. The ideas and techniques of this paper may be starting point for further research in these areas. Journal of Orissa Mathematical Society 35 2 Preliminary Results Let K be a nonempty closed set in a real Hilbert space H. We denote by ⟨·,·⟩and ∥· ∥be the inner product and norm, respectively. Let F : K →R be a continuous function. Deﬁnition 2.1. .The set K in H is said to be convex set, if u+t(v−u) ∈K, ∀u,v ∈K,t ∈[0,1]. Deﬁnition 2.2. A function F is said to be convex, if F((1−t)u+tv) ≤(1−t)F(u)+tF(v), ∀u,v ∈K, t ∈[0,1]. (2.1) We now consider the concept of the exponentially convex function, which is mainly due to Noor and Noor [14, 15] and Rashid et al[21, 22] as: Deﬁnition 2.3. A function F is said to be exponentially convex function, if eF((1−t)u+tv) ≤(1−t)eF(u) +teF(v), ∀u,v ∈K, t ∈[0,1]. We remark that Deﬁnition 2.5 can be rewritten in the following equivalent way, which is due to Antczak . Deﬁnition 2.4. A function F is said to be exponentially convex function, if F((1−t)a+tb) ≤log[(1−t)eF(a) +teF(b)], ∀a,b ∈K, t ∈[0,1]. (2.2) A function is called the exponentially concave function f, if −f is exponentially convex function. It is obvious that two concepts are equivalent. This equivalent have been used to discuss various aspects of the exponentially convex functions. It is worth mentioning that one can also deduce the concept of exponentially convex functions from r-convex functions, which were considered by Avriel and Bern-stein . 36 M. A. Noor and K. I. Noor For the applications of the exponentially convex functions in the mathematical programming and in-formation theory, see Antczak ,Alirezaei and Mathar and Pal et al.. For the applications of the exponentially concave function in the communication and information theory, we have the following example. Example : The error function er f(x) = 2 √π Z x 0 e−t2dt, becomes an exponentially concave function in the form er f(√x), x ≥0, which describes the bit/symbol erorr probability of communication systems depending on the square root of the underlying signal-to-noise ratio. This shows that the exponentially concave functions can play important part in communica-tion theory and information theory. Deﬁnition 2.5. A function F is said to be exponentially afﬁne convex function, if eF((1−t)u+tv) = (1−t)eF(u) +teF(v), ∀u,v ∈K, t ∈[0,1]. Deﬁnition 2.6. The function F on the convex set K is said to be exponentially quasi convex, if eF(u+t(v−u)) ≤max{eF(u),eF(v)}, ∀u,v ∈K,t ∈[0,1]. From the above deﬁnitions, we have eF(u+t(v−u)) ≤ (1−t)eF(u) +teF(v)) ≤ max{eF(u),eF(v)}. This shows that every exponentially convex function is a exponentially quasi-convex function. However, the converse is not true. Journal of Orissa Mathematical Society 37 Let K = I = [a,b] be the interval. We now deﬁne the exponentially convex functions on I. Deﬁnition 2.7. Let I = [a,b]. Then F is exponentially convex function, if and only if, 1 1 1 a x b eF(a) eF(x) eF(b) ≥0; a ≤x ≤b. One can easily show that the following are equivalent: 1. F is exponentially convex function. 2. eF(x) ≤eF(a) + eF(b)−eF(a) b−a (x−a). 3. eF(x)−eF(a) x−a ≤eF(b)−eF(a) b−a . 4. (b−x)eF(a) +(a−b)eF(x) +(x−a)eF(b)) ≥0. 5. F(a) (b−a)(a−x) + eF(x) (x−b)(a−x) + eF(b (b−a)(x−b) ≤0, where x = (1−t)a+tb ∈[0,1]. 3 Main Results In this section, we consider some basic properties of generalized strongly convex functions. Theorem 3.1. Let F be a strictly exponentially convex function. Then any local minimum of F is a global minimum. Proof. Let the exponentially convex function F have a local minimum at u ∈K. Assume the contrary, that is, F(v) < F(u) for some v ∈K. Since F is exponentially convex, so eF(u+t(v−u)) < teF(v) +(1−t)eF(u), for 0 < t < 1. 38 M. A. Noor and K. I. Noor Thus eF(u+t(v−u)) −eF(u) < t[eF(v) −eF(u)] < 0, from which it follows that eF(u+t(v−u)) < eF(u), for arbitrary small t > 0, contradicting the local minimum. Theorem 3.2. If the function F on the convex set K is exponentially convex, then the level set Lα = {u ∈ K : eF(u) ≤α, α ∈R} is a convex set. Proof. Let u,v ∈Lα. Then eF(u) ≤α and eF(v) ≤α. Now, ∀t ∈(0,1), w = v+t(u−v) ∈K, since K is a convex set. Thus, by the exponentially convexity of F, we have Fe(v+t(u−v)) ≤ (1−t)eF(v) +teF(u) ≤ (1−t)α +tα = α, from which it follows that v+t(u−v) ∈Lα Hence Lα is convex set. Theorem 3.3. The function F is exponentially convex, if and only if epi(F) = {(u,α) : u ∈K : eF(u) ≤α,α ∈R} is a convex set. Proof. Assume thatF is exponentially convex. Let (u,α), (v,β) ∈epi(F). Then it follows that eF(u) ≤ α and eF(v) ≤β. Thus, ∀t ∈[0,1], u,v ∈K, we have eF(u+t(v−u)) ≤ (1−t)eF(u) +teF(v) ≤ (1−t)α +tβ, which implies that (u+t(v−u),(1−t)α +tβ) ∈epi(F). Journal of Orissa Mathematical Society 39 Thus epi(F) is a convex set. Conversely, let epi(F) be a convex set. Let u,v ∈K. Then (u,eF(u)) ∈epi(F) and (v,eF(v)) ∈epi(F). Since epi(F) is a convex set, we must have (u+t(v−u),(1−t)eF(u) +teF(v)) ∈epi(F), which implies that eF(u+t(v−u)) ≤(1−t)eF(u) +teF(u). This shows that F is exponentially convex function. Theorem 3.4. The function F is exponentially quasi convex, if and only if, the level set Lα = {u ∈K,α ∈ R : eF(u) ≤α} is a convex set. Proof. Let u,v ∈Lα. Then u,v ∈K and max(eF(u),eF(v)) ≤α. Now for t ∈(0,1),w = u+t(v−u) ∈K, We have to prove that u+t(v−u) ∈Lα. By the exponentially quasi convexity of F, we have eF(u+t(v−u)) ≤max(eF(u),eF(v)) ≤α, which implies that u+t(v−u) ∈Lα, showing that the level set Lα is indeed a convex set. Conversely, assume that Lα is a convex set. Then for any u,v ∈Lα,t ∈[0,1], u +t(v −u) ∈Lα. Let u,v ∈Lα for α = max(eF(u),eF(v)) and eF(v) ≤eF(u). Then from the deﬁnition of the level set Lα, it follows that eF(u+t (v,u)) ≤max(eF(u),eF(v)) ≤α. Thus F is an exponentially quasi convex function. This completes the proof. Theorem 3.5. Let F be an exponentially convex function.. Let µ = infu∈K F(u). Then the set E = {u ∈ K : eF(u) = µ} is a convex set of K. If F is strictly exponentially , then E is a singleton. 40 M. A. Noor and K. I. Noor Proof. Let u,v ∈E. For 0 < t < 1, let w = u+t(v−u). Since F is a exponentially convex function, then F(w) = eF(u+t(v−u)) ≤ (1−t)eF(u) +teF(v) = tµ +(1−t)µ = µ, which implies that to w ∈E. and hence E is a convex set. For the second part, assume to the contrary that F(u) = F(v) = µ. Since K is a convex set, then for 0 < t < 1,u +t(v −u) ∈K. Further, since F is strictly exponentially convex, eF(u+t(v−u)) < (1−t)eF(u) +teF(v) = (1−t)µ +tµ = µ. This contradicts the fact that µ = infu∈K F(u) and hence the result follows. Theorem 3.6. If F is an exponentially convex function such that eF(v) < eF(u),∀u,v ∈K, then F is a strictly exponentially quasi convex function. Proof. By the exponentially convexity of the function F, ∀u,v ∈K,t ∈[0,1], we have eF(u+t(v−u)) ≤(1−t)eF(u) +teF(v) < eF(u), since eF(v) < eF(u), which shows that the function F is strictly exponentially quasi convex. We now derive some properties of the differentiable exponentially convex functions. Theorem 3.7. Let F be a differentiable function on the convex set K. Then the function F is exponentially convex function, if and only if, eF(v) −eF(u) ≥⟨eF(u)F′(u),v−u⟩, ∀v,u ∈K. (3.1) Journal of Orissa Mathematical Society 41 Proof. Let F be a exponentially convex function. Then eF(u+t(v−u)) ≤(1−t)eF(u) +teF(v), ∀u,v ∈K, which can be written as eF(v) −eF(u) ≥{eF(u+t(v−u)) −eF(u) t }. Taking the limit in the above inequality as t →0 , we have eF(v) −eF(u) ≥⟨eF(u)F′(u),v−u)⟩, which is (3.1), the required result. Conversely, let (3.1) hold. Then ∀u,v ∈K,t ∈[0,1], vt = u+t(v−u) ∈K, we have eF(v) −eF(vt) ≥ ⟨eF(vt)F′(vt),v−vt)⟩ = (1−t)⟨eF(vt)F′(vt),v−u⟩. (3.2) In a similar way, we have eF(u) −eF(vt) ≥ ⟨eF(vt)F′(vt),u−vt)⟩ = −t⟨eF(vt)F′(vt),v−u⟩. (3.3) Multiplying (3.2) by t and (3.3) by (1−t) and adding the resultant, we have eF(u+t(v−u)) ≤(1−t)eF(u) +teF(v), showing that F is a exponentially convex function. 42 M. A. Noor and K. I. Noor Remark 3.1. From (3.1), we have eF(v)−F(u) −1 ≥⟨F′(u),v−u⟩, ∀v,u ∈K, which can be written as F(v)−F(u) ≥log{1+⟨F′(u),v−u⟩} ∀v,u ∈K, (3.4) Changing the role of u and v in (3.4), we also have F(u)−F(v) ≥log{1+⟨F′(v),u−v⟩} ∀v,u ∈K, (3.5) Adding (3.4) and (3.5), we have ⟨F′(u)−F′(v),u−v⟩≥(⟨F′(u)u−v⟩)(⟨F′(v),u−v⟩) which express the monotonicity of the differential F′(.) of the exponentially convex function. Theorem 3.7 enables us to introduce the concept of the exponentially monotone operators, which appears to be new ones. Deﬁnition 3.1. The differential F′(.) is said to be exponentially monotone, if ⟨eF(u)F′(u)−eF(v)F′(v),u−v⟩≥0, ∀u,v ∈H. Deﬁnition 3.2. The differential F′(.) is said to be exponentially pseudo-monotone, if ⟨eF(u)F′(u),v−u⟩≥0, ⇒⟨eF(v)F′(v),v−u⟩≥0, ∀u,v ∈H. From these deﬁnitions, it follows that exponentially monotonicity implies exponentially pseudo-monotonicity, but the converse is not true. Theorem 3.8. Let F be differentiable on the convex set K. Then (3.1) holds, if and only if, F′ satisﬁes ⟨eF(u)F′(u)−eF(v)F′(v),u−v⟩≥0, ∀u,v ∈K. (3.6) Journal of Orissa Mathematical Society 43 Proof. Let F be a exponentially convex function on the convex set K. Then, from Theorem 3.1, we have eF(v) −eF(u) ≥⟨eF(u)F′(u),v−u⟩, ∀u,v ∈K. (3.7) Changing the role of u and v in (3.7), we have eF(u) −eF(v) ≥⟨eF(v)F′(v),u−v)⟩, ∀u,v ∈K. (3.8) Adding (3.7) and (3.8), we have ⟨eF(u)F′(u)−eF(v)F′(v),u−v⟩≥0, which shows that F′ is exponentially monotone. Conversely, from (3.6), we have ⟨eF(v)F′(v),u−v⟩≤⟨eF(u)F′(u),u−v)⟩. (3.9) Since K is an convex set, ∀u,v ∈K, t ∈[0,1] vt = u+t(v−u) ∈K. Taking v = vt in (3.9), we have ⟨eF(vt)F′(vt),u−vt⟩ ≤ ⟨eF(u)F′(u),u−vt⟩ = −t⟨eF(u)F′(u),v−u⟩, which implies that ⟨eF(vt)F′(vt),v−u⟩≥⟨eF(u)F′(u),v−u⟩. (3.10) Consider the auxiliary function g(t) = eF(u+t(v−u)), from which, we have g(1) = eF(v), g(0) = eF(u). 44 M. A. Noor and K. I. Noor Then, from (3.10), we have g′(t) = ⟨eF(vt)F′(vt,v−u⟩≥⟨eF(u)F′(u),v−u⟩. (3.11) Integrating (3.11) between 0 and 1, we have g(1)−g(0) = Z 1 0 g′(t)dt ≥⟨eF(u)F′(u),v−u⟩. Thus it follows that eF(v) −eF(u) ≥⟨eF(u)F′(u),v−u⟩, which is the required (3.1). We now give a necessary condition for exponentially pseudo-convex function. Theorem 3.9. Let F′ be exponentially pseudomonotone. Then F is a exponentially pseudo-convex func-tion. Proof. Let F′ be a exponentially pseudomonotone. Then, ∀u,v ∈K, ⟨eF(u)F′(u),v−u⟩≥0. implies that ⟨eF(v)F′(v),v−u⟩≥0. (3.12) Since K is an convex set, ∀u,v ∈K, t ∈[0,1], vt = u+t(v−u) ∈K. Taking v = vt in (3.12), we have ⟨eF(vt)F′(vt),v−u⟩≥0. (3.13) Consider the auxiliary function g(t) = eF(u+t(v−u)) = eF(vt), ∀u,v ∈K,t ∈[0,1], Journal of Orissa Mathematical Society 45 which is differentiable, since F is differentiable function. Then, using (3.13), we have g′(t) = ⟨eF(vt)F′(vt),v−u)⟩≥0. Integrating the above relation between 0 to 1, we have g(1)−g(0) = Z 1 0 g′(t)dt ≥0, that is, eF(v) −eF(u) ≥0, showing that F is a exponentially pseudo-convex function. Deﬁnition 3.3. The function F is said to be sharply exponentially pseudo convex, if there exists a constant µ > 0 such that ⟨eF(u)F′(u),v−u⟩ ≥ 0 ⇒ F(v) ≥ eF(v+t(u−v)), ∀u,v ∈K,t ∈[0,1]. Theorem 3.10. Let F be a sharply exponentially pseudo convex function on K. Then ⟨eF(v)F′(v),v−u⟩≥0, ∀u,v ∈K. Proof. Let F be a sharply exponentially pesudo convex function on K. Then eF(v) ≥eF(v+t(u−v)), ∀u,v ∈K,t ∈[0,1]. from which we have 0 ≤ lim t→0{eF(v+t(u−v)) −eF(v) t } = ⟨eF(v)F′(v),v−u⟩, the required result. 46 M. A. Noor and K. I. Noor Deﬁnition 3.4. A function F is said to be a pseudo convex function, if there exists a strictly positive bifunction b(.,.), such that eF(v) < eF(u) ⇒ eF(u+t(v−u)) < eF(u) +t(t −1)b(v,u),∀u,v ∈K,t ∈[0,1]. Theorem 3.11. If the function F is exponentially convex function such that eF(v) < eF(u), then the function F is exponentially pseudo convex. Proof. Since eF(v) < eF(u) and F is exponentially convex function, then ∀u,v ∈K, t ∈[0,1], we have eF(u+t l(v,u)) ≤ eF(u) +t(eF(v) −eF(u)) < eF(u) +t(1−t)(eF(v) −eF(u)) = eF(u) +t(t −1)(eF(u) −eF(v))) < eF(u) +t(t −1)b(u,v), where b(u,v) = eF(u) −eF(v) > 0, the required result. This shows that the function F is exponentially convex function. We now discuss the optimality condition for the differentiable exponentially convex functions, which is the main motivation of our next result. Theorem 3.12. Let F be a differentiable exponentially convex function. Then u ∈K is the minimum of the function F, if and only if, u ∈K satisﬁes the inequality ⟨eF(u)F′(u),v−u⟩≥0, ∀u,v ∈K. (3.14) Proof. Let u ∈K be a minimum of the function F. Then F(u) ≤F(v),∀v ∈K. Journal of Orissa Mathematical Society 47 from which, we have eF(u) ≤eF(v),∀v ∈K. (3.15) Since K is a convex set, so, ∀u,v ∈K, t ∈[0,1], vt = (1−t)u+tv ∈K. Taking v = vt in (3.15), we have 0 ≤ lim t→0{eF(u+t(v−u)) −eF(u) t } = ⟨eF(u)F′(u),v−u⟩. (3.16) Since F is differentiable exponentially convex function, so eF(u+t(v−u)) ≤eF(u) +t(eF(v) −eF(u), u,v ∈K,t ∈[0,1], from which, using (3.16), we have eF(v) −eF(u) ≥ lim t→0{eF(u+t(v−u)) −eF(u) t } = ⟨eF(u)F′(u),v−u⟩≥0, from which , we have eF(v) −eF(u) ≥0, which implies that F(u) ≤F(v), ∀v ∈K. This shows that u ∈K is the minimum of the differentiable exponentially convex function, the required result. 48 M. A. Noor and K. I. Noor Remark 3.2. The inequality of the type (3.14) is called the exponentially variational inequality and appears to be new one. For the applications, formulations, numerical methods and other aspects of variational inequalities, see Noor [12, 13]. We now show that the difference of exponentially convex function and exponentially afﬁne convex function is again an exponentially convex function. Theorem 3.13. Let f be a exponentially afﬁne convex function. Then F is a exponentially convex function, if and only if, g = F −f is a exponentially convex function. Proof. Let f be exponentially afﬁne convex function. Then ef((1−t)u+tv) = (1−t)ef(u) +tef(v), ∀u,v ∈K, t ∈[0,1]. (3.17) From the exponentially convexity of F, we have eF((1−t)u+tv) ≤(1−t)eF(u) +teF(v), ∀u,v ∈K, t ∈[0,1]. (3.18) From (3.17 ) and (3.18), we have eF((1−t)u+tv) −e f((1−t)u+tv) ≤(1−t)(eF(u) −ef(u))+t(eF(v) −ef(v)), (3.19) from which it follows that eg((1−t)u+tv) = eF((1−t)u+tv) −ef((1−t)u+tv) ≤ (1−t)(eF(u) −ef(u))+t(eF(v) −ef(v)), which show that g = F −f is an exponentially convex function. The inverse implication is obvious. Journal of Orissa Mathematical Society 49 4 Conclusion In this paper, we have introduced and studied a new class of convex functions, which is called the exponentially convex function. It has been shown that exponentially convex functions enjoy several properties which convex functions have. We have shown that the minimum of the differentiable expo-nentially convex functions can be characterized by a new class of variational inequalities, which is called the exponential variational inequality. One can explore the applications of the exponentially variational inequalities This may stimulate further research. References 1. N. I. 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17555	https://www.ijord.com/index.php/ijord/article/download/1677/940/8305	International Journal of Research in Dermatology \| May -June 20 23 \| Vol 9 \| Issue 3 Page 123 International Journa l of Research in Dermatology Reddy PG et al. Int J Res Dermatol . 20 23Ma y;9(3): 123 -125 Original Research Article Encountering steroid treatment induced tinea incognito: a case report Pallavi Gaddam Reddy 1, Maryam Maqsood 2 INTRODUCTION The term dermatophytosis or tinea reflects a wide range of fungal infections that can affect the skin, hair, and nails. Tinea infections can be differentiated by the area of the body affected, and the degree of the infections caused. Patients with decrease d immune response, related co - morbidities like diabetes mellitus, old age as well as children, patients with altered or poor circulation, and corticosteroid use possess more risk of developing a dermatophyte infection. When these infections are superficia l in nature, they can be symptomatically diagnosed based on their typical presentation signs. Nowadays, there have been numerous reports regarding the presentations of tinea being quite atypical like clinical signs and symptoms, which more often than not may lead to mis diagnosis or a late diagnosis. 1-3 Tinea incognito refers to a type of dermatophyte infection that affects the skin, and its signs and symptoms upon presentation tend to appear permutated due to the misuse of corticostero ids or calcineurin inhibitors. 4,5 The emergence of tinea incognita may resemble the clinical presentations of rosacea, eczema, lichen, psoriasis, lupus, viral infections, impetigo, or seborrheic dermatitis. Moreover, it has been mostly seen to manifest itself in patients wit h previous skin conditions such as psoriasis which was managed using mostly topical medication regimens. Mostly prevailing amongst psoriatic patients, the onychomycosis and tinea pedis often present as tinea incognito. The cause of their atypical presentat ion is mostly attributed to the DOI: http s://dx.doi.org /10.18203/issn.2455 -4529.IntJResDermatol20231165 ABSTRACT Tinea incognito refers to a type of dermatophyte infection that affects the skin, and its signs and symptoms upon presentation tend to appear permutated due to the prior use of immunosuppressants, corticosteroids or calcineurin inhi bitors. We present a 42 -year -old Asian (Indian) male patient with scaly erythematous rashes, mimicking annular erythema, and developing post -corticosteroid usage after an elective hair transplant procedure. The results of the biopsy reported that erythema annulare centrifugum (EAC) was absent, and the sample was suggestive of potential infective folliculitis. By day 3 of the presentation, with fungal stain tested positive for the presence of fungal infection. Based o n this, the final diagnosis of t inea inco gnito was made. The final treatment prescribed for tinea incognito was an anti -fungal tablet of Itraconazole at a dose of 100 mg twice a day for 4 weeks, topical luliconazole, ciclopirox and anti -histamines for itching. Topical corticosteroids can change t he clinical appearance of tinea by reducing erythema and scaling while enabling the fungus to grow freely without presenting the typical clinical indications of tinea . Practitioners should follow patients on corticosteroid treatment to alert them to potent ial cutaneous problems due to possible fungal infections, whenever warranted . Keywords: Tinea incognito, Anti -fungal therapy, Corticosteroids, Immunosuppressant 1 Department of Dermatology and Venereal Diseases, 2Department of Infectious Diseases, Apollo Hospitals, Jubilee Hills, Hyderabad, Telangana, India Received: 07 Febr uary 2023 Accepted: 03 March 2023 Correspondence: Dr. Maryam Maqsood , E-mail: maryammaqsood99@gmail.com Copyright: © the author(s), publisher and licensee Medip Academy . This is an open -access article distributed under the terms of the Creative Commons Attribution Non -Commercial License, which permits unrestricted non -commercial use, distribution, and reproduction in any medium, provided the original work is properly cit ed. Reddy PG et al. Int J Res Dermatol . 20 23 Ma y;9( 3): 123 -125 International Journal of Research in Dermatology \|May -June 20 23 \| Vol 9 \| Issue 3 Page 124 administration of immunosuppressive drugs and corticosteroids as a part of the drug regimen used for the treatment of psoriasis. 6 Despite the fact that the interrelationship between a few non -dermatological diseases and psor iasis is well elucidated scientifically, there haven’t been extensive reports on dermatological comorbidities themselves and their rel ationship with tinea incognito. 7 CA SE REPORT A 42 -year -old male presented to the dermatology department with the chief co mplaints of red eruptions with itching on the neck, anterior and posterior aspects of the trunk, axillary region, groin, and buttocks, extending towards distal aspects of all four limbs since 2 days. The patient did not have any known allergies to medicine s or food; and had no recent contact with known allergens, travel, or known sick contacts. The patient is a known case of bronchial asthma on deriphyllin, using inhalers for budesonide and levosalbutamol since more than a year ago, for the management of hi s symptoms. The patient has been recently diagnosed with hypertension for which he took tablet telmisartan for one day, followed by a change to tablet olmesartan medoxomil 10mg taken for 4 days prior to presenting with rashes. His HbA1c was recorded as 5.7 %. Figur e 1: Eruptions with itching as seen on the back . History of the presentation includes the patient electively undergoing a hair transplant procedure at another facility 7 days back, after which he was prescribed the following medications: oral pantoprazole 40 mg once a day for 5 days, Oral Amoxicillin and clavulanic acid 625 mg twice a day for 5 days, an oral tablet of a fixed -dose formulation containing diclofenac 50 mg + paracetamol 325 mg + serratiopeptidase 15 mg to be taken SOS, oral cortic osteroid betamethasone sodium phosphate 2 mg, pre and pro -biotic capsules fortified with lactobacillus to be taken once daily, and topical formulation of mupirocin for external application. A hair vitamin tablet consisting of biotin, amino acids, minerals, vitamins, soya isoflavones, and grape seed extracts were given for 6 months, along with topical minoxidil 5%. Betadine solution and normal saline were to be used for wound care. This regimen was followed by the patient for a period of 7 days prior to pres enting with the complaint. Because the patient’s scaly skin lesions were erythematous and patches exhibited ring -like morphology with a few having central clearing, with the presence of pustules or vesicles, the working diagnosis was presumed to be annular erythema. Our differential diagnosis included a flare of psoriasis with annular lesions, tinea corporis or tinea incognita, and erythema annulare. Blood tests did not reveal any abnormalities. A sample of the skin was sent for examination via biopsy, for a definitive diagnosis. The other working differential diagnoses included erythema annulare centrifugum (EAC), erythema perstans, and erythema multiforme. A fungal stain was also conducted on the scaling obtained superficially from the annular lesions in o rder to rule out the possibility of fungal infection. Meanwhile while awaiting biopsy results, the patient was empirically initiated on tablet hydroxyzine hydrochloride 25 mg HS for 2 weeks, tablet fexofenadine hydrochloride 180mg twice a day for 1 week, t ablet ranitidine 300 mg before breakfast for 10 days, tablet prednisolone 30 mg after breakfast for 10 days, cream halobetasone (0.05% w/w) + fusidic acid (2% w/w) topical application on the affected area for 15 days, and calamine lotion for local application on SOS basis. The infection prophylaxis for biopsy sutures was tablet azithromycin 250mg twice daily to be taken for 5 days. The results of the biopsy reported that EAC was absent, and the sample was suggestive of pote ntial folliculitis. By day 3 of the presentation, the fungal stain tested positive for the presence of fungal elements. Based on this, the final diagnosis of tinea incognito was made. The corticosteroid dose was prescribed at a 2 -day dose reduction/ taper ing regimen and the patient was advised to stop the use of halobetasone and fusidic acid. The final treatment prescribed for tinea incognito was an anti - fungal tablet of itraconazole at a dose of 100 mg twice a day for 4 weeks, topical luliconazole lotion (1% w/w) for external application at night, a ciclopirox (1% w/w) ointment to be applied during the day for 4 weeks. Oral administration of 25 mg hydroxyzine was advised to be continued once daily for the following 2 weeks. The patient improved on treatmen t and clinical cure was observed on subsequent follow -ups . DISCUSSION Tinea incognito is difficult to diagnose since it is similar to other non -infectious skin diseases and rashes. To avoid opportunistic cutaneous infections, patients on corticosteroid med ication must be closely monitored. Misdiagnosis or delayed diagnosis of tinea can pose a practical and epidemiologic issue. Reddy PG et al. Int J Res Dermatol . 20 23 Ma y;9( 3): 123 -125 International Journal of Research in Dermatology \|May -June 20 23 \| Vol 9 \| Issue 3 Page 125 When patients on any form of immunosuppressive agent present to clinics with increasing widespread scaly erythematous skin eruptions and pruritus, concurrent tinea incognito should be investigated, and a potassium hydroxide (KOH) study and/or biopsy should be sent to obtain a specific diagnosis. The usage of corticosteroids can change the clinical appearance of tinea by reducing eryth ema and scaling while enabling the fungus to grow freely without presenting the typical clinical indications of tinea. Tinea incognito has a wide range of clinical symptoms, including rosacea, eczema, lichen, psoriasis, lupus erythematosus, viral infection s, neurodermatitis, impetigo, contact dermatitis, and seborrheic dermatitis; and hence must be carefully evaluated to avoid exacerbation o f the main underlying disease. 8,9 CONCLUSION Dermatophyte infections are prevalent these days due to the use of cortic osteroids or immunosuppressive therapies which lead them to have atypical presentations, often about a predisposing factor. To recapitulate, an adult male patient with a history of steroid use with regards to hair transplantation procedure presented with annular lesions that were diagnosed as tinea incognita when the mycology testing was positive for fungal infection, and the skin biopsy was negative for suspected EAC. The antifungal treatment was administered, and the lesions resolved . Funding: No funding sources Conflict of interest: None declared Ethical approval: Not required REFERENCES Ansar A, Farshchian M, Nazeri H, Ghiasian SA. Clinico -epidemiological and mycological aspects of tinea incognito in Iran: a 16 -year study. Japanese J Med Mycol. 2011;52:25 -32. Kim WJ, Kim TW, Mun JH, Song M, Kim HS, Ko HC, et al. Tinea incognito in Korea and its risk factors: nine -year multicenter survey. J Korean Med Sci. 2013;28:145 -51. Atzori L, Pau M, Aste N, Aste N. Dermatophyte infections mimicking other skin diseases: a 154 - person case survey of tinea atypica in the district of Cagliari (Italy). Int J Dermatol. 2012;51:410 -5. Crawford KM, Bostrom P, Russ B, Boyd J. Pimecrolimus -induced tinea incognito. Skinmed. 2004;3:352 -3. Rallis E, Koumantaki -Mathioudaki E. Pimecrolimus induced tinea incognito masquerading as intertriginous psoriasis. Mycoses. 2008;51:71 -3. Đorđević Betetto L, Žgavec B, Bergant Suhodolčan A. Psoriasis -like tinea incognita: a case report and literature review. Acta Dermatovenerol Alp Pannonica Adriat. 2020;29(1):43 -5. Zander N, Schäfer I, Radtke M, Jacobi A, Heigel H, Augustin M. Dermatological comorbidity in psoriasis: results from a large -scale cohort of employees. Arch Der matol Res. 2017;309:349 -56. Diruggiero D. Successful Management of Psoriasis and Treatment -induced Tinea Incognito: A Case Report. J Clin Aesthet Dermatol. 2020;13(9):S21 -5. Arenas R, Moreno -Coutiño G, Vera L, Welsh O. Tinea incognito. Clin Dermatol. 2010; 28(2):137 -9. Cite this article as: Reddy PG, Maqsood M . Encountering steroid treatment induced tinea incognito: a case report . Int J Res Dermatol 20 23;9:123 -5.
17556	https://artofproblemsolving.com/articles/files/SatoNT.pdf?srsltid=AfmBOorUYLcu2lunCTO7QcRmFIp2hxCAmWVPUbneb_UIjKDr3B9WV3dV	Number Theory Naoki Sato sato@artofproblemsolving.com 0 Preface This set of notes on number theory was originally written in 1995 for students at the IMO level. It covers the basic background material that an IMO student should be familiar with. This text is meant to be a reference, and not a replacement but rather a supplement to a number theory textbook; several are given at the back. Proofs are given when appropriate, or when they illustrate some insight or important idea. The problems are culled from various sources, many from actual contests and olympiads, and in general are very difficult. The author welcomes any corrections or suggestions. 1 Divisibility For integers a and b, we say that a divides b, or that a is a divisor (or factor ) of b, or that b is a multiple of a, if there exists an integer c such that b = ca , and we denote this by a \| b. Otherwise, a does not divide b, and we denote this by a - b. A positive integer p is a prime if the only divisors of p are 1 and p. If pk \| a and pk+1 - a where p is a prime, i.e. pk is the highest power of p dividing a, then we denote this by pk‖a.Useful Facts • If a, b > 0, and a \| b, then a ≤ b. • If a \| b1, a \| b2, . . . , a \| bn, then for any integers c1, c2, . . . , cn, a \| n ∑ i=1 bici. Theorem 1.1 . The Division Algorithm . For any positive integer a and integer b, there exist unique integers q and r such that b = qa + r and 0 ≤ r < a , with r = 0 iff a \| b.1Theorem 1.2 . The Fundamental Theorem of Arithmetic . Every integer greater than 1 can be written uniquely in the form pe1 1 pe2 2 · · · pek k , where the pi are distinct primes and the ei are positive integers. Theorem 1.3 . (Euclid) There exist an infinite number of primes. Proof . Suppose that there are a finite number of primes, say p1, p2, . . . , pn. Let N = p1p2 · · · pn + 1. By the fundamental theorem of arithmetic, N is divisible by some prime p. This prime p must be among the pi, since by assumption these are all the primes, but N is seen not to be divisible by any of the pi, contradiction. Example 1.1 . Let x and y be integers. Prove that 2 x + 3 y is divisible by 17 iff 9 x + 5 y is divisible by 17. Solution . 17 \| (2 x + 3 y) ⇒ 17 \| [13(2 x + 3 y)], or 17 \| (26 x + 39 y) ⇒ 17 \| (9 x + 5 y), and conversely, 17 \| (9 x + 5 y) ⇒ 17 \| [4(9 x + 5 y)], or 17 \| (36 x + 20 y) ⇒ 17 \| (2 x + 3 y). Example 1.2 . Find all positive integers d such that d divides both n2 +1 and ( n + 1) 2 + 1 for some integer n. Solution . Let d \| (n2 + 1) and d \| [( n + 1) 2 + 1], or d \| (n2 + 2 n + 2). Then d \| [( n2 + 2 n + 2) − (n2 + 1)], or d \| (2 n + 1) ⇒ d \| (4 n2 + 4 n + 1), so d \| [4( n2+2 n+2) −(4 n2+4 n+1)], or d \| (4 n+7). Then d \| [(4 n+7) −2(2 n+1)], or d \| 5, so d can only be 1 or 5. Taking n = 2 shows that both of these values are achieved. Example 1.3 . Suppose that a1, a2, . . . , a2n are distinct integers such that the equation (x − a1)( x − a2) · · · (x − a2n) − (−1) n(n!) 2 = 0 has an integer solution r. Show that r = a1 + a2 + · · · + a2n 2n . (1984 IMO Short List) Solution . Clearly, r 6 = ai for all i, and the r − ai are 2 n distinct integers, so \|(r − a1)( r − a2) · · · (r − a2n)\| ≥ \| (1)(2) · · · (n)( −1)( −2) · · · (−n)\| = ( n!) 2, 2with equality iff {r − a1, r − a2, . . . , r − a2n} = {1, 2, . . . , n, −1, −2, . . . , −n}. Therefore, this must be the case, so (r − a1) + ( r − a2) + · · · + ( r − a2n)= 2 nr − (a1 + a2 + · · · + a2n)= 1 + 2 + · · · + n + ( −1) + ( −2) + · · · + ( −n) = 0 ⇒ r = a1 + a2 + · · · + a2n 2n . Example 1.4 . Let 0 < a 1 < a 2 < · · · < a mn +1 be mn + 1 integers. Prove that you can select either m + 1 of them no one of which divides any other, or n + 1 of them each dividing the following one. (1966 Putnam Mathematical Competition) Solution . For each i, 1 ≤ i ≤ mn + 1, let ni be the length of the longest sequence starting with ai and each dividing the following one, among the integers ai, ai+1 , . . . , amn +1 . If some ni is greater than n then the problem is solved. Otherwise, by the pigeonhole principle, there are at least m + 1 values of ni that are equal. Then, the integers ai corresponding to these ni cannot divide each other. Useful Facts • Bertrand’s Postulate . For every positive integer n, there exists a prime p such that n ≤ p ≤ 2n. • Gauss’s Lemma . If a polynomial with integer coefficients factors into two polynomials with rational coefficients, then it factors into two poly-nomials with integer coefficients. Problems 1. Let a and b be positive integers such that a \| b2, b2 \| a3, a3 \| b4, b4 \| a5,. . . . Prove that a = b.2. Let a, b, and c denote three distinct integers, and let P denote a poly-nomial having all integral coefficients. Show that it is impossible that P (a) = b, P (b) = c, and P (c) = a.(1974 USAMO) 33. Show that if a and b are positive integers, then ( a + 12 )n + ( b + 12 )n is an integer for only finitely many positive integers n.(A Problem Seminar , D.J. Newman) 4. For a positive integer n, let r(n) denote the sum of the remainders when n is divided by 1, 2, . . . , n respectively. Prove that r(k) = r(k − 1) for infinitely many positive integers k.(1981 K¨ ursch´ ak Competition) 5. Prove that for all positive integers n,0 < n ∑ k=1 g(k) k − 2n 3 < 23, where g(k) denotes the greatest odd divisor of k.(1973 Austrian Mathematics Olympiad) 6. Let d be a positive integer, and let S be the set of all positive integers of the form x2 + dy 2, where x and y are non-negative integers. (a) Prove that if a ∈ S and b ∈ S, then ab ∈ S.(b) Prove that if a ∈ S and p ∈ S, such that p is a prime and p \| a,then a/p ∈ S.(c) Assume that the equation x2 + dy 2 = p has a solution in non-negative integers x and y, where p is a given prime. Show that if d ≥ 2, then the solution is unique, and if d = 1, then there are exactly two solutions. 2 GCD and LCM The greatest common divisor of two positive integers a and b is the great-est positive integer that divides both a and b, which we denote by gcd( a, b ), and similarly, the lowest common multiple of a and b is the least positive 4integer that is a multiple of both a and b, which we denote by lcm( a, b ). We say that a and b are relatively prime if gcd( a, b ) = 1. For integers a1, a2,. . . , an, gcd( a1, a 2, . . . , a n) is the greatest positive integer that divides all of a1, a2, . . . , an, and lcm( a1, a 2, . . . , a n) is defined similarly. Useful Facts • For all a, b, gcd( a, b ) · lcm( a, b ) = ab . • For all a, b, and m, gcd( ma, mb ) = m gcd( a, b ) and lcm( ma, mb ) = mlcm( a, b ). • If d \| gcd( a, b ), then gcd (ad , bd ) = gcd( a, b ) d . In particular, if d = gcd( a, b ), then gcd( a/d, b/d ) = 1; that is, a/d and b/d are relatively prime. • If a \| bc and gcd( a, c ) = 1, then a \| b. • For positive integers a and b, if d is a positive integer such that d \| a, d \| b, and for any d′, d′ \| a and d′ \| b implies that d′ \| d, then d =gcd( a, b ). This is merely the assertion that any common divisor of a and b divides gcd( a, b ). • If a1a2 · · · an is a perfect kth power and the ai are pairwise relatively prime, then each ai is a perfect kth power. • Any two consecutive integers are relatively prime. Example 2.1 . Show that for any positive integer N , there exists a multiple of N that consists only of 1s and 0s. Furthermore, show that if N is relatively prime to 10, then there exists a multiple that consists only of 1s. Solution . Consider the N + 1 integers 1, 11, 111, . . . , 111...1 ( N + 1 1s). When divided by N , they leave N + 1 remainders. By the pigeonhole princi-ple, two of these remainders are equal, so the difference in the corresponding integers, an integer of the form 111...000, is divisible by N . If N is relatively prime to 10, then we may divide out all powers of 10, to obtain an integer of the form 111...1 that remains divisible by N .5Theorem 2.1 . For any positive integers a and b, there exist integers x and y such that ax + by = gcd( a, b ). Furthermore, as x and y vary over all integers, ax + by attains all multiples and only multiples of gcd( a, b ). Proof . Let S be the set of all integers of the form ax +by , and let d be the least positive element of S. By the division algorithm, there exist integers q and r such that a = qd + r, 0 ≤ r < d . Then r = a − qd = a − q(ax + by ) = (1 − qx )a − (qy )b, so r is also in S. But r < d , so r = 0 ⇒ d \| a, and similarly, d \| b, so d \| gcd( a, b ). However, gcd( a, b ) divides all elements of S,so in particular gcd( a, b ) \| d ⇒ d = gcd( a, b ). The second part of the theorem follows. Corollary 2.2 . The positive integers a and b are relatively prime iff there exist integers x and y such that ax + by = 1. Corollary 2.3 . For any positive integers a1, a2, . . . , an, there exist integers x1, x2, . . . , xn, such that a1x1+a2x2+· · · +anxn = gcd( a1, a 2, . . . , a n). Corollary 2.4 . Let a and b be positive integers, and let n be an integer. Then the equation ax + by = n has a solution in integers x and y iff gcd( a, b ) \| n. If this is the case, then all solutions are of the form (x, y ) = ( x0 + t · bd , y 0 − t · ad ) , where d = gcd( a, b ), ( x0, y 0) is a specific solution of ax + by = n, and t is an integer. Proof . The first part follows from Theorem 2.1. For the second part, as stated, let d = gcd( a, b ), and let ( x0, y 0) be a specific solution of ax + by = n,so that ax 0 + by 0 = n. If ax + by = n, then ax + by − ax 0 − by 0 = a(x − x0) + b(y − y0) = 0, or a(x − x0) = b(y0 − y), and hence (x − x0) · ad = ( y0 − y) · bd. Since a/d and b/d are relatively prime, b/d must divide x − x0, and a/d must divide y0 − y. Let x − x0 = tb/d and y0 − y = ta/d . This gives the solutions described above. 6Example 2.2 . Prove that the fraction 21 n + 4 14 n + 3 is irreducible for every positive integer n. (1959 IMO) Solution . For all n, 3(14 n + 3) − 2(21 n + 4) = 1, so the numerator and denominator are relatively prime. Example 2.3 . For all positive integers n, let Tn = 2 2n Show that if m 6 = n, then Tm and Tn are relatively prime. Solution . We have that Tn − 2 = 2 2n − 1 = 2 2n−1·2 − 1= ( Tn−1 − 1) 2 − 1 = T 2 n−1 − 2Tn−1 = Tn−1(Tn−1 − 2) = Tn−1Tn−2(Tn−2 − 2) = · · · = Tn−1Tn−2 · · · T1T0(T0 − 2) = Tn−1Tn−2 · · · T1T0, for all n. Therefore, any common divisor of Tm and Tn must divide 2. But each Tn is odd, so Tm and Tn are relatively prime. Remark . It immediately follows from this result that there are an infinite number of primes. The Euclidean Algorithm . By recursive use of the division algorithm, we may find the gcd of two positive integers a and b without factoring either, and the x and y in Theorem 2.1 (and so, a specific solution in Corollary 2.4). For example, for a = 329 and b = 182, we compute 329 = 1 · 182 + 147 , 182 = 1 · 147 + 35 , 147 = 4 · 35 + 7 , 35 = 5 · 7, and stop when there is no remainder. The last dividend is the gcd, so in our example, gcd(329,182) = 7. Now, working through the above equations 7backwards, 7 = 147 − 4 · 35 = 147 − 4 · (182 − 1 · 147) = 5 · 147 − 4 · 182 = 5 · (329 − 182) − 4 · 182 = 5 · 329 − 9 · 182 . Remark . The Euclidean algorithm also works for polynomials. Example 2.4 . Let n be a positive integer, and let S be a subset of n + 1 elements of the set {1, 2, . . . , 2n}. Show that (a) There exist two elements of S that are relatively prime, and (b) There exist two elements of S, one of which divides the other. Solution . (a) There must be two elements of S that are consecutive, and thus, relatively prime. (b) Consider the greatest odd factor of each of the n + 1 elements in S. Each is among the n odd integers 1, 3, . . . , 2 n − 1. By the pigeon-hole principle, two must have the same greatest odd factor, so they differ (multiplication-wise) by a power of 2, and so one divides the other. Example 2.5 . The positive integers a1, a2, . . . , an are such that each is less than 1000, and lcm( ai, a j ) > 1000 for all i, j, i 6 = j. Show that n ∑ i=1 1 ai < 2. (1951 Russian Mathematics Olympiad) Solution . If 1000 m+1 < a ≤ 1000 m , then the m multiples a, 2 a, . . . , ma do not exceed 1000. Let k1 the number of ai in the interval ( 1000 2 , 1000], k2 in (1000 3 , 1000 2 ], etc. Then there are k1 + 2 k2 + 3 k3 + · · · integers, no greater than 1000, that are multiples of at least one of the ai. But the multiples are distinct, so k1 + 2 k2 + 3 k3 + · · · < 1000 ⇒ 2k1 + 3 k2 + 4 k3 + · · · = ( k1 + 2 k2 + 3 k3 + · · · ) + ( k1 + k2 + k3 + · · · ) < 1000 + n< 2000 . 8Therefore, n ∑ i=1 1 ai ≤ k1 21000 + k2 31000 + k3 41000 + · · · = 2k1 + 3 k2 + 4 k3 + · · · 1000 < 2. Note: It can be shown that n ≤ 500 as follows: Consider the greatest odd divisor of a1, a2, . . . , a1000 . Each must be distinct; otherwise, two differ, multiplication-wise, by a power of 2, which means one divides the other, contradiction. Also, there are only 500 odd numbers between 1 and 1000, from which the result follows. It also then follows that n ∑ i=1 1 ai < 32. Useful Facts • Dirichlet’s Theorem . If a and b are relatively prime positive integers, then the arithmetic sequence a, a + b, a + 2 b, . . . , contains infinitely many primes. Problems 1. The symbols ( a, b, . . . , g ) and [ a, b, . . . , g ] denote the greatest common divisor and lowest common multiple, respectively of the positive inte-gers a, b, . . . , g. Prove that [a, b, c ]2 [a, b ][ a, c ][ b, c ] = (a, b, c )2 (a, b )( a, c )( b, c ). (1972 USAMO) 2. Show that gcd( am − 1, a n − 1) = agcd( m,n ) − 1 for all positive integers a > 1, m, n.93. Let a, b, and c be positive integers. Show that lcm( a, b, c ) = abc · gcd( a, b, c )gcd( a, b ) · gcd( a, c ) · gcd( b, c ). Express gcd( a, b, c ) in terms of abc , lcm( a, b, c ), lcm( a, b ), lcm( a, c ), and lcm( b, c ). Generalize. 4. Let a, b be odd positive integers. Define the sequence ( fn) by putting f1 = a, f2 = b, and by letting fn for n ≥ 3 be the greatest odd divisor of fn−1 + fn−2. Show that fn is constant for n sufficiently large and determine the eventual value as a function of a and b.(1993 USAMO) 5. Let n ≥ a1 > a 2 > · · · > a k be positive integers such that lcm( ai, a j ) ≤ n for all i, j. Prove that ia i ≤ n for i = 1, 2, . . . , k. 3 Arithmetic Functions There are several important arithmetic functions, of which three are pre-sented here. If the prime factorization of n > 1 is pe1 1 pe2 2 · · · pek k , then the number of positive integers less than n, relatively prime to n, is φ(n) = ( 1 − 1 p1 ) ( 1 − 1 p2 ) · · · ( 1 − 1 pk ) n = pe1−11 pe2−12 · · · pek −1 k (p1 − 1)( p2 − 1) · · · (pk − 1) , the number of divisors of n is τ (n) = ( e1 + 1)( e2 + 1) · · · (ek + 1) , and the sum of the divisors of n is σ(n) = ( pe1 1 pe1−11 + · · · + 1)( pe2 2 pe2−12 + · · · + 1) · · · (pek k pek −1 k · · · + 1) = (pe1+1 1 − 1 p1 − 1 ) ( pe2+1 2 − 1 p2 − 1 ) · · · (pek +1 k − 1 pk − 1 ) . Also, φ(1), τ (1), and σ(1) are defined to be 1. We say that a function f is multiplicative if f (mn ) = f (m)f (n) for all relatively prime positive 10 integers m and n, and f (1) = 1 (otherwise, f (1) = 0, which implies that f (n) = 0 for all n). Theorem 3.1 . The functions φ, τ , and σ are multiplicative. Hence, by taking the prime factorization and evaluating at each prime power, the formula above are found easily. Example 3.1 . Find the number of solutions in ordered pairs of positive integers ( x, y ) of the equation 1 x + 1 y = 1 n, where n is a positive integer. Solution . From the given, 1 x + 1 y = 1 n ⇔ xy = nx + ny ⇔ (x − n)( y − n) = n2. If n = 1, then we immediately deduce the unique solution (2,2). For n ≥ 2, let n = pe1 1 pe2 2 · · · pek k be the prime factorization of n. Since x, y > n ,there is a 1-1 correspondence between the solutions in ( x, y ) and the factors of n2, so the number of solutions is τ (n2) = (2 e1 + 1)(2 e2 + 1) · · · (2 ek + 1) . Example 3.2 . Let n be a positive integer. Prove that ∑ d\|n φ(d) = n. Solution . For a divisor d of n, let Sd be the set of all a, 1 ≤ a ≤ n, such that gcd( a, n ) = n/d . Then Sd consists of all elements of the form b · n/d ,where 0 ≤ b ≤ d, and gcd( b, d ) = 1, so Sd contains φ(d) elements. Also, it is clear that each integer between 1 and n belongs to a unique Sd. The result then follows from summing over all divisors d of n.Problems 1. Let n be a positive integer. Prove that n ∑ k=1 τ (k) = n ∑ k=1 ⌊nk ⌋ . 11 2. Let n be a positive integer. Prove that ∑ d\|n τ 3(d) = ∑ d\|n τ (d)  2 . Prove that if σ(N ) = 2 N + 1, then N is the square of an odd integer. (1976 Putnam Mathematical Competition) 4 Modular Arithmetic For a positive integer m and integers a and b, we say that a is congruent to b modulo m if m \| (a − b), and we denote this by a ≡ b modulo m, or more commonly a ≡ b (mod m). Otherwise, a is not congruent to b modulo m,and we denote this by a 6 ≡ b (mod m) (although this notation is not used often). In the above notation, m is called the modulus , and we consider the integers modulo m. Theorem 4.1 . If a ≡ b and c ≡ d (mod m), then a + c ≡ b + d (mod m)and ac ≡ bd (mod m). Proof . If a ≡ b and c ≡ d (mod m), then there exist integers k and l such that a = b + km and c = d + lm . Hence, a + c = b + d + ( k + l)m, so a + c ≡ b + d (mod m). Also, ac = bd + dkm + blm + klm 2 = bd + ( dk + bl + klm )m, so ac ≡ bd (mod m). Useful Facts • For all integers n, n2 ≡ { 01 } (mod 4) { if n is even , if n is odd . • For all integers n, n2 ≡  041  (mod 8)  if n ≡ 0 (mod 4) , if n ≡ 2 (mod 4) , if n ≡ 1 (mod 2) . 12 • If f is a polynomial with integer coefficients and a ≡ b (mod m), then f (a) ≡ f (b) (mod m). • If f is a polynomial with integer coefficients of degree n, not identically zero, and p is a prime, then the congruence f (x) ≡ 0 (mod p)has at most n solutions modulo p, counting multiplicity. Example 4.1 . Prove that the only solution in rational numbers of the equation x3 + 3 y3 + 9 z3 − 9xyz = 0 is x = y = z = 0. (1983 K¨ ursch´ ak Competition) Solution . Suppose that the equation has a solution in rationals, with at least one non-zero variable. Since the equation is homogeneous, we may obtain a solution in integers ( x0, y 0, z 0) by multiplying the equation by the cube of the lowest common multiple of the denominators. Taking the equa-tion modulo 3, we obtain x30 ≡ 0 (mod 3). Therefore, x0 must be divisible by 3, say x0 = 3 x1. Substituting, 27 x31 + 3 y30 + 9 z30 − 27 x1y0z0 = 0 ⇒ y30 + 3 z30 + 9 x31 − 9x1y0z0 = 0 . Therefore, another solution is ( y0, z 0, x 1). We may then apply this reduction recursively, to obtain y0 = 3 y1, z0 = 3 z1, and another solution ( x1, y 1, z 1). Hence, we may divide powers of 3 out of our integer solution an arbitrary number of times, contradiction. Example 4.2 . Does one of the first 10 8 + 1 Fibonacci numbers terminate with 4 zeroes? Solution . The answer is yes. Consider the sequence of pairs ( Fk, F k+1 )modulo 10 4. Since there are only a finite number of different possible pairs (10 8 to be exact), and each pair is dependent only on the previous one, this sequence is eventually periodic. Also, by the Fibonacci relation, one can find the previous pair to a given pair, so this sequence is immediately periodic. But F0 ≡ 0 (mod 10 4), so within 10 8 terms, another Fibonacci number divisible by 10 4 must appear. 13 In fact, a computer check shows that 10 4 \| F7500 , and ( Fn) modulo 10 4 has period 15000, which is much smaller than the upper bound of 10 8.If ax ≡ 1 (mod m), then we say that x is the inverse of a modulo m,denoted by a−1, and it is unique modulo m. Theorem 4.2 . The inverse of a modulo m exists and is unique iff a is relatively prime to m. Proof . If ax ≡ 1 (mod m), then ax = 1+ km for some k ⇒ ax −km = 1. By Corollary 2.2, a and m are relatively prime. Now, if gcd( a, m ) = 1, then by Corollary 2.2, there exist integers x and y such that ax + my = 1 ⇒ ax =1 − my ⇒ ax ≡ 1 (mod m). The inverse x is unique modulo m, since if x′ is also an inverse, then ax ≡ ax ′ ≡ 1 ⇒ xax ≡ xax ′ ≡ x ≡ x′. Corollary 4.3 . If p is a prime, then the inverse of a modulo p exists and is unique iff p does not divide a. Corollary 4.4 . If ak ≡ bk (mod m) and k is relatively prime to m, then a ≡ b (mod m). Proof . Multiplying both sides by k−1, which exists by Theorem 4.2, yields the result. We say that a set {a1, a 2, . . . , a m} is a complete residue system modulo m if for all i, 0 ≤ i ≤ m−1, there exists a unique j such that aj ≡ i (mod m). Example 4.3 . Find all positive integers n such that there exist complete residue systems {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} modulo n for which {a1 + b1, a 2 + b2, . . . , a n + bn} is also a complete residue system. Solution . The answer is all odd n. First we prove necessity. For any complete residue system {a1, a 2, . . . , a n} modulo n, we have that a1 + a2 + · · · + an ≡ n(n + 1) /2 (mod n). So, if all three sets are complete residue systems, then a1 +a2 +· · · +an +b1 +b2 +· · · +bn ≡ n2 +n ≡ 0 (mod n)and a1 + b1 + a2 + b2 + · · · + an + bn ≡ n(n + 1) /2 (mod n), so n(n + 1) /2 ≡ 0(mod n). The quantity n(n + 1) /2 is divisible by n iff ( n + 1) /2 is an integer, which implies that n is odd. Now assume that n is odd. Let ai = bi = i for all i. Then ai + bi = 2 i for all i, and n is relatively prime to 2, so by Corollary 4.4, {2, 4, . . . , 2n} is a complete residue system modulo n. Theorem 4.5 . Euler’s Theorem . If a is relatively prime to m, then aφ(m) ≡ 1 (mod m). 14 Proof . Let a1, a2, . . . , aφ(m) be the positive integers less than m that are relatively prime to m. Consider the integers aa 1, aa 2, . . . , aa φ(m). We claim that they are a permutation of the original φ(m) integers ai, modulo m. For each i, aa i is also relatively prime to m, so aa i ≡ ak for some k. Since aa i ≡ aa j ⇔ ai ≡ aj (mod m), each ai gets taken to a different ak under multiplication by a, so indeed they are permuted. Hence, a1a2 · · · aφ(m) ≡ (aa 1)( aa 2) · · · (aa φ(m)) ≡ aφ(m)a1a2 · · · aφ(m) ⇒ 1 ≡ aφ(m) (mod m). Remark . This gives an explicit formula for the inverse of a modulo m: a−1 ≡ aφ(m)−2 (mod m). Alternatively, one can use the Euclidean algorithm to find a−1 ≡ x as in the proof of Theorem 4.2. Corollary 4.6 . Fermat’s Little Theorem (FLT) . If p is a prime, and p does not divide a, then ap−1 ≡ 1 (mod p). Example 4.4 . Show that if a and b are relatively prime positive integers, then there exist integers m and n such that am + bn ≡ 1 (mod ab ). Solution . Let S = am + bn, where m = φ(b) and n = φ(a). Then by Euler’s Theorem, S ≡ bφ(a) ≡ 1 (mod a), or S − 1 ≡ 0 (mod a), and S ≡ aφ(b) ≡ 1 (mod b), or S − 1 ≡ 0 (mod b). Therefore, S − 1 ≡ 0, or S ≡ 1(mod ab ). Example 4.5 . For all positive integers i, let Si be the sum of the products of 1, 2, . . . , p − 1 taken i at a time, where p is an odd prime. Show that S1 ≡ S2 ≡ · · · ≡ Sp−2 ≡ 0 (mod p). Solution . First, observe that (x − 1)( x − 2) · · · (x − (p − 1)) = xp−1 − S1xp−2 + S2xp−3 − · · · − Sp−2x + Sp−1. This polynomial vanishes for x = 1, 2, . . . , p − 1. But by Fermat’s Little Theorem, so does xp−1 − 1 modulo p. Taking the difference of these two polynomials, we obtain another polynomial of degree p − 2 with p − 1 roots modulo p, so it must be the zero polynomial, and the result follows from comparing coefficients. 15 Remark . We immediately have that ( p − 1)! ≡ Sp−1 ≡ − 1 (mod p), which is Wilson’s Theorem. Also, xp − x ≡ 0 (mod p) for all x, yet we cannot compare coefficients here. Why not? Theorem 4.7 . If p is a prime and n is an integer such that p \| (4 n2 + 1), then p ≡ 1 (mod 4). Proof . Clearly, p cannot be 2, so we need only show that p 6 ≡ 3 (mod 4). Suppose p = 4 k + 3 for some k. Let y = 2 n, so by Fermat’s Little Theorem, yp−1 ≡ 1 (mod p), since p does not divide n. But, y2 + 1 ≡ 0, so yp−1 ≡ y4k+2 ≡ (y2)2k+1 ≡ (−1) 2k+1 ≡ − 1 (mod p), contradiction. Therefore, p ≡ 1 (mod 4). Remark . The same proof can be used to show that if p is a prime and p \| (n2 + 1), then p = 2 or p ≡ 1 (mod 4). Example 4.6 . Show that there are an infinite number of primes of the form 4 k + 1 and of the form 4 k + 3. Solution . Suppose that there are a finite number of primes of the form 4k + 1, say p1, p2, . . . , pn. Let N = 4( p1p2 · · · pn)2 + 1. By Theorem 4.7, N is only divisible by primes of the form 4 k + 1, but clearly N is not divisible by any of these primes, contradiction. Similarly, suppose that there are a finite number of primes of the form 4k + 3, say q1, q2, . . . , qm. Let M = 4 q1q2 · · · qm − 1. Then M ≡ 3 (mod 4), so M must be divisible by a prime of the form 4 k + 3, but M is not divisible by any of these primes, contradiction. Example 4.7 . Show that if n is an integer greater than 1, then n does not divide 2 n − 1. Solution . Let p be the least prime divisor of n. Then gcd( n, p − 1) = 1, and by Corollary 2.2, there exist integers x and y such that nx +( p−1) y = 1. If p \| (2 n − 1), then 2 ≡ 2nx +( p−1) y ≡ (2 n)x(2 p−1)y ≡ 1 (mod p) by Fermat’s Little Theorem, contradiction. Therefore, p - (2 n − 1) ⇒ n - (2 n − 1). Theorem 4.8 . Wilson’s Theorem . If p is a prime, then ( p − 1)! ≡ − 1(mod p). (See also Example 4.5.) Proof . Consider the congruence x2 ≡ 1 (mod p). Then x2 − 1 ≡ (x − 1)( x + 1) ≡ 0, so the only solutions are x ≡ 1 and −1. Therefore, for each i,2 ≤ i ≤ p − 2, there exists a unique inverse j 6 = i of i, 2 ≤ j ≤ p − 2, modulo 16 p. Hence, when we group in pairs of inverses, (p − 1)! ≡ 1 · 2 · · · (p − 2) · (p − 1) ≡ 1 · 1 · · · 1 · (p − 1) ≡ − 1 (mod p). Example 4.8 . Let {a1, a 2, . . . , a 101 } and {b1, b 2, . . . , b 101 } be complete residue systems modulo 101. Can {a1b1, a 2b2, . . . , a 101 b101 } be a complete residue system modulo 101? Solution . The answer is no. Suppose that {a1b1, a 2b2, . . . , a 101 b101 } is a complete residue system modulo 101. Without loss of generality, assume that a101 ≡ 0 (mod 101). Then b101 ≡ 0 (mod 101), because if any other bj was congruent to 0 modulo 101, then aj bj ≡ a101 b101 ≡ 0 (mod 101), contradiction. By Wilson’s Theorem, a1a2 · · · a100 ≡ b1b2 · · · b100 ≡ 100! ≡−1 (mod 101), so a1b1a2b2 · · · a100 b100 ≡ 1 (mod 101). But a101 b101 ≡ 0(mod 101), so a1b1a2b2 · · · a100 b100 ≡ 100! ≡ − 1 (mod 101), contradiction. Theorem 4.9 . If p is a prime, then the congruence x2 + 1 ≡ 0 (mod p)has a solution iff p = 2 or p ≡ 1 (mod 4). (Compare to Theorem 7.1) Proof . If p = 2, then x = 1 is a solution. If p ≡ 3 (mod 4), then by the remark to Theorem 4.7, no solutions exist. Finally, if p = 4 k + 1, then let x = 1 · 2 · · · (2 k). Then x2 ≡ 1 · 2 · · · (2 k) · (2 k) · · · 2 · 1 ≡ 1 · 2 · · · (2 k) · (−2k) · · · (−2) · (−1) (multiplying by 2 k −1s) ≡ 1 · 2 · · · (2 k) · (p − 2k) · · · (p − 2) · (p − 1) ≡ (p − 1)! ≡ − 1 (mod p). Theorem 4.10 . Let p be a prime such that p ≡ 1 (mod 4). Then there exist positive integers x and y such that p = x2 + y2. Proof . By Theorem 4.9, there exists an integer a such that a2 ≡ − 1(mod p). Consider the set of integers of the form ax − y, where x and y are integers, 0 ≤ x, y < √p. The number of possible pairs ( x, y ) is then (b√pc + 1) 2 > (√p)2 = p, so by pigeonhole principle, there exist integers 0 ≤ x1, x 2, y 1, y 2 < √p, such that ax 1−y1 ≡ ax 2−y2 (mod p). Let x = x1−x2 and y = y1 − y2. At least one of x and y is non-zero, and ax ≡ y ⇒ a2x2 ≡ 17 −x2 ≡ y2 ⇒ x2 + y2 ≡ 0 (mod p). Thus, x2 + y2 is a multiple of p, and 0 < x 2 + y2 < (√p)2 + ( √p)2 = 2 p, so x2 + y2 = p. Theorem 4.11 . Let n be a positive integer. Then there exist integers x and y such that n = x2 + y2 iff each prime factor of n of the form 4 k + 3 appears an even number of times. Theorem 4.12 . The Chinese Remainder Theorem (CRT) . If a1, a2, . . . , ak are integers, and m1, m2, . . . , mk are pairwise relatively prime integers, then the system of congruences x ≡ a1 (mod m1),x ≡ a2 (mod m2), ... x ≡ ak (mod mk)has a unique solution modulo m1m2 · · · mk. Proof . Let m = m1m2 · · · mk, and consider m/m 1. This is relatively prime to m1, so there exists an integer t1 such that t1 · m/m 1 ≡ 1 (mod m1). Accordingly, let s1 = t1 · m/m 1. Then s1 ≡ 1 (mod m1) and s1 ≡ 0(mod mj ), j 6 = 1. Similarly, for all i, there exists an si such that si ≡ 1(mod mi) and si ≡ 0 (mod mj ), j 6 = i. Then, x = a1s1 + a2s2 + · · · + aksk is a solution to the above system. To see uniqueness, let x′ be another solution. Then x − x′ ≡ 0 (mod mi) for all i ⇒ x − x′ ≡ 0 (mod m1m2 · · · mk). Remark . The proof shows explicitly how to find the solution x. Example 4.9 . For a positive integer n, find the number of solutions of the congruence x2 ≡ 1 (mod n). Solution . Let the prime factorization of n be 2 epe1 1 pe2 2 · · · pek k . By CRT, x2 ≡ 1 (mod n) ⇔ x2 ≡ 1 (mod pei i ) for all i, and x2 ≡ 1 (mod 2 e). We consider these cases separately. We have that x2 ≡ 1 (mod pei i ) ⇔ x2 − 1 = ( x − 1)( x + 1) ≡ 0 (mod pei i ). But pi cannot divide both x − 1 and x + 1, so it divides one of them; that is, x ≡ ± 1 (mod pei i ). Hence, there are two solutions. Now, if ( x − 1)( x + 1) ≡ 0 (mod 2 e), 2 can divide both x − 1 and x + 1, but 4 cannot divide both. For e = 1 and e = 2, it is easily checked that there are 1 and 2 solutions respectively. For e ≥ 3, since there is at most one factor 18 of 2 in one of x − 1 and x + 1, there must be at least e − 1 in the other, for their product to be divisible by 2 e. Hence, the only possibilities are x − 1 or x + 1 ≡ 0, 2 e−1 (mod 2 e), which lead to the four solutions x ≡ 1, 2 e−1 − 1, 2e−1 + 1, and 2 e − 1. Now that we know how many solutions each prime power factor con-tributes, the number of solutions modulo n is simply the product of these, by CRT. The following table gives the answer: e Number of solutions 0, 1 2k 2 2k+1 ≥ 3 2k+2 Theorem 4.11 . Let m be a positive integer, let a and b be integers, and let k = gcd( a, m ). Then the congruence ax ≡ b (mod m) has k solutions or no solutions according as k \| b or k - b.Problems 1. Prove that for each positive integer n there exist n consecutive positive integers, none of which is an integral power of a prime. (1989 IMO) 2. For an odd positive integer n > 1, let S be the set of integers x,1 ≤ x ≤ n, such that both x and x + 1 are relatively prime to n. Show that ∏ x∈S x ≡ 1 (mod n). Find all positive integer solutions to 3 x + 4 y = 5 z .(1991 IMO Short List) 4. Let n be a positive integer such that n + 1 is divisible by 24. Prove that the sum of all the divisors of n is divisible by 24. (1969 Putnam Mathematical Competition) 5. (Wolstenholme’s Theorem) Prove that if 1 + 12 + 13 + · · · + 1 p − 119 is expressed as a fraction, where p ≥ 5 is a prime, then p2 divides the numerator. 6. Let a be the greatest positive root of the equation x3 − 3x2 + 1 = 0. Show that ba1788 c and ba1988 c are both divisible by 17. (1988 IMO Short List) 7. Let {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} be complete residue systems modulo n, such that {a1b1, a 2b2, . . . , a nbn} is also a complete residue system modulo n. Show that n = 1 or 2. 8. Let m, n be positive integers. Show that 4 mn − m − n can never be a square. (1984 IMO Proposal) 5 Binomial Coefficients For non-negative integers n and k, k ≤ n, the binomial coefficient (nk ) is defined as n! k!( n − k)! , and has several important properties. By convention, (nk ) = 0 if k > n .In the following results, for polynomials f and g with integer coefficients, we say that f ≡ g (mod m) if m divides every coefficient in f − g. Theorem 5.1 . If p is a prime, then the number of factors of p in n! is ⌊np ⌋ + ⌊ np2 ⌋ + ⌊ np3 ⌋ · · · . It is also n − sn p − 1 , where sn is the sum of the digits of n when expressed in base p. Theorem 5.2 . If p is a prime, then (pi ) ≡ 0 (mod p)20 for 1 ≤ i ≤ p − 1. Corollary 5.3 . (1 + x)p ≡ 1 + xp (mod p). Lemma 5.4 . For all real numbers x and y, bx + yc ≥ b xc + byc. Proof . x ≥ b xc ⇒ x + y ≥ b xc + byc ∈ Z, so bx + yc ≥ b xc + byc. Theorem 5.5 . If p is a prime, then (pk i ) ≡ 0 (mod p)for 1 ≤ i ≤ pk − 1. Proof . By Lemma 5.4, k ∑ j=1 (⌊ ipj ⌋ + ⌊pk − ipj ⌋) ≤ k ∑ j=1 ⌊pk pj ⌋ , where the LHS and RHS are the number of factors of p in i!( pk − i)! and pk! respectively. But, ⌊ ipk ⌋ = ⌊pk −ipk ⌋ = 0 and ⌊pk pk ⌋ = 1, so the inequality is strict, and at least one factor of p divides (pk i ). Corollary 5.6 . (1 + x)pk ≡ 1 + xpk (mod p). Example 5.1 . Let n be a positive integer. Show that the product of n consecutive positive integers is divisible by n!. Solution . If the consecutive integers are m, m + 1, . . . , m + n − 1, then m(m + 1) · · · (m + n − 1) n! = (m + n − 1 n ) . Example 5.2 . Let n be a positive integer. Show that (n + 1) lcm (( n 0 ) , (n 1 ) , . . . , (nn )) = lcm(1 , 2, . . . , n + 1) . (AMM E2686) Solution . Let p be a prime ≤ n + 1 and let α (respectively β) be the highest power of p in the LHS (respectively RHS) of the above equality. Choose r so that pr ≤ n + 1 < p r+1 . Then clearly β = r. We claim that if pr ≤ m < p r+1 , then pr+1 - (mk ) for 0 ≤ k ≤ m. (∗)21 Indeed, the number of factors of p in (mk ) is γ = r ∑ s=1 (⌊ mps ⌋ − ⌊ kps ⌋ − ⌊m − kps ⌋) . Since each summand in this sum is 0 or 1, we have γ ≤ r; that is, () holds. For 0 ≤ k ≤ n, let ak = ( n + 1) (nk ) = ( n − k + 1) (n + 1 k ) = ( k + 1) (n + 1 k + 1 ) . By ( ∗), pr+1 does not divide any of the integers (nk ), (n+1 k ), or (n+1 k+1 ). Thus, pr+1 can divide ak only if p divides each of the integers n + 1, n − k + 1, and k + 1. This implies that p divides ( n + 1) − (n − k + 1) − (k + 1) = −1, contradiction. Therefore, pr+1 - ak. On the other hand, for k = pr − 1, we have that k ≤ n and ak = ( k + 1) (n+1 k+1 ) is divisible by pr. Therefore, β = r = α. Theorem 5.7 . Lucas’s Theorem . Let m and n be non-negative integers, and p a prime. Let m = mkpk + mk−1pk−1 + · · · + m1p + m0, and n = nkpk + nk−1pk−1 + · · · + n1p + n0 be the base p expansions of m and n respectively. Then (mn ) ≡ (mk nk )( mk−1 nk−1 ) · · · (m1 n1 )( m0 n0 ) (mod p). Proof . By Corollary 5.6, (1 + x)m ≡ (1 + x)mk pk +mk−1pk−1+··· +m1p+m0 ≡ (1 + x)pk mk (1 + x)pk−1mk−1 · · · (1 + x)pm 1 (1 + x)m0 ≡ (1 + xpk )mk (1 + xpk−1 )mk−1 · · · (1 + xp)m1 (1 + x)m0 (mod p). By base p expansion, the coefficient of xn on both sides is (mn ) ≡ (mk nk )( mk−1 nk−1 ) · · · (m1 n1 )( m0 n0 ) (mod p). 22 Corollary 5.8 . Let n be a positive integer. Let A(n) denote the number of factors of 2 in n!, and let B(n) denote the number of 1s in the binary expansion of n. Then the number of odd entries in the nth row of Pascal’s Triangle, or equivalently the number of odd coefficients in the expansion of (1 + x)n, is 2 B(n). Furthermore, A(n) + B(n) = n for all n.Useful Facts • For a polynomial f with integer coefficients and prime p,[f (x)] pn ≡ f (xpn ) (mod p). Problems 1. Let a and b be non-negative integers, and p a prime. Show that (pa pb ) ≡ (ab ) (mod p). Let an be the last non-zero digit in the decimal representation of the number n!. Is the sequence a1, a2, a3, . . . eventually periodic? (1991 IMO Short List) 3. Find all positive integers n such that 2 n \| (3 n − 1). 4. Find the greatest integer k for which 1991 k divides 1990 1991 1992 1992 1991 1990 . (1991 IMO Short List) 5. For a positive integer n, let a(n) and b(n) denote the number of binomial coefficients in the nth row of Pascal’s Triangle that are congruent to 1 and 2 modulo 3 respectively. Prove that a(n) − b(n) is always a power of 2. 6. Let n be a positive integer. Prove that if the number of factors of 2 in n! is n − 1, then n is a power of 2. 23 7. For a positive integer n, let Cn = 1 n + 1 (2nn ) , and Sn = C1 + C2 + · · · + Cn.Prove that Sn ≡ 1 (mod 3) if and only if there exists a 2 in the base 3 expansion of n + 1. 6 Order of an Element We know that if a is relatively prime to m, then there exists a positive integer n such that an ≡ 1 (mod m). Let d be the smallest such n. Then we say that d is the order of a modulo m, denoted by ord m(a), or simply ord( a) if the modulus m is understood. Theorem 6.1 . If a is relatively prime to m, then an ≡ 1 (mod m) iff ord( a) \| n. Furthermore, an0 ≡ an1 iff ord( a) \| (n0 − n1). Proof . Let d = ord( a). It is clear that d \| n ⇒ an ≡ 1 (mod m). On the other hand, if an ≡ 1 (mod m), then by the division algorithm, there exist integers q and r such that n = qd + r, 0 ≤ r < d . Then an ≡ aqd +r ≡ (ad)qar ≡ ar ≡ 1 (mod m). But r < d , so r = 0 ⇒ d \| n. The second part of the theorem follows. Remark . In particular, by Euler’s Theorem, ord( a) \| φ(m). Example 6.1 . Show that the order of 2 modulo 101 is 100. Solution . Let d = ord(2). Then d \| φ(101), or d \| 100. If d < 100, then d divides 100/2 or 100/5; that is, d is missing at least one prime factor. However, 250 ≡ 1024 5 ≡ 14 5 ≡ 196 · 196 · 14 ≡ (−6) · (−6) · 14 ≡ − 1 (mod 101) , and 220 ≡ 1024 2 ≡ 14 2 ≡ − 6 (mod 101) , so d = 100. Example 6.2 . Prove that if p is a prime, then every prime divisor of 2p − 1 is greater than p.24 Solution . Let q \| (2 p − 1), where q is a prime. Then 2 p ≡ 1 (mod q), so ord(2) \| p. But ord(2) 6 = 1, so ord(2) = p. And by Fermat’s Little Theorem, ord(2) \| (q − 1) ⇒ p ≤ q − 1 ⇒ q > p .In fact, for p > 2, q must be of the form 2 kp + 1. From the above, ord(2) \| (q − 1), or p \| (q − 1) ⇒ q = mp + 1. Since q must be odd, m must be even. Example 6.3 . Let p be a prime that is relatively prime to 10, and let n be an integer, 0 < n < p . Let d be the order of 10 modulo p.(a) Show that the length of the period of the decimal expansion of n/p is d.(b) Prove that if d is even, then the period of the decimal expansion of n/p can be divided into two halves, whose sum is 10 d/ 2 − 1. For example, 1/7 = 0 .142857, so d = 6, and 142 + 857 = 999 = 10 3 − 1. Solution . (a) Let m be the length of the period, and let n/p =0.a 1a2 . . . a m. Then 10 mnp = a1a2 . . . a m.a 1a2 . . . a m ⇒ (10 m − 1) np = a1a2 . . . a m, an integer. Since n and p are relatively prime, p must divide 10 m − 1, so d divides m. Conversely, p divides 10 d −1, so (10 d −1) n/p is an integer, with at most d digits. If we divide this integer by 10 d − 1, then we obtain a rational number, whose decimal expansion has period at most d. Therefore, m = d.(b) Let d = 2 k, so n/p = 0 .a 1a2 . . . a kak+1 . . . a 2k. Now p divides 10 d −1 = 10 2k − 1 = (10 k − 1)(10 k + 1). However, p cannot divide 10 k − 1 (since the order of 10 is 2 k), so p divides 10 k + 1. Hence, 10 knp = a1a2 . . . a k.a k+1 . . . a 2k ⇒ (10 k + 1) np = a1a2 . . . a k + 0 .a 1a2 . . . a k + 0 .a k+1 . . . a 2k is an integer. This can occur iff a1a2 . . . a k +ak+1 . . . a 2k is a number consisting only of 9s, and hence, equal to 10 k − 1. Problems 25 1. Prove that for all positive integers a > 1 and n, n \| φ(an − 1). 2. Prove that if p is a prime, then pp−1 has a prime factor that is congruent to 1 modulo p.3. For any integer a, set na = 101 a−100 ·2a. Show that for 0 ≤ a, b, c, d ≤ 99 , n a + nb ≡ nc + nd (mod 10100) implies {a, b } = {c, d }.(1994 Putnam Mathematical Competition) 4. Show that if 3 ≤ d ≤ 2n+1 , then d - (a2n 1) for all positive integers a. 7 Quadratic Residues Let m be an integer greater than 1, and a an integer relatively prime to m. If x2 ≡ a (mod m) has a solution, then we say that a is a quadratic residue of m. Otherwise, we say that a is a quadratic non-residue . Now, let p be an odd prime. Then the Legendre symbol (ap ) is assigned the value of 1 if a is a quadratic residue of p. Otherwise, it is assigned the value of −1. Theorem 7.1 . Let p be an odd prime, and a and b be integers relatively prime to p. Then (a) (ap ) ≡ a(p−1) /2 (mod p), and (b) (ap )( bp ) = (ab p ) . Proof . If the congruence x2 ≡ a (mod p) has a solution, then a(p−1) /2 ≡ xp−1 ≡ 1 (mod p), by Fermat’s Little Theorem. If the congruence x2 ≡ a (mod p) has no solutions, then for each i, 1 ≤ i ≤ p − 1, there is a unique j 6 = i, 1 ≤ j ≤ p − 1, such that ij ≡ a. Therefore, all the integers from 1 to p − 1 can be arranged into ( p − 1) /2 such pairs. Taking their product, a(p−1) /2 ≡ 1 · 2 · · · (p − 1) ≡ (p − 1)! ≡ − 1 (mod p), 26 by Wilson’s Theorem. Part (b) now follows from part (a). Remark . Part (a) is known as Euler’s criterion. Example 7.1 . Show that if p is an odd prime, then (1 p ) + (2 p ) · · · + (p − 1 p ) = 0 . Solution . Note that 1 2, 2 2, . . . , (( p − 1) /2) 2 are distinct modulo p,and that (( p + 1) /2) 2, . . . , ( p − 1) 2 represent the same residues, simply in reverse. Hence, there are exactly ( p−1) /2 quadratic residues, leaving ( p−1) /2quadratic non-residues. Therefore, the given sum contains ( p − 1) /2 1s and (p − 1) /2 −1s. Theorem 7.2 . Gauss’s Lemma . Let p be an odd prime and let a be relatively prime to p. Consider the least non-negative residues of a, 2 a, . . . , (( p − 1) /2) a modulo p. If n is the number of these residues that are greater than p/ 2, then (ap ) = ( −1) n. Theorem 7.3 . If p is an odd prime, then (−1 p ) = ( −1) (p−1) /2; that is, (−1 p ) = { 1 if p ≡ 1 (mod 4) , −1 if p ≡ 3 (mod 4) . Proof . This follows from Theorem 4.9 (and Theorem 7.1). Theorem 7.4 . If p is an odd prime, then (2 p ) = ( −1) (p2−1) /8; that is, (2 p ) = { 1 if p ≡ 1 or 7 (mod 8) , −1 if p ≡ 3 or 5 (mod 8) . 27 Proof . If p ≡ 1 or 5 (mod 8), then 2(p−1) /2 (p − 12 ) ! ≡ 2 · 4 · 6 · · · (p − 1) ≡ 2 · 4 · 6 · · · (p − 12 ) · ( −p − 32 ) · · · (−5) · (−3) · (−1) ≡ (−1) (p−1) /4 (p − 12 ) ! ⇒ 2(p−1) /2 ≡ (−1) (p−1) /4 (mod p). By Theorem 7.1, (2 p ) = ( −1) (p−1) /4. Hence, (2 p ) = 1 or −1 according as p ≡ 1 or 5 (mod 8). Similarly, if p ≡ 3 or 7 (mod 8), then 2(p−1) /2 (p − 12 ) ! ≡ 2 · 4 · 6 · · · (p − 32 ) · ( −p − 12 ) · · · (−5) · (−3) · (−1) ≡ (−1) (p+1) /4 (p − 12 ) ! ⇒ 2(p−1) /2 ≡ (−1) (p+1) /4 (mod p). Hence, (2 p ) = 1 or −1 according as p ≡ 7 or 3 (mod 8). Example 7.2 . Prove that if n is an odd positive integer, then every prime divisor of 2 n − 1 is of the form 8 k ± 1. (Compare to Example 6.2) Solution . Let p \| (2 n − 1), where p is prime. Let n = 2 m + 1. Then 2n ≡ 22m+1 ≡ 2(2 m)2 ≡ 1 (mod p) ⇒ (2 p ) = 1 ⇒ p is of the form 8 k ± 1. Theorem 7.5 . The Law of Quadratic Reciprocity . For distinct odd primes p and q, (pq ) ( qp ) = ( −1) p−12 · q−12 . Example 7.3 . For which primes p > 3 does the congruence x2 ≡ − 3(mod p) have a solution? Solution . We seek p for which (−3 p ) = (−1 p ) ( 3 p ) = 1. By quadratic reciprocity, (3 p ) ( p 3 ) = ( −1) (p−1) /2 = (−1 p ) , 28 by Theorem 7.3. Thus, in general, (−3 p ) = (−1 p ) ( 3 p ) = (p 3 ) ( −1 p )2 = (p 3 ) . And, ( p 3 ) = 1 iff p ≡ 1 (mod 3). Since p 6 ≡ 4 (mod 6), we have that x2 ≡ − 3(mod p) has a solution iff p ≡ 1 (mod 6). Example 7.4 . Show that if p = 2 n + 1, n ≥ 2, is prime, then 3 (p−1) /2 + 1 is divisible by p. Solution . We must have that n is even, say 2 k, for otherwise p ≡ 0(mod 3). By Theorem 7.1, (3 p ) ≡ 3(p−1) /2 (mod p). However, p ≡ 1 (mod 4), and p ≡ 4k + 1 ≡ 2 (mod 3) ⇒ (p 3 ) = −1, and by quadratic reciprocity, (3 p ) ( p 3 ) = ( −1) (p−1) /2 = 1 , so (3 p ) = −1 ⇒ 3(p−1) /2 + 1 ≡ 0 (mod p). Useful Facts • (a) If p is a prime and p ≡ 1 or 3 (mod 8), then there exist positive integers x and y such that p = x2 + 2 y2.(b) If p is a prime and p ≡ 1 (mod 6), then there exist positive integers x and y such that p = x2 + 3 y2.Problems 1. Show that if p > 3 is a prime, then the sum of the quadratic residues among the integers 1, 2, . . . , p − 1 is divisible by p.2. Let Fn denote the nth Fibonacci number. Prove that if p > 5 is a prime, then Fp ≡ (p 5 ) (mod p). 29 3. Show that 16 is a perfect 8 th power modulo p for any prime p.4. Let a, b, and c be positive integers that are pairwise relatively prime, and that satisfy a2 − ab + b2 = c2. Show that every prime factor of c is of the form 6 k + 1. 5. Let p be an odd prime and let ζ be a primitive pth root of unity; that is, ζ is a complex number such that ζp = 1 and ζk 6 = 1 for 1 ≤ k ≤ p − 1. Let Ap and Bp denote the set of quadratic residues and non-residues modulo p, respectively. Finally, let α = ∑ k∈Ap ζk and β = ∑ k∈Bp ζk.For example, for p = 7, α = ζ + ζ2 + ζ4 and β = ζ3 + ζ5 + ζ6. Show that α and β are the roots of x2 + x +1 − (−1 p ) p 4 = 0 . 8 Primitive Roots If the order of g modulo m is φ(m), then we say that g is a primitive root modulo m, or simply of m. Example 8.1 . Show that 2 is a primitive root modulo 3 n for all n ≥ 1. Solution . The statement is easily verified for n = 1, so assume the result is true for some n = k; that is, 2 φ(3 k ) ≡ 22·3k−1 ≡ 1 (mod 3 k). Now, let d be the order of 2 modulo 3 k+1 . Then 2 d ≡ 1 (mod 3 k+1 ) ⇒ 2d ≡ 1 (mod 3 k), so 2 · 3k−1 \| d. However, d \| φ(3 k+1 ), or d \| 2 · 3k. We deduce that d is either 2 · 3k−1 or 2 · 3k. Now we require the following lemma: Lemma. 2 2·3n−1 ≡ 1 + 3 n (mod 3 n+1 ), for all n ≥ 1. This is true for n = 1, so assume it is true for some n = k. Then by assumption, 22·3k−1 = 1 + 3 k + 3 k+1 m for some integer m ⇒ 22·3k = 1 + 3 k+1 + 3 k+2 M for some integer M (obtained by cubing) ⇒ 22·3k ≡ 1 + 3 k+1 (mod 3 k+2 ). By induction, the lemma is proved. Therefore, 2 2·3k−1 ≡ 1 + 3 k 6 ≡ 1 (mod 3 k+1 ), so the order of 2 modulo 3 k+1 is 2 · 3k, and again by induction, the result follows. 30 Corollary 8.2 . If 2 n ≡ − 1 (mod 3 k), then 3 k−1 \| n. Proof . The given implies 2 2n ≡ 1 (mod 3 k) ⇒ φ(3 k) \| 2n, or 3 k−1 \| n. Theorem 8.3 . If m has a primitive root, then it has φ(φ(m)) (distinct) primitive roots modulo m. Theorem 8.4 . The positive integer m has a primitive root iff m is one of 2, 4, pk, or 2 pk, where p is an odd prime. Theorem 8.5 . If g is a primitive root of m, then gn ≡ 1 (mod m) iff φ(m) \| n. Furthermore, gn0 ≡ gn1 iff φ(m) \| (n0 − n1). Proof . This follows directly from Theorem 6.1. Theorem 8.6 . If g is a primitive root of m, then the powers 1, g, g2,. . . , gφ(m)−1 represent each integer relatively prime to m uniquely modulo m.In particular, if m > 2, then gφ(m)/2 ≡ − 1 modulo m. Proof . Clearly, each power gi is relatively prime to m, and there are φ(m) integers relatively prime to m. Also, if gi ≡ gj (mod m), then gi−j ≡ 1 ⇒ φ(m) \| (i − j) by Theorem 8.6, so each of the powers are distinct modulo m. Hence, each integer relatively prime to m is some power gi modulo m. Furthermore, there is a unique i, 0 ≤ i ≤ φ(m) − 1, such that gi ≡ − 1 ⇒ g2i ≡ 1 ⇒ 2i = φ(m), or i = φ(m)/2. Proposition 8.7 . Let m be a positive integer. Then the only solutions to the congruence x2 ≡ 1 (mod m) are x ≡ ± 1 (mod m) iff m has a primitive root. Proof . This follows from Example 4.9. Example 8.2 . For a positive integer m, let S be the set of positive integers less than m that are relatively prime to m, and let P be the product of the elements in S. Show that P ≡ ± 1 (mod m), with P ≡ − 1 (mod m)iff m has a primitive root. Solution . We use a similar strategy as in the proof of Wilson’s Theorem. The result is clear for m = 2, so assume that m ≥ 3. We partition S as follows: Let A be the elements of S that are solutions to the congruence x2 ≡ 1 (mod m), and let B be the remaining elements. The elements in B can be arranged into pairs, by pairing each with its distinct multiplicative inverse. Hence, the product of the elements in B is 1 modulo m.The elements in A may also be arranged into pairs, by pairing each with 31 its distinct additive inverse, i.e. x and m − x. These must be distinct, because otherwise, x = m/ 2, which is not relatively prime to m. Note that their product is x(m − x) ≡ mx − x2 ≡ − 1 (mod m). Now if m has a primitive root, then by Proposition 8.7, A consists of only the two elements 1 and −1, so P ≡ − 1 (mod m). Otherwise, by Example 4.9, the number of elements of A is a power of two that is at least 4, so the number of such pairs in A is even, and P ≡ 1 (mod m). Remark . For m prime, this simply becomes Wilson’s Theorem. Theorem 8.8 .(1) If g is a primitive root of p, p a prime, then g or g + p is a primitive root of p2, according as gp−1 6 ≡ 1 (mod p2) or gp−1 ≡ 1 (mod p2). (2) If g is a primitive root of pk, where k ≥ 2 and p is prime, then g is a primitive root of pk+1 .By Theorem 8.6, given a primitive root g of m, for each a relatively prime to m, there exists a unique integer i modulo φ(m) such that gi ≡ a (mod m). This i is called the index of a with respect to the base g, denoted by ind g(a)(i is dependent on g, so it must be specified). Indices have striking similarity to logarithms, as seen in the following properties: (1) ind g(1) ≡ 0 (mod φ(m)), ind g(g) ≡ 1 (mod φ(m)), (2) a ≡ b (mod m) ⇒ ind g(a) ≡ ind g(b) (mod φ(m)), (3) ind g(ab ) ≡ ind g(a) + ind g(b) (mod φ(m)), (4) ind g(ak) ≡ k ind g(a) (mod φ(m)). Theorem 8.9 . If p is a prime and a is not divisible by p, then the con-gruence xn ≡ a (mod p) has gcd( n, p − 1) solutions or no solutions according as a(p−1) / gcd( n,p −1) ≡ 1 (mod p) or a(p−1) / gcd( n,p −1) 6 ≡ 1 (mod p). Proof . Let g be a primitive root of p, and let i be the index of a with respect to g. Also, any solution x must be relatively prime to p, so let u be the index of x. Then the congruence xn ≡ a becomes gnu ≡ gi (mod p) ⇔ nu ≡ i (mod p − 1). Let k = gcd( n, p − 1). Since g is a primitive root of p, k \| i ⇔ gi(p−1) /k ≡ a(p−1) /k ≡ 1. The result now follows from Theorem 4.11. 32 Remark . Taking p to be an odd prime and n = 2, we deduce Euler’s criterion. Example 8.3 Let n ≥ 2 be an integer and p = 2 n + 1. Show that if 3(p−1) /2 + 1 ≡ 0 (mod p), then p is a prime. (The converse to Example 7.4.) Solution . From 3 (p−1) /2 ≡ 32n−1 ≡ − 1 (mod p), we obtain 3 2n ≡ 1(mod p), so the order of 3 is 2 n = p−1, but the order also divides φ(p) ≥ p−1. Therefore, φ(p) = p − 1, and p is a prime. Example 8.4 . Prove that if n = 3 k−1, then 2 n ≡ − 1 (mod 3 k). (A partial converse to Corollary 8.2.) Solution . By Example 8.1, 2 is a primitive root of 3 k. Therefore, 2 has order φ(3 k) = 2 · 3k−1 = 2 n ⇒ 22n ≡ 1 ⇒ (2 n − 1)(2 n + 1) ≡ 0 (mod 3 k). However, 2 n − 1 ≡ (−1) 3k−1 − 1 ≡ 1 6 ≡ 0 (mod 3), so 2 n + 1 ≡ 0 (mod 3 k). Example 8.5 . Find all positive integers n > 1 such that 2n + 1 n2 is an integer. (1990 IMO) Solution . Clearly, n must be odd. Now assume that 3 k‖n; that is, 3 k is the highest power of 3 dividing n. Then 3 2k \| n2 \| (2 n + 1) ⇒ 2n ≡ − 1(mod 3 2k) ⇒ 32k−1 \| n, by Corollary 8.2 ⇒ 2k − 1 ≤ k ⇒ k ≤ 1, showing that n has at most one factor of 3. We observe that n = 3 is a solution. Suppose that n has a prime factor greater than 3; let p be the least such prime. Then p \| (2 n +1) ⇒ 2n ≡ − 1 (mod p). Let d be the order of 2 modulo p. Since 2 2n ≡ 1, d \| 2n. If d is odd, then d \| n ⇒ 2n ≡ 1, contradiction, so d is even, say d = 2 d1. Then 2 d1 \| 2n ⇒ d1 \| n. Also, d \| (p − 1), or 2d1 \| (p − 1) ⇒ d1 ≤ (p − 1) /2 < p . But d1 \| n, so d1 = 1 or d1 = 3. If d1 = 1, then d = 2, and 2 2 ≡ 1 (mod p), contradiction. If d1 = 3, then d = 6, and 26 ≡ 1 (mod p), or p \| 63 ⇒ p = 7. However, the order of 2 modulo 7 is 3, which is odd, again contradiction. Therefore, no such p can exist, and the only solution is n = 3. Useful Facts • All prime divisors of the Fermat number 2 2n 1, n > 1, are of the form 2n+2 k + 1. 33 Problems 1. Let p be an odd prime. Prove that 1i + 2 i + · · · + ( p − 1) i ≡ 0 (mod p)for all i, 0 ≤ i ≤ p − 2. 2. Show that if p is an odd prime, then the congruence x4 ≡ − 1 (mod p)has a solution iff p ≡ 1 (mod 8). 3. Show that if a and n are positive integers with a odd, then a2n ≡ 1(mod 2 n+2 ). 4. The number 142857 has the remarkable property that multiplying it by 1, 2, 3, 4, 5, and 6 cyclically permutes the digits. What are other numbers that have this property? Hint: Compute 142857 × 7. 9 Dirichlet Series Despite the intimidating name, Dirichlet series are easy to work with, and can provide quick proofs to certain number-theoretic identities, such as Example 3.2. Let α be a function taking the positive integers to the integers. Then we say that f (s) = ∞ ∑ n=1 α(n) ns = α(1) + α(2) 2s + α(3) 3s + · · · is the Dirichlet series generating function (Dsgf) of the function α,which we denote by f (s) ↔ α(n). Like general generating functions, these generating functions are used to provide information about their correspond-ing number-theoretic functions, primarily through manipulation of the gen-erating functions. Let 1 denote the function which is 1 for all positive integers; that is, 1( n) = 1 for all n. Let δ1(n) be the function defined by δ1(n) = { 1 if n = 1 , 0 if n > 1. It is easy to check that 1 and δ1 are multiplicative. 34 Now, let α and β be functions taking the positive integers to the integers. The convolution of α and β, denoted α ∗ β, is defined by (α ∗ β)( n) = ∑ d\|n α(d)β(n/d ). Note that convolution is symmetric; that is, α ∗ β = β ∗ α. Theorem 9.1 . Let f (s) ↔ α(n) and g(s) ↔ β(n). Then ( f · g)( s) ↔ (α ∗ β)( n). We now do three examples. The Dsgf of 1( n) is the well-known Riemann Zeta function ζ(s): ζ(s) = ∞ ∑ n=1 1 ns = 1 + 12s + 13s + · · · , so ζ(s) ↔ 1( n). This function will play a prominent role in this theory. What makes this theory nice to work with is that we may work with these functions at a purely formal level; no knowledge of the analytic properties of ζ(s) or indeed of any other Dsgf is required. By Theorem 9.1, the number-theoretic function corresponding to ζ2(s) is ∑ d\|n 1( d)1( n/d ) = ∑ d\|n 1 = τ (n). Hence, ζ2(s) ↔ τ (n). Finally, it is clear that 1 ↔ δ1(n). If α is a multiplicative function, then we can compute the Dsgf corre-sponding to α using the following theorem. Theorem 9.2 . Let α be a multiplicative function. Then ∞ ∑ n=1 α(n) ns = ∏ p ∞ ∑ k=0 α(pk) pks = ∏ p [1 + α(p)p−s + α(p2)p−2s + α(p3)p−3s + · · · ], where the product on the right is taken over all prime numbers. 35 As before, if we take α = 1, then we obtain ζ(s) = ∏ p (1 + p−s + p−2s + p−3s + · · · )= ∏ p ( 11 − p−s ) = 1 ∏ p (1 − p−s), an identity that will be useful. We say that a positive integer n > 1 is square-free if n contains no repeated prime factors; that is, p2 - n for all primes p. With this in mind, we define the M¨ obius function μ as follows: μ(n) =  1 if n = 1 , 0 if n is not square-free, and (−1) k if n is square-free and has k prime factors . It is easy to check that μ is multiplicative. By Theorem 9.2, the corresponding Dsgf is given by ∏ p (1 − p−s) = 1 ζ(s). Hence, 1 /ζ (s) ↔ μ(n), and this property makes the the seemingly mysterious function μ very important, as seen in the following theorem. Theorem 9.3 . (M¨ obius Inversion Formula) Let α and β be functions such that β(n) = ∑ d\|n α(d). Then α(n) = ∑ d\|n μ(n/d )β(d). Proof . Let f (s) ↔ α(n) and g(s) ↔ β(n). The condition is equivalent to β = α ∗ 1, or g(s) = f (s)ζ(s), and the conclusion is equivalent to α = β ∗ μ,or f (s) = g(s)/ζ (s). Theorem 9.4 . Let f (s) ↔ α(n). Then for any integer k, f (s − k) ↔ nkα(n). 36 For more on Dirichlet series, and generating functions in general, see H. Wilf, Generatingfunctionology .Problems 1. Let α, β, and γ be functions taking the positive integers to the integers. (a) Prove that α ∗ δ1 = α.(b) Prove that ( α ∗ β) ∗ γ = α ∗ (β ∗ γ). (c) Prove that if α and β are multiplicative, then so is α ∗ β.2. Prove that the following relations hold: ζ(s − 1) ζ(s) ↔ φ(n),ζ(s)ζ(s − 1) ↔ σ(n),ζ(s) ζ(2 s) ↔ \| μ(n)\|. Let the prime factorization of a positive integer n > 1 be pe1 1 pe2 2 · · · pek k .Define the functions λ and θ by λ(n) = ( −1) e1+e2+··· +ek and θ(n) = 2 k.Set λ(1) = θ(1) = 1. Show that λ and θ are multiplicative, and that ζ(2 s) ζ(s) ↔ λ(n) and ζ2(s) ζ(2 s) ↔ θ(n). For all positive integers n, let f (n) = n ∑ m=1 n gcd( m, n ). (a) Show that f (n) = ∑ d\|n dφ (d). (b) Let n = pe1 1 pe2 2 · · · pek k 1 be the prime factorization of n. Show that f (n) = (p2e1+1 1 + 1 p1 + 1 ) ( p2e2+1 2 + 1 p2 + 1 ) · · · (p2ek +1 1 + 1 pk + 1 ) . Verify Example 3.2 in one calculation. 37 6. Let id denote the identity function; that is, id( n) = n for all n. Verify each of the following identities in one calculation: (a) φ ∗ τ = σ.(b) μ ∗ 1 = δ1.(c) μ ∗ id = φ.(d) φ ∗ σ = id · τ .(e) σ ∗ id = 1 ∗ (id · τ ). 7. Let a1, a2, . . . , be the sequence of positive integers satisfying ∑ d\|n ad = 2 n for all n. Hence, a1 = 2, a2 = 2 2 − 2 = 2, a3 = 2 3 − 2 = 6, a4 =24 − 2 − 2 = 12, and so on. Show that for all n, n \| an.Hint: Don’t use the Dsgf of ( an)∞ 1 ; use the M¨ obius Inversion Formula. Bigger Hint: Consider the function f : [0 , 1] → [0 , 1] defined by f (x) = {2x}, where {x} = x − b xc is the fractional part of x. Find how the formula in the problem relates to the function f (n) = f ◦ f ◦ · · · ◦ f ︸︷︷︸ n .8. For all non-negative integers k, let σk be the function defined by σk(n) = ∑ d\|n dk. Thus, σ0 = τ and σ1 = σ. Prove that ζ(s)ζ(s − k) ↔ σk(n). 10 Miscellaneous Topics 10.1 Pell’s Equations Pell’s equations (or Fermat’s equations, as they are rightly called) are diophantine equations of the form x2 − dy 2 = N , where d is a positive non-square integer. There always exist an infinite number of solutions when N = 1, which we characterize. 38 Theorem 10.1.1 . If ( a, b ) is the lowest positive integer solution of x2 − dy 2 = 1, then all positive integer solutions are of the form (xn, y n) = ( (a + b√d)n + ( a − b√d)n 2 , (a + b√d)n − (a − b√d)n 2√d ) . We will not give a proof here, but we will verify that every pair indicated by the formula is a solution. The pair ( xn, y n) satisfy the equations xn + yn √d = ( a + b√d)n, and xn − yn √d = ( a − b√d)n. Therefore, x2 n − dy 2 n = ( xn + yn √d)( xn − yn √d)= ( a + b√d)n(a − b√d)n = ( a2 − db 2)n = 1 , since ( a, b ) is a solution. Remark . The sequences ( xn), ( yn) satisfy the recurrence relations xn =2ax n−1 − xn−2, yn = 2 ay n−1 − yn−2.For x2 − dy 2 = −1, the situation is similar. If ( a, b ) is the least positive solution, then the ( xn, y n) as above for n odd are the solutions of x2 − dy 2 = −1, and the ( xn, y n) for n even are the solutions of x2 − dy 2 = 1. Example 10.1.1 Find all solutions in pairs of positive integers ( x, y ) to the equation x2 − 2y2 = 1. Solution . We find that the lowest positive integer solution is (3,2), so all positive integer solutions are given by (xn, y n) = ( (3 + 2 √2) n + (3 − 2√2) n 2 , (3 + 2 √2) n − (3 − 2√2) n 2√2 ) . The first few solutions are (3,2), (17,12), and (99,70). 39 Example 10.1.2 . Prove that the equation x2 −dy 2 = −1 has no solution in integers if d ≡ 3 (mod 4). Solution . It is apparent that d must have a prime factor of the form 4 k +3, say q. Then x2 ≡ − 1 (mod q), which by Theorem 4.9 is a contradiction. Problems 1. In the sequence 12, 53, 11 8 , 27 19 , . . . , the denominator of the nth term ( n > 1) is the sum of the numerator and the denominator of the ( n − 1) th term. The numerator of the nth term is the sum of the denominators of the nth and ( n−1) th term. Find the limit of this sequence. (1979 Atlantic Region Mathematics League) 2. Let x0 = 0, x1 = 1, xn+1 = 4 xn − xn−1, and y0 = 1, y1 = 2, yn+1 =4yn − yn−1. Show for all n ≥ 0 that y2 n = 3 x2 n (1988 Canadian Mathematical Olympiad) 3. The polynomials P , Q are such that deg P = n, deg Q = m, have the same leading coefficient, and P 2(x) = ( x2 − 1) Q2(x) + 1. Show that P ′(x) = nQ (x). (1978 Swedish Mathematical Olympiad, Final Round) 10.2 Farey Sequences The nth Farey sequence is the sequence of all reduced rationals in [0,1], with both numerator and denominator no greater than n, in increasing order. Thus, the first 5 Farey sequences are: 0/1, 1/1, 0/1, 1/2, 1/1, 0/1, 1/3, 1/2, 2/3, 1/1, 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1, 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1. Properties of Farey sequences include the following: 40 (1) If a/b and c/d are consecutive fractions in the same sequence, in that order, then ad − bc = 1. (2) If a/b , c/d , and e/f are consecutive fractions in the same sequence, in that order, then a + eb + f = cd. (3) If a/b and c/d are consecutive fractions in the same sequence, then among all fractions between the two, ( a + c)/(b + d) (reduced) is the unique fraction with the smallest denominator. For proofs of these and other interesting properties, see Ross Honsberger, “Farey Sequences”, Ingenuity in Mathematics .Problems 1. Let a1, a2, . . . , am be the denominators of the fractions in the nth Farey sequence, in that order. Prove that 1 a1a2 1 a2a3 · · · + 1 am−1am = 1 . 10.3 Continued Fractions Let a0, a1, . . . , an be real numbers, all positive, except possibly a0. Then let 〈a0, a 1, . . . , a n〉 denote the continued fraction a0 + 1 a1 + · · · + 1 an−1 + 1 an . If each ai is an integer, then we say that the continued fraction is simple .Define sequences ( pk) and ( qk) as follows: p−1 = 0 , p0 = a0, and pk = akpk−1 + pk−2,q−1 = 0 , q0 = 1 , and qk = akpk−1 + qk−2, for k ≥ 1. Theorem 10.3.1 . For all x > 0 and k ≥ 1, 〈a0, a 1, . . . , a k−1, x 〉 = xp k−1 + pk−2 xq k−1 + qk−2 . 41 In particular, 〈a0, a 1, . . . , a k〉 = pk qk . Theorem 10.3.2 . For all k ≥ 0, (1) pkqk−1 − pk−1qk = ( −1) k−1,(2) pkqk−2 − pk−2qk = ( −1) kak.Define ck to be the kth convergence 〈a0, a 1, . . . , a k〉 = pk/q k. Theorem 10.3.3 . c0 < c 2 < c 4 < · · · < c 5 < c 3 < c 1.For a nice connection between continued fractions, linear diophantine equations, and Pell’s equations, see Andy Liu, “Continued Fractions and Diophantine Equations”, Volume 3, Issue 2, Mathematical Mayhem .Problems 1. Let a = 〈1, 2, . . . , 99 〉 and b = 〈1, 2, . . . , 99 , 100 〉. Prove that \|a − b\| < 199!100! . (1990 Tournament of Towns) 2. Evaluate 8 √√√√√2207 − 12207 − 12207 − · · · . Express your answer in the form a+b√cd , where a, b, c, d are integers. (1995 Putnam) 10.4 The Postage Stamp Problem Let a and b be relatively prime positive integers greater than 1. Consider the set of integers of the form ax + by , where x and y are non-negative integers. The following are true: (1) The greatest integer that cannot be written in the given form is ( a − 1)( b − 1) − 1 = ab − a − b.42 (2) There are 12 (a − 1)( b − 1) positive integers that cannot be written in the given form. (3) For all integers t, 0 ≤ t ≤ ab − a − b, t can be written in the given form iff ab − a − b − t cannot be. (If you have not seen or attempted this enticing problem, it is strongly suggested you have a try before reading the full solution.) Before presenting the solution, it will be instructive to look at an example. Take a = 12 and b = 5. The first few non-negative integers, in rows of 12, with integers that cannot be written in the given form in bold, are shown: 0 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 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 With this arrangement, one observation should become immediately ap-parent, namely that bold numbers in each column end when they reach a multiple of 5. It should be clear that when reading down a column, once one hits an integer that can be written in the given form, then all successive integers can be as well, since we are adding 12 for each row we go down. It will turn out that this one observation is the key to the solution. Proof . Define a grapefruit to be an integer that may be written in the given form. For each i, 0 ≤ i ≤ a − 1, let mi be the least non-negative integer such that b \| (i + am i). It is obvious that for k ≥ mi, i + ak is a grapefruit. We claim that for 0 ≤ k ≤ mi − 1, i + ak is not a grapefruit. It is sufficient to show that i + a(mi − 1) is not a grapefruit, if mi ≥ 1. Let i + am i = bn i, ni ≥ 0. Since i + a(mi − b) = b(ni − a), mi must be strictly less than b; otherwise, we can find a smaller mi. Then i + a(mi − b) ≤ a−1−a = −1, so ni < a , or ni ≤ a−1. Suppose that ax +by = i+a(mi −1) = bn i − a, for some non-negative integers x and y. Then a(x + 1) = b(ni − y), so ni − y is positive. Since a and b are relatively prime, a divides ni − y.However, ni ≤ a − 1 ⇒ ni − y ≤ a − 1, contradiction. Therefore, the greatest non-grapefruit is of the form bn i − a, ni ≤ a − 1. The above argument also shows that all positive integers of this form are also non-grapefruits. Hence, the greatest non-grapefruit is b(a−1) −a = ab −a−b,proving (1). 43 Now, note that there are mi non-grapefruits in column i. The above tells us the first grapefruit appearing in column i is nib. Since 0, b, 2 b, . . . , ( a−1) b appear in different columns (because a and b are relatively prime), and there are a columns, we conclude that as i varies from 0 to a − 1, ni takes on 0, 1, . . . , a − 1, each exactly once. Therefore, summing over i, 0 ≤ i ≤ a − 1, ∑ i (i + am i) = ∑ i i + ∑ i am i = a(a − 1) 2 + a ∑ i mi = ∑ i bn i = a(a − 1) b 2 ⇒ a ∑ i mi = a(a − 1)( b − 1) 2 ⇒ ∑ i mi = (a − 1)( b − 1) 2 , proving (2). Finally, suppose that ax 1 + by 1 = t, and ax 2 + by 2 = ab − a − b − t, for some non-negative integers x1, x2, y1, and y2. Then a(x1 + x2) + b(y1 + y2) = ab − a − b, contradicting (1). So, if we consider the pairs ( t, ab − a − b − t), 0 ≤ t ≤ (a − 1)( b − 1) /2 − 1, at most one element in each pair can be written in the given form. However, we have shown that exactly ( a − 1)( b − 1) /2 integers cannot be written in the given form, which is the number of pairs. Therefore, exactly one element of each pair can be written in the given form, proving (3). Remark . There is a much shorter proof using Corollary 2.4. Can you find it? For me, this type of problem epitomizes problem solving in number theory, and generally mathematics, in many ways. If I merely presented the proof by itself, it would look artificial and unmotivated. However, by looking at a specific example, and finding a pattern, we were able to use that pattern as a springboard and extend it into a full proof. The algebra in the proof is really nothing more than a translation of observed patterns into formal notation. (Mathematics could be described as simply the study of pattern.) Note also that we used nothing more than very elementary results, showing how powerful basic concepts can be. It may have been messy, but one should never be afraid to get one’s hands dirty; indeed, the deeper you go, the 44 more you will understand the importance of these concepts and the subtle relationships between them. By trying to see an idea through to the end, one can sometimes feel the proof almost working out by itself. The moral of the story is: A simple idea can go a long way. For more insights on the postage stamp problem, see Ross Honsberger, “A Putnam Paper Problem”, Mathematical Gems II .Problems 1. Let a, b, and c be positive integers, no two of which have a common divisor greater than 1. Show that 2 abc − ab − bc − ca is the largest integer that cannot be expressed in the form xab + yca + zab , where x, y, and z are non-negative integers. (1983 IMO) References A. Adler & J. Coury, The Theory of Numbers , Jones and Bartlett I. Niven & H. Zuckerman, An Introduction to the Theory of Numbers ,John Wiley & Sons c© First Version October 1995 c© Second Version January 1996 c© Third Version April 1999 c© Fourth Version May 2000 Thanks to Ather Gattami for an improvement to the proof of the Postage Stamp Problem. This document was typeset under L ATEX, and may be freely distributed provided the contents are unaltered and this copyright notice is not removed. Any comments or corrections are always welcomed. It may not be sold for profit or incorporated in commercial documents without the express permis-sion of the copyright holder. So there. 45
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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in College Algebra 2e 6.5 Logarithmic Properties College Algebra 2e6.5 Logarithmic Properties Contents Contents Highlights Table of contents Preface 1 Prerequisites 2 Equations and Inequalities 3 Functions 4 Linear Functions 5 Polynomial and Rational Functions 6 Exponential and Logarithmic Functions Introduction to Exponential and Logarithmic Functions 6.1 Exponential Functions 6.2 Graphs of Exponential Functions 6.3 Logarithmic Functions 6.4 Graphs of Logarithmic Functions 6.5 Logarithmic Properties 6.6 Exponential and Logarithmic Equations 6.7 Exponential and Logarithmic Models 6.8 Fitting Exponential Models to Data Chapter Review Exercises 7 Systems of Equations and Inequalities 8 Analytic Geometry 9 Sequences, Probability, and Counting Theory Answer Key Index Search for key terms or text. Close Learning Objectives In this section, you will: Use the product rule for logarithms. Use the quotient rule for logarithms. Use the power rule for logarithms. Expand logarithmic expressions. Condense logarithmic expressions. Use the change-of-base formula for logarithms. Figure 1 The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan) In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be basic. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is basic, consider the following pH levels of some common substances: Battery acid: 0.8 Stomach acid: 2.7 Orange juice: 3.3 Pure water: 7 (at 25° C) Human blood: 7.35 Fresh coconut: 7.8 Sodium hydroxide (lye): 14 To determine whether a solution is acidic or basic, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where H+H+H+ is the concentration of hydrogen ion in the solution pH=−log([H+])=log(1[H+])pH=−log([H+])=log(1[H+])pH=−log([H+])=log(1[H+]) The equivalence of −log([H+])−log([H+])−log([H+]) and log(1[H+])log(1[H+])log(1[H+]) is one of the logarithm properties we will examine in this section. Using the Product Rule for Logarithms Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove. log b 1=0 log b b=1 log b 1=0 log b b=1 log b 1=0 log b b=1 For example, log 5 1=0 log 5 1=0 log 5 1=0 since 5 0=1.5 0=1.5 0=1. And log 5 5=1 log 5 5=1 log 5 5=1 since 5 1=5.5 1=5.5 1=5. Next, we have the inverse property. log b(b x)=x b log b x=x,x>0 log b(b x)=x b log b x=x,x>0 log b(b x)=x b log b x=x,x>0 For example, to evaluate log(100),log(100),log(100), we can rewrite the logarithm as log 10(10 2),log 10(10 2),log 10(10 2), and then apply the inverse property log b(b x)=x log b(b x)=x log b(b x)=x to get log 10(10 2)=2.log 10(10 2)=2.log 10(10 2)=2. To evaluate e ln(7),e ln(7),e ln(7), we can rewrite the logarithm as e log e 7,e log e 7,e log e 7, and then apply the inverse property b log b x=x b log b x=x b log b x=x to get e log e 7=7.e log e 7=7.e log e 7=7. Finally, we have the one-to-one property. log b M=log b N if and only if M=N log b M=log b N if and only if M=N log b M=log b N if and only if M=N We can use the one-to-one property to solve the equation log 3(3 x)=log 3(2 x+5)log 3(3 x)=log 3(2 x+5)log 3(3 x)=log 3(2 x+5) for x.x.x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x:x:x: 3 x=2 x+5 x=5 Set the arguments equal.Subtract 2 x.3 x=2 x+5 Set the arguments equal.x=5 Subtract 2 x.3 x=2 x+5 Set the arguments equal.x=5 Subtract 2 x. But what about the equation log 3(3 x)+log 3(2 x+5)=2?log 3(3 x)+log 3(2 x+5)=2?log 3(3 x)+log 3(2 x+5)=2? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation. Recall that we use the product rule of exponents to combine the product of powers by adding exponents: x a x b=x a+b.x a x b=x a+b.x a x b=x a+b. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below. Given any real number x x x and positive real numbers M,N,M,N,M,N, and b,b,b, where b≠1,b≠1,b≠1, we will show log b(M N)=log b(M)+log b(N).log b(M N)=log b(M)+log b(N).log b(M N)=log b(M)+log b(N). Let m=log b M m=log b M m=log b M and n=log b N.n=log b N.n=log b N. In exponential form, these equations are b m=M b m=M b m=M and b n=N.b n=N.b n=N. It follows that log b(M N)=log b(b m b n)=log b(b m+n)=m+n=log b(M)+log b(N)Substitute for M and N.Apply the product rule for exponents.Apply the inverse property of logs.Substitute for m and n.log b(M N)=log b(b m b n)Substitute for M and N.=log b(b m+n)Apply the product rule for exponents.=m+n Apply the inverse property of logs.=log b(M)+log b(N)Substitute for m and n.log b(M N)=log b(b m b n)Substitute for M and N.=log b(b m+n)Apply the product rule for exponents.=m+n Apply the inverse property of logs.=log b(M)+log b(N)Substitute for m and n. Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider log b(w x y z).log b(w x y z).log b(w x y z). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors: log b(w x y z)=log b w+log b x+log b y+log b z log b(w x y z)=log b w+log b x+log b y+log b z log b(w x y z)=log b w+log b x+log b y+log b z The Product Rule for Logarithms The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms. log b(M N)=log b(M)+log b(N)for b>0 log b(M N)=log b(M)+log b(N)for b>0 log b(M N)=log b(M)+log b(N)for b>0 How To Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms. Factor the argument completely, expressing each whole number factor as a product of primes. Write the equivalent expression by summing the logarithms of each factor. Example 1 Using the Product Rule for Logarithms Expand log 3(30 x(3 x+4)).log 3(30 x(3 x+4)).log 3(30 x(3 x+4)). Solution We begin by factoring the argument completely, expressing 30 30 30 as a product of primes. log 3(30 x(3 x+4))=log 3(2⋅3⋅5⋅x⋅(3 x+4))log 3(30 x(3 x+4))=log 3(2⋅3⋅5⋅x⋅(3 x+4))log 3(30 x(3 x+4))=log 3(2⋅3⋅5⋅x⋅(3 x+4)) Next we write the equivalent equation by summing the logarithms of each factor. log 3(30 x(3 x+4))=log 3(2)+log 3(3)+log 3(5)+log 3(x)+log 3(3 x+4)log 3(30 x(3 x+4))=log 3(2)+log 3(3)+log 3(5)+log 3(x)+log 3(3 x+4)log 3(30 x(3 x+4))=log 3(2)+log 3(3)+log 3(5)+log 3(x)+log 3(3 x+4) Try It #1 Expand log b(8 k).log b(8 k).log b(8 k). Using the Quotient Rule for Logarithms For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: x a x b=x a−b.x a x b=x a−b.x a x b=x a−b. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule. Given any real number x x x and positive real numbers M,M,M,N,N,N, and b,b,b, where b≠1,b≠1,b≠1, we will show log b(M N)=log b(M)−log b(N).log b(M N)=log b(M)−log b(N).log b(M N)=log b(M)−log b(N). Let m=log b M m=log b M m=log b M and n=log b N.n=log b N.n=log b N. In exponential form, these equations are b m=M b m=M b m=M and b n=N.b n=N.b n=N. It follows that log b(M N)=log b(b m b n)=log b(b m−n)=m−n=log b(M)−log b(N)Substitute for M and N.Apply the quotient rule for exponents.Apply the inverse property of logs.Substitute for m and n.log b(M N)=log b(b m b n)Substitute for M and N.=log b(b m−n)Apply the quotient rule for exponents.=m−n Apply the inverse property of logs.=log b(M)−log b(N)Substitute for m and n.log b(M N)=log b(b m b n)Substitute for M and N.=log b(b m−n)Apply the quotient rule for exponents.=m−n Apply the inverse property of logs.=log b(M)−log b(N)Substitute for m and n. For example, to expand log(2 x 2+6 x 3 x+9),log(2 x 2+6 x 3 x+9),log(2 x 2+6 x 3 x+9), we must first express the quotient in lowest terms. Factoring and canceling we get, log(2 x 2+6 x 3 x+9)=log(2 x(x+3)3(x+3))=log(2 x 3)Factor the numerator and denominator.Cancel the common factors.log(2 x 2+6 x 3 x+9)=log(2 x(x+3)3(x+3))Factor the numerator and denominator.=log(2 x 3)Cancel the common factors.log(2 x 2+6 x 3 x+9)=log(2 x(x+3)3(x+3))Factor the numerator and denominator.=log(2 x 3)Cancel the common factors. Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule. log(2 x 3)=log(2 x)−log(3)=log(2)+log(x)−log(3)log(2 x 3)=log(2 x)−log(3)=log(2)+log(x)−log(3)log(2 x 3)=log(2 x)−log(3)=log(2)+log(x)−log(3) The Quotient Rule for Logarithms The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms. log b(M N)=log b M−log b N log b(M N)=log b M−log b N log b(M N)=log b M−log b N How To Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely. Example 2 Using the Quotient Rule for Logarithms Expand log 2(15 x(x−1)(3 x+4)(2−x)).log 2(15 x(x−1)(3 x+4)(2−x)).log 2(15 x(x−1)(3 x+4)(2−x)). Solution First we note that the quotient is factored and in lowest terms, so we apply the quotient rule. log 2(15 x(x−1)(3 x+4)(2−x))=log 2(15 x(x−1))−log 2((3 x+4)(2−x))log 2(15 x(x−1)(3 x+4)(2−x))=log 2(15 x(x−1))−log 2((3 x+4)(2−x))log 2(15 x(x−1)(3 x+4)(2−x))=log 2(15 x(x−1))−log 2((3 x+4)(2−x)) Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5. log 2(15 x(x−1))−log 2((3 x+4)(2−x))=[log 2(3)+log 2(5)+log 2(x)+log 2(x−1)]−[log 2(3 x+4)+log 2(2−x)]=log 2(3)+log 2(5)+log 2(x)+log 2(x−1)−log 2(3 x+4)−log 2(2−x)log 2(15 x(x−1))−log 2((3 x+4)(2−x))=[log 2(3)+log 2(5)+log 2(x)+log 2(x−1)]−[log 2(3 x+4)+log 2(2−x)]=log 2(3)+log 2(5)+log 2(x)+log 2(x−1)−log 2(3 x+4)−log 2(2−x)log 2(15 x(x−1))−log 2((3 x+4)(2−x))=[log 2(3)+log 2(5)+log 2(x)+log 2(x−1)]−[log 2(3 x+4)+log 2(2−x)]=log 2(3)+log 2(5)+log 2(x)+log 2(x−1)−log 2(3 x+4)−log 2(2−x) Analysis There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for x=−4 3 x=−4 3 x=−4 3 and x=2.x=2.x=2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that x>0,x>0,x>0,x>1,x>1,x>1,x>−4 3,x>−4 3,x>−4 3, and x<2.x<2.x<2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises. Try It #2 Expand log 3(7 x 2+21 x 7 x(x−1)(x−2)).log 3(7 x 2+21 x 7 x(x−1)(x−2)).log 3(7 x 2+21 x 7 x(x−1)(x−2)). Using the Power Rule for Logarithms We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as x 2?x 2?x 2? One method is as follows: log b(x 2)=log b(x⋅x)=log b x+log b x=2 log b x log b(x 2)=log b(x⋅x)=log b x+log b x=2 log b x log b(x 2)=log b(x⋅x)=log b x+log b x=2 log b x Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example, 100=10 2 3–√=3 1 2 1 e=e−1 100=10 2 3=3 1 2 1 e=e−1 100=10 2 3=3 1 2 1 e=e−1 The Power Rule for Logarithms The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base. log b(M n)=n log b M log b(M n)=n log b M log b(M n)=n log b M How To Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm. Express the argument as a power, if needed. Write the equivalent expression by multiplying the exponent times the logarithm of the base. Example 3 Expanding a Logarithm with Powers Expand log 2 x 5.log 2 x 5.log 2 x 5. Solution The argument is already written as a power, so we identify the exponent, 5, and the base, x,x,x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log 2(x 5)=5 log 2 x log 2(x 5)=5 log 2 x log 2(x 5)=5 log 2 x Try It #3 Expand ln x 2.ln x 2.ln x 2. Example 4 Rewriting an Expression as a Power before Using the Power Rule Expand log 3(25)log 3(25)log 3(25) using the power rule for logs. Solution Expressing the argument as a power, we get log 3(25)=log 3(5 2).log 3(25)=log 3(5 2).log 3(25)=log 3(5 2). Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log 3(5 2)=2 log 3(5)log 3(5 2)=2 log 3(5)log 3(5 2)=2 log 3(5) Try It #4 Expand ln(1 x 2).ln(1 x 2).ln(1 x 2). Example 5 Using the Power Rule in Reverse Rewrite 4 ln(x)4 ln(x)4 ln(x) using the power rule for logs to a single logarithm with a leading coefficient of 1. Solution Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression 4 ln(x),4 ln(x),4 ln(x), we identify the factor, 4, as the exponent and the argument, x,x,x, as the base, and rewrite the product as a logarithm of a power: 4 ln(x)=ln(x 4).4 ln(x)=ln(x 4).4 ln(x)=ln(x 4). Try It #5 Rewrite 2 log 3 4 2 log 3 4 2 log 3 4 using the power rule for logs to a single logarithm with a leading coefficient of 1. Expanding Logarithmic Expressions Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example: log b(6 x y)=log b(6 x)−log b y=log b 6+log b x−log b y log b(6 x y)=log b(6 x)−log b y=log b 6+log b x−log b y log b(6 x y)=log b(6 x)−log b y=log b 6+log b x−log b y We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power: log b(A C)=log b(A C−1)=log b(A)+log b(C−1)=log b A+(−1)log b C=log b A−log b C log b(A C)=log b(A C−1)=log b(A)+log b(C−1)=log b A+(−1)log b C=log b A−log b C log b(A C)=log b(A C−1)=log b(A)+log b(C−1)=log b A+(−1)log b C=log b A−log b C We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product. With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm. Example 6 Expanding Logarithms Using Product, Quotient, and Power Rules Rewrite ln(x 4 y 7)ln(x 4 y 7)ln(x 4 y 7) as a sum or difference of logs. Solution First, because we have a quotient of two expressions, we can use the quotient rule: ln(x 4 y 7)=ln(x 4 y)−ln(7)ln(x 4 y 7)=ln(x 4 y)−ln(7)ln(x 4 y 7)=ln(x 4 y)−ln(7) Then seeing the product in the first term, we use the product rule: ln(x 4 y)−ln(7)=ln(x 4)+ln(y)−ln(7)ln(x 4 y)−ln(7)=ln(x 4)+ln(y)−ln(7)ln(x 4 y)−ln(7)=ln(x 4)+ln(y)−ln(7) Finally, we use the power rule on the first term: ln(x 4)+ln(y)−ln(7)=4 ln(x)+ln(y)−ln(7)ln(x 4)+ln(y)−ln(7)=4 ln(x)+ln(y)−ln(7)ln(x 4)+ln(y)−ln(7)=4 ln(x)+ln(y)−ln(7) Try It #6 Expand log(x 2 y 3 z 4).log(x 2 y 3 z 4).log(x 2 y 3 z 4). Example 7 Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression Expand log(x−−√).log(x).log(x). Solution log(x−−√)=log x(1 2)=1 2 log x log(x)=log x(1 2)=1 2 log x log(x)=log x(1 2)=1 2 log x Try It #7 Expand ln(x 2−−√3).ln(x 2 3).ln(x 2 3). Q&A Can we expandln(x 2+y 2)?ln(x 2+y 2)?ln(x 2+y 2)? No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Example 8 Expanding Complex Logarithmic Expressions Expand log 6(64 x 3(4 x+1)(2 x−1)).log 6(64 x 3(4 x+1)(2 x−1)).log 6(64 x 3(4 x+1)(2 x−1)). Solution We can expand by applying the Product and Quotient Rules. log 6(64 x 3(4 x+1)(2 x−1))=log 6 64+log 6 x 3+log 6(4 x+1)−log 6(2 x−1)=log 6 2 6+log 6 x 3+log 6(4 x+1)−log 6(2 x−1)=6 log 6 2+3 log 6 x+log 6(4 x+1)−log 6(2 x−1)Apply the Quotient Rule.Simplify by writing 64 as 2 6.Apply the Power Rule.log 6(64 x 3(4 x+1)(2 x−1))=log 6 64+log 6 x 3+log 6(4 x+1)−log 6(2 x−1)Apply the Quotient Rule.=log 6 2 6+log 6 x 3+log 6(4 x+1)−log 6(2 x−1)Simplify by writing 64 as 2 6.=6 log 6 2+3 log 6 x+log 6(4 x+1)−log 6(2 x−1)Apply the Power Rule.log 6(64 x 3(4 x+1)(2 x−1))=log 6 64+log 6 x 3+log 6(4 x+1)−log 6(2 x−1)Apply the Quotient Rule.=log 6 2 6+log 6 x 3+log 6(4 x+1)−log 6(2 x−1)Simplify by writing 64 as 2 6.=6 log 6 2+3 log 6 x+log 6(4 x+1)−log 6(2 x−1)Apply the Power Rule. Try It #8 Expand ln((x−1)(2 x+1)2√(x 2−9)).ln((x−1)(2 x+1)2(x 2−9)).ln((x−1)(2 x+1)2(x 2−9)). Condensing Logarithmic Expressions We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing. How To Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient. Example 9 Using the Product and Quotient Rules to Combine Logarithms Write log 3(5)+log 3(8)−log 3(2)log 3(5)+log 3(8)−log 3(2)log 3(5)+log 3(8)−log 3(2) as a single logarithm. Solution Using the product and quotient rules log 3(5)+log 3(8)=log 3(5⋅8)=log 3(40)log 3(5)+log 3(8)=log 3(5⋅8)=log 3(40)log 3(5)+log 3(8)=log 3(5⋅8)=log 3(40) This reduces our original expression to log 3(40)−log 3(2)log 3(40)−log 3(2)log 3(40)−log 3(2) Then, using the quotient rule log 3(40)−log 3(2)=log 3(40 2)=log 3(20)log 3(40)−log 3(2)=log 3(40 2)=log 3(20)log 3(40)−log 3(2)=log 3(40 2)=log 3(20) Try It #9 Condense log 3−log 4+log 5−log 6.log 3−log 4+log 5−log 6.log 3−log 4+log 5−log 6. Example 10 Condensing Complex Logarithmic Expressions Condense log 2(x 2)+1 2 log 2(x−1)−3 log 2((x+3)2).log 2(x 2)+1 2 log 2(x−1)−3 log 2((x+3)2).log 2(x 2)+1 2 log 2(x−1)−3 log 2((x+3)2). Solution We apply the power rule first: log 2(x 2)+1 2 log 2(x−1)−3 log 2((x+3)2)=log 2(x 2)+log 2(x−1−−−−√)−log 2((x+3)6)log 2(x 2)+1 2 log 2(x−1)−3 log 2((x+3)2)=log 2(x 2)+log 2(x−1)−log 2((x+3)6)log 2(x 2)+1 2 log 2(x−1)−3 log 2((x+3)2)=log 2(x 2)+log 2(x−1)−log 2((x+3)6) Next we apply the product rule to the sum: log 2(x 2)+log 2(x−1−−−−√)−log 2((x+3)6)=log 2(x 2 x−1−−−−√)−log 2((x+3)6)log 2(x 2)+log 2(x−1)−log 2((x+3)6)=log 2(x 2 x−1)−log 2((x+3)6)log 2(x 2)+log 2(x−1)−log 2((x+3)6)=log 2(x 2 x−1)−log 2((x+3)6) Finally, we apply the quotient rule to the difference: log 2(x 2 x−1−−−−√)−log 2((x+3)6)=log 2 x 2 x−1−−−−√(x+3)6 log 2(x 2 x−1)−log 2((x+3)6)=log 2 x 2 x−1(x+3)6 log 2(x 2 x−1)−log 2((x+3)6)=log 2 x 2 x−1(x+3)6 Try It #10 Rewrite log(5)+0.5 log(x)−log(7 x−1)+3 log(x−1)log(5)+0.5 log(x)−log(7 x−1)+3 log(x−1)log(5)+0.5 log(x)−log(7 x−1)+3 log(x−1) as a single logarithm. Example 11 Rewriting as a Single Logarithm Rewrite 2 log x−4 log(x+5)+1 x log(3 x+5)2 log x−4 log(x+5)+1 x log(3 x+5)2 log x−4 log(x+5)+1 x log(3 x+5) as a single logarithm. Solution We apply the power rule first: 2 log x–4 log(x+5)+1 x log(3 x+5)=log(x 2)−log(x+5)4+log((3 x+5)x−1)2 log x–4 log(x+5)+1 x log(3 x+5)=log(x 2)−log(x+5)4+log((3 x+5)x−1)2 log x–4 log(x+5)+1 x log(3 x+5)=log(x 2)−log(x+5)4+log((3 x+5)x−1) Next we rearrange and apply the product rule to the sum: log(x 2)−log(x+5)4+log((3 x+5)x−1)log(x 2)−log(x+5)4+log((3 x+5)x−1)log(x 2)−log(x+5)4+log((3 x+5)x−1) =log(x 2)+log((3 x+5)x−1)−log(x+5)4=log(x 2)+log((3 x+5)x−1)−log(x+5)4=log(x 2)+log((3 x+5)x−1)−log(x+5)4 =log(x 2(3 x+5)x−1)−log(x+5)4=log(x 2(3 x+5)x−1)−log(x+5)4=log(x 2(3 x+5)x−1)−log(x+5)4 Finally, we apply the quotient rule to the difference: =log(x 2(3 x+5)x−1)−log(x+5)4=log x 2(3 x+5)x−1(x+5)4=log(x 2(3 x+5)x−1)−log(x+5)4=log x 2(3 x+5)x−1(x+5)4=log(x 2(3 x+5)x−1)−log(x+5)4=log x 2(3 x+5)x−1(x+5)4 Try It #11 Condense 4(3 log(x)+log(x+5)−log(2 x+3)).4(3 log(x)+log(x+5)−log(2 x+3)).4(3 log(x)+log(x+5)−log(2 x+3)). Example 12 Applying of the Laws of Logs Recall that, in chemistry, pH=−log[H+].pH=−log[H+].pH=−log[H+]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH? Solution Suppose C C C is the original concentration of hydrogen ions, and P P P is the original pH of the liquid. Then P=–log(C).P=–log(C).P=–log(C). If the concentration is doubled, the new concentration is 2 C.2 C.2 C. Then the pH of the new liquid is pH=−log(2 C)pH=−log(2 C)pH=−log(2 C) Using the product rule of logs pH=−log(2 C)=−(log(2)+log(C))=−log(2)−log(C)pH=−log(2 C)=−(log(2)+log(C))=−log(2)−log(C)pH=−log(2 C)=−(log(2)+log(C))=−log(2)−log(C) Since P=–log(C),P=–log(C),P=–log(C), the new pH is pH=P−log(2)≈P−0.301 pH=P−log(2)≈P−0.301 pH=P−log(2)≈P−0.301 When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301. Try It #12 How does the pH change when the concentration of positive hydrogen ions is decreased by half? Using the Change-of-Base Formula for Logarithms Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or e,e,e, we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs. To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms. Given any positive real numbers M,b,M,b,M,b, and n,n,n, where n≠1 n≠1 n≠1 and b≠1,b≠1,b≠1, we show log b M=log n M log n b log b M=log n M log n b log b M=log n M log n b Let y=log b M.y=log b M.y=log b M.By exponentiating both sides with base b b b, we arrive at an exponential form, namely b y=M.b y=M.b y=M. It follows that log n(b y)y log n b y log b M=log n M=log n M=log n M log n b=log n M log n b Apply the one-to-one property.Apply the power rule for logarithms.Isolate y.Substitute for y.log n(b y)=log n M Apply the one-to-one property.y log n b=log n M Apply the power rule for logarithms.y=log n M log n b Isolate y.log b M=log n M log n b Substitute for y.log n(b y)=log n M Apply the one-to-one property.y log n b=log n M Apply the power rule for logarithms.y=log n M log n b Isolate y.log b M=log n M log n b Substitute for y. For example, to evaluate log 5 36 log 5 36 log 5 36 using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log. log 5 36=log(36)log(5)≈2.2266 Apply the change of base formula using base 10.Use a calculator to evaluate to 4 decimal places.log 5 36=log(36)log(5)Apply the change of base formula using base 10.≈2.2266 Use a calculator to evaluate to 4 decimal places.log 5 36=log(36)log(5)Apply the change of base formula using base 10.≈2.2266 Use a calculator to evaluate to 4 decimal places. The Change-of-Base Formula The change-of-base formula can be used to evaluate a logarithm with any base. For any positive real numbers M,b,M,b,M,b, and n,n,n, where n≠1 n≠1 n≠1 and b≠1,b≠1,b≠1, log b M=log n M log n b.log b M=log n M log n b.log b M=log n M log n b. It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs. log b M=ln M ln b log b M=ln M ln b log b M=ln M ln b and log b M=log M log b log b M=log M log b log b M=log M log b How To Given a logarithm with the form log b M,log b M,log b M, use the change-of-base formula to rewrite it as a quotient of logs with any positive basen,n,n, where n≠1.n≠1.n≠1. Determine the new base n,n,n, remembering that the common log, log(x),log(x),log(x), has base 10, and the natural log, ln(x),ln(x),ln(x), has base e.e.e. Rewrite the log as a quotient using the change-of-base formula The numerator of the quotient will be a logarithm with base n n n and argument M.M.M. The denominator of the quotient will be a logarithm with base n n n and argument b.b.b. Example 13 Changing Logarithmic Expressions to Expressions Involving Only Natural Logs Change log 5 3 log 5 3 log 5 3 to a quotient of natural logarithms. Solution Because we will be expressing log 5 3 log 5 3 log 5 3 as a quotient of natural logarithms, the new base, n=e.n=e.n=e. We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5. log b M log 5 3=ln M ln b=ln 3 ln 5 log b M=ln M ln b log 5 3=ln 3 ln 5 log b M=ln M ln b log 5 3=ln 3 ln 5 Try It #13 Change log 0.5 8 log 0.5 8 log 0.5 8 to a quotient of natural logarithms. Q&A Can we change common logarithms to natural logarithms? Yes. Remember that log 9 log 9 log 9 means log 10 9.log 10 9.log 10 9. So, log 9=ln 9 ln 10.log 9=ln 9 ln 10.log 9=ln 9 ln 10. Example 14 Using the Change-of-Base Formula with a Calculator Evaluate log 2(10)log 2(10)log 2(10) using the change-of-base formula with a calculator. Solution According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e.e.e. log 2 10=ln 10 ln 2≈3.3219 Apply the change of base formula using base e.Use a calculator to evaluate to 4 decimal places.log 2 10=ln 10 ln 2 Apply the change of base formula using base e.≈3.3219 Use a calculator to evaluate to 4 decimal places.log 2 10=ln 10 ln 2 Apply the change of base formula using base e.≈3.3219 Use a calculator to evaluate to 4 decimal places. Try It #14 Evaluate log 5(100)log 5(100)log 5(100) using the change-of-base formula. Media Access these online resources for additional instruction and practice with laws of logarithms. The Properties of Logarithms Expand Logarithmic Expressions Evaluate a Natural Logarithmic Expression 6.5 Section Exercises Verbal 1. How does the power rule for logarithms help when solving logarithms with the form log b(x−−√n)?log b(x n)?log b(x n)? What does the change-of-base formula do? Why is it useful when using a calculator? Algebraic For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs. 3. log b(7 x⋅2 y)log b(7 x⋅2 y)log b(7 x⋅2 y) ln(3 a b⋅5 c)ln(3 a b⋅5 c)ln(3 a b⋅5 c) 5. log b(13 17)log b(13 17)log b(13 17) log 4(x z w)log 4(x z w)log 4(x z w) 7. ln(1 4 k)ln(1 4 k)ln(1 4 k) log 2(y x)log 2(y x)log 2(y x) For the following exercises, condense to a single logarithm if possible. 9. ln(7)+ln(x)+ln(y)ln(7)+ln(x)+ln(y)ln(7)+ln(x)+ln(y) log 3(2)+log 3(a)+log 3(11)+log 3(b)log 3(2)+log 3(a)+log 3(11)+log 3(b)log 3(2)+log 3(a)+log 3(11)+log 3(b) 11. log b(28)−log b(7)log b(28)−log b(7)log b(28)−log b(7) ln(a)−ln(d)−ln(c)ln(a)−ln(d)−ln(c)ln(a)−ln(d)−ln(c) 13. −log b(1 7)−log b(1 7)−log b(1 7) 1 3 ln(8)1 3 ln(8)1 3 ln(8) For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs. 15. log(x 15 y 13 z 19)log(x 15 y 13 z 19)log(x 15 y 13 z 19) ln(a−2 b−4 c 5)ln(a−2 b−4 c 5)ln(a−2 b−4 c 5) 17. log(x 3 y−4−−−−−√)log(x 3 y−4)log(x 3 y−4) ln(y y 1−y−−−√)ln(y y 1−y)ln(y y 1−y) 19. log(x 2 y 3 x 2 y 5−−−−√3)log(x 2 y 3 x 2 y 5 3)log(x 2 y 3 x 2 y 5 3) For the following exercises, condense each expression to a single logarithm using the properties of logarithms. log(2 x 4)+log(3 x 5)log(2 x 4)+log(3 x 5)log(2 x 4)+log(3 x 5) 21. ln(6 x 9)−ln(3 x 2)ln(6 x 9)−ln(3 x 2)ln(6 x 9)−ln(3 x 2) 2 log(x)+3 log(x+1)2 log(x)+3 log(x+1)2 log(x)+3 log(x+1) 23. log(x)−1 2 log(y)+3 log(z)log(x)−1 2 log(y)+3 log(z)log(x)−1 2 log(y)+3 log(z) 4 log 7(c)+log 7(a)3+log 7(b)3 4 log 7(c)+log 7(a)3+log 7(b)3 4 log 7(c)+log 7(a)3+log 7(b)3 For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base. 25. log 7(15)log 7(15)log 7(15) to base e e e log 14(55.875)log 14(55.875)log 14(55.875) to base 10 10 10 For the following exercises, suppose log 5(6)=a log 5(6)=a log 5(6)=a and log 5(11)=b.log 5(11)=b.log 5(11)=b. Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of a a a and b.b.b. Show the steps for solving. 27. log 11(5)log 11(5)log 11(5) log 6(55)log 6(55)log 6(55) 29. log 11(6 11)log 11(6 11)log 11(6 11) Numeric For the following exercises, use properties of logarithms to evaluate without using a calculator. log 3(1 9)−3 log 3(3)log 3(1 9)−3 log 3(3)log 3(1 9)−3 log 3(3) 31. 6 log 8(2)+log 8(64)3 log 8(4)6 log 8(2)+log 8(64)3 log 8(4)6 log 8(2)+log 8(64)3 log 8(4) 2 log 9(3)−4 log 9(3)+log 9(1 729)2 log 9(3)−4 log 9(3)+log 9(1 729)2 log 9(3)−4 log 9(3)+log 9(1 729) For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places. 33. log 3(22)log 3(22)log 3(22) log 8(65)log 8(65)log 8(65) 35. log 6(5.38)log 6(5.38)log 6(5.38) log 4(15 2)log 4(15 2)log 4(15 2) 37. log 1 2(4.7)log 1 2(4.7)log 1 2(4.7) Extensions Use the product rule for logarithms to find all x x x values such that log 12(2 x+6)+log 12(x+2)=2.log 12(2 x+6)+log 12(x+2)=2.log 12(2 x+6)+log 12(x+2)=2. Show the steps for solving. 39. Use the quotient rule for logarithms to find all x x x values such that log 6(x+2)−log 6(x−3)=1.log 6(x+2)−log 6(x−3)=1.log 6(x+2)−log 6(x−3)=1. Show the steps for solving. Can the power property of logarithms be derived from the power property of exponents using the equation b x=m?b x=m?b x=m? If not, explain why. If so, show the derivation. 41. Prove that log b(n)=1 log n(b)log b(n)=1 log n(b)log b(n)=1 log n(b) for any positive integers b>1 b>1 b>1 and n>1.n>1.n>1. Does log 81(2401)=log 3(7)?log 81(2401)=log 3(7)?log 81(2401)=log 3(7)? 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17558	https://math.stackexchange.com/questions/2198358/how-to-know-when-to-put-calculator-in-radian-or-degree-mode	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How to know when to put calculator in radian or degree mode? Ask Question Asked Modified 7 years, 6 months ago Viewed 173k times 2 $\begingroup$ For a take home test that I have, I have some questions that say Find the exact function value, if it exists. $\sin\left(60^\circ\right)$ $\tan\left(-45^\circ\right)$ $\cos (5\pi/2)$ $\sec (3\pi)$ The second part says Find the function values. Round to four decimals places.¨: $\cos \left(111.4^\circ\right)$ $\sin \left(-18^\circ\right)$ $\sec (9Ï/10)$ $\tan (42.5)$ How do I know when my calculator should be in radian or degree mode? Hopefully this makes sense... trigonometry Share edited Mar 22, 2017 at 15:04 gt6989b 55k33 gold badges4040 silver badges7575 bronze badges asked Mar 22, 2017 at 14:54 Ria CarRia Car 2111 gold badge11 silver badge33 bronze badges $\endgroup$ 11 8 $\begingroup$ If there is a degree symbol, you should have your calculator in degree mode. $\endgroup$ Arthur – Arthur 2017-03-22 14:57:26 +00:00 Commented Mar 22, 2017 at 14:57 1 $\begingroup$ If the input is in degrees (the circle symbol), use degree mode, otherwise use radian mode. $\endgroup$ quasi – quasi 2017-03-22 14:57:47 +00:00 Commented Mar 22, 2017 at 14:57 1 $\begingroup$ Do you know the difference between degrees and radians? It seems to me that there might be more fundamental problem than typing a value into calculator. $\endgroup$ Ennar – Ennar 2017-03-22 14:58:50 +00:00 Commented Mar 22, 2017 at 14:58 1 $\begingroup$ @uniquesolution, that is rather ill-advised. For example, $\tan 42.5$ is not in degrees, while $\tan \pi^\circ$ is. $\endgroup$ Ennar – Ennar 2017-03-22 15:00:38 +00:00 Commented Mar 22, 2017 at 15:00 1 $\begingroup$ @uniquesolution, I don't understand what you are disagreeing with. Are you saying that $42.5$ is in degrees, while $\pi^\circ$ is in radians? $\endgroup$ Ennar – Ennar 2017-03-22 15:03:15 +00:00 Commented Mar 22, 2017 at 15:03 \| Show 6 more comments 2 Answers 2 Reset to default 12 $\begingroup$ If there is a degree symbol, ${ \ }^\circ$, then use degree mode. If there is no degree symbol, then use radian mode. Even if there is no $\pi$ in the number. In your examples, assuming there are no typos: Sin 60 ° degree mode because there is a degree symbol Tan(-45 °) degree mode because there is a degree symbol Cos 5Ï/2 radian mode because there is no degree symbol Sec 3Ï radian mode because there is no degree symbol I should mention that for those first 4 problems, I think the point is actually not to use a calculator. The second part says ¨Find the function values. Round to four decimals places.¨: Cos 111.4° degree mode because there is a degree symbol Sin(-18°) degree mode because there is a degree symbol Sec 9Ï/10 radian mode because there is no degree symbol tan 42.5 radian mode because there is no degree symbol Share answered Mar 22, 2017 at 14:59 user307169user307169 $\endgroup$ Add a comment \| 4 $\begingroup$ The ° symbol means "degrees." Any answer marked with that is definitely in degrees. The technically correct thing to do is to assume that everything is in radians unless otherwise specified. However, humans tend to be bad at being technically correct, so if you haven't been told to use radians unless otherwise specified I would consider making contextual judgement calls. If you have been taught the technically correct rule, definitely use it. This interpretation agrees with the rules of thumb that I am about to give everywhere that it's applicable, leaving the last problem. I would guess that $42.5$ is supposed to be in radians, because everywhere else in the problem the professor has been careful to use the degree symbol, making me think its omission is deliberate. If none of the problems had been marked with a degree symbol, I might think otherwise since $42.5$ is much bigger than $2\pi$. In contexts where you think your professor has simplified by opting to not use the degree symbol, some general rules of thumb can be applied. If it's unspecified and a $\pi$ shows up, you should assume radians. If it's unspecified and of the form $360/n$ for some integer $n$, use degrees. Another good rule of thumb is that if one interpretation gives an algebraic answer, use that interpretation. I'm unsure if that solves the last problem, though plugging it into Wolfram Alpha will tell you. Share edited Mar 22, 2017 at 15:19 answered Mar 22, 2017 at 14:59 Stella BidermanStella Biderman 31.5k66 gold badges4949 silver badges9494 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 1 Finding exact value of trigonometric functions 1 Finding the exact values of trig functions in a quadrant What is the value for $\tan^{-1}(3)+\tan^{-1}(3)$? 1 Solve the equations for $0\leq x<2\pi$, note $x$ is in radians. 1 Finding an angle of a triangle with three given sides (radian) 1 Antiderivative of $\sin{x}$ when $x$ is given in degrees 0 If $x < 90^\circ$, such that $\sin x = \frac {2}{\sqrt{5}}$, find $\cos (x)$ 2 Simplifying $\frac{\sin x+\sin x\tan^2x}{\tan x}$ to $\sec x$ 8 How to find sin of any fraction-angles, and how do you find them in fraction forms and not in decimal forms? 0 Let $a=5^{\large \log_5(\sin x)}$ and $b=7^{\large \log_7(\cos x)}$ then $\frac ab$ can be equal to : Hot Network Questions Why weren’t Prince Philip’s sisters invited to his wedding to Princess Elizabeth? Switch between math versions but without math versions ICC in Hague not prosecuting an individual brought before them in a questionable manner? Quantizing EM field by imposing canonical commutation relations Numbers Interpreted in Smallest Valid Base Identifying a thriller where a man is trapped in a telephone box by a sniper Can I use the TEA1733AT for a 150-watt load despite datasheet saying 75 W? Data lost/Corrupted on iCloud The geologic realities of a massive well out at Sea Space Princess Space Tours: Black Holes merging - what would you visually see? Why do universities push for high impact journal publications? How to design a circuit that outputs the binary position of the 3rd set bit from the right in an 8-bit input? Riffle a list of binary functions into list of arguments to produce a result How do trees drop their leaves? What happens when the jewels run out? How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? 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17559	https://math.stackexchange.com/questions/1330516/problem-from-olympiad-book-by-arthur-engel-invariant-problem	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Problem from olympiad book by Arthur Engel (invariant problem) Ask Question Asked Modified 9 years, 9 months ago Viewed 969 times 2 $\begingroup$ There are $a$ white, $b$ black, and $c$ red chips on a table. In one step, you may choose two chips of different colors and replace them by a chip of the third color. If just one chip will remains at the end, its color will not depend on the evolution of the game. When can this final state be reached? This the solution given in the book: Solution: All three numbers $a$, $b$, $c$ change their parity in one step. If one of the numbers has different parity from the other two, it will retain this property to the end. This will be the one that remains. I don't understand this given solution. contest-math Share edited Dec 6, 2015 at 16:33 Xoque55 4,50944 gold badges3030 silver badges4747 bronze badges asked Jun 18, 2015 at 18:11 BlueyBluey 2,1242121 silver badges3737 bronze badges $\endgroup$ 5 $\begingroup$ Do you understand what is meant by the parity of an integer? $\endgroup$ Brian M. Scott – Brian M. Scott 2015-06-18 18:17:30 +00:00 Commented Jun 18, 2015 at 18:17 $\begingroup$ whether number is even or odd $\endgroup$ Bluey – Bluey 2015-06-18 18:19:35 +00:00 Commented Jun 18, 2015 at 18:19 1 $\begingroup$ Right. Now check that at each move, the parities of the numbers of chips of each color change. For instance, if you take a white and a black chip, you lose one of each of those colors and gain one red chip, so the parities of all three numbers change. Now suppose that two of $a,b$ and $c$ are even and the third is odd, or two are odd and the third is even: one pile has a different parity from the other two. At each move you change the parity of every pile, so the pile that had a different parity to begin with still has a different parity. This must continue until the game is over. ... $\endgroup$ Brian M. Scott – Brian M. Scott 2015-06-18 18:23:17 +00:00 Commented Jun 18, 2015 at 18:23 $\begingroup$ ... Now when theres only one chip left, the pile that contains it has odd parity, and the two empty piles have even parity. Thus, the pile with the remaining chip must be the one whose parity was different from the other two all the way through the game. $\endgroup$ Brian M. Scott – Brian M. Scott 2015-06-18 18:25:04 +00:00 Commented Jun 18, 2015 at 18:25 $\begingroup$ Youre welcome. $\endgroup$ Brian M. Scott – Brian M. Scott 2015-06-18 18:30:29 +00:00 Commented Jun 18, 2015 at 18:30 Add a comment \| 0 Reset to default 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 contest-math See similar questions with these tags. 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17560	https://resources.finalsite.net/images/v1721669618/oleanschoolsorg/pp8vpytsr3yuhfeneujo/Unit_06_-_Exponential_Equations_and_Functions.pdf	UNIT 6: EXPONENTIAL EQUATIONS & FUNCTIONS Part A: Video Tutorial Section Video 1: (Review how to use a graphic calculator to find exponents and square roots) Video 2 and 3: (Properties of Square Roots) (More Examples of Properties of Square Roots) Video 4 and 5: (Simplifying Square Roots) (More Examples of Simplifying Square Roots) Video 6 and 7: (Factoring Square Roots with Variables) (Factoring Square Roots with Variables) Video 8 (BASIC Review of Rules of Exponents) Videos 9 and 10: (Product of Powers Property Exponents) (More Examples of Powers Property Exponents) Videos 11 and 12: (Quotient of Powers Property Exponents) (More Examples of Quotient of Powers Property) Exponents) Videos 13 and 14: (Writing Expressions in Radical Form) (Writing Expressions in Rational Form) Video 15: (Using a Graphing Calculator to Solve) Video 16: (Introduction to Exponential Functions) Video 17: (Basics You Need to Understand About Functions) Videos 18 and 19: (Linear Vs. Exponential Functions) (More Examples of Linear Vs. Exponential Functions) Video 20: (Solving for X, When X is an Exponent) Videos 21 and 22: (Graphing Exponential Functions) (Graphing Exponential Functions Using Graphing Calculator) Videos 23, 24, 25: (Exponential Growth and Decay) Exponential Growth: (Exponential Decay) Videos 26 and 27: Geometric Sequences) (More Examples for geometric Sequences) Video 28: (Extending Geometric Sequences) Videos 29 and 30 (Graphing Geometric Sequences) (More Examples of Graphing a Geometric Sequence) Video 31: (Recursively Defined Sequences) Videos 32 and 33: (Recursive Formulas for Sequences) (More Examples of Recursive Formulas for Sequences) Video 34 and 35: Between Recursive Rules & Explicit Rules) (More Examples of Translating Between Recursive Rules & Explicit Rules) Video 36: (Writing Recursive Rules for Other Sequences) (More Examples for Writing Recursive Rules for Other Sequences) Part B : Vocabulary, Hints and Explanations Important Vocabulary That Students need to Understand! closed when the operation performed on any two numbers in the set results in a number that is also in the set Nth root when bn = a for an integer n greater than 1 The nth root of a number may be real or imaginary numbers. Exponential function A function of the form y = abx, where a exponential growth occurs when a quantity increases by the same factor over equal intervals of time exponential groth function A function in the form of: a(1+ r)1, where a > 0 and r > 0 compound interest interest earned on the principal and on previously earned interest. exponential decay when a quantity decreases by the same factor over equal intervals of time exponential decay function A function of the form: y = a(1-r)1 where a > 0 and 0< r<1 geometric sequence The ratio between consecutive terms common ratio The ratio found in a geometric sequence recursive rule Gives the beginning term (s0 of a sequence and an equation that indicates how any term an in the sequence relates to the previous term Some students struggle to recall that exponents mean repeated multiplication of the base number and NOT the base number and exponent multiplied together. A student many think that 52 = 5(2) 52 = 25 not 5(2) = 10 Hint: Tell the student that exponents may be tiny but they are powerful (like bees), they make numbers grow big fast (just like a bee sting makes your hand swell up fast) Making sure that the student understands the concept of exponents before working with exponents in functions is important. Hint: A student may relate or understand exponents for squares if he sees a multiplication table and draws a line through the perfect squares. This can also help when factoring square roots. Properties of square roots: A student needs to understand that when working with square roots the “hard operations” apply but not the “easy operations”. If a student thinks of addition and subtraction as the “easy operations” (because they have been doing those since Elementary School), they can say that those operations do not apply to something “hard” like square roots. (Note- on line version of word will not put in the square root or radical sign. When creating examples, I had to use the words “square root”. When teaching you will want to use the symbol.) Ex: square root of 36 + 64 ≠ square root of 36 + square root of 64 Square root of 100 ≠ 6 + 8 10 ≠ 14 With that same reasoning, the “harder operations” which are multiplication and division do apply to “hard math” such as square roots. Ex: Ex. Square root of 4 • 9 = square root of 4 • square root of 9 Square root of 36 = 2 • 3 6 = 6 Simplifying a Square Root Expression: A student needs to think in terms of factoring a radical (another name for square root) into perfect squares whenever possible. Ex: = square root of 150 = square root of 25 • square root of 6 (25 is a perfect square - 5 5 = 25) Square root of 150 = 5 • square root of 6 Properties of Exponents: Multiplying with powers (exponents) - when you multiply powers with the same base number (or variable), keep the base and add the exponents Ex: 43 • 42 = 43+2 = 45 A student may want to multiply the bases and add the exponents. Have the student complete 43 = 64 42 = 16 64 + 16 = 80 165 = 1,048,576 When a student breaks the equation into the component parts, he may see the error in multiplying the bases. Dividing with powers (exponents) - when you divide powers with the same base number (or variable), keep the base and subtract. Hint: A student is pairing up operations. The “get larger” operations of multiplication and addition go together. The “get smaller” operations of division and subtraction go together. Negative Powers: A negative power “flips” you out! Ex: 4-2 = 1/42 = 1/16 To the power of one: Remind students that a base without an exponent is that base to the power of one. Ex: 52 • 5 = 5(2+1) = 53 To the power of zero: Any base to the power of zero equals one Ex: 123,4560 = 1 Radicals and Rational Exponents The rule states WHEN bn = a for an integer >1, b is the nth root of a (say what??!!??) The following examples may help you understand. Also, this is a calculator job or a guess and check! Ex: 43 = 64 SO - 3square root of 64 = square root of 4•4•4 The nth root can be expressed as fractional power. Ex: 641/3 = 3square root of 64 Hint: Fractional exponents make answers smaller while whole number exponents make things larger. T 42= 4• 4 = 16 41/2 = (square root) 4 = 2 A student can use the properties of exponents (from above) to simplify expressions involving rational exponents. Ex: 163/4 = (161/4)3 rewrite the exponent such that one over the denominator becomes the exponent (base and exponent in parenthesis) to the power of the numerator = 23 find the fourth root of 16 = 8 solve for the cube of 2 Hint: the numerator navigates to outside Exponential Functions An exponential function is a non-linear function where the y increases by exponentially (repeated multiplication) versus addition. Think of bacteria growing on a surface! Formula: y = abx To find the y-intercept in exponential function, substitute zero for the x (which is the exponent). In y=abx the y value changes by factors of b as the x values increase by 1. Ex: x 0 1 2 3 y 2 10 50 1250 In this example x is increasing by 1, while y is changing by a factor of times 5. The equation for this example is: y = 2(5)x The graph of an exponential function is a smooth, curved line. Exponential Growth and Decay Exponential growth occurs when a quantity increases by the same factor over equal intervals of time. Please note the formula on the reference sheet for the Algebra 1 Regents is NOT the formula that Algebra 1 students should use! Students will need to memorize the formula for exponential growth and decay. Exponential growth formula: y = a(1 + r) t y = final amount a = initial amount r = rate of growth (in decimal form) t = time (1 + r) = the growth factor Exponential decay formula: y = a(1 - r) t y = final amount a = initial amount r = rate of growth (in decimal form) t = time (1 - r) = the decay factor Compound Interest Simple interest is interest earned on principal (or initial amount) only. Compound interest is interest earned on principal AND previously earned interest. Hint: explain a savings account to a student by demonstrating a simple interest versus a compound interest. Demonstrate that an initial savings account (or investment) of $500 at 5% interest for 5 years using both simple interest and compound interest. Simple interest: I=prt I = 500(.05)(5) I = $125 500 + 125 = $625 Compound interest (without the formula) Year one I = 500 (.05) = 25 Year two I = 525 (.05) = 26.25 Year three I = 551.25 (.05) = 27.56 Year three I = 578.81 (.05) = 28.94 Year four I = 607.75 (.05) = 30.39 Year five I = 638.14 (.05) = 31.91 With compound interest – after five years you have $670.05 The formula for compound interest is: y= P(I+ r/n)nt P = principal (initial amount) r = annual interest rate t= time in years n = number of times interest is compounded per year In the above example the interest was compounded annually. If interest is compounded every 6 months the n would equal 2 times per year. Geometric Sequence In Unit 5 students learned to solve arithmetic sequences where each term in the sequence was found by adding the same intervals. In a geometric sequence, each term in the sequence is found by multiplying the previous term by a common ratio. Ex: Position 1 2 3 4 5 6 7 Term 3 (3• 2) = 6 (6•2)= 12 (12•2) = 24 (24•2) = 48 (48•2) = 96 (96•2) = 192 Formula for geometric sequence: This formula IS in the Algebra 1 Reference Sheet and CAN be used by Algebra 1 students to solve geometric sequence for a desired term. Formula: an = a1r(n-1) an is the desired term in the sequence a1 is the first term in the sequence r is the common ration (what you are multiplying by) Recursively Defined Sequences The formulas learned above for arithmetic and geometric sequences were explicit equations. A recursive rule gives the beginning term and an equation that indicates how many terms an in the sequence relates to the previous term. When writing a recursive rule, the student writes the first term, the rule. Ex; a1 = 3, an = an-1 + 3 for arithmetic sequences a1 = 1, an = 3an-1 for geometric sequences The recursive equation for an arithmetic sequence is: an = an-1 + d , where d is the common difference A student may be required to write the terms in a sequence and graph. Ex: Write terms in a sequence : Given: a1 = 2 , an = an-1 + 3 Each term that follows would be: a1 = 2 a2 = a1 + 3 = 5 a3 = a2 + 3 = 8 a4 = a3 + 3 = 11 The student then graphs the coordinates: (1,2) (2,5) (3,8) (4,11) A student may be required to write the equation in recursive format from a chart. Ex: Given a chart position 1 2 3 4 5 term 3 8 13 18 23 The sequence is arithmetic with the first term 3 and the common difference 5. Recursive equation for an arithmetic sequence is: an = an-1 + d (substitute 5 for the d) The equation for this table is: an = an-1 +5 Therefore, the recursive rule is: a1 = 3, an = an-1 + 5 Translating Recursive Rules into Explicit Equations for Arithmetic Sequences: Given a recursive rule: a1 = 25, an = an-1 + 10 The first stated term is 25 and the common difference is 10 Recall the equation for an arithmetic sequence is: an = a1 + (n-1) d Substitute 25 for the first term, and 10 for d (common difference) an = 25 + (n-1)10 an = 25 + 10n – 10 distribute an = 10n + 15 simplify Translating Explicit Equations into Recursive Rules for Arithmetic Sequences: Given an explicit equation: an = -2n + 3 The equation represents a sequence with the first term –2(1) + 3 = 1, with a common difference of –2 Recall the recursive equation is: an = an-1 + d Substitute –2 for d an = an-1 + (-2) The recursive rule becomes: an = 1, an = an-1 -2 The recursive equation for a geometric sequence is: an = r• an-1 , where r is the common ratio The student may be required to write terms in a sequence and graph. Ex: Write the terms in a sequence Given a1 = 1, an = 3an-1 Each term that follows would be: a1 = 1 a2 = 3an-1 = 3(1) = 3 a3 = 3a2 = 3(3) = 9= a4 = 3(9) = 27 a5 = 3(27) = 81 The student then graphs the coordinates: (1,1) (2,3) (3,9) (4,27), (5,81) A student may be required to write the equation in recursive format from a chart. Ex: Given a chart position 1 2 3 4 5 term 5 10 20 40 80 The sequence is geometric with the first term 5 and the common ratio 2. Recall the recursive equation for a geometric sequence is: an = r•an-1 Substitute 2 for the common ratio , r an =2an-1 Therefore, the recursive rule is: an = 5, an =2an-1 Translating Recursive Rules into Explicit Equations for Geometric Sequences: Given a recursive rule: a1 =3, an = 2an-1 The first term is 3 and the common ratio is 2. Recall the equation for a geometric sequence is: an = a1rn-1 Substitute 3 for a1 and 2 for r: an = 3(2)n-1 Translating Explicit Equations into Recursive Rules for Geometric Sequences: Given an explicit equation: an =-3(2)n-1 The equation represents a geometric sequence with the first term of –3 and a common ration of 2. Recall the recursive equation is: an = r•an-1 Substitute 2 for the common ratio, r an = 2an-1 Therefore, the recursive rule is: a1 = -3, an =2an-1
17561	https://en.wikipedia.org/wiki/KOH_test	Jump to content Search Contents (Top) 1 Procedure 2 Evaluation 3 References 4 External links KOH test العربية 日本語 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Test to differentiate between skin fungi \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "KOH test" – news · newspapers · books · scholar · JSTOR (April 2011) (Learn how and when to remove this message) \| The KOH test, also known as a potassium hydroxide preparation or KOH prep, is a quick, inexpensive fungal test to differentiate dermatophytes and Candida albicans symptoms from other skin disorders like psoriasis and eczema. Dermatophytes are a type of fungus that invades the top layer of the skin, hair, or nails. There are three genera of fungi commonly implicated: Trichophyton (found in skin, nail, and hair infections), Epidermophyton (skin and nail infections), and Microsporum (skin and hair infections).[citation needed] Dermatophytes produce an infection commonly known as ringworm or tinea. It can appear as "jock itch" in the groin or inner thighs (tinea cruris); on the scalp and hair (tinea capitis) resulting in brittle hair shafts that fall out easily. Tinea unguium affects the nails and athlete's foot (tinea pedis) affects the feet. Tinea versicolor refers to a fungal infection of the skin caused by Malassezia furfur. It appears anywhere on the skin and produces red or gray, scaly patches of itchy skin. Deeper infections may be discoloured, ulcerative and purulent. A Candida yeast infection can also be identified by a KOH test by taking scrapings from the mouth (oral thrush), vagina (vaginitis) and skin (candidiasis). There are over 40 different fungus species known to cause disease in humans, of which Candida albicans is the most common and most frequently tested for. Procedure [edit] The KOH test for fungus is conducted on an outpatient basis and patients do not need to prepare in advance. Results are usually available while the patient waits or the next day if sent to a clinical laboratory. The KOH test procedure may be performed by a physician, nurse practitioner, physician associate, medical assistant, nurse, midwife or medical laboratory technician. If fungal cultures are required, the test is performed by a technologist who specializes in microbiology.[citation needed] Collection: Skin, nail, or hair samples are collected from the infected area on the patient. For skin samples, a scalpel or edge of a glass slide is used to gently scrape skin scales from the infected area. For hair samples, a forceps is used to remove hair shafts and follicles from the infected site. If the test is being sent to a laboratory, the scrapings are placed in a sterile covered container. The scrapings are placed directly onto a microscope slide and are covered with 10% or 20% potassium hydroxide. The slide is left to stand until clear, normally between five and fifteen minutes, in order to dissolve skin cells, hair, and debris. To enhance clearing dimethyl sulfoxide can be added to the slide. To make the fungi easier to see lactophenol cotton blue stain can be added. The slide is gently heated to speed up the action of the KOH. Adding calcofluor-white stain to the slide will cause the fungi to become fluorescent, making them easier to identify under a fluorescence microscope. Place the slide under a microscope to read. Evaluation [edit] Dermatophytes are easily recognized under the microscope by their long branch-like tubular structures called hyphae. Fungi causing ringworm infections produce septate (segmented) hyphae. Some show the presence of spores formed directly from the hyphae (arthroconidia). Under the microscope Tinea versicolor is recognized by curved hyphae and round yeast forms that give it a spaghetti-and-meatball appearance. Yeast cells appear round or oval and budding forms may be seen. The KOH prep cannot identify the specific organism; the specimen can be submitted for fungal culture to identify the organism. A normal, or negative, KOH test shows no fungi (no dermatophytes or yeast). Dermatophytes or yeast seen on a KOH test indicate the person has a fungal infection. Follow-up tests are usually unnecessary. The skin may be sore after the test because of the tissue being scraped off the top of the surface of the skin. References [edit] ^ MedlinePlus Encyclopedia: Skin lesion KOH exam ^ Birnbaum, PS (May 1985). "Cost containment: freestanding emergency centers and the emergency department". The American Journal of Emergency Medicine. 3 (3): 259. doi:10.1016/0735-6757(85)90105-6. PMID 3994805. ^ "R.R.O. 1990, Reg. 682: LABORATORIES". Government of Ontario. 2014-07-24. Retrieved March 4, 2017. ^ Frances Talaska Fischbach; Marshall Barnett Dunning (2004). A manual of laboratory and diagnostic tests. Williams & Wilkins. ISBN 0781741807.[page needed] External links [edit] Medical-dictionary.thefreedictionary.com Medicalhealthtests.com \| v t e Techniques in clinical microbiology \| \| Isolationand culture \| \| \| \| --- \| \| Isolation techniques \| Asepsis Streak plate Selective media \| \| Cultures by body site \| Blood culture Genital culture Sputum culture Throat culture Urine culture Wound culture \| \| Cultures by organism \| Bacterial culture Fungal culture Viral culture \| \| \| Identificationand testing \| \| \| \| --- \| \| Manual testing: basic techniques \| Colonial morphology + Hemolysis Staining + Gram stain + Acid-fast stain + Giemsa stain + India ink stain + Ziehl–Neelsen stain Wet prep Rapid tests + Oxidase + Catalase + Indole + PYR \| \| Manual testing:biochemical and immunologic tests \| ALA test Amino acid decarboxylase test Bile solubility test CAMP test Citrate test Coagulase test DNAse test IMViC KOH test Methyl red test Nitrite test ONPG test Oxidative/fermentation glucose test Phenylalanine deaminase test Reverse CAMP test Salt tolerance test Sulfide indole motility test Triple sugar iron test Urease test + rapid Voges–Proskauer test X and V factor test Bacitracin susceptibility test Optochin susceptibility test Novobiocin susceptibility test Lancefield grouping RPR test \| \| Automated and point-of-care testing \| Analytical profile index MALDI-TOF Polymerase chain reaction VITEK Rapid strep test Monospot test \| \| Antibiotic susceptibility testing \| Beta-lactamase test Disk diffusion test Etest McFarland standards Minimum inhibitory concentration \| \| \| Equipment \| Agar plate + Growth medium Anaerobic jar + Gas-pak Durham tube Biosafety cabinet Incubator Inoculation loop Inoculation needle \| Retrieved from " Category: Medical tests Hidden categories: Wikipedia articles needing page number citations from August 2015 Articles with short description Short description is different from Wikidata Articles needing additional references from April 2011 All articles needing additional references All articles with unsourced statements Articles with unsourced statements from December 2024 Articles with unsourced statements from July 2023 KOH test Add topic
17562	https://www.expii.com/t/write-systems-of-linear-inequalities-from-a-graph-4426	Expii Write Systems of Linear Inequalities from a Graph - Expii We can write a system of inequalities from a graph. Use the line to determine the equation. Use the shaded area and type of line to determine sign. Explanations (3) Hannah Bonville Text 5 When writing an inequality from a graph, there are a few things we need to do: Determine the equation of the line. This gives us the general form of our inequality once we remove the equal sign. Note whether the line is dotted or solid. Recall that dotted lines indicate < or > in an inequality and solid lines indicate ≤ or ≥. Use the graph's shading to determine which way your inequality sign will face. Image source: By Caroline Kulczycky This works for single inequalities as well as systems of inequalities. If you have more than one line on the graph, you'll need to use these steps to write more than one inequality. To go over how to graph inequalities on a coordinate plane, check out this lesson. Let's look at an example: Made using Desmos STEP 1: Determine the equation of the line Report Share 5 Like Related Lessons Systems of Linear Inequalities, Word Problems - Examples Quadratic Equations - Definition & Examples Graphing Systems of Linear Inequalities - Process & Examples Square Roots of Quadratic Equations - Definition & Examples View All Related Lessons Alex Federspiel Text 3 Writing a System of Linear Inequalities from a Graph We can graph systems of inequalities to visualize the solutions. We can also look at a graph of a system and reverse engineer the inequalities. Created using Desmos Take the graph above for example. There are two lines so you know the system has two equations. Use whatever method you prefer to find the equations each line. I like to take two sets of points for each line and use y−y1=m(x−x1) where m=y2−y1x2−x1. So the equation for the red (vertical) line is: x=1 And the equation for the blue (horizontal) line is: y=2 Once we have our equations, we look at the highlighted region and determine if the equations are < or >. The highlighted purple region has x points that are larger than 1 which is the equation of the line. So we can determine: x>1 But the red line isn't dashed. This means the equation is greater than or equal to instead of just greater than. So the equation is actually: x≥1 For the second equation we notice that the y points for the highlighted region are larger than 2. So we can determine: y>2 Since the blue line is dashed, we know that its only greater than and not greater than or equal to. Given the above graph we can determine the system of linear inequalities to be: x≥1 y>2 Report Share 3 Like Anusha Rahman Video 1 (Video) Writing a System of Linear Inequalities from a Graph by Rick Faulk In this video by Rick Faulk, he goes over how to go from a graph to a system of linear inequalities. Summary In the graph below, we are working to figure out what the system of inequalities is that is represented by the graph. Image source: by Rick Faulk We want to tackle this problem one inequality at a time. We are first going to look at the dotted line. We can see that it's shaded below the line, so we can assume a less than sign: y< We can see that the y-intercept is +5. So, we add that to our inequality: y<+5 We need a slope. We figure out that the slope is −12. y<−x2+5 We will do something really similar for the second line, the one going diagonally from the bottom-left to the top-right. We know that there is shading above the line, so we will use a greater than or equal to sign. y≥ We can see that the y-intercept is −1. Our next inequality becomes: y≥−1 We can calculate the slope to be 1. So, our second inequality is: y≥x−1 Therefore, the system of linear inequalities that is represented by the graph are: y<−12+5 y≥x−1 Try the next one on your own! Then follow along with the video to see how you did. You've reached the end How can we improve? General Bug Feature Send Feedback
17563	https://www.khanacademy.org/math/arithmetic/x18ca194a:intro-to-division	Intro to division \| Arithmetic \| Math \| 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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Arithmetic 19 units · 203 skillsUnit 1 Intro to multiplicationUnit 2 1-digit multiplicationUnit 3 Intro to divisionUnit 4 Understand fractionsUnit 5 Place value through 1,000,000Unit 6 Add and subtract through 1,000,000Unit 7 Multiply 1- and 2-digit numbersUnit 8 Divide with remaindersUnit 9 Add and subtract fraction (like denominators)Unit 10 Multiply fractionsUnit 11 Decimals and place valueUnit 12 Add and subtract decimalsUnit 13 Add and subtract fractions (different denominators)Unit 14 Multiply and divide multi-digit numbersUnit 15 Divide fractionsUnit 16 Multiply and divide decimalsUnit 17 Exponents and powers of tenUnit 18 Add and subtract negative numbersUnit 19 Multiply and divide negative numbers Course challenge Test your knowledge of the skills in this course.Start Course challenge Math Arithmetic Unit 3: Intro to division 1,700 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test Division with groups of objects Divide with visuals Relate division to multiplication Relate multiplication and division equations Relate division to multiplication word problems Intro to division: Quiz 1 Divide by 1 Divide by 2 Divide by 4 Divide by 5 Divide by 10 Intro to division: Quiz 2 Divide by 3 Divide by 6 Divide by 7 Divide by 8 Divide by 9 Intro to division: Quiz 3 Basic division Multiplication and division word problems (within 100) Intro to division: Quiz 4 Intro to division: Unit test About this unit We all know that multiplication is just repeated addition, but what about division? Think of it as repeated subtraction! In this unit, you'll learn how to divvy up numbers with ease and expand your mathematical toolkit even further. Division intro Learn Division as equal groups (Opens a modal) Visualizing division with arrays (Opens a modal) Practice Up next for you:Division with groups of objectsGet 5 of 7 questions to level up! Start Not started Divide with visualsGet 5 of 7 questions to level up! Practice Not started Relating multiplication and division Learn Relating division to multiplication (Opens a modal) Multiplication word problem: parking lot (Opens a modal) Division word problem: school building (Opens a modal) Practice Relate division to multiplicationGet 5 of 7 questions to level up! Practice Not started Relate multiplication and division equationsGet 5 of 7 questions to level up! Practice Not started Relate division to multiplication word problemsGet 3 of 4 questions to level up! Practice Not started Quiz 1 Level up on the above skills and collect up to 400 Mastery points Start quiz Divide by 1, 2, or 4 Learn No videos or articles available in this lesson Practice Divide by 1Get 8 of 11 questions to level up! Practice Not started Divide by 2Get 7 of 10 questions to level up! Practice Not started Divide by 4Get 7 of 10 questions to level up! Practice Not started Divide by 5 or 10 Learn No videos or articles available in this lesson Practice Divide by 5Get 7 of 10 questions to level up! Practice Not started Divide by 10Get 7 of 10 questions to level up! Practice Not started Quiz 2 Level up on the above skills and collect up to 400 Mastery points Start quiz Divide by 3 or 6 Learn No videos or articles available in this lesson Practice Divide by 3Get 7 of 10 questions to level up! Practice Not started Divide by 6Get 7 of 10 questions to level up! Practice Not started Divide by 7, 8, or 9 Learn No videos or articles available in this lesson Practice Divide by 7Get 7 of 10 questions to level up! Practice Not started Divide by 8Get 7 of 10 questions to level up! Practice Not started Divide by 9Get 7 of 10 questions to level up! Practice Not started Quiz 3 Level up on the above skills and collect up to 400 Mastery points Start quiz 1-digit division Learn No videos or articles available in this lesson Practice Basic divisionGet 5 of 7 questions to level up! Practice Not started Multiplication and division word problems Learn Multiplication word problem: soda party (Opens a modal) Division word problem: blueberries (Opens a modal) Practice Multiplication and division word problems (within 100)Get 3 of 4 questions to level up! Practice Not started Quiz 4 Level up on the above skills and collect up to 160 Mastery points Start quiz Unit test Level up on all the skills in this unit and collect up to 1,700 Mastery points!Start Unit test Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization. Donate or volunteer today! 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17564	https://www.idexot.com/media/wysiwyg/02_Gaussian_Beam_Optics.pdf	Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings www.cvimellesgriot.com 2.1 Gaussian Beam Optics Gaussian Beam Optics Gaussian Beam Propagation 2.2 Transformation and Magnification by Simple Lenses 2.6 Real Beam Propagation 2.10 Lens Selection 2.13 Gaussian Beam Optics 2 BEAM WAIST AND DIVERGENCE In order to gain an appreciation of the principles and limitations of Gaussian beam optics, it is necessary to understand the nature of the laser output beam. In TEM00 mode, the beam emitted from a laser begins as a perfect plane wave with a Gaussian transverse irradiance profile as shown in figure 2.1. The Gaussian shape is truncated at some diameter either by the internal dimensions of the laser or by some limiting aperture in the optical train. To specify and discuss the propagation characteristics of a laser beam, we must define its diameter in some way. There are two commonly accepted definitions. One definition is the diameter at which the beam irradiance (intensity) has fallen to 1/e2 (13.5 percent) of its peak, or axial value and the other is the diameter at which the beam irradiance Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.2 In most laser applications it is necessary to focus, modify, or shape the laser beam by using lenses and other optical elements. In general, laser-beam propagation can be approximated by assuming that the laser beam has an ideal Gaussian intensity profile, which corresponds to the theoretical TEM00 mode. Coherent Gaussian beams have peculiar transformation properties which require special consideration. In order to select the best optics for a particular laser application, it is important to understand the basic properties of Gaussian beams. Unfortunately, the output from real-life lasers is not truly Gaussian (although the output of a single mode fiber is a very close approximation). To accom-modate this variance, a quality factor, M2 (called the “M-squared” factor), has been defined to describe the deviation of the laser beam from a theoretical Gaussian. For a theoretical Gaussian, M2 = 1; for a real laser beam, M2>1. The M2 factor for helium neon lasers is typically less than 1.1; for ion lasers, the M2factor typically is between 1.1 and 1.3. Collimated TEM00 diode laser beams usually have an M2 ranging from 1.1 to 1.7. For high-energy multimode lasers, the M2 factor can be as high as 25 or 30. In all cases, the M2 factor affects the characteristics of a laser beam and cannot be neglected in optical designs. In the following section, Gaussian Beam Propagation, we will treat the characteristics of a theoretical Gaussian beam (M2=1); then, in the section Real Beam Propagation we will show how these characteristics change as the beam deviates from the theoretical. In all cases, a circularly symmetric wavefront is assumed, as would be the case for a helium neon laser or an argon-ion laser. Diode laser beams are asymmetric and often astigmatic, which causes their transformation to be more complex. Although in some respects component design and tolerancing for lasers is more critical than for conventional optical components, the designs often tend to be simpler since many of the constraints associated with imaging systems are not present. For instance, laser beams are nearly always used on axis, which eliminates the need to correct asymmetric aberration. Chromaticaberrations are of no concern in single-wavelength lasers, although they are critical for some tunable and multiline laser applications. In fact, the only significant aberration in most single-wavelength applications is primary (third-order) spherical aberration. Scatter from surface defects, inclusions, dust, or damaged coatings is of greater concern in laser-based systems than in incoherent systems. Speckle content arising from surface texture and beam coherence can limit system performance. Because laser light is generated coherently, it is not subject to some of the limitations normally associated with incoherent sources. All parts of the wavefront act as if they originate from the same point; consequently, the emergent wavefront can be precisely defined. Starting out with a well-defined wavefront permits more precise focusing and control of the beam than otherwise would be possible. For virtually all laser cavities, the propagation of an electromagnetic field, E(0), through one round trip in an optical resonator can be described mathematically by a propagation integral, which has the general form where K is the propagation constant at the carrier frequency of the opti-cal signal, pis the length of one period or round trip, and the integral is over the transverse coordinates at the reference or input plane. The function K is commonly called the propagation kernel since the field E(1)(x, y), after one propagation step, can be obtained from the initial field E (0)(x0, y0) through the operation of the linear kernel or “propagator” K(x, y, x0, y0). By setting the condition that the field, after one period, will have exactly the same transverse form, both in phase and profile (amplitude variation across the field), we get the equation where Enm represents a set of mathematical eigenmodes, and gnm a corresponding set of eigenvalues. The eigenmodes are referred to as transverse cavity modes, and, for stable resonators, are closely approx-imated by Hermite-Gaussian functions, denoted by TEMnm. (Anthony Siegman, Lasers) The lowest order, or “fundamental” transverse mode, TEM00 has a Gaussian intensity profile, shown in figure 2.1, which has the form In this section we will identify the propagation characteristics of this low-est-order solution to the propagation equation. In thenext section, Real Beam Propagation, we will discuss the propagation characteristics of higher-order modes, as well as beams that have been distorted by diffraction or various anisotropic phenomena. Gaussian Beam Propagation E x y e K x y x y E x y dx dy jkp InputPlane ( ) , , , , , 1 0 0 0 0 0 0 0 ( ) = ( ) ( ) − ( ) ∫∫ gnm nm nm E x y K x y x y E x y dx dy InputPlane , , , , , ( ) ≡ ( ) ( ) ∫∫ 0 0 0 0 0 0 I x y e k x y , ( )∝ − + ( ) 2 2 (2.1) (2.2) (2.3) Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings www.cvimellesgriot.com 2.3 Gaussian Beam Optics Gaussian Beam Optics 13.5 CONTOUR RADIUS 41.5w 20 40 60 80 100 4 PERCENT IRRADIANCE 0 w 1.5w 1/e2 diameter 13.5% of peak FWHM diameter 50% of peak direction of propagation Figure 2.1 Irradiance profile of a Gaussian TEM00 mode Figure 2.2 Diameter of a Gaussian beam toward infinity as z is further increased, asymptotically approaching the value of z itself. The plane z = 0 marks the location of a Gaussian waist, or a place where the wavefront is flat, and w0 is called the beam waist radius. The irradiance distribution of the Gaussian TEM00 beam, namely, where w = w(z) and P is the total power in the beam, is the same at all cross sections of the beam. The invariance of the form of the distribution is a special consequence of the presumed Gaussian distribution at z = 0. If a uniform irradiance distribution had been presumed at z = 0, the pattern at z = ∞would have been the familiar Airy disc pattern given by a Bessel function, whereas the pattern at intermediate zvalues would have been enormously complicated. Simultaneously, as R(z) asymptotically approaches z for large z, w(z) asymptotically approaches the value where z is presumed to be much larger than pw0 /l so that the 1/e2 irradiance contours asymptotically approach a cone of angular radius (intensity) has fallen to 50 percent of its peak, or axial value, as shown in figure 2.2. This second definition is also referred to as FWHM, or full width at half maximum. For the remainder of this guide, we will be using the 1/e2 definition. Diffraction causes light waves to spread transversely as they propagate, and it is therefore impossible to have a perfectly collimated beam. The spreading of a laser beam is in precise accord with the predictions of pure diffraction theory; aberration is totally insignificant in the present context. Under quite ordinary circumstances, the beam spreading can be so small it can go unnoticed. The following formulas accurately describe beam spread-ing, making it easy to see the capabilities and limitations of laser beams. Even if a Gaussian TEM00 laser-beam wavefront were made perfectly flat at some plane, it would quickly acquire curvature and begin spreading in accordance with where z is the distance propagated from the plane where the wavefront is flat, l is the wavelength of light, w0 is the radius of the 1/e2 irradiance contour at the plane where the wavefront is flat, w(z) is the radius of the 1/e2 contour after the wave has propagated a distance z, and R(z) is the wavefront radius of curvature after propagating a distance z. R(z) is infinite at z=0, passes through a minimum at some finite z, and rises again R z z w z w z w z w ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ 1 1 0 2 2 0 0 2 2 p l l p and ⎤ ⎦ ⎥ ⎥ / 1 2 (2.4) I r I e P w e r w r w ( ) / / = = − − 0 2 2 2 2 2 2 2 2 p , (2.6) w z z w ( ) = l p 0 (2.7) (2.5) v l p = ( ) = w z z w0 . (2.8) and Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings This value is the far-field angular radius (half-angle divergence) of the Gaussian TEM00 beam. The vertex of the cone lies at the center of the waist, as shown in figure 2.3. It is important to note that, for a given value of l, variations of beam diameter and divergence with distance z are functions of a single parameter, w0, the beam waist radius. www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.4 w w0 w0 z w0 1 e2 irradiance surface v asymptotic cone Figure 2.3 Growth in 1/e2 radius with distance propagated away from Gaussian waist laser 2w0 v Gaussian profile z = 0 planar wavefront 2w0 2 z = zR maximum curvature Gaussian intensity profile z = q planar wavefront Figure 2.4 Changes in wavefront radius with propagation distance curvature is a maximum. Far-field divergence (the number quoted in laser specifications) must be measured at a distance much greater than zR (usually >10#zR will suffice). This is a very important distinction because calculations for spot size and other parameters in an optical train will be inaccurate if near- or mid-field divergence values are used. For a tightly focused beam, the distance from the waist (the focal point) to the far field can be a few millimeters or less. For beams coming directly from the laser, the far-field distance can be measured in meters. Typically, one has a fixed value for w0 and uses the expression to calculate w(z) for an input value of z. However, one can also utilize this equation to see how final beam radius varies with starting beam radius at a fixed distance, z. Figure 2.5 shows the Gaussian beam propagation equation plotted as a function of w0, with the particular values of l = 632.8 nm and z = 100 m. The beam radius at 100 m reaches a minimum value for a starting beam radius of about 4.5 mm. Therefore, if we wanted to achieve the best combination of minimum beam diameter and minimum beam spread (or best collimation) over a distance of 100 m, our optimum starting beam radius would be 4.5 mm. Any other starting value would result in a larger beam at z = 100 m. We can find the general expression for the optimum starting beam radius for a given distance, z. Doing so yields Using this optimum value of w0 will provide the best combination of minimum starting beam diameter and minimum beam spread [ratio of w z 0 optimum ( ) = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ l p 1 2 / (2.10) w z w z w ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 0 0 2 2 1 2 1 l p / Near-Field vs Far-Field Divergence Unlike conventional light beams, Gaussian beams do not diverge linearly. Near the beam waist, which is typically close to the output of the laser, the divergence angle is extremely small; far from the waist, the divergence angle approaches the asymptotic limit described above. The Raleigh range (zR), defined as the distance over which the beam radius spreads by a factor of √ _ 2, is given by At the beam waist (z = 0), the wavefront is planar [R(0) = ∞]. Likewise, at z = ∞, the wavefront is planar [R(∞) = ∞]. As the beam propagates from the waist, the wavefront curvature, therefore, must increase to a maximum and then begin to decrease, as shown in figure 2.4. The Raleigh range, considered to be the dividing line between near-field divergence and mid-range divergence, is the distance from the waist at which the wavefront z w R = p l 0 2 (2.9) . Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings www.cvimellesgriot.com 2.5 Gaussian Beam Optics Gaussian Beam Optics STARTING BEAM RADIUS w0 (mm) FINAL BEAM RADIUS (mm) 0 1 2 3 4 5 6 7 8 9 10 0 20 40 60 80 100 Figure 2.5 Beam radius at 100 m as a function of starting beam radius for a HeNe laser at 632.8 nm beam expander w(–zR) = 2w0 beam waist 2w0 zR zR w(zR) = 2w0 Figure 2.6 Focusing a beam expander to minimize beam radius and spread over a specified distance Location of the beam waist The location of the beam waist is required for most Gaussian-beam calculations. CVI Melles Griot lasers are typically designed to place the beam waist very close to the output surface of the laser. If a more accurate location than this is required, our applications engineers can furnish the precise location and tolerance for a particular laser model. APPLICATION NOTE Do you need . . . BEAM EXPANDERS CVI Melles Griot offers a range of precision beam expanders for better performance than can be achieved with the simple lens combinations shown here. w(z) to w0] over the distance z. For z = 100 m and l = 632.8 nm, w0 (optimum) = 4.48 mm (see example above). If we put this value for w0 (optimum) back into the expression for w(z), Thus, for this example, By turning this previous equation around, we find that we once again have the Rayleigh range (zR), over which the beam radius spreads by a factor of √ _ 2 as If we use beam-expanding optics that allow us to adjust the position of the beam waist, we can actually double the distance over which beam divergence is minimized, as illustrated in figure 2.6. By focusing the beam-expanding optics to place the beam waist at the midpoint, we can restrict beam spread to a factor of √ _ 2 over a distance of 2zR, as opposed to just zR. This result can now be used in the problem of finding the starting beam radius that yields the minimum beam diameter and beam spread over 100 m. Using 2(zR) = 100 m, or zR = 50 m, and l = 632.8 nm, we get a value of w(zR) = (2l/p)½ = 4.5 mm, and w0 = 3.2 mm. Thus, the optimum starting beam radius is the same as previously calculated. However, by focusing the expander we achieve a final beam radius that is no larger than our starting beam radius, while still maintaining the √ _ 2 factor in overall variation. Alternately, if we started off with a beam radius of 6.3 mm, we could focus the expander to provide a beam waist of w0 = 4.5 mm at 100 m, and a final beam radius of 6.3 mm at 200 m. z w w z w R R with = ( ) = p l 0 2 0 2 . w 100 2 4 48 6 3 ( ) = ( ) = . . mm w z w ( ) = ( ) 2 0 (2.11) with Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings Self recommends calculating zR, w0, and the position of w0 for each optical element in the system in turn so that the overall transformation of the beam can be calculated. To carry this out, it is also necessary to consider magnification: w0″/w0. The magnification is given by The Rayleigh range of the output beam is then given by All the above formulas are written in terms of the Rayleigh range of the input beam. Unlike the geometric case, the formulas are not symmetric with respect to input and output beam parameters. For back tracing beams, it is useful to know the Gaussian beam formula in terms of the Rayleigh range of the output beam: Transformation and Magnification by Simple Lenses www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.6 It is clear from the previous discussion that Gaussian beams transform in an unorthodox manner. Siegman uses matrix transformations to treat the general problem of Gaussian beam propagation with lenses and mirrors. A less rigorous, but in many ways more insightful, approach to this problem was developed by Self (S.A.Self, “Focusing of Spherical Gaussian Beams”). Self shows a method to model transformations of a laser beam through simple optics, under paraxial conditions, by calculating the Rayleigh range and beam waist location following each individual optical element. These parameters are calculated using a formula analogous to the well-known standard lens-maker’s formula. The standard lens equation is written as where sis the object distance, s″is the image distance, and fis the focal length of the lens. For Gaussian beams, Self has derived an analogous formula by assuming that the waist of the input beam represents the object, and the waist of the output beam represents the image. The formula is expressed in terms of the Rayleigh range of the input beam. In the regular form, In the far-field limit as z/R approaches 0 this reduces to the geometric optics equation. A plot of s/f versus s″/f for various values of z/R/f is shown in figure 2.7. For a positive thin lens, the three distinct regions of interest correspond to real object and real image, real object and virtual image, and virtual object and real image. The main differences between Gaussian beam optics and geometric optics, highlighted in such a plot, can be summarized as follows: $ There is a maximum and a minimum image distance for Gaussian beams. $ The maximum image distance occurs at s = f=z/R, rather than at s = f. $ There is a common point in the Gaussian beam expression at s/f = s″/f = 1. For a simple positive lens, this is the point at which the incident beam has a waist at the front focus and the emerging beam has a waist at the rear focus. $ A lens appears to have a shorter focal length as z/R/f increases from zero (i.e., there is a Gaussian focal shift). 1 1 1 s f s f / / . + ″ = (2.12) z m z R 2 R ′′ = . (2.16) 1 1 1 2 s s z s f f + ″ + ″ ″ − = R /( ) . (2.17) m w w s f z f = ″ = −( ) ⎡ ⎣ ⎤ ⎦+ ( ) { } 0 0 2 2 1 1 / / . R (2.15) 1 1 1 1 2 2 s z s f s f s f z f + − ( ) + ″ = ( ) + ( R R or, in dimensionless form, / / / ) − ( ) + ″ ( ) = / / / . s f s f 1 1 1 (2.13) (2.14) 0 1 2 4 5 41 0 1 2 3 4 5 3 0 0.25 1 2 parameter zR f ( ) IMAGE DISTANCE (s″/f ) OBJECT DISTANCE (s/f ) 0.50 42 43 44 45 41 42 43 44 Figure 2.7 Plot of lens formula for Gaussian beams with normalized Rayleigh range of the input beam as the parameter or, in dimensionless form, Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings BEAM CONCENTRATION The spot size and focal position of a Gaussian beam can be determined from the previous equations. Two cases of particular interest occur when s = 0 (the input waist is at the first principal surface of the lens system) and s = f (the input waist is at the front focal point of the optical system). For s = 0, we get For the case of s=f, the equations for image distance and waist size reduce to the following: Substituting typical values into these equations yields nearly identical results, and for most applications, the simpler, second set of equations can be used. In many applications, a primary aim is to focus the laser to a very small spot, as shown in figure 2.8, by using either a single lens or a combination of several lenses. If a particularly small spot is desired, there is an advantage to using a well-corrected high-numerical-aperture microscope objective to concen-trate the laser beam. The principal advantage of the microscope objective over a simple lens is the diminished level of spherical aberration. Although microscope objectives are often used for this purpose, they are not always designed for use at the infinite conjugate ratio. Suitably optimized lens systems, known as infinite conjugate objectives, are more effective in beam-concentration tasks and can usually be identified by the infinity symbol on the lens barrel. DEPTH OF FOCUS Depth of focus (8Dz), that is, the range in image space over which the focused spot diameter remains below an arbitrary limit, can be derived from the formula The first step in performing a depth-of-focus calculation is to set the allowable degree of spot size variation. If we choose a typical value of 5 percent, or w(z)0 = 1.05w0, and solve for z = Dz, the result is Since the depth of focus is proportional to the square of focal spot size, and focal spot size is directly related to f-number (f/#), the depth of focus is proportional to the square of the f/# of the focusing system. TRUNCATION In a diffraction-limited lens, the diameter of the image spot is where K is a constant dependent on truncation ratio and pupil illumination, l is the wavelength of light, and f/# is the speed of the lens at truncation. The intensity profile of the spot is strongly dependent on the intensity profile of the radiation filling the entrance pupil of the lens. For uniform pupil illumination, the image spot takes on the Airy disc intensity profile shown in figure 2.9. www.cvimellesgriot.com 2.7 Gaussian Beam Optics Gaussian Beam Optics D p l z w ≈± 0 32 0 2 . . d K = × × l f /# (2.20) s f w f w ″ = = and l p / . 0 w z w z w ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 0 0 2 2 1 2 1 l p / . w Dbeam 1 e2 2w 0 Figure 2.8 Concentration of a laser beam by a laser-line focusing singlet .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 INTENSITY 2.44 l (f-number) 13.5% intensity 50% intensity Figure 2.9 Airy disc intensity distribution at the image plane and ′′ = +( ) = +( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ s f f w w f w f w 1 1 0 2 2 0 0 2 2 1 2 l p l p l p / / / / and (2.18) (2.19) and Calculation of spot diameter for these or other truncation ratios requires that K be evaluated. This is done by using the formulas and The K function permits calculation of the on-axis spot diameter for any beam truncation ratio. The graph in figure 2.11 plots the K factor vs T(Db/Dt). The optimal choice for truncation ratio depends on the relative importance of spot size, peak spot intensity, and total power in the spot as demon-strated in the table below. The total power loss in the spot can be calculated by using for a truncated Gaussian beam. A good compromise between power loss and spot size is often a truncation ratio of T=1. When T=2 (approximately uniform illumination), fractional power loss is 60 percent. When T = 1, d1/e 2 is just 8.0 percent larger than when T = 2, whereas fractional power loss is down to 13.5 percent. Because of this large savings in power with relatively little growth in the spot diameter, truncation ratios of 0.7 to 1.0 are typically used. Ratios as low as 0.5 might be employed when laser power must be conserved. However, this low value often wastes too much of the available clear aperture of the lens. Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings If the pupil illumination is Gaussian in profile, the result is an image spot of Gaussian profile, as shown in figure 2.10. When the pupil illumination is between these two extremes, a hybrid intensity profile results. In the case of the Airy disc, the intensity falls to zero at the point dzero = 2.44#l#f/#, defining the diameter of the spot. When the pupil illumination is not uniform, the image spot intensity never falls to zero making it necessary to define the diameter at some other point. This is commonly done for two points: dFWHM = 50-percent intensity point and d1/e 2 = 13.5% intensity point. It is helpful to introduce the truncation ratio where Db is the Gaussian beam diameter measured at the 1/e2 intensity point, and Dt is the limiting aperture diameter of the lens. If T = 2, which approximates uniform illumination, the image spot intensity profile approaches that of the classic Airy disc. When T = 1, the Gaussian profile is truncated at the 1/e2 diameter, and the spot profile is clearly a hybrid between an Airy pattern and a Gaussian distribution. When T = 0.5, which approximates the case for an untruncated Gaussian input beam, the spot intensity profile approaches a Gaussian distribution. www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.8 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 INTENSITY 1.83 l (f-number) 13.5% intensity 50% intensity Figure 2.10 Gaussian intensity distribution at the image plane T D D = b t (2.21) K T T FWHM = + − ( ) − − ( ) 1 029 0 7125 0 2161 0 6445 0 2161 2.179 2.221 . . . . . (2.22) K T T e 1 1.821 1.891 2 1 6449 0 6460 0 2816 0 5320 0 2816 / . . . . . . = + − ( ) − − ( ) (2.23) P e D D L t b = −( ) 2 2 / (2.24) 0.5 1.5 1.0 2.0 2.5 3.0 K FACTOR 0 T(Db/Dt) 1.0 2.0 3.0 4.0 spot diameter = K ! l ! f-number spot measured at 50% intensity level spot measured at 13.5% intensity level Figure 2.11 K factors as a function of truncation ratio Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings SPATIAL FILTERING Laser light scattered from dust particles residing on optical surfaces may produce interference patterns resembling holographic zone planes. Such patterns can cause difficulties in interferometric and holographic applications where they form a highly detailed, contrasting, and confusing background that interferes with desired information. Spatial filtering is a simple way of suppressing this interference and maintaining a very smooth beam irradiance distribution. The scattered light propagates in different direc-tions from the laser light and hence is spatially separated at a lens focal plane. By centering a small aperture around the focal spot of the direct beam, as shown in figure 2.12, it is possible to block scattered light while allowing the direct beam to pass unscathed. The result is a cone of light that has a very smooth irradiance distribution and can be refocused to form a collimated beam that is almost uniformly smooth. As a compromise between ease of alignment and complete spatial filtering, it is best that the aperture diameter be about two times the 1/e2 beam contour at the focus, or about 1.33 times the 99% throughput contour diameter. www.cvimellesgriot.com 2.9 Gaussian Beam Optics Gaussian Beam Optics focusing lens pinhole aperture Figure 2.12 Spatial filtering smoothes the irradiance distribution Truncation Ratio dFWHM d1/e2 dzero PL(%) Spot Diameters and Fractional Power Loss for Three Values of Truncation Infinity 1.03 1.64 2.44 100 2.0 1.05 1.69 — 60 1.0 1.13 1.83 — 13.5 0.5 1.54 2.51 — 0.03 Do you need . . . SPATIAL FILTERS CVI Melles Griot offers 3-axis spatial filters with precision micrometers (07 SFM 201 and 07 SFM 701). These devices feature an open-design that provides access to the beam as it passes through the instrument. The spatial filter consists of a precision, differential-micrometer y-z stage, which controls the pinhole location, and a single-axis translation stage for the focusing lens. The spatial filter mount accepts LSL-series focusing optics, OAS-series microscope objectives, and PPM-series mounted pinholes. The precision individual pinholes are for general-purpose spatial filtering. The high-energy laser precision pinholes are constructed specifically to withstand irradiation from high-energy lasers. Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings Real Beam Propagation www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.10 In the real world, truly Gaussian laser beams are very hard to find. Low-power beams from helium neon lasers can be a close approximation, but the higher the power of the laser is, the more complex the excitation mechanism (e.g., transverse discharges, flash-lamp pumping), and the higher the order of the mode is, the more the beam deviates from the ideal. To address the issue of non-Gaussian beams, a beam quality factor, M2, has come into general use. For a typical helium neon laser operating in TEM00 mode, M2 <1.1. Ion lasers typically have an M2 factor ranging from 1.1 to 1.7. For high-energy multimode lasers, the M2 factor can be as high as 10 or more. In all cases, the M2 factor affects the characteristics of a laser beam and cannot be neglected in optical designs, and truncation, in general, increases the M2 factor of the beam. In Laser Modes, we will illustrate the higher-order eigensolutions to the propagation equation, and in The Propagation Constant, M2will be defined. The section Incorporating M2 into the Propagation Equations defines how non-Gaussian beams propagate in free space and through optical systems. THE PROPAGATION CONSTANT The propagation of a pure Gaussian beam can be fully specified by either its beam waist diameter or its far-field divergence. In principle, full characterization of a beam can be made by simply measuring the waist diameter, 2w0, or by measuring the diameter, 2w(z), at a known and specified distance (z) from the beam waist, using the equations and where l is the wavelength of the laser radiation, and w(z) and R(z) are the beam radius and wavefront radius, respectively, at distance z from the beam waist. In practice, however, this approach is fraught with problems— it is extremely difficult, in many instances, to locate the beam waist; relying on a single-point measurement is inherently inaccurate; and, most important, pure Gaussian laser beams do not exist in the real world. The beam from a well-controlled helium neon laser comes very close, as does the beam from a few other gas lasers. However, for most lasers (even those LASER MODES The fundamental TEM00 mode is only one of many transverse modes that satisfy the round-trip propagation criteria described in Gaussian Beam Propagation. Figure 2.13 shows examples of the primary lower-order Hermite-Gaussian (rectangular) solutions to the propagation equation. Note that the subscripts n and m in the eigenmode TEMnm are correlated to the number of nodes in the x and y directions. In each case, adjacent lobes of the mode are 180 degrees out of phase. The propagation equation can also be written in cylindrical form in terms of radius (r) and angle (f). The eigenmodes (Erf) for this equation are a series of axially symmetric modes, which, for stable resonators, are closely approximated by Laguerre-Gaussian functions, denoted by TEMrf. For the lowest-order mode, TEM00, the Hermite-Gaussian and Laguerre-Gaussian functions are identical, but for higher-order modes, they differ significantly, as shown in figure 2.14. TEM00 TEM01 TEM10 TEM11 TEM02 Figure 2.13 Low-order Hermite-Gaussian resonator modes The mode, TEM01, also known as the “bagel” or “doughnut” mode, is considered to be a superposition of the Hermite-Gaussian TEM10 and TEM01 modes, locked in phase quadrature. In real-world lasers, the Hermite-Gaussian modes predominate since strain, slight misalignment, or contamination on the optics tends to drive the system toward rectan-gular coordinates. Nonetheless, the Laguerre-Gaussian TEM10 “target” or “bulls-eye” mode is clearly observed in well-aligned gas-ion and helium neon lasers with the appropriate limiting apertures. w z w z w ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 0 0 2 2 1 2 1 l p / R z z w z ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 0 2 2 p l Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings specifying a fundamental TEM00 mode), the output contains some com-ponent of higher-order modes that do not propagate according to the formula shown above. The problems are even worse for lasers operating in high-order modes. The need for a figure of merit for laser beams that can be used to determine the propagation characteristics of the beam has long been recognized. Specifying the mode is inadequate because, for example, the output of a laser can contain up to 50 percent higher-order modes and still be considered TEM00. www.cvimellesgriot.com 2.11 Gaussian Beam Optics Gaussian Beam Optics TEM00 TEM01 TEM10 Figure 2.14 Low-order axisymmetric resonator modes 2wR(R) = M[2w (R)] = M(2w0√2) 2v z embedded Gaussian mixed mode 2w0R = M(2w0) 2w0 z = 0 z = R M(2v) Figure 2.15 The embedded Gaussian The concept of a dimensionless beam propagation parameter was developed in the early 1970s to meet this need, based on the fact that, for any given laser beam (even those not operating in the TEM00 mode) the product of the beam waist radius (w0) and the far-field divergence (v) are constant as the beam propagates through an optical system, and the ratio where w0Rand vR, the beam waist and far-field divergence of the real beam, respectively, is an accurate indication of the propagation characteristics of the beam. For a true Gaussian beam, M2 = 1. EMBEDDED GAUSSIAN The concept of an “embedded Gaussian,” shown in figure 2.15, is useful as a construct to assist with both theoretical modeling and laboratory measurements. A mixed-mode beam that has a waist M (not M2) times larger than the embedded Gaussian will propagate with a divergence M times greater than the embedded Gaussian. Consequently the beam diameter of the mixed-mode beam will always be M times the beam diameter of the embedded Gaussian, but it will have the same radius of curvature and the same Rayleigh range (z = R). M w w 2 0 0 = R R v v (2.25) In a like manner, the lens equation can be modified to incorporate M2. The standard equation becomes and the normalized equation transforms to Incorporating M2 Into the Propagation Equations In the previous section we defined the propagation constant M2 where w0R and vR are the beam waist and far-field divergence of the real beam, respectively. For a pure Gaussian beam, M2 = 1, and the beam-waist beam-divergence product is given by It follows then that for a real laser beam, The propagation equations for a real laser beam are now written as where wR(z) and RR(z) are the 1/e2 intensity radius of the beam and the beam wavefront radius at z, respectively. The equation for w0(optimum) now becomes The definition for the Rayleigh range remains the same for a real laser beam and becomes For M2 = 1, these equations reduce to the Gaussian beam propagation equations . Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.12 w0v l p = / w M 0 2 R R v l p l p = > (2.26) w z w z M w R z z w z M R R R R R ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ( ) = + 0 2 0 2 2 1 2 0 2 1 1 l p p l / and 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ (2.27) R z z w z w z w z w ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ 1 1 0 2 2 0 0 2 2 p l l p and ⎤ ⎦ ⎥ ⎥ / 1 2 w zM 0 2 optimum ( ) = l p 1/2 (2.29) (2.28) M w w 2 0 0 = R R v v 1 1 1 2 2 s z M s f s f +( ) − ( ) + ′′ = R / / (2.31) 1 1 1 1 2 2 s f z M f s f s f / / / / / . ( ) + ( ) − ( ) + ′′ ( ) = R (2.32) Do you need . . . CVI Melles Griot Lasers and Laser Accessories CVI Melles Griot manufactures many types of lasers and laser systems for laboratory and OEM applications. Laser types include helium neon (HeNe) and helium cadmium (HeCd) lasers, argon, krypton, and mixed gas (argon/krypton) ion lasers; diode lasers and diode-pumped solid-state (DPSS) lasers. CVI Melles Griot also offers a range of laser accessories including laser beam expanders, generators, laser-line collimators, spatial filters and shear-plate collimation testers. z w R R = p M l 0 2 2 (2.30) and and EXAMPLE: OBTAIN AN 8-MM SPOT AT 80 M Using the CVI Melles Griot HeNe laser 25 LHR 151, produce a spot 8 mm in diameter at a distance of 80 m, as shown in figure 2.16 The CVI Melles Griot 25 LHR 151 helium neon laser has an output beam radius of 0.4 mm. Assuming a collimated beam, we use the propagation formula to determine the spot size at 80 m: or 80.6-mm beam diameter. This is just about exactly a factor of 10 larger than we wanted. We can use the formula for w0(optimum) to determine the smallest collimated beam diameter we could achieve at a distance of 80 m: Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings Lens Selection www.cvimellesgriot.com 2.13 Gaussian Beam Optics Gaussian Beam Optics The most important relationships that we will use in the process of lens selection for Gaussian-beam optical systems are focused spot radius and beam propagation. Focused Spot Radius where wF is the spot radius at the focal point, and wL is the radius of the collimated beam at the lens. M2 is the quality factor (1.0 for a theoretical Gaussian beam). Beam Propagation and where w0R is the radius of a real (non-Gaussian) beam at the waist, and wR (z) is the radius of the beam at a distance z from the waist. For M2 = 1, the formulas reduce to that for a Gaussian beam. w0(optimum) is the beam waist radius that minimizes the beam radius at distance z, and is obtained by differentiating the previous equation with respect to distance and setting the result equal to zero. Finally, where zR is the Raleigh range. w fM w F L = l p 2 (2.33) w z w z M w R z z w z M R R R R R ( ) = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ( ) = + 0 2 0 2 2 1 2 0 2 1 1 l p p l / and 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ w zM 0 2 optimum ( ) = l p 1/2 z w R = p l 0 2 0.8 mm 45 mm 80 m 8 mm Figure 2.16 Lens spacing adjusted empirically to achieve an 8-mm spot size at 80 m We can also utilize the equation for the approximate on-axis spot size caused by spherical aberration for a plano-convex lens at the infinite conjugate: This formula is for uniform illumination, not a Gaussian intensity profile. However, since it yields a larger value for spot size than actually occurs, its use will provide us with conservative lens choices. Keep in mind that this formula is for spot diameter whereas the Gaussian beam formulas are all stated in terms of spot radius. w zM 0 2 optimum ( ) = l p 1/2 w( ) . . , . 80 0 4 1 0 6328 10 80 000 0 4 3 2 2 1 m -= + × × ( ) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ p / 2 = 40.3-mm beam radius spot diameter (3 -order spherical aberration) f rd = ( ) 0 067 . /# f 3 w0 3 1 2 0 6328 10 80 000 4 0 optimum mm. ( ) = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − . , . / p and In order to determine necessary focal lengths for an expander, we need to solve these two equations for the two unknowns. In this case, using a negative value for the magnification will provide us with a Galilean expander. This yields values of f2 = 55.5 mm] and f1 = 45.5 mm. Ideally, a plano-concave diverging lens is used for minimum spherical aberration, but the shortest catalog focal length available is -10 mm. There is, however, a biconcave lens with a focal length of 5 mm (LDK-5.0-5.5-C). Even though this is not the optimum shape lens for this application, the extremely short focal length is likely to have negligible aberrations at this f-number. Ray tracing would confirm this. Now that we have selected a diverging lens with a focal length of 45 mm, we need to choose a collimating lens with a focal length of 50 mm. To determine whether a plano-convex lens is acceptable, check the spherical aberration formula. Clearly, a plano-convex lens will not be adequate. The next choice would be an achromat, such as the LAO-50.0-18.0. The data in the spot size charts indicate that this lens is probably diffraction limited at this f-number. Our final system would therefore consist of the LDK-5.0-5.5-C spaced about 45 mm from the LAO-50.0-18.0, which would have its flint element facing toward the laser. This tells us that if we expand the beam by a factor of 10 (4.0 mm/0.4 mm), we can produce a collimated beam 8 mm in diameter, which, if focused at the midpoint (40 m), will again be 8 mm in diameter at a distance of 80 m. This 10# expansion could be accomplished most easily with one of the CVI Melles Griot beam expanders, such as the 09 LBX 003 or 09 LBM 013. However, if there is a space constraint and a need to perform this task with a system that is no longer than 50 mm, this can be accomplished by using catalog components. Figure 2.17 illustrates the two main types of beam expanders. The Keplerian type consists of two positive lenses, which are positioned with their focal points nominally coincident. The Galilean type consists of a negative diverging lens, followed by a positive collimating lens, again positioned with their focal points nominally coincident. In both cases, the overall length of the optical system is given by and the magnification is given by where a negative sign, in the Galilean system, indicates an inverted image (which is unimportant for laser beams). The Keplerian system, with its internal point of focus, allows one to utilize a spatial filter, whereas the Galilean system has the advantage of shorter length for a given magnification. Fundamental Optics Gaussian Beam Optics Optical Specifications Material Properties Optical Coatings www.cvimellesgriot.com Gaussian Beam Optics Gaussian Beam Optics 2.14 Keplerian beam expander f1 f2 Galilean beam expander f1 f2 Figure 2.17 Two main types of beam expanders overall length = + f f 1 2 magnification = f f 2 1 f f f f 1 2 2 1 50 10 + = = − mm and . The spot diameter resulting from spherical aberration is m The spot diameter resulting from d 0 067 50 6 25 14 3 . . . × = m iffraction (2 ) is m w0 3 2 0 6328 10 50 4 0 5 ( . ) . . × = − p m References A. Siegman. Lasers (Sausalito, CA: University Science Books, 1986). S. A. Self. “Focusing of Spherical Gaussian Beams.” Appl. Opt. 22, no. 5 (March 1983): 658. H. Sun. “Thin Lens Equation for a Real Laser Beam with Weak Lens Aperture Truncation.” Opt. Eng. 37, no. 11 (November 1998). R. J. Freiberg, A. S. Halsted. “Properties of Low Order Transverse Modes in Argon Ion Lasers.” Appl. Opt. 8, no. 2 (February 1969): 355-362. W. W. Rigrod. “Isolation of Axi-Symmetric Optical-Resonator Modes.”Appl. Phys. Let. 2, no. 3 (February 1963): 51-53. M. Born, E. Wolf. Principles of Optics Seventh Edition (Cambridge, UK: Cam-bridge University Press, 1999). and The spot diameter resulting from spherical aberration is The spot diameter resulting from diffraction (2wo) is
17565	https://www.sciencedirect.com/science/article/abs/pii/S0196885822001427	Cycle structure of random parking functions - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (34) Cited by (1) Advances in Applied Mathematics Volume 144, March 2023, 102458 Cycle structure of random parking functions Author links open overlay panel J.E.Paguyo Show more Add to Mendeley Share Cite rights and content Abstract We initiate the study of the cycle structure of uniformly random parking functions. Using the combinatorics of parking completions, we compute the asymptotic expected value of the number of cycles of any fixed length. We obtain an upper bound on the total variation distance between the joint distribution of cycle counts and independent Poisson random variables using a multivariate version of Stein's method via exchangeable pairs. Under a mild condition, the process of cycle counts converges in distribution to a process of independent Poisson random variables. Introduction Consider n parking spots placed sequentially on a one-way street. A line of n cars enter the street one at a time, with each car having a preferred parking spot. The i th car drives to its preferred spot, π i, and parks if the spot is available. If the spot is already occupied, the car parks in the first available spot after π i. If the car is unable to find any available spots, the car exits the street without parking. A sequence of preferences π=(π 1,…,π n) is a parking function if all n cars are able to park. More precisely, let [n]:={1,…,n}. A sequence π=(π 1,…,π n)∈[n]n is a parking function of size n if and only if \|{k:π k≤i}\|≥i for all i∈[n]. This follows from the pigeonhole principle, since a parking function must have at least one coordinate with value equal to 1, at least two coordinates with value at most 2, and so on. Equivalently, π=(π 1,…,π n)∈[n]n is a parking function if and only if π(i)≤i for all i∈[n], where (π(1),…,π(n)) is π sorted in a weakly increasing order π(1)≤⋯≤π(n). Let PF n denote the set of parking functions of size n. The total number of parking functions of size n is\|PF n\|=(n+1)n−1.An elegant proof using a circle argument is due to Pollak and can be found in . Parking functions were introduced by Konheim and Weiss in their study of the hash storage structure. It has since found many applications to combinatorics, probability, and computer science. Observe that the expression for \|PF n\| is reminiscent of Cayley's formula for labeled trees. Indeed, Foata and Riordan established a bijection between parking functions and trees on n+1 labeled vertices. Connections to other combinatorial objects have since been established. For example, Stanley found connections to noncrossing set partitions and hyperplane arrangements , and Pitman and Stanley found a relation to volume polynomials of certain polytopes. The literature on the combinatorics of parking functions is vast and we refer the reader to Yan for an accessible survey. There are several generalizations of the classical parking functions. In , Yan considered x-parking functions, where parking functions are associated with a vector x, and constructed a bijection with rooted forests. Kung and Yan then introduced [a,b]-parking functions in and u-parking functions in , and for both generalizations, they gave explicit formulas for moments of sums of these parking functions. In , Postnikov and Shapiro introduced G-parking functions, where G is a digraph on [n]; the classical parking functions correspond to the case where G=K n+1. Gorsky, Mazin, and Vazirani studied rational parking functions and found connections to affine permutations and representation theory. Returning to the classical notion of parking on a one-way street, Ehrenborg and Happ studied parking in which cars have different sizes, and in they extended this to study parking cars behind a trailer of fixed size. Much of the recent combinatorial work on parking functions concerns parking completions. Suppose that ℓ of the n spots are already occupied, denoted by v=(v 1,…,v ℓ), where the entries are in increasing order, and that we want to find parking preferences for the remaining n−ℓ cars so that they can all successfully park. The set of successful preference sequences are the parking completions for the sequence v=(v 1,…,v ℓ). Gessel and Seo obtained a formula for parking completions where the occupied spots consist of a contiguous block starting from the first spot, v=(1,…,ℓ). Diaconis and Hicks introduced the parking function shuffle to count parking completions with one spot arbitrarily occupied. Extending this result, Aderinan et al. used join and split operations to count parking completions where ℓ≤n occupied spots are arbitrarily occupied. Probabilistic questions about parking functions have also been considered, but probabilistic problems tend to be more complicated than enumeration results. In their study of the width of rooted labeled trees, Chassaing and Marckert discovered connections between parking functions, empirical processes, and the Brownian bridge. The asymptotic distribution for the cost construction for hash tables with linear probing (which is equivalent to the area statistic of parking functions) was studied by Flajolet, Poblete, and Viola and Janson , where it was shown to converge to normal, Poisson, and Airy distributions depending on the ratio between the number of cars and spots. More recently, Diaconis and Hicks studied the distribution of coordinates, descent pattern, area, and other statistics of random parking functions. Yao and Zeilberger used an experimental mathematics approach combined with some probability to study the area statistic. In , Kenyon and Yin explored links between combinatorial and probabilistic aspects of parking functions. Extending previous work, Yin developed the parking function multi-shuffle to obtain formulas for parking completions, moments of multiple coordinates, and all possible covariances between two coordinates for (m,n)-parking functions (where there are m≤n cars and n spots) and u-parking functions . In this paper, we continue the probabilistic study of parking functions by establishing the asymptotic distribution of the cycle counts of parking functions, partially answering a question posed by Diaconis and Hicks in . Let C k(π) be the number of k-cycles in the parking function π∈PF n. Our first result shows that the expected number of k-cycles in a random parking function is asymptotically 1 k. Theorem 1.1 Let π∈PF n be a parking function chosen uniformly at random. Then E(C k(π))∼1 k. Let μ and ν be probability distributions. The total variation distance between μ and ν is d T V(μ,ν):=sup A⊆Ω⁡\|μ(A)−ν(A)\|,where Ω is a measurable space. If X and Y are random variables with distributions μ and ν, respectively, then we write d T V(X,Y) in place of d T V(μ,ν). Our main result gives an upper bound on the total variation distance between the joint distribution of cycle counts (C 1,…,C d) of a random parking function and a Poisson process (Z 1,…,Z d), where {Z k} are independent Poisson random variables with rate λ k=E(C k). Theorem 1.2 Let π∈PF n be a parking function chosen uniformly at random. Let C k=C k(π)be the number of k-cycles in π and let W=(C 1,C 2,…,C d). Let Z=(Z 1,Z 2,…,Z d), where{Z k}are independent Poisson random variables with rate λ k=E(C k). Then d T V(W,Z)=O(d 4 n−d). Moreover if d=o(n 1/4), then the process of cycle counts converges in distribution to a process of independent Poisson random variables(C 1,C 2,…)→D(Y 1,Y 2,…)as n→∞, where{Y k}are independent Poisson random variables with rate 1 k . The proof uses a multivariate Stein's method with exchangeable pairs. Stein's method via exchangeable pairs has previously been used to prove limit theorems in a wide range of settings. We refer the reader to for an accessible survey. In particular, Judkovich used this approach to prove a Poisson limit theorem for the joint distribution of cycle counts in uniformly random permutations without long cycles. We remark that our limit theorem parallels the result of Arratia and Tavare on the cycle structure of uniformly random permutations. There is a vast probabilistic literature on the cycle structure of random permutations, which include the works of Shepp and Lloyd on ordered cycle lengths and DeLaurentis and Pittel on a functional central limit theorem for cycle lengths with connections to Brownian motion. Our paper initiates a parallel study of the cycle structure in random parking functions, but further study is fully warranted. The paper is organized as follows. Section 2 introduces definitions and notation that we use throughout the paper, and gives the necessary background and relevant results that we use to prove our main theorems. In Section 3, we compute the exact values for the expected number of fixed points and transpositions in a random parking function, and obtain the asymptotic formula, Theorem 1.1, for the expected number of k-cycles, for general k. We then apply Stein's method via exchangeable pairs in Section 4 to establish the Poisson limit theorem for joint cycle counts, Theorem 1.2. We conclude with some final remarks and open problems in Section 5. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Preliminaries In this section we introduce the definitions and notation that we use throughout the paper, and give necessary background, tools, and techniques that we use to prove our main results. Expected number of cycles of a fixed length In this section we compute the expected number of k-cycles in a random parking function. We consider the cases of fixed points and transpositions separately, as we are able to compute their expected values exactly. For general k, we compute the asymptotic expected number of k-cycles. Poisson limit theorem for cycles In this section, we introduce a useful directed graph representation for parking functions and construct our exchangeable pair. We then use these constructions along with the multivariate Stein's method to prove Theorem 1.2. Parking functions are a subset of a more general class of functions, F n, called random mappings, which are functions f:[n]→[n] from the set [n] to itself. Random mappings are extensively used in computer science and computational mathematics, for example in random number generators, cycle detection, and integer factorization. There is an extensive literature on the probabilistic properties of random mappings and we only cite a few here. Harris initiated the classical theory of random Acknowledgments We thank Jason Fulman for suggesting this problem and Peter Kagey for a helpful discussion. We also thank an anonymous referee for many helpful comments and suggestions. Recommended articles References (34) P. Diaconis et al. Probabilizing parking functions Adv. Appl. Math. (2017) J.P.S. Kung et al. Exact formulas for moments of sums of classical parking functions Adv. Appl. Math. (2003) J. Pitman Forest volume decompositions and Abel-Cayley-Hurwitz multinomial expansions J. Comb. Theory, Ser. A (2002) C.H. Yan Generalized parking functions, tree inversions, and multicolored graphs Adv. Appl. Math. (2001) A. Adeniran et al. Enumerating parking completions using join and split Electron. J. Comb. (2020) D.J. Aldous et al. Brownian bridge asymptotics for random mappings Random Struct. Algorithms (1994) R. Arratia et al. Total variation asymptotics for Poisson process approximations of logarithmic combinatorial assemblies Ann. Probab. (1995) R. Arratia et al. The cycle structure of random permutations Ann. Probab. (1992) P. Chassaing et al. Parking functions, empirical processes, and the width of rooted labeled trees Electron. J. Comb. (2001) S. Chatterjee et al. Exchangeable pairs and Poisson approximation Probab. Surv. (2005) J.M. DeLaurentis et al. Random permutations and Brownian motion Pac. J. Math. (1985) R. Ehrenborg et al. Parking cars of different sizes Am. Math. Mon. (2016) R. Ehrenborg et al. Parking cars after a trailer Australas. J. Comb. (2018) P. Flajolet et al. Random mapping statistics P. Flajolet et al. On the analysis of linear probing hashing Algorithmica (1998) D. Foata et al. Mappings of acyclic and parking functions Aequ. Math. (1974) I. Gessel et al. A refinement of Cayley's formula for trees Electron. J. Comb. (2006) View more references Cited by (1) The Foata–Fuchs proof of Cayley’s formula, and its probabilistic uses 2023, Electronic Communications in Probability View full text © 2022 Elsevier Inc. All rights reserved. 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17566	https://math.stackexchange.com/questions/3369742/looking-for-references-to-pythagorean-triple-subsets	elementary number theory - Looking for references to Pythagorean triple subsets - 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 Looking for references to Pythagorean triple subsets Ask Question Asked 6 years ago Modified4 years, 2 months ago Viewed 417 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I knew nothing about generating Pythagorean triples in 2009 so I looked for them in a spreadsheet. Millions of formulas later, I found a pattern of sets shown in the sample below. S e t n S e t 1 S e t 2 S e t 3 S e t 4 T r i p l e 1 3,4,5 15,8,17 35,12,37 63,16,65 T r i p l e 2 5,12,13 21,20,29 45,28,53 77,36,85 T r i p l e 3 7,24,25 27,36,45 55,48,73 91,60,109 T r i p l e 4 9,40,41 33,56,65 65,72,97 105,88,137 S e t n T r i p l e 1 T r i p l e 2 T r i p l e 3 T r i p l e 4 S e t 1 3,4,5 5,12,13 7,24,25 9,40,41 S e t 2 15,8,17 21,20,29 27,36,45 33,56,65 S e t 3 35,12,37 45,28,53 55,48,73 65,72,97 S e t 4 63,16,65 77,36,85 91,60,109 105,88,137 In each S e t n S e t n, (C−B)=(2 n−1)2(C−B)=(2 n−1)2, the increment between consecutive values of A A is 2(2 n−1)k 2(2 n−1)k where k k is the member number or count within the set, and A=(2 n−1)2+2(2 n−1)k A=(2 n−1)2+2(2 n−1)k. I solved the Pythagorean theorem for B B and C C, substituted now-known the expressions for A A and (C−B)(C−B), and got B=2(2 n−1)k+2 k 2 C=(2 n−1)2+2(2 n−1)k+2 k 2 B=2(2 n−1)k+2 k 2 C=(2 n−1)2+2(2 n−1)k+2 k 2. I have since learned the my formula is the equivalent of replacing (m,n)(m,n) in Euclid's formula with ((2 n−1+k),k)((2 n−1+k),k). I found ways of using either my formula or Euclid's to find triples given only sides, perimeters, ratios, and areas as well as polygons and pyramids constructed of dissimilar primitive triples. I found that the first member of each set (k=1)(k=1) and all members of S e t 1(n=1)S e t 1(n=1) are primitive. I found that, if (2 n−1)(2 n−1) is prime, only primitives will be generated in S e t n S e t n if A=(2 n−1)2+2(2 n−1)k+⌊k−1 2 n−2⌋A=(2 n−1)2+2(2 n−1)k+⌊k−1 2 n−2⌋ and I found that, if (2 n−1)(2 n−1) is composite, I could obtain only primitives in S e t n S e t n by generating and subtracting the set of [multiple] triples generated when k k is a 1 1-or-more multiple of any factor of (2 n−1)(2 n−1). The primitive count in the former is obtained directly; the count for the latter is obtained by combinatorics. I'm trying to write a paper "On Finding Pythagorean Triples". Surely someone has discovered these sets in the 2300 2300 years since Euclid but I haven't found and reference to them or any subsets of Pythagorean triples online or in the books I've bought and read. So my question is: "Where have these distinct sets of triples been mentioned before?" I would like to cite the work if I can find it. The bounty just expired and neither of the two answers has been helpful. I have not quite a day to award the bounty. Any takers? Where and when have these sets been discovered before? elementary-number-theory reference-request soft-question pythagorean-triples Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 2, 2020 at 19:05 poetasispoetasis asked Sep 25, 2019 at 17:56 poetasispoetasis 6,895 2 2 gold badges 15 15 silver badges 42 42 bronze badges 11 3 Well, why not take a primitive triple and multiply each term by an odd square?lulu –lulu 2019-09-25 18:03:26 +00:00 Commented Sep 25, 2019 at 18:03 I know the subset contains odd square multiples of primitives. I don't need to find them. I'm looking for what has been studied about their properties. I have my own observations and I would like to compare them to what's been done. I even developed a formula that generates the entire subset but I'm sure it must have been done before.poetasis –poetasis 2019-09-25 18:40:13 +00:00 Commented Sep 25, 2019 at 18:40 1 I don't know, if Maor, Eli, 2007: The Pythagorean Theorem: a 4000-year History. Princeton Univ. Press. ISBN 9-780-691-14823-6 helps (I haven't read it), but generally it's useful to read books about the history of the Pythagorean triple.user90369 –user90369 2019-10-11 12:28:24 +00:00 Commented Oct 11, 2019 at 12:28 1 Your splitting into sets is very special, it's not clear for which purpose (although I've read the explanation for Nilotpal Kanti Sinha). Perhaps you are the first one who is splitting into sets. ;) If you don't really find what you are looking for, it might be better to broaden your topic (as long as the core of the topic is maintained) and allow more ideas. Then more readers can be reached and the probability of finding suitable literature increases. ;) --- For the bounty a grace period of 24 hours is left. :)user90369 –user90369 2019-10-11 16:14:14 +00:00 Commented Oct 11, 2019 at 16:14 1 Yes, I have already noticed that here. But very special questions don't usually have many readers. If I find suitable literature, I write it of course, but the hope is unfortunately small.user90369 –user90369 2019-10-11 20:01:33 +00:00 Commented Oct 11, 2019 at 20:01 \|Show 6 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. In L. E. Dickson, History of the Theory of Numbers, Volume II, page 167 T. Fantet de Lagny 18 18 replaced m m by d+n d+n in (1)(1) and obtained x=2 n(d+n),y=d(d+2 n),z=x+d 2=y+2 n 2.x=2 n(d+n),y=d(d+2 n),z=x+d 2=y+2 n 2. The footnote 18 is briefly "Hist. Acad. Sc. Paris, 1729, 318." Your formulas are A=(2 n−1)2+2(2 n−1)k,B=2(2 n−1)k+2 k 2,C=(2 n−1)2+2(2 n−1)k+2 k 2.A=(2 n−1)2+2(2 n−1)k,B=2(2 n−1)k+2 k 2,C=(2 n−1)2+2(2 n−1)k+2 k 2. Get this from Lagny's formulas if d d is replaced by 2 n−1 2 n−1 and n n is replaced with k.k. Thus, your formula is equivalent to de Lagny's except 2 n−1=d 2 n−1=d is always odd, however, if d d is even, the triple has a common factor of 2 2 and can not be primitive. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 18, 2021 at 23:33 answered Jul 18, 2021 at 22:33 SomosSomos 37.6k 3 3 gold badges 35 35 silver badges 85 85 bronze badges 1 It will take me a while to understand your post but it looks like a step in the right direction. None of the other posts were useful and I downvoted them both. I lost my bounty points but at least they did not get them.poetasis –poetasis 2021-07-18 22:44:08 +00:00 Commented Jul 18, 2021 at 22:44 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. This paper defines the 'height' of a triple as C−B C−B and classifies Pythagorean triples in terms of their height and a parameter k k. Height and excess of Pythagorean triples, D McCullough - Mathematics Magazine, 2005 - Taylor & Francis,  Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 9, 2019 at 6:50 Angela PretoriusAngela Pretorius 10.3k 2 2 gold badges 30 30 silver badges 44 44 bronze badges 1 I'm sorry, I did not see anything related to the sets I discovered or the formula I developed in the link. If someone has found these sets, surely the formula would follow. e.g. F(1,1)=(3,4,5),F(1,2)=(5,12,13),F(1,3)=(7,24,25),F(1,4)=(9,40,41)F(1,1)=(3,4,5),F(1,2)=(5,12,13),F(1,3)=(7,24,25),F(1,4)=(9,40,41)F(2,1)=(15,8,17),F(2,2)=(21,20,29),F(2,3)=(27,36,45)F(2,1)=(15,8,17),F(2,2)=(21,20,29),F(2,3)=(27,36,45) and so on. Can you, perhaps, find a reference to a variation of Euclid's formula where A=((2 m−1+n)2−n 2)B=(2(2 m−1+n)n)C=((2 m−1+n)2+n 2)A=((2 m−1+n)2−n 2)B=(2(2 m−1+n)n)C=((2 m−1+n)2+n 2)? It produces the sets shown in the example.poetasis –poetasis 2019-10-09 18:22:51 +00:00 Commented Oct 9, 2019 at 18:22 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let A 2+B 2=C 2 A 2+B 2=C 2 be a Pythagoream triplet. The formula that you have mentioned is a special case of the general formula which gives all Pythagoream triplets. A=n(r 2−s 2),B=2 n r s),C=n(r 2+s 2)A=n(r 2−s 2),B=2 n r s),C=n(r 2+s 2) where n,r,s n,r,s are some positive integers. In case you want to generate all primitive Pythagorean triplets where a,b,c a,b,c have no common factors then take gcd(r,s)=n=1 gcd(r,s)=n=1. Every other special type of triplets can be generated from this general formula so there is actually nothing left. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 11, 2019 at 12:07 Nilotpal SinhaNilotpal Sinha 24.5k 5 5 gold badges 38 38 silver badges 111 111 bronze badges 1 I know my formula is a special case. It generates only and all triples where G C D(A,B,C)=(2 m−1)2,m∈N G C D(A,B,C)=(2 m−1)2,m∈N which includes all primitives. One advantage is that it eliminates the trivials, the doubles, the even squares and other non-odd-square multiples of primitives generated by Euclid's formula. What I'm looking for is a reference to any earlier discovery of the s e t s s e t s shown in the example. My discovery appears to be original but I know that is vanity so I'm looking to give credit if prior art can be found.poetasis –poetasis 2019-10-11 15:12:16 +00:00 Commented Oct 11, 2019 at 15:12 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 elementary-number-theory reference-request soft-question pythagorean-triples See similar questions with these tags. 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17567	https://peverati.github.io/pchem1/GibbsFree.html	9 Gibbs Energy In this chapter, we will concentrate on chemical processes that happen at constant TT and constant PP.41 As such, we will focus our attention on the Gibbs energy. 9.1 Gibbs Equation Recalling from chapter 8, the definition of GG is: G=U−TS+PV=H−TS,G=U−TS+PV=H−TS,(9.1) which, taking the differential at constant TT and PP, becomes: dG=dH=0⏞−SdT−TdS=dH−TdS.dG=dH=0−SdT−TdS=dH−TdS.(9.2) Integrating eq. (9.2) between the initial and final states of a process results in: ∫fidG=∫fidH−T∫fidSΔG=ΔH−TΔS∫fidG=∫fidH−T∫fidSΔG=ΔH−TΔS(9.3) which is the famous Gibbs equation for ΔGΔG. Using Definition 8.2, we can use ΔGΔG to infer the spontaneity of a chemical process that happens at constant TT and PP using ΔG≤0ΔG≤0. If we set ourselves at standard conditions, we can calculate the standard Gibbs energy of formation, ΔrxnG−⊖−ΔrxnG−⊖−, for any reaction as: ΔrxnG−⊖−=ΔrxnH−⊖−−TΔrxnS−⊖−=∑iνiΔfH−⊖−i−T∑iνiS−⊖−i,ΔrxnG−⊖−=ΔrxnH−⊖−−TΔrxnS−⊖−=∑iνiΔfH−⊖−i−T∑iνiS−⊖−i,(9.4) where ΔfH−⊖−iΔfH−⊖−i are the standard enthalpies of formation, S−⊖−iS−⊖−i are the standard entropies, and νiνi are the stoichiometric coefficients for every species ii involved in the reaction. All these quantities are commonly available, and we have already discussed their usage in chapters 4 and 7, respectively.42 The following four options are possible for ΔG−⊖−ΔG−⊖− of a chemical reaction: \| ΔG−⊖−ΔG−⊖− \| \| ΔH−⊖−ΔH−⊖− \| ΔS−⊖−ΔS−⊖− \| Spontaneous? \| \| – \| if \| – \| + \| Always \| \| + \| if \| + \| – \| Never \| \| –/+ \| if \| – \| – \| Depends on TT: spontaneous at low Tspontaneous at low T \| \| +/– \| if \| + \| + \| Depends on TT: spontaneous at high Tspontaneous at high T \| Or, in other words: Exothermic reactions that increase the entropy are always spontaneous. Endothermic reactions that reduce the entropy are always non-spontaneous. For the other two cases, the spontaneity of the reaction depends on the temperature: Exothermic reactions that reduce the entropy are spontaneous at low TT. Endothermic reactions that increase the entropy are spontaneous at high TT. A simple criterion to evaluate the entropic contribution of a reaction is to look at the total number of moles of the reactants and the products (as the sum of the stoichiometric coefficients). If the reaction is producing more molecules than it destroys (\|∑productsνi\|>\|∑reactantsνi\|)(∣∣∑productsνi∣∣>∣∣∑reactantsνi∣∣), it will increase the entropy. Vice versa, if the total number of moles in a reaction is reducing (\|∑productsνi\|<\|∑reactantsνi\|)(∣∣∑productsνi∣∣<∣∣∑reactantsνi∣∣), the entropy will also reduce. As we saw in section 8.2, the natural variables of the Gibbs energy are the temperature, TT, the pressure, PP, and chemical composition, as the number of moles {ni}{ni}. The Gibbs energy can therefore be expressed using the total differential as (see also, last formula in eq. (8.16)): dG(T,P,{ni})=(∂G∂T)P,{ni}⏟temperature dependencedT+(∂G∂P)T,{ni}⏟pressure dependencedP+∑i(∂G∂ni)T,P,{nj≠i}⏟composition dependencedni.dG(T,P,{ni})=(∂G∂T)P,{ni}temperature dependencedT+(∂G∂P)T,{ni}pressure dependencedP+∑i(∂G∂ni)T,P,{nj≠i}composition dependencedni.(9.5) If we know the behavior of GG as we vary each of the three natural variables independently of the other two, we can reconstruct the total differential dGdG. Each of these terms represents a coefficient in eq. (9.5), which are given in eq. (8.17). 9.2 Temperature Dependence of ΔGΔG (∂G∂T)P,{ni}=−S(∂G∂T)P,{ni}=−S Let’s analyze the first coefficient that gives the dependence of the Gibbs energy on temperature. Since this coefficient is equal to −S−S and the entropy is always positive, GG must decrease when TT increases at constant PP and {ni}{ni}, and vice versa. If we replace this coefficient for −S−S in the Gibbs equation, eq. (9.3), we obtain: ΔG=ΔH+T(∂ΔG∂T)P,{ni},ΔG=ΔH+T(∂ΔG∂T)P,{ni},(9.6) and since eq. (9.6) includes both ΔGΔG and its partial derivative with respect to temperature (∂ΔG∂T)P,{ni}(∂ΔG∂T)P,{ni} we need to rearrange it to include the temperature derivative only. To do so, we can start by evaluating the partial derivative of (ΔGT)(ΔGT) using the chain rule: [∂(ΔGT)∂T]P,{ni}=1T(∂ΔG∂T)P,{ni}−1T2ΔG,⎡⎢ ⎢⎣∂(ΔGT)∂T⎤⎥ ⎥⎦P,{ni}=1T(∂ΔG∂T)P,{ni}−1T2ΔG,(9.7) which, replacing ΔGΔG from eq. (9.6) into eq. (9.7), becomes: [∂(ΔGT)∂T]P,{ni}=1T(∂ΔG∂T)P,{ni}−1T2[ΔH+T(∂ΔG∂T)P,{ni}]=1T(∂ΔG∂T)P,{ni}−ΔHT2−1T(∂ΔG∂T)P,{ni},⎡⎢ ⎢⎣∂(ΔGT)∂T⎤⎥ ⎥⎦P,{ni}=1T(∂ΔG∂T)P,{ni}−1T2[ΔH+T(∂ΔG∂T)P,{ni}]=1T(∂ΔG∂T)P,{ni}−ΔHT2−1T(∂ΔG∂T)P,{ni},(9.8) which simplifies to: [∂(ΔGT)∂T]P,{ni}=−ΔHT2.⎡⎢ ⎢⎣∂(ΔGT)∂T⎤⎥ ⎥⎦P,{ni}=−ΔHT2.(9.9) Equation (9.9) is known as the Gibbs–Helmholtz equation, and is useful in its integrated form to calculate the Gibbs energy for a chemical reaction at any temperature TT by knowing just the standard Gibbs energy of formation and the standard enthalpy of formation for the individual species, which are usually reported at T=298KT=298K. The integration is performed as follows: ∫Tf=TTi=298K∂(ΔrxnGT)∂T=−∫Tf=TTi=298KΔrxnHT2ΔrxnG−⊖−(T)T=ΔrxnG−⊖−298K+ΔrxnH−⊖−(1T−1298K),∫Tf=TTi=298K∂(ΔrxnGT)∂T=−∫Tf=TTi=298KΔrxnHT2ΔrxnG−⊖−(T)T=ΔrxnG−⊖−298K+ΔrxnH−⊖−(1T−1298K),(9.10) giving the integrated Gibbs–Helmholtz equation: ΔrxnG−⊖−(T)T=∑iνiΔfG−⊖−i298K+∑iνiΔfH−⊖−i(1T−1298K)ΔrxnG−⊖−(T)T=∑iνiΔfG−⊖−i298K+∑iνiΔfH−⊖−i(1T−1298K)(9.11) 9.3 Pressure Dependence of ΔGΔG (∂G∂P)T,{ni}=V(∂G∂P)T,{ni}=V We can now turn the attention to the second coefficient that gives how the Gibbs energy changes when the pressure change. To do this, we put the system at constant TT and {ni}{ni}, and then we consider infinitesimal variations of GG. From eq. (8.8): dG=VdP−SdT+∑iμidniconstantT,{ni}→dG=VdP,dG=VdP−SdT+∑iμidniconstantT,{ni}−−−−−−−−−→dG=VdP,(9.12) which is the differential equation that we were looking for. To study changes of G for macroscopic changes in P, we can integrate eq. (9.12) between initial and final pressures, and considering an ideal gas, we obtain: ∫fidG=∫fiVdPΔG=nRT∫fidPP=nRTlnPfPi. If we take Pi=P−⊖−=1bar, we can rewrite eq. (9.13) as: G=G−⊖−+nRTlnPfP−⊖−, which is useful to convert standard Gibbs energies of formation at pressures different than standard pressure, using: ΔfG=ΔfG−⊖−+nRTlnPfP−⊖−⏟=1bar=ΔfG−⊖−+nRTlnPf For liquids and solids, V is essentially independent of P (liquids and solids are incompressible), and eq. (9.12) can be integrated as: ΔG=∫fiVdP=V∫fidP=VΔP. The plots in Figure 9.1 show the remarkable difference in the behaviors of ΔfG for a gas and for a liquid, as obtained from eqs. (9.13) and (9.16). Figure 9.1: Dependence of the Gibbs Energy of Formation of Liquid and Gaseous Ethanol at T = 310 K. The Curves Cross at the Vapor Pressure of Liquid Ethanol at this Temperature, which is 0.1 bar. 9.4 Composition Dependence of ΔG (∂G∂ni)T,P=μi The third and final coefficient gives the chemical potential as the dependence of G on the chemical composition at constant T and P. Similarly to the previous cases, we can take the definition of the coefficient and integrate it directly between the initial and final stages of a reaction. If we consider a reaction product, pure substance i, at the beginning of the reaction there will be no moles of it ni=0, and consequently G=0.43 We can then integrate the left-hand side between zero and the number of moles of product at the end of the reaction, n, and the right-hand side between zero and the Gibbs energy of the product, G. The integral will become: ∫G0dG=∫n0μ∗dn, where μ∗ indicates the chemical potential of a pure substance, which is independent on the number of moles by definition. As such, eq. (9.17) becomes: ∫G0dG=μ∗∫n0dn→G=μ∗n→μ∗=Gn, which gives a straightforward interpretation of the chemical potential of a pure substance as the molar Gibbs energy. We can start from eq. (9.14) and write for a pure substance that is brought from Pi=P−⊖− to Pf=P at constant T: G−G−⊖−=nRTlnPP−⊖−, dividing both sides by n, we obtain: Gn−G−⊖−n=RTlnPP−⊖−, which, for a pure substance at P−⊖−=1bar, becomes: μ∗−μ−⊖−=RTlnP. Notice that, while we use the pressure of the gas inside the logarithm in eq. (9.21), the quantity is formally divided by the standard pressure P−⊖−=1bar, and therefore it is a dimensionless quantity, as it should be. For simplicity of notation, however, we will omit the division by P−⊖− in the remaining of this textbook, especially wherever it does not create confusion. Let’s now consider a mixture of ideal gases, and let’s try to find out whether the chemical potential of a pure gas inside the mixture, μmixturei, is the same as its chemical potential outside the mixture, μ∗. To do so, we can use eq. (9.21) and replace the pressure P with the partial pressure Pi: μmixturei=μ−⊖−i+RTlnPi, where the partial pressure Pi can be obtained from the simple relation that is known as Dalton’s Law: Pi=yiP, with yi being the concentration of gas i measured as a mole fraction in the gas phase yi=ninTOT<1. Replacing eq. (9.23) into eq. (9.22), we obtain: μmixturei=μ−⊖−i+RTln(yiP)=μ−⊖−i+RTlnP⏟μ∗i+RTlnyi, which then reduces to the following equation: μmixturei=μ∗i+RTlnyi. Analyzing eq. (9.25), we can immediately see that, since yi<1: μmixturei<μ∗i, or, in other words, the chemical potential of a substance in the mixture is always lower than the chemical potential of the pure substance. If we consider a process where we start from two separate pure ideal gases and finish with a mixture of the two, we can calculate the change in Gibbs energy due to the mixing process with: ΔmixingG=∑ni(μmixturei−μ∗i)<0, or, in other words, the process is spontaneous under all circumstances, and pure ideal gases will always mix. 9.5 Chapter Review 9.5.1 Practice Problems Problem 1: Calculating Gibbs Energy Change A chemical reaction occurs at constant temperature and pressure. The enthalpy change of the reaction is −50kJ/mol and the entropy change is −0.15kJ/(mol·K). Calculate the Gibbs energy change at 298K. Is the reaction spontaneous? Solution: We can use Gibbs equation, eq. (9.3), to solve this problem: ΔrxnG=ΔrxnH−TΔrxnS. Given: ΔrxnH=−50kJ/mol, ΔrxnS=−0.15kJ/(mol·K), T=298K. Then: ΔrxnG=−50kJ/mol−(298K)(−0.15kJ/(mol·K)) ΔrxnG=−50kJ/mol+44.7kJ/mol Answer: the Gibbs energy change for the reaction is −5.3kJ/mol. The reaction is spontaneous. Problem 2: Calculating ΔG of Reaction from Standard Free Energies of Formation Calculate ΔrxnG−⊖− at 298K for the reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l) Using the standard Gibbs energies of formation reported in the tables at the end of the book. Solution: We can use the equation: ΔrxnG−⊖−=∑νiΔfG−⊖−i Substituting the values: ΔrxnG−⊖−=[1(−394.4)+2(−237.1)]−[1(−50.7)+2(0)] ΔrxnG−⊖−=−817.9kJ/mol WHich gives us the final answer, after fixing the significant digits. Answer: The change in Gibbs energy for the reported reaction at 298K is −818kJ/mol. 9.5.2 Study Questions Which of the following is the correct definition of the Gibbs energy? G=U−TS. G=H+TS. G=H−TS. G=H−PV+TS. G=H+PV−TS. For a spontaneous process at constant T and P, which of the following is true? ΔG>0. ΔG<0. ΔG=0. ΔG≥0. ΔG is undefined. Which of the following is NOT a natural variable of the Gibbs energy? Temperature. Pressure. Chemical composition. Volume. Number of moles. What is the temperature dependence coefficient of G? (∂G∂T)P,{ni}=S. (∂G∂T)P,{ni}=−S. (∂G∂T)P,{ni}=H. (∂G∂T)P,{ni}=−H. (∂G∂T)P,{ni}=V. What is the pressure dependence coefficient of G? (∂G∂P)T,{ni}=−V. (∂G∂P)T,{ni}=V. (∂G∂P)T,{ni}=S. (∂G∂P)T,{ni}=−S. (∂G∂P)T,{ni}=H. For an ideal gas, how does G change with pressure at constant T? G=G−⊖−−nRTlnPP−⊖−. G=G−⊖−+nRTlnPP−⊖−. G=G−⊖−+nRTPP−⊖−. G=G−⊖−−nRTPP−⊖−. G=G−⊖−+RTlnPP−⊖−. What is the composition dependence coefficient of G? (∂G∂ni)T,P,{nj≠i}=−μi. (∂G∂ni)T,P,{nj≠i}=μi. (∂G∂ni)T,P,{nj≠i}=RTlnμi. (∂G∂ni)T,P,{nj≠i}=RTμi. (∂G∂ni)T,P,{nj≠i}=μiRT. For a pure substance, what is the relationship between chemical potential and molar Gibbs energy? μ∗=Gn2. μ∗=GRT. μ∗=G. μ∗=Gn. μ∗=nG. How does the chemical potential of a substance in an ideal gas mixture compare to its chemical potential as a pure substance? μmixturei>μ∗i. μmixturei<μ∗i. μmixturei=μ∗i. μmixturei=RTlnμ∗i. μmixturei=μ∗iRT. What is the equation for the change in Gibbs energy due to mixing of ideal gases? ΔmixingG=∑ni(μmixturei+μ∗i). ΔmixingG=∑ni(μmixturei−μ∗i)>0. ΔmixingG=∑ni(μmixturei−μ∗i)=0. ΔmixingG=∑ni(μmixturei−μ∗i)<0. ΔmixingG=RT∑nilnμmixturei. Answers: Click to reveal 1.c, 2.b, 3.d, 4.b, 5.b, 6.b, 7.b, 8.d, 9.b, 10.d The majority of chemical reactions in a lab happens at those conditions, and all biological functions happen at those conditions as well.↩︎ It is not uncommon to see values of ΔfG−⊖− tabulated alongside ΔfH−⊖− and S−⊖−i, which simplifies even further the calculation. In fact, a comprehensive list of standard Gibbs energy of formation of inorganic and organic compounds is reported in the appendix of this book 15. For cases where ΔfG−⊖− are not reported, they can always be calculated by their constituents.↩︎ For reactants, the same situation usually applies but in reverse. More complicated cases where the reaction does not consume all reactants are possible, but insignificant for the following treatment.↩︎
17568	https://pmc.ncbi.nlm.nih.gov/articles/PMC2663759/	Auxin and Ethylene Regulate Elongation Responses to Neighbor Proximity Signals Independent of Gibberellin and DELLA Proteins in Arabidopsis - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Plant Physiol . 2009 Apr;149(4):1701–1712. doi: 10.1104/pp.108.133496 Search in PMC Search in PubMed View in NLM Catalog Add to search Auxin and Ethylene Regulate Elongation Responses to Neighbor Proximity Signals Independent of Gibberellin and DELLA Proteins in Arabidopsis1,[C],[W],[OA] Ronald Pierik Ronald Pierik 1 Plant Ecophysiology, Institute of Environmental Biology, Utrecht University, 3584 CA Utrecht, The Netherlands Find articles by Ronald Pierik 1,, Tanja Djakovic-Petrovic Tanja Djakovic-Petrovic 1 Plant Ecophysiology, Institute of Environmental Biology, Utrecht University, 3584 CA Utrecht, The Netherlands Find articles by Tanja Djakovic-Petrovic 1,2,3, Diederik H Keuskamp Diederik H Keuskamp 1 Plant Ecophysiology, Institute of Environmental Biology, Utrecht University, 3584 CA Utrecht, The Netherlands Find articles by Diederik H Keuskamp 1,2, Mieke de Wit Mieke de Wit 1 Plant Ecophysiology, Institute of Environmental Biology, Utrecht University, 3584 CA Utrecht, The Netherlands Find articles by Mieke de Wit 1, Laurentius ACJ Voesenek Laurentius ACJ Voesenek 1 Plant Ecophysiology, Institute of Environmental Biology, Utrecht University, 3584 CA Utrecht, The Netherlands Find articles by Laurentius ACJ Voesenek 1 Author information Article notes Copyright and License information 1 Plant Ecophysiology, Institute of Environmental Biology, Utrecht University, 3584 CA Utrecht, The Netherlands Corresponding author; e-mail r.pierik@uu.nl. 2 These authors contributed equally to the article. 3 Present address: Seminis, Westeinde 161, 1601 BM Enkhuizen, The Netherlands. Received 2008 Dec 4; Accepted 2009 Feb 5. Copyright © 2009, American Society of Plant Biologists PMC Copyright notice PMCID: PMC2663759 PMID: 19211699 See commentary "Light quality controls shoot elongation through regulation of multiple hormones" in Plant Signal Behav, volume 4 on page 755. Abstract Plants modify growth in response to the proximity of neighbors. Among these growth adjustments are shade avoidance responses, such as enhanced elongation of stems and petioles, that help plants to reach the light and outgrow their competitors. Neighbor detection occurs through photoreceptor-mediated detection of light spectral changes (i.e. reduced red:far-red ratio [R:FR] and reduced blue light intensity). We recently showed that physiological regulation of these responses occurs through light-mediated degradation of nuclear, growth-inhibiting DELLA proteins, but this appeared to be only part of the full mechanism. Here, we present how two hormones, auxin and ethylene, coregulate DELLAs but regulate shade avoidance responses through DELLA-independent mechanisms in Arabidopsis (Arabidopsis thaliana). Auxin appears to be required for both seedling and mature plant shoot elongation responses to low blue light and low R:FR, respectively. Auxin action is increased upon exposure to low R:FR and low blue light, and auxin inhibition abolishes the elongation responses to these light cues. Ethylene action is increased during the mature plant response to low R:FR, and this growth response is abolished by ethylene insensitivity. However, ethylene is also a direct volatile neighbor detection signal that induces strong elongation in seedlings, possibly in an auxin-dependent manner. We propose that this novel ethylene and auxin control of shade avoidance interacts with DELLA abundance but also controls independent targets to regulate adaptive growth responses to surrounding vegetation. Plants respond to competing neighbors in a variety of ways. Among these are an upward leaf movement and enhanced shoot elongation to consolidate light capture in dense stands (Aphalo et al., 1999; Ballaré, 1999; Vandenbussche et al., 2005; Franklin, 2008). These so-called shade avoidance responses can be initiated early on in canopy development upon sensing the reduced red:far-red ratio (R:FR) in the canopy light by the phytochrome family of photoreceptors (Morgan and Smith, 1976; Ballaré et al., 1990; Franklin et al., 2003). Plant neighbor detection also involves blue light (Aphalo et al., 1999; Ballaré, 1999; Vandenbussche et al., 2005), which, like red light, is strongly reduced in well-developed canopies as a result of absorption by chlorophyll. When applied individually, both light signals can induce functional shade avoidance responses, such as hypocotyl, stem, and petiole elongation and upward leaf movement (Ballaré et al., 1991; Casal and Sánchez, 1994; Pierik et al., 2004b; Franklin and Whitelam, 2005; Franklin, 2008). Many downstream signal transduction components involving several plant hormones operate to induce the growth responses upon detection of canopy signals. A reduction of the R:FR can either sensitize plants to GA (Weller et al., 1994; López-Juez et al., 1995) or enhance the production of bioactive GAs (Beall et al., 1996). Absence of GA or proper GA signaling consequently results in strongly attenuated elongation responses to low R:FR (Reid et al., 1990; López-Juez et al., 1995; Pierik et al., 2004a). Recent work on Arabidopsis (Arabidopsis thaliana) has greatly enhanced our understanding of the mechanisms underpinning this regulation mechanism. Key factors in GA responses are DELLA proteins. The Arabidopsis genome encodes five DELLAs (GA-INSENSITIVE [GAI], REPRESSOR OF GA1-3 [RGA], RGA-LIKE1 [RGL1], RGL2, and RGL3). DELLAs are negative regulators of GA responses such as elongation growth (primarily GAI and RGA; Dill and Sun, 2001; King et al., 2001), are targeted for degradation by GA (Schwechheimer, 2008), and are degraded upon low R:FR detection through phytochromes (Djakovic-Petrovic et al., 2007). This may in part be through enhanced GA biosynthesis, as suggested by the enhanced expression of the GA biosynthesis gene GA 20-OXIDASE (Hisamatsu et al., 2005) but can also follow from more direct interactions between phytochromes and DELLA proteins. Phytochromes act in part through their direct interaction with the bHLH family of phytochrome-interacting factors (PIFs), and one of these (PIF3-LIKE5) has been shown to control the expression of DELLA genes (Oh et al., 2007). Furthermore, DELLA proteins can bind to PIF4, thus preventing PIF4-induced transcriptional regulation of target genes associated with cell elongation (de Lucas et al., 2008). Interestingly, shade avoidance responses can occur normally in the absence of GA if DELLA proteins are not present, such as in DELLA knockouts (Djakovic-Petrovic et al., 2007). Furthermore, DELLA absence alone (e.g. through GA application or DELLA gene knockouts) does not suffice to induce the full shade avoidance response in Arabidopsis. Thus, although GA is required to degrade DELLAs, this is not the only route that is engaged to regulate shade avoidance. Therefore, we are investigating alternative mechanisms that regulate shade avoidance responses. The plant hormone auxin has been suggested to be important for shade avoidance (Morelli and Ruberti, 2000), although rigorous experimental evidence is still limited. Transcript levels of several auxin-related genes are regulated by reduced R:FR, such as various AUX/IAA genes, auxin efflux-associated PIN genes (Devlin et al., 2003), and SMALL AUXIN UPREGULATED15 (SAUR15) and SAUR68 (Roig-Villanova et al., 2007). A newly characterized route for auxin biosynthesis from l-Trp to indole-3-pyruvic acid using a Trp aminotransferase (TAA1; Stepanova et al., 2008; Tao et al., 2008) is also rapidly enhanced upon far-red enrichment in a low-light background (Tao et al., 2008). Furthermore, the auxin-resistant axr1-12 mutant displays an attenuated hypocotyl elongation response to low R:FR (Steindler et al., 1999). In addition, the volatile hormone ethylene has been associated with shade avoidance, both as a primary neighbor detection signal (through atmospheric accumulation) and as a downstream target for photoreceptor signaling (Pierik et al., 2004b, 2007). Therefore, ethylene is another pathway involved in the regulation of shade avoidance responses. Both auxin and ethylene, however, have been suggested to affect DELLA stability, resulting in DELLA regulation by GA, auxin, and ethylene (Achard et al., 2003; Fu and Harberd, 2003). Furthermore, ethylene can also stimulate auxin biosynthesis through the TAA1 route (Stepanova et al., 2008) that is also enhanced during shade treatment, and ethylene can affect auxin responses, as was shown for AUXIN REPONSE FACTOR2 expression during apical hook formation (Li et al., 2004). Ethylene-auxin interactions are also known for root growth control, where ethylene appears to stimulate auxin production and transport, thus controlling root growth (Ruzicka et al., 2007; Stepanova et al., 2007; Swarup et al., 2007). Rigorous studies are now required to shed light on the roles and interactions of these two hormones during the control of shade avoidance responses to neighbor-derived light signals. Here, we investigated how interactions between auxin, ethylene, and DELLA proteins regulate shade avoidance responses induced by reduced R:FR and reduced blue light photon fluence rates. We show that ethylene and auxin are important regulators of shade avoidance in Arabidopsis, where ethylene at least partly acts through auxin action. This pathway affects DELLA abundance, but this interaction appears to have only limited functionality during shade avoidance. We conclude that the ethylene-auxin pathway is an obligatory signaling route that is functionally parallel to the earlier identified GA-DELLA signaling system controlling shade avoidance responses. RESULTS Low R:FR-Induced Petiole Elongation Depends on Phytochrome B and Low Blue Light-Induced Hypocotyl Elongation Depends on Cryptochromes The R:FR and the blue light photon fluence rate are reduced in plant canopies, and individually both signals can induce shade avoidance responses. Reduced R:FR led to a fast increase of petiole elongation. This response was abolished in the phyB mutant (Fig. 1), confirming that R:FR-mediated shade avoidance occurs primarily through PhyB signaling. Arabidopsis petioles respond only weakly to reduced blue light photon fluence rates (data not shown), but hypocotyls of light-grown seedlings show a very dramatic elongation response to this signal. As shade avoidance responses to reduced levels of blue light have hardly been studied in Arabidopsis, we first tested which photoreceptors contribute to this response. Our data (Fig. 1) indicate that the two cryptochromes CRY1 and CRY2 are redundantly involved in the induction of hypocotyl elongation upon low blue treatment. Single mutants for these two photoreceptors displayed no severe reduction of the response, whereas the double mutant cry1 cry2 was essentially unresponsive to a reduction of blue light. Even double phototropin mutants were hardly disturbed for this response, suggesting that this family of blue light photoreceptors does not play an important role in the hypocotyl elongation response studied here. Phytochromes also do not seem to affect the cryptochrome-mediated responses to low blue light, as even the quadruple phytochrome mutant phya phyb phyd phye and the chromophore mutant hy2 displayed clear low blue light responses, despite constitutively elongated phenotypes. Figure 1. Open in a new tab Photoreceptor involvement in shade avoidance responses of petioles to low R:FR (A) and hypocotyl responses to low blue light photon fluence rates (B). Data are means ± se (n = 10 for petioles, n = 30–50 for hypocotyls). Different letters indicate significant differences. Low R:FR and Low Blue Light Signaling Result in Enhanced Auxin Activity In order to establish the involvement of auxin in growth responses to low R:FR and low blue light, auxin action was visualized in pIAA19∷GUS-expressing lines (Tatematsu et al., 2004) and quantified with quantitative reverse transcription (qRT)-PCR for this IAA19 auxin reporter gene. These data confirm that low R:FR leads to increased auxin action in petioles (almost 3-fold increase of IAA19 expression quantified with qRT-PCR), especially in the more lateral tissues (Fig. 2). Low blue light gave only a slight increase in auxin activity in petioles (data not shown), which is consistent with the very weak petiole elongation response upon this light cue (data not shown). However, low blue light led to strongly enhanced GUS staining of pIAA19∷GUS, corresponding to a 5-fold increase of this gene compared with control light measured with qRT-PCR in hypocotyls, where also the elongation response was very pronounced. Similar data were obtained with another auxin reporter, DR5∷GFP (data not shown). Low R:FR, which can also stimulate hypocotyl elongation, also led to enhanced pIAA19∷GUS activity in hypocotyls (data not shown). Taken together, these data confirm the occurrence of increased auxin action during shade avoidance, and our next step was to establish whether this is functionally relevant to shade avoidance. Figure 2. Open in a new tab Auxin involvement in shade avoidance responses of petioles to low R:FR and hypocotyl responses to low blue light photon fluence rates. A to E, Petioles in control and low R:FR light. F to I, Hypocotyls in control and low blue light. A to C and F to H, Auxin activity, shown with the auxin-responsive pIAA19∷GUS reporter (A, B, F, and G) and IAA19 gene expression (C and H), is increased upon low R:FR and low blue exposure. D, E, and I, Undisturbed auxin signaling and transport are required for shade avoidance responses to low R:FR (D and E) and low blue light photon fluence rates (I). Data are means ± se (n = 8–12 for petioles, n = 30–50 for hypocotyls). Different letters indicate significant differences (P< 0.05). [See online article for color version of this figure.] The petiole elongation response to low R:FR was inhibited by treatment with the auxin transport inhibitor naphthylphthalamic acid (NPA; Fig. 2). In accordance with this, low blue light-induced hypocotyl elongation was much reduced upon NPA treatment as well (Fig. 2). As could be expected, NPA led to a reduction of the increased pIAA19∷GUS activity of low blue light-exposed hypocotyls and restricted the GUS staining in low blue light to a faint staining in the central cylinder of the upper 30% of the hypocotyl (Supplemental Fig. S1), consistent with the fact that NPA disturbs auxin transport (Petrasek et al., 2003). Auxin involvement was further suggested by the lack of low R:FR-induced petiole elongation in the axr1-12 and axr2-1 auxin signaling mutants (Fig. 2). Likewise, these mutants displayed a much reduced hypocotyl elongation response to low blue light (Fig. 2). We also tested the iaa19/msg2-3 mutant, which showed a somewhat reduced elongation response to low blue light and low R:FR (data not shown), but this effect was much less severe than shown for axr1-12 and axr2-1. Hypocotyl lengths for axr1-12 and axr2-1 under control light conditions were not notably different from those for wild-type accession Columbia (Col-0). This is in agreement with some other reports (Steindler et al., 1999) showing similar hypocotyl lengths for axr1-12 and Col-0 (around 1 mm) but in contrast to others. For example, Collett et al. (2000) report a somewhat reduced hypocotyl length for axr1-12 (measured under low-light conditions, which probably induce low blue light-mediated shade avoidance) and a more strongly reduced length for axr1-3 in higher light. Timpte et al. (1994) report reduced hypocotyl length for axr2-1 relative to Col-0 under continuous light. In those two studies, Col-0 hypocotyls were much longer than those under control light conditions in our experiments, whereas mutant hypocotyl lengths were very similar to what we found. Tentatively, the very different light conditions (both photoperiod and intensity), in combination with the addition of sugars to their medium (which were not added in these experiments), may explain why the constitutive lengths differ between our experiments and those by Collett et al. (2000) and Timpte et al. (1994). We conclude that auxin action is enhanced during light-mediated shade avoidance responses in petioles and hypocotyls, particularly in the more lateral regions of these organs. Auxin action appears to be important for petiole and hypocotyl elongation in response to low R:FR and reduced blue light fluence rates, as these responses are diminished when auxin transport or signaling is disrupted. Auxin-Mediated Shade Avoidance Is DELLA Independent As auxin can affect GA action by reducing DELLA stability (Fu and Harberd, 2003), we tested if auxin involvement in shade avoidance is through interaction with DELLA proteins. We showed recently that the DELLA protein RGA is degraded upon exposure to low R:FR in petioles and low blue light in hypocotyls (Djakovic-Petrovic et al., 2007). Figure 3 shows that inhibition of auxin transport by NPA leads to higher abundance of this protein (visualized through confocal microscopy of the GFP-RGA fusion protein) in low R:FR-treated petioles as compared with non-NPA-treated plants. These data indicate that auxin can interact with GA signaling during shade avoidance and suggest that auxin facilitates the degradation of DELLA proteins, subsequently allowing growth. We tested this hypothesis by studying whether low R:FR-induced petiole elongation is still inhibited by NPA in the quadruple DELLA knockout mutant rga24 gait6 rgl1 rgl2, where DELLA accumulation can obviously not occur. Figure 3E shows that NPA strongly inhibited low R:FR-induced petiole elongation in the wild type. However, the quadruple DELLA knockout mutant also showed a complete inhibition of shade avoidance by NPA. These data suggest that auxin acts functionally independent of DELLA proteins. In accordance with this, GA addition to the auxin-resistant mutant axr2-1 could not rescue low R:FR-induced petiole elongation of this mutant (Fig. 3F). Very similar interactions were found for low blue light-induced hypocotyl elongation (Fig. 4), where NPA partly prevented low blue light-induced degradation of the DELLA protein RGA (Fig. 4, compare B and D). However, NPA very strongly inhibited low blue light-induced hypocotyl elongation in the quadruple DELLA knockout, similar to what was observed for low R:FR-induced petiole elongation. Interestingly, and different from petioles, NPA reduced hypocotyl length in control light, but this was not observed in plants with low DELLA abundance: quadruple DELLA knockouts and GA-treated seedlings (Fig. 4G). As shown previously, enhanced DELLA stability in the gai gain-of-function mutant inhibits shade avoidance, and the reduction of hypocotyl length was even more severe in the presence of NPA (Fig. 4G). In conclusion, auxin-mediated shade avoidance likely represents an alternative route toward shade avoidance, functionally parallel to the earlier described GA-DELLA route. Figure 3. Open in a new tab Interactions between auxin action and DELLA abundance during low R:FR-induced petiole elongation. A to D, The pRGA∷GFP:RGA reporter shows the abundance of the DELLA protein RGA (green speckles on a background of red-fluorescing chloroplasts). Note that RGA abundance is enhanced upon auxin transport inhibition with the NPA. E, Petiole elongation responses to low R:FR are inhibited by NPA treatment, also in the quadruple DELLA knockout mutant. F, GA addition cannot rescue the lack of shade avoidance in auxin-resistant mutants. Data are means ± se (n = 8–12). Different letters indicate significant differences. Figure 4. Open in a new tab Auxin-DELLA interaction during low blue light-induced hypocotyl elongation. A to F, Inhibition of auxin transport (NPA) prevents low blue light-induced DELLA degradation, and this is overcome by the addition of GA, as evidenced by the pRGA∷GFP:RGA reporter. G, NPA effects under standard and low blue light conditions on hypocotyl length in the quadruple DELLA knockout mutant and the DELLA gain-of-function mutant gai. Data are means ± se (n = 30–50). Different letters indicate significant differences. Ethylene-Induced Shade Avoidance Does Not Act through DELLA Regulation The gaseous hormone ethylene is known to play an important role in the regulation of shade avoidance in addition to auxin (Pierik et al., 2004b). Furthermore, interactions between ethylene and GA, and more specifically DELLA abundance, and between ethylene and auxin are well known. Low R:FR treatment gave the classic stimulation of ethylene production (Fig. 5A) and a consistent up-regulation of ERS2 expression (data not shown), an ethylene receptor-encoding gene that is a marker for ethylene signaling (Millenaar et al., 2005). Low R:FR-induced petiole elongation was absent in the ethylene-insensitive mutants ein2-1 and ein3-1 eil1-3 (ein3 ein3-like1) and in plants pretreated with the ethylene action inhibitor 1-methylcyclopropane (1-MCP; Fig. 5, B and C), confirming that ethylene is a positive regulator of shade avoidance. Low R:FR-induced petiole elongation was also inhibited by 1-MCP in quadruple DELLA knockout plants, suggesting that ethylene-mediated petiole elongation in low R:FR does not require DELLA proteins (Fig. 5D). Notably, the inhibitory effect of 1-MCP on low R:FR-induced petiole elongation was stronger in the Col-0 accession as compared with Landsberg erecta (L er; Fig. 5), suggesting that there could be genetic variation for ethylene involvement in these responses. Despite the importance of endogenous ethylene signaling for low R:FR-induced petiole elongation, exogenous ethylene application has only little effect on petiole elongation in Arabidopsis (data not shown; Millenaar et al., 2005). Hypocotyl elongation of light-grown seedlings, on the other hand, is stimulated by ethylene (Smalle et al., 1997). Consistently, application of the ethylene precursor 1-aminocyclopropane-1-carboxylic acid (ACC), known to yield high ethylene levels, also stimulates hypocotyl length (Fig. 6B, compare control Col-0 with control Col-0 + ACC). Low blue light exposure, however, did not affect ethylene production (Fig. 6A), nor did it affect the expression of the ethylene marker gene ERS2 (data not shown). Accordingly, low blue light-induced hypocotyl elongation is only partly dependent upon endogenous ethylene levels, as evidenced by the reduced but still very clearly present response of ethylene-insensitive mutants (Fig. 6C). An additional experiment on the hypocotyl elongation response to low R:FR also indicated no involvement of ethylene (data not shown), suggesting that ethylene involvement in light-mediated elongation growth is more prominent in petioles than in hypocotyls. Figure 5. Open in a new tab Ethylene involvement in low R:FR-induced petiole elongation. A, Low R:FR stimulates ethylene production. FW, Fresh weight. B and C, Petiole elongation under standard and low R:FR light conditions in ethylene-insensitive mutants (B) or upon exposure to the ethylene action inhibitor 1-MCP (C). D, The effect of inhibition of ethylene action does not depend on DELLAs. Data are means ± se (n = 8–12). Different letters indicate significant differences. Figure 6. Open in a new tab Ethylene regulation of hypocotyl elongation in control and low blue light. A, Ethylene evolution from control light-exposed and low blue light-exposed seedlings is similar. FW, Fresh weight. B, Exogenous application of the ethylene precursor ACC stimulates hypocotyl elongation. C, Ethylene-insensitive mutants have a slightly reduced hypocotyl elongation response to low blue light conditions. Data are means ± se (n = 3 for A, n = 30–50 for B and C). Different letters indicate significant differences. Ethylene, however, is not only a putative downstream component in light-mediated shade avoidance but can also act as a primary volatile neighbor detection signal that can induce shade avoidance even in the absence of light quality changes (Pierik et al., 2004b). As this may occur through an interaction with GA, we tested whether ACC application affects DELLA abundance in light-grown hypocotyls. Figure 7 shows that under control light conditions a small (30%) ACC-induced stimulation of the GFP-RGA signal was found, whereas ACC under these light conditions gave a pronounced stimulation of hypocotyl length. ACC, on the other hand, gave a pronounced (more than 10-fold) increase of RGA-GFP abundance under low blue light conditions compared with non-ACC-exposed seedlings under low blue light (Fig. 7, compare B and D). However, under these low blue light conditions, no effect of ACC on the hypocotyl length of L er seedlings was found (Fig. 7E), whereas in Col-0 there was even a slight stimulation under these conditions (Fig. 6B). To further understand if interactions of ethylene with DELLAs are relevant to the effects of ethylene on hypocotyl length, a range of ACC concentrations were tested on wild-type L er as well as on the quadruple DELLA knockout in which DELLA stabilization by ACC cannot occur. These knockouts appeared to respond similarly to wild-type L er, although DELLA knockout hypocotyls were constitutively elongated due to the absence of the DELLA growth repressors. Further evidence for the hypothesis that ethylene-induced hypocotyl elongation does not act through DELLA regulation came from experiments on the severe GA-deficient ga1-3 mutant and the DELLA gain-of-function mutant gai that expresses a mutant DELLA protein, gai, that is irresponsive to GA. Although ga1-3 is constitutively dwarfed, it still showed an approximately 60% stimulation of hypocotyl length by ACC, which is only slightly less than the 80% increase in wild-type L er under control light conditions (Fig. 7E). The gai mutant showed no disturbance of the ethylene response at all under control light conditions and under low blue light conditions; this mutant even gained a response where there was none in L er. These data together indicate that stimulation of hypocotyl elongation by ethylene can occur independently of GA in Arabidopsis, despite the finding that ACC seems to enhance DELLA abundance under low blue light photon fluence rates. Figure 7. Open in a new tab Ethylene-DELLA interactions during hypocotyl elongation. A to D, The pRGA∷GFP:RGA reporter shows the abundance of the DELLA protein RGA (green speckles on a background of red-fluorescing chloroplasts). Note that RGA abundance is enhanced by the ethylene precursor ACC under low blue light conditions. E and F, ACC dose-response curves in wild-type L er (E) and the quadruple DELLA knockout mutant (F) under normal and low blue light conditions. G, ACC-induced stimulation of hypocotyl length does not depend on GA, as evidenced by the GA-deficient ga1-3 and the GA-insensitive gai (DELLA gain-of-function) mutants. Data are means ± se (n = 30–50). Different letters indicate significant differences. Ethylene-Induced Hypocotyl Elongation Is Reduced in Two Auxin-Resistant Mutants The final step to be elucidated is if auxin and ethylene regulate hypocotyl elongation separately or if these two hormones are in fact part of one pathway. To address this question, we studied the importance of ethylene for auxin-induced hypocotyl elongation and vice versa. To this end, ACC was applied to the auxin-resistant axr1-12 and axr2-1 mutants and indole-3-acetic acid (IAA; a plant-produced auxin) was applied to the ethylene-insensitive ein2-1 and double ein3-1 eil1-3 mutants under control light conditions (Fig. 8). We found that axr1-12 still showed a weak response, albeit much reduced compared with Col-0, to ACC, but this mutant also appeared to still respond weakly to auxin. The more severely auxin-resistant axr2-1 mutant, on the other hand, showed virtually no response to ACC, suggesting that intact auxin signaling could be required for ethylene to stimulate hypocotyl elongation. The ein2-1 and ein3-1 eil1-3 mutants were entirely unresponsive to ACC, confirming their ethylene insensitivity, but still responded properly to exogenous auxin. These data suggest that auxin signaling can be required for ethylene effects on hypocotyl elongation. It thus appears that auxin controls hypocotyl elongation in a pathway that is parallel to GA and that ethylene might stimulate elongation growth through this auxin pathway. Figure 8. Open in a new tab Hypocotyl elongation responses to applied auxin (IAA) or ethylene (applied as the ethylene precursor ACC) of wild-type (Col-0) plants and auxin-resistant (axr1-12 and axr2-1) and ethylene-insensitive (ein2-1 and ein3-1 eil3-1) mutants. Seedlings were in control light conditions. Data are means ± se (n = 30–50). Different letters indicate significant differences. CT, Control. DISCUSSION Reaching out for light is essential to plant survival in dense stands. Shade avoidance responses are induced upon neighbor detection through various signals, among which are a reduced R:FR and a low blue light photon fluence rate. It is well known that low R:FR is primarily signaled by phytochrome B (Franklin, 2008). Here, we show that hypocotyl elongation in response to reduced blue light is mediated by the blue light receptors cryptochromes 1 and 2 (Fig. 1B). This indicates that cryptochrome photoreceptors are involved in plant neighbor detection in addition to phytochromes. Although much is known about photoreceptor signaling itself, little is known about how these signals are translated into an adaptive growth response. We showed recently that GA regulation of shade avoidance acts through DELLA proteins. It was shown that DELLA degradation is essential to allow for shade avoidance responses in hypocotyls and petioles in response to low blue light and low R:FR, respectively (Djakovic-Petrovic et al., 2007). However, it was also noted that DELLA degradation alone is not sufficient to induce these responses, as multiple DELLA knockouts showed normal shade avoidance responses in both the presence and absence of GA. This suggests that additional signal transduction pathways have to be engaged to induce light-mediated shade avoidance responses in Arabidopsis. Here, we investigated if auxin and ethylene could be involved in those alternative pathways, as these hormones have been implicated in shade avoidance previously (Morelli and Ruberti, 2000; Pierik et al., 2004b). Using an auxin-responsive promoter-GUS fusion (pIAA19∷GUS) reporter and qRT-PCR for this auxin-responsive IAA19 gene, we show that low R:FR and low blue light conditions lead to enhanced auxin action in Arabidopsis petioles and hypocotyls (Fig. 2). This is in agreement with a model for auxin action in shade avoidance that was posed a number of years ago, where enhanced lateral auxin distribution in stems or hypocotyls was suggested to regulate cell elongation during shade avoidance (Morelli and Ruberti, 2000). The patterns we found for this auxin action reporter are not only consistent with the predicted auxin distribution pattern but are also disrupted when the auxin transport inhibitor NPA is applied (Supplemental Fig. S1). The enhanced auxin action, therefore, is likely brought about by regulated auxin transport toward the lateral regions of the hypocotyl and petiole. In addition, auxin biosynthesis is also likely to be enhanced under these shade-avoiding conditions, as was recently shown for Arabidopsis seedlings (Tao et al., 2008). As this additional auxin would also be transported toward the more lateral regions of the elongating petioles and hypocotyls, this would further contribute to the observed auxin-reporter staining patterns. Next, we tested the importance of auxin for shade avoidance responses. The disrupted auxin transport caused by NPA is associated with strongly reduced shade avoidance responses in both hypocotyls and petioles. Furthermore, genetic evidence confirms that auxin signaling is important for shade avoidance, as both of the auxin-resistant mutants, axr1-12 and axr2-1, show much reduced responses. This appears to apply to elongation responses in both petioles and hypocotyls to low R:FR and low blue light, respectively (Fig. 2). In addition, stimulation of hypocotyl elongation upon low R:FR has also been shown to be impaired in the axr1-12 mutant (Steindler et al., 1999). Therefore, we conclude that enhanced auxin action, indicated by the pIAA19∷GUS reporter, is required for shade avoidance, which had been suggested before (Morelli and Ruberti, 2000) but for which little causal evidence existed so far. Auxin is well known to affect GA biosynthesis (Ross et al., 2000) and DELLA protein stability (Fu and Harberd, 2003). Therefore, we investigated whether auxin accumulation during shade avoidance requires GA signaling. First, we confirmed that auxin affects GA signaling by studying DELLA protein abundance in NPA-treated and non-NPA-treated plants (Figs. 3 and 4). NPA inhibits shade avoidance to low blue light and low R:FR in hypocotyls and petioles, respectively; accordingly, it leads to enhanced DELLA abundance under these light conditions. This would be an indication that NPA-induced inhibition of shade avoidance may be related to an enhanced abundance of growth-inhibiting DELLA proteins under these conditions. This would be in agreement with the earlier observed stabilizing effects of auxin transport inhibition on DELLA proteins in Arabidopsis roots (Fu and Harberd, 2003). These data on Arabidopsis roots have led to the idea that DELLA protein stability can be affected by several signals, among which is auxin, and that these proteins can thus be seen as a molecular mechanism for cross talk. As enhanced DELLA stability (e.g. in the gai mutant) leads to reduced shade avoidance (Djakovic-Petrovic et al., 2007; Fig. 4G), it would have seemed a likely option that NPA-mediated DELLA stabilization would explain the reduced shade avoidance upon NPA treatment. However, although we show here that this auxin-DELLA cross talk may occur during shade avoidance, we also show that this is not fundamental for shade avoidance to occur. This is most clearly indicated by the novel finding that low R:FR-induced elongation is abolished by NPA treatment in the quadruple DELLA knockout gait6 rga24 rgl1 rgl2 to the same extent as in wild-type plants (Fig. 4E). In other words, DELLA proteins are most likely not essential for the reduction of shade avoidance during auxin inhibition. Although DELLA proteins are more abundant during NPA treatment, this does not explain the inhibition of shade avoidance under these conditions, as the same growth inhibition occurs when these DELLAs are not present (Figs. 3 and 4). These data indicate that the shade avoidance response mediated by auxin does not require GA signaling but rather constitutes a separate hormonal pathway regulating shade avoidance. This is further confirmed by the fact that the addition of GA does not rescue shade avoidance responses in the auxin-resistant axr2-1 mutant. The volatile hormone ethylene can also be a player in shade avoidance, both as a hormone required to regulate petiole elongation responses to low R:FR and as a direct neighbor detection signal (Pierik et al., 2004b). We demonstrate that ethylene production increases upon low R:FR signaling. Furthermore, we show that an intact ethylene signaling pathway is required for low R:FR-induced petiole elongation in Arabidopsis (Fig. 5). Unlike ethylene involvement in petiole elongation, low blue light-induced hypocotyl elongation in Arabidopsis appeared not to rely heavily on intact ethylene signaling. However, ethylene application to light-grown Arabidopsis seedlings does induce strong hypocotyl elongation under control light conditions (Figs. 6 and 7, E–G; Smalle et al., 1997; Vandenbussche et al., 2003; Pierik et al., 2006). We show that this hypocotyl elongation response to ethylene does not occur in the axr2-1 mutant, which is also not responsive to IAA (Fig. 8). The axr1-12 mutant, however, does still show a weak but significant response to IAA and accordingly also shows a weak response to ACC. In line with these observations, ACC-induced hypocotyl elongation is also abolished upon treatment with the polar auxin transport inhibitor NPA (Vandenbussche et al., 2003). Therefore, we hypothesize that auxin may be a downstream regulator of ethylene-induced hypocotyl elongation. This would be consistent with the fact that in order to control root growth, ethylene also acts through auxin by enhancing auxin production and transport in roots (Ruzicka et al., 2007; Stepanova et al., 2007; Swarup et al., 2007). Similar to what was shown for auxin, the interaction of ethylene with GA and DELLA proteins does not seem to explain the involvement of ethylene in shoot elongation growth. In fact, ethylene-mediated hypocotyl elongation in Arabidopsis seems to be independent of GA altogether, since the GA-insensitive gai1 and GA-deficient ga1-3 mutants both retain a substantial hypocotyl elongation response to ACC (Fig. 7E). This is consistent with findings by De Grauwe et al. (2007), who also showed that gai is still ACC responsive. Those authors also show that GA-induced hypocotyl elongation does not require ethylene, despite the fact that their transcript-profiling experiments suggested that some interactions were present (De Grauwe et al., 2007). The lack of GA involvement in ethylene-induced hypocotyl elongation in Arabidopsis, however, is not general for all species. For example, tobacco (Nicotiana tabacum) stem elongation to low blue light photon fluence rates requires ethylene, which in turn can stimulate stem elongation only if sufficient GA is present (Pierik et al., 2004a). Furthermore, ethylene-induced elongation of internodes in rice (Oryza sativa) and petioles in Rumex palustris, two flooding-tolerant species, is entirely diminished by inhibition of GA production (for review, see Bailey-Serres and Voesenek, 2008; Jackson, 2008). Ethylene itself can already induce very different, and sometimes opposite, growth responses in different species (Pierik et al., 2006), and even simple responses, like stimulation of elongation growth, can be regulated in different ways among different plant species. In summary, we propose that enhanced lateral distribution of auxin activity in elongating shoot organs constitutes an essential regulatory mechanism to adaptively modulate elongation growth upon light-mediated neighbor detection. The volatile hormone ethylene may exert its effects in the shade avoidance response by acting through the auxin pathway in Arabidopsis. While this novel route can affect the stability of the nuclear growth-suppressing DELLA proteins, ethylene- and auxin-mediated regulation of shade avoidance appears to act predominantly through DELLA-independent mechanisms. MATERIALS AND METHODS Plant Growth For experiments on petioles of full-grown plants, Arabidopsis (Arabidopsis thaliana) plants were grown essentially as described (Millenaar et al., 2005; Djakovic-Petrovic et al., 2007). In short, seeds were put on moist filter paper, stratified at 4°C in the dark, germinated in 200 μ mol m−2 s−1 photosynthetically active radiation (PAR; 9 h of light, 15 h of dark) for 4 d, transferred to pots after germination, and put at 200 μ mol m−2 s−1 PAR (9 h of light, 15 h of dark, 21°C, 70% relative humidity). The petioles of the third youngest leaves of plants at 36 to 38 d after sowing were used for experiments and measured at the start of the experiment (time 0) and after 24 h of treatment. For hypocotyl experiments, seeds were surface sterilized in hypoclorite (0.4%) for 10 min and rinsed three times with ethanol and then two times with sterile demineralized water. Seeds were then transferred to sterile low-nutrient (0.4% Murashige and Skoog medium) agar (0.8%, w/v) plates and stratified for 4 d in the dark (4°C). Thereafter, plates were placed in the light for 2 h and then kept in the dark for 24 h to synchronize germination. After this period, the seeds were placed under standard light conditions (described in the next section) or in light conditions with the same total photon fluence rate but depleted in the blue light region. Seedlings were allowed to grow for 7 d in the low blue light treatment before photographs were taken through a stereo microscope. From these photographs, hypocotyl lengths were determined digitally with ImageJ software ( Involvement of auxin in shade avoidance responses was tested using the auxin-resistant axr1-12 (Lincoln et al., 1990) and axr2-1 (Wilson et al., 1990) gain-of-function mutants and the pIAA19∷GUS auxin reporter (Tatematsu et al., 2004). Ethylene involvement was tested using the ethylene-insensitive mutants ein2-1 (Guzman and Ecker, 1990) and ein3-1 eil1-3 (Alonso et al., 2003). As these mutants are in a Col-0 background, this accession served as the wild-type control. Interactions between auxin and GA signaling were tested with a quadruple DELLA knockout mutant (rga24 gait6 rgl1-1 rgl2-1; Achard et al., 2007), the GA-insensitive gai gain-of-function mutant (Koornneef et al., 1985), and the pRGA∷GFP:RGA reporter, all in the L er background and with L er as the wild-type control. The involvement of photoreceptors in the induction of shade avoidance was tested with the following photoreceptor mutants (background in parentheses): cry1-304 (Col-0), cry2-1 (Col-0; Guo et al., 1998); cry1 cry2 (= hy4-2 fha-1; L er), hy2-1 (L er), phyb-1 (L er; Koornneef et al., 1980); phya-201 phyb-1 phyd-1 phye-1 (L er; Franklin et al., 2003); phot1-101 (L er; Liscum and Briggs, 1995); phot2-5 (Wassilewskija [Ws]), phot1-101 phot2-5 (Ws/L er; Sakai et al., 2001); and phot1-101 phot2-5 cry1 cry2 (Ws/L er; Ohgishi et al., 2004). Light Treatments Control light conditions were obtained by filtering standard growth chamber light (Philips HPI 400 W + Philips Halogen 150 W) through spectrally neutral shading cloth, achieving a total light intensity of 147 μ mol m−2 s−1 PAR (400–700 nm), which contained 25 μ mol m−2 s−1 blue light (400–500 nm) and had a R:FR (655–665 nm:725–735 nm) of 1.1. Low blue light conditions were obtained by filtering the standard growth chamber light through a double layer of blue light-absorbing filter paper (Medium Yellow 010; Lee Filters), yielding 0.7 μ mol m−2 s−1 blue light, R:FR of 1.1, and 147 μ mol m−2 s−1 PAR. The R:FR was lowered in the low R:FR treatment by adding far-red light (730 nm far-red light-emitting diodes; Shinkoh Electronics) to a control light background. As a result, R:FR was lowered to 0.28 in the low R:FR treatment, whereas PAR was 140 μ mol m−2 s−1 and blue light photon fluence rate was 24 μ mol m−2 s−1. Full spectra are available in Supplemental Figure S2. Pharmacological Experiments The involvement of auxin was not only investigated genetically but also by the use of the auxin transport inhibitor NPA (Petrasek et al., 2003) and the auxin IAA. NPA was brushed onto the leaves (25 μ m NPA, 0.1% ethanol, and 0.1% Tween) or added to the agar nutrient medium (25 μ m). NPA concentrations were based on a dose-response curve for NPA in control and low blue light (Supplemental Fig. S3). For petioles of mature plants, this concentration was substantially lower than what has been used in other species (Cox et al., 2004). Controls received similar amounts and concentrations of dissolvent without NPA. IAA treatments occurred in a similar way, but with a concentration of 10 μ m (this is the lowest concentration to saturate the hypocotyl elongation response to IAA in light-grown seedlings without having negative effects; Vandenbussche et al., 2003), and a similar 10 μ m concentration was used for GA 3 (Djakovic-Petrovic et al., 2007). The ethylene precursor ACC was tested at a range of concentrations (0, 0.01, 0.1, 1, 10, and 20 μ m) and appeared to give an almost saturating effect at 1 μ m, but its effect continued to increase, particularly in the quadruple DELLA knockout mutant, until 20 μ m (Fig. 7, F and G). Therefore, in the other experiments, ACC was applied at a final concentration of 20 μ m, a representative ACC concentration for studies on hypocotyl elongation that gives saturated responses without noticeable specific side effects (Smalle et al., 1997; Vandenbussche et al., 2003). Ethylene perception was inhibited with 1-MCP gas (Sisler and Serek, 2003). 1-MCP was applied 3 h prior to the start of light treatment at a final concentration of 10 μ L L−1, obtained from SmartFresh powder (Rohm and Haas). Prior to light treatment, the plants were incubated for 3 h with this concentration of 1-MCP, and this rendered the plants insensitive to ethylene during the experiment in a standard growth room environment (Sisler and Serek, 2003; Millenaar et al., 2005). Ethylene Emission The effect of light quality on ethylene production in mature plants (low R:FR) and seedlings (low blue light) was determined. Ethylene measurements were made in triplicate for the same treatment duration as for all growth and molecular reporter studies (i.e. 1 d of low R:FR treatment for mature plants and 7 d of low blue light exposure for seedlings). Measurements at earlier time points gave very similar data as those obtained from these final time points (data not shown). For mature plants, a 300-mg sample of shoot material was incubated in a small closed air volume for 20 min. This incubation time was found to be long enough for ethylene to accumulate to detectable levels but short enough to prevent wounding-derived ethylene production. Then, 1 mL of air sample was analyzed for ethylene with a gas chromatograph that was equipped with a Photo Ionization Detector (Syntech Spectras Analyzer GC955-100; Synspec). From these values, ethylene production was calculated in pmol g−1 fresh weight h−1. Ethylene release from seedlings was measured by growing 35 seedlings in a 10-mL cap flask that was filled with 5 mL of agar-containing (0.8%, v/v) low-nutrient growth medium (0.4% Murashige and Skoog medium). After 6 d, the cap flask was closed and ethylene was allowed to accumulate. After 24 h, the head space was sampled and analyzed for ethylene as described above for mature plants. GUS Assay In order to visualize auxin action in control and low R:FR-treated petioles and control and low blue light-treated hypocotyls, GUS abundance was studied in transgenic pIAA19∷GUS lines expressing the GUS enzyme driven by the IAA19 promoter (Tatematsu et al., 2004). This has been shown to be a good indicator of auxin action. The GUS assay for seedlings was performed by overnight incubation of freshly harvested material in the staining solution [1 m m 5-bromo-4-chloro-3-indolyl β-d-glucuronide in 100 m m NaPi buffer, pH 7, 0.1 m m EDTA, 0.1% Triton X-100, 1 m m K 4 Fe(CN)6, 1 m m K 3 Fe(CN)6, and 0.52 mg mL−1 dimethyl formamide]. The GUS assay for leaf rosettes was performed after a pretreatment of 20 s in acetone and a fixative treatment (0.3% formaldehyde, 10 m m MES, and 0.3 m mannitol) of 45 min. The rosettes were then washed with 100 m m NaPi (pH 7.0). The histochemical reaction was performed by incubating the rosette for 24 h with 1 m m 5-bromo-4-chloro-3-indolyl β-d-glucuronide in 100 m m NaPi buffer (pH 7.0) with 0.1 m m EDTA. The staining was followed by bleaching with an ethanol series from 50% to 90%, after which the material was photographed. qRT-PCR In order to give an independent quantitative estimation of the auxin-responsive IAA19 gene used in the GUS assay above, we analyzed its expression in low blue light-exposed hypocotyls (3 d of exposure) and low R:FR-exposed petioles (24 h of exposure). To this end, total RNA was extracted from petioles (two petioles from each of five plants were pooled per extraction, with three replicate extractions) and seedlings (up to 100 seedlings per extraction, with three replicate extractions) using the RNeasy Plant Mini Kit (Qiagen), including on-column DNase digestion to eliminate genomic DNA from the samples. RNA transcripts at 1 μ g (hypocotyls) or 3 μ g (petioles) were reverse transcribed to cDNA with the SuperScript III Reverse Transcriptase kit (Invitrogen) and random hexamer primers. qRT-PCR was performed using a Bio-Rad MyiQ single-color detection system on a 20-μ L reaction mix containing 40 ng (hypocotyls) or 30 ng (petioles) of cDNA, 10 μ L of SYBR Green Supermix (Bio-Rad), and gene-specific primers: IAA19-F (At3g15540), 5′-GGCTTGAGATAACGGAGCTG-3′; IAA19-R, 5′-ACCATCTTTCAAGGCCACAC-3′. 18S ribosomal RNA was used as an internal standard to normalize for differences in cDNA concentration between samples: 18S-F, 5′-CGTTGCTCTGATGATTCATGA-3′; 18S-R, 5′-GTTGATAGGGCAGAAATTTGAATGAT-3′. Threshold cycle values were obtained from PCR with an efficiency of approximately 2, and gene expression values were calculated according to Livak and Schmittgen (2001), with control light plants as the final reference with expression levels set at 1. GFP Visualization and Quantification To study DELLA protein abundance, GFP fluorescence was studied in pRGA∷GFP:RGA transgenic plants, as described (Djakovic-Petrovic et al., 2007). Essentially, DELLA-GFP fluorescence was visualized with confocal laser scanning microscopy (40× magnification) using a 488-nm excitation wavelength, a 505- to 530-nm band-path filter to separate GFP, and a 560-nm long-pass filter to determine chlorophyll fluorescence. Z-stacks were made for 149.5 μ m tissue thickness from the basal end of petioles and hypocotyls. It has repeatedly been shown that the RGA-GFP signal shown through confocal imaging shows a good correspondence with the signal being studied through western blotting using an anti-GFP antibody (Achard et al., 2007; Navarro et al., 2008). GFP fluorescence was quantified on at least three replicate images from independent specimens, with a macro developed in house using KS400 (version 3.0) software (Carl Zeiss Vision). Fluorescence values were calculated relative to control light conditions, which were set at 100%. Statistical Analyses Data were analyzed with one-way ANOVA and Tukey's post-hoc comparisons (SPSS version 14) to allow for comparisons among all means. When necessary, data were log transformed to meet the requirement of homogenic variances. Supplemental Data The following materials are available in the online version of this article. Supplemental Figure S1. Inhibition of auxin transport with 25 μ m NPA leads to reduced auxin activity in the hypocotyl, as shown with the auxin-responsive pIAA19∷GUS reporter. Supplemental Figure S2. Spectral composition of the different light conditions used throughout these studies. Supplemental Figure S3. Dose-response relationship for hypocotyl length and applied NPA concentrations in control and low blue light-exposed seedlings. Supplementary Material [Supplemental Data] pp.108.133496_index.html (1KB, html) Acknowledgments We thank Diederik van Bentum and Rashmi Sasidharan for help with experiments and two anonymous reviewers for their very helpful comments on an earlier version of the manuscript. Seeds were obtained from the Nottingham Arabidopsis Stock Centre or provided by N.P. Harberd (gai, gait6 rga24 rgl1-1 rgl2-1, and pRGA∷GFP:RGA reporter), K.T. Yamamoto (pIAA19∷GUS), T. Sakai (phot1, phot2, phot1 phot2, and cry1 cry2 phot1 phot2), K.A. Franklin and G.C. Whitelam (phyABDE and cry2), M. Koornneef (hy2), C. Lin (cry1 cry2), and J.R. Ecker (ein3-1 eil1-3). 1 This work was supported by the Netherlands Organization for Scientific Research (VENI grant no. 86306001 to R.P.). The author responsible for distribution of materials integral to the findings presented in this article in accordance with the policy described in the Instructions for Authors (www.plantphysiol.org) is: Ronald Pierik (r.pierik@uu.nl). 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Supplementary Materials [Supplemental Data] pp.108.133496_index.html (1KB, html) pp.108.133496_1.pdf (209.2KB, pdf) pp.108.133496_2.pdf (167.3KB, pdf) pp.108.133496_3.pdf (46.9KB, pdf) Articles from Plant Physiology are provided here courtesy of Oxford University Press ACTIONS View on publisher site PDF (936.0 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract RESULTS DISCUSSION MATERIALS AND METHODS Supplementary Material Acknowledgments References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
17569	https://www.youtube.com/watch?v=6bQ6HJsAv4A	Unit conversion: centimeters to meters \| Measurement and data \| 5th grade \| Khan Academy Khan Academy 9090000 subscribers 944 likes Description 581623 views Posted: 23 Sep 2013 Which are smaller: centimeters or meters? If you know the answer to that question, then solving this problem gets a little easier. Let's convert centimeters to meters. Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? Grade 5th on Khan Academy: Whether you are a robot, possum, magic unicorn, or just a normal human being, 5th grade is incredible! You will deepen your knowledge of the decimals that you were first exposed to in 4th grade. In particular, you'll learn to do all the arithmetic operations--addition, subtraction, multiplication, division--with them and really understand what a decimal represents. This will be complemented by building your fraction muscles where you'll learn to add, subtract, and multiply any fractions (and begin to divide them). On the measurement front, you'll explore volume (or how much "space" something takes up) and continue the unit conversion journey begun in 4th grade. 5th grade is also the first time that you get exposed to the fundamental concept of graphing points on a coordinate plane, and you'll get a very brief exposure to the concept of exponents (when exploring "powers" of 10)! (Content was selected for this grade level based on a typical curriculum in the United States.) About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan AcademyÂ Ã Âªs 5th grade channel: Subscribe to Khan Academy: Transcript: 0 seconds Convert 37 centimeters to meters. Let me write this over. 37, and I'll write centi in a different color to emphasize the prefix. 37 centimeters, and we want to convert it to meters. So we really just have to remind ourselves what the prefix centi means. Centi literally means 1/100th, or one hundredth of a meter. So this is really saying 37 hundredth meters, so let's write it that way. This is literally 37/100 of a meter. So 1/100 meters. These are equivalent statements. So what are 37/100 of a meter? Well, that's going to be 37/100 of a meter, which can be rewritten as, if you wanted to write it as a decimal, that's 0. you could view it as 3/10 and 7/100, or 37/100. So 0.37, 0.37 meters. Another way that you could have thought about this is look, I'm going to go from centimeters to meters. I have 100. I need 100 centimeters to get to one meter, so I'm going to have to divide by 100 in order to figure out how many meters I have. And you should always do a reality check. If I convert centimeters to meters, should I get a larger number or a smaller number? Well, however many centimeters I have, I'm going to have a fewer number of meters. Meters is a larger unit. So you should have a smaller value here, and it should be a smaller value by a factor of 100. So you literally could have started off with 37 centimeters. 1 minute, 58 seconds Let me write it this way, 37 centimeters. 2 minutes, 3 seconds And actually, let's go through a couple of the units right over here. Now, if you wanted to turn it to decimeters, so this is 1/100. Decimeters, one over 1/10. So you would divide. This would be 3.7 decimeters. Let me write this as 3.7 decimeters. 2 minutes, 24 seconds And a decimeter is 1/10 of a meter, so you would divide by 10 again. So this is 0.37 meters. So one way to think about it is to go from centimeters to meters, you're going to divide by 100. Dividing by 100, you would move the decimal space over to the left two times. Doing it once divides by 10. Doing it twice divides by 100. So you get to 0.37 meters. 3 minutes, 1 second
17570	https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem?srsltid=AfmBOopMcQPBKvBrXbA4yDxVy5BMKhVCLh8IfVVroPJuRO8pswD5tx6T	Art of Problem Solving Power of a Point Theorem - 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 Power of a Point Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Power of a Point Theorem The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two linesintersect a circle and each other. Contents [hide] 1 Statement 1.1 Case 1 (Inside the Circle): 1.2 Case 2 (Outside the Circle): 1.2.1 Classic Configuration 1.2.2 Tangent Line 1.3 Case 3 (On the Border/Useless Case): 1.4 Alternate Formulation 1.5 Hint for Proof 2 Notes 3 Proof 3.1 Case 1 (Inside the Circle) 3.2 Case 2 (Outside the Circle) 3.3 Case 3 (On the Circle Border) 4 Problems 4.1 Introductory 4.2 Intermediate 4.3 Olympiad 5 See Also 5.1 External Links Statement There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application. Case 1 (Inside the Circle): If two chords and intersect at a point within a circle, then Case 2 (Outside the Circle): Classic Configuration Given lines and originate from two unique points on the circumference of a circle ( and ), intersect each other at point , outside the circle, and re-intersect the circle at points and respectively, then Tangent Line Given Lines and with tangent to the related circle at , lies outside the circle, and Line intersects the circle between and at , Case 3 (On the Border/Useless Case): If two chords, and , have on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is so no matter what, the constant product is . Alternate Formulation This alternate formulation is much more compact, convenient, and general. Consider a circle and a point in the plane where is not on the circle. Now draw a line through that intersects the circle in two places. The power of a point theorem says that the product of the length from to the first point of intersection and the length from to the second point of intersection is constant for any choice of a line through that intersects the circle. This constant is called the power of point . For example, in the figure below Hint for Proof Draw extra lines to create similar triangles (Draw on all three figures. Draw another line as well.) Notice how this definition still works if and coincide (as is the case with ). Consider also when is inside the circle. The definition still holds in this case. Notes One important result of this theorem is that both tangents from any point outside of a circle to that circle are equal in length. The theorem generalizes to higher dimensions, as follows. Let be a point, and let be an -sphere. Let two arbitrary lines passing through intersect at , respectively. Then Proof. We have already proven the theorem for a -sphere (a circle), so it only remains to prove the theorem for more dimensions. Consider the plane containing both of the lines passing through . The intersection of and must be a circle. If we consider the lines and with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds. Proof Case 1 (Inside the Circle) Join and . In (Angles subtended by the same segment are equal) (Vertically opposite angles) (Corresponding sides of similar triangles are in the same ratio) Case 2 (Outside the Circle) Join and (Why?) Now, In (shown above) (common angle) (Corresponding sides of similar triangles are in the same ratio) Case 3 (On the Circle Border) Length of a point is zero so no proof needed:) Problems Introductory Find the value of in the following diagram: Solution Find the value of in the following diagram: Solution (ARML) In a circle, chords and intersect at . If and , find the ratio . Solution (ARML) Chords and of a given circle are perpendicular to each other and intersect at a right angle at point . Given that , , and , find . Solution Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is (Source) Intermediate Two tangents from an external point are drawn to a circle and intersect it at and . A third tangent meets the circle at , and the tangents and at points and , respectively (this means that T is on the minor arc ). If , find the perimeter of . (Source) Square of side length has a circle inscribed in it. Let be the midpoint of . Find the length of that portion of the segment that lies outside of the circle. (Source) is a chord of a circle such that and Let be the center of the circle. Join and extend to cut the circle at Given find the radius of the circle. (Source) Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find (Source) Let be a triangle inscribed in circle . Let the tangents to at and intersect at point , and let intersect at . If , , and , can be written as the form , where and are relatively prime integers. Find . (Source) Olympiad Given circles and intersecting at points and , let be a line through the center of intersecting at points and and let be a line through the center of intersecting at points and . Prove that if and lie on a circle then the center of this circle lies on line . (Source) Let be a point interior to triangle (with ). The lines , and meet again its circumcircle at , , respectively . The tangent line at to meets the line at . Show that from follows . (Source) See Also Geometry Planar figures External Links Handout on AoPS Forums This article is a stub. Help us out by expanding it. Retrieved from " Categories: Geometry Theorems 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.
17571	https://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/TangentsAndOrthogonalCircle.shtml	Typesetting math: 100% Site... What's new Content page Front page Index page About Privacy policy Help with math Subjects... Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles... Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections... Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math... Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry Poles, Polars and Orthogonal Circles Given circles A(BD) and point C(BD), with centers at A and C, meeting in points B and D, let E be any point on BD. Draw tangents EF, EG from E to A(BD). Then FG passes through C if and only if the two circles are orthogonal. Solution Below is an applet that is supposed to illustrate the assertion: When you drag points or lines in the applet, the objects leave a trace. To remove the traces press Reset. If you do, drag point E along BD. You will be able to observe that the lines FG all pass through a fixed point. This point is the pole of chord BD in A(BD). You may surmise that that pole coincides with C only if the two circles are orthogonal. \|Contact\| \|Front page\| \|Content\| \|Geometry\| Copyright © 1996-2018 Alexander Bogomolny Given circles A(BD) and point C(BD), with centers at A and C, meeting in points B and D, let E be any point on BD. Draw tangents EF, EG from E to A(BD). Then FG passes through C if and only if the two circles are orthogonal. The polar of a point outside a circle with respect to that circle is defined by the two points of tangency of the tangents from the point to the circle. Point E, therefore, is known to be the polar of chord FG. Since E is chosen on BD, the pole of the latter, by La Hire's theorem, lies necessarily on the polar of the former. Let's P stands for the polar of BD. Then PF is tangent to A(BD) only if PF⊥AF, implying that P is the center of circle C(BD), or that the two circles are orthogonal. Poles and Polars Brianchon's Theorem Complete Quadrilateral Harmonic Ratio Harmonic Ratio in Complex Domain Inversion Joachimsthal's Notations La Hire's Theorem La Hire's Theorem, a Variant La Hire's Theorem in Ellipse Nobbs' Points, Gergonne Line Polar Circle Pole and Polar with Respect to a Triangle Poles, Polars and Quadrilaterals Straight Edge Only Construction of Polar Tangents and Diagonals in Cyclic Quadrilateral Secant, Tangents and Orthogonality Poles, Polars and Orthogonal Circles Seven Problems in Equilateral Triangle, Solution to Problem 1 \|Contact\| \|Front page\| \|Content\| \|Geometry\| Copyright © 1996-2018 Alexander Bogomolny
17572	https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-arithmetic-operations/cc-6th-dividing-fractions/v/conceptual-understanding-of-dividing-fractions-by-fractions	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
17573	https://english.stackexchange.com/questions/585906/a-few-chairs-or-few-chairs	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. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. 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. A few chairs or few chairs? [duplicate] I was wondering why we use the article "a" before few? For example, why we say a few chairs instead of few chairs? Is it similar to the word "a lot"? We say a lot of chairs instead of lot of chairs. 1 Answer 1 “A few chairs” would mean some chairs, not many but not none. The quantity is notable for being non-zero. “Few chairs” would mean not as many chairs as you expected, or as you needed. The quantity is notable for being low. “A few people” have been to the Moon. “Few people” do not know a few people have been to the Moon. Start asking to get answers Find the answer to your question by asking. Explore related questions See similar questions with these tags. Linked Related Hot Network Questions English Language & Usage Company Stack Exchange Network Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.9.26.34547
17574	https://m.youtube.com/watch?v=XzVj_4PsH0E	If t1 & t2 are the ends of a focal \| important concept of parabola #projecteducation #jeeshorts #jee project education 5310 subscribers 5 likes Description 79 views Posted: 6 Feb 2025 If t1 & t2 are the ends of a focal chord of the parabola y² = 4ax then t1 t2 = −1. Hence co−ordinates at the extremities of a focal chord can be taken as (at², 2at) & (a/t², -2a/t) If t1 & t2 are the ends of a focal \| important concept of parabola #projecteducation #jeeshorts #jee #parabolashorts important highlight of parabola Hello Learner , I am glad that i will able to share my knowledge with you. let me introduce myself . I am Anurag Kasaudhan graduated from IIT(ISM) Dhanbad hoping for, at the end of the video you will be amazed ! please visit : telegram channel link : For previous year topic wise solution please navigate the playlist i have made all the arrangement . ✍JEE advanced 2024 math paper 1 solution:- ✍JEE advanced 2024 math paper 2 solution:- ✍JEE Advanced 2023 Math Paper 1 solutions: ✍JEE Advanced 2023 Math Paper 2 solutions: ✍ JEE-main 2023:- 31 January shift-1 math's solution:- ✍ JEE- main 2023:- 29 January shift-1 math's solution:- ✍JEE Advanced 2022 Math Paper 1 solutions: ✍JEE Advanced 2022 Math paper 2 solution: ✍JEE-main 2022:- 24 June shift-2 math's solution: ✍JEE-main 2022:- 25 June shift-1 math's solution: ✍JEE Advanced 2021 Math Paper 1 solutions: ✍JEE Advanced 2021 Math Paper 2 solutions: ✍JEE Advanced 2020 Math Paper 1 solutions: ✍JEE Advanced 2020 Math Paper 2 solutions: ✍JEE advanced 2019 math paper 1 solution: ✍JEE advanced 2019 math paper 2 solution: ✍JEE advanced 2016 math paper 1 solution: ✍important question of differential equation : ✍relation b/w AM & GM: ✍important questions of matrix: 🖖thank you for watching🖖 if you like the content please like ,share and subscribe jeeshorts #parabolashorts #impconcept #projecteducation #parabolaproperty #jeemain #jeeadvanced #reels #jeereels #conceptshorts 2 comments Transcript: पैराबोला के फोकल कॉर्ड का अगर एक एंड का पैरामीटर t होता है दूसरे एंड का पैरामीटर -1 बा t होता है बहुत ध्यान से सुनना मैं इसको प्रूफ करता हूं और एक्चुअली यह बहुत कॉमन सा प्रॉपर्टी है तुम्हें पता होगा इसका यूज केस बहुत तगड़ा है तो उसका मैं अभी यूज केस दिखाता हूं प्रॉपर्ली सुनना हर चीज देखते हैं है ना तो हमने क्या बोला है कि अगर फोकल कॉर्ड है पैराबोला का और अगर एक एंड का पैरामीटर टी है मीस हम पैराबोला के ओबवियस बात है y स् 4x की बात करें ठीक है ना तो अगर फोकल मतलब फोकस से पास 0 से पास हो रहा है इसका पैरामीटर t है तो हमारे पास ये पॉइंट बन जाएगा ए स् 2at तो इस मुझे इसका निकालना है तो मान लो इसको मैंने t1 मान तो बनेगा ये at1 स् 21 अब बस t1 और t में रिलेशन निकालना है तो इसको हम सॉल्व करते हैं प्रॉपर्ली देखना तो हमारे पास इसका जो स्लोप हो जाएगा ब हमारा 28 अपन ए स्क माइ ए और इसका जो स्लोप बन जाएगा बगा 21 अपन at1 स् - तो 2 कैंसिल हो जाएगा क्रॉस मल्टीप्लाई कर लो तुम्हारे पास बन जाएगा ओके a से a भी उड़ जाएगा तुम्हारा t / t स् - 1 इक्वल टू हो जाएगा t1 / t1 स् - 1 क्रॉस मल्टीप्लाई तो बचेगा tt1 का स्क्वा - t = t1 t स् - t1 सबको एक तरफ लेके जाओ तो जाएगा tt1 का स्क्वा - t1 t स् - t + t1 = 0 यहां से t1 t कॉमन ले लो तो हमारा बन जाएगा t1 - t और यहां से प्व कॉमन ले लो तो बनेगा t1 - t इ 0 क्लियर ना तो हमारे पास ये हो जाएगा t1 - t टाइम्स ऑफ t1 t ् 1 = 0 तो इसको जीरो करते हमारा t1 इ t आ जाएगा जो इंपॉसिबल है दो दो डिफरेंट पॉइंट है ठीक है ना तो हमारे पास t1 का वल हो जाएगा t1 इल इसको जीरो करने से आएगा इसको रो t1 वैल्यू हो जाएगा / t जो मैंने बोला यानी अगर एक एंड का पैरामीटर हमारा t है अगर यहां का पैरामीटर t है तो यहां का हमारा t1 वलू हो जाएगा t1 हमने निकाल लिया t1 जो है वो मानस बाटी है इसका जो यूज केस है इसका यूज केस बहुत तगड़ा है इसको मैं नेक्स्ट डील में शेयर करता हूं तो यह बहुत बेरी मतलब ये बहुत इंपोर्टेंट रिजल्ट है ले इसका यूज इसमें डायरेक्टर सर्कल भी आता है सब कुछ आता है तो डियर हम लोग इसका नेक्स्ट पार्ट में सीखते हैं थैंक यू फॉर वाचिंग टेक केयर
17575	https://datavizproject.com/data-type/choropleth-map-2/	by Ferdio Choropleth Map Also called: Choropleth Heat Map A Choropleth Map is a thematic map in which areas are shaded or patterned in proportion to the measurement of the statistical variable being displayed on the map, such as population density or per-capita income. The choropleth map provides an easy way to visualize how a measurement varies across a geographic area or it shows the level of variability within a region. Family Function Shape Input Examples 1 dataset. 100 visualizations. Discover data viz applied to real data in our latest project! Hire us Interested in your own custom designed Choropleth Map? Ferdio is a leading infographic and data visualization agency specialized in transforming data and information into captivating visuals. Ferdio applies unique competencies of creativity, insight and experience throughout every project with a wide range of services. Do you have any feedback or suggestions for Choropleth Map? Let us know
17576	http://mathcentral.uregina.ca/QQ/database/QQ.02.06/ken1.html	Cutting off the top of a triangle Dear Math Central, I'm a creative type who needs to find how you can determine the horizontal, relative percentages within a triangle. Attached is a pdf showing approximately the 50% line, but I'd like to know how to determine any percentage (horizontally) within a triangle . I know that if I put a horizontal line though the height of the triangle I won't be showing 50% above and 50% below. The bottom part has to be larger (correct me if I'm wrong). I'm not concerned about height as I am wanting to break down the percentages into 60% at the bottom, 25% in the middle and 15% at the top. Thanks in advance for your help. Sincerely, Ken Hi Ken, I produced a diagram much like the one you sent and then I put some labels on it. The initial triangle has base of length b and height h so the area is given by > A =1/2 b h The horizontal line cuts off a certain proportion of the area above the line. I want to call this proportion p. (I would rather work with proportions than percentages.) The horizontal cut-off line is of length b' and the triangle above this line has height h'. Thus the area of the triangle above the cut-off line is given by > pA =1/2 b' h' Substitute A from the equation above to get > p 1/2 b h =1/2 b' h' Thus > p b h = b' h' or > > p =b'/bh'/h The original triangle and the triangle above the cut-off line are similar so > b'/b = h'/h and hence > p =h'/hh'/h= (h'/h)2 Finally > h' = √p h Thus in your example you want 15% of the area above the cut-off line so p = 0.15 and √0.15 = 0.39. Thus draw the cut-off line so that the height of the triangle above the cut-off line is 39% of the height of the original triangle. The next cut-off line is to leave 40% of the area of the original triangle above it so here p = 0.40 and √0.40 = 0.63. Thus draw the next cut-off line so that the height of the triangle above this cut-off line is 63% of the height of the original triangle. Penny
17577	https://math.stackexchange.com/questions/31893/equation-for-getting-the-length-of-the-minor-axis-of-an-ellipse	geometry - Equation for getting the length of the minor axis (of an ellipse) - 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 Equation for getting the length of the minor axis (of an ellipse) Ask Question Asked 14 years, 5 months ago Modified2 years, 4 months ago Viewed 5k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I'm looking for an equation that can help me determine the length of the minor axis. I know the length of the major axis and have the Cartesian coordinates of a point somewhere on the ellipse. How can I use these to get the length of the minor axis? geometry analytic-geometry conic-sections Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 14, 2011 at 17:19 Américo Tavares 39.2k 14 14 gold badges 110 110 silver badges 252 252 bronze badges asked Apr 9, 2011 at 11:50 user9348 user9348 5 Does the major and minor axis lies respectively on the x and y-axis?Américo Tavares –Américo Tavares 2011-04-09 12:04:59 +00:00 Commented Apr 9, 2011 at 12:04 Actually it's the reverse. The major axis is on the y-axis and the minor axis is on the x-axis. The center is 0,0.user9348 –user9348 2011-04-09 12:06:47 +00:00 Commented Apr 9, 2011 at 12:06 Then you can write the equation of the ellipse, and you also know that the coordinates of the given point satisfies that equation.Américo Tavares –Américo Tavares 2011-04-09 12:15:31 +00:00 Commented Apr 9, 2011 at 12:15 Could you give me an example? I'm ashamed to admit that my Math is rather rusty.user9348 –user9348 2011-04-09 12:22:00 +00:00 Commented Apr 9, 2011 at 12:22 please add the information that "The major axis is on the y-axis and the minor axis is on the x-axis" to the question.Américo Tavares –Américo Tavares 2011-04-09 14:01:39 +00:00 Commented Apr 9, 2011 at 14:01 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Added: In a comment OP states that "The major axis is on the y-axis and the minor axis is on the x-axis." The equation of an ellipse whose major and minor axis are respectively on the y y and x x-axis is x 2 b 2+y 2 a 2=1,(∗)x 2 b 2+y 2 a 2=1,(∗) where a a is the semimajor axe and b b is the semiminor axe. You are given 2 a 2 a and you need to find 2 b 2 b. Let the coordinates of the given point be (x 1,y 1)(x 1,y 1). Since it is on the ellipse, its coordinates must satisfy (∗)(∗) x 2 1 b 2+y 2 1 a 2=1.(∗∗)x 1 2 b 2+y 1 2 a 2=1.(∗∗) Clearing denominators and then dividing by y 2 1−a 2 y 1 2−a 2 we get a 2 x 2 1+b 2 y 2 1=a 2 b 2⇔(y 2 1−a 2)b 2=−a 2 x 2 1⇔b 2=−a 2 x 2 1 y 2 1−a 2=a 2 x 2 1 a 2−y 2 1.a 2 x 1 2+b 2 y 1 2=a 2 b 2⇔(y 1 2−a 2)b 2=−a 2 x 1 2⇔b 2=−a 2 x 1 2 y 1 2−a 2=a 2 x 1 2 a 2−y 1 2. Since a 2−y 2 1≥0 a 2−y 1 2≥0 and b>0 b>0, we obtain b=a\|x 1\|a 2−y 2 1−−−−−−√.(∗∗∗)b=a\|x 1\|a 2−y 1 2.(∗∗∗) The length of the minor axe is 2 b 2 b. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 5, 2014 at 18:22 answered Apr 9, 2011 at 13:55 Américo TavaresAmérico Tavares 39.2k 14 14 gold badges 110 110 silver badges 252 252 bronze badges 5 Thanks, that did the trick. I wasn't sure what the "\|" symbols around x1 were supposed to represent so I just wrote it as a x1 and it worked. The point is now consistently on the ellipse regardless of changes to its position.user9348 –user9348 2011-04-09 15:32:54 +00:00 Commented Apr 9, 2011 at 15:32 @Joe: \|x 1\|\|x 1\| is the absolute value of x 1 x 1 (en.wikipedia.org/wiki/Absolute_value). You need to get a positive value for b b.Américo Tavares –Américo Tavares 2011-04-09 17:28:31 +00:00 Commented Apr 9, 2011 at 17:28 Ah yes. That cleared thing up. The days of my schooling are far behind me so it's no wonder I forgot some of the rarer mathematical symbols.user9348 –user9348 2011-04-09 20:47:05 +00:00 Commented Apr 9, 2011 at 20:47 Could you please elaborate on how you divide by x 2−a 2 x 2−a 2? I tried long division and I got something else. For example a 2 x 2/(x 2−a 2)a 2 x 2/(x 2−a 2) is what?user3680 –user3680 2014-02-05 09:23:09 +00:00 Commented Feb 5, 2014 at 9:23 1 @Calle Thanks! It's a typo, it should be "dividing by y 2 1−a 2 y 1 2−a 2": x 2 1 b 2+y 2 1 a 2=⇔⇔⇔1⇔a 2 b 2(x 2 1 b 2+y 2 1 a 2)=a 2 b 2 a 2 x 2 1+b 2 y 2 1=a 2 b 2(y 2 1−a 2)b 2=−a 2 x 2 1 b 2=−a 2 x 2 1 y 2 1−a 2=a 2−y 2 1.x 1 2 b 2+y 1 2 a 2=1⇔a 2 b 2(x 1 2 b 2+y 1 2 a 2)=a 2 b 2⇔a 2 x 1 2+b 2 y 1 2=a 2 b 2⇔(y 1 2−a 2)b 2=−a 2 x 1 2⇔b 2=−a 2 x 1 2 y 1 2−a 2=a 2−y 1 2. Corrected.Américo Tavares –Américo Tavares 2014-02-05 18:21:23 +00:00 Commented Feb 5, 2014 at 18:21 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. So you know the length of the semimajor axis, and it's along y. Let's call this axis 'a'. We'll call the length of the semiminor axis 'b'. x^2 / b^2 + y^2 / a^2 = 1. You also have another point (x1, y1). Simply sub this into the equation and solve for b! x1^2 / b^2 + y1^2 / a^2 = 1 (a^2 - y^1)/ a^2 = x1^2 / b^2 x1^2 a^2 / (a^2 - y1^2)`= b^2 Of course, this approach won't work if a^2 = y1^2 (as you'll be dividing by 0), but a point on the ellipse should mean this will never be the case. I may have made an algebraic mistake somewhere there, but the approach should still be good. :) Hope this helps. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 9, 2011 at 13:26 answered Apr 9, 2011 at 12:55 Ben StottBen Stott 377 2 2 silver badges 12 12 bronze badges 2 'a' is the semimajor axis (and 'b' the semiminor axis).Américo Tavares –Américo Tavares 2011-04-09 13:21:18 +00:00 Commented Apr 9, 2011 at 13:21 Yes, of course. Sorry for that - my brain usually melts at this time of night! Corrected in my post.Ben Stott –Ben Stott 2011-04-09 13:26:12 +00:00 Commented Apr 9, 2011 at 13:26 Add a comment\| You must log in to answer this question. 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17578	https://www.titance.com/documents/Open%20channel%20flow%201%20-%20update%20nov%202013.pdf	1 Open Channel Flow I - The Manning Equation and Uniform Flow Harlan H. Bengtson, PhD, P.E. COURSE CONTENT 1. Introduction Flow of a liquid may take place either as open channel flow or pressure flow. Pressure flow takes place in a closed conduit such as a pipe, and pressure is the primary driving force for the flow. For open channel flow, on the other hand the flowing liquid has a free surface at atmospheric pressure and the driving force is gravity. Open channel flow takes place in natural channels like rivers and streams. It also occurs in manmade channels such as those used to transport wastewater and in circular sewers flowing partially full. In this course several aspects of open channel flow will be presented, discussed and illustrated with examples. The main topic of this course is uniform open channel flow, in which the channel slope, liquid velocity and liquid depth remain constant. First, however, several ways of classifying open channel flow will be presented and discussed briefly. Open Channel Flow Examples: A River and an Irrigation Canal 2 2. Topics Covered in this Course I. Methods of Classifying Open Channel Flow A. Steady State of Unsteady State Flow B. Laminar or Turbulent Flow C. Uniform or Non-uniform Flow D. Supercritical, Subcritical or Critical Flow II. Calculations for Uniform Open Channel Flow A. The Manning Equation B. The Manning Roughness Coefficient C. The Reynold's Number D. The Hydraulic Radius E. The Manning Equation in S.I. Units F. The Manning Equation in Terms of V Instead of Q G. The Easy Parameters to Calculate with the Manning Equation H. The Hard Parameter to Calculate - Determination of Normal Depth I. Circular Pipes Flowing Full or Partially Full J. Uniform Flow in Natural Channels III. Summary IV. References and Websites 3. Methods of Classifying Open Channel Flow Open Channel flow may be classified is several ways, including i) steady state or unsteady state, ii) laminar or turbulent, iii) uniform or nonuniform, and iv) subcritical, critical or supercritical flow. Each of these will be discussed briefly in the rest of this section, and then uniform open channel flow will be covered in depth in the rest of the course. Steady State or Unsteady State Flow: The meanings of the terms steady state and unsteady state are the same for open channel flow as for a variety of other flowing fluid applications. For steady state flow, there are no changes in velocity 3 patterns and magnitude with time at a given channel cross section. Unsteady state flow, on the other hand, does have changing velocity with time at a given cross section. Unsteady state open channel flow takes place when there is a changing flow rate, as for example in a river after a rain storm. Steady state open channel flow takes place when there is a constant flow rate of liquid is passing through the channel. Steady state or nearly steady state conditions are present for many practical open channel flow situations. The equations and calculations in this course will be for steady state flow. Laminar or Turbulent Flow: Classification of a given flow as either laminar or turbulent is important in several fluid flow applications, such as pipe flow and flow past a flat plate, as well as in open channel flow. In each case a Reynold’s number is the criterion used to predict whether a given flow will be laminar or turbulent. Open channel flow is typically laminar for a Reynold’s number below 500 and turbulent for a Reynold’s number greater than 12,500. A flow with Reynold’s number between 500 and 12,500 may be either laminar or turbulent, depending on other conditions, such as the upstream channel conditions and the roughness of the channel walls. More details on the Reynold’s number for open channel flow and its calculation will be given in Section 3, Calculations for Uniform Open Channel Flow. The Reynolds number is greater than 12,500, and thus the flow is turbulent for most practical cases of water transportation in natural or manmade open channels. A notable example of laminar open channel flow is flow of a thin liquid layer on a large flat surface, such as rainfall runoff from a parking lot, highway, or airport runway. This type of flow is often called sheet flow. Laminar and Turbulent Flow Background: The difference between laminar and turbulent flow in pipes and the quantification of the conditions for each of them, was first observed and reported on by Osborne Reynolds in . His classic experiments utilized injection of dye into a transparent pipe containing a flowing fluid. When the flow was laminar he observed that they dye flowed in a streamline and didn’t mix with the rest of the fluid. Under turbulent flow conditions, however, the net velocity of the fluid is in the direction of flow, but there are eddy currents in all directions that cause mixing of the fluid, so that the entire fluid became colored in Reynold’s experiments. Laminar and turbulent flow are illustrated for open channel flow in figure 1. 4 Figure 1. Dye injection into laminar & turbulent open channel flow Laminar flow is also sometimes called streamline flow. It occurs for flows with high viscosity fluids and/or low velocity and/or high viscosity. Turbulent flow, on the other hand, occurs for fluid flows with low viscosity and/or high velocity. Uniform or Non-Uniform Flow: Uniform flow will be present in a portion of open channel (called a reach of channel) with a constant flow rate of liquid passing through it, constant bottom slope, and constant cross-section shape & size. With these conditions present, the average velocity of the flowing liquid and the depth of flow will remain constant in that reach of channel. For reaches of channel where the bottom slope, cross-section shape, and/or cross-section size change, non-uniform flow will occur. Whenever the bottom slope and channel cross-section shape and size become constant in a downstream reach of channel, another set of uniform flow conditions will occur there. This is illustrated in Figure 2. Figure 2. Uniform and Non-uniform Open Channel Flow 5 Supercritical, Subcitical, or Critical Flow: Any open channel flow will be supercritical, subcritical or critical flow. The differences among these three classifications of open channel flow, however, are not as obvious or intuitive as with the other classifications (steady or unsteady state, laminar or turbulent, and uniform or non-uniform). Your intuition will probably not lead you to expect some of the behaviors for subcritical and supercritical flow and the transitions between them. Subcritical flow occurs with relatively low liquid velocity and relatively deep flow, while supercritical flow occurs with relatively high liquid velocity and relatively shallow flow. The Froude number (Fr = V/(gl)1/2) can be used to determine whether a given flow is supercritical, subcritical or critical. Fr is less than one for subcritical flow, greater than one for supercritical flow and equal to one for critical flow. Further discussion of subcritical, supercritical and critical flow is beyond the scope of this course. 4. Calculations for Uniform Open Channel Flow Uniform open channel flow takes place in a channel reach that has constant channel cross-section size and shape, constant surface roughness, and constant bottom slope. With a constant flow rate of liquid moving though the channel, these conditions lead to flow at a constant liquid velocity and depth, as illustrated in Figure 2. The Manning Equation is a widely used empirical equation that relates several uniform open channel flow parameters. This equation was developed in 1889 by the Irish engineer, Robert Manning. In addition to being empirical, the Manning Equation is a dimensional equation, so the units must be specified for a given constant in the equation. For commonly used U.S. units the Manning Equation and the units for its parameters are as follows: Q = (1.49/n)A(Rh 2/3)S1/2 (1) Where: Q is the volumetric flow rate passing through the channel reach in ft3/sec. 6 A is the cross-sectional area of flow perpendicular to the flow direction in ft2. S is the bottom slope of the channel in ft/ft (dimensionless). n is a dimensionless empirical constant called the Manning Roughness coefficient. Rh is the hydraulic radius = A/P. Where: A is the cross-sectional area as defined above in ft2, and P is the wetted perimeter of the cross-sectional area of flow in ft. Actually, S is the slope of the hydraulic grade line. For uniform flow, however, the depth of flow is constant, so the slope of the hydraulic grade line is the same as that for the liquid surface and the same as the channel bottom slope, so the channel bottom slope is typically used for S in the Manning Equation. The Manning Roughness Coefficient, n, was noted above to be a dimensionless, empirical constant. Its value is dependent on the nature of the channel and its surfaces. Many handbooks and textbooks have tables with values of n for a variety of channel types and surfaces. A typical table of this type is given as Table 1 below. It gives n values for several man-made open channel surfaces. What is HYDRAULIC RADIUS ? Does it need to be round ? 7 Table 1. Manning Roughness Coefficient, n, for Selected Surfaces The Reynold’s number is defined as Re = VRh/ for open channel flow, where Rh is the hydraulic radius, as defined above, V is the liquid velocity (= Q/A), and  and  are the density and viscosity of the flowing fluid, respectively. Any consistent set of units can be used for RhV, and , because the Reynold’s number is dimensionless. In order to use the Manning equation for uniform open channel flow, the flow must be in the turbulent regime. Forunately, Re is greater than 12,500 for nearly all practical cases of water transport through an open channel, so the flow is turbulent and the Manning equation can be used. Sheet flow, as mentioned above, is a rather unique type of open channel flow, and is the primary example of laminar flow with a free water surface. 8 The Manning equation doesn't contain any properties of water, however, in order to calculate a value for Reynold's number, values of density and viscosity for the water in question are needed. Many handbooks, textbooks, and websites have tables of density and viscosity values for water as a function of temperature. Table 2 below summarizes values of density and viscosity of water from 32oF to 70oF. Table 2. Density and Viscosity of Water Example #1: Water is flowing 1.5 feet deep in a 4 foot wide, open channel of rectangular cross section, as shown in the diagram below. The channel is made of concrete (made with steel forms), with a constant bottom slope of 0.003. a) Estimate the flow rate of water in the channel. b) Was the assumption of turbulent flow correct ? 9 Solution: a) Based on the description, this will be uniform flow. Assume that the flow is turbulent in order to be able to use equation (1), the Manning equation. All of the parameters on the right side of equation (1) are known or can be calculated: From Table 1, n = 0.011. The bottom slope is given as: S = 0.003. From the diagram, it can be seen that the cross-sectional area perpendicular to flow is 1.5 ft times 4 ft = 6 ft2. Also from the figure, it can be seen that the wetted perimeter is 1.5 + 1.5 + 4 ft = 7 ft. The hydraulic radius can now be calculated: Rh = A/P = 6 ft2/7 ft = 0.8571 ft Substituting values for all of the parameters into Equation 1: Q = (1.49/0.011)(6)(0.85712/3)(0.0031/2) = 40.2 ft3/sec = Q b) Since no temperature was specified, assume a temperature of 50o F. From Table 2,  = 1.94 slugs/ft3, and  = 2.730 x 10-5 lb-s/ft2. Calculate average velocity, V: V = Q/A = 40.2/6 ft/sec = 6.7 ft/sec Reynold’s number (Re = VRh/) can now be calculated: Re = VRh/ = (1.94)(6.7)(0.8571)/( 2.730 x 10-5) = 4.08 x 105 Since Re > 12,500, this is turbulent flow The Hydraulic Radius is an important parameter in the Manning Equation. Some common cross-sectional shapes used for open channel flow calculations are rectangular, circular, semicircular, trapezoidal, and triangular. Example #1 has already illustrated calculation of the hydraulic radius for a rectangular open channel. Calculations for the other four shapes will now be considered briefly. Common examples of gravity flow in a circular open channel are the flows in storm sewers, sanitary sewers and circular culverts. Culverts and storm and 10 sanitary sewers usually flow only partially full, however the "worst case" scenario of full flow is often used for hydraulic design calculations. The diagram below shows a representation of a circular channel flowing full and one flowing half full. Figure 3. Circular and Semicircular Open Channel Cross-Sections For a circular conduit with diameter, D, and radius, R, flowing full, the hydraulic radius can be calculated as follows: The x-sect. area of flow is: A = R2 = (D/2)2 = D2/4 The wetted perimeter is: P = 2R = D Hydraulic radius = Rh = A/P = (D2/4)/(D), simplifying: For a circular conduit flowing full: Rh = D/4 (2) If a circular conduit is flowing half full, there will be a semicircular cross-sectional area of flow, the area and perimeter are each half of the value shown above for a circle, so the ratio remains the same, D/4. Thus: For a semicircular x-section: Rh = D/4 (3) 11 A trapezoidal shape is sometimes used for manmade channels and it is also often used as an approximation of the cross-sectional shape for natural channels. A trapezoidal open channel cross-section is shown in Figure 3 along with the parameters used to specify its size and shape. Those parameters are b, the bottom width; B, the width of the liquid surface; l, the wetted length measured along the sloped side, y; the liquid depth; and , the angle of the sloped side from the vertical. The side slope is also often specified as: horiz: vert = z:1. Figure 4. Trapezoidal Open Channel Cross-section The hydraulic radius for the trapezoidal cross-section is often expressed in terms of liquid depth, bottom width, & side slope (y, b, & z) as follows: The cross-sectional area of flow = the area of the trapezoid = A = y(b + B)/2 = (y/2)(b + B) From Figure 4, one can see that B is greater than b by the length, zy at each end of the liquid surface. Thus: B = b + 2zy Substituting into the equation for A: A = (y/2)(b + b + 2zy) = (y/2)(2b + 2zy) Simplifying: A = by + zy2 As seen in Figure 4, the wetted perimeter for the trapezoidal cross-section is: 12 P = b + 2l By Pythagoras’ Theorem: l2 = y2 + (yz)2 or l = (y2 + (yz)2)1/2 Substituting into the above equation for P and simplifying: P = b + 2y(1 + z2)1/2 Thus for a trapezoidal cross-section the hydraulic radius is found by substituting equations (2) & (3) into Rh = A/P, yielding the following equation: For a trapezoid: Rh = (by + zy2)/( b + 2y(1 + z2)1/2) (4) A triangular open channel cross-section is shown in Figure 5. As would be typical, this cross-section has both sides sloped from vertical at the same angle. Several parameters that are typically used to specify the size and shape of a triangular cross-section are shown in the figure as follows: y, the depth of flow; B, the width of the liquid surface; l, the wetted length measured along the sloped side; and the side slope specified as: horiz : vert = z : 1. Figure 5. Triangular Open Channel Cross-section 13 The wetted perimeter and cross-sectional area of flow for a triangular open channel of the configuration shown in Figure 5, can be expressed in terms of the depth of flow, y, and the side slope, z, as follows: The area of the triangular area of flow is: A = ½ By, but from Figure 5: B = 2yz, Thus: A = ½ (2yz)y or simply: A = y2z The wetted perimeter is: P = 2l and l2 = y2 + (yz)2 , solving for l and substituting: P = 2[y2(1 + z2)]1/2 Hydraulic radius: Rh = A/P For a trianglular x-section: Rh = y2z/(2[y2(1 + z2)]1/2 ) Or simplifying: Rh = yz/[2(1 + z2)1/2 ] (5) Example #2: A triangular flume has 10 ft3/sec of water flowing at a depth of 2 ft above the vertex of the triangle. The side slopes of the flume are: horiz : vert = 1 : 1. The bottom slope of the flume is 0.004. What is the Manning roughness coefficient, n, for this flume? Solution: From the problem statement: y = 2 ft and z = 1, substituting into Equation (5): Rh = 2(1)/(2[2(1 + 12)]1/2 ) = 0.707 ft The cross-sectional area of flow is: A = y2z = (22)(1) = 4 ft2 Substituting these values for Rh and A along with given values for Q and S into equation (1) gives: 10 = (1.49/n)(4)(0.7072/3)(0.0041/2) Solving for n: n = 0.030 14 The Manning Equation in SI Units has the constant equal to 1.00 instead of 1.49. The equation and units are as shown below: Q = (1.00/n)A(Rh 2/3)S1/2 (6) Where: Q is the volumetric flow rate passing through the channel reach in m3/sec. A is the cross-sectional area of flow perpendicular to the flow direction in m2. S is the bottom slope of the channel in m/m (dimensionless). n is the dimensionless empirical Manning Roughness coefficient Rh is the hydraulic radius = A/P. Where: A is the cross-sectional area as defined above in m2 and P is the wetted perimeter of the cross-sectional area of flow in m. The Manning Equation in terms of V instead of Q: Sometimes it's convenient to have the Manning Equation expressed in terms of average velocity, V, rather than volumetric flow rate, Q, as follows for U.S. units (The constant would be 1.00 for S.I. units.): V = (1.49/n)(Rh 2/3)S1/2 (7) Where the definition of average velocity, V, is the volumetric flow rate divided by the cross-sectional area of flow: V = Q/A (8) 15 The Easy Parameters to Calculate with the Manning Equation: Q, V, S, and n are the easy parameters to calculate. If any of these is the unknown, with adequate known information, the Manning equation can be solved for that unknown parameter and then used to calculate the unknown by calculating Rh and substituting known parameters into the equation. This is illustrated for calculation of Q in Example #1 and calculation of n in Example #2. Another example here illustrates bottom slope, S, as the unknown. Then in the next section, we’ll take a look at the hard parameter to calculate, normal depth. Example #3: Determine the bottom slope required for a 12 inch diameter circular storm sewer made of centrifugally spun concrete, if must have an average velocity of 3.0 ft/sec when it’s flowing full. Solution: Solving Equation (6) for S, gives: S = {(nV)/[1.49(Rh 2/3)]}2. The velocity, V, was specified as 3 ft/sec. From Table 1, n = 0.013 for centrifugally spun concrete. For the circular, 12 inch diameter sewer, Rh = D/4 = ¼ ft. Substituting into the equation for S gives: S = {(0.013)(3.0)/[1.49(1/4)2/3]}2 = 0.00435 = S The Hard Parameter to Calculate - Determination of Normal Depth: For a given flow rate through a channel reach of known shape size & material and known bottom slope, there will be a constant depth of flow, called the normal depth, sometimes represented by the symbol, yo. Determination of the unknown normal depth, yo, for given values of Q, n, S, and channel size and shape, is more difficult than determination of Q, V, n, or S, as discussed in the previous section. It will be possible to get an equation with yo as the only unknown, however, in most cases it isn’t possible to solve the equation explicitly for yo, so an iterative or “trial and error” solution is needed. Example #4 illustrates this type of problem and solution. Example #4: Determine the normal depth for a water flow rate of 15 ft3/sec, through a rectangular channel with a bottom slope of 0.0003, bottom width of 3 ft, and Manning roughness coefficient of 0.013. 16 Solution: Substituting specified values into the Manning equation [ Q = (1.49/n)A(Rh 2/3)S1/2 ] gives: 15 = (1.49/0.013)(3yo)(( 3yo/(3 + 2yo))2/3)(0.00031/2) Rearranging this equation gives: 3 yo(3yo/(3 + 2yo))2/3 = 7.5559 There’s a unique value of yo that satisfies this equation, even though the equation can’t be solved explicitly for yo. The solution can be found by an iterative process, that is, by trying different values of yo until you find the one that makes the left hand side of the equation equal to 7.5559, to the degree of accuracy needed. A spreadsheet such as Excel helps a great deal in carrying out such an iterative solution. The table below shows an iterative solution to Example #4. Trying values of 1, 2, & 3 for yo, shows that the correct value for yo lies between 2 and 3. Then the next four trials for yo, shows that it is between 2.6 and 2.61. The next two entries show that the right hand column is closest to 7.5559 for yo = 2.60, thus yo = 2.60 to three significant figures. yo 3yo[3yo/(3 + 2yo)]2/3 1 2.134 2 5.414 3 9.000 2.5 7.184 2.7 7.906 2.6 7.555 2.61 7.580 For a trapezoidal or triangular channel, the procedure for determining normal depth would be the same. In those cases the equations for Rh are a bit more complicated, and the side slope, z, must be specified, but the overall procedure would be like that used in the example above. 17 Circular Pipes Flowing Full or Parially Full: For a circular pipe, flowing full under gravity flow, such as a storm sewer, Rh = D/4, and A = D2/4 can be substituted into the Manning equation to give the following simplified forms: Q = (1.49/n)(D2/4)((D/4)2/3)S1/2 (8) V = (1.49/n)((D/4)2/3)S1/2 (9) The diameter required for a given velocity or given flow rate at full pipe flow, with known slope and pipe material can be calculated directly, by solving the above equations for D, giving the following two equations: D = 4[Vn/(1.49S1/2)]3/2 (10) D = {[45/3/(1.49)]3/8}Qn/S1/2 = 1.33Qn/S1/2 (11) Calculations for the hydraulic design of storm sewers are typically made on the basis of the circular pipe flowing full under gravity. Storm sewers actually flow less than full much of the time, however, due to storms less intense than the design storm, so there is sometimes interest in finding the flow rate or velocity for a specified depth of flow in a storm sewer of known diameter, slope and n value. Equations are available for these calculations, but they are rather awkward to use, so a convenient to use graph correlating V/Vfull and Q/Qfull to d/D (depth of flow/diameter of pipe), has been prepared and is widely available in handbooks, textbooks and on the internet. That graph is given in Figure 7 below. The depth of flow, d, and pipe diameter, D, are shown in Figure 6, and Figure 7 gives the correlation between V/Vfull, Q/Qfull, and d/D. 18 Figure 6. Depth of Flow, d, and Diameter, D, for Partially Full Pipe Flow 19 Figure 7. Flow Rate and Velocity Ratios in Pipes Flowing Partially Full Example #5: Calculate the velocity and flow rate in a 24 inch diameter storm sewer with slope = 0.0018 and n = 0.012, when it is flowing full under gravity. Solution: From Equation (9): V = (1.49/n)((D/4)2/3)S1/2 = (1.49/0.012)((2/4)2/3)(0.00181/2) = 3.32 ft/sec = Vfull Then: Q = VA = (3.32)(22/4) = 3.32 = 10.43 cfs = Qfull Example #6: What would be the velocity and flow rate of water in the storm sewer from Example #5, when it is flowing at a depth of 18 inches? Solution: d/D = 18/24 = 0.75 From Figure 7: for d/D = 0.75: Q/Qfull = 0.80 and V/Vfull = 0.97 Thus: Q = 0.8 Qfull = (0.80)(10.43) = 8.34 cfs = Q and: V = 0.97 Vfull = (0.97)(3.32) = 3.22 ft/sec = V Uniform Flow in Natural Channels: The Manning equation is widely applied to flow in natural channels as well as manmade channels. One of the main differences for application to natural channels is less precision in estimating a value for the Manning roughness coefficient, n, due to the great diversity in the type of channels. Another difference is less likelihood of truly constant slope and channel shape and size over an extended reach of channel. One way of handling the problem of determining a value for n is the experimental approach. The depth of flow, channel shape and size, bottom slope and volumetric flow rate are each measured for a channel reach with reasonably constant values for those parameters. Then an empirical value for n is calculated. The value of n can then be used to calculate depth for a given flow or velocity, or to calculate velocity and flow rate for a given depth for that reach of channel. 20 There are many tables of n values for natural channels in handbooks, textbooks and on the internet. An example is the table on the next two pages from the Indiana Department of Transportation Design Manual, available on the internet at: 21 22 23 Similar tables are available on many state agency websites. Note that this table gives minimum, normal and maximum values of the Manning Roughness coefficient, n, for a wide range of natural and excavated or dredged channel descriptions. Example #6: A reach of channel for a stream on a plain is described as clean, straight, full stage, no rifts or deep pools. The bottom slope is reasonably constant at 0.00025 for a reach of this channel. Its cross-section is also reasonably constant for this reach, and can be approximated by a trapezoid with bottom width equal to 7 feet, and side slopes, with horiz : vert equal to 3:1. Using the minimum and maximum values of n in the above table for this type of stream, find the range of volumetric flow rates represented by a 4 ft depth of flow. Solution to Example #6: From the problem statement, b = 7 ft, S = 0.00025, z = 3, and y = 4 ft. From the above table, item 1. a. (1) under “Natural Stream”, the minimum expected value of n is 0.025 and the maximum is 0.033. Substituting values for b, z, and y into equation (4) for a trapezoidal hydraulic radius gives: Rh = [(7)(4) + 3(42)]/[7 + (2)(4)(1 + 32)1/2 ] = 2.353 ft Also A = (7)(4) + 3(42) = 76 ft2 Substituting values into the Manning Equation [Q = (1.49/n)A(Rh 2/3)S1/2] gives the following results: Minimum n (0.025): Qmax = (1.49/0.025)(76)(2.3532/3)(0.00025)1/2 Qmax = 126.7 ft3/sec Maximum n (0.033): Qmin = (1.49/0.033)(76)(2.3532/3)(0.00025)1/2 Qmin = 95.99 ft3/sec 24 5. Summary Open channel flow, which has a free liquid surface at atmospheric pressure, occurs in a variety of natural and man-made settings. Open channel flow may be classified as i) laminar or turbulent, ii) steady state or unsteady state, iii) uniform or non-uniform, and iv) critical, subcritical, or supercritical flow. Many practical cases of open channel flow can be treated as turbulent, steady state, uniform flow. Several open channel flow parameters are related through the empirical Manning Equation, for turbulent, uniform open channel flow (Q = (1.49/n)A(Rh 2/3)S1/2). The use of the Manning equation for uniform open channel flow calculations and for the calculation of parameters in the equation, such as cross-sectional area and hydraulic radius, are illustrated in this course through worked examples. 6. References and Websites 1. Bengtson, H.H., “Manning Equation/Open Channel Flow Calculations with Excel Spreadsheets,” an online article at www.EngineeringExcelSpreadsheets.com. 2. Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4th Ed., New York: John Wiley and Sons, Inc, 2002. 3. Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959. 4. Camp, T.R., “Design of sewers to facilitate flow,” Sewage Works Journal, 18 (3), 1946 5. Steel, E. W. & McGhee, T. J., Water Supply and Sewerage, 5th Ed. New York, McGraw-Hill Book Company, 1979 Websites: 1. Indiana Department of Transportation Design Manual, available on the internet at: 2. Illinois Department of Transportation Drainage Manual, available on the internet at:
17579	https://www.surveymonkey.com/learn/research-and-analysis/p-value-calculator/	Home Survey best practices How to analyze survey data P-value calculator: How to calculate p-value P-value calculator: How to calculate p-value Is your data sufficient to reject the null hypothesis? Calculate p-value with our calculator. Get started P-value calculator Z Score A z-score indicates a data point's distance from the mean in standard deviations. Find it in a standard normal distribution table or software. Test Type Use a two-tailed test for differences in any direction. Choose left- or right-tailed if you expect much lower or higher results. Choose... Significance Level Typically set at 0.05, this is your threshold to accept the statistically significance of the results. The p-value is 0 The result is not significant at p ≥ 0.05 Picture this: you're diving deep into the world of analytics and statistics, trying to make sense of all those numbers and data points. Suddenly, you stumble upon a little gem called the p-value. It's like a secret code that researchers use to unlock the mysteries of hypothesis testing and significance. The primary use of the p-value is for decision-making in hypothesis testing. It helps researchers assess if the observed data is enough to reject the null hypothesis for an alternative hypothesis. Researchers also use the p-value to compare groups or test for correlations. Gather answers using the SurveyMonkey p-value calculator above. What is a p-value? The p-value stands for probability value. It measures the likelihood of a result, assuming the null hypothesis is true. It's a probability gauge showing how likely your result is, assuming no real difference (the null hypothesis). The p-value quantifies the strength of evidence against the null hypothesis. It is typically compared to a predetermined level of significance, such as 0.05. When the p-value is low, it tells you, "This result probably didn't happen by chance!" This gives you the green light to reject the null hypothesis and consider that your hypothesis might be true. The p-value is important because researchers use it to decide whether to accept or reject the null hypothesis. Some examples of research questions that can use the p-value are: “Do men and women differ in customer satisfaction?” “Is satisfaction with training programs associated with employee satisfaction?” A low p-value suggests there are differences among the groups you tested. It also indicates that real, predictable relationships among variables may exist. Researchers can then interpret the significance of their findings and communicate the strength of evidence to stakeholders and peers. How to calculate p-value To calculate a p-value, first determine the probability of obtaining your data if the null hypothesis were true. Then, compare this probability to your chosen significance level (usually 0.05) to decide if your results are statistically significant. Calculate the p-value from z-score To calculate a p-value from a z-score, look up the z-score in a standard normal distribution table. Alternatively, use software to find the corresponding probability. This probability represents the likelihood of observing a value as extreme as the z-score under the null hypothesis. The following formulas give the p-value: Left-tailed z-test: p-value = P(Zscore) Right-tailed z-test: p-value = 1 - P(Zscore) Two-tailed z-test: p-value = 2 × P(−\|Zscore\|) or 2 - 2 × P(\|Zscore\|) Here’s the step-by-step guide on how to calculate the p-value from a z-score: Understand the problem:You have data and want to know how likely it is to get that result. You also want to see how likely something more extreme would be, assuming the null hypothesis is true. Find the z-score:Start by finding the z-score of your data. This tells you how many standard deviations away from the mean your data point is. Find your z-score by leveraging statistical software (like R or SPSS) or look up the deviation in a table (like this one). Determine the direction:Choose a one-tailed test (extreme values in one direction) or a two-tailed test (in both directions). If you expect the difference to be significantly smaller or larger, use a one-tailed test—the left- or right-tailed test. If you don’t have a hypothesis about which direction the difference will be, use a two-tailed test. Look up the z-score:Using a standard normal table, software, or a p-value calculator, find the cumulative probability. Calculate the p-value by leveraging the p-value calculator above or: For a one-tailed test: If the z-score is positive (right-tailed test), subtract the cumulative probability from 1. If the z-score is negative (left-tailed test), use the cumulative probability directly. For a two-tailed test: Double the cumulative probability to account for both tails. Then, subtract it from 1 if the z-score is positive. Interpret the p-value: If the p-value is very small (usually less than 0.05), it suggests that your data is unlikely under the null hypothesis, indicating statistical significance. You can also use our p-value calculator above to interpret the p-value based on the confidence level. Calculate the p-value from t-score To calculate a p-value from a t-score, first, determine the t-score representing the difference between your sample mean and the population mean. Then, use a t-distribution table or software to find the probability of observing that t-score. This indicates the likelihood of obtaining your sample results under the null hypothesis. The following formulas gives the p-value from the t-score. Left-tailed t-test: p-value = cdft,d(tscore) Right-tailed t-test: p-value = 1 - cdft,d(tscore) Two-tailed t-test: p-value = 2 × cdft,d(−\|tscore\|) or p-value = 2 - 2 × cdft,d(\|tscore\|) Where cdft,d represents the cumulative distribution function of the t-Student distribution with d degrees of freedom. Here’s the step-by-step guide on how to calculate the p-value from a t-score: Understand the situation:You have sample data and want to know how likely it is to get your results. This assumes there's no actual difference in the population. Calculate the t-score: This measurement tells you how different your sample mean is from the population mean. Determine degrees of freedom:This is based on your sample size. It helps you look up the correct probability in the t-distribution table. Check the t-distribution table:Look up your calculated t-score in the table. This provides the probability of observing that difference or more if there is no actual difference in the population. Interpret the result:If the p-value is very small, your sample results are unlikely under the null hypothesis. This suggests your results might be significant. Calculate the p-value from the Pearson Correlation To obtain the p-value for a Pearson correlation coefficient, first use the calculated coefficient to derive a t-statistic. Then, you can find its associated p-value using the t-distribution with degrees of freedom (n - 2). The formula to get the t-statistic from a Pearson correlation coefficient is below: Where: r is the Pearson correlation coefficient. n is the sample size. After obtaining the t-statistic, you can calculate the p-value using the cumulative distribution function of the t-distribution. This uses n - 2 degrees of freedom, where n is the sample size. Here's the general process: Understand the situation: You have some sample data and want to see if two variables are correlated. Calculate the t-statistic:Convert the correlation coefficient (r) to a t-statistic using the formula above. Determine the degrees of freedom: Calculate the degrees of freedom (df). Use the formula 𝑑𝑓 = n - 2, where n is the sample size. Find the p-value: Once you have the t-statistic and degrees of freedom, you can use a t-distribution table or a statistical software package to find the p-value associated with the calculated t-statistic. Interpret the result: If the p-value is less than your chosen significance level (commonly 0.05), you reject the null hypothesis and conclude that there is a statistically significant correlation between the two variables. Otherwise, you fail to reject the null hypothesis. Calculate the p-value from the chi-square score To calculate the p-value from a chi-square score, determine the degrees of freedom associated with the chi-square distribution. Then, use statistical tables or software to find the probability of obtaining a chi-square value as extreme as the observed one. You can get the p-value with the help of the following formula: p-value=1− cdfχ² (x; df) Where: x is the chi-square test statistic. cdfχ² is the cumulative distribution function of the chi-square distribution. df is the degrees of freedom. You subtract the cumulative probability from 1 because the chi-square distribution is right-skewed, so the tail area to the right of the observed chi-square value corresponds to the p-value. Here are the step-by-step instructions to calculate the p-value from a chi-square score: Understand the situation: You have categorical data and want to see whether the variables have significant associations. Compute the chi-square score. Determine degrees of freedom (df): Degrees of freedom are determined by the number of categories minus 1 for a simple chi-square test. For a chi-square test of independence, it's calculated as (rows−1)×(columns−1). Find the p-value: Use statistical tables or software to find the cumulative probability for the chi-square and degrees of freedom. This is the area to the right of the chi-square value under the chi-square distribution curve. Interpret the result: Compare the obtained p-value with your chosen significance level (commonly 0.05). If the p-value is less than the significance level, you reject the null hypothesis and conclude that there's a significant association between the variables. Otherwise, you fail to reject the null hypothesis. How to interpret p-value If the p-value is less than or equal to 0.05 (or any chosen significance level), it suggests the result is statistically significant. This means the observed result is significant at the α level. This means the probability of obtaining an extreme result, assuming the null hypothesis is true, is very low. Typically, this probability is less than 5%. Therefore, you reject the null hypothesis in favor of the alternative hypothesis. This indicates some evidence to support the claim made by the alternative hypothesis. If the p-value is greater than 0.05, it suggests that the observed result is not statistically significant at the chosen significance level. In other words, there’s insufficient evidence to reject the null hypothesis. This means we cannot conclude that the observed result differs from what would be expected under the null hypothesis. Related reading:How to analyze survey data Common p-value mistakes Using the p-value to represent real-world probabilities Some people believe that a p-value of 0.05 means there is a 95% chance the test hypothesis is true and a 5% chance it is false. This is a misinterpretation of the p-value. P-values indicate the likelihood of observing the data, assuming the null hypothesis is true. They are not direct measures of the probability of hypotheses being true or false. Treating p-value as effect size or importance Treating the p-value as synonymous with effect size or importance is a common misconception. This blurs the line between statistical significance and practical significance. A small p-value indicates the observed result is unlikely due to random chance. However, it does not convey the effect's magnitude. Additionally, it does not reflect the practical relevance of that effect. For instance, even tiny deviations from the null hypothesis may yield statistically significant p-values in large datasets despite being practically insignificant. Also, if an experiment yields significant differences multiple times, it’s likely to observe nonsignificant results sometimes because this is based on probability. Conversely, a large p-value doesn't necessarily imply that the observed effect is trivial. Instead, it suggests that the data do not provide convincing evidence against the null hypothesis. To accurately assess the practical importance of findings, it's essential to complement p-values with measures of effect size. Effect size quantifies the magnitude of the observed effect. It helps researchers contextualize results within the broader scope of the research question or application. This distinction ensures that statistical significance aligns with meaningful real-world implications. It guides informed decision-making and interpretation of research outcomes. Failing to account for multiple testing The multiple testing problem arises when researchers conduct numerous hypothesis tests on the same dataset without appropriately adjusting the significance level. This practice significantly inflates the likelihood of encountering false positives, also known as Type I errors. In these situations, the null hypothesis is incorrectly rejected. Imagine a scenario where several independent tests are conducted simultaneously. Even if each test maintains a low significance level (e.g., α = 0.05), the cumulative probability of observing at least one significant result by chance alone increases. This occurs as the number of tests increases. Researchers employ statistical correction techniques such as the Bonferroni correction to make it more difficult to reject the null hypothesis. These solutions will help maintain stringent control over the overall false positive rate. They ensure the probability of false positives across all tests remains below the specified threshold. P-value best practices Interpret results in context Consider the practical implications of your findings within the broader context of your research question or application. Avoid overinterpreting statistically significant results or dismissing nonsignificant results without careful consideration. Suppose you find a statistically significant improvement in test scores among students taught with a new method. This improvement is compared to those taught using the traditional method. You should avoid overinterpreting the results. Instead, consider factors such as the size of the effect. Is the score improvement substantial enough to justify implementing the new teaching method on a large scale? Would this finding be replicated in other studies with similar conditions? Are there other factors, such as cost, that need to be considered? Conversely, nonsignificant results could be due to other factors, such as small sample size or measurement error. Therefore, it's important to critically evaluate the study design, data quality, and potential sources of bias before drawing conclusions. Report all p-values Regardless of their significance, all p-values for all variables in a study should be included. This provides a comprehensive picture of the analysis. It enables readers to assess the robustness of the findings. By reporting all p-values, researchers convey the complete range of statistical analyses, including those with nonsignificant results. This transparency allows readers to evaluate the consistency and reliability of the findings across different variables and analyses. It also promotes integrity in research by presenting the data in its entirety, without bias or distortion. Be skeptical of small p-values Interpreting small p-values requires caution. They can sometimes be misleading indicators of the significance of observed effects. Recognizing that small p-values can arise from genuine effects and large sample sizes is crucial. Large sample sizes increase the statistical power to detect trivial deviations from the null hypothesis. Therefore, small p-values in studies with large sample sizes may not necessarily reflect meaningful or practically significant effects. P-value FAQs What is a z-score? A z-score, or standard score, measures how many standard deviations a data point is above or below the mean. A positive z-score indicates that the data point is above the mean. A negative z-score indicates that it is below the mean. What is a t-score? A t-score, or t-value, measures a data point's distance from a t-distribution's mean. It assesses whether the means of the two groups are significantly different from each other. A large t-score indicates that groups are different, while a small t-score indicates that groups are similar. What is a Pearson score? The Pearson correlation coefficient, often called the Pearson correlation or correlation coefficient, is a statistical measure. It quantifies the strength and direction of the linear relationship between two continuous variables. It evaluates how closely the data points in a scatterplot align along a straight line. What is chi-square score? The chi-square score, also known as the chi-square statistic, is a statistical test. It assesses the independence or association between categorical variables in a contingency table. It quantified the difference between the observed and expected frequencies under the assumption of independence. Discover more resources Explore our toolkits Discover our toolkits, designed to help you leverage feedback in your role or industry. Learn more SurveyMonkey 101: Sending your first survey Learn how to send your first survey with confidence Learn more What 100 billion questions taught us about asking better ones Learn how to ask better questions using a helpful framework Learn more How to analyze survey data in Excel Learn how to analyze survey data in Excel and gain insights with our quick and easy-to-follow guide. Learn more See how SurveyMonkey can power your curiosity Get startedView plans
17580	https://math.mit.edu/~poonen/notes02.pdf	18.02 LECTURE NOTES, FALL 2021 BJORN POONEN These are an approximation of what was covered in lecture. (Please clear your browser’s cache before reloading this ﬁle to make sure you are getting the current version.) Thursday, September 9 We do not live in a 1-dimensional world. And 3 dimensions is not enough either, since applications in engineering, machine learning, economics, etc. routinely involve quantities depending on lots of parameters. 18.02 provides the background for understanding functions with multivariable input and output. In Fall 2021 we are modernizing the content of 18.02, in particular to incorporate more linear algebra that is essential for many of the applications above. Time to turn on the ﬁrehose! 1. Vectors 1.1. Vector notation. A vector v in R3 is an ordered triple of real numbers. (In print, use bold face letters like v; in handwriting, use ⃗ v.) In 18.02, vectors are often1 written as column vectors like    2 3 5   ; Edwards & Penney write the same vector as ⟨2, 3, 5⟩. The 3-dimensional space R3 is the set of all such triples. (You can have vectors in R2 or R95786 too, if you want.) Geometrically, a vector is an arrow with a length and a direction; its position does not matter. The standard basis vectors for R3 are e1 :=    1 0 0   , e2 :=    0 1 0   , e3 :=    0 0 1   . (The symbol := means “is deﬁned to be”.) (Alternative notations: i, j, k, or ˆ x, ˆ y, ˆ z to indicate that these are in the directions of the x-, y-, and z-axes.2) Deﬁne 0 :=    0 0 0   . 1One reason is to facilitate matrix-vector multiplication later on. 2One advantage of the notation e1, e2, . . . is that it generalizes to arbitrary dimension. One disadvantage of i, j, k is that i can be confused with i := √−1. 1 If P is a point in space, then the position vector P is the vector pointing from (0, 0, 0) to P. (Alternative notations: OP or − → OP.) Sometimes we’ll think of the point (2, 3, 5) and its position vector    2 3 5   as being the same. The length of v =    a b c   is \|v\| := √ a2 + b2 + c2 (also called magnitude). This formula can be explained by using the Pythagorean theorem twice. A unit vector is a vector of length 1. 1.2. Vector operations. Addition: 3 1 ! + 1 4 ! = 4 5 ! . Subtraction: 1 4 ! − 3 1 ! = −2 3 ! . Geometrically: parallelogram law for +, triangle law for −: a b a + b A A B B − → AB = B −A Important: If A and B are two points, and A and B are their position vectors, then the vector from A to B is − → AB = B −A, because this is what you have to add to A to get to B. Scalar multiplication: −10 scalar 3 1 ! vector = −30 −10 ! vector . Scalar means number (use this word to emphasize that it is not a vector). Scalar multiplication is a scalar times a vector, and the result is a vector. Geometrically: cv has the same (or opposite) direction as v, but possibly a diﬀerent length. Two vectors v and w are parallel if one of them is a scalar multiple of the other. Problem 1.1. Fill in the blanks: 3 4 ! = positive scalar unit vector . 2 Solution: The positive scalar must be the length, 5. So the answer is 3 4 ! = 5 positive scalar 3/5 4/5 ! unit vector . In general, for any nonzero v: v = \|v\| length v \|v\| unit vector in the direction of v . Question 1.2. Does the zero vector 0 := 0 0 ! have a direction? It is best to say that it has every direction, and hence to say that it is parallel to every other vector, and perpendicular to every other vector. Another example, involving addition and scalar multiplication: 2 e1 + 3 e2 + 5 e3 linear combination of e1, e2, e3 =    2 3 5   . In general, a linear combination of vectors v1, . . . , vn is any vector obtained by multiplying the vectors by (possibly diﬀerent) scalars and adding the results, i.e., an expression of the form c1v1 + · · · + cnvn for some scalars c1, . . . , cn. 1.3. Solving geometry problems with vectors. Question 1.3. Suppose that M is the midpoint of segment AB. In terms of the position vectors A and B, what is the position vector M? A 1 2(B −A) O A M B To get to M from the origin, ﬁrst go to A and then go halfway from A to B. The position vector A gets you to A. The vector from A to B is B −A, so the vector from A to M is 1 2(B −A). Thus M = A + 1 2(B −A) = A + B 2 . Question 1.4. Explain why the midpoints of the sides of a space quadrilateral form a parallelogram. 3 How should one approach a problem like this? 1. Give variable names to the objects given in the problem. Let A, B, C, D be the vertices of the quadrilateral in order. Let A′ be the midpoint of AB, let B′ be the midpoint of BC, let C′ be the midpoint of CD, and let D′ be the midpoint of DA. Let A, B, C, D, A′, B′, C′, D′ be the corresponding position vectors. 2. Write down known equations relating the variables. We know that A′ = A + B 2 , B′ = B + C 2 , C′ = C + D 2 , D′ = D + A 2 . 3. See if these equations imply the desired conclusion. We compute B′ −A′ = C −A 2 C′ −D′ = C −A 2 . Thus segments B′A′ and C′D′ are parallel and have the same length. This means that A′B′C′D′ is a parallelogram. 1.4. Some advice for success in 18.02. • Read the Information pages on Canvas. • Reading assignments are posted on Canvas. Do the reading before lecture! • Come to oﬃce hours! (Oﬃce hours generally consist of a small group of students discussing additional examples not covered in lecture or recitation, asking questions, getting started on diﬃcult homework problems, etc. The recitation leaders for this class are some of the best math postdocs and grad students worldwide, and oﬃce hours are your best chance to learn from them.) • Homework: – It’s long and has diﬃcult problems, so start now! – The problems indicate the date after which you should have the knowledge to do them. – Work together in groups! It’s OK if other people tell you how to solve a problem, but don’t be looking at their solution as you write your own. – Do what it takes (come to oﬃce hours, discuss problems with others) so that when you submit an assignment you are pretty sure that it is complete and correct. 4 1.5. Dot product. Do we multiply vectors coordinate-wise? No! Why not? This does not give a notion with useful geometric meaning. Instead: Dot product (also called scalar product or inner product):    2 3 5    vector ·    7 8 9    vector = (2)(7) + (3)(8) + (5)(9) = 14 + 24 + 45 = 83 scalar. Important special case: a · a = \|a\|2. Theorem 1.5 (Geometric interpretation of the dot product). If θ is the angle between nonzero vectors a and b, then a · b = \|a\| \|b\| cos θ. (If a or b is 0, then θ can be taken to be any real number, and the formula still holds, with both sides being 0.) a b a −b θ Why? Proof (=explanation). Let a = \|a\|, b = \|b\|, c = \|a −b\|. Then c2 = (a −b) · (a −b) = a · a −a · b −b · a + b · b (since · satisﬁes the distributive law) = a2 + b2 −2 a · b. On the other hand, the law of cosines says c2 = a2 + b2 −2ab cos θ. Comparing shows that a · b = ab cos θ. □ Corollary 1.6. Vectors a and b are perpendicular ⇐ ⇒a · b = 0. 5 1.6. Scalar component of a vector in the direction of another vector. Suppose that a is a vector, and b is a nonzero vector. The question we want to answer is “How much of a is in the direction of b?” Start by dropping a perpendicular from a to the line spanned by b. This gives a decomposition a = p parallel to b + q perpendicular to b . Then p is in the same direction as in b b a p q θ or p is in the opposite direction b a p q θ or p = 0. The vector p is called the projection of a onto the line spanned by b. The vector p is also called the vector component of a in the direction of b. To get the scalar component, take its length with an appropriate sign: Deﬁnition 1.7. The scalar component of a in the direction of b is compb a := ±\|p\|, where the sign is + or −according to whether p is in the same or opposite direction as b. (If p = 0, either sign gives 0.) Although p is a vector, compb a is a scalar, as the name suggests! Another geometric interpretation: Set up a new coordinate system in which the x-axis is in the direction of b; then compb a is the new x-coordinate of a. Computing the scalar component: The diagrams show that compb a = \|a\| cos θ. (1) 6 But the easiest way to compute compb a is usually to dot a with the direction unit vector b \|b\|: compb a = a · b \|b\| = a · b \|b\| . (This is equivalent to formula (1), since a · b = \|a\| \|b\| cos θ.) Example 1.8. compe2    −2 −3 −5   =    −2 −3 −5   · e2 = −3. (In this example, the b was already a unit vector, so there was no need to compute b \|b\|.) 2. Matrices 2.1. Matrix notation. Deﬁnition 2.1. An m × n matrix is a rectangular array of numbers with m rows and n columns. Example: A :=      3 5 π 4 0 0 6 7 9 1 −2 3      is a 4 × 3 matrix. (The dimensions are always given in the order “height × width”.) The notation aij means the number in the ith row and jth column. In the example above, a32 = 7. (One could write a3,2 but people often omit the comma.) Two matrices A and B are equal if A has the same dimensions as B and aij = bij for all i and j. An m × 1 matrix is the same as a (column) vector in Rm. Example:    2 3 5   ∈R3. 2.2. Matrix times a vector. Example 2.2. The product 1 2 3 2 3 5 ! matrix    100 1 10    vector 7 is deﬁned by taking the dot product of each matrix row with the vector: ⟨1, 2, 3⟩·    100 1 10   = 132 ⟨2, 3, 5⟩·    100 1 10   = 253, so the ﬁnal result is 1 2 3 2 3 5 ! matrix    100 1 10    vector = 132 253 ! vector . In general, a matrix-vector product Ax is deﬁned when the dot products are deﬁned, which is when the width of A equals the height of x. Example 2.3. 6 7 8 2 3 5 !    1 0 0   = 6 2 ! . In general: Ae1 = (1st column of A) , Ae2 = (2nd column of A) , and so on. Example 2.4. 6 7 8 2 3 5 !    x y z   = 6x + 7y + 8z 2x + 3y + 5z ! = x 6 2 ! + y 7 3 ! + z 8 5 ! . In general: Ax = (some linear combination of the columns of A) . 2.3. Matrix operations. Addition, subtraction, and scalar multiplication are deﬁned entrywise, as for vectors. (For addition and subtraction, the two matrices have to have the same dimensions.) Transpose: 2 3 5 7 11 13 !T =    2 7 3 11 5 13   . Each row of the original matrix corresponds to a column of the transpose matrix. The ij-entry of A equals the ji-entry of AT. Multiplication: If A is m × n and B is n × p, then AB is the m × p matrix whose ij-entry is the dot product (ith row of A) · (jth column of B). 8 (If the second dimension of A does not match the ﬁrst dimension of B, then the dot products are not deﬁned, so AB is not deﬁned.) Example 2.5. Multiplying a 2 × 3 matrix by a 3 × 3 matrix gives a 2 × 3 matrix: 2 3 7 0 5 1 !    10 0 9 1 2 0 0 1 0   = 23 13 18 5 11 0 ! . For example, the 2, 1 entry is computed as follows: ⟨0, 5, 1⟩ 2nd row ·    10 1 0    1st column = 0(10) + 5(1) + 1(0) = 5 2,1 entry. Warning 2.6. Even when AB and BA both make sense, they might be unequal. (In other words, matrix multiplication is not commutative.) Friday, September 10 2.4. Determinants. To each square matrix A is associated a number called its determinant. det a := a = ± length of segment in the real line R determined by a det a b c d ! := ad −bc = ± area of parallelogram in R2 formed by ⟨a, b⟩and ⟨c, d⟩ (the sign is + if ⟨c, d⟩is counterclockwise from ⟨a, b⟩) det    a1 a2 a3 b1 b2 b3 c1 c2 c3   := a1b2c3 + a2b3c1 + a3b1c2 −c1b2a3 −c2b3a1 −c3b1a2 = ± volume of parallelepiped in R3 formed by the rows a, b, c (the sign is + if a, b, c agree with the right hand rule) (In the formula for the 3 × 3 determinant, each + term is the product along a “southeast” diagonal with wraparound and each −term is the product along a “northeast” diagonal with wraparound.) The geometric interpretations are true also for the segment/parallelogram/parallelepiped formed by the columns instead of the rows. 9 Alternative notation for determinant: \|A\| = a b c d . (This is a scalar, not a matrix!) Laplace expansion (along the ﬁrst row) for a 3 × 3 determinant: a1 a2 a3 b1 b2 b3 c1 c2 c3 = +a1 b2 b3 c2 c3 −a2 b1 b3 c1 c3 + a3 b1 b2 c1 c2 . The general rule leading to the formula above is this: (1) Move your ﬁnger along the entries in a row. (2) At each position, compute the minor, deﬁned as the smaller determinant obtained by crossing out the row and the column through your ﬁnger; then multiply the minor by the number you are pointing at, and adjust the sign according to the checkerboard pattern + − + − + − + − + (the pattern always starts with + in the upper left corner). (3) Add up the results. There is a similar expansion for a determinant of any size, computed along any row or column. Properties of determinants: D-1: Interchanging two rows changes the sign of det A. D-2: If one of the rows is all 0, then det A = 0. D-3: Multiplying an entire row by a scalar c multiples det A by c. D-4: Adding a multiple of a row to another row does not change det A. These properties can all be interpreted geometrically. The same properties hold for column operations. Example 2.7. Let A := 2 3 5 7 ! . If we add 10 times the ﬁrst row to the second row (and leave the ﬁrst row unchanged), we get a new matrix B := 2 3 25 37 ! . Property D-4 says that det B = det A. Question 2.8. Suppose that A is a 3 × 3 matrix such that det A = 5. Doubling every entry of A gives a matrix 2A. What is det(2A)? 10 Solution. Each time we multiply a row by 2, the determinant gets multiplied by 2. We need to do this three times to double the whole matrix A, so the determinant gets multiplied by 2 · 2 · 2 = 8. Thus det(2A) = 8 det(A) = 40. □ 2.5. Cross product. 2.5.1. Deﬁnition. Deﬁnition 2.9 (Cross product, also called vector product). The cross product of vectors a and b in R3 is another vector in R3: a × b := e1 e2 e3 a1 a2 a3 b1 b2 b3 := + a2 a3 b2 b3 e1 − a1 a3 b1 b3 e2 + a1 a2 b1 b2 e3 = D a2b3 −a3b2, a3b1 −a1b3, a1b2 −a2b1 E . The cross product is deﬁned for vectors in R3 only! 2.5.2. Geometric interpretation of the cross product. If a and b are nonzero vectors forming an angle θ, then a × b is the vector determined by the following three conditions: • It is perpendicular to a and b. • Its length is \|a\|\|b\| sin θ (which equals the area of the parallelogram formed by a and b). • Its direction is given by the right hand rule: if you point the ﬁngers of your right hand in the direction of a so that bending your ﬁngers makes them point in the direction of b, then your thumb shows the direction of a × b. (We won’t explain in 18.02 why the determinant and cross product have the geometric interpretations claimed.) 2.5.3. Properties of cross products. For vectors a and b, a and b are parallel ⇐ ⇒a × b = 0 (these happen when θ is 0 or π) a × a = 0 b × a = −a × b. 2.5.4. Cross products involving e1, e2, e3. Problem 2.10. What is e1 × e3? Solution. It is a vector perpendicular to e1 and e3, and its length is (1)(1) sin π 2 = 1, so it must be e2 or −e2. The right hand rule tells you which: e1 × e3 = −e2. □ 11 If you like, you can use the following mnemonic: Start by drawing e1 e2 e3 The cross product of any one of these with the next one is the third vector in the sequence: for example, e3 × e1 = e2. But if the two input vectors are in the order opposite to the order speciﬁed by the arrows, then insert a negative sign: e1 × e3 = −e2. (And if the two input vectors are the same, the cross product is 0.) 3. Matrices as linear transformations 3.1. Functions with vector input and/or vector output. Scalar input, scalar output: Alternative notation for the function f(x) := sin x: f : R − →R x 7− →sin x. The ﬁrst line speciﬁes that • the set of allowable inputs is R (the domain), and • every output is an element of R (the codomain). The second line shows a typical input x in the domain, and shows which element of the codomain it is mapped to. (Note: The codomain speciﬁes only the type of output; every output is an element of the codomain, but there might be other elements of the codomain that are not outputs. In contrast, the range of f (also called the image of f) is the set of all actual outputs of f, which in this example is the interval [−1, 1]. In general, the range is a subset of the codomain.) Scalar input, vector output (vector-valued function): r: R − →R2 t 7− → cos t sin t ! . (We use a bold letter r instead of r since its values are vectors.) For example, r(π/3) = 1/2 √ 3/2 ! . 12 Vector input, scalar output (multivariable function): f : R3 − →R    x y z   7− →x2 + y2 −z2. It could also be written f(x, y, z) := x2 + y2 −z2. For example, f(1, 2, 3) = 1 + 4 −9 = −4. Vector input, vector output: Can we have functions with vector input and vector output? Yes! See Example 3.1 below. 3.2. Going from a matrix to a linear transformation. Warm-up: Given a number, say 3, we get a function f : R − →R x 7− →3x that multiplies each input by 3. What is the higher-dimensional analogue? Example 3.1. Given the 2 × 3 matrix 6 7 8 2 3 5 ! , we get a function f : R3 − →R2 x 7− → 6 7 8 2 3 5 ! x with vector input and vector output! Explicitly, f    x y z   = 6x + 7y + 8z 2x + 3y + 5z ! . Each coordinate of the output is a linear combination of the input variables. What does this function do to e2? Answer: f(e2) = f    0 1 0   = 7 3 ! . In general, an m × n matrix A gives rise to a function f : Rn − →Rm x 7− →Ax (note that m and n get reversed). Functions arising in this way are called linear transformations. Sometimes we use the matrix A itself instead of f to denote the linear transformation. 13 3.3. Depicting a linear transformation with a domain-codomain diagram. 3.3.1. A single-variable function. To visualize a function like f(x) := 3x, we would usually draw its graph in R2, the line y = 3x. But there is another way: Draw the domain and the codomain (two copies of the real line), and show what certain features in the domain get transformed to: 0 0 1 3 2 6 f domain R codomain R For example, f maps the point 2 to the point 6. The diagram shows how f expands everything by a factor of 3. 3.3.2. Higher-dimensional analogue. Consider the matrix 2 0 0 1 ! and the associated linear transformation f : R2 − →R2 x y ! 7− → 2 0 0 1 ! x y ! = 2x y ! . Drawing a graph of f would require 4 dimensions (2 for the input, 2 fo the output), so instead let’s draw a domain-codomain diagram. How does f transform the van Gogh unit smiley? Tuesday, September 14 2 0 0 1 ! 14 For example, the ear at −1 0 ! is mapped to −2 0 ! . The linear transformation f stretches in the horizontal direction only. 3.4. Going from a linear transformation to a matrix. Given a linear transformation f, how do we reconstruct the matrix A? Answer 1: If you know a formula for f, just read oﬀthe entries of the matrix. For example, if f    x y z   = 6x + 7y 2x + 5z ! , then A = 6 7 0 2 0 5 ! . Answer 2: If you know only a geometric description of the linear transformation f, then get A as the matrix whose columns are f(e1), f(e2), etc. Here is an example illustrating Answer 2: Question 3.2. Given θ, there is a 2 × 2 matrix R that rotates each vector in R2 counter-clockwise by the angle θ. What is it? Solution. ⟨cos θ, sin θ⟩ ⟨−sin θ, cos θ⟩ θ e1 e2 R Thus (ﬁrst column of R) = Re1 = cos θ sin θ ! (second column of R) = Re2 = −sin θ cos θ ! , so R = cos θ −sin θ sin θ cos θ ! . □ 15 3.5. Area scaling factor. Example 3.3. What does the dilation given by 3 0 0 3 ! do to area? (Hint: Consider what it does to a 1 × 1 square.) Answer: It multiplies area by 9. Notice that det 3 0 0 3 ! = 9 too. In general, for any 2 × 2 matrix A, the associated linear transformation has area scaling factor = \|det A\|. The area scaling factor is always nonnegative, while det A could be negative, so it is necessary to take the absolute value. Why is the formula correct? If A = a b c d ! , then the square formed by 1 0 ! and 0 1 ! is transformed into the parallelogram formed by a c ! and b d ! ; the square has area 1 and the parallelogram has area \|det A\|. 3.6. Composition. Matrix multiplication corresponds to composition of the linear transfor-mations. More explicitly, if • f : Rn →Rm corresponds to a matrix A, and • g: Rp →Rn corresponds to a matrix B, then the composition Rp g / f◦g 5 Rn f / Rm x / f(g(x)) corresponds to the matrix product AB. In other words, if you multiply x by B (apply the inner function g ﬁrst!) and then multiply the result by A, you get the same output vector as if you multiplied x by AB: A(Bx) = (AB)x. Matrix multiplication is deﬁned as it is to make this true. Example 3.4. Let α and β be any two angles. Let Rα be the 2 × 2 rotation matrix for α, and so on. Then Rα Rβ = Rα+β 16 because rotating by β and then rotating by α has the same eﬀect as rotating by α + β. (In this example the order did not matter, but in other examples it does.) What famous formulas do you get if you compare entries in this identity? 3.7. Identity matrix. The identity transformation f : R3 − →R3    x y z   7− →    x y z   =    1x + 0y + 0z 0x + 1y + 0z 0x + 0y + 1z   . is associated to the 3 × 3 identity matrix I :=    1 0 0 0 1 0 0 0 1   . (You can guess what the 4 × 4 identity matrix looks like. The positions of the 1s form what is called the diagonal of the matrix.) The matrix I acts like the number 1: IA = A and AI = A whenever the multiplication makes sense. 3.8. Inverse matrices. 3.8.1. Motivation: solving linear systems. To solve 3x = 5, multiply both sides by 3−1. Similarly, one way to solve 2x1 + 3x2 = 4 4x1 + 5x2 = 6, is to rewrite as 2 3 4 5 ! x1 x2 ! = 4 6 ! , which has the shape Ax = b, and left multiply both sides by A−1 to get x = A−1b. 3.8.2. Deﬁnition. Deﬁnition 3.5. The inverse of an n × n matrix A is another n × n matrix, A−1, such that AA−1 = I and A−1A = I. It exists if and only if det A ̸= 0; in that case, A is called invertible, or nonsingular. If A corresponds to the linear transformation f, then A−1 corresponds to the inverse function f −1 (if it exists). For example, the inverse of the rotation matrix Rθ is R−θ. 17 3.8.3. Formula for 2 × 2 matrices. Theorem 3.6. Let A = a b c d ! . • If det A ̸= 0, then A−1 = 1 det A d −b −c a ! . • If det A = 0, then A−1 does not exist. For square matrices larger than 2×2, there is still a formula, but it is much more complicated (see M.2), and there are better algorithms for computing A−1, and anyway one would typically use a computer for this. 3.9. Equations of planes. Question 3.7. What does the set of vectors perpendicular to ⟨1, 2, 3⟩look like? Solution. It’s a plane through the origin. Its equation is ⟨1, 2, 3⟩· ⟨x, y, z⟩= 0, which is x + 2y + 3z = 0. □ The vector n := ⟨1, 2, 3⟩is called a normal vector to the plane. (Normal is another word for perpendicular.) Question 3.8. What is the plane with normal vector ⟨1, 2, 3⟩passing through (4, 5, 6)? Solution. A point (x, y, z) lies on this plane if the vector from (4, 5, 6) to (x, y, z) (not the position vector of (x, y, z)!) is perpendicular to ⟨1, 2, 3⟩, so its equation is ⟨1, 2, 3⟩· (⟨x, y, z⟩−⟨4, 5, 6⟩) = 0, which is (x −4) + 2(y −5) + 3(z −6) = 0. □ Question 3.9. (Followup question) What is the distance from (2, 3, 5) to that plane? Solution. If we choose any point on the plane, say (4, 5, 6), and form the vector v := ⟨2, 3, 5⟩−⟨4, 5, 6⟩= ⟨−2, −2, −1⟩ between the two points, then the desired distance is not the length of v, because the straight line path from (2, 3, 5) to (4, 5, 6) is not the shortest path from (2, 3, 5) to the plane. Instead we want “the amount of v in the direction parallel to the normal vector n := ⟨1, 2, 3⟩”, taking 18 the absolute value if necessary. Thus the desired distance is the absolute value of the scalar component of v in the direction of n. That scalar component is compn v = v · n \|n\| = ⟨−2, −2, −1⟩· ⟨1, 2, 3⟩ √ 12 + 22 + 32 = −2 −4 −3 √ 14 = −9 √ 14, so the distance is 9/ √ 14. □ Question 3.10. Are the vector ⟨−5, 1, 1⟩and the plane x + 2y + 3z = 6 • parallel, • perpendicular, • both, • or neither? Hint: ⟨−5, 1, 1⟩· ⟨1, 2, 3⟩= 0. Solution. The vector ⟨−5, 1, 1⟩is perpendicular not to the plane, but to a normal vector of the plane. So ⟨−5, 1, 1⟩is parallel to the plane. The vectors perpendicular to the plane are the scalar multiples of the normal vector, so ⟨−5, 1, 1⟩is not like this. So the answer is that the vector and plane are parallel. □ Thursday, September 16 4. Linear algebra 4.1. Linear combinations. Suppose that you have a list of vectors. To get a linear combi-nation of them, 1. multiply each vector by a scalar and 2. add the results. vectors typical linear combination v −3v v, w 2v + (−5)w v, w, x 4v + 0w + 7x 19 4.2. Span. Deﬁnition 4.1. Given one vector v, the span of v is the set of all linear combinations of v, which is the set of all scalar multiples of v: Span(v) := {all vectors cv, where c ranges over all real numbers}. Example 4.2. Suppose that v = 2 1 ! . What are some scalar multiples of v? Well, there’s 2 2 1 ! = 4 2 ! , 3 2 1 ! = 6 3 ! , −1 2 2 1 ! = −1 −1/2 ! , and so on. Then Span(v) is an inﬁnite set whose elements are all of these vectors. Geometric interpretation: Think of each of these vectors as a point in the plane: (4, 2), (6, 3), (−1, −1/2), and so on; then Span(v) is the set of all these points. So Span(v) is a line through the origin, speciﬁcally, the line y = 1 2x. Deﬁnition 4.3. Given two vectors, v and w, the span of v and w is the set of all linear combinations of v and w: Span(v, w) := {all vectors c1v + c2w, where c1 and c2 range over all pairs of real numbers}. Example 4.4. Consider e1 =    1 0 0   and e2 =    0 1 0   . What are some linear combinations of e1 and e2? Well, there’s 3    1 0 0   + 4    0 1 0   =    3 4 0   , 5    1 0 0   + (−7)    0 1 0   =    5 −7 0   , and so on. Then Span(e1, e2) is the set of all such vectors: this is the set of all the vectors in R3 whose 3rd coordinate is 0. Geometric interpretation: Span(e1, e2) is the xy-plane. It is deﬁned by the equation z = 0. 20 Question 4.5. Is the span of two vectors in R3 always a plane? Answer: Usually it is, but not always. For example, the span of    1 2 3   and    2 4 6   is a line because the linear combinations are all scalar multiples of the ﬁrst vector. An even more degenerate case: if both vectors are 0, then their span is the set {0} containing only the one vector 0. Problem 4.6. Let a =    2 1 0   and b =    0 1 3   . What is the equation of the plane Span(a, b)? Solution. To give the equation of a plane through the origin, all we need is a normal vector. It needs to be perpendicular to both a and b. One possible normal vector is a × b =    3 −6 2   . The equation of the plane is then    3 −6 2   · x = 0, which is 3x −6y + 2z = 0. (To check the answer, verify that a and b satisfy the equation.) □ One can also take the span of more than two vectors: again it is deﬁned to be the set of all linear combinations that one can form with the given vectors. Example 4.7. Span(e1, e2, e3) is the whole space R3. 4.3. Linearly dependent vectors. Deﬁnition 4.8. Vectors v1, . . . , vn are linearly dependent if one of them is a linear combination of the others. Equivalent deﬁnition: Vectors v1, . . . , vn are linearly dependent if there exist scalars c1, . . . , cn not all zero such that c1v1 + · · · + cnvn = 0. Example 4.9. The vectors 2 1 ! and 6 3 ! are linearly dependent because the second is a scalar multiple of the ﬁrst: 6 3 ! = 3 2 1 ! . 21 According to the equivalent deﬁnition, they are linearly dependent because 3 2 1 ! + (−1) 2 1 ! = 0. The span of 2 1 ! and 6 3 ! is only a line. 6 3 ! 2 1 ! Example 4.10. The vectors 2 1 ! and 1 3 ! are linearly independent. Their span is the whole plane R2. 1 3 ! 2 1 ! In general, dim Span(v1, . . . , vn) is    n, if v1, . . . , vn are linearly independent, less than n, if v1, . . . , vn are linearly dependent. (We haven’t deﬁned dimension mathematically, but you probably have an intuitive sense of what it is.) Example 4.11. The vectors    1 1 1   ,    2 3 5   ,    4 5 7   , are linearly dependent since the 3rd vector is a linear combination of the ﬁrst two:    4 5 7   = 2    1 1 1   + 1    2 3 5   . 22 The span of the three vectors is the same as the span of the ﬁrst two, which is a plane, only 2-dimensional. Example 4.12. The vectors e1, e2, e3 are linearly independent and their span is the whole space R3, which is 3-dimensional. 4.4. Basis. Deﬁnition 4.13. A list of vectors is a basis of R2 if 1. the span of the vectors is the whole plane R2, and 2. they are linearly independent. Example 4.14. 2 1 ! , 1 3 ! is a basis of R2. Every basis of R2 has exactly 2 vectors, as we’ll explain later. Given a basis b1, b2, we can form the 2 × 2 basechange matrix B := b1 b2 whose columns are b1 and b2. In the example, B = 2 1 1 3 ! . 4.5. Coordinates with respect to a basis. A basis b1, b2 sets up a new coordinate system on R2 (this is why bases are important). Speciﬁcally, given any vector v in R2, the coordinates of v with respect to the basis b1, b2 are the numbers c1, c2 such that c1b1 + c2b2 = v. Problem 4.15. Find the coordinates of 7 6 ! with respect to the basis 2 1 ! , 1 3 ! . Solution. We need to ﬁnd numbers c1, c2 such that c1 2 1 ! + c2 1 3 ! = 7 6 ! . Solving the resulting system 2c1 + c2 = 7 c1 + 3c2 = 6 yields c1 = 3 and c2 = 1. □ The two conditions in the deﬁnition of basis are what guarantee that the coordinates exist and are unique: • First, since Span(b1, b2) = R2, every vector v in R2 is a linear combination c1b1+c2b2. 23 • If some v could be so expressed in two diﬀerent ways, say, 10b1 + 40b2 = v = 2b1 + 3b2 then subtracting would give 8b1 + 37b2 = 0, which is impossible since b1 and b2 are linearly independent. 4.6. Changing coordinates. The plane R2 has its original coordinate system given by e1, e2 and the new coordinate system given by b1, b2. Any given point can have coordinates (x1, x2) in the original system and diﬀerent coordinates (c1, c2) in the new system: x1e1 + x2e2 = c1b1 + c2b2. Let’s rewrite the equation in terms of x = x1 x2 ! , B := b1 b2 , and c = c1 c2 ! : x = Bc. This shows how, given c, to get x. Because it is also possible to convert the other way, B is automatically invertible, and c = B−1x. Friday, September 17 4.7. Basis of Rn. All the statements about R2 have analogues for Rn. For a list of vectors in Rn, the small-text label on each sloped equality in dimension of the span linearly independent span is Rn number of vectors n is equivalent to the equality holding. If any two of the three equalities hold, then so does the third. Conclusion: To check that a list of vectors is a basis of Rn, it is enough to check any two of the following three conditions: • the span of the vectors is the whole space Rn, • the vectors are linearly independent, • the number of vectors is n; 24 then the other one will hold too. (The deﬁnition of basis is that the ﬁrst two hold.) Another way to test whether vectors b1, . . . , bn ∈Rn form a basis: 1. Form the matrix B whose columns are b1, . . . , bn. 2. Calculate det B. 3. Then b1, . . . , bn is a basis of Rn ⇐ ⇒det B ̸= 0 4.8. Square systems of linear equations. A square system of linear equations is one with the same number of equations as variables. 4.8.1. Homogeneous systems. The system x + 2y + 3z = 0 8x + 4z = 0 7x + 6y + 5z = 0 is called homogeneous3 since all the right sides are 0. In matrix form, it is Ax = 0 for some square matrix A. 4.8.2. Inhomogeneous systems. The system x + 2y + 3z = 0 8x + 4z = 37 7x + 6y + 5z = 23 is inhomogeneous since some of the right sides are nonzero. It has the form Ax = b for some square matrix A and column vector b. 4.8.3. Behavior of solution set. What are the solutions to a square system? The behavior depends on whether the system is homogeneous, and on whether det A is nonzero: if det A ̸= 0 if det A = 0 homogeneous system Ax = 0 only x = 0 inﬁnitely many inhomogeneous system Ax = b only x = A−1b inﬁnitely many or none In the det A ̸= 0 column, why is there only one solution? Answer: If det A ̸= 0, then A−1 exists, so the system can be solved for x by multiplying by A−1. 3Both of the letters e in homogeneous are pronounced ee, and the stress is on the third syllable! 25 Example 4.16. The homogeneous system x + 2y = 0 2x + 4y = 0 has inﬁnitely many solutions (they form a line). Example 4.17. The inhomogeneous system x + 2y = 3 2x + 4y = 6 has inﬁnitely many solutions (again a line). Example 4.18. The inhomogeneous system x + 2y = 3 2x + 4y = 7 has no solutions. 4.8.4. Geometric interpretation in the 3 × 3 case. For any 3 × 3 matrix A, the solution set to the homogeneous system Ax = 0 is the intersection of 3 planes passing through 0 (assuming that no row of A is all 0). • If det A ̸= 0, the intersection contains only 0 (this is what usually happens). • If det A = 0, the intersection is either a line or a plane (through 0). For a 3 × 3 inhomogeneous system Ax = b, the solution set is again the intersection of 3 planes, but they may be shifted away from 0. 5. Parametric lines and curves 5.1. Lines. Two ways to describe lines in R3: • intersection of two planes • parametric equation (to be introduced now) Think of the trajectory of an airplane moving at constant velocity. Let r0 be the position vector of the airplane at time t = 0. Let v be the velocity. Where is the airplane at time t = 1? Answer: r0 + v. At time t = 2? Answer: r0 + 2v. In general, at time t the airplane is at r(t) := r0 + tv. 26 t 0 1 2 airplane v v r(0) = r0 r(1) = r0 + v r(2) = r0 + 2v r domain R codomain R3 This is a function with scalar input, vector output: r: R − →R3 t 7− →r0 + tv. Each input real number t gives one output point r(t) on the line, and as t varies, these points trace out the whole line. (The codomain is R3, but the range is a line in R3.) The input variable t is called the parameter. Problem 5.1. Parametrize the line L through (1, 2, 3) and (4, 1, 3). Solution. Use initial position vector r0 :=    1 2 3   and velocity vector v :=    4 1 3   −    1 2 3   =    3 −1 0   so that at time t = 1 the point reaches (4, 1, 3). So L is given by r :=    1 2 3   + t    3 −1 0   . Another way to express the answer is to give the parametric equations of L in terms of coordinate functions: x = 1 + 3t, y = 2 −t, z = 3. □ Question 5.2. The lines x = 1 + 3t, y = 2 −t, z = 3 and x = 2t, y = −1 + t, z = 1 + t 27 (1) are the same, (2) are parallel, (3) intersect in one point, or (4) are skew (i.e., do not intersect, but are not parallel either)? Answer. The velocity vectors    3 −1 0   and    2 1 1   are not scalar multiples of each other, so the lines are not the same, and are not parallel. They intersect if the system 1 + 3t = 2s 2 −t = −1 + s 3 = 1 + s. is solvable. Why did we use a diﬀerent variable on the right side? Imagine airplanes moving along the lines. If we used the same t on both sides, a solution would be a time when both airplanes are at the same place. If we use t on the left and s on the right, a solution would mean that airplane #1 at some time is at the same point as airplane #2 at a possibly diﬀerent time, meaning that their paths still cross, and this is what we’re trying to test! OK, let’s now solve the system. The last equation implies s = 2. The ﬁrst equation then implies t = 1. These values make all three equations true. So the lines intersect, at the point r1(1) = r2(2) =    4 1 3   . □ 5.2. Parametric equations of curves. Question 5.3. As t ranges through all real numbers, x = 2 cos t, y = sin t describes 1) A circle 2) An ellipse 3) A line of slope 1/2 4) A point. Answer. Without the 2, it would be the unit circle. The 2 stretches it in the x-direction, to make an ellipse. It is given by the implicit equation x2 + 4y2 = 4. To ﬁnd the implicit equation, one must eliminate t from the parametric equations; how to do this depends on the shape of the parametric equations, and may require some guesswork. In 28 this problem, we know that sin2 t + cos2 t = 1, and this can be rewritten as y2 + (x/2)2 = 1 in which t does not appear, which is equivalent to x2 + 4y2 = 4. □ The parametric equations and the implicit equation are completely diﬀerent ways of describing the same curve: the parametric equations tell you the curve one point at a time, one point for each value of t; the implicit equation gives you a test for when a given point in R2 lies on the curve. Slope of tangent line to this ellipse at a given time? Use the chain rule dy dt known = dy dx ??? dx dt known to get y′ = dy dx = dy/dt dx/dt = cos t −2 sin t = −1 2 cot t. The following question was not discussed in lecture. What is d2y dx2? It’s d2y dx2 = dy′ dx = dy′/dt dx/dt = 1 2 csc2 t −2 sin t = − 1 4 sin3 t. 5.3. Parametric equations of curves: an example. This example will be done in recita-tion on Monday, September 20. Problem 5.4 (Supplementary notes, Exercise 1I-4). A roll of plastic tape of outer radius a is held in a ﬁxed position while the tape is being unwound counterclockwise. The end P of the unwound tape is always held so that the unwound portion is perpendicular to the roll. Taking the center of the roll to be the origin O, and the end P to be initially at (a, 0), write parametric equations for the motion of P. P = (a, 0) Initial position aθ a θ P After unwinding θ radians 29 Solution. Parameter: The radian measure θ of the amount of tape unwound so far. The length of tape unwound so far is aθ. The position of the point where the unwound tape meets the roll: (a cos θ, a sin θ). Final answer: r = ⟨a cos θ, a sin θ⟩+ aθ⟨cos θ, sin θ⟩ = ⟨a(1 + θ) cos θ, a(1 + θ) sin θ⟩. □ Tuesday, September 21 6. Eigenvalues and eigenvectors 6.1. Introduction. Recall that the matrix 2 0 0 1 ! stretches in the horizontal direction by a factor of 2. 2 0 0 1 ! In particular, 2 0 0 1 ! e1 eigenvector = 2 eigenvaluee1. The stretching factor (here 2) is called an eigenvalue, and the vector in the stretching direction (here e1) is called an eigenvector. Deﬁnition 6.1. Suppose that A is a square matrix. • An eigenvalue of A is a scalar λ such that Av = λv for some nonzero vector v. (The scalar λ may be 0.) • An eigenvector of A associated to a given λ is a vector v such that Av = λv. (Warning: Some authors require an eigenvector to be not the zero vector.) 30 Try the “Matrix Vector” mathlet Problem 6.2. Let A = 1 −2 −1 0 ! and let v = 2 −1 ! . Is v an eigenvector of A? Solution. The calculation Av = 1 −2 −1 0 ! 2 −1 ! = 4 −2 ! = 2v, shows that v is an eigenvector, and that the associated eigenvalue is 2. □ Given A, how can we ﬁnd the eigenvalues and eigenvectors? For this we need some more concepts of linear algebra. 6.2. Trace. Deﬁnition 6.3. The trace of an n × n matrix A is the sum of the entries along the diagonal: tr A := a11 + a22 + · · · + ann Example 6.4. If A =    4 6 9 1 7 8 2 3 5   , then tr A = 4 + 7 + 5 = 16. 6.3. Characteristic polynomial. Use λ to denote a scalar-valued variable. Deﬁnition 6.5. The characteristic polynomial of an n × n matrix A is det(λI −A). The reason for this deﬁnition will be clear in the next section when we show how to compute eigenvalues. Remark 6.6. We often calculate the characteristic polynomial using det(A−λI) instead. This turns out to be the same as det(λI −A), except negated when n is odd. (The reason is that changing the signs of all n rows of the matrix A −λI ﬂips the sign of the determinant n times.) Usually we care only about the roots of the polynomial, so negating the whole polynomial doesn’t make a diﬀerence. In any case, det(A −λI) = det(λI −A) for 2 × 2 matrices (since 2 is even). Problem 6.7. What is the characteristic polynomial of A := 7 2 3 5 ! ? Solution. We have A −λI = 7 2 3 5 ! − λ 0 0 λ ! = 7 −λ 2 3 5 −λ ! det(A −λI) = (7 −λ)(5 −λ) −2(3) = λ2 −12λ + 29 . □ 31 Here is a shortcut for 2 × 2 matrices: Theorem 6.8. If A is a 2 × 2 matrix, then the characteristic polynomial of A is λ2 −(tr A)λ + (det A). Proof. Write A = a b c d ! . Then the characteristic polynomial is det(A −λI) = det a −λ b c d −λ ! = (a −λ)(d −λ) −bc = λ2 −(a + d)λ + (ad −bc) = λ2 −(tr A)λ + (det A). □ We can solve Problem 6.7 again, using this shortcut: the matrix A := 7 2 3 5 ! has tr A = 12 and det A = 29, so the characteristic polynomial of A is λ2 −12λ + 29. For an n × n matrix, the characteristic polynomial is a degree n polynomial in the variable λ and its leading coeﬃcient is 1, so the polynomial looks like λn + . . . . Remark 6.9. Suppose that n > 2. Then, for an n × n matrix A, the characteristic polynomial has the form λn −(tr A)λn−1 + · · · ± det A where the ± is + if n is even, and −if n is odd. So knowing tr A and det A determines some of the coeﬃcients of the characteristic polynomial, but not all of them. 6.4. Computing all the eigenvalues. Warm-up problem: Given a square matrix A, how can we test if 5 is an eigenvalue? Solution. The following are equivalent: • 5 is an eigenvalue. • There exists a nonzero solution to Av = 5v 5v −Av = 0 5Iv −Av = 0 (5I −A)v = 0. • det(5I −A) = 0. • Evaluating the characteristic polynomial det(λI −A) at 5 gives 0. 32 • 5 is a root of the characteristic polynomial. □ The same test works for any number in place of 5. (Now that we know how this works, we never again have to go through the argument above.) Conclusion: eigenvalues = roots of the characteristic polynomial . Steps to ﬁnd all the eigenvalues of a square matrix A: 1. Calculate the characteristic polynomial det(λI −A) or det(A −λI). 2. The roots of this polynomial are all the eigenvalues of A. Problem 6.10. Find all the eigenvalues of A := 1 −2 −1 0 ! . Solution. We have tr A = 1 + 0 = 1 and det A = 0 −2 = −2, so the characteristic polynomial is p(λ) = λ2 −(tr A)λ + (det A) = λ2 −λ −2 = (λ −2)(λ + 1). Its roots are 2 and −1; these are the eigenvalues. □ 6.5. Computing eigenvectors. Problem 6.11. Find all the eigenvectors of A := 1 −2 −1 0 ! associated with the eigenvalue 2. Solution. By deﬁnition, an eigenvector associated to the eigenvalue 2 is a vector v = v w ! satisfying Av = 2v Av −2v = 0 (A −2I)v = 0 1 −2 −1 0 ! − 2 0 0 2 !! v = 0 −1 −2 −1 −2 ! v w ! = 0 0 ! , which is equivalent to −v −2w = 0. 33 One solution is −2 1 ! ; the other solutions are the scalar multiples of this one. Conclusion: The eigenvectors with eigenvalue 2 are all the scalar multiples of −2 1 ! . □ Remark 6.12. In this example, the matrix equation became two copies of the same equation −v −2w = 0. More generally, for any 2 × 2 matrix A and eigenvalue λ, one of the two equations will be a scalar multiple of the other, so again we need to consider only one of them. In particular, the system of two equations will always have a nonzero solution (as there must be, by deﬁnition of eigenvalue). A similar calculation shows that the eigenvectors of A associated with the eigenvalue −1 are the scalar multiples of 1 1 ! . To summarize: Steps to ﬁnd all the eigenvectors associated to a given eigenvalue λ of a 2 × 2 matrix A: 1. Calculate A −λI. 2. Expand (A −λI)v = 0 using v = v w ! ; this gives a system of two equations in x and y. 3. Solve the system; one of the equations will be redundant, so nonzero solutions will exist. 4. The solution vectors v w ! are the eigenvectors associated to λ. Remark 6.13. Let A be a 2 × 2 matrix. • If A is the matrix 3I = 3 0 0 3 ! , then the only eigenvalue is 3, and every vector is an eigenvector with eigenvalue 3. If A = aI for some other scalar a, a similar thing happens. • Otherwise, for each eigenvalue λ, the system (A −λI)v = 0 amounts to one nontrivial equation (the other is redundant), so the eigenvectors associated to λ will be the scalar multiples of a single nonzero vector. In this case, if λ is real, then the set of all real eigenvectors forms a line through the origin, called the eigenline of λ. Thursday, September 23 34 7. Derivatives and integrals of functions with scalar input, vector output 7.1. Derivative of a vector-valued function. 7.1.1. Deﬁnition and physical interpretation. Deﬁnition 7.1. The derivative of a vector-valued function r(t) is the vector-valued function r′(t) := lim h→0 r(t + h) −r(t) h . (Alternative notation: dr dt .) It measures the rate of change of r at each time t. For example, if r(t) is the position of a moving particle at time t, then r′(t) is its velocity at time t. 7.1.2. Calculating the derivative. Derivatives can be calculated coordinate-wise: For example, if r(t) = t3 cos t ! , then r′(t) = 3t2 −sin t ! . This holds because all the ingredients used in the deﬁnition of derivative (vector subtraction, scalar multiplication by 1/h, and limits) can be calculated coordinate-wise. It’s the same in 3D. Advice: Use the deﬁnition to understand the derivative physically, but work coordinate-wise to calculate it. 7.2. Integration of vector-valued functions. Similarly, the deﬁnite integral Z b a f(t) dt := lim ∆t→0 n X i=1 f(t∗ i ) ∆t (where [a, b] is subdivided into n subintervals of width ∆t, and t∗ 1, . . . , t∗ n are sample points, one in each subinterval) can be computed coordinate-wise: for example, For example, if f(t) = t3 cos t ! , then Z 1 0 f(t) dt = t4/4\|1 0 sin t\|1 0 ! = 1/4 sin 1 ! . Also, there is a fundamental theorem of calculus for vector-valued functions: if f(t) = F′(t), then Z b a f(t) dt integral of a rate of change = F(b) −F(a) total change . It follows from the scalar function version. 35 7.3. Acceleration. Recall: If r(t) is the position of an object at time t, then its velocity at time t is v(t) := r′(t). Next, its acceleration at time t is a(t) := v′(t) = r′′(t). All of these are vector-valued functions. On the other hand, its speed is \|v(t)\|; this is a scalar function whose values are nonnegative. Question 7.2. What does it mean if a particle’s position vector r(t) satisﬁes d\|r\| dt = 0 for all t? Answer: The length of the position vector is constant, so the particle is staying on a sphere centered at the origin. 7.4. Diﬀerentials. In 18.01, for a function u(x), one deﬁnes du := u′(x) dx Here du and dx are diﬀerentials (not numbers, not vectors, not matrices, but a new kind of object — it does act like a tiny number, though). Similarly, for a vector-valued function r(t) =    x(t) y(t) z(t)   , one deﬁnes the “vector-valued diﬀerential” dr = r′(t) dt, which is the same as    dx dy dz   =    x′(t) dt y′(t) dt z′(t) dt   . Informally, one imagines dt as being a very tiny number, and dr is a tiny vector measuring the change in position during the short time interval from time t to time t + dt. 7.5. Arc length. Next deﬁne ds = \|dr\| := \|r′(t)\| dt = p x′(t)2 + y′(t)2 + z′(t)2 dt. Informally, ds measures the distance travelled during the same time interval [t, t + dt]. To get the total distance s(t) travelled along the trajectory from a starting point r(0) to a variable end point r(t), one “adds up” the distance travelled over all the tiny time subintervals, by integrating: s(t) = Z t 0 ds = Z t 0 \|r′(u)\| du. (The u in the integral can be thought of as a variable number increasing from 0 up to t; we couldn’t just use t, because the name t is already being used, as the upper limit of the integration.) This scalar-valued function s(t) is called the arc length function. Question 7.3. What are the physical meanings of Z t 0 \|dr\| and Z t 0 dr ? 36 Solution. The ﬁrst one is the deﬁnition of arc length s(t), which measures distance along the trajectory from the starting point r(0) to the end point r(t). On the other hand, “adding up” the inﬁnitesimal changes in position vector gives the total change in position vector, Z t 0 dr = r(t) −r(0), by the fundamental theorem of calculus, and the length of this vector is the distance as the crow ﬂies. □ 7.6. Unit tangent vector. Suppose that v(t) ̸= 0. Then v(t) is a tangent vector to the curve. Deﬁnition 7.4. The unit tangent vector at r(t) is the unit vector in the direction of v(t): T := v(t) \|v(t)\|. Informally, T is also the unit vector in the direction of dr = v(t) dt; then the identity dr = T ds is like the expression of a nonzero vector as a unit vector multiplied by a length. 7.7. Foci of an ellipse. (This section is just to help make sense of Kepler’s ﬁrst law, and to help with one of the homework problems.) One way to write down an ellipse is to write an equation x2 a2 + y2 b2 = 1. Here a is half the width, and b is half the height. But if you were to ask an ancient Greek what an ellipse is, the answer would be: Fix two points P and Q and ﬁx a number ℓgreater than PQ. Then the locus (possible positions) of a point R such that PR + RQ = ℓis an ellipse. The points P and Q are the foci (plural of focus) of the ellipse. Another property of the foci: Inside an elliptical room with mirror walls, if you place a light at one focus, the rays will reﬂect and meet again at the other focus. (Same for sound.) 7.8. Kepler’s second law. In the early 1600s, Johannes Kepler noticed that Tycho Brahe’s data on planetary motion was consistent with three laws: (1) The orbit of a planet is an ellipse with the sun at one focus. (2) A planet moves in a plane containing the sun, and the line segment connecting the sun to the planet sweeps out area at a constant rate. (3) The square of the period of revolution of a planet about the sun is proportional to the cube of the major semiaxis of its elliptical orbit. Another law: “Gravitational force F is central”, which in mathematical terms says 37 (2′) Acceleration is central, that is, the planet’s acceleration vector a is always parallel to the vector from the sun to the planet. Theorem 7.5. Kepler’s second law (2) is equivalent to the “acceleration is central” law (2′). Proof. Let the origin be where the sun is. Let r = r(t) be the position vector of the planet. Let A = A(t) be the area swept out from time 0 to time t. Between time t and t + dt, dA ≈Area(triangle) = 1 2 \|r × dr\|. Divide by dt to get a rate: dA dt = 1 2 r × dr dt = 1 2\|r × v\|. So far, this was just setting things up. Now, to prove equivalence, we prove that each law implies the other. = ⇒: Suppose that Kepler’s second law holds: dA dt is constant. Then \|r × v\| is constant. But the direction of r × v is also constant since it is perpendicular to the plane of motion (and by continuity cannot suddenly switch to the opposite direction). Thus r × v is constant. So d dt(r × v) = 0. On the other hand, by a rule for diﬀerentiating a cross product (from the textbook reading for today), d dt(r × v) = dr dt × v + r × dv dt (important: keep the r and v in order) = v × v + r × a = r × a. Combining the previous two equations gives r × a = 0. This means that a is parallel to r. ⇐ = : The converse, that the acceleration being central implies Kepler’s second law, can be proved by reversing the steps of the previous paragraph. □ Friday, September 24 38 8. Derivatives of multivariable functions 8.1. Graphs and level curves of two-variable functions. Three ways to depict a 2-variable function f(x, y): (1) Map of its values: At many points (x, y) in the plane, write the value f(x, y). For example, if f(x, y) := y(y + 1)/2 −x + 10, then the values at integer points are 30 29 28 27 26 25 24 23 22 21 20 25 24 23 22 21 20 19 18 17 16 15 21 20 19 18 17 16 15 14 13 12 11 18 17 16 15 14 13 12 11 10 9 8 16 15 14 13 12 11 10 9 8 7 6 15 14 13 12 11 10 9 8 7 6 5 15 14 13 12 11 10 9 8 7 6 5 16 15 14 13 12 11 10 9 8 7 6 18 17 16 15 14 13 12 11 10 9 8 21 20 19 18 17 16 15 14 13 12 11 25 24 23 22 21 20 19 18 17 16 15 (the colored entry is f(0, 0)). (2) Graph: Above each point (x, y) in the plane, plot a point in R3 whose z-coordinate is the value f(x, y). Taken together, these points form a surface in R3 called the graph of f. It is the set of points in space satisfying the equation z = f(x, y). (3) Level curves: For each number h, the level curve at height h is the set of points (x, y) in the xy-plane such that f(x, y) = h. In the example above, each level curve is a parabola. Question 8.1. Consider f(x, y) = x2 + y2. Its graph is a paraboloid. Draw level curves for equally spaced values of h; these are circles in R2 centered at (0, 0), namely x2 + y2 = h. Then as one goes out, are the circles (1) getting closer together (2) occurring at equally spaced radii (3) getting farther apart? Answer: Getting closer together. The farther out you are, the steeper the paraboloid is, so the shorter you have to go horizontally to get a given increase in height. 8.2. Partial derivatives. 8.2.1. Introduction via an example. 39 Here is a map showing values of a function f(x, y) at integer points, with the colored value at (0, 0): 20 19 18 17 16 15 16 16 16 16 16 16 12 13 14 15 16 17 8 10 12 14 16 18 4 7 10 13 16 19 If one starts at (0, 0) and moves to the right, the value increases by 2 for each increase of x by 1; thus the rate of change, denoted ∂f ∂x(0, 0), equals 2. (Note: ∂f ∂x is often pronounced “partial f partial x”.) Similarly, ∂f ∂y (0, 0) = 3, and ∂f ∂x(2, −1) = 3. Since ∂f ∂x has potentially diﬀerent values at diﬀerent points, it is itself a function deﬁned on the plane. 8.2.2. Deﬁnition. Deﬁnition 8.2. The partial derivative of f(x, y) with respect to x is a function ∂f ∂x whose value at (x0, y0) is • the rate of change of f(x, y) when x is varying near x = x0 and y is held constant at the value y0, or, • more precisely, ∂f ∂x(x0, y0) := lim h→0 f(x0 + h, y0) −f(x0, y0) h (if the limit exists). Other notations: fx(x0, y0), ∂f ∂x 0 . The 0 subscript on the derivative means “evaluate me at (x0, y0)”. Question 8.3. What is the diﬀerence between d f dx and ∂f ∂x? • d f dx is used if f is a function of x alone. • ∂f ∂x is used if f is a function of several variables but we are measuring the rate of change of f arising from a change in only the variable x. The deﬁnition of ∂f ∂y is similar to the deﬁnition of ∂f ∂x. 8.2.3. How to compute ∂f ∂x. View y as a constant and diﬀerentiate with respect to x. (And then evaluate at (x0, y0) if desired.) Example: If f(x, y) = x3y5, then fx = 3x2y5 and fx(2, 1) = 12. 40 Question 8.4. Let f(x, y) = xy for (x, y) in the half-plane x > 0. What is ∂f ∂y ? Possible answers: (1) xy ln x (2) yxy−1 (3) 0 (4) None of the above Answer: (1). Computing ∂f ∂y is like computing d dy2y = d dyey ln 2 = ey ln 2 ln 2 = 2y ln 2 except with a “constant” x in place of the 2. So the answer is xy ln x. 8.3. Second partial derivatives. To calculate the second partial derivative fxy = (fx)y = ∂ ∂yfx = ∂ ∂y ∂ ∂xf = ∂2f ∂y ∂x, ﬁrst take the x-derivative of f, and then take the y-derivative of the result. The other second partial derivatives are fxx, fyx, and fyy. For example, fxx = ∂2f ∂x2 . For most functions arising in practice, including any function for which all the second partial derivatives are continuous, fxy = fyx. So usually you don’t have to worry about the order in which you take derivatives. 8.4. Total derivative. If f : R2 →R is a function of x and y, then the total derivative is the 1 × 2 matrix of functions f ′(x, y) := ∂f ∂x ∂f ∂y , whose value at each input point (x0, y0) is the 1 × 2 matrix of partial derivatives at that point. (Alternative notation: (Df)(x).) The dimensions of the matrix are the same as they would be for a linear transformation R2 →R (but f here doesn’t have to be a linear transformation) Example 8.5. Let f(x, y) := x3y5. This is a function f : R2 →R, but it is not a linear transformation since x3y5 is not just a linear combination of the input variables x and y. The total derivative of f is f ′(x, y) = 3x2y5 5x3y4 and its value at (2, 1) is the 1 × 2 matrix of numbers f ′(2, 1) = 12 40 . 41 In general, for a function f : Rn − →Rm    x1 . . . xn   7− →    f1(x1, . . . , xn) . . . fm(x1, . . . , xn)   , the total derivative of f is the m × n matrix of functions f ′(x) :=       ∂f1 ∂x1 · · · ∂f1 ∂xn . . . . . . ∂fm ∂x1 · · · ∂fm ∂xn       , whose value at any input point in Rn is an m × n matrix of numbers. The m rows correspond to the coordinate functions f1, . . . , fm of f, and the n columns correspond to partial derivatives with respect to diﬀerent input variables ∂ ∂x1 , . . . , ∂ ∂xn . Example 8.6. For the function f : R2 − →R2 x y ! 7− → x2y x sin y ! the total derivative is f ′(x, y) = 2xy x2 sin y x cos y ! , and its value at any input point (x0, y0) like (2, 3) is a 2 × 2 matrix of numbers. 8.5. Linear approximation. 8.5.1. Warm-up: single-variable function. Question 8.7. How do you estimate f(x) for x near 3? Another way to state the question: Question 8.8. How does f(3 + ∆x) compare to f(3), when ∆x is a small number? (Below we’ll use green for constants that do not depend on ∆x.) 42 Answer (assuming that f is diﬀerentiable). The change ∆x in the input gets approximately magniﬁed by the derivative f ′(3): ∆f ≈f ′(3) ∆x f(3 + ∆x) −f(3) ≈f ′(3) ∆x f(3 + ∆x) ≈f(3) + f ′(3) ∆x. □ More generally, for any initial input number x0, there is the relative change formula ∆f ≈f ′(x0) ∆x. and the linear approximation formula f(x0 + ∆x) ≈ f(x0) starting value + f ′(x0) ∆x adjustment that gives the best linear approximation to f(x) for x := x0 + ∆x near x0. Another way to write the linear approximation formula (substitute x = x0 + ∆x, so ∆x = x −x0): When x is close to the number x0, f(x) ≈f(x0) + f ′(x0) (x −x0). The following problem was not discussed in lecture. Problem 8.9. Let f(x) := √x + 1. Estimate f(3.1). Solution. We have f(3) = 2, f ′(x) = 1 2(x+1)−1/2, f ′(3) = 1/4, so f(3.1) ≈f(3)+f ′(3)(0.1) = 2 + (1/4)(0.1) = 2.025. □ 8.5.2. Two-variable function. For f(x, y), the boxed formulas are the same except that x0 and ∆x are vectors, and f ′(x0) is now the total derivative. For example, the relative change formula for f(x, y) says ∆f ≈f ′(x0) ∆x = ∂f ∂x 0 ∂f ∂y 0 ∆x ∆y ! = ∂f ∂x 0 ∆x + ∂f ∂y 0 ∆y, so the relative change formula expands to become ∆f ≈ ∂f ∂x 0 ∆x + ∂f ∂y 0 ∆y (it tells you approximately how much f changes in response to changes in x and y). The new value f(x0 + ∆x, y0 + ∆y) is the old value f(x0, y0) plus the change ∆f; this gives the linear approximation formula 43 f(x0 + ∆x, y0 + ∆y) ≈f(x0, y0) starting value + ∂f ∂x 0 ∆x adjustment from ∆x + ∂f ∂y 0 ∆y adjustment from ∆y . Another way to write the linear approximation formula: f(x, y) ≈f(x0, y0) + ∂f ∂x 0 (x −x0) + ∂f ∂y 0 (y −y0). The approximation can be expected to be reasonably good if (x, y) is close to (x0, y0) and the partial derivatives are continuous in a neighborhood of (x0, y0), so that they don’t change too suddenly. Tuesday, September 28 Problem 8.10. A point P in R2 is near (−12, 5), but its coordinates could be oﬀby as much as 0.1 each. How far is P from the origin? Estimate the maximum error. Solution. Let f(x, y) = p x2 + y2. Then fx = x √ x2+y2 and fy = y √ x2+y2, so f(x, y) ≈f(−12, 5) + fx(−12, 5) ∆x + fy(−12, 5) ∆y ≈13 −12 13 ∆x + 5 13 ∆y, which is 13 with maximum error of absolute value ≈12 13 (0.1) + 5 13 (0.1) ≈0.13. (The worst error occurs when (∆x, ∆y) = (−0.1, 0.1) or (∆x, ∆y) = (0.1, −0.1).) □ 8.5.3. Three-variable function and beyond. This section was not explicitly covered in lecture. Approximation can be done for functions of more than two variables too. Question 8.11. What is the best linear polynomial approximation to a 3-variable function f(x, y, z) for inputs near a starting point (x0, y0, z0)? As one moves from (x0, y0, z0) to (x, y, z), the relative change formula says that ∆f is caused by ∆x, ∆y, ∆z, each magniﬁed by the corresponding partial derivative: ∆f change in f ≈(fx)0 ∆x + (fy)0 ∆y + (fz)0 ∆z. This gives the linear approximation formula f(x, y, z) = f(x0, y0, z0) starting value + ∆f change in f f(x, y, z) ≈f(x0, y0, z0) + (fx)0 (x −x0) + (fy)0 (y −y0) + (fz)0 (z −z0). 44 9. Maximum and minimum 9.1. Review of 1-variable max/min. 1. Identify the function f(x) to be maximized. 2. Identify the domain I on which f is to be maximized. This is the set of inputs of f that are under consideration. It is not necessarily the whole real line R. For a 1-variable function, usually the domain I is an interval in the real line, such as (−∞, 3] or (−4, −3) or . . . . 3. Check all of the following to ﬁnd potential maxima: A. critical points in I (points where f ′(x) = 0) B. points in I where f ′(x) is undeﬁned C. behavior of f(x) as x →∞or x →−∞(if applicable) D. boundary behavior (value/limit at the endpoints of I). The global max is to be found among these points, so evaluate f at these points to ﬁnd the maximum value. (Warning: If the values of f become larger and larger as x approaches ±∞or an endpoint of I, then the global max does not exist.) Here are some examples illustrating A–D above: • The global max of f(x) := −x2 on R occurs at x = 0 where the derivative f ′(x) = −2x becomes zero. • The function f(x) := 5−\|x\| on R has a global max at x = 0, where f ′(x) is undeﬁned. • Consider the function f(x) := x3 −3x on R. Its derivative f ′(x) = 3x2 −3 becomes 0 at x = ±1. But neither x = −1 nor x = 1 is a global max; in fact, f(x) →+∞as x →+∞, so a global max does not exist. • Consider f(x) := x3 −3x again, but this time restricted to the interval [−10, 10]. As before, f ′(x) = 0 if and only if x = ±1. We have f(−1) = 2 and f(1) = −2, but the global max on [−10, 10] actually occurs on the boundary of the interval, where f(10) = 970. 9.2. Constraint inequalities and constraint equations. Not covered at this point in lecture, though it should have been. For a function in more than one variable, the domain is typically more complicated than an interval. Example 9.1. The set of points (x, y) in R2 satisfying the constraint inequalities x ≥0, y ≥0, and 2x + 3y < 6 is a solid right triangle T (with the hypotenuse missing). A real-valued function f could have domain T, in which case one would write f : T →R. Example 9.2. The set of points (x, y, z) in R3 satisfying the constraint equation x2+y2+z2 = 4 is a sphere of radius 2. In contrast, x2 + y2 + z2 ≤4 deﬁnes a solid ball in R3. Similarly, x2 + y2 = 9 deﬁnes a circle in R2, while x2 + y2 ≤9 deﬁnes a disk. 45 9.3. Dimension. Not covered at this point in lecture, though it should have been. The dimension of a set is how many parameters are needed to specify a point in the set. (To make this deﬁnition precise would require more care; we won’t worry about this in 18.02.) Examples: • Any given circle is 1-dimensional (need only one parameter θ to distinguish diﬀerent points on it). • A disk is 2-dimensional (need r and θ). • The plane R2 itself is 2-dimensional (need x and y). • A point is 0-dimensional. Usually, each constraint equation reduces the dimension by 1. This leads to the following: Rule of thumb: Usually, if a set deﬁned by e constraint equations in n variables, it will be (n −e)-dimensional. Example 9.3. The constraint equation x + 2y + 3z = 5 deﬁnes a 2-dimensional set (a plane). Example 9.4. The constraint equations x2 + y2 + z2 = 100 x + 2y + 3z = 5 deﬁne a 1-dimensional set (a circle, arising as the intersection of a sphere and a plane). Warning 9.5. It is not always true that each constraint equation reduces the dimension by 1. Constraint inequalities usually do not aﬀect the dimension. 9.4. Boundary. Not covered at this point in lecture, though it should have been. Rather than deﬁne the boundary of a set precisely, let’s give some examples. Question 9.6. What are the boundaries of the following sets? • The disk x2 + y2 ≤1. Answer: The circle x2 + y2 = 1. • The ﬁrst quadrant, where x ≥0 and y ≥0. Answer: The nonnegative parts of the x- and y-axes. • The interval consisting of numbers x such that 3 ≤x ≤5. Answer: The two points x = 3 and x = 5. • The circle x2 + y2 = 1. Answer: No constraint inequalities, so no boundary! 46 • The arc of the circle x2 + y2 = 1 given by 0 ≤θ ≤π/4. Answer: The points (1, 0) (where θ = 0) and ( √ 2/2, √ 2/2) (where θ = π/4). General rule of thumb: To ﬁnd the boundary, take one of the constraint inequalities and change it to =; this gives one piece of the boundary. Then do this for each constraint inequality to get all the pieces of the boundary. In particular, if there are no constraint inequalities, there is no boundary. Another rule of thumb: For a set R, dim(boundary of R) = dim R −1 (if there is a boundary at all). Example 9.7. A disk in R2 is 2-dimensional. Its boundary is a circle, which is 1-dimensional. 9.5. Critical points. Deﬁnition 9.8. A critical point for f(x, y) is a point (a, b) such that fx(a, b) = 0 and fy(a, b) = 0. Question 9.9. What are the critical points of f(x, y) := x4 + y3 −3y? Solution: Compute the partial derivatives, and set them equal to 0: 4x3 = 0 3y2 −3 = 0. Solving this system gives x = 0 and y = ±1. So the critical points are (0, 1) and (0, −1). 9.6. Global and local extrema. Deﬁnition 9.10. The function f(x, y) has a global maximum at the point (a, b) if f(x, y) ≤ f(a, b) for all points (x, y) in the domain of f (that is, for all points (x, y) where f(x, y) is deﬁned). Global minimum is similar. Example 9.11. If f(x, y) = (x −y)2 + 5, then every point along the line x = y is a global minimum. Deﬁnition 9.12. The function f(x, y) has a local maximum at the point (a, b) if there exists some number ϵ > 0 such that at every point (x, y) in the domain whose distance to (a, b) is less than ϵ, the inequality f(x, y) ≤f(a, b) holds. In other words, (a, b) might not be a global maximum, but it becomes a maximum if one restricts the domain to a small neighborhood of (a, b). 47 Local minimum is similar. Every global max is automatically a local max. Drew a level curve diagram, waved the chalk over it, and asked the students to yell whenever it crossed a local min or local max. Theorem 9.13. Every local max (or local min) is a critical point, assuming that the partial derivatives exist at the point being tested. Proof. The point has to be a local max for the partial functions obtained by plugging a number into one of the variables, so by one-variable calculus, the derivatives of the partial functions are 0 there if they exist. □ Question 9.14. True or false: Every critical point of a function f(x, y) is either a local min or a local max. (Hint: What happens for functions of 1 variable?) Answer: False. Here are two counterexamples. (1) f(x, y) := x3 has a critical point at (0, 0), but it is not a local max (because there are nearby points to the right where f(x, y) > 0), and not a local min (because there are nearby points to the left where f(x, y) < 0). −27 −8 −1 0 1 8 27 −27 −8 −1 0 1 8 27 −27 −8 −1 0 1 8 27 −27 −8 −1 0 1 8 27 −27 −8 −1 0 1 8 27 −27 −8 −1 0 1 8 27 −27 −8 −1 0 1 8 27 (2) f(x, y) := x2 −y2 has a critical point at (0, 0), but it is neither a local max nor a local min, because there are nearby points (on one axis or the other) where the value is larger or smaller than 0. 0 −5 −8 −9 −8 −5 0 5 0 −3 −4 −3 0 5 8 3 0 −1 0 3 8 9 4 1 0 1 4 9 8 3 0 −1 0 3 8 5 0 −3 −4 −3 0 5 0 −5 −8 −9 −8 −5 0 9.7. Solving unconstrained max/min problems. General method for solving a max/min problem (without constraint equations): 1. Identify the function f(x, y) to be maximized. 48 2. Identify the domain R on which f is to be maximized. • Easy case: only constraint inequalities or no constraints, so that the dimension of R equals the number of variables, which for this function is 2. (Example: constraint inequalities x, y > 0 deﬁning the ﬁrst quadrant.) • Hard case: constraint equation(s), so that the dimension of R is less than the number of variables. (Example: constraint equation x2 + y2 = 1 deﬁning the unit circle, which is of dimension only 1.) Assume that we are in the easy case. (In the hard case, a more sophisticated method is needed: Lagrange multipliers.) 3. Check all of the following to ﬁnd potential maxima: A. critical points in R (points where fx = 0 and fy = 0 simultaneously) B. points in R where fx or fy is undeﬁned C. behavior at ∞(what happens to f as (x, y) →∞in R?) D. boundary behavior (if there are constraint inequalities) — this may lead to another, lower-dimensional max/min problem (and, even in the “easy” case, this might require Lagrange multipliers if you can’t parametrize the boundary). The global max (a, b), if it exists, is to be found among these points, so evaluate f at these points to ﬁnd the maximum value. (Warning: If the values of f become larger and larger as (x, y) approaches the boundary or ∞, then the global max does not exist.) The method also works for ﬁnding max/min of functions in 3 variables on a 3-dimensional region, and so on. Terminology: the global maximum is the location (a, b), but the maximum value is f(a, b). In any max/min problem, you will need to determine which is being asked for. Midterm 1 on Thursday, September 30. Friday, October 1 Example 9.15. Find the point on the surface xyz2 = 2 closest to the origin. Solution: There is a constraint equation (3 variables, but only 2-dimensional domain), but fortunately we can eliminate z by solving for z. Exploit symmetry: The surface lies in the regions where xy > 0, which means that x, y > 0 or x, y < 0. Because of the symmetries (x, y, z) 7→(−x, −y, z) (x, y, z) 7→(x, y, −z) 49 that preserve the equation of the surface, it’s enough to consider the parts with x, y > 0 and z > 0: this is the set of points of the form x, y, r 2 xy for x, y > 0. 1. What function do we want to minimize? Shortcut: The point where distance is minimized is the same as the point where distance2 is minimized; let’s use distance2 since it has a simpler formula, namely f(x, y) := x2 + y2 + 2 xy. 2. What is the region R? Since we are considering x, y > 0 only, R is the ﬁrst quadrant. (It’s 2-dimensional, so we don’t need Lagrange multipliers.) 3. A. Critical points: solve fx = 2x − 2 x2y = 0 fy = 2y − 2 xy2 = 0. These lead to x3y = 1 xy3 = 1, which imply x3y = xy3, and we may divide by xy to get x2 = y2, so x = y or x = −y. Since x, y > 0, we must have x = y. Then x4 = 1, and x > 0 so x = 1. Thus the only critical point is (x, y) = (1, 1). B. Points in R where fx or fy is undeﬁned: none. (The points where x = 0 or y = 0 are not part of R.) C. Behavior as (x, y) →∞: as x or y grows, the function f(x, y) tends to +∞(because of the x2 and y2 terms in f(x, y)), so it’s not approaching a minimum out there. D. Boundary behavior: As x →0 (from the right, while y is bounded), the function f(x, y) tends to +∞(because of the 2 xy term in f(x, y)), so it’s not approaching a minimum there. Same if y →0. Conclusion: f(x, y) is minimized at (x, y) = (1, 1), and z = √ 2 there, so the point is (1, 1, √ 2). The other symmetric points are (1, 1, − √ 2) and (−1, −1, √ 2) and (−1, −1, − √ 2). (And the minimum value of the distance is p D(1, 1) = √1 + 1 + 2 = 2.) The following good question came from a student: 50 Question 9.16. The point (1, 1, √ 2) we computed seems to be the point with minimum height, closest to the xy-plane (among those on the piece of the surface above R), but how do we know that there isn’t another point on the surface that is even closer to the origin? Answer: The function we minimized was not the z-coordinate of a point on the surface, but the squared distance to the origin, so the point we computed is truly the one closest to the origin. Another way of saying this: we found the lowest point on the graph z = f(x, y) above R, but this graph is diﬀerent from the piece of the original surface, which is given by z = r 2 xy. If we had wanted the point closest to the xy-plane, we would have instead minimized the z-coordinate, which is given by the function r 2 xy; in that case, the minimum does not exist since r 2 xy can be made arbitrarily close to 0 by taking x and y to be large positive numbers. 9.8. Least squares interpolation. Problem: Given data points (x1, y1), . . . , (xn, yn) that approximately lie on an unknown line y = ax + b, ﬁnd the line. What are the unknowns here? a and b! Given a candidate line y = ax + b, how do we measure how good of an approximation it is? For each input xi, the line predicts an output of axi + b, but the actual output was yi, so the error in the prediction is \|yi −(axi + b)\|. Then the total error from all the data points would be n X i=1 \|yi −(axi + b)\|, and we want to ﬁnd a, b that make this small. Because of the absolute values, the partial derivatives of this function do not exist everywhere, which complicates the minimization problem, so instead we try to minimize the sum of the squares of the errors. Deﬁnition 9.17. The least squares line (the “best” line) is the one for which D := n X i=1 (yi −(axi + b))2 is minimum. This D is a function D(a, b). 51 The minimum occurs where ∂D ∂a and ∂D ∂b are both 0. Instead of expanding out D, use the chain rule: ∂D ∂a = n X i=1 2(yi −(axi + b))(−xi) = 0 ∂D ∂b = n X i=1 2(yi −(axi + b))(−1) = 0. This is a system of two linear equations in a and b (remember: the xi and yi are given numbers). Solving for a and b gives the best line. The same method can be used to approximate data points by the graph of an unknown quadratic function y = ax2 + bx + c. How do you know what kind of function to use? Maybe the source of the data suggests that a particular shape of function is the right answer. Or maybe the data themselves look as if they can be ﬁtted with a parabola, or . . . . If you suspect that x and y are related by a power law, a function y = cxd, then plot ln x versus ln y so that the relationship is linear again: ln y = C + d ln x, where C := ln c and d are constants to be solved for. In other words, ﬁnd the line that best approximates the points (ln xi, ln yi); this gives C and d. 9.9. Second derivative test for two-variable functions. 9.9.1. Introduction: estimating a two-variable polynomial near (0, 0). The degree of a mono-mial is the sum of the exponents; for example, the degree of x2y3 is 5. The monomials in a two-variable polynomial f(x, y) can be grouped by degree, and in estimating f(x, y) near (0, 0), it is the lowest degree terms that matter most (x is much larger than x2, for instance). Example 9.18. Suppose that f(x, y) = 5 (the constant term is f(0, 0)) + 2x + 7y (these coeﬃcients are fx(0, 0), fy(0, 0)) + x2 + 6xy + 11y2 + · · · . After the 5, the terms that matter most for (x, y) near (0, 0) are the 2x and 7y. The best linear approximation is 5 + 2x + 7y. As one moves from (0, 0) to the right, f(x, y) increases; as one moves to the left f(x, y) decreases; and so on — the monomials of degree ≥2 are too small to interfere with this when (x, y) is close to (0, 0). 52 A linear combination of the variables, like 2x + 7y, is called a linear form. A linear combination of the degree 2 monomials, like x2 + 6xy + 11y2, is called a quadratic form. Example 9.19. Suppose that f(x, y) = 5 (the constant term is f(0, 0)) + 0x + 0y (fx(0, 0) and fy(0, 0) are 0) + x2 + 6xy + 11y2 + · · · . This time f(x, y) has a critical point at (0, 0). To determine whether nearby values are greater or less than the value f(0, 0) = 5, what matters most is whether the nearby values of the quadratic form x2 + 6xy + 11y2 are positive or negative. To understand functions like this, we’ll develop a general method for testing whether a quadratic form is positive/negative near (0, 0). 9.9.2. Behavior of quadratic forms. Examples of critical point behavior at (0, 0): • x2 + y2 has a local min, • −x2 −y2 has a local max, • x2 −y2 has a saddle point. (A saddle point is a critical point that is neither a local min nor a local max.4) Example 9.20. f(x, y) = x2 + 6xy + 11y2. Complete the square: it’s (x + 3y)2 + 2y2, so it has a local min at (0, 0). A general quadratic form can be written in the form f(x, y) := 1 2Ax2 + Bxy + 1 2Cy2. (The reason for putting 1 2 in the formula is so that fxx = A, fxy = fyx = B, fyy = C at (0, 0).) Let’s suppose that A ̸= 0. Completing the square rewrites f(x, y) as A 2 x + B Ay 2 + AC −B2 2A y2, and its behavior depends on the signs of the coeﬃcients in front of the squares. 4Sometimes people require more before they call a critical point a “saddle point”: they may require the graph to curve upward along one line through the point, and downwards along a line in a diﬀerent direction, like a saddle on a horse or like a Pringles potato chip. 53 Case Conclusion AC −B2 > 0 and A > 0 local min AC −B2 > 0 and A < 0 local max AC −B2 < 0 saddle point AC −B2 = 0 inconclusive The formula AC −B2 can be remembered as a determinant: AC −B2 = A B B C = fxx fxy fyx fyy . Tuesday, October 5 9.9.3. Second derivative test. The same table lets one analyze functions that are more complicated than quadratic forms: Second derivative test: Suppose that f(x, y) has a critical point at (a, b) (that is, fx(a, b) = 0 and fy(a, b) = 0), and that all second derivatives exist and are continuous in a neighborhood of (a, b). Deﬁne A := fxx(a, b), B := fxy(a, b), C := fyy(a, b). Then the type of the critical point is given by the table above. This works even when A = 0. “Inconclusive” means that it could be anything: a local min, a local max, or a saddle point (possibly a weird one). Warning 9.21. It’s not OK to use the second derivative test when there is a constraint equation! 9.9.4. An example. Question 9.22. What kind of critical point does f(x, y) := xy(x −y) = x2y −xy2 have at (0, 0)? Calculate all the partial derivatives up to second order: fx = 2xy −y2 fy = x2 −2xy fxx = 2y fxy = fyx = 2x −2y fyy = −2x. 54 Since fx and fy are 0 at (0, 0), there is a critical point there. Next, A := fxx(0, 0) = 0 B := fxy(0, 0) = 0 C := fyy(0, 0) = 0. Since AC −B2 = 0, the second derivative test is inconclusive. So, what next? The lines x = 0, y = 0 and x −y = 0 divide the plane into six regions on which f(x, y) is alternately positive or negative (to see which are which, evaluate f at a point in each region). So f(x, y) has neither a local min nor a local max at (0, 0). In fact, it has what is called a monkey saddle: there are three negative regions, two for the legs and one for the tail. x y + − + − + − left leg right leg tail Challenge: Find a surface that has an octopus saddle! 10. More on derivatives of a multivariable function 10.1. Diﬀerentials. The total diﬀerential of f(x, y) is d f := f ′(x) total derivative dx = fx fy dx dy ! . In other words, d f := fx dx + fy dy. This should remind you of the relative change formula ∆f ≈fx ∆x + fy ∆y. 55 10.2. Chain rule. In 18.01, if f is a function of x, and x is a function of t, then the chain rule says that d dtf(x(t)) = f ′(x(t)) x′(t). If instead f = f(x, y), where x = x(t) and y = y(t), then in terms of the vector quantity x = x y ! one has f = f(x) and x = x(t), and the analogous formula is d dtf(x(t)) = f ′(x(t)) total derivative x′(t) = fx fy dx/dt dy/dt ! . Thus the (multivariable) chain rule says d f dt = fx dx dt + fy dy dt . One way to remember this: “divide d f := fx dx + fy dy by dt”. The chain rule can also be written as d f dt = ∂f ∂x dx dt + ∂f ∂y dy dt . One can also use a dependency diagram to remember the chain rule: f ∂f ∂x ∂f ∂y x dx dt y dy dt t This shows that f is a function of x, y, that x is a function of t, and that y is a function of t. A change in t is magniﬁed by the factor dx/dt to produce a change in x, which is magniﬁed by ∂f/∂x to produce a change in f; simultaneously, the change in t is magniﬁed by the factor dy/dt to produce a change in y, which is magniﬁed by ∂f/∂y to produce a change in f. Thus the total magniﬁcation factor by which changes in t produce changes in f is d f dt = ∂f ∂x dx dt + ∂f ∂y dy dt . 56 Example 10.1 (more complicated version). Suppose that Q = Q(u, v, w) where u = u(x, y), v = v(x, y), and w = w(x, y). The dependency diagram Q ∂Q ∂u ∂Q ∂v ∂Q ∂w u ∂u ∂x v ∂v ∂x w ∂w ∂x x y shows that ∂Q ∂x = ∂Q ∂u ∂u ∂x + ∂Q ∂v ∂v ∂x + ∂Q ∂w ∂w ∂x and ∂Q ∂y is similar. Most general chain rule: If f and g are diﬀerentiable functions with vector input, vector output, such that f(g(x)) is deﬁned, then the total derivative of f(g(x)) is the matrix product f ′(g(x)) g′(x). 10.3. Using the chain rule to derive the product and quotient rules. Example 10.2 (Product rule). If f = uv where u = u(x) and v = v(x), then d f dx = fu du dx + fv dv dx. In other words, d dx(uv) = vu′ + uv′. The quotient rule can also be obtained this way. 10.4. Review: deﬁnitions of cos and sin. What is the meaning of cos θ and sin θ? θ θ P = ⟨cos θ, sin θ⟩ ⟨1, 0⟩ 57 Deﬁnition 10.3. Draw a unit circle centered at the origin. Let P be the point reached by going θ units counterclockwise from (1, 0). (If θ is negative, this means going clockwise.) Then cos θ := x-coordinate of P sin θ := y-coordinate of P. Example 10.4. cos π = (x-coordinate of (−1, 0)) = −1. 10.5. Review: polar coordinates. A point P in the plane can be speciﬁed in rectangular coordinates x, y or in polar coordinates r, θ. Here r means OP, the distance from P to the origin, and θ means the angle measured counterclockwise from the positive x-axis to the ray − → OP. The value of θ is not completely determined by P, since one can add any integer times 2π. Then P = x y ! = r cos θ sin θ ! , and we get the conversion formulas x = r cos θ y = r sin θ. for going from r, θ to x, y. Problem 10.5. In reverse, given x, y, how can we ﬁnd r, θ? Solution. Finding r is easy: r = p x2 + y2. Finding θ is trickier. If x = 0 or y = 0, then (x, y) is on one of the axes and θ will be an appropriate integer multiple of π/2. So assume that x and y are nonzero. The correct θ satisﬁes tan θ = y/x, but there are also other angles that satisfy this equation, namely θ + kπ for any integer k. Some of these other angles point in the opposite direction. In particular, tan−1(y/x) might be in the opposite direction. By deﬁnition, the angle tan−1(y/x) always lies in (−π/2, π/2), pointing into the right half plane, so it will be wrong when (x, y) lies in the left half plane; in that case, adjust tan−1(y/x) by adding or subtracting π to get a possible θ. Finally, if desired, add an integer multiple of 2π to get the principal value, which is the θ in the interval (−π, π]. The “2-variable arctangent function” in Mathematica and MATLAB looks not only at y/x, but also at the point (x, y), to calculate a correct θ: θ = ArcTan[x,y] Mathematica = atan2(y,x) MATLAB . Warning: Some people require θ ∈[0, 2π) instead. Warning: In MATLAB, be careful to use (y, x) and not (x, y). An alternative approach to ﬁnding θ is to use cos θ = x/r or sin θ = y/r, but again one may need to adjust to get the quadrant right. □ 58 Thursday, October 7 Problem 10.6. Convert (x, y) = (−1, −1) to polar coordinates. Solution. First, r = p (−1)2 + (−1)2 = √ 2. Evaulating tan−1(y/x) at (−1, −1) gives tan−1(1) = π/4, pointing in the direction opposite to (−1, −1). Subtracting π gives −3π/4 as a possible θ. (The other possible angles θ are the numbers −3π/4 + 2πk, where k can be any integer.) □ Question 10.7. At a time when a particle is at (4, 3) and has velocity vector 3 −1 ! (in rectangular coordinates), what is dr dt ? Solution 1. View r = r(x, y) and x = x(t), y = y(t). Namely, start with r = p x2 + y2, and apply the chain rule dr dt = ∂r ∂x dx dt + ∂r ∂y dy dt . Here ∂r ∂x = 1 2(x2 + y2)−1/2(2x) = x p x2 + y2 ∂r ∂y = 1 2(x2 + y2)−1/2(2y) = y p x2 + y2, which at (4, 3) are 4/5 and 3/5. Also, at the given time dx/dt dy/dt ! = 3 −1 ! , so dr dt = ∂r ∂x dx dt + ∂r ∂y dy dt = 4 5(3) + 3 5(−1) = 9 5. □ Solution 2 (mostly avoiding square roots). Start with r2 = x2 + y2. Apply d dt: 2rdr dt = 2xdx dt + 2ydy dt . At the given time, r = 5, so this becomes 2(5)dr dt = 2(4)(3) + 2(3)(−1), 59 which again leads to dr dt = 9 5. □ 11. Complex numbers Complex numbers are expressions of the form a + bi, where a and b are real numbers, and i is a new symbol. Multiplication of complex numbers will eventually be deﬁned so that i2 = −1. (Electrical engineers sometimes write j instead of i, because they want to reserve i for current, but everybody else thinks that’s weird.) Just as the set of all real numbers is denoted R, the set of all complex numbers is denoted C. The notation “α ∈C” means literally that α is an element of the set of complex numbers, so it is a short way of saying “α is a complex number”. Question 11.1. Is 9 a real number or a complex number? Possible answers: 1. real number 2. complex number 3. both 4. neither Answer: Both, because 9 can be identiﬁed with 9 + 0i. 11.1. Operations on complex numbers. real part Re(a + bi) := a imaginary part Im(a + bi) := b (Note: It is b, not bi, so Im(a + bi) is real!) complex conjugate a + bi := a −bi (negate the imaginary component) Question 11.2. What is Im(17 −83i)? Possible answers: • 17 • 17i • 83 • −83 • 83i • −83i Answer: The imaginary part is −83, without the i. (In lecture there was a joke about the Greek letter Ξ; you had to be there.) 60 One can add, subtract, multiply, and divide complex numbers (except for division by 0). Addition, subtraction, and multiplication are deﬁned as for polynomials, except that after multiplication one simpliﬁes by using i2 = −1; for example, (2 + 3i)(1 −5i) = 2 −7i −15i2 = 17 −7i. To divide z by w, multiply z/w by w/w so that the denominator becomes real; for example, 2 + 3i 1 −5i = 2 + 3i 1 −5i · 1 + 5i 1 + 5i = 2 + 13i + 15i2 1 −25i2 = −13 + 13i 26 = −1 2 + 1 2i. The arithmetic operations on complex numbers satisfy the same properties as for real numbers (zw = wz and so on). The mathematical jargon for this is that C, like R, is a ﬁeld. In particular, for any complex number z and integer n, the nth power zn can be deﬁned in the usual way (need z ̸= 0 if n < 0); e.g., z3 := zzz, z0 := 1, z−3 := 1/z3. (Warning: Although there is a way to deﬁne zn also for a complex number n, when z ̸= 0, it turns out that zn has more than one possible value for non-integral n, so it is ambiguous notation.) If you change every i in the universe to −i (that is, take the complex conjugate everywhere), then all true statements remain true. For example, i2 = −1 becomes (−i)2 = −1. Another example: If z = v + w, then z = v + w; in other words, v + w = v + w for any complex numbers v and w. Similarly, v w = v w. These two identities say that complex conjugation respects addition and multiplication. 11.2. The complex plane. Just as real numbers can be plotted on a line, complex numbers can be plotted on a plane: plot a + bi at the point (a, b). 61 Addition and subtraction of complex numbers have the same geometric interpretation as for vectors. The same holds for scalar multiplication by a real number. (The geometric interpretation of multiplication by a complex number is diﬀerent; we’ll explain it soon.) Complex conjugation reﬂects a complex number in the real axis. The absolute value (also called magnitude or modulus) \|z\| of a complex number z = a + bi is its distance to the origin: \|a + bi\| := √ a2 + b2 (this is a real number). For a complex number z, inequalities like z < 3 do not make sense, but inequalities like \|z\| < 3 do, because \|z\| is a real number. The complex numbers satisfying \|z\| < 3 are those in the open disk of radius 3 centered at 0 in the complex plane. (Open disk means the disk without its boundary.) 62 11.3. Some useful identities. The following are true for all complex numbers z: Re z = z + z 2 , Im z = z −z 2i , z = z, zz = \|z\|2. Also, for any real number c and complex number z, Re(cz) = c Re z, Im(cz) = c Im z. (These can fail if c is not real.) Proof of the ﬁrst identity. Write z as a + bi. Then Re z = a, z + z 2 = (a + bi) + (a −bi) 2 = a, so Re z = z + z 2 . □ The proofs of the others are similar. Many identities have a geometric interpretation too. For example, Re z = z + z 2 says that Re z is the midpoint between z and its reﬂection z. 11.4. Complex roots of polynomials. real polynomial: polynomial with real coeﬃcients complex polynomial: polynomial with complex coeﬃcients Example 11.3. How many roots does the polynomial z3 −3z2 + 4 have? It factors as (z −2)(z −2)(z + 1), so it has only two distinct roots (2 and −1). But if we count 2 twice, then the number of roots counted with multiplicity is 3, equal to the degree of the polynomial. 63 Some real polynomials, like z2 + 9, cannot be factored completely into degree 1 real polynomials, but do factor into degree 1 complex polynomials: (z + 3i)(z −3i). In fact, every complex polynomial factors completely into degree 1 complex polynomials — this is proved in advanced courses in complex analysis. This implies the following: Fundamental theorem of algebra. Every degree n complex polynomial f(z) has exactly n complex roots, if counted with multiplicity. Since real polynomials are special cases of complex polynomials, the fundamental theorem of algebra applies to them too. For real polynomials, the non-real roots can be paired oﬀ with their complex conjugates. Example 11.4. The degree 3 polynomial z3+z2−z+15 factors as (z+3)(z−1−2i)(z−1+2i), so it has three distinct roots: −3, 1 + 2i, and 1 −2i. Of these roots, −3 is real, and 1 + 2i and 1 −2i form a complex conjugate pair. Example 11.5. Want a fourth root of i? The fundamental theorem of algebra guarantees that z4 −i = 0 has a complex solution (in fact, four of them). The fundamental theorem of algebra is useful for understanding all the eigenvalues, and this is used in 18.03 for constructing solutions to linear ordinary diﬀerential equations (ODEs) with constant coeﬃcients. 11.5. Real and imaginary parts of complex-valued functions. Suppose that y(t) is a complex-valued function of a real variable t. It can be expressed as y: R − →C t 7− →y(t) = f(t) + i g(t) for some real-valued functions f(t) and g(t), namely f(t) := Re y(t) and g(t) := Im y(t). Diﬀerentiation and integration can be done component-wise, as for vector-valued functions of a single real variable: y′(t) = f ′(t) + i g′(t) Z y(t) dt = Z f(t) dt + i Z g(t) dt. Example 11.6. Suppose that y(t) = 2 + 3i 1 + it . Then y(t) = 2 + 3i 1 + it = 2 + 3i 1 + it · 1 −it 1 −it = (2 + 3t) + i(3 −2t) 1 + t2 = 2 + 3t 1 + t2 \| {z } f(t) + i 3 −2t 1 + t2 \| {z } g(t) . The functions in parentheses labelled f(t) and g(t) are real-valued, so these are the real and imaginary parts of the function y(t). 2 64 Friday, October 8 11.6. The complex exponential function. Raising e to a complex number has no a priori meaning; it needs to be deﬁned. People long ago tried to deﬁne it so that the key properties of the function et for real numbers t would be true for complex numbers too. They succeeded, and we will too! Deﬁnition 11.7 (Euler’s formula). For each real number t, eit := cos t + i sin t. (The reason for this deﬁnition will be explained in 18.03. Brieﬂy, it is that a function y(t) = eit should have the properties y′(t) = i y(t) and y(0) = 1, and the function cos t + i sin t is the unique function with these properties.) Remark 11.8. Some older books use the awful abbreviation cis t := cos t + i sin t, but this belongs in a cispool [sic], since eit is a more useful expression for the same thing. As t increases, the complex number eit = cos t + i sin t travels counterclockwise around the unit circle. Deﬁnition 11.9. For any complex number a + bi, where a and b are real numbers, ea+bi := eaebi = ea(cos b + i sin b). Properties: • e0 = 1. • ezew = ez+w for all complex numbers z and w. • 1 ez = e−z for every complex number z. • (ez)n = enz for every complex number z and integer n. • e−it = cos t −i sin t = eit for every real number t. • \|eit\| = 1 for every real number t. 65 11.7. Polar forms of a complex number. Given a nonzero complex number z = x + yi, we can express the point (x, y) in polar coordinates r and θ: x = r cos θ, y = r sin θ. Then x + yi = (r cos θ) + (r sin θ)i = r(cos θ + i sin θ). In other words, z = reiθ . The expression reiθ is called a polar form of the complex number z. Here r is required to be a positive real number (assuming z ̸= 0), so r = \|z\|. Any possible θ for z (a possible value for the angle or argument of z) may be called arg z, but this is dangerously ambiguous notation since there are many values of θ for the same z: this means that arg z is not a function. The θ in (−π, π] is called the principal value of the argument and is denoted in various ways: θ = Arg z = Arg[z] Mathematica = ArcTan[x,y] Mathematica = atan2(y,x) MATLAB . Example 11.10. Suppose that z = −3i. So z corresponds to the point (0, −3). Then r = \|z\| = 3, but there are inﬁnitely many possibilities for the angle θ. One possibility is −π/2; all the others are obtained by adding integer multiples of 2π: arg z = . . . , −5π/2, −π/2, 3π/2, 7π/2, . . . . 66 So z has many polar forms: · · · = 3ei(−5π/2) = 3ei(−π/2) = 3ei(3π/2) = 3ei(7π/2) = · · · . 2 Test for equality of two nonzero complex numbers in polar form: r1eiθ1 = r2eiθ2 ⇐ ⇒ r1 = r2 and θ1 = θ2 + 2πk for some integer k. This assumes that r1 and r2 are positive real numbers, and that θ1 and θ2 are real numbers, as you would expect for polar coordinates. Question 11.11. How do you convert a nonzero complex number z = x + yi to polar form? Solution. Convert (x, y) to polar coordinates (r, θ). Then z = reiθ. □ 11.8. Operations in polar form. Some arithmetic operations on complex numbers are easy in polar form: multiplication: (r1eiθ1)(r2eiθ2) = r1r2ei(θ1+θ2) (multiply absolute values, add angles) reciprocal: 1 reiθ = 1 re−iθ division: r1eiθ1 r2eiθ2 = r1 r2 ei(θ1−θ2) (divide absolute values, subtract angles) nth power: (reiθ)n = rneinθ for any integer n complex conjugation: reiθ = re−iθ. Taking absolute values gives identities: \|z1z2\| = \|z1\| \|z2\|, 1 z = 1 \|z\|, z1 z2 = \|z1\| \|z2\|, \|zn\| = \|z\|n, \|z\| = \|z\|. Question 11.12. What happens if you take a smiley in the complex plane and multiply each of its points by 3i? 67 Solution: Since i = eiπ/2, multiplying by i adds π/2 to the angle of each point; that is, it rotates counterclockwise by 90◦(around the origin). Next, multiplying by 3 does what you would expect: dilate by a factor of 3. Doing both leads to. . . For example, the nose was originally on the real line, a little less than 2, so multiplying it by 3i produces a big nose close to (3i)2 = 6i. 2 68 Question 11.13. How do you trap a lion? Answer: Build a cage in the shape of the unit circle \|z\| = 1. Get inside the cage. Make sure that the lion is outside the cage. Apply the function 1/z to the whole plane. Voila! The lion is now inside the cage, and you are outside it. (Only problem: There’s a lot of other stuﬀinside the cage too. Also, don’t stand too close to z = 0 when you apply 1/z.) Question 11.14. Why not always write complex numbers in polar form? Answer: Because addition and subtraction are diﬃcult in polar form! 12. Gradient and its applications 12.1. Gradient. Deﬁnition 12.1. The gradient of a scalar-valued function f(x, y, z) is the vector-valued function ∇f :=           ∂f ∂x ∂f ∂y ∂f ∂z           . 69 It contains the same information as the total derivative f ′(x) := ∂f ∂x ∂f ∂y ∂f ∂z ; in fact, ∇f is just the transpose of the total derivative f ′(x). There is a similar deﬁnition for a function in any number of variables. Problem 12.2. Let f(x, y, z) = x2y + 7z, and let P = (2, 3, 5). What is ∇f(P)? Solution. ∇f =    2xy x2 7   , so its value at (2, 3, 5) is    12 4 7   . □ We can restate certain theorems and deﬁnitions in terms of ∇: • Chain rule: If f = f(x, y, z) and r(t) = (x(t), y(t), z(t)), then d dtf(r(t)) = f ′(r(t)) r′(t) = ∂f ∂x dx dt + ∂f ∂y dy dt + ∂f ∂z dz dt = ∇f(r(t)) · r′(t). • Critical points of f(x, y): These are points P where ∇f(P) = 0. A geometric property of ∇: Theorem 12.3. (a) In 2D, ∇f(P) is perpendicular to the level curve f(x, y) = c through P. (b) In 3D, ∇f(P) is perpendicular to the level surface f(x, y, z) = c through P. Proof. Both statements follow from the following claim: If r(t) is any parametric curve on which f is constant, then at each time t, the vector ∇f(r(t)) is perpendicular to the tangent vector (velocity vector) r′(t). Let’s prove this claim. To say that f is constant on the curve means that there is a constant c such that f(r(t)) = c for all t. Take d dt and use the chain rule: ∇f(r(t)) · r′(t) = 0. This says that ∇f(r(t)) is perpendicular to r′(t). □ Tuesday, October 12 70 Examples 12.4. • If f(x, y, z) = 2x + 3y + 5z, then at every point P in R3, the vector ∇f =    2 3 5   is perpendicular to the level surface 2x + 3y + 5z = c through P (which is a plane). • If f(x, y) = x2+y2, then ∇f = 2x 2y ! , which points out from the origin, perpendicular to the level curves (which are circles). 12.2. Tangent plane. Deﬁnition 12.5. The tangent plane to the surface f(x, y, z) = c at P = (x0, y0, z0) is the plane with normal vector ∇f(P) through P: ∂f ∂x(P) (x −x0) + ∂f ∂y (P) (y −y0) + ∂f ∂z (P) (z −z0) = 0. (This makes sense only if ∇f(P) ̸= 0.) Example 12.6. If f(x, y, z) = x2 + y2 −z2 then ∇f =    2x 2y −2z   , so the cone x2 + y2 −z2 = 0 has a tangent plane at each point except (0, 0, 0). At P = (3, 4, 5), the gradient vector ∇f(P) is    6 8 −10   , so the tangent plane is 6(x −3) + 8(y −4) −10(z −5) = 0. (Check the answer: (3, 4, 5) is on this plane, and the normal vector is correct.) 12.3. Directional derivatives. Recall: Given f(x, y), the partial derivatives fx and fy measure the rate of change of f as one moves in the direction of e1 or e2. What about other directions? Deﬁnition 12.7. Let u be any unit vector. The directional derivative of f(x) in the direction u is Duf(x) := d dsf(x + su) s=0 = lim s→0 f(x + su) −f(x) s . This is because r(s) := x + su describes the position of a point starting at x and moving at speed 1 in the direction u. Why s, and not t? Answer: speed is 1, so distance traveled s equals time t. 71 Alternative notation: d f ds u (P). Sometimes also d f ds P or just d f ds (this presumes that P and u are understood). Question 12.8. If f(x, y) := x2 + y2 and x = 6 8 ! , in which directions u is the directional derivative maximum? Possible answers: (1) e1 (2) e2 (3) 3/5 4/5 ! (4) −3/5 −4/5 ! (5) 4/5 −3/5 ! or −4/5 3/5 ! . (6) None of these. To ﬁgure out the answer, we’ll use the following: Theorem 12.9 (Formula for calculating directional derivatives). Duf(x) = ∇f(x) · u. Proof. Duf(x) = d dsf(x + su) s=0 = ∇f(x + su) · d ds(x + su) s=0 = ∇f(x) · u. □ Given f and x, we maximize Duf(x) by choosing u to point in the direction of ∇f(x), so that the cos θ factor in the geometric formula for a dot product is cos 0 = 1. And if u is so chosen, then Duf(x) = \|∇f(x)\|, by that geometric formula. Hence: Geometric interpretation of ∇f: • direction of ∇f = direction in which f is increasing the fastest (perpendicular to level curve/surface) • length of ∇f = directional derivative of f in that direction 72 Example 12.10. Back to f(x, y) := x2 + y2 and P = (6, 8). We have ∇f(x, y) = 2x 2y ! , which at P is 12 16 ! , which is 20 times the unit vector u := 3/5 4/5 ! . So the function is increasing the fastest in the direction of 3/5 4/5 ! (and its rate of increase in that direction is 20). So the answer to Question 12.8 is (3). Question 12.11. In Question 12.8, in which directions v is the directional derivative 0? Solution. These are the directions v such that ∇f(P) · v = 0. We already calculated that ∇f(P) = 12 16 ! , so we want the unit vectors v satisfying 12 16 ! · v = 0. These are the unit vectors v perpendicular to the vector u := 3/5 4/5 ! in the previous question, so v = 4/5 −3/5 ! or v = −4/5 3/5 ! . (Whereas u is perpendicular to the level curve, these vectors v are tangent to the level curve. This makes sense, since the value of f is constant along the level curve, so the directional derivative of f should be 0 in either direction along the level curve.) □ 12.4. Constraint equations vs. constraint inequalities. (1) Find the minimum value of f(x, y) := 2x2 −7xy subject to the constraint inequalities x ≥1, y ≥1, and x + y ≤4. This is a standard max/min problem with boundaries to check. (2) Find the minimum value of f(x, y, z) := x2 + y2 + z2 subject to the constraint equation 3x + 5y + z = 9. Eliminate z to get an equivalent standard max/min problem: Find the minimum value of F(x, y) := x2 + y2 + (9 −3x −5y)2 with no constraint equation. (For other such problems, however, you might need Lagrange multipliers to study the boundary.) (3) Find the minimum value of f(x, y, z) := x2 + y2 + z2 subject to the constraint equation ex+y+2z = x9yz + 2. 73 When there is a constraint equation, and you can’t or don’t want to eliminate it, you need Lagrange multipliers! 12.5. Constraint equations and dimension. (This was discussed in lecture here, but I have moved it earlier in these notes, to a more logical place.) 12.6. Lagrange multipliers. Lagrange multipliers — a method for ﬁnding max/min of f(x, y) when x and y are required to satisfy a constraint g(x, y) = c (the “hard” case): 1. Identify the function f(x, y) to be maximized (or minimized). 2. Identify the constraint equation g(x, y) = c and constraint inequalities that deﬁne the domain R of f. Usually the presence of the constraint equation means that the dimension of R is one less than the number of variables (but constraint inequalities do not reduce the dimension). The constraint inequalities will be useful later, to determine what boundaries need to be checked. 3. Compute ∇f and ∇g. 4. Solve the system g = c ∇f = λ ∇g in (x, y, λ) to ﬁnd the possible pairs (x, y) (we don’t care about the value of λ, so a good strategy is to eliminate it). 5. Check also A. points on g = c where ∇g = 0 B. points on g = c where f or g is not diﬀerentiable C. behavior at ∞(what happens to f as (x, y) approaches ∞along the constraint curve g = c?) D. boundary points (if there are constraint inequalities) The global max (a, b) is to be found among these points from steps 4 and 5, so evaluate f at these points to ﬁnd the maximum value. (Warning: If the values of f become larger and larger as (x, y) approaches the boundary or ∞along g(x, y) = c, then the global max does not exist.) Thursday, October 14 We drew a level diagram for a function f(x, y) representing height of a point on a mountain, and a level curve g(x, y) = c representing a smooth train track, and we explained why, when the train is at its highest point on the track, ∇f (pointing up the mountain) is a scalar times ∇g (pointing perpendicular to the track). 74 Why does this work? Why should ∇f be a multiple of ∇g at a maximum (when ∇g ̸= 0)? Answer. 1. The vector ∇g is perpendicular to the level curve g = c. 2. At the maximum, ∇f is perpendicular to g = c too. (Reason: If not, then for one of the two directions u along g = c, the value of ∇f · u would be positive, which says that the directional derivative Duf would be positive, so one could increase f by moving in that direction along g = c, meaning that we weren’t really at a maximum of f.) Since at the maximum both ∇f and ∇g are perpendicular to the level curve, ∇f must be a multiple of ∇g there (if ∇g ̸= 0). □ Sample problem: Find the highest point on the intersection of the cylinder x2 + y2 = 13 and the plane 2x + 3y −z = 8 in R3. Solution. Steps 1 and 2: Identify the function f(x, y) to be maximized/minimized, and the constraint equation g(x, y) = c and constraint inequalities that deﬁne the domain R of f. We want to maximize z subject to two constraint equations, x2+y2 = 13 and 2x+3y−z = 8. But we can use the second equation to eliminate z and express everything in terms of x and y; namely, z = 2x + 3y −8. Now we want to maximize 2x + 3y −8 subject to the constraint x2 + y2 = 13. So take f(x, y) := 2x + 3y −8 and g(x, y) := x2 + y2 and c := 13. No constraint inequalities. Step 3: Compute ∇f and ∇g. We get ∇f = 2 3 ! and ∇g = 2x 2y ! . Step 4: Solve the system g = c ∇f = λ∇g in (x, y, λ) to ﬁnd the possible pairs (x, y) (we don’t care about the value of λ, so a good strategy is to eliminate it). The system to be solved is x2 + y2 = 13 2 = λ(2x) 3 = λ(2y). Multiply the second equation by y and the third equation by x and equate to get 2y = 3x. 75 Solve for y and substitute into the ﬁrst equation: x2 + 3 2x 2 = 13 13 4 x2 = 13 x2 = 4 x = ±2. So we get (x, y) = (2, 3) or (−2, −3). Step 5: Check also A. points on g = c where ∇g = 0 No such points, since ∇g = 0 only at (0, 0), which is not on x2 + y2 = 13. B. points on g = c where f or g is not diﬀerentiable No such points. C. behavior at ∞(what happens to f as (x, y) approaches ∞along the constraint curve g = c?) Not applicable — the constraint curve is bounded. D. boundary points (if there are constraint inequalities) Not applicable. So the only points to check are (2, 3) and (−2, −3). We have f(2, 3) = 5, and f(−2, −3) = −21, so the maximum is at (x, y) = (2, 3). There, z = f(2, 3) = 5. So the highest point is (2, 3, 5). □ Lagrange multipliers apply also to max/min of functions of more than 2 variables. (Lagrange multipliers also can be used when there is more than one constraint — this is discussed in the textbook, but we won’t study such problems in this course.) The second derivative test as we stated it cannot be used when there is a constraint equation. Sample problem 2: Find the point on the surface x2 + y2 = (z −1)3 closest to the origin. Solution. (We are going to make some mistakes below in red, and then correct them in green.) We want to minimize f(x, y, z) := x2 + y2 + z2 subject to the constraint g = 0 where g(x, y, z) := x2 + y2 −(z −1)3. We need to solve the Lagrange multiplier system x2 + y2 −(z −1)3 = 0    2x 2y 2z   = λ    2x 2y −3(z −1)2   , 76 which, when written out in components, says x2 + y2 −(z −1)3 = 0 2x = λ(2x) 2y = λ(2y) 2z = λ(−3(z −1)2). Either the second or third equation implies λ = 1. Substituting λ = 1 in the last equation leads to 2z = −3(z −1)2 3(z −1)2 + 2z = 0 3z2 −4z + 3 = 0 but 42 −4(3)(3) < 0, so there are no solutions. Oops, we divided by either x or y to obtain λ = 1, so that was valid only if x ̸= 0 or y ̸= 0. Thus to ﬁnish ﬁnding all solutions to the Lagrange multiplier system, We also need to consider the case in which x = y = 0. Then the constraint equation implies z = 1, and substituting this into the last of the four equations in the Lagrange multiplier system says 2 = λ(0) which is impossible. So there is no minimum. Oops, we forgot to check points where ∇g = 0. This says    2x 2y −3(z −1)2   =    0 0 0   , which leads to (x, y, z) = (0, 0, 1). (The other parts of Step 5 do not give anything additional.) By the geometry, there has to be a minimum somewhere, and (0, 0, 1) is the only candidate, so (0, 0, 1) is the closest point. □ The surface could be sketched by considering the slices obtained by intersecting with horizontal planes z = c. The z = 1 slice is a point, and the z = c slice for each c > 1 is a circle. The point (0, 0, 1) turns out to be what is called a singularity (it turns out to be a sharp point on the surface). Friday, October 15 In general, if you see something that is aﬀecting the climate at MIT and you don’t feel comfortable discussing it directly with the people involved, there are resources at MIT that you can reach out to. Here are some of them: 77 • Student Support Services (S3), studentlife.mit.edu/s3 • Oﬃce of Minority Education (OME), ome.mit.edu • Institute Discrimination & Harassment Response (IDHR), idhr.mit.edu 12.7. Review: gradient and directional derivatives. Geometric interpretation of ∇f: • direction of ∇f = direction in which f is increasing the fastest (perpendicular to level curve/surface) • length of ∇f = directional derivative of f in that direction Formula for directional derivative of f(x), starting at x and moving in the direction of the unit vector u: Duf(x) = ∇f(x) · u. Just as the value of a partial derivative ∂f/∂x at a point is a scalar, the value of a directional derivative is a scalar. What does Duf mean? If you travel a distance s in the direction of u, then the value of f increases by approximately (Duf)s. Example 12.12. Suppose that f(x, y) = x2 + 5y. What is (∇f)(6, 1) and what does it say geometrically? Well, ∇f = ⟨2x, 5⟩, so (∇f)(6, 1) = ⟨12, 5⟩. This has length √ 122 + 52 = 13 and direction 12 13, 5 13 . This says that f is increasing the fastest in the direction of 12 13, 5 13 , and that the directional derivative (rate of change) in that direction is 13. We have f(6, 1) = 41. What is the nearest point to (6, 1) where f takes the value 41.26, approximately? If from (6, 1) we move a distance ϵ in the direction of fastest increase, the value of f increases by 13ϵ, so we need 13ϵ = 0.26, so ϵ = 0.02. The point reached from (6, 1) by moving 0.02 units in the direction of 12 13, 5 13 is ⟨6, 1⟩+ 0.02 12 13, 5 13 ≈⟨6, 1⟩+ 0.02 ⟨1, 1/2⟩= ⟨6.02, 1.01⟩. (The approximation of 5/13 by 1/2 was pretty rough, but good enough for blackboard work.) What is the equation of the level curve through (6, 1)? The value of f at (6, 1) is 41, so the level curve has equation x2 + 5y = 41. What is the slope of the level curve at (6, 1)? Solution 1: It is perpendicular to ∇f, so the slope is −12/5. Solution 2: The level curve is y = −1 5x2 + 41 5 , i.e., the graph of ℓ(x) := −1 5x2 + 41 5 , and ℓ′(x) = −2 5x, so the slope is ℓ′(6) = −12 5 . 78 In which direction(s) u is the directional derivative of f at (6, 1) equal to 56/5? If u = ⟨a, b⟩, then we need a2 +b2 = 1 (so that u is a unit vector), and (∇f)(6, 1)·u = 56/5, which says 12a + 5b = 56/5. Solving this system (many algebra details skipped) leads to two possibilities 3 5, 4 5 , 837 845, −116 845 . In which direction(s) u is the directional derivative of f at (6, 1) equal to 14? Solution: None, because the maximum directional derivative was 13! In which direction(s) u is the directional derivative of f at (6, 1) equal to −13? Solution: Since (∇f)(6, 1) had length 13, the only unit vector u that dots with it to give −13 is the unit vector in the opposite direction, namely −12/13 −5/13 ! . What does the graph of f look like? The part above the horizontal line y = 0 is the parabola z = x2. The part above the horizontal line y = 1 is the higher parabola z = x2 + 5. All these parabolas together form the biggest slide ever. Example 12.13. Let f(x, y) = x2 + 2y2. What is Duf at (1, 3), where u is the direction that is 45◦counterclockwise from ∇f at (1, 3)? Solution. At (1, 3), ∇f = 2x 4y ! = 2 12 ! , so Duf = (∇f) · u = \|∇f\| \|u\| cos 45◦ = √ 22 + 122 (1) ( √ 2/2) = √ 12 + 62 √ 2 = √ 74. □ 12.8. Review: level curves and graph. Problem 12.14. Draw the level curves of f(x, y) := x2 −y2. What does the graph look like? Solution. The hyperbola x2 −y2 = 4 is the level curve corresponding to the value 4, the union of two lines x2 −y2 = 0 is the level curve corresponding to the value 0, and so on. To draw the graph, by visualizing a point above (or below) each point on the xy-plane, with height given by the value of the function there: the result is a saddle, with critical point 79 at (0, 0). The point (0, 0) is not a local max since there are larger values in any neighborhood of (0, 0) (for example, just to the right). It is also not a local min since there are larger values in any neighborhood (for example, just above). A critical point that is neither a local max nor a local min is called a saddle point, so that is what we have. □ 12.9. Review: complex exponential. Problem 12.15. Sketch the trajectory of f : R − →C t 7− →e(−3+2i)t in the complex plane. Solution. To understand f(t) geometrically, write it in polar form: f(t) = e(−3+2i)t = e−3tei(2t). Thus the radius is the exponentially decreasing function e−3t, and the angle is the steadily increasing function 2t. This means that the trajectory is spiralling in counterclockwise towards 0, going around inﬁnitely many times but never reaching 0. Going backwards in time, the trajectory spirals outward clockwise to ∞. □ 12.10. Review: roots of unity. Problem 12.16. Find all complex solutions to z8 = 1, and plot them in the complex plane. (These are called the 8th roots of unity.) Solution. Write both sides in polar form. Thus z = reiθ, with the unknowns r and θ to be solved for instead of z, and 1 = 1ei(0). Now we can rewrite the equation: z8 = 1 (reiθ)8 = 1ei(0) r8ei(8θ) = 1ei(0). Does this mean that r8 = 1 and 8θ = 0? Not quite, since it could be that 8θ = 0 + 2πk for some integer k, and we need to consider all values of k in order to get all solutions. The equation r8 = 1 implies r = 1, since the distance r in polar form is always a nonnegative real number. The equation 8θ = 0 + 2πk implies θ = kπ/4, so the possibilities for θ are . . . , −2π/4, −π/4, 0, π/4, 2π/4, . . . . Reassembling r and θ into z shows that z = ei(kπ/4), so the possibilities for z are . . . , ei(−2π/4), ei(−π/4), ei(0), ei(π/4), ei(2π/4), . . . . 80 These are points along the unit circle in C, equally separated by an angle π/4. But they are not all diﬀerent! They repeat after every 8 terms, so there are only 8 distinct solutions. In other words, adding 8 to k does not change ei(kπ/4), so one gets all the solutions by taking the ones with k = 0, 1, 2, . . . , 7, say. These are ei(0) = cos 0 + i sin 0 = 1 ei(π/4) = cos(π/4) + i sin(π/4) = √ 2 2 + i √ 2 2 ei(2π/4) = cos(2π/4) + i sin(2π/4) = i ei(3π/4) = cos(3π/4) + i sin(3π/4) = − √ 2 2 + i √ 2 2 ei(4π/4) = cos(4π/4) + i sin(4π/4) = −1 ei(5π/4) = cos(5π/4) + i sin(5π/4) = − √ 2 2 −i √ 2 2 ei(6π/4) = cos(6π/4) + i sin(6π/4) = −i ei(7π/4) = cos(7π/4) + i sin(7π/4) = √ 2 2 −i √ 2 2 . □ Remark 12.17. One could have predicted in advance that z8 = 1 would have 8 solutions, since 1. they are the roots of the degree 8 polynomial z8 −1, and 2. none of the roots are multiple roots, because at a multiple root the derivative would have to be 0, but the derivative is 8z7 which is nonzero at every root (0 is not a root). Midterm 2 on Tuesday, October 19. Thursday, October 21 13. Integrals of multivariable functions 13.1. Double integrals. Let R be a region in R2, cut into tiny regions R1, . . . , Rn. Choose (x1, y1) in R1, . . . , (xn, yn) in Rn. Then x R f(x, y) dA ≈f(x1, y1) Area(R1) + · · · + f(xn, yn) Area(Rn). It’s a number. (The actual deﬁnition involves a limit as the maximum size of the Ri goes to 0.) If f ≥0 everywhere on R, then s R f(x, y) dA can be interpreted geometrically as the volume under the graph of f (above R). 81 13.1.1. Computing double integrals as iterated integrals. Suppose that R is a rectangle [a, b] × [c, d]. Partition [a, b] into tiny subintervals, so R gets sliced into thin rectangles, and the volume above it gets sliced like cheese, into slabs. Then x R f(x, y) dA = Z x=b x=a Z y=d y=c f(x, y) dy dx =: Z b a Z d c f(x, y) dy dx. Think of the inner integral (with x treated as a constant) as the area of a slab; multiplying it by the width “dx” of a slab gives the volume of the slab and we sum these (“integrate”) to get the total volume. Similarly, x R f(x, y) dA = Z d c Z b a f(x, y) dx dy. 13.1.2. Non-rectangular regions. Problem 13.1. Let R be the bounded region between y −x = 2 and y = x2. Find s R(2x + 4y) dA. Solution: We’ll compute it as s (2x + 4y) dy dx. What are the limits of integration? Step 1: Sketch the region. Solving y −x = 2 and y = x2 shows that the line and the parabola intersect at (−1, 1) and (2, 4). Step 2: The outer integral goes from the smallest x-coordinate of a point in R to the largest x-coordinate. The smallest x-coordinate is −1 and the largest x-coordinate is 2. So the outer integral will look like Z x=2 x=−1 Step 3: Hold x ﬁxed, and increase y; look at the y-values where the line enters and leaves R — usually these depend on x. The inner integral goes from the smaller y-value to the larger y-value. It enters at y = x2 and leaves at y = x + 2, so the iterated integral is Z x=2 x=−1 Z y=x+2 y=x2 (2x + 4y) dy dx. Then how do you evaluate the iterated integral? Step 4: Evaluate the inner integral ﬁrst, treating x as constant. The result will be a function of x. 82 It’s Z y=x+2 y=x2 (2x + 4y) dy = 2xy + 2y2 y=x+2 y=x2 = (2x(x + 2) + 2(x + 2)2) −(2x3 + 2x4) = −2x4 −2x3 + 4x2 + 12x + 8. Step 5: Then evaluate the outer integral, which is just a deﬁnite integral of a 1-variable function of x. Answer: Z x=2 x=−1 Z y=x+2 y=x2 (2x + 4y) dy dx = Z 2 −1 (−2x4 −2x3 + 4x2 + 12x + 8) dx = 333 10 . 13.1.3. Dividing the region into pieces. Sometimes to compute a double integral, the region needs to divided into two or more pieces. The integral is then the sum of the integrals over the pieces. Example: rectangle with a smaller rectangle removed from its center. 13.1.4. Volume between two surfaces. If f(x, y) and g(x, y) are functions on a region R and f(x, y) ≥g(x, y) for all (x, y) in R, then the volume of the solid between the graphs of f and g is s R(f(x, y) −g(x, y)) dA. Problem (p. 960, #44): Set up an iterated integral that computes the volume of the region bounded by the surfaces z = x2 + 3y2 and z = 4 −y2. Solution: The intersection of the two surfaces is deﬁned by the system z = x2 + 3y2 z = 4 −y2. The projection to the xy-plane of this is obtained by eliminating z: x2 + 3y2 = 4 −y2. So R is the region bounded by the ellipse x2 + 4y2 = 4. Inside this ellipse the function 4 −y2 is larger than x2 + 3y2 (as you can see by comparing their values at (0, 0)). So we want x R ((4 −y2) −(x2 + 3y2)) dA = Z 1 −1 Z √ 4−4y2 −√ 4−4y2(4 −x2 −4y2) dx dy. 83 (Note: it is OK to write R 1 −1 instead of R y=1 y=−1, since the variable is determined by the matching diﬀerential: the integrals are nested, so the leftmost R corresponds to the rightmost diﬀerential dy, and so on.) Example 13.2. What’s wrong with the following? Z 5 2 Z x+7 x+3 (x2 + 2y) dx dy Since the outer diﬀerential on the right is dy, this would mean Z y=5 y=2 Z x=x+7 x=x+3 (x2 + 2y) dx dy, but the inner limits make no sense. In the inner integral, y is constant, and the range for x could depend on y, but the range for x cannot be expressed in terms of x! 13.2. Area in polar coordinates. Question: What is the area of the region described in polar coordinates by r ∈[3, 3.2], θ ∈[π/4, π/4 + 0.1], to the nearest hundredth? Answer: 0.06. Why? In general, the region with polar coordinates in [r, r + ∆r] and [θ, θ + ∆θ] is approximately a rectangle with sides ∆r and r∆θ (the latter is the length of an arc of radius r and measure ∆θ, by deﬁnition of radian). So its area is approximately r ∆r ∆θ, which in our case is 3(0.2)(0.1) = 0.06. (To be more precise, the area is between the areas of two rectangles, between 0.2(3(0.1)) = 0.06 and 0.2(3.2(0.1)) = 0.064.) To remember: dA = dx dy = r dr dθ. Friday, October 22 13.3. Double integrals in polar coordinates. Problem 13.3. Re-express Z √ 2 1 Z √ 2−y2 −√ 2−y2 x p x2 + y2 dx dy as an iterated integral in polar coordinates. Solution. Substitute x = r cos θ y = r sin θ dx dy = r dr dθ 84 but also re-compute the limits of integration. The inequalities − p 2 −y2 ≤x ≤ p 2 −y2 are equivalent to x2 ≤2 −y2, and to x2 + y2 ≤2. Thus the region is deﬁned by x2 + y2 ≤2 1 ≤y ≤ √ 2. In the circle of radius √ 2 centered at (0, 0), our region is the upper segment obtained by cutting the circle with the chord from (−1, 1) to (1, 1). (Draw it!) So θ ∈[π/4, 3π/4], and the upper limit for r is √ 2, but what is the lower limit for r? The inequality y ≥1 becomes r sin θ ≥1 in polar coordinates, so r ≥1/ sin θ. Final answer: Z 3π/4 π/4 Z √ 2 1/ sin θ (r cos θ)r r dr dθ. □ 13.4. Applications of double integrals. 13.4.1. Average value. Warm-up: The average of numbers x1, . . . , xn is x1 + · · · + xn n . Deﬁnition 13.4. The average value of f(x, y) on a region R is s R f dA Area(R) 13.4.2. Mass and centroid. For a metal plate in the xy-plane, mass = (mass per unit area \| {z } density )(area) But if its density is not constant, δ = δ(x, y) (in g/cm2, say), then each piece of area must be multiplied by the density there. Deﬁnition 13.5. The mass of a 2-dimensional object is m := x R δ(x, y) dA \| {z } dm . The x-coordinate of the centroid is the average of the x-coordinates of the points in the region, weighted by density. Same for the y-coordinate. So: 85 Deﬁnition 13.6. The centroid of a 2-dimensional object is the point (¯ x, ¯ y) where ¯ x := s R x dm m = s R x δ(x, y) dA s R δ(x, y) dA ¯ y := s R y dm m = s R y δ(x, y) dA s R δ(x, y) dA . (If density is not speciﬁed, assume δ(x, y) ≡1.) The centroid is also called the center of mass or the center of gravity. Sometimes symmetry gives a shortcut. For example, the centroid of an equilateral triangle (of constant density) has reﬂectional symmetry in each of its three altitudes, so the centroid must lie on all three altitudes, hence at the center. Another example: A parallelogram has 180◦rotational symmetry around the point where the two diagonals meet, so that point must be the centroid. 13.4.3. Moment of inertia. The moment of inertia of an object with respect to an axis measures how diﬃcult it is to rotate it (around that axis). For a point mass: I = (distance to axis)2 m In general (since diﬀerent pieces are at diﬀerent distances to the axis): I = x R (distance to axis)2 dm. Special case: The polar moment of inertia of an object in the xy-plane is its moment of inertia around the z-axis. The distance from (x, y, z) to the nearest point on the z-axis (namely, (0, 0, z)) is p x2 + y2, so (distance to axis)2 = x2 + y2 in this special case. Example 13.7. Polar moment of inertia of a triangle R of constant density 1 with vertices at (±3, 0) and (0, 2)? Answer: Since density is 1, we have dm = dA. The polar moment of inertia is x R (x2 + y2) dA = Z 2 0 Z 3−3y/2 −(3−3y/2) (x2 + y2) dx dy = · · · 13.4.4. Volume of revolution. Theorem 13.8 (First theorem of Pappus). A plane region R lies on one side of an axis, and is then rotated 360◦around the axis. Let A = Area(R), and let D be the distance traveled by the centroid of R. Then the solid of revolution has volume V = AD. 86 Why is this true? Set up a coordinate system so that the axis is the y-axis, and R is in the half-plane x ≥0. A little rectangle of area dA sweeps out a ring of volume approximately 2πx ∆A since the ring can be cut into approximate rectangular parallelepipeds with base dA and heights summing to the length of the ring (2πx). Summing over all rectangles and taking a limit as their size goes to 0 yields V = x R 2πx dA = (2πA) s R x dA A = A(2π¯ x) = AD. Tuesday, October 26 13.5. Change of variables in double integrals. 13.5.1. Review of change of variables in one-variable integrals. For example, Z 1 1/2 1 √ 1 −x2 dx = Z π/2 π/6 1 p 1 −sin2 u (cos u du) . where x = sin u dx = cos u du. The boundary points were re-expressed in terms of u: x = 1 corresponds to u = π/2 x = 1/2 corresponds to u = π/6. Today: The 2-variable analogue. 13.5.2. Transformations. View (u, v) as input variables, and (x, y) as output variables: (x, y) = F(u, v). Example 13.9 (from long ago). The linear transformation F(u, v) := (2u, v) (that is, x = 2u, y = v) stretches in the horizontal direction. The smiley becomes wider. 87 u v x y 2 0 0 1 ! Example 13.10. The nonlinear transformation F(u, v) := (u, v −3u2) (that is, x = u, y = v −3u2) does something worse. One way to visualize F: Plug in various input points (u, v) and see where they get mapped to. Another way: Draw images of the lines u = c and v = c. The smiley is not smiling anymore. But if you had to do an integral over the deformed smiley, you could convert it into an integral over the original smiley, as we’ll explain. u v x y F S R Important: From now on, F should give a bijective transformation from a region S in the uv-plane to a region R in the xy-plane. Bijective means that F matches the points in S with the points in R perfectly: each point in R comes from exactly one point in S. In particular, diﬀerent points in S are mapped to diﬀerent points inside R (this last condition is called one-to-one in Edwards & Penney). Without this condition, the uv-integral might double-count some parts of the xy-integral. If the condition is violated only the boundary, it’s still OK. 88 13.5.3. Jacobian determinant and area scaling factor. The Jacobian determinant of the trans-formation F is the determinant of the total derivative: ∂(x, y) ∂(u, v) := det F′(u) = det xu xv yu yv ! . It is a scalar-valued function of (u, v). The area scaling factor is the absolute value of the Jacobian determinant: ∂(x, y) ∂(u, v) = det xu xv yu yv ! . This too is a scalar-valued function of (u, v). A linear transformation has a constant area scaling factor, but a nonlinear transformation can stretch diﬀerent parts of S diﬀerently. Remark 13.11. If for some reason you instead have u, v as functions of x, y, and don’t feel like solving for x, y in terms of u, v, then ﬁrst compute the upside-down Jacobian determinant ∂(u, v) ∂(x, y) = det ux uy vx vy ! and then use the identity ∂(x, y) ∂(u, v) = ∂(u, v) ∂(x, y) −1 . The exponent −1 here denotes the reciprocal of the function (1 divided by the function), not the inverse function. 13.5.4. Steps for changing variables. Goal: Re-express x R f(x, y) dx dy as a double integral in new variables u, v. (1) Choose a transformation x = x(u, v), y = y(u, v) in order to make the integrand simpler or the region simpler. (2) Find the equations of the boundary curves of R in terms of x, y. (3) Rewrite these in terms of u, v to ﬁnd the corresponding region S in the uv-plane. (4) Compute the Jacobian determinant ∂(x, y) ∂(u, v) = det xu xv yu yv ! 89 (5) Take the absolute value of the Jacobian determinant to get the area scaling factor: ∂(x, y) ∂(u, v) . Area is always nonnegative! (6) Substitute x = x(u, v) y = y(u, v) dx dy = ∂(x, y) ∂(u, v) du dv. into the integral, and remember to use the equations of the boundary curves in terms of u, v to describe the new region S in the uv-plane, and hence determine the new limits of integration. Remark 13.12. For the change of variable x = r cos θ, y = r sin θ (with r, θ playing the roles of u, v), the area scaling factor ∂(x, y) ∂(r, θ) turns out to be r. (Check this yourself!) This gives another explanation of the conversion formula dx dy = r dr dθ. Problem 13.13. Let R be the square with corners (±1, 0) and (0, ±1). Evaluate I := x R sin2(x −y) x + y + 2 dx dy. Solution. It’s often a good idea to choose u, v so that the sides of R correspond to u = constant, v = constant. In this problem: Try u = x + y, v = x −y, so x = (u + v)/2, y = (u −v)/2. Boundary curves: u = 1, u = −1, v = 1, v = −1. Jacobian: ∂(x, y) ∂(u, v) = det 1/2 1/2 1/2 −1/2 ! = −1/2 so dx dy = −1 2 du dv = 1 2 du dv. So I = Z 1 −1 Z 1 −1 sin2 v u + 2 1 2 du dv. To ﬁnish, one should evaluate the inside integral (with variable u), and then the outside integral (with variable v). □ 90 13.5.5. Why is the area scaling factor equal to ∂(x, y) ∂(u, v) ? The rectangle [u, u+du]×[v, v +dv] maps to (approximately) a parallelogram whose sides are given by the vectors xu du yu du ! and xv dv yv dv ! In other words, as the input moves du in the u-direction, the x-coordinate of the output changes by approximately xu du, and so on, because xu is the rate of change of x with respect to u (as v is held constant). The area of the parallelogram formed by those two vectors is the absolute value of the determinant: det xu du xv dv yu du yv dv ! = det xu xv yu yv ! du dv. (Here we used twice that if one multiplies an entire column of a matrix by a “number”, the determinant gets multiplied by that number.) This explains why dx dy = ∂(x, y) ∂(u, v) du dv. Problem 13.14. Let R be the region in the ﬁrst quadrant inside the ellipse x2 + 9y2 = 12 but outside the circle x2 + y2 = 4. Evaluate x R xyey2 1 −y2 dx dy. Solution. Since x2 and y2 appear in the equations of the boundary curves, and since y2 appears in ey2 in the integrand, let’s use the change of variables u = x2 and v = y2, so x = √u and y = √v. The original region R is described by x2 + 9y2 ≤12 and x2 + y2 ≥4 and x, y ≥0. The corresponding region S in the uv-plane is described by u + 9v ≤12 and u + v ≥4 and u, v ≥0; this is a triangle with vertices at (4, 0), (12, 0), and (3, 1). We have ∂(u, v) ∂(x, y) = det ux uy vx vy ! = det 2x 0 0 2y ! = 4xy, so ∂(x, y) ∂(u, v) = 1 4xy, which is positive, so its absolute value is also 1 4xy, which is the area scaling factor: dx dy = 1 4xy du dv. Thus the integral becomes x S xyey2 1 −y2 1 4xy du dv = 1 4 x S ev 1 −v du dv. 91 (We re-expressed everything in terms of u and v.) At this point, we have the option of doing the integration in the order dv du instead of du dv, but the order dv du would require dividing the triangle into two triangles, so it is easier to keep v as the outer variable. The range for v is [0, 1]. For each v, we get the lower and upper limits for u by solving u + v = 4 and u + 9v = 12 for u in terms of v. This leads to 1 4 Z v=1 v=0 Z u=12−9v u=4−v ev 1 −v du dv = 1 4 Z v=1 v=0 ev 1 −vu u=12−9v u=4−v dv = 1 4 Z v=1 v=0 ev 1 −v(8 −8v) dv = 2 Z v=1 v=0 ev dv = 2 ev\|v=1 v=0 = 2e −2. □ Thursday, October 28 13.6. Triple integrals. Triple integral: y T f(x, y, z) dV where f(x, y, z) is continuous on a 3-dimensional region T. (Think of dividing T into tiny blocks, multiply the value of f at a sample point in each block by the volume of the block, and add the results to get an approximation.) Problem 13.15. Let T be the region in R3 where x2 + y2 + z2 ≤1 and z ≥0. So T is the upper half of the unit ball in R3. Find the centroid of T. (Assume constant density.) Solution. By symmetry, it must be of the form (0, 0, ¯ z), where ¯ z is the average value of z on the half-ball: ¯ z := t T z dV 1 2 4 3π13. At a given height z, the cross-section is the disk Dz deﬁned by x2 + y2 ≤1 −z2 (with z viewed as constant), a disk of radius √ 1 −z2. So y T z dV = Z z=1 z=0 x Dz z dx dy ! dz. 92 At this point we could write the inner double integral as an iterated integral and obtain the following: Z z=1 z=0 Z y= √ 1−z2 y=− √ 1−z2 Z x=√ 1−y2−z2 x=−√ 1−y2−z2 z dx dy dz. But it is easier instead to factor the z out of the inner double integral (z is a constant for the inner integral, in which x and y are the variables), and to recognize what remains as an area: y T z dV = Z z=1 z=0 z Area(Dz) dz = Z z=1 z=0 z π √ 1 −z22 dz = π Z 1 0 (z −z3) dz = π 4 , so ¯ z = π/4 1 2 4 3π13 = 3 8. □ 13.7. Cylindrical coordinates. In R3, replace x, y by polar coordinates, but keep z: this leads to cylindrical coordinates. To convert from cylindrical coordinates (r, θ, z) to rectangular coordinates (x, y, z): x = r cos θ y = r sin θ z = z. The other way: Given (x, y, z), compute r = p x2 + y2 tan θ = y x (check quadrant) z = z. What happens to volume? dV = dx dy dz = dz r dr dθ. This can also be visualized geometrically. 93 Let’s redo the half-ball integral: In cylindrical coordinates, the half-ball is given by r2 + z2 ≤1 and z ≥0. For each ﬁxed point (r, θ) in the plane, look at the vertical segment above/below it contained in T to get the range of integration for z: y T z dV = Z θ=2π θ=0 Z r=1 r=0 Z z= √ 1−r2 z=0 z dz r dr dθ = Z 2π 0 Z 1 0 1 −r2 2 r dr dθ = Z 2π 0 1 8 dθ = π 4 , which agrees with what we got before. 13.8. Spherical coordinates. Spherical coordinates are (ρ, φ, θ) (pronounced “rho, phi, theta”), where ρ := distance to (0, 0, 0) φ := angle down from z-axis (angle between position vector and positive z-axis) θ := same θ as in cylindrical coordinates, “longitude”, depends only on x, y. Warning: Some books use diﬀerent Greek letters here, so when reading another book, check the deﬁnitions of the variables. Imagine an r × z swinging door hinged along the z-axis, with one corner at (0, 0, 0) and opposite corner at (x, y, z), so ρ is the length of the diagonal of the door, and φ is the angle the diagonal forms with the positive z-axis. By trigonometry in the upper half of the door (above the diagonal), r = ρ sin φ is the width, and z = ρ cos φ is the height (or −height if φ > π/2). Finally, θ controls how far counterclockwise the door has swung. 94 x y z (x, y, z) ρ r φ θ z Range of possible values: ρ ∈[0, ∞) φ ∈[0, π] θ ∈[0, 2π] (or [−π, π] or . . . ) (Along the z-axis, θ is indeterminate — it can be any value. At the origin, φ is indeterminate too.) 13.8.1. Conversion between spherical coordinates and rectangular coordinates. • Given (ρ, φ, θ), we have r = ρ sin φ, so x = r cos θ = ρ sin φ cos θ y = r sin θ = ρ sin φ sin θ z = ρ cos φ. • Given (x, y, z), compute ρ := p x2 + y2 + z2, then convert (x, y) to polar coordinates (r, θ) to get θ, and ﬁnally get φ from cos φ = z ρ. 95 (That is better than using sin φ = r ρ since the latter cannot distinguish between φ and π −φ.) 13.8.2. Relationship to latitude and longitude. A great circle on a sphere S is the intersection of S with a plane through the center. If P, Q are in S, the shortest path along the sphere connecting P and Q is an arc of a great circle. (To visualize this, rotate the sphere so that P and Q are on the equator.) Set up an xyz-coordinate system with the origin at the center of the earth, with the positive z-axis passing through the North Pole, with the xy-plane containing the Equator, and with Greenwich (near London) contained in the xz-plane. Then θ◦(i.e., θ measured in degrees) is longitude (east if θ > 0). And 90◦−φ◦is latitude (number of degrees up from the Equator). Friday, October 29 13.8.3. Volume in spherical coordinates. The region with spherical coordinates in the intervals [ρ, ρ+dρ], [φ, φ+dφ], [θ, θ+dθ] deﬁnes a tiny box of volume approximately dρ times the surface area of a little “rectangle” on the sphere of sides ρ dφ and r dθ (these “sides” are actually tiny arcs of circles, and the formula for the length of such an arc is (radius)(arc measure)). The volume of the box is approximately (dρ)(ρ dφ)(r dθ) = ρ2 sin φ dρ dφ dθ, so dV = dx dy dz = ρ2 sin φ dρ dφ dθ. Another way to ﬁnd this formula: use the change of variable formula dx dy dz = ∂(x, y, z) ∂(ρ, φ, θ) dρ dφ dθ. (The volume scaling factor is the absolute value of the 3 × 3 Jacobian determinant.) 13.9. Gravitation. 13.9.1. Gravitational force exerted by a point mass. Let F be the gravitational force of a point of mass M with position vector r acting on a point of mass m at (0, 0, 0), so F is in the same direction as r, which points from m to M. Let ρ = \|r\|, which is the distance between the two masses. Then Newton says F = GmM ρ2 magnitude r ρ direction . (2) 96 13.9.2. Gravitational force exerted by a solid body. Now suppose that instead of a point of mass M, there is a solid body T whose density is given by δ = δ(x, y, z). Now what is the total force F that T exerts on a point mass m at (0, 0, 0)? Each little piece of T will be pulling m, in diﬀerent directions, and we need to add up (integrate) these little force vectors to compute the total force. A little piece of volume dV at position r has mass dM = δ dV , so by formula (2), the force it exerts on m is Gm dM ρ2 r ρ. To integrate this would involve a triple integral of a vector-valued function but we’ve only talked about triple integrals of scalar-valued functions in lecture so far, so let’s instead just compute the z-component of F: (z-component of F) = compe3 F = F · e3 = y T Gm dM ρ2 r ρ · e3 = y T Gm dM ρ2 z ρ = Gm y T cos φ ρ2 dM (since z = ρ cos φ) = Gm y T cos φ ρ2 δ dV = Gm y T δ cos φ sin φ dρ dφ dθ (since dV = ρ2 sin φ dρ dφ dθ). Example 13.16. Suppose that T is a solid sphere of radius a centered at (0, 0, a), of constant density δ = 1, and suppose that m = 1. Warning: The (0, 0, 0) for our spherical coordinates is not at the center of T, but at its south pole, where m is! This means that if r is the position vector of a point P inside T, then r is pointing from the south pole to P, and the angle φ that r and the positive z-axis form will be in the range [0, π/2], not [0, π]. 97 By symmetry, the gravitational force F is pointing straight up, so \|F\| = (z-component of F) = G y T cos φ sin φ dρ dφ dθ (by the previous calculation) = G Z 2π θ=0 Z π/2 φ=0 Z 2a cos φ ρ=0 cos φ sin φ dρ dφ dθ = G Z 2π 0 Z π/2 0 2a cos2 φ sin φ dφ dθ = 2Ga Z 2π 0 −1 3 cos3 φ π/2 0 dθ = 2Ga Z 2π 0 1 3 dθ = 4πGa 3 . On the other hand, since δ = 1, the total mass of T is M = Volume(T) = 4 3πa3, so if all the mass of T were concentrated at its center, then Newton’s law of gravitation for a point mass would give \|F\| = GM a2 = 4πGa 3 . Same answer! Remark 13.17. Newton proved more generally that a solid sphere of constant density exerts the same force on any external mass m (not necessarily on the surface) as would a point mass M placed at its center. Tuesday, November 2 14. Vector fields Deﬁnition 14.1. A vector ﬁeld is a function whose value at each point of a region is a vector. It could be showing, for example, • the wind velocity at each point, • the velocity of a ﬂuid at each point, • the strength and direction of an electric ﬁeld or magnetic ﬁeld, • or the gradient of a function at each point. 98 Mathematically, a 2-dimensional vector ﬁeld has the form F: R2 − →R2 (x, y) 7− → P(x, y) Q(x, y) ! = P(x, y) e1 + Q(x, y) e2, where P : R2 →R and Q: R2 →R are functions. Example 14.2. Consider the vector ﬁeld F(x, y) := −y 0 ! . Its values in the upper half plane are vectors pointing to the left, and its values in the lower half plane are vectors pointing to the right. (The students in class demonstrated this by forming one big vector ﬁeld with their hands.) Example 14.3. The gradient ﬁeld of f(x, y) := x2 + y2 is F(x, y) := 2x 2y ! , which is always pointing outward. Think of the direction of fastest increase, looking at a top view of a paraboloid. The gradient ﬁeld is everywhere perpendicular to the level curves, which are circles. F is continuous ⇐ ⇒P and Q are continuous. F is diﬀerentiable ⇐ ⇒P and Q are diﬀerentiable. F is continuously diﬀerentiable ⇐ ⇒P and Q are diﬀerentiable, and their partial derivatives are continuous. 14.1. Line integrals. Let C be an oriented curve in R2. (Oriented means that one of the two directions along the curve has been chosen.) Three kinds of line integrals: • Line integral with respect to arc length: R C f(x, y) ds. (Remember: s = distance traveled = arc length.) • Line integral with respect to coordinate variables: R C f(x, y) dx or R C f(x, y) dy or R C P(x, y) dx + Q(x, y) dy. • Line integral of a vector ﬁeld: R C F · dr. Which are vectors and which are scalars? They are all scalars! 14.1.1. What do these mean? Divide C into n pieces by labelling points P0, P1, . . . , Pn in order along C where P0 and Pn are the endpoints. Choose a sample point P ∗ i on the arc from 99 Pi−1 to Pi. Let ∆si be the arc length of that arc. Then Z C f(x, y) ds := lim max ∆si→0 (f(P ∗ 1 )∆s1 + · · · + f(P ∗ n)∆sn) Z C f(x, y) dx := lim max ∆si→0 (f(P ∗ 1 )(x(P1) −x(P0)) + · · · + f(P ∗ n)(x(Pn) −x(Pn−1))) Z C F · dr := lim max ∆si→0 F(P ∗ 1 ) · − − → P0P1 + · · · + F(P ∗ n) · − − − − → Pn−1Pn . Example 14.4. Z C 1 ds = arc length of C. 14.1.2. How do you compute these? (1) Choose a parametrization of C, say r(t) := x(t) y(t) ! for t ∈[a, b]. Now x, y, s, r are all functions of t. (2) Substitute x = x(t) and y = y(t) in the integrand. (3) Also substitute dx = x′(t) dt dy = y′(t) dt ds = p dx2 + dy2 = p x′(t)2 + y′(t)2 dt dr = dx dy ! = x′(t) y′(t) ! dt. (4) Change R C to R b a . (5) Evaluate the resulting 1-variable integral in t. Problem 14.5. Let C be the upper half of the circle x2 + y2 = 4, oriented counterclockwise. Let F(x, y) = −y 0 ! . Compute R C F · dr. Is the answer going to be positive or negative? Positive, because everywhere along the curve, F and dr form an angle less than 90◦. 100 Solution. Choose the parametrization 2 cos t 2 sin t ! for t ∈[0, π]. Then Z C F · dr = Z C −y 0 ! · dx dy ! = Z π 0 −2 sin t 0 ! · −2 sin t 2 cos t ! dt = Z π 0 4 sin2 t dt = Z π 0 (2 −2 cos 2t) dt = 2π. □ 14.1.3. Geometric interpretation. Recall that dr = T ds = dx dy ! , where T is the unit tangent vector. So if F = P Q ! , then Z C F · dr = Z C F · T tangential component of F ds = Z C P dx + Q dy. 14.2. Applications of line integrals. 14.2.1. Applications of line integrals with respect to arc length. If C is a wire whose density (mass per unit length) is δ(x, y), then dm = δ ds, so the quantities mass m := Z C dm mass-weighted average ¯ f := R C f dm m centroid := (¯ x, ¯ y) moment of inertia I := Z C (distance to axis)2 dm all involve line integrals with respect to arc length. 14.2.2. Work. Given a force ﬁeld that is constant (independent of position), work := − − → force · − − − − − − − − − → displacement := F · ∆r, 101 More generally, for an object moving along C, the work done by a not necessarily constant force ﬁeld F is W := Z C F · dr. (Think of adding up the work done over each little piece of C.) 14.3. Adding and reversing paths. If the endpoint of path C1 equals the start point of C2, then concatenating gives a path C1 + C2, and Z C1+C2 · · · = Z C1 · · · + Z C2 · · · . Let C be a piecewise diﬀerentiable oriented curve. Then −C denotes the same curve with the reverse orientation. For each piece of C, the “tiny change in position” dr changes sign, so its components dx and dy change sign too, but the “tiny length” ds = \|dr\| stays the same. Thus Z −C f ds = Z C f ds Z −C P dx + Q dy = − Z C P dx + Q dy Z −C F · dr = − Z C F · dr. Thursday, November 4 14.4. Fundamental theorem of calculus for line integrals. 1st FTC: If f(x) is a continuous function on [a, b], and F(x) := Z x a f(t) dt then F ′(x) = f(x). Main purpose: Construction of an antiderivative of f(x). 2nd FTC: If G(x) is such that G′(x) is continuous on an interval [a, b], then Z b a G′(t) dt = G(b) −G(a). Main purpose: Evaluation of a deﬁnite integral R b a G′(t) dt. FTC for line integrals: If C starts at A and ends at B, and f is a function that is continuously diﬀerentiable at each point of C, then Z C ∇f · dr = f(B) −f(A). 102 The left side is adding up the change in f along each piece of the curve, and the right side is the total change in f. The formula is the same for a function f in any number of variables. Proof. Let r(t) for t ∈[a, b] be a parametrization of C. Let G(t) = f(r(t)). Then Z C ∇f · dr = Z b a ∇f · r′(t) dt = Z b a G′(t) dt (by the gradient form of the chain rule) = G(b) −G(a) (by the 2nd FTC) = f(r(b)) −f(r(a)) = f(B) −f(A). □ Consequences for gradient ﬁelds: • It is very easy to compute a line integral of a gradient ﬁeld. There’s no need to parametrize the curve — the FTC gives the answer directly! • R B A ∇f · dr is path independent for each pair of points A and B: this says that the value of R C ∇f · dr is the same for every curve C from A to B! • If C is a closed curve (starts and ends at the same point), then H C ∇f · dr = 0. (There is no diﬀerence between H and R : the former is just a notation to help remind you that the curve is closed.) These are properties special to gradient vector ﬁelds. Example 14.6. Let F = y 0 ! . Let C be the upper half of the unit circle, and let C′ be the lower half, both oriented from (1, 0) to (−1, 0). The geometry shows that R C F · dr is negative, but R C′ F · dr is positive. So line integrals of F are not path independent. Thus F cannot be ∇f for any function f. 14.5. The operator ∇. Introduce the notation ∇:=           ∂ ∂x ∂ ∂y ∂ ∂z           . The symbol ∇is sometimes pronounced “nabla” (ancient Greek word for harp). It is not an actual vector, since its entries are not numbers (or functions). But it helps us remember the 103 deﬁnition of gradient of a function f(x, y, z), ∇f :=           ∂f ∂x ∂f ∂y ∂f ∂z           , which is a kind of derivative of f (it is the transpose of the total derivative f ′(x)). 14.6. Divergence of a 3D vector ﬁeld. For a diﬀerentiable 3D vector ﬁeld F(x, y, z) =    P(x, y, z) Q(x, y, z) R(x, y, z)   , we can now use ∇to deﬁne two new fancy kinds of derivative, called divergence and curl. Deﬁnition 14.7. The divergence of F is the function div F := ∇· F :=           ∂ ∂x ∂ ∂y ∂ ∂z           ·        P Q R        := Px + Qy + Rz. Vector or scalar? The divergence of F is a scalar (at each point), just like any dot product. So don’t make the mistake of thinking that div F is    Px Qy Rz   ! Problem 14.8. Let F(x, y, z) :=    x y z   , a 3D vector ﬁeld that is pointing radially outward at each point. What is div F? 104 Solution. By deﬁnition, div F = ∇· F =           ∂ ∂x ∂ ∂y ∂ ∂z           ·        x y z        = ∂ ∂x(x) + ∂ ∂y(y) + ∂ ∂z(z) = 3. (Usually div F is a function taking diﬀerent values at diﬀerent points in R3, but in this problem, div F turned out to be a constant function.) □ Physical interpretation of divergence: Let F be the velocity ﬁeld of a ﬂuid of density 1. Then, at each point, think of div F as measuring the net amount of ﬂuid exiting a tiny cube per unit volume and per unit time. (If more ﬂuid is entering than exiting, then the net amount will be negative.) It is not obvious from the formula for div F why div F measures this. We’ll explain this later, using the concept of ﬂux. Example 14.9. Suppose that the velocity ﬁeld F is the vector ﬁeld of Problem 14.8: F(x, y, z) :=    x y z   . For a tiny cube centered at (0, 0, 0), the net ﬂow is outward. For a tiny cube centered at (0, 0, 1), the net ﬂow is again outward, since the rate of ﬂow out the top is slightly greater than the rate of ﬂow into the bottom (and there is also ﬂuid ﬂowing out the sides). These observations are consistent with div F being positive at these points. 14.7. Curl of a 3D vector ﬁeld. Again assume that F(x, y, z) =    P(x, y, z) Q(x, y, z) R(x, y, z)   . 105 Deﬁnition 14.10. The curl of F is the function curl F := ∇× F := e1 e2 e3 ∂ ∂x ∂ ∂y ∂ ∂z P Q R := (Ry −Qz)e1 + (Pz −Rx)e2 + (Qx −Py)e3. Vector or scalar? The curl of F is a vector (at each point), just like any cross product. Theorem 14.11 (Curl of a gradient ﬁeld is zero). For any function f : R3 →R with continuous second partial derivatives, curl(∇f) = 0. Proof. By deﬁnition, curl(∇f) = ∇× ∇f = e1 e2 e3 ∂ ∂x ∂ ∂y ∂ ∂z fx fy fz = (fzy −fyz)e1 + (fxz −fzx)e2 + (fyx −fxy)e3 = 0. □ Physical interpretation of curl: Let F be the velocity ﬁeld of a ﬂuid again. Through any given point, there is an axis around which a tiny paddlewheel in the ﬂuid would rotate the fastest. Let ω be the angular speed in radians per second. • The direction of curl F is along this axis, with direction given by the right hand rule (if your ﬁngers are in the direction of rotation, your thumb gives the direction of curl F). • The length of curl F is 2ω. If instead F is a force ﬁeld (measuring force exerted per unit mass), then curl F measures the torque per unit of moment of inertia exerted on a tiny dumbbell. An analogy: − − → force mass = − − − − − − − − → acceleration := d dt(− − − − − → velocity) − − − − → torque moment of inertia = − − − − − − − − − − − − − − → angular acceleration := d dt(− − − − − − − − − − − → angular velocity). 106 (The direction of rotational quantities like torque or angular velocity is along the axis of rotation, given by the right hand rule.) Friday, November 5 14.8. Simply connected regions. Suppose that T is a connected region in R2 or R3. Let’s say that a closed curve C inside T is shrinkable in T if it is possible to continuously shrink C to a point within T. Example 14.12. If T is R2 with the unit disk removed, then a circle of radius 2 centered at the origin is contained in T but is not shrinkable in T. (To shrink it, the curve would have to enter the unit disk, at least temporarily, and that is not allowed.) Deﬁnition 14.13. Call T simply connected if every closed curve C in T is shrinkable in T. Question 14.14. Which of the following are simply connected? • The annulus in R2 deﬁned by 1 < x2 + y2 < 9? No, because the circle x2 + y2 = 4 is not shrinkable inside the annulus. • R2 with a point removed? No, for the same reason. • R3? Yes. • a solid ball? Yes. • a solid torus (donut)? No, because a curve going all the way around the donut is not shrinkable. • R3 with a solid ball removed? Yes, since a curve going around the ball can be slipped around the ball. • R3 with a point removed? Yes, for the same reason. • R3 with an inﬁnite cylinder removed? No, because a curve going around the cylinder is not shrinkable. 15. Conservative vector fields All vector ﬁelds F today are assumed to be continuous in a 3D region T. If considering curl F, then F is assumed to be continuously diﬀerentiable. 15.1. Equivalent conditions. Here are four conditions that F might or might not satisfy: (1) F is a gradient ﬁeld (equal to ∇f for some f = f(x, y, z) on T). (2) R B A F · dr is path independent for every A and B in T (for paths inside T). (3) H C F · dr = 0 for every closed curve C inside T. (4) curl F = 0 at every point of T. Theorem 15.1 (Equivalence of conditions). 107 • Conditions (1), (2), (3) are equivalent. • Any of (1), (2), (3) implies (4). • If T is simply connected, then all four conditions are equivalent. Example 15.2. Suppose that F is a vector ﬁeld that satisﬁes condition (2), the path independence condition. Then the equivalence of (1), (2), (3) means that F also is a gradient ﬁeld, and F also satisﬁes the closed curve condition. Moreover, any of those conditions implies (4), so curl F = 0 too. Example 15.3. Suppose that F is a vector ﬁeld that satisﬁes condition (4), that curl F = 0 at every point of T. There is no guarantee that F satisﬁes conditions (1), (2), (3) too (the implication goes the wrong direction). But if we also know that T is simply connected, then F satisﬁes all four conditions. Deﬁnition 15.4. Say that F is conservative if it F satisﬁes any of conditions (1), (2), (3) (it doesn’t matter which, since they are all equivalent!). Any diﬀerentiable function f such that ∇f = F is called a potential function for F — the multivariable analogue of an antiderivative. Problem 15.5. Consider a solid spherical planet of mass M. Let the region T be all space outside the planet. The gravitational ﬁeld F at each point of T represents the force that the planet would exert on a unit mass at that point. In terms of spherical coordinates (ρ, φ, θ) with origin at the planet’s center, Newton’s inverse square law says F = −GM ρ2 b ρ. Is F conservative? Solution. It is enough to check any of conditions (1), (2), (3). In fact, since T is simply connected, we can check any of (1), (2), (3), (4). Usually (2) and (3) are hard to check, because they involve the integrals along many paths. Usually (4) is the easiest to check since it just involves calculating a curl, but we didn’t ever work out the formula for curl in spherical coordinates. So let’s instead check (1), that F is a gradient ﬁeld. To do this, we’ll guess f and then verify that F = ∇f. Let f = GM ρ . Is it true that ∇f = F? Yes, because • ∇f is radially inward, in the direction of −ˆ ρ, just like F, by symmetry; and • ∇f has the same ˆ ρ-component as F: ∇f · ˆ ρ = Dˆ ρf = ∂ ∂ρ GM ρ = −GM ρ2 = F · ˆ ρ. Thus F is a gradient ﬁeld. In other words, F is conservative. □ 108 Remark 15.6. Another way to verify that ∇f = F would have been to use the general formula for ∇f in spherical coordinates: ∇f = ∂f ∂ρ ˆ ρ + 1 ρ ∂f ∂φ ˆ φ + 1 ρ sin φ ∂f ∂θ ˆ θ = −GM ρ2 ˆ ρ + 0ˆ φ + 0ˆ θ = F. 15.2. Proof of equivalence. Why are conditions (1), (2), (3) equivalent? Proof. (1) = ⇒(3): We already showed that a line integral of a gradient ﬁeld along a closed curve is 0. (3) = ⇒(2): Suppose that H C F · dr = 0 for every closed curve C. We need to prove path independence. So suppose that C1 and C2 are two paths from A to B. Then C1 + (−C2) is a closed loop, so 0 = I C1+(−C2) F · dr = Z C1 F · dr − Z C2 F · dr. So R C1 F · dr = R C2 F · dr. (2) = ⇒(1): Suppose that line integrals of F are independent of path. Choose a start point a in T. Deﬁne f(x) := Z x a F · dr. (This makes sense since the value is independent of the path.) We will check ∇f = F coordinate by coordinate. To calculate the ﬁrst coordinate fx at a point b, we need the values of f along the horizontal line through b, so deﬁne g(x) := f(x, b2, b3) = Z b a F · dr + Z (x,b2,b3) b F · dr. Along that line, fx = dg dx = d dx Z (x,b2,b3) b F · dr (the ﬁrst term of g(x) is a constant) = d dx Z x b1    P Q R   ·    dt 0 0    (using r(t) := (t, b2, b3)) = d dx Z x b1 P dt = P by the one-variable 1st FTC). 109 Similarly, fy = Q and fz = R. Thus ∇f = F. □ Also, Theorem 14.11 shows that if F is a gradient ﬁeld, then curl F = 0, so (1) = ⇒(4). To ﬁnish the proof of Theorem 15.1, we need to explain why if T is simply connected, (4) implies one of the ﬁrst three conditions. This requires Stokes’ theorem, so we’ll postpone this. 15.3. Finding the potential. We had time only to sketch the ﬁrst solution to Problem 2 below. The details of this section will be covered in recitation on Monday, November 8. Consider the vector ﬁeld F =    3x2y + 5yz x3 + 7z + 5xz 7y + 5xy + ez    on R3. Problem 1: Is there a function f such that ∇f = F? Solution: Since R3 is simply connected, we can check any of the four conditions. Let’s test (4): curl F := ∇× F = e1 e2 e3 ∂ ∂x ∂ ∂y ∂ ∂z 3x2y + 5yz x3 + 7z + 5xz 7y + 5xy + ez = e1((7 + 5x) −(7 + 5x)) −e2(5y −5y) + e3((3x2 + 5z) −(3x2 + 5z)) = 0. So the answer is yes. Problem 2: Can you ﬁnd such a potential function f? First solution (FTC in reverse): If ∇f = F, then f(a, b, c) −f(0, 0, 0) = Z C ∇f · dr = Z C F · dr = Z (a,b,c) (0,0,0) (3x2y + 5yz) dx + (x3 + 7z + 5xz) dy + (7y + 5xy + ez) dz. Choose the path C1 + C2 + C3 where C1 goes from (0, 0, 0) to (a, 0, 0), C2 goes from (a, 0, 0) to (a, b, 0), and C3 goes from (a, b, 0) to (a, b, c). Add up the three integrals. For C1, use 110 the parametrization r(t) := ⟨t, 0, 0⟩for t ∈[0, a]; then x = t, y = 0, z = 0, dx = dt, dy = 0, dz = 0, so Z C1 F · dr = Z a 0 0 dt = 0. Similarly, for C2 use ⟨a, t, 0⟩for t ∈[0, b]; then x = a, y = t, z = 0, dx = 0, dy = dt, dz = 0, so Z C2 F · dr = Z b 0 a3 dt = a3b. Finally, for C3, use the parametrization ⟨a, b, t⟩for t ∈[0, c], so x = a, y = b, z = t, dx = 0, dy = 0, dz = dt, so Z C3 F · dr = Z c 0 (7b + 5ab + et) dt = 7bc + 5abc + ec −1. Summing yields Z C F · dr = a3b + 7bc + 5abc + ec −1. Thus f(x, y, z) = x3y + 7yz + 5xyz + ez −1 is one possibility. (The others are obtained by adding any constant.) Second solution (antiderivative method): We know fx = 3x2y + 5yz fy = x3 + 7z + 5xz fz = 7y + 5xy + ez. The fx equation implies f = x3y + 5xyz + g(y, z) (3) for some g(y, z). Taking ∂ ∂y gives fy = x3 +5xz +gy and comparing with the given fy equation shows that gy = 7z, so g = 7yz +h(z) for some function h(z). Substituting back into equation (3) gives f = x3y + 5xyz + 7yz + h(z). (4) Taking ∂ ∂z gives fz = 5xy + 7y + hz and comparing with the given fz equation shows that hz = ez + c for some constant c. Substituting back into equation (4) gives f = x3y + 5xyz + 7yz + ez + c. We check that this really has the right gradient. 111 Third solution (guess and check): Just guess one possible potential function f, and check that it has the right gradient; then the complete set of solutions is the set of functions of the form f + c, where c is any number. Midterm 3 covers up to here. Tuesday, November 9 15.4. Conservative force ﬁelds in physics. In physics, kinetic energy is 1 2m\|v\|2 = 1 2mv·v, and if F is a force ﬁeld (e.g., gravitational force) and F = −∇V , then the scalar-valued function V is called potential energy. If an object moves from A to B along a path C, 1 2mv · v B A increase in kinetic energy FTC = Z mv′(t) · v(t) dt (by the product rule for the dot product) = Z ma(t) · r′(t) dt = Z C F · dr work done by F on the object FTC = −V \|B A decrease in potential energy , so kinetic energy + potential energy is constant: conservation of energy! That’s why F is called conservative. And this also explains why physicists like to write F as −∇V instead of just ∇V . 16. Review 16.1. Equivalent conditions for a vector ﬁeld to be conservative. A gradient ﬁeld (one that is ∇f for some diﬀerentiable function f(x, y, z)) is a special kind of vector ﬁeld. What is another name for gradient ﬁelds? Conservative vector ﬁelds. Most 3D vector ﬁelds are not conservative, not gradient ﬁelds. If F is a gradient ﬁeld, any function f such that ∇f = F is called a potential function. Question 16.1. Can a vector ﬁeld F have more than one potential function? Answer. Yes, if there is any potential function at all, say f, then f + c for any number c is another one, since ∇(f + c) = ∇f. In fact, these are all the potential functions for F: if f1, f2 are both potential functions for F, then the function g := f1 −f2 satisﬁes ∇g = ∇f1 −∇f2 = F −F = 0 everywhere, so all partial derivatives of g are 0, which means that g is constant. □ 112 Given a 3D vector ﬁeld F, why is it helpful to know whether F is conservative? • It makes computing line integrals R C F · dr easy, at least if one can ﬁnd a potential function f, because of the FTC for line integrals. • It has other nice consequences, such as path independence and the closed curve property. Here are the four conditions that F might or might not satisfy: (1) F is a gradient ﬁeld (equal to ∇f for some f = f(x, y, z) on T). (2) R B A F · dr is path independent for every A and B in T (for paths inside T). (3) H C F · dr = 0 for every closed curve C inside T. (4) curl F = 0 at every point of T. Case 1: T is simply connected. Then all four conditions are equivalent. Does this mean that every F satisﬁes the the four conditions? No. What it means is that if F satisﬁes any one of the conditions, then F satisﬁes all four conditions. The easiest one to check is usually (4) (because one just has to compute curl F) or sometimes (1) (if one can ﬁnd a potential function f). Case 2: T is not simply connected. Then the ﬁrst three conditions are equivalent, and they imply the fourth. If one knows only that the fourth condition holds, then the ﬁrst three might hold, or they might fail. 16.2. Example in which equivalence fails. This example was rewritten from what was said in class, to hopefully make it clearer. Consider R2 −{(x, 0) : x ≥0}, which is the plane with a slit along the nonnegative x-axis. Let θ1(x, y) be the scalar-valued function with domain R2 −{(x, 0) : x ≥0} that gives the value of the polar coordinate θ in the range (0, 2π). It cannot be extended to a continuous function on the punctured plane R2 −{(0, 0)} because its values jump as one crosses the positive x-axis. Let θ2(x, y) be the function with domain R2 −{(x, 0) : x ≤0} that gives the value of θ in the range (−π, π). The functions θ1 and θ2 do not come from a single function deﬁned on R2 −{(0, 0)} because they do not agree in the lower half-plane. On the other hand, they diﬀer there only by a constant (namely, 2π), so ∇θ1 and ∇θ2 do agree at every point where both are deﬁned. Thus there is a 2D vector ﬁeld F deﬁned on R2 −{(0, 0)} that agrees with each of these. We drew F: it is pointing in the counterclockwise direction, and is stronger at points closer to origin, like a whirlpool. What is the formula for F? At a point in R2 −{(0, 0)} with position vector r, • the direction of F is the direction in which θ increases the fastest, which is 90◦ counterclockwise from r, and • the length of F is the rate of increase in that direction, which is the directional derivative dθ ds, which equals 1 r, since the little piece of arc length ds is r dθ. 113 Thus F = 1 r 90◦counterclockwise rotation of the outward unit vector r r = 1 r2 (90◦counterclockwise rotation of r) = 1 x2 + y2 0 −1 1 0 ! x y ! = 1 x2 + y2 −y x ! =    − y x2 + y2 x x2 + y2   . (Another way to get the formula for F would be to take the gradient of tan−1(y/x) + constant or π/2 −tan−1(x/y) + constant; at least one of these formulas makes sense at each point (x, y) ̸= (0, 0).) Now let us add one dimension, and think of θ1, θ2, F as functions of (x, y, z) that happen not to depend on z, with F having a horizontal value at each point. The new F is a 3D vector ﬁeld deﬁned on the region T := R3 −(z-axis), given by F(x, y, z) :=       − y x2 + y2 x x2 + y2 0       . Question 16.2. Is T simply connected? Answer. No, because a closed curve that goes around the z-axis is not shrinkable inside T. □ Question 16.3. Does curl F = 0 hold at every point of T? Answer. Yes, because in a neigborhood of each point of T, the vector ﬁeld F agrees with either ∇θ1 or ∇θ2, and the curl of any gradient ﬁeld is 0. □ Question 16.4. Is F conservative on T? This does not follow automatically from curl F = 0 since (4) does not necessarily imply (1), (2), (3) on a region that is not simply connected. Answer. No. Let us check that the closed curve condition (3) fails. If C is the counterclockwise unit circle, then F and dr point in the same direction along C, so the geometric interpretation 114 of the dot product shows that R C F · dr > 0. (In fact, the value of the integral is 2π, because it is adding up the change in θ along C.) Thus it is not true that R C F · dr = 0 for every closed curve C in T. This means that (3) fails. Since (1), (2), (3) are equivalent, they all fail, and F is not conservative. □ Conclusion: F: T →R3 is a vector ﬁeld for which conditions (1), (2), (3) fail, but (4) holds. 16.3. Cylinder in spherical coordinates. Problem 16.5. Let T be the 3D region deﬁned by x2 + y2 ≤3 and 0 ≤z ≤1. Find limits of integration for an iterated integral over T in spherical coordinates. Solution: The iterated integral should have the shape Z θ= Z φ= Z ρ= · · · ρ2 sin φ dρ dφ dθ The range for θ is [0, 2π]. For each θ, the range for φ is [0, π/2] (from straight up to horizontal). For each θ and φ, the range for ρ is from 0 up to something, but the upper limit has a diﬀerent formula depending on whether the ray outward from the origin hits the top surface or the lateral surface of the cylinder, which depends on how φ compares to π/3 (an angle of a right triangle of width √ 3 and height 1). This means the integral must be broken into two. If φ ≤π/3, then the upper limit for ρ satisﬁes z = 1, which is ρ cos φ = 1, so ρ = 1/ cos φ. If φ ≥π/3, then the upper limit for ρ satisﬁes r = √ 3, which is ρ sin φ = √ 3, so ρ = √ 3/ sin φ. Thus the answer is Z 2π θ=0 Z π/3 φ=0 Z 1/ cos φ ρ=0 · · · ρ2 sin φ dρ dφ dθ + Z 2π θ=0 Z π/2 φ=π/3 Z √ 3/ sin φ ρ=0 · · · ρ2 sin φ dρ dφ dθ. Tuesday, November 16 17. Surfaces 17.1. Parametrized curves again. Problem 17.1. Parametrize the upper half of the circle x2 + y2 = 9, counterclockwise. To parametrize a curve in R2, the parameter needs to be one variable t that has a diﬀerent value at each point of the curve, so that the diﬀerent values of the parameter correspond to the diﬀerent points on the curve. Then express the x-coordinate and y-coordinate of the point on the curve in terms of t to get r(t) := x(t) y(t) ! . 115 Solution 1. Let t be the counterclockwise angle from the x-axis, so t is the θ of polar coordinates. The range for t is the interval [0, π], and each number t corresponds to the point r(t) := 3 cos t 3 sin t ! . The function r maps [0, π] in the t-line to the semicircle in the xy-plane. □ Solution 2. Let t = −x; we use −x instead of x so that as the parameter increases, the point moves counterclockwise. The range for t is the interval [−3, 3]. What is the point on the curve with a given t-value? Well, it has x = −t and then solving x2 + y2 = 9 for y gives y = √ 9 −x2 = √ 9 −t2; we use the positive square root since we are parametrizing the upper half of the circle. Thus the parametrization is r(t) := −t √ 9 −t2 ! for t ∈[−3, 3]. □ 17.2. Examples of parametrized surfaces. To parametrize a surface S in R3, one needs two parameters that together specify a point on the surface, and then one needs to express the x-, y-, and z-coordinates of the point in terms of the two parameters. Problem 17.2. Let S be the lateral surface of the cylinder x2 + y2 = 9 for 0 ≤z ≤5. (Lateral means “on the side”. In cylindrical coordinates, the equation would be r = 3; that’s why it’s a cylinder.) What is a parametrization of S? Solution. We can try to use cylindrical coordinates (r, θ, z), but r is not helping to distinguish diﬀerent points on the cylinder because it has the same value 3 at every point of S. So use θ and z as the parameters. What are x, y, z of a point on S in terms of the parameters? Answer: x = 3 cos θ, y = 3 sin θ, and z = z! So the parametrization is r(θ, z) :=    3 cos θ 3 sin θ z   . The range for θ is [0, 2π] and the range for z is [0, 5]. The function r maps the rectangle R := [0, 2π]×[0, 5] onto S. (Strictly speaking, the points with θ = 0 duplicate the points with θ = 2π, so the vertical segment on the right of the cylinder has been doubly parametrized, but we won’t worry about this, because we are interested in surface area, and the doubly-counted segment has zero area.) □ Problem 17.3. Parametrize the surface S of the Earth, assuming that it is perfectly spherical. 116 Solution. One could use latitude and longitude as the two parameters, but let’s use φ and θ. One would not include ρ because it does not help specify the position of a point already known to be on the surface. The x-, y-, and z-coordinates can be expressed in terms of φ and θ, using formulas we have seen before: x = r cos θ = ρ sin φ cos θ y = r sin θ = ρ sin φ sin θ z = ρ cos φ so r(φ, θ) :=    ρ sin φ cos θ ρ sin φ sin θ ρ cos φ,    for φ ∈[0, π] and θ ∈[0, 2π]. (In these formulas, ρ is not a variable: ρ is a constant, namely the radius of the Earth.) The function r maps the rectangle R := [0, π] × [0, 2π] onto S. □ Problem 17.4. Let S be the lateral surface of the cone x2 + y2 = z2 for 0 ≤z ≤5. (In cylindrical coordinates, the equation would be r = z; that’s why it’s a cone.) What is a parametrization of S? Solution. Use the r and θ of cylindrical coordinates. (Using θ and z would also work.) The parametrization is r(r, θ) =    r cos θ r sin θ r   , for r ∈[0, 3] and θ ∈[0, 2π]. Thus r maps the rectangle R := [0, 3] × [0, 2π] onto S. □ 17.3. Parametrized surfaces in general. A region R in R2 can be mapped to a curved surface S in R3. Example: A disk can be mapped to a Salvador Dal´ ı watch. In general, a parametrized surface S is the image of a region R in the uv-plane under a function r: R − →R3 (u, v) 7− →r(u, v) =    x(u, v) y(u, v) z(u, v)   . (One could use any variable names in place of u and v.) Form ru :=    xu yu zu   , rv :=    xv yv zv    117 (these are the two columns of the total derivative of r). At each point (u, v) in R, one can evaluate ru and rv there to get two vectors in R3. The parametrized surface S is smooth at the point corresponding to (u, v) if these two vectors are linearly independent (i.e., nonzero and nonparallel). This condition guarantees that tiny rectangles in R get mapped to tiny “parallelograms” in S (instead of being compressed into a 1-dimensional object, for example). The quotation marks are there because the “parallelograms” in S could be slightly warped and hence are not actual parallelograms. Problem 17.5. Parametrize the lateral surface S of the cone x2 + y2 = z2 for 0 ≤z ≤5 as before. Is S smooth? Solution. Our parametrization was r(r, θ) =    r cos θ r sin θ r   , Then rr =    cos θ sin θ 1   , rθ =    −r sin θ r cos θ 0   . At a point where r = 0, we get rθ = 0. But at a point where r ̸= 0, we have that rθ is a nonzero vector in the xy-plane while rr has a nonzero vertical component so rr and rθ are nonzero and nonparallel. Thus S is smooth except at the points where r = 0, which corresponds to r = 0. This makes sense: the cone is smooth except at its vertex. □ 17.4. Surface area. Suppose that S is a surface parametrized by r: R →S. To approximate the surface area of S, we cut R into tiny rectangles, look at their images under r, and add up the areas of these tiny “parallelograms” covering S. Imagine a tiny rectangle [u, u + du] × [v, v + dv] in the uv-plane. It will be mapped to a “parallelogram” whose sides are the vectors ru du and rv dv. Deﬁne dS := (ru du) × (rv dv) = ru × rv du dv, which we imagine as a tiny vector. What is the geometric meaning of dS? • The length of dS is the area of the parallelogram, which should be thought of as a piece of surface area, denoted dS := \|ru × rv\| du dv. • The direction of dS is a unit normal vector n to the surface. 118 So dS = n dS. Deﬁne the surface area of S to be Area(S) := x R dS = x R \|ru × rv\| du dv. Thursday, November 18 17.4.1. Surface area of a hemisphere. Problem 17.6. Let S be the upper half of the unit sphere centered at the origin. What is its surface area? (There are many ways to do this; our purpose here is to use the general method for calculating surface area via a parametrization.) Solution. We already know a parametrization of S: the function r(φ, θ) :=    sin φ cos θ sin φ sin θ cos φ    maps the rectangle R := [0, π/2] × [0, 2π] in the φθ-plane onto S. Then rφ =    cos φ cos θ cos φ sin θ −sin φ   , rθ =    −sin φ sin θ sin φ cos θ 0   , 119 so rφ × rθ = e1 e2 e3 cos φ cos θ cos φ sin θ −sin φ −sin φ sin θ sin φ cos θ 0 = (sin2 φ cos θ)e1 + (sin2 φ sin θ)e2 + (sin φ cos φ cos2 θ + sin φ cos φ sin2 θ)e3 = (sin2 φ cos θ)e1 + (sin2 φ sin θ)e2 + (sin φ cos φ)e3 = (sin φ)r \|rφ × rθ\| = sin φ dS = sin φ dφ dθ Area(S) = Z 2π 0 Z π/2 0 sin φ dφ dθ = Z 2π 0 dθ = 2π. □ Note: The geometric interpretation of the cross product shows that the vector rφ × rθ is perpendicular to the sphere at each point; this explains why it turned out to be a scalar multiple of r. 17.5. Surface integrals. Recall: For a curve C, Z C f ds = Z b a f(r(t)) \|r′(t)\| dt. (That is, we compute the left side by choosing a parametrization r(t) of C, for t ∈[a, b].) Similarly: If S is a surface, and f = f(x, y, z), then x S f dS = x R f(r(u, v)) \|ru × rv\| du dv, where r(u, v) is a parametrization of S, mapping the points (u, v) ∈R to the points of S. To be remembered: 1. To compute a 1-dimensional integral on a curve C, you must choose a parametrization r(t) to convert it to an integral on a straight interval [a, b] in the real line. 2. Generally, to compute a 2-dimensional integral on a curved surface S, you must choose a parametrization r(u, v) to convert it to an integral over a ﬂat region R in the uv-plane. The only exceptions to this are: 120 • If the integrand is a constant c, then Z C c ds = c Length(C) x S c dS = c Area(S), and you might know Length(C) or Area(S) already by geometry. • Sometimes you can use the fundamental theorem of calculus to calculate a line integral. Similarly, sometimes you can use fancy versions of the fundamental theorem of calculus called the divergence theorem and Stokes’ theorem (coming soon), to convert the integral to some other kind of integral. 17.6. Applications of surface integrals. Deﬁnition 17.7. The average value of f on a surface S is ¯ f := s S f dS Area(S) If S is a metal surface whose density (mass per unit area) at each point is given by a function δ: S →R, then dm = δ dS, so the quantities mass m := x S dm mass-weighted average ¯ f := s S f dm m centroid := (¯ x, ¯ y, ¯ z) moment of inertia I := x S (distance to axis)2 dm all involve surface integrals. 17.7. Flux. Suppose that F is a continuous 3D vector ﬁeld, and S is a parametrized surface. Then one gets dS = n dS, where n is a unit normal at each point of S. The ﬂux of F across S is a special kind of surface integral: ﬂux of F across S := x S F · n normal component of F dS = x S F · dS. There are actually two choices for n at each point, and the parametrization is specifying one of them; if one wants ﬂux for the opposite direction of ﬂow, negate the integral. If S bounds a 3D region (like a sphere enclosing a ball), then one usually chooses n to be the outward unit normal. 121 17.7.1. Physical meaning of ﬂux. Intuitive explanation: Imagine a 3D ﬂuid with constant velocity ﬁeld F. The amount of ﬂuid that ﬂows across a tiny parallelogram of area dS in unit time is the ﬂuid in a parallelepiped of base dS and height F · n. Summing over all tiny parallelepipeds comprising a surface S gives x S F · n dS. That’s ﬂux! This intuition works more generally for any continuous F and smooth S, since F is almost constant on small regions, and S is well approximated by tiny parallelograms. Flux measures the rate of ﬂow (volume per unit time, measured in m3/s, say). 17.7.2. Computing 3D ﬂux. To compute a ﬂux integral x S F · dS : (1) Choose a parametrization of S, say r(u, v) for (u, v) in the ﬂat 2D region R. (The whole point is to convert the surface integral over the curved surface S into a double integral over the ﬂat region R.) (2) Compute the partial derivatives ru, rv, and their cross product ru × rv. (3) Make sure that ru × rv is in the same direction as the desired unit normal n. (If it is in the opposite direction, negate your answer at the end.) (4) Substitute dS = ru × rv du dv. to get x S F · dS = x R F(r(u, v)) · (ru × rv) du dv. (5) Evaluate the double integral over the ﬂat 2-dimensional region R; usually this will be done by converting it to an iterated integral. Example 17.8. Let S in R3 be deﬁned by x2 + y2 = 4 and 0 ≤z ≤3. (So S is the lateral surface of a cylinder.) Let F(x, y, z) =    yz x2 + 2 5   . What is the outward ﬂux of F across S? The solution below was ﬁnished on the next day. Solution. Use the θ and z of cylindrical coordinates as parameters (r is the constant 2, hence not usable as a parameter). The parametrization is r(θ, z) :=    2 cos θ 2 sin θ z   , 122 for 0 ≤θ ≤2π and 0 ≤z ≤3. Then rθ =    −2 sin θ 2 cos θ 0   , rz =    0 0 1   , so rθ × rz = e1 e2 e3 −2 sin θ 2 cos θ 0 0 0 1 =    2 cos θ 2 sin θ 0   . Notice that rθ × rz is in the same direction as the outward unit normal n. Next dS = rθ × rz dθ dz =    2 cos θ 2 sin θ 0   dθ dz. So the ﬂux is x S F · dS = Z 3 0 Z 2π 0    yz x2 + 2 5   ·    2 cos θ 2 sin θ 0   dθ dz = Z 3 0 Z 2π 0 yz(2 cos θ) + (x2 + 2)(2 sin θ) dθ dz = Z 3 0 Z 2π 0 (2 sin θ)z(2 cos θ) + ((2 cos θ)2 + 2)(2 sin θ) dθ dz = Z 3 0 Z 2π 0 2z sin 2θ + 8 cos2 θ sin θ + 4 sin θ dθ dz = 0, because R 2π 0 sin 2θ dθ is the integral over two complete cycles, which is 0, and the last two terms in the integrand are negated by θ 7→θ + π so their integrals over [0, π] cancel with the integrals over [π, 2π]. □ Friday, November 19 123 18. The divergence theorem 18.1. Flux across the faces of a box. Let us explain the connection between ﬂux and divergence. Imagine a rectangular box [x0, x1] × [y0, y1] × [z0, z1] whose dimensions ∆x := x1 −x0 ∆y := y1 −y0 ∆z := z1 −z0 are very small. Let F(x, y, z) :=    P(x, y, z) Q(x, y, z) R(x, y, z)    be a continuously diﬀerentiable 3D vector ﬁeld; this means that all nine partial derivatives, like Py and Rz, are continuous functions. Question 18.1. What is the ﬂux of F across the top face of the box? Solution. On the top face, z takes the constant value z1, and x and y can be used as parameters. The geometry shows that dS = dx dy and n = e3 (no need to calculate a cross product). Thus the ﬂux across the top face is Z top face F · n dS = Z y1 y=y0 Z x1 x=x0 F(x, y, z1) · e3 dx dy = Z y1 y=y0 Z x1 x=x0 R(x, y, z1) dx dy. □ Question 18.2. What is the ﬂux of F across the bottom face of the box? Solution. On the bottom face, z takes the constant value z0, and x and y can again be used as parameters. The geometry shows that dS = dx dy and n = −e3. Thus the ﬂux across the bottom face is Z bottom face F · n dS = Z y1 y=y0 Z x1 x=x0 F(x, y, z0) · (−e3) dx dy = Z y1 y=y0 Z x1 x=x0 −R(x, y, z0) dx dy. □ Question 18.3. If the box is very small, what is an estimate for the net ﬂux through both the top and bottom faces? Solution. Adding the previous two answers gives Z y1 y=y0 Z x1 x=x0 (R(x, y, z1) −R(x, y, z0)) dx dy. 124 If the height ∆z is very small, then we can use the relative change formula R(x, y, z1) −R(x, y, z0) ≈Rz ∆z (there is only one term instead of three, since x and y are not changing). Moreover, if the box is very small, then the continuous function Rz is approximately constant on the box, so Z y1 y=y0 Z x1 x=x0 (R(x, y, z1) −R(x, y, z0)) dx dy ≈ Z y1 y=y0 Z x1 x=x0 Rz ∆z dx dy ≈Rz ∆z Z y1 y=y0 Z x1 x=x0 dx dy = Rz ∆z (∆x ∆y) = Rz Vol(box). □ Question 18.4. If the box is very small, what is an estimate for the net outward ﬂux across all six faces of the box? Solution. Using similar arguments for each opposite pair of faces gives Faces Approximate ﬂux through them top and bottom Rz Vol(box) front and back Px Vol(box) left and right Qy Vol(box) total div F Vol(box) □ Conclusion: div F ≈net outward ﬂux across the sides the box volume of the box . Thus, at each point, div F measures the ﬂux out of a tiny box, per unit volume. 18.2. Divergence as a source rate. Suppose that F is the velocity ﬁeld of an incompressible 3D ﬂuid. Suppose also that ﬂuid is being created at each point of the region (maybe there are tiny pipes pumping ﬂuid into each location). The source rate at a point is the rate at which ﬂuid is being created per unit volume. Incompressible means that the amount of ﬂuid within any region is constant, so if ﬂuid is being created inside a tiny box at some rate, then ﬂuid must overﬂow out of the box at the same rate. In other words, source rate = outward ﬂux per unit volume = div F at each point. 125 18.3. Bounded regions. What about the outward ﬂux across the boundary of a larger region? That is what the divergence theorem is about. But ﬁrst we have to discuss what kind of regions are allowed. Deﬁnition 18.5. A region T in R3 is called bounded if there is a ball that contains it. The big mouth test: Does eating T in one bite require just a big mouth, or an inﬁnitely big mouth? If a big mouth is enough, then T is bounded. Question 18.6. Which of the following regions are bounded? • a solid cube? Yes. • the orthant deﬁned by x, y, z ≥0? No, because it has points arbitrarily far from the origin — no ball can contain them all. • a solid torus? Yes. • the solid inﬁnite cylinder in R3 deﬁned by x2 + y2 ≤4? No. • R3 itself? No! 18.4. Closed surfaces. Deﬁnition 18.7. A closed surface S is a (piecewise smooth) surface that is the entire boundary of a bounded region T in R3. To say that it is positively-oriented means that at each (smooth) point of S we choose the outward unit normal n. Question 18.8. Which of the following are closed surfaces? (1) the sphere x2 + y2 + z2 = 9? Yes, it is the boundary of a ball. (2) (the outside of) a cube? Yes, it is the boundary of the solid cube. (3) (the outside of) a torus? Yes, it is the boundary of the solid torus. (4) the lateral part of a cylinder? No, since the boundary of a solid cylinder would have to include the top and bottom disks too. (5) the inﬁnite cylinder x2 + y2 = 4 in R3? No — it is the boundary of a solid inﬁnite cylinder, but the solid inﬁnite cylinder is not bounded. An integral over a closed surface S is sometimes written using the notation v S, although s S is also correct. 18.5. Divergence theorem. Theorem 18.9 (Divergence theorem, also known as Gauss’s theorem or Ostrogradsky’s theorem — actually discovered earlier by Lagrange). Let S be a positively-oriented closed surface bounding a region T in R3. Let F be a vector ﬁeld that is continuously diﬀerentiable not only on S but also on T. Then { S F · dS = y T div F dV. 126 18.6. Physical interpretation of the divergence theorem. Suppose that F is the ve-locity ﬁeld of an incompressible 3D ﬂuid. As we saw before, div F is the source rate (the rate at which ﬂuid is being created per unit volume). Thus div F dV is the rate of ﬂuid creation inside the little piece of volume dV . Integrating shows that the right side of the divergence theorem is the rate of ﬂuid creation inside all of T. But the ﬂuid is incompressible, so ﬂuid must be ﬂowing out of T at the same rate, to keep the amount of ﬂuid inside T constant. The rate at which ﬂuid is ﬂowing out of T is the outward ﬂux of F across the boundary of T, which is the left side of the divergence theorem. Tuesday, November 23 18.7. Another explanation of the divergence theorem. Suppose that two tiny boxes are stacked next to each other to form a larger box. Then the ﬂux out of box 1 plus the ﬂux out of box 2 equals the ﬂux out of the combined box, because the ﬂuxes across the now-interior face are computed in opposite directions and hence cancel. The same principle holds for stacking many tiny boxes: adding up the ﬂuxes out of each box gives the total ﬂux out of their union U. Now T can be approximated by such a union U of tiny boxes. So ﬂux out of U ≈ the sum of the ﬂuxes out of the tiny boxes (5) What happens in the limit as the boxes get smaller, so that U approximates T better and better? The left side of (5) tends to the ﬂux out of T, which is v S F · dS, the outward ﬂux across the boundary of T. The ﬂux out of each tiny box of volume dV is div F dV , and we are adding these up. Thus in the limit, (5) becomes { S F · dS = y T div F dV. This should make the divergence theorem believable even if we did not justify the details. Good enough for physics! 18.8. The extended divergence theorem. Sometimes it happens that the boundary of a region T has more than one connected component. Example 18.10. If T is the solid spherical shell deﬁned by 4 ≤x2 + y2 + z2 ≤9, then the boundary of T with outward unit normal consists of an outer sphere S of radius 3 with outward unit normal and an inner sphere S′ of radius 2 with inward unit normal (because outward from the point of view of T is towards the center). 127 The extended divergence theorem says that if the boundary of T consists of positively-oriented closed surfaces S1, . . . , Sn, then { S1 F · dS + · · · + { Sn F · dS = y T div F dV. In the example above, the extended divergence theorem becomes { S F · dS − { S′ F · dS = y T div F dV (6) if each sphere is oriented so that n points away from the origin. The minus sign is there because the orientation of S′ in the extended divergence theorem (from the point of view of T) is given by the unit normal pointing towards the origin, the opposite of what we are using in (6). 18.9. Divergence and gravitation. Lemma 18.11. Let F be a 3D vector ﬁeld pointing radially outward and whose magnitude is 1/ρ2 (deﬁned everywhere except the origin). Then div F = 0 at every point except the origin. Proof. Since the radially outward unit vector is r \|r\| = r ρ, an explicit formula for F is F = 1 ρ2 r ρ = ρ−3    x y z   =    ρ−3x ρ−3y ρ−3z   . To calculate div F, we need to calculate partial derivatives of the three coordinate functions, while remembering that ρ is really a function of x, y, z. We could just substitute ρ = p x2 + y2 + z2 and calculate the partial derivatives explicitly. Alternatively, taking ∂ ∂x of ρ2 = x2 + y2 + z2 gives 2ρ∂ρ ∂x = 2x, so ∂ρ ∂x = x ρ. Now ∂ ∂x(ρ−3x) = ρ−3 + (−3ρ−4)∂ρ ∂xx = ρ−3 −3ρ−5x2. 128 Summing this with the corresponding equations for y and z gives div F = 3ρ−3 −3ρ−5(x2 + y2 + z2) = 3ρ−3 −3ρ−5ρ2 = 0. □ Let F = F(x, y, z) be the gravitational ﬁeld of a point mass M at (0, 0, 0) (i.e., the force that it would exert on a unit mass at (x, y, z)). By Newton’s inverse square law, F = −GM ρ2 r ρ. This is just a constant times the vector ﬁeld in Lemma 18.11, so div F = 0 at every point in R3 except (0, 0, 0). Question 18.12. Let F be the gravitational ﬁeld of a point of mass M at the origin. Let Sa be the sphere of radius a centered at (0, 0, 0). Let S2a be the sphere of twice the radius. Let’s compare the ﬂux across Sa with the ﬂux across S2a. Which of the following is correct? (1) The ﬂux across S2a is 4 times as much, because the integral is over a surface area that is 4 times bigger. (2) The ﬂux across S2a is 1/4 as much, because the gravitational ﬁeld is 1/4 as strong. (3) The ﬂuxes are equal and nonzero. (4) The ﬂuxes are both 0, because the divergence theorem says { Sa F · dS = y T div F dV = 0, since div F = 0 everywhere. Answer: The ﬂuxes are equal and nonzero. (The two eﬀects in the ﬁrst two answers cancel each other out. The application of the divergence theorem in the last answer is wrong: F is not deﬁned at (0, 0, 0), so the right side of the divergence theorem does not even make sense.) Question 18.13. What is the ﬂux across the sphere Sa of radius a centered at (0, 0, 0)? 129 Answer: At every point, F and n are in opposite directions, so ﬂux = { Sa F · n dS = { Sa −\|F\| dS = { Sa −GM a2 dS = −GM a2 Area(Sa) = −GM a2 (4πa2) = −4πGM. It is independent of the radius a! Even better, if S is any closed surface enclosing the point mass, and T is the 3D region between S and a small sphere Sa centered at the point mass (so T has a bubble inside), then the extended divergence theorem shows that { S F · dS − { Sa F · dS = y T div F dV = 0, so { S F · dS = { Sa F · dS = −4πGM. The same claim is true if there are many point masses inside S, or even some planets inside S, because the force ﬁelds add. This proves Gauss’s law: gravitational ﬂux across S = −4πGM, where M is the mass enclosed by S. Problem 18.14. What is the gravitational ﬁeld inside a hollow spherical planet? Solution. Inside a centered sphere S of radius r inside the hollow part, symmetry implies that F = cn for some c depending only on r. Then the ﬂux across S is 4πr2c, but Gauss’s law says that it equals 0, so c = 0. Thus the gravitational ﬁeld is 0 everywhere inside the hollow part. □ 130 Challenge question 18.15. For a donut-shaped planet, if you are standing on the inner circle of the planet, is gravity pulling you towards the center of mass or is it pulling you towards the planet under your feet? Hint: Imagine ﬁlling in the donut hole with a cylinder of very small height. The total ﬂux through the cylinder is 0 by Gauss’s law. On the other hand, is the outward ﬂux through the top and bottom disks positive or negative? If you ﬁgure that out, that can help answer the question, because it must be cancelled by the ﬂux through the lateral surface of the cylinder. Tuesday, November 30 18.10. Application of the divergence theorem to an electric ﬁeld. Place a point charge Q at (0, 0, 0). Let E = E(x, y, z) be the electric ﬁeld it creates. Coulomb’s law: E = Q/4πϵ0 ρ2 r ρ, where ϵ0 is a constant. The physics is diﬀerent, but the math is the same as for gravitation, with the constant Q/4πϵ0 in place of −GM. The gravitational ﬂux was −4πGM, which is 4π times the constant appearing in the inverse square law for gravitation. Similarly, the electric ﬂux is 4π(Q/4πϵ0) = Q ϵ0 . Summary: ﬁeld ﬂux gravitational F = −GM ρ2 r ρ −4πGM electric E = Q/4πϵ0 ρ2 r ρ Q/ϵ0 So we get Gauss’s law for an electric ﬁeld (also called the Gauss–Coulomb law): The electric ﬂux across a closed surface S equals the charge enclosed divided by ϵ0, { S E · dS = Q ϵ0 . 131 19. Stokes’ theorem 19.1. Curves bounding surfaces. In R3, let C be a closed curve, and let S be a bounded surface whose boundary is C. Roughly speaking, S “ﬁlls in” C. (Here S is usually not a closed surface; it usually does not enclose a 3-dimensional region.) For example, C could be the Equator on the Earth, and S could be the Northern Hemisphere. 1. An orientation of C is a choice of direction along C. 2. An orientation of S is a (continuously varying) choice of unit normal vector n at each point of S. For Stokes’ theorem to hold, these two choices must be compatible. There are two equivalent ways to say what compatible means: • If you walk along C in the chosen direction, with S to your left, then n is pointing up. • If your right thumb is pointing in the direction of n, then your ﬁngers point in the chosen direction along C. In the example above, if the orientation of the Equator C is east, then the compatible orientation of the Northern Hemisphere S is the one that has n pointing up into space at each point. 19.2. The theorem. Theorem 19.1 (Stokes’ theorem5). In R3, let C be an oriented piecewise smooth curve, and let S be an oriented, bounded, piecewise smooth surface with boundary C. Assume that the orientations of C and S are compatible. Let F be a 3D vector ﬁeld that is continuously diﬀerentiable everywhere on S. Then I C F · dr = x S (curl F) · dS. Remark 19.2. The orientation of C is needed to deﬁne H C F · dr, and the orientation of S is needed to deﬁne dS = n dS in s S(curl F)·dS. If these orientations are not chosen compatibly, the two sides in Stokes’ theorem will diﬀer by a sign. 19.3. Example. Problem 19.3. Let S be the boundary of the cylinder x2 + y2 ≤9, 0 ≤z ≤2 excluding the base in the xy-plane. Let F =    y −x y3   . Compute the outward ﬂux of curl F across S in as many ways as you can. 5Actually ﬁrst proved by Lord Kelvin, who mailed it to Stokes, who included it as a question in a physics competition for students. 132 Solution 1: Use Stokes’ theorem to convert it to a line integral on the circle C given by x2 + y2 = 9 in the xy-plane. ﬂux of curl F across S = x S (curl F) · dS Stokes’ = I C F · dr. To compute the latter, choose a parametrization of C (counterclockwise, so as to agree with the orientation of S). Let’s use r(t) =    3 cos t 3 sin t 0    for 0 ≤t ≤2π, so dr =    −3 sin t 3 cos t 0   dt I C F · dr = Z 2π 0    3 sin t −3 cos t (3 sin t)3   ·    −3 sin t 3 cos t 0   dt = Z 2π 0 −9 sin2 t −9 cos2 t dt = Z 2π 0 −9 dt = −18π. Solution 2: Compute the ﬂux directly from the deﬁnition. The surface S consists of the top disk S1 and the lateral surface S2 of the cylinder. We have curl F = e1 e2 e3 ∂ ∂x ∂ ∂y ∂ ∂z y −x y3 = 3y2e1 −2e3. On S1, the unit normal is n = e3, so (curl F) · n = −2 x S1 (curl F) · n dS = −2 Area(S1) = −18π. 133 For S2, we use the z and θ of cylindrical coordinates as parameters: r(z, θ) :=    3 cos θ 3 sin θ z    for z ∈[0, 2] and θ ∈[0, 2π]. Then rz =    0 0 1    rθ =    −3 sin θ 3 cos θ 0    rz × rθ = e1 e2 e3 0 0 1 −3 sin θ 3 cos θ 0 = (−3 cos θ) e1 + (−3 sin θ) e2. This gives the opposite of the desired orientation for n (the vector rz × rθ is horizontally inward instead of outward), so insert a minus sign to get n dS = −rz × rθ dz dθ, so that the outward ﬂux is x S1 (curl F) · n dS = − Z 2π 0 Z 2 0 (curl F) · (rz × rθ) dz dθ = − Z 2π 0 Z 2 0    3(3 sin θ)2 0 −2   ·    −3 cos θ −3 sin θ 0   dz dθ = − Z 2π 0 Z 2 0 −81 sin2 θ cos θ dz dθ = Z 2π 0 162 sin2 θ cos θ dθ = 54 sin3 θ 2π 0 = 0. So the total ﬂux of curl F across S is −18π + 0 = −18π. 134 Thursday, December 2 Solution 3: Use the divergence theorem to compute the ﬂux of curl F across the entire boundary of the cylinder, and then subtract the ﬂux across the bottom disk. We already calculated that curl F = 3y2 e1 −2 e3 so div(curl F) = ∂ ∂x3y2 + ∂ ∂y0 + ∂ ∂z(−2) = 0 + 0 + 0 = 0. (In fact, it turns out that div(curl F) = 0 holds for any vector ﬁeld whose component functions have continuous second partial derivatives.) By the divergence theorem, the outward ﬂux of curl F across the entire boundary of the cylinder is { entire boundary (curl F) · dS = y solid cylinder div(curl F) dV = y solid cylinder 0 dV = 0. On the other hand, the outward ﬂux across the bottom disk of the cylinder is x bottom disk (curl F) · dS = x bottom disk (curl F) · (−e3) dS (since the outward unit normal is downward) = x bottom disk (3y2e1 −2e3) · (−e3) dS = x bottom disk 2 dS = 2 Area(bottom disk) = 2(π · 32) = 18π. Subtracting gives the ﬂux of curl F through S only: x S (curl F) · dS = 0 −18π = −18π. 19.4. Line integral around a tiny rectangle. Remember how we explained the divergence theorem by approximating the 3D region T by a union of little boxes and by showing that the ﬂux out of each little box was approximated by div F dV ? We are now going to explain Stokes’ theorem in a similar way by approximating the surface S by a union of tiny rectangles hinged together (like a disco ball) and by showing that the line integral around each little rectangle of area dS is approximated by (curl F) · n dS. 135 Imagine a horizontal rectangle R = [x0, x1] × [y0, y1] × {z0} whose dimensions ∆x := x1 −x0 ∆y := y1 −y0 are very small. Its area, which we’ll call dS, is ∆x ∆y. Its boundary Ctiny is a closed curve consisting of four line segments, oriented as shown: Cbottom Cright Ctop Cleft (x0, y0, z0) (x1, y0, z0) (x1, y1, z0) (x0, y1, z0) R Question 19.4. What is the line integral R Cbottom F · dr along the bottom? Answer. The path Cbottom is parametrized by (x, y0, z0) for x ∈[x0, x1], so Z Cbottom F · dr = Z x1 x=x0 F(x, y0, z0) · e1 dx = Z x1 x=x0 P(x, y0, z0) dx. □ Question 19.5. What is the line integral R Ctop F · dr along the top? Answer. The path Ctop is parametrized backwards by (x, y1, z0) for x ∈[x0, x1], so Z Ctop F · dr = − Z x1 x=x0 F(x, y1, z0) · e1 dx = − Z x1 x=x0 P(x, y1, z0) dx. □ Question 19.6. What is an estimate for the line integrals along the top and bottom combined? (They almost cancel.) Answer. Z Cbottom F · dr + Z Ctop F · dr = Z x1 x=x0 (P(x, y0, z0) −P(x, y1, z0)) dx ≈− Z x1 x=x0 Py(x, y0, z0) ∆y dx (linear approximation) ≈−Py ∆y Z x1 x=x0 dx (Py is nearly constant on the rectangle) = −Py ∆y ∆x = −Py dS. □ 136 Question 19.7. What is an estimate for the line integral H Ctiny F · dr around the whole tiny rectangle? Answer. A similar calculation shows that Z Cright F · dr + Z Cleft F · dr ≈ Qx dS. (7) Combining and answer to Question 19.6 and (7) gives I Ctiny F · dr ≈ (Qx −Py) dS = (curl F) · n dS, (8) since curl F =    Ry −Qz Pz −Rx Qx −Py   and n =    0 0 1   since R is horizontal. Moreover, the smaller R is, the better the approximation. □ The same will hold for a tiny rectangle in any position in space, because of the geometric invariance of work and curl under rotations of space (they have a meaning that is independent of the coordinate system you are working in). 19.5. Explaining Stokes’ theorem for a general surface. Suppose that two tiny rect-angles are hinged along one side. Then adding the line integrals around each gives the line integral around the combined surface, since the line integrals along the hinge segment cancel. The same holds for a surface built out of more than two rectangles. Therefore, when equation (8) is summed over all these rectangles, the left side sum becomes the line integral H C F · dr around the boundary of the whole surface; meanwhile, in the limit as the rectangles’ sizes tend to 0, the right side sum tends to a surface integral and the approximation becomes an equality: I C F · dr = x S (curl F) · n dS. This should make Stokes’ theorem believable even if we did not justify the details. 19.6. Conservative vector ﬁelds and Stokes’ theorem. Do you remember the four conditions (1) F is a gradient ﬁeld (equal to ∇f for some f = f(x, y, z) on T). (2) R B A F · dr is path independent for every A and B in T (for paths inside T). (3) H C F · dr = 0 for every closed curve C inside T. (4) curl F = 0 at every point of T. on a continuously diﬀerentiable 3D vector ﬁeld F? We explained why (1), (2), (3) are equivalent, and why any of (1), (2), (3) implies (4). We also claimed that when T is simply connected, then all four conditions are equivalent, but we never explained why (4) implied any of (1), (2), (3). Now that we have Stokes’ theorem, we can ﬁnally explain this. 137 Proposition 19.8. When T is simply connected, (4) implies (3). Sketch of proof. Suppose that T is simply connected and that (4) holds, so curl F = 0 at every point of T. We need to prove that (3) holds. Since (3) is a statement about every closed curve in T, the proof begins by letting C be any closed curve in T, and we need to show that H C F · dr = 0. Because T is simply connected, the curve C is shrinkable to a point in T. A time lapse photograph of the shrinking would show C tracing out a surface S as it shrinks, and S is contained in T. It turns out that if C is piecewise smooth, then S can be arranged to be piecewise smooth too, so that Stokes’ theorem applies: I C F · dr = x S (curl F) · dS = 0, since curl F = 0 everywhere on S by the assumption (4). □ 19.7. Extended Stokes’ theorem. Sometimes the boundary of a surface S in R3 may consist of several closed curves C1, . . . , Cn. In this situation, the extended Stokes’ theorem says I C1 F · dr + · · · + I Cn F · dr = x S (curl F) · dS. It can also be written I ∂S F · dr = x S curl F · dS where ∂S is an abbreviation for the boundary of S. (The ∂is the same symbol used for partial diﬀerentiation, but its meaning here is diﬀerent!) Example 19.9. If S is a pair of pants, oriented by the outward unit normal, then ∂S consists of three curves: the waistline, and the two cuﬀs at the bottom. The waistline is oriented clockwise when viewed from the top. At this point, all the material on the ﬁnal exam has been covered. The remaining topics are partly for review, and partly to show how the material is used in physics and elsewhere. Friday, December 3 138 20. Green’s theorem Stokes’ theorem and the divergence theorem have 2D versions: I C F · dr ﬂux 2D Green’s theorem Green’s theorem for ﬂux 3D Stokes’ theorem divergence theorem The 2D versions are less important for physics, but they come up sometimes. 20.1. Green’s theorem is Stokes’ theorem in 2D. 20.1.1. Setup. R: a bounded region in the xy-plane C: the boundary of R, assumed to be a piecewise smooth closed curve F(x, y) = P(x, y) Q(x, y) ! : a continuously diﬀerentiable 2D vector ﬁeld on R. Viewing R as a surface in R3, orient R by choosing n = e3 (up, out of the xy-plane). What is the compatible orientation of C? Counterclockwise, by the right hand rule. Stokes’ theorem is for 3D vector ﬁelds, not 2D vector ﬁelds. So build from F a 3D vector ﬁeld ˜ F(x, y, z) :=    P(x, y) Q(x, y) 0   . Then curl ˜ F = e1 e2 e3 ∂ ∂x ∂ ∂y ∂ ∂z P(x, y) Q(x, y) 0 = (Qx −Py)e3. (9) The expression Qx −Py is sometimes called the 2D scalar curl of the 2D vector ﬁeld F. 20.1.2. Application of Stokes’ theorem. Apply Stokes’ theorem to ˜ F on R. Left side of Stokes’ theorem: I C ˜ F · dr = I C    P Q 0   ·    dx dy dz    139 is the same as I C F · dr = I C P dx + Q dy . Right side of Stokes’ theorem: x R (curl ˜ F) · dS = x R (curl ˜ F) · n dS = x R (Qx −Py)e3 · e3 dA = x R (Qx −Py) dA . So the result of applying Stokes’ theorem is Theorem 20.1 (Green’s theorem). For the setup above, I C P dx + Q dy = x R (Qx −Py) dA. Another way to write the same equation: I C F · dr = x R curl F 2D scalar curl dA. 20.1.3. Computing area with Green’s theorem. Let R and C be as in the setup for Green’s theorem. If we choose P = 0 and Q = x, then Qx −Py = 1, so Green’s theorem (with the two sides reversed) gives Area(R) = I C x dy. If we choose P = −y and Q = 0, then Qx −Py = 1 again, so Area(R) = I C −y dx. These formulas give the area of R in terms of a calculation along the boundary! In the 1800s, people invented the planimeter, a mechanical device with an arm attached to a pencil such that if the pencil traces out a closed curve, the device calculates the area enclosed. It works by eﬀectively calculating H C x dy. 20.2. Green’s theorem for ﬂux is the divergence theorem in 2D. 140 20.2.1. Setup. R: a bounded region in the xy-plane C: the boundary of R, assumed to be a piecewise smooth closed curve n: the outward unit normal in R2 at each point of C F(x, y) = M(x, y) N(x, y) ! : a continuously diﬀerentiable 2D vector ﬁeld on R. The divergence theorem requires a 3D vector ﬁeld on a 3D region T. So build from F a 3D vector ﬁeld ˜ F(x, y, z) :=    M(x, y) N(x, y) 0   and thicken R into a slab T := R × [0, 1]. The boundary S of T consists of the lateral surface C × [0, 1], the bottom R × {0}, and the top R × {1}. Equip S with the outward unit normal vector ˜ n at each point. 20.2.2. Application of the divergence theorem. Apply the divergence theorem to ˜ F on T. Left side of the divergence theorem: First, what is ˜ F · ˜ n at a boundary point (x, y, z)? • along the lateral surface, ˜ F · ˜ n at (x, y, z) has the same value as F · n at (x, y); • along the top, ˜ F · ˜ n = 0 since ˜ F is horizontal while ˜ n is vertical; • along the bottom, ˜ F · ˜ n = 0 since ˜ F is horizontal while ˜ n is vertical. Thus ﬂux across S = ﬂux across the lateral surface C × [0, 1] = x C×[0,1] ˜ F · ˜ n dS = I C Z 1 z=0 F · n dz ds = I C (F · n) Z 1 z=0 dz ds (since F · n does not depend on z) = I C F · n ds ﬂux of 2D ﬁeld F across C . 141 Right side of the divergence theorem: First, div ˜ F = Mx + Ny, which we also call the 2D divergence of F, denoted div F. So the right side is y T div ˜ F dV = x R Z 1 z=0 div F dz dA = x R (div F) Z 1 z=0 dz dA = x R div F dA . So the result of applying the divergence theorem is Theorem 20.2 (Green’s theorem for ﬂux, also called Green’s theorem in normal form). For the setup above, I C F · n ds ﬂux of 2D ﬁeld F across C = x R div F 2D divergence dA . (10) Now n ds is the 90◦clockwise rotation of T ds = dr = dx dy ! in the plane, so on the left of (10) we have F · n ds = M N ! · dy −dx ! = −N dx + M dy, while on the right of (10) we have div F = Mx + Ny, so (10) can be rewritten as I C −N dx + M dy = x R (Mx + Ny) dA . (11) Mathematically, Green’s theorem and Green’s theorem for ﬂux are essentially the same: Taking P = −N and Q = M in Green’s theorem gives Green’s theorem for ﬂux. Tuesday, December 7 21. Generalized Stokes’ theorem The fundamental theorem of calculus, the fundamental theorem of calculus for line integrals, the divergence theorem, and Stokes’ theorem all ﬁt the template “ Z ∂R F = Z R dF”, where ∂R is the boundary of R and dF is some kind of derivative/diﬀerential of F. 142 FTC: The boundary of an interval [a, b] is a set of two points {a, b}. f(b) −f(a) = Z b a f ′(x) dx. FTC for line integrals: The boundary of an oriented curve C from A to B is {A, B}. f(B) −f(A) = Z C ∇f · dr. Divergence theorem: The boundary of a 3D region T is a closed surface S (or many closed surfaces). Let F be a 3D vector ﬁeld. { S F · dS = y T div F dV. Stokes’ theorem: The boundary of a surface S in R3 is a closed curve C (or many closed curves). Let F be a 3D vector ﬁeld. I C F · dr = x S curl F · dS. Can you make Green’s theorem and Green’s theorem for ﬂux also ﬁt the template? Yes, try it! In fact, all six theorems are special cases of a generalized Stokes’ theorem that applies to diﬀerential forms on n-dimensional manifolds for any n! 22. Maxwell’s equations Maxwell’s equations relate electric and magnetic ﬁelds. They can be expressed in diﬀerential form (involving the derivative-like operators div and curl), or in integrated form (involving line integrals and surface integrals). Mathematics proves none of them. Instead, the role of mathematics is to show that the diﬀerential form is mathematically equivalent to the integrated form. 22.1. Deﬁnitions of the physical quantities. Introduce the following quantities (in SI units): E := electric ﬁeld (newtons/coulomb = volt/m) B := magnetic ﬁeld (teslas) ρ := charge density (coulomb/m3) J := current density (amp/m2) t := time (s) µ0, ϵ0 are constants. 143 22.2. The two forms of Maxwell’s equations. Here are Maxwell’s equations in diﬀerential form: div E = ρ ϵ0 div B = 0 curl E = −dB dt curl B = µ0ϵ0 dE dt + µ0J. And here are Maxwell’s equations in integrated form: { S E · dS = Q ϵ0 (Gauss-Coulomb law) { S B · dS = 0 (Gauss’s law for magnetism) I C E · dr = −d dt x S B · dS (Faraday’s law) I C B · dr = µ0ϵ0 d dt x S E · dS + µ0IS (Amp ere’s law). In the ﬁrst two equations, S is a closed surface, and Q is the charge enclosed by S. In the last two equations, C is the boundary curve of a surface S, and IS is the current ﬂowing through S (that is, the ﬂux of J across S). As you can see, the diﬀerential form of the equations is a little simpler. But the integrated form tells us about the macroscopic quantities that we ultimately care about. (A similar thing happens with gravitation: Newton’s law for gravitation is simplest when expressed as a law about force or acceleration, but integrating it twice gives information about position, which is ultimately what we care about.) 22.3. Equivalence of the two forms. Let’s prove the equivalence of the two forms of the third Maxwell equation (Faraday’s law). The key to the proof will be Stokes’ theorem. Proof that the diﬀerential form implies the integrated form. First suppose that curl E = −dB dt at every point. Take the surface integral of both sides over S: x S curl E · dS = x S −dB dt · dS. 144 Apply Stokes’ theorem to the left side, and on the right side convert the integral of a derivative to the derivative of an integral (usually it is OK to do this, just as the sum of derivatives is the derivative of the sum): I C E · dr = −d dt x S B · dS. □ Proof that the integrated form implies the diﬀerential form. Suppose that I C E · dr = −d dt x S B · dS. holds for every surface S, with C being the boundary of S. Reversing the steps above leads to x S curl E · dS = x S −dB dt · dS for every surface S. By Lemma 22.1 below, this implies that curl E = −dB dt everywhere. □ Lemma 22.1. If F and G are continuous 3D vector ﬁelds such that x S F · dS = x S G · dS for every surface S, then F = G everywhere. Proof. Suppose not. Then we can choose a point P where F ̸= G. Let S be a tiny disk perpendicular to F −G at P. If S is small enough, then x S (F −G) · dS ̸= 0. Thus x S F · dS ̸= x S G · dS, a contradiction. □ 23. Review 23.1. Flux. Flux quantiﬁes how much a vector ﬁeld F is “ﬂowing” across a surface S: ﬂux := x S F · n normal component of F dS = x S F · dS. The part of F that matters is the part going across S — this is F · n, which is the scalar component of F in the direction of the unit normal vector to S. Because of the geometric interpretation of the dot product F · n in terms of the angle θ between F and n, the ﬂux through a tiny piece of surface area dS will be 145 • positive if F is roughly in the same direction as n (that is, θ < π/2), • negative if F and n are roughly in opposite directions (that is, θ > π/2), , and • zero if F is perpendicular to n (that is, θ = π/2, so F is parallel to S, so that F is not really ﬂowing across S). Also, the deﬁnition of ﬂux is a surface integral because one needs to add up the ﬂux through each tiny piece of S. Question 23.1. How do you compute ﬂux? Is it necessary to parametrize S? Answer. Parametrizing S will work, but it is not always necessary: • If F · n is constant on S, then pull it out of s S F · n dS so that it remains to compute the surface area of S. • Otherwise, ﬁnd a parametrization r(u, v) of S and use n dS = dS = ru × rv du dv. □ Thursday, December 9 23.2. A parade of diﬀerentials. What is the diﬀerence between ds, dS, dS, etc.? • dr can be thought of as a tiny vector measuring change of position along a curve. Use: Line integrals of vector ﬁelds (for example, work) have the shape R C F · dr. dr =    dx dy dz   = T ds. • ds can be thought of as the length of a tiny piece of a curve. Uses: Length of a curve, line integrals of scalar functions. ds = \|dr\| = p dx2 + dy2 + dz2 = p x′(t)2 + y′(t)2 + z′(t)2 dt. • dA can be thought of as the area of a tiny piece of R2: dA = dx dy (rectangular) = r dr dθ (polar) = ∂(x, y) ∂(u, v) du dv (change of variable). (The last formula is for a change of variable x = x(u, v) and y = y(u, v).) 146 • dS can be thought of as the area of a tiny piece of a surface. Uses: Surface area, surface integrals of scalar functions. If the surface is parametrized by r(u, v), then dS = \|ru × rv\| du dv (That’s the area of a tiny parallelogram of sides ru du and rv dv in R3.) For a sphere of radius ρ parametrized by φ, θ, this becomes dS = ρ2 sin φ dφ dθ. • dS means n dS, where n is a unit normal to the surface. Use: Flux of a 3D vector ﬁeld F across a surface S is s S F · dS. In terms of r(u, v): dS = ru × rv du dv. The length of dS is dS. The direction of dS is the unit normal n = ru × rv \|ru × rv\|. • dV can be thought of as the volume of a tiny region in R3: dV = dx dy dz (rectangular) = dz r dr dθ (cylindrical) = ρ2 sin φ dρ dφ dθ (spherical) = ∂(x, y, z) ∂(u, v, w) du dv dw (change of variable). • dm can be thought of as a tiny bit of mass. It is δ ds, or δ dS, or δ dV , depending on whether the object is of dimension 1, 2, or 3 (such as a wire, a warped metal plate, or a planet). The density δ is mass per unit length, or mass per unit area, or mass per unit volume. Uses: Mass, mass-weighted average, centroid, moment of inertia, gravitational ﬁeld. 23.3. Table of diﬀerentials and their uses. n-dim integral Diﬀerential n-forms Uses Z ds, dr arc length, work x dS, dS surface area, ﬂux across surface y dV volume it depends dm mass, average, centroid, moment of inertia Make sure that the dimension of the integral matches what you are trying to compute! 147 Example 23.2. Flux of a 3D vector ﬁeld F is ﬂux across a surface, so it should be a 2D integral. (In fact, it is the integral of the normal component F · n with respect to surface area, so it is s S F · n dS.) 23.4. Ice cream in rectangular, cylindrical, and spherical coordinates. Problem 23.3. Let T be an ice cream cone with vertex (0, 0, 0), height 4, and ﬂat top of radius 3. (It’s the disappointing kind of ice cream cone in which the ice cream does not bulge at the top.) Find triple integrals expressing the volume in rectangular, cylindrical, and spherical coordinates. Solution. Cylindrical: Above a point in the xy-plane with polar coordinates (r, θ) (with r ≤3) the height of the bottom of the cone is 4 3r by similar triangles, and the height of the top of the cone is 4. In other words, the cone is described by 4 3r ≤z ≤4 (which implies r ≤3). So the volume is Z 2π θ=0 Z 3 r=0 Z z=4 z= 4 3 r dz r dr dθ. We did not have time for the rest of this solution, but it will be discussed in my review session. Rectangular: The inequality r ≤3 corresponds to x2 + y2 ≤9, and the range for z above the point (x, y) in this disk is 4 3 p x2 + y2 ≤z ≤4. So the volume is Z 3 x=−3 Z √ 9−x2 y=− √ 9−x2 Z 4 z= 4 3 √ x2+y2 dz dy dx. Spherical: Either argue geometrically to ﬁnd the ranges for ρ, φ, θ, or just substitute the change of coordinates into the inequalities. Let’s do the latter: 4 3r ≤z ≤4 becomes 4 3ρ sin φ ≤ρ cos φ ≤4. The ﬁrst inequality says tan φ ≤3/4, so φ ≤tan−1(3/4). The second inequality says ρ ≤4/ cos φ. So the volume is Z 2π θ=0 Z tan−1(3/4) φ=0 Z 4/ cos φ ρ=0 ρ2 sin φ dρ dφ dθ. □ Problem 23.4. Same cone without the ice cream. What is a parametrization for the lateral surface of the cone? Solution. First, choose your parameters. How do you describe a position on the cone with two numbers? One possible answer: give the angle θ and the height z, i.e., the θ and z of 148 cylindrical coordinates. The radius at height z is 3 4z, so x = 3 4z cos θ y = 3 4z sin θ z = z. In other words, r(θ, z) =    3 4z cos θ 3 4z sin θ z   . □ 24. Life after 18.02 18.01 Calculus 18.03 Diﬀeqs 18.02 Multivar calc 18.04 Complex 18.05 Prob/stat 18.06(1) Linear algebra 18.062/6.042 Math for CS 18.S096 Intro to math reasoning 18.300 Applied math 18.200(A) Discrete math 18.600 Probability 18.700 Linear algebra 18.100 Real analysis 18.701 Algebra applied o / theoretical For sciences/engineering, 18.03, 18.05, 18.06 are useful. For computer science, 18.06 and 18.062/6.042. To prepare for more advanced math subjects (18.100 and greater), or if considering a math major (or double major with math), take 18.S096. This is a new subject developed by me and Profs. Dyatlov and Seidel that uses selected topics from combinatorics, algebra, and analysis (including fun topics like permutations and diﬀerent sizes of inﬁnity) to teach how to understand abstract concepts and prove theorems. The subject 18.062/6.042 is similar, but it uses topics aimed towards computer science subjects. 149 Many 18.02 students become math majors, or do a double major with math. If you are considering this, taking 18.S096 is a great way to get a sense of what it is like to be a math major — it is designed for students with your background. 25. Thank you Thank you to • Jennifer French for implementing my Part A problems on the MITx site; • Theresa Cummings and the Math Academic Services staﬀfor handling many admin-istrative details (such as printing and proctoring exams, helping with students with special situations, etc.); • Jean-Michel Claus and the rest of the team that developed the eigenvalue–eigenvector mathlet, among many others; • the math department faculty who helped develop 18.02 over decades; • the professors I consulted in other departments whose advice helped me modernize the 18.02 content this fall to make it more relevant to their subjects; • the MIT audio-visual staﬀon hand at each lecture; • the MIT Facilities staﬀfor improving the ventilation in 34-101 upon our request; • the board cleaners at each lecture; • and especially the recitation instructors (Zongchen Chen, Yuqiu Fu, Duncan Levear, Hyunki Min, Yilin Wang, Pu Yu) for their hard work over the semester! And thank you to all the 18.02 students for making this class fun to teach! I hope that you all ace the ﬁnal! 150
17581	https://pubmed.ncbi.nlm.nih.gov/15985474/	Adrenal insufficiency in meningococcal sepsis: bioavailable cortisol levels and impact of interleukin-6 levels and intubation with etomidate on adrenal function and mortality - 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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Adrenal insufficiency in meningococcal sepsis: bioavailable cortisol levels and impact of interleukin-6 levels and intubation with etomidate on adrenal function and mortality Marieke den Brinker1,Koen F M Joosten,Olivia Liem,Frank H de Jong,Wim C J Hop,Jan A Hazelzet,Marije van Dijk,Anita C S Hokken-Koelega Affiliations Expand Affiliation 1 Department of Pediatrics, Division of Endocrinology and Division of Pediatric Intensive Care, Erasmus Medical Center--Sophia Children's Hospital, P.O. Box 2060, 3000 CB Rotterdam, The Netherlands. m.denbrinker@erasmusmc.nl PMID: 15985474 DOI: 10.1210/jc.2005-1107 Item in Clipboard Adrenal insufficiency in meningococcal sepsis: bioavailable cortisol levels and impact of interleukin-6 levels and intubation with etomidate on adrenal function and mortality Marieke den Brinker et al. J Clin Endocrinol Metab.2005 Sep. Show details Display options Display options Format J Clin Endocrinol Metab Actions Search in PubMed Search in NLM Catalog Add to Search . 2005 Sep;90(9):5110-7. doi: 10.1210/jc.2005-1107. Epub 2005 Jun 28. Authors Marieke den Brinker1,Koen F M Joosten,Olivia Liem,Frank H de Jong,Wim C J Hop,Jan A Hazelzet,Marije van Dijk,Anita C S Hokken-Koelega Affiliation 1 Department of Pediatrics, Division of Endocrinology and Division of Pediatric Intensive Care, Erasmus Medical Center--Sophia Children's Hospital, P.O. Box 2060, 3000 CB Rotterdam, The Netherlands. m.denbrinker@erasmusmc.nl PMID: 15985474 DOI: 10.1210/jc.2005-1107 Item in Clipboard Cite Display options Display options Format Abstract Context: Adequate adrenal function is pivotal to survive meningococcal sepsis. Objectives: The objective of the study was to evaluate adrenocortical function in meningococcal disease. Design: This was an observational cohort study. Setting: The study was conducted at a university-affiliated pediatric intensive care unit. Patients: Sixty children with meningococcal sepsis or septic shock participated in the study. Main outcome measures: The differences in adrenal function between nonsurvivors (n = 8), shock survivors (n = 43), and sepsis survivors (n = 9) on pediatric intensive care unit admission were measured. Results: Nonsurvivors had significantly lower median cortisol to ACTH ratio than shock survivors and sepsis survivors. Because cortisol binding globulin and albumin levels did not significantly differ among the groups, bioavailable cortisol levels were also significantly lower in nonsurvivors than sepsis survivors. Nonsurvivors had significantly lower cortisol to 11-deoxycortisol ratios but not lower 11-deoxycortisol to 17-hydroxyprogesterone ratios than survivors. Using multiple regression analysis, decreased cortisol to ACTH ratio was significantly related to higher IL-6 levels and intubation with etomidate (one single bolus), whereas decreased cortisol to 11-deoxycortisol ratio was significantly related only to intubation with etomidate. Aldosterone levels tended to be higher in nonsurvivors than shock survivors, whereas plasma renin activity did not significantly differ. Conclusions: Our study shows that the most severely ill children with septic shock had signs of adrenal insufficiency. Bioavailable cortisol levels were not more informative on adrenal function than total cortisol levels. Besides disease severity, one single bolus of etomidate during intubation was related to decreased adrenal function and 11beta-hydroxylase activity. Decreased adrenal function was not related to decreased 21-hydroxylase activity. Based on our results, it seems of vital importance to take considerable caution using etomidate and consider combining its administration with glucocorticoids during intubation of children with septic shock. PubMed Disclaimer Similar articles One single dose of etomidate negatively influences adrenocortical performance for at least 24h in children with meningococcal sepsis.den Brinker M, Hokken-Koelega AC, Hazelzet JA, de Jong FH, Hop WC, Joosten KF.den Brinker M, et al.Intensive Care Med. 2008 Jan;34(1):163-8. doi: 10.1007/s00134-007-0836-3. Epub 2007 Aug 21.Intensive Care Med. 2008.PMID: 17710382 Free PMC article. Adrenal response in patients with septic shock of abdominal origin: relationship to survival.Riché FC, Boutron CM, Valleur P, Berton C, Laisné MJ, Launay JM, Chappuis P, Peynet J, Vicaut E, Payen D, Cholley BP.Riché FC, et al.Intensive Care Med. 2007 Oct;33(10):1761-6. doi: 10.1007/s00134-007-0770-4. Epub 2007 Jul 6.Intensive Care Med. 2007.PMID: 17618417 Adrenal status in children with septic shock using low-dose stimulation test.Sarthi M, Lodha R, Vivekanandhan S, Arora NK.Sarthi M, et al.Pediatr Crit Care Med. 2007 Jan;8(1):23-8. doi: 10.1097/01.pcc.0000256622.63135.90.Pediatr Crit Care Med. 2007.PMID: 17251878 Adrenal dysfunction in critically ill children.Karagüzel G, Cakir E.Karagüzel G, et al.Minerva Endocrinol. 2014 Dec;39(4):235-43. Epub 2014 Jul 29.Minerva Endocrinol. 2014.PMID: 25069846 Review. [Etomidate can not be recommended to patients with septic shock].Toft P, Jensen AG.Toft P, et al.Ugeskr Laeger. 2011 May 16;173(20):1421-3.Ugeskr Laeger. 2011.PMID: 21586246 Review.Danish. See all similar articles Cited by The effect of etomidate on adrenal function in critical illness: a systematic review.Albert SG, Ariyan S, Rather A.Albert SG, et al.Intensive Care Med. 2011 Jun;37(6):901-10. doi: 10.1007/s00134-011-2160-1. Epub 2011 Mar 4.Intensive Care Med. 2011.PMID: 21373823 24-Hour protein, arginine and citrulline metabolism in fed critically ill children - A stable isotope tracer study.de Betue CTI, Garcia Casal XC, van Waardenburg DA, Schexnayder SM, Joosten KFM, Deutz NEP, Engelen MPKJ.de Betue CTI, et al.Clin Nutr. 2017 Jun;36(3):876-887. doi: 10.1016/j.clnu.2016.12.023. Epub 2017 Jan 4.Clin Nutr. 2017.PMID: 28089618 Free PMC article. Differential effects of etomidate and its pyrrole analogue carboetomidate on the adrenocortical and cytokine responses to endotoxemia.Pejo E, Feng Y, Chao W, Cotten JF, Le Ge R, Raines DE.Pejo E, et al.Crit Care Med. 2012 Jan;40(1):187-92. doi: 10.1097/CCM.0b013e31822d7924.Crit Care Med. 2012.PMID: 21926608 Free PMC article. Long-term health-related quality of life in survivors of meningococcal septic shock in childhood and their parents.Buysse CM, Raat H, Hazelzet JA, Vermunt LC, Utens EM, Hop WC, Joosten KF.Buysse CM, et al.Qual Life Res. 2007 Dec;16(10):1567-76. doi: 10.1007/s11136-007-9271-8. Epub 2007 Oct 19.Qual Life Res. 2007.PMID: 17952627 Rare case of meningococcal sepsis-induced testicular failure, primary hypothyroidism and hypoadrenalism: Is there a link?Bachmeier CAE, Malabu U.Bachmeier CAE, et al.BMJ Case Rep. 2018 Sep 15;2018:bcr2018224437. doi: 10.1136/bcr-2018-224437.BMJ Case Rep. 2018.PMID: 30219775 Free PMC article. 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17582	https://zh.wikipedia.org/zh-hans/%E6%9C%80%E5%B0%8F%E5%85%AC%E5%80%8D%E6%95%B8	最小公倍数 - 维基百科，自由的百科全书跳转到内容 [x] 主菜单主菜单移至侧栏隐藏导航首页分类索引特色内容新闻动态最近更改随机条目帮助帮助维基社群方针与指引互助客栈知识问答字词转换 IRC即时聊天联络我们关于维基百科特殊页面搜索搜索 [x] 外观资助维基百科创建账号登录 [x] 个人工具资助维基百科创建账号登录目录移至侧栏隐藏序言 1 与最大公因数之关系 2 计算方法开关计算方法子章节 2.1 递归计算多个整数的最小公倍数 3 程式代码开关程式代码子章节 3.1 C 3.2 C++ 3.3 C# 3.4 Go 3.5 Java 3.6 Pascal 3.7 Python 3.8 Ruby 3.9 Swift 4 应用 5 参见 6 参考来源 [x] 开关目录最小公倍数 [x] 72种语言 العربية Asturianu Azərbaycanca Беларуская Български বাংলা Català کوردی Čeština Чӑвашла Dansk Deutsch Ελληνικά English Esperanto Español Eesti Euskara فارسی Suomi Français Galego עברית हिन्दी Magyar Հայերեն Bahasa Indonesia Italiano 日本語 Қазақша ಕನ್ನಡ 한국어 Lombard Lietuvių Latviešu Македонски മലയാളം Монгол मराठी Bahasa Melayu Malti Plattdüütsch Nederlands Norsk nynorsk Norsk bokmål ଓଡ଼ିଆ Polski Piemontèis پنجابی Português Română Русский Simple English Slovenčina سرائیکی Slovenščina Shqip Српски / srpski Svenska தமிழ் తెలుగు ไทย Tagalog Türkçe Українська اردو Oʻzbekcha / ўзбекча Tiếng Việt 吴语 ייִדיש 閩南語 / Bân-lâm-gí 粵語编辑链接条目讨论 [x] 简体不转换简体繁體大陆简体香港繁體澳門繁體大马简体新加坡简体臺灣正體阅读编辑查看历史 [x] 工具工具移至侧栏隐藏操作阅读编辑查看历史常规链入页面相关更改上传文件固定链接页面信息引用此页获取短链接下载二维码打印/导出下载为PDF 打印版本在其他项目中维基共享资源 Wikifunctions 维基数据项目外观移至侧栏隐藏维基百科，自由的百科全书描述12、18之间的因倍数关系的文氏图最小公倍数（英语：least common multiple，lcm）是数论中的一个概念。若有一个数 X{\displaystyle X}，可以被另外两个数 A{\displaystyle A}、B{\displaystyle B}整除，且 X{\displaystyle X}同时大于或等于 A{\displaystyle A}和 B{\displaystyle B}，则 X{\displaystyle X}为 A{\displaystyle A}和 B{\displaystyle B}的公倍数。A{\displaystyle A}和 B{\displaystyle B}的公倍数有无限个，而所有正的公倍数中，最小的公倍数就叫做最小公倍数。同样地，若干个整数公有的倍数中最小的正整数称为它们的最小公倍数。n{\displaystyle n}个整数 a 1,a 2,⋯,a n{\displaystyle a_{1},a_{2},\cdots ,a_{n}}的最小公倍数一般记作：[a 1,a 2,⋯,a n]{\displaystyle [a_{1},a_{2},\cdots ,a_{n}]}，或者参照英文记法记作 lcm⁡(a 1,a 2,⋯,a n){\displaystyle \operatorname {lcm} (a_{1},a_{2},\cdots ,a_{n})}。对分数进行加减运算时，要求两数的分母相同才能计算，故需要通分；标准的计算步骤是将两个分数的分母通分成它们的最小公倍数，然后将通分后的分子相加。与最大公因数之关系 [编辑] 两个整数的最小公倍数与最大公因数之间有如下的关系： lcm⁡(a,b)=\|a⋅b\|gcd⁡(a,b){\displaystyle \operatorname {lcm} (a,b)={\frac {\|a\cdot b\|}{\operatorname {gcd} (a,b)}}} 计算方法 [编辑] 最小公倍数可以通过多种方法得到，最直接的方法是列举法，从小到大列举出其中一个数（如最大数）的倍数，当这个倍数也是另一个数的倍数时，就求得最小公倍数。另一个方法是利用公式 lcm⁡(a 1,a 2)=a 1 a 2 gcd(a 1,a 2){\displaystyle \operatorname {lcm} (a_{1},a_{2})={\frac {a_{1}a_{2}}{\gcd(a_{1},a_{2})}}}来求解，这时首先要知道它们的最大公因数。而最大公因数可以通过短除法得到。利用整数的唯一分解定理，还可以用质因数分解法。将每个整数进行质因数分解。对每个质数，在质因数分解的表达式中寻找次数最高的乘幂，最后将所有这些质数乘幂相乘就可以得到最小公倍数。譬如求216、384和210的最小公倍数。对216、384和210来说： 216=2 3×3 3{\displaystyle 216=2^{3}\times 3^{3}}，384=2 7×3 1{\displaystyle 384=2^{7}\times 3^{1}}，210=2 1×3 1×5 1×7 1{\displaystyle 210=2^{1}\times 3^{1}\times 5^{1}\times 7^{1}}。其中 2{\displaystyle 2}对应的最高次乘幂为 2 7{\displaystyle 2^{7}}；3{\displaystyle 3}对应的最高次乘幂为 3 3{\displaystyle 3^{3}}；5{\displaystyle 5}和 7{\displaystyle 7}对应的最高次乘幂分别是 5 1{\displaystyle 5^{1}}与 7 1{\displaystyle 7^{1}}。将这些乘幂乘起来，就可以得到最小公倍数：[216,384,210]=2 7×3 3×5 1×7 1=120960{\displaystyle [216,384,210]=2^{7}\times 3^{3}\times 5^{1}\times 7^{1}=120960}。短除法利用短除法，可以快速计算出多个整数的最小公倍数。以下为例子: 假设我们要求12、20和42的最小公倍数。 a: 6 \|12 18 42 b: 2 3 7 最小公倍数=a×b 因此，12、18和42和最小公倍数=6×2×3×7 所以，6×2×3×7=252，12、18和42的最小公倍数是252 递归计算多个整数的最小公倍数 [编辑] 可以递归求出多个整数的最小公倍数：欲求 lcm⁡(a 1,...,a n)(n≥3){\displaystyle \operatorname {lcm} (a_{1},...,a_{n})(n\geq 3)}，只需求 lcm⁡(a 1,...,a n−2,lcm⁡(a n−1,a n)){\displaystyle \operatorname {lcm} (a_{1},...,a_{n-2},\operatorname {lcm} (a_{n-1},a_{n}))}。这利用了性质 lcm⁡(a 1,a 2,a 3)=lcm⁡(lcm⁡(a 1,a 2),a 3){\displaystyle \operatorname {lcm} (a_{1},a_{2},a_{3})=\operatorname {lcm} (\operatorname {lcm} (a_{1},a_{2}),a_{3})}。该性质证明如下：记 a 1,a 2,a 3{\displaystyle a_{1},a_{2},a_{3}} 的质因数分解分别为∏i=1 n p i e 1 i,∏i=1 n p i e 2 i,∏i=1 n p i e 3 i{\displaystyle \prod {i=1}^{n}p{i}^{e_{1i}},\prod {i=1}^{n}p{i}^{e_{2i}},\prod {i=1}^{n}p{i}^{e_{3i}}}，其中 p i{\displaystyle p_{i}} 是第 i{\displaystyle i} 个质数。那么根据最小公倍数的定义，lcm⁡(a 1,a 2,a 3)=∏i=1 n p i max(e 1 i,e 2 i,e 3 i){\displaystyle \operatorname {lcm} (a_{1},a_{2},a_{3})=\prod {i=1}^{n}p{i}^{\max(e_{1i},e_{2i},e_{3i})}}， lcm⁡(lcm⁡(a 1,a 2),a 3)=lcm⁡(∏i=1 n p i max(e 1 i,e 2 i),a 3)=∏i=1 n p i max(max(e 1 i,e 2 i),e 3 i)=∏i=1 n p i max(e 1 i,e 2 i,e 3 i){\displaystyle \operatorname {lcm} (\operatorname {lcm} (a_{1},a_{2}),a_{3})=\operatorname {lcm} (\prod {i=1}^{n}p{i}^{\max(e_{1i},e_{2i})},a_{3})=\prod {i=1}^{n}p{i}^{\max(\max(e_{1i},e_{2i}),e_{3i})}=\prod {i=1}^{n}p{i}^{\max(e_{1i},e_{2i},e_{3i})}}，证毕。程式代码 [编辑] 以下使用辗转相除法求得最大公因数，之后再求最小公倍数。 C [编辑] int GCD(int a, int b) { if(b) while((a %= b) && (b %= a)); return a + b; } int LCM(int a, int b) { return a b / GCD(a, b); } C++ [编辑] template T GCD(T a, T b) { if (b) while((a %= b) && (b %= a)); return a + b; } template T LCM(T a, T b) { return a b / GCD(a, b); } C [编辑] int GCD(int a, int b) { return a % b == 0 ? b : GCD(b, a % b); } int LCM(int a, int b) { return a b / GCD(a, b); } Go [编辑] func GCD(a, b int) int { if b == 0 { return a } return GCD(b, a%b) } func LCM(a, b int) int { return a b / GCD(a, b) } Java [编辑] int GCD(int a, int b) { return a % b == 0 ? b : GCD(b, a % b); } int LCM(int a, int b) { return a b / GCD(a, b); } Pascal [编辑] function gcd(a,b:integer):longint; begin if b=0 then gcd:=a else gcd:=gcd(b,a mod b); end; function lcm(a,b:integer):longint; begin lcm:=(ab) div gcd(a,b); end; Python [编辑] def gcd(a, b): return a if b == 0 else gcd(b, a % b) def lcm(a, b): return a b / gcd(a, b) Ruby [编辑] def gcd(a, b) b.zero? ? a : gcd(b, a % b) end def lcm(a, b) a b / gcd(a, b) end Swift [编辑] func gcd(_ a: Int, _ b: Int) -> Int { return b == 0 ? a : gcd(b, a % b) } func lcm(_ a: Int, _ b: Int) -> Int { return a b / gcd(a, b) } 应用 [编辑] 求最小公倍数是进行分数加减法时的步骤之一。将分母通分时，会把所有分数的分母通分为它们的最小公倍数，然后将分子相加。例如： 2 21+1 6=4 42+7 42=11 42{\displaystyle {2 \over 21}+{1 \over 6}={4 \over 42}+{7 \over 42}={11 \over 42}} 其中分母42就是21与6的最小公倍数。参见 [编辑] 公倍数公因数最大公因数参考来源 [编辑] 柯召，孙绮，孙琦. 《数论讲义》. 高等教育出版社. 2005. ISBN 753205473X. 阿尔伯特·H·贝勒著谈祥柏译. 《数论妙趣：数学女王的盛情款待》. 上海教育出版社. 1998. ISBN 7040091909. 检索自“ 分类：四则运算数字运算数论本页面最后修订于2025年5月8日 (星期四) 23:04。本站的全部文字在知识共享署名-相同方式共享 4.0协议之条款下提供，附加条款亦可能应用。（请参阅使用条款） Wikipedia®和维基百科标志是维基媒体基金会的注册商标；维基™是维基媒体基金会的商标。维基媒体基金会是按美国国内税收法501(c)(3)登记的非营利慈善机构。隐私政策关于维基百科免责声明行为准则开发者统计 Cookie声明手机版视图搜索搜索 [x] 开关目录最小公倍数 72种语言添加话题
17583	https://bingweb.binghamton.edu/~suzuki/ThermoStatFIles/6.7%20%20PD%20Cosmic%20neutrino%20background.pdf	Cosmic neutrino background (CB) Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: December 23, 2018) The cosmic neutrino background (CνB)is the universe's background particle radiation composed of neutrinos. They are sometimes known as relic neutrinos. The CνB is a relic of the big bang; while the cosmic microwave background radiation (CMB) dates from when the universe was 379,000 years old, the CνB decoupled (separated) from matter when the universe was just one second old. It is estimated that today, the CνB has a temperature of roughly 1.95 K. As neutrinos rarely interact with matter, these neutrinos still exist today. They have a very low energy, around 10−4 to 10−6 eV. Even high energy neutrinos are notoriously difficult to detect, and the CνB has energies around 10−10 times smaller, so the CνB may not be directly observed in detail for many years, if at all. However, Big Bang cosmology makes many predictions about the CνB, and there is very strong indirect evidence that the CνB exists. 1. Neutrino () A neutrino (denoted by the Greek letter ν) is a fermion (an elementary particle with half-integer spin) that interacts only via the weak subatomic force and gravity. The mass of the neutrino is much smaller than that of the other known elementary particles. Although only differences of squares of the three mass values are known as of 2016, cosmological observations imply that the sum of the three masses must be less than one millionth that of the electron. The neutrino is so named because it is electrically neutral and because its rest mass is so small (-ino) that it was long thought to be zero. Weak interactions create neutrinos in one of three leptonic flavors: electron neutrinos, muon neutrinos or tau neutrinos, in association with the corresponding charged lepton. A neutrino created with a specific flavor is in an associated specific quantum superposition of all three mass states. As a result, neutrinos oscillate between different flavors in flight. For each neutrino, there also exists a corresponding antiparticle, called an antineutrino, which also has half-integer spin and no electric charge. They are distinguished from the neutrinos by having opposite signs of lepton number and chirality. To conserve total lepton number, in nuclear beta decay, electron neutrinos appear together with only positrons (anti-electrons) or electron-antineutrinos, and electron antineutrinos with electrons or electron neutrinos. 2. Chemical potential A pair production (Huang): the grand canonical ensemble The grand canonical ensemble includes systems with different particle numbers, with a mean value N determined by the chemical potential. This makes sense only if N is a conserved quantity, for otherwise the chemical potential would be zero, as in the case of photon. We consider a reaction e e      , 2      The reaction establishes an average value for the conserved quantum number  N N . The grand partition function is given by N N G N N N N Z z Z Z         where z e  ,     N B N Z T k F ln or     N F N e Z      N B N Z T k F ln or     N F N e Z  Then we have exp[ ( ) ( )] N N N N N N z Z Z N N F F                We put ) ( ) (        N N F F N N f  We find the condition that 0 ,                   V T N N F N f  or 0      . 0 ,                    V T N N F N f  or 0      . where                V T N N F , , and                V T N N F , . Thus we have 0       . When 0  , 0       3. ((Problem and solution)) D.V. Schroeder, An Introduction to Thermal Physics (Addison Wesley, 1999). Problem 7-48 In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos () and antineutrinos (), currently at an effective temperature of 1.95 K. There are three species of neutrinos (electron neutrino, muon neutrino, and tau neutrino), each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless. (a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of neutrinos, so trhat their chemical potentials are equal;      . Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction 2      where  is photon. Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that 0  for both the neutrinos and the antineutrinos. (b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. (Hint: there are very few differences between this “neutrino gas” and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a mathematical table. (c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K. ((Solution)) The condition for equilibrium is the same as the reaction equation, but with the name of each species replaced by its chemical potential. So for the reaction 2      the equilibrium condition would be 2         But the chemical potential for the photons is zero, while the chemical potentials of the neutrinos and ant-neutrinos are equal to each other, if they are equally abundant. Therefore we must have 0       For neutrino, we use the dispersion relation as cp c k  ℏ There are three kinds of neutrinos; electron neutrinos, muon neutrinos or tau neutrinos. The internal energy is 2 3 0 3(2 1) 4 (2 ) 1 c k V c k U s k dk e          ℏ ℏ where s=1/2. We put x c k   ℏ. Thus we get 2 3 2 0 4 4 3 2 3 3 0 4 4 2 3 3 6 4 8 1 3 1 7 40 x B x B V x dx x U c c e k T V x dx c e k T V c                       ℏ ℏ ℏ ℏ or 4 4 2 4 3 3 7 21 40 2 B SB k T U V V T c c      ℏ with 2 4 2 4 3 2 3 3 4 4 60 15 SB B B k k c c c c       ℏ ℏ , 2 4 3 2 60 B SB k c    ℏ , and 3 4 0 7 1 120 x x dx e      . The number of neutrinos is 2 3 0 1 3(2 1) 4 (2 ) 1 c k V N s k dk e          ℏ where s=1/2. We put x c k   ℏ. Thus we get 3 3 2 2 3 3 0 3 2 3 1 3(1.80309) B x B k T x N V dx c e k T V c               ℏ ℏ where 2 0 1.80309 1 x x dx e     The number density of neutrino at T = 1.95 K (cosmic neutrino background) 3 2 3 (1.80309) B N k T n V c            ℏ =3.38457 x 108/m3. The entropy is 3 4 42 3 3 SB U S V T T c      ___________ 4. Photon () We note that the internal energy of photon is 4 2 4 4 4 3 3 4 15 SB B T k T U V T V V c c        ℏ . The entropy: 4 3 4 16 4 3 3 3 SB U S V T V T T c        The number of photons is 2 2 3 2 2 4 (2 ) 1 k k V V k dk N k dkn e           The zero-momentum state is ignored in the continuum approximation used. The energy dispersion for photon; ck ℏ   Since cdk d ℏ   , we have 2 2 3 1 ( ) 1 V d N c e         ℏ We put   x Then we get 3 2 2 3 0 ( ) ( ) 1 B x k T V x dx N c e       ℏ We note that 2 3 0 2 ( 1) 2.40412 1 x x dx z e        , 20206 . 1 ) 1 ( 3   z  Then we have 3 2 3 ( ) 2.40416 ( ) B k T V N c    ℏ The number density of photon at T = 2.72 K (cosmic microwave background) 3 8 3 ( ) 0.2435926 4.08255 10 / m B N n V k T c      ℏ ____________ 5. Electron-positron pair For positron and electron, we use the dispersion relation as cp c k  ℏ The internal energy is 2 3 0 2(2 1) 4 (2 ) 1 p e c k V c k U s k dk e        ℏ ℏ where s=1/2. We put x c k   ℏ. Thus we get 2 3 2 0 4 4 3 2 3 3 0 4 4 2 3 3 4 4 8 1 2 1 7 60 p e x B x B V x dx x U c c e k T V x dx c e k T V c                       ℏ ℏ ℏ ℏ or 4 4 2 4 3 3 7 7 60 B p e SB k T U V V T c c     ℏ with 3 4 0 7 1 120 x x dx e      The entropy: 3 4 28 3 3 p e SB p e U S V T T c     The number density: 2 e 3 0 1 2(2 1) 4 (2 ) 1 p c k V N s k dk e        ℏ where s=1/2. We put x c k   ℏ. Thus, we get 3 3 2 2 3 3 0 3 2 2 1 1.80309 2 B p e x B k T x N V dx c e k T V c              ℏ ℏ where 2 0 1.80309 1 x x dx e     ____________ The total entropy of positron-electron, neutrino, and photon, 3 3 28 16 42 86 ( ) 3 3 3 3 SB SB tot S V T V T c c       3 3 28 16 44 ( ) 3 3 3 SB SB e p S S V T V T c c        3 28 3 SB e p S V T c   3 42 ( ) 3 SB S V T c    3 16 3 SB S V T c    3 3 3 3 42 ( ) ( ) 3 16 ( ) 3 43 21 0.75 SB SB V T S T c S T V T c T T                    3 3 86 ( ) 3 42 3 43 21 SB total SB T V S c T S V c            3 3 86 ( ) 3 16 3 21 4 SB total SB T V S c T S V c            6. Cosmic neutrino background (CB) Given the temperature of the CMB, the temperature of the CνB can be estimated. Before neutrinos decoupled from the rest of matter, the universe primarily consisted of neutrinos, electrons, positrons, and photons, all in thermal equilibrium with each other. Once the temperature dropped to approximately 2.5 MeV, the neutrinos decoupled from the rest of matter. Despite this decoupling, neutrinos and photons remained at the same temperature as the universe expanded. However, when the temperature dropped below the mass of the electron, most electrons and positrons annihilated, transferring their heat and entropy to photons, and thus increasing the temperature of the photons. So the ratio of the temperature of the photons before and after the electron-positron annihilation is the same as the ratio of the temperature of the neutrinos and the photons today. To find this ratio, we assume that the entropy of the universe was approximately conserved by the electron-positron annihilation. Then using where S is the entropy, g is the effective degrees of freedom and T is the temperature, we find that where T denotes the temperature before the electron-positron annihilation and T denotes after the electron-positron annihilation: 16/3 for photons, since they are massless bosons 28/3 for electrons and positrons, since they are fermions. Given the current value of Tγ = 2.725 K, it follows that Tν = 1.945 K. __________ Neutrino  Neutrino  Neutrino Electron Positron  Electron annihilation Positron      Photon Photon Photon T  T _______________ REFERENCES R.K. Pathria and P.D. Beale, Statistical Mechanics, third edition (Elsevier, 2011). K. Huang, Introduction to Statistical Physics, second edition (CRC Press, 2010). L.D. Landau and E.M. Lifshitz, Statistical Physics (Pergamon, 1980). V. Barger, D. Marfatia, and K. Whisnant, The Physics of Neutrinos (Princeton Univ. Press, 2012). D.V. Schroeder, An Introduction to Thermal Physics (Addison Wesley, 1999).
17584	https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/2%3A_Higher_order_linear_ODEs/2.2%3A_Constant_coefficient_second_order_linear_ODEs	Published Time: 2013-11-07T23:23:20Z 2.2: Constant coefficient second order linear ODEs - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 2: Higher order linear ODEs Differential Equations for Engineers (Lebl) { } { "2.1:Second_order_linear_ODEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.2:_Constant_coefficient_second_order_linear_ODEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.3:_Higher_order_linear_ODEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.4:_Mechanical_Vibrations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.5:_Nonhomogeneous_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.6:_Forced_Oscillations_and_Resonance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.E:_Higher_order_linear_ODEs(Exercises)" : "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", "0:_Introduction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1:_First_order_ODEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2:_Higher_order_linear_ODEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3:_Systems_of_ODEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4:_Fourier_series_and_PDEs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5:_Eigenvalue_problems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6:_The_Laplace_Transform" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7:_Power_series_methods" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8:_Nonlinear_Systems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Appendix_A:_Linear_Algebra" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Appendix_B:_Table_of_Laplace_Transforms" : "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, 24 Feb 2025 03:36:28 GMT 2.2: Constant coefficient second order linear ODEs 350 350 Delmar Larsen { } Anonymous Anonymous 2 false false [ "article:topic", "authorname:lebl", "Euler\u2019s formula", "license:ccbysa", "showtoc:no", "autonumheader:yes2", "licenseversion:40", "source@ ] [ "article:topic", "authorname:lebl", "Euler\u2019s formula", "license:ccbysa", "showtoc:no", "autonumheader:yes2", "licenseversion:40", "source@ ] 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. Bookshelves 3. Differential Equations 4. Differential Equations for Engineers (Lebl) 5. 2: Higher order linear ODEs 6. 2.2: Constant coefficient second order linear ODEs Expand/collapse global location Differential Equations for Engineers (Lebl) Front Matter 0: Introduction 1: First order ODEs 2: Higher order linear ODEs 3: Systems of ODEs 4: Fourier series and PDEs 5: Eigenvalue problems 6: The Laplace Transform 7: Power series methods 8: Nonlinear Systems Appendix A: Linear Algebra Appendix B: Table of Laplace Transforms Back Matter 2.2: Constant coefficient second order linear ODEs Last updated Feb 24, 2025 Save as PDF 2.1: Second order linear ODEs 2.3: Higher order linear ODEs Page ID 350 Jiří Lebl Oklahoma State University ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Solving Constant Coefficient Equations 1. Exercise 2.2.1 2.2.1 2. Theorem 2.2.1 2.2.1 3. Example 2.2.1 2.2.1 1. Solution 4. Example 2.2.2 2.2.2:/2:_Higher_order_linear_ODEs/2.2:_Constant_coefficient_second_order_linear_ODEs#Example_.5C(.5CPageIndex.7B2.7D.5C):) 1. Solution/2:_Higher_order_linear_ODEs/2.2:_Constant_coefficient_second_order_linear_ODEs#Solution_2) 5. Exercise 2.2.2 2.2.2: Linear Independence/2:_Higher_order_linear_ODEs/2.2:_Constant_coefficient_second_order_linear_ODEs#Exercise_.5C(.5CPageIndex.7B2.7D.5C):_Linear_Independence) 2.2.2 Complex numbers and Euler’s formula Exercise 2.2.3 2.2.3 Theorem 2.2.2 2.2.2 Exercise 2.2.4 2.2.4: Exercise 2.2.5 2.2.5 2.2.3 Complex roots Theorem 2.2.3 2.2.3 Example 2.2.3 2.2.3 Solution Example 2.2.4 2.2.4 Solution Footnotes Solving Constant Coefficient Equations Suppose we have the problem y′′−6 y′+8 y=0,y(0)=−2,y′(0)=6 y″−6 y′+8 y=0,y(0)=−2,y′(0)=6 This is a second order linear homogeneous equation with constant coefficients. Constant coefficients means that the functions in front of y′′y″, y′y′, and y y are constants and do not depend on x x. To guess a solution, think of a function that you know stays essentially the same when we differentiate it, so that we can take the function and its derivatives, add some multiples of these together, and end up with zero. Let us try1 1 a solution of the form y=e r x y=e r x. Then y′=r e r x y′=r e r x and y′′=r 2 e r x y″=r 2 e r x. Plug in to get y′′−6 y′+8 y r 2 e r xy′′−6 r e r xy′+8 e r xy r 2−6 r+8(r−2)(r−4)=0,=0,=0(divide through by e r x),=0.(2.2.1)(2.2.1)y″−6 y′+8 y=0,r 2 e r x⏟y″−6 r e r x⏟y′+8 e r x⏟y=0,r 2−6 r+8=0(divide through by e r x),(r−2)(r−4)=0. Hence, if r=2 r=2 or r=4 r=4, then e r x e r x is a solution. So let y 1=e 2 x y 1=e 2 x and y 2=e 4 x y 2=e 4 x. Exercise 2.2.1 2.2.1 Check that y 1 y 1 and y 2 y 2 are solutions. Solution The functions e 2 x e 2 x and e 4 x e 4 x are linearly independent. If they were not linearly independent we could write e 4 x=C e 2 x e 4 x=C e 2 x for some constant C C, implying that e 2 x=C e 2 x=C for all x x, which is clearly not possible. Hence, we can write the general solution as y=C 1 e 2 x+C 2 e 4 x y=C 1 e 2 x+C 2 e 4 x We need to solve for C 1 C 1 and C 2 C 2. To apply the initial conditions we first find y′=2 C 1 e 2 x+4 C 2 e 4 x y′=2 C 1 e 2 x+4 C 2 e 4 x. We plug in x=0 x=0 and solve. −2 6=y(0)=C 1+C 2=y′(0)=2 C 1+4 C 2(2.2.2)(2.2.2)−2=y(0)=C 1+C 2 6=y′(0)=2 C 1+4 C 2 Either apply some matrix algebra, or just solve these by high school math. For example, divide the second equation by 2 to obtain 3=C 1+2 C 2 3=C 1+2 C 2, and subtract the two equations to get 5=C 2 5=C 2. Then C 1=−7 C 1=−7 as −2=C 1+5−2=C 1+5. Hence, the solution we are looking for is y=−7 e 2 x+5 e 4 x y=−7 e 2 x+5 e 4 x Let us generalize this example into a method. Suppose that we have an equation a y′′+b y′+c y=0,(2.2.3)(2.2.3)a y″+b y′+c y=0, where a,b,c a,b,c are constants. Try the solution y=e r x y=e r x to obtain a r 2 e r x+b r e r x+c e r x=0 a r 2 e r x+b r e r x+c e r x=0 Divide by e r x e r x to obtain the so-called characteristic equation of the ODE: a r 2+b r+c=0 a r 2+b r+c=0 Solve for the r r by using the quadratic formula. r 1,r 2=−b±b 2−4 a c−−−−−−−√2 a r 1,r 2=−b±b 2−4 a c 2 a Therefore, we have e r 1 x e r 1 x and e r 2 x e r 2 x as solutions. There is still a difficulty if r 1=r 2 r 1=r 2, but it is not hard to overcome. Theorem 2.2.1 2.2.1 Suppose that r 1 r 1 and r 2 r 2 are the roots of the characteristic equation. If r 1 r 1 and r 2 r 2 are distinct and real (when b 2−4 a c>0 b 2−4 a c>0), then(2.2.3)(2.2.3)has the general solution y=C 1 e r 1 x+C 2 e r 2 x y=C 1 e r 1 x+C 2 e r 2 x If r 1=r 2 r 1=r 2(happens when b 2−4 a c=0 b 2−4 a c=0), then(2.2.3)(2.2.3)has the general solution y=(C 1+C 2 x)e r 1 x y=(C 1+C 2 x)e r 1 x For another example of the first case, take the equation y′′−k 2 y=0 y″−k 2 y=0. Here the characteristic equation is r 2−k 2=0 r 2−k 2=0 or (r−k)(r+k)=0(r−k)(r+k)=0. Consequently, e−k x e−k x and e k x e k x are the two linearly independent solutions. Example 2.2.1 2.2.1 Solve y′′−k 2 y=0.y″−k 2 y=0. Solution The characteristic equation is r 2−k 2=0 r 2−k 2=0 or (r−k)(r+k)=0(r−k)(r+k)=0. Consequently, e−k x e−k x and e k x e k x are the two linearly independent solutions, and the general solution is y=C 1 e k x+C 2 e−k x.y=C 1 e k x+C 2 e−k x. Since cosh s=e s+e−s 2 cosh⁡s=e s+e−s 2 and sinh s=e s−e−s 2 sinh⁡s=e s−e−s 2, we can also write the general solution as y=D 1 cosh(k x)+D 2 sinh(k x).y=D 1 cosh⁡(k x)+D 2 sinh⁡(k x). Example 2.2.2 2.2.2: Find the general solution of y′′−8 y′+16 y=0 y″−8 y′+16 y=0 Solution The characteristic equation is r 2−8 r+16=(r−4)2=0 r 2−8 r+16=(r−4)2=0. The equation has a double root r 1=r 2=4 r 1=r 2=4. The general solution is, therefore, y=(C 1+C 2 x)e 4 x=C 1 e 4 x+C 2 x e 4 x y=(C 1+C 2 x)e 4 x=C 1 e 4 x+C 2 x e 4 x Exercise 2.2.2 2.2.2: Linear Independence Check that e 4 x e 4 x and x e 4 x x e 4 x are linearly independent. Answer That e 4 x e 4 x solves the equation is clear. If x e 4 x x e 4 x solves the equation, then we know we are done. Let us compute y′=e 4 x+4 x e 4 x y′=e 4 x+4 x e 4 x and y′′=8 e 4 x+16 x e 4 x y″=8 e 4 x+16 x e 4 x. Plug in y′′−8 y′+16 y=8 e 4 x+16 x e 4 x−8(e 4 x+4 x e 4 x)+16 x e 4 x=0 y″−8 y′+16 y=8 e 4 x+16 x e 4 x−8(e 4 x+4 x e 4 x)+16 x e 4 x=0 We should note that in practice, doubled root rarely happens. If coefficients are picked truly randomly we are very unlikely to get a doubled root. Let us give a short proof for why the solution x e r x x e r x works when the root is doubled. This case is really a limiting case of when the two roots are distinct and very close. Note that e r 2 x−e x 1 x r 2−r 1 e r 2 x−e x 1 x r 2−r 1 is a solution when the roots are distinct. When we take the limit as r 1 r 1 goes to r 2 r 2, we are really taking the derivative of e r x e r x using r r as the variable. Therefore, the limit is x e r x x e r x, and hence this is a solution in the doubled root case. 2.2.2 Complex numbers and Euler’s formula It may happen that a polynomial has some complex roots. For example, the equation r 2+1=0 r 2+1=0 has no real roots, but it does have two complex roots. Here we review some properties of complex numbers. Complex numbers may seem a strange concept, especially because of the terminology. There is nothing imaginary or really complicated about complex numbers. A complex number is simply a pair of real numbers, (a,b)(a,b). We can think of a complex number as a point in the plane. We add complex numbers in the straightforward way, (a,b)+(c,d)=(a+c,b+d)(a,b)+(c,d)=(a+c,b+d). We define multiplication by (a,b)×(c,d)=def(a c−b d,a d+b c).(a,b)×(c,d)=def(a c−b d,a d+b c). It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further, and most importantly (0,1)×(0,1)=(−1,0)(0,1)×(0,1)=(−1,0). Generally we just write (a,b)(a,b) as (a+i b)(a+i b), and we treat i i as if it were an unknown. We do arithmetic with complex numbers just as we would with polynomials. The property we just mentioned becomes i 2=−1 i 2=−1. So whenever we see i 2 i 2, we replace it by −1−1. The numbers i i and −i−i are the two roots of r 2+1=0 r 2+1=0. Note that engineers often use the letter j j instead of i i for the square root of −1−1. We will use the mathematicians’ convention and use i i. Exercise 2.2.3 2.2.3 Make sure you understand (that you can justify) the following identities: i 2=−1,i 3=−1,i 4=1 i 2=−1,i 3=−1,i 4=1, 1 i=−i 1 i=−i, (3−7 i)(−2−9 i)=⋯=−69−13 i(3−7 i)(−2−9 i)=⋯=−69−13 i, (3−2 i)(3+2 i)=3 2−(2 i)2=3 2+2 2=13(3−2 i)(3+2 i)=3 2−(2 i)2=3 2+2 2=13, 1 3−2 i=1 3−2 i 3+2 i 3+2 i=3+2 i 13=3 13+2 13 i 1 3−2 i=1 3−2 i 3+2 i 3+2 i=3+2 i 13=3 13+2 13 i. We can also define the exponential e a+i b e a+i b of a complex number. We do this by writing down the Taylor series and plugging in the complex number. Because most properties of the exponential can be proved by looking at the Taylor series, these properties still hold for the complex exponential. For example the very important property: e x+y=e x e y e x+y=e x e y. This means that e a+i b=e a e i b e a+i b=e a e i b. Hence if we can compute e i b e i b, we can compute e a+i b e a+i b. For e i b e i b we use the so-called Euler’s formula. Theorem 2.2.2 2.2.2 Euler's Formula e i θ=cos θ+i sin θ a n d e−i θ=cos θ−i sin θ e i θ=cos⁡θ+i sin⁡θ a n d e−i θ=cos⁡θ−i sin⁡θ In other words, e a+i b=e a(cos(b)+i sin(b))=e a cos(b)+i e a sin(b)e a+i b=e a(cos⁡(b)+i sin⁡(b))=e a cos⁡(b)+i e a sin⁡(b). Exercise 2.2.4 2.2.4: Using Euler’s formula, check the identities: cos θ=e i θ+e−i θ 2 and sin θ=e i θ−e−i θ 2 i cos⁡θ=e i θ+e−i θ 2 and sin⁡θ=e i θ−e−i θ 2 i Exercise 2.2.5 2.2.5 Double angle identities: Start with e i(2 θ)=(e i θ)2 e i(2 θ)=(e i θ)2. Use Euler on each side and deduce: Answer cos(2 θ)=cos 2 θ−sin 2 θ and sin(2 θ)=2 sin θ cos θ cos⁡(2 θ)=cos 2 θ−sin 2 θ and sin⁡(2 θ)=2 sin⁡θ cos⁡θ For a complex number a+i b a+i b we call a a the real part and b b the imaginary part of the number. Often the following notation is used, Re(a+i b)=a and Im(a+i b)=b Re(a+i b)=a and Im(a+i b)=b 2.2.3 Complex roots Suppose that the equation a y′′+b y′+c y=0 a y″+b y′+c y=0 has the characteristic equation a r 2+b r+c=0 a r 2+b r+c=0 that has complex roots. By the quadratic formula, the roots are −b±b 2−4 a c−−−−−−−√2 a−b±b 2−4 a c 2 a. These roots are complex if b 2−4 a c<0 b 2−4 a c<0. In this case the roots are r 1,r 2=−b 2 a±i 4 a c−b 2−−−−−−−√2 a r 1,r 2=−b 2 a±i 4 a c−b 2 2 a As you can see, we always get a pair of roots of the form α±i β α±i β. In this case we can still write the solution as y=C 1 e(α+i β)x+C 2 e(α−i β)x y=C 1 e(α+i β)x+C 2 e(α−i β)x However, the exponential is now complex valued. We would need to allow C 1 C 1 and C 2 C 2 to be complex numbers to obtain a real-valued solution (which is what we are after). While there is nothing particularly wrong with this approach, it can make calculations harder and it is generally preferred to find two real-valued solutions. Here we can use Euler’s formula. Let y 1=e(α+i β)x and y 2=e(α−i β)x y 1=e(α+i β)x and y 2=e(α−i β)x Then note that y 1 y 2=e a x cos(β x)+i e a x sin(β x)=e a x cos(β x)−i e a x sin(β x)(2.2.4)(2.2.4)y 1=e a x cos⁡(β x)+i e a x sin⁡(β x)y 2=e a x cos⁡(β x)−i e a x sin⁡(β x) Linear combinations of solutions are also solutions. Hence, y 3 y 4=y 1+y 2 2=e a x cos(β x)=y 1−y 2 2 i=e a x sin(β x)(2.2.5)(2.2.5)y 3=y 1+y 2 2=e a x cos⁡(β x)y 4=y 1−y 2 2 i=e a x sin⁡(β x) are also solutions. Furthermore, they are real-valued. It is not hard to see that they are linearly independent (not multiples of each other). Therefore, we have the following theorem. Theorem 2.2.3 2.2.3 For the homegneous second order ODE a y′′+b y′+c y=0 a y″+b y′+c y=0 If the characteristic equation has the roots α±i β α±i β(when b 2−4 a c<0 b 2−4 a c<0), then the general solution is y=C 1 e a x cos(β x)+C 2 e a x sin(β x)y=C 1 e a x cos⁡(β x)+C 2 e a x sin⁡(β x) Example 2.2.3 2.2.3 Find the general solution of y′′+k 2 y=0 y″+k 2 y=0, for a constant k>0 k>0. Solution The characteristic equation is r 2+k 2=0 r 2+k 2=0. Therefore, the roots are r=±i k r=±i k and by the theorem we have the general solution y=C 1 cos(k x)+C 2 sin(k x)y=C 1 cos⁡(k x)+C 2 sin⁡(k x) Example 2.2.4 2.2.4 Find the solution of y′′−6 y′+13 y=0,y(0)=0,y′(0)=10.y″−6 y′+13 y=0,y(0)=0,y′(0)=10. Solution The characteristic equation is r 2−6 r+13=0 r 2−6 r+13=0. By completing the square we get (r−3)2+2 2=0(r−3)2+2 2=0 and hence the roots are r=3±2 i r=3±2 i. By the theorem we have the general solution y=C 1 e 3 x cos(2 x)+C 2 e 3 x sin(2 x)y=C 1 e 3 x cos⁡(2 x)+C 2 e 3 x sin⁡(2 x) To find the solution satisfying the initial conditions, we first plug in zero to get 0=y(0)=C 1 e 0 cos 0+C 2 e 0 sin 0=C 1 0=y(0)=C 1 e 0 cos⁡0+C 2 e 0 sin⁡0=C 1 Hence C 1=0 C 1=0 and y=C 2 e 3 x sin(2 x)y=C 2 e 3 x sin⁡(2 x). We differentiate y′=3 C 2 e 3 x sin(2 x)+2 C 2 e 3 x cos(2 x)y′=3 C 2 e 3 x sin⁡(2 x)+2 C 2 e 3 x cos⁡(2 x) We again plug in the initial condition and obtain 10=y′(0)=2 C 2 10=y′(0)=2 C 2, or C 2=5 C 2=5. Hence the solution we are seeking is y=5 e 3 x sin(2 x)y=5 e 3 x sin⁡(2 x) Footnotes Making an educated guess with some parameters to solve for is such a central technique in differential equations, that people sometimes use a fancy name for such a guess: ansatz, German for “initial placement of a tool at a work piece.” Yes, the Germans have a word for that. This page titled 2.2: Constant coefficient second order linear ODEs is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform. Back to top 2.1: Second order linear ODEs 2.3: Higher order linear ODEs Was this article helpful? Yes No Recommended articles 14.4: Constant coefficient second order linear ODEs 4.4: Homogeneous Linear Second Order Equations with Constant Coefficients 2.2: Constant coefficient second order linear ODEs 4.1b: Constant coefficient second order linear ODEs 2.E: Higher order linear ODEs (Exercises)These are homework exercises to accompany Libl's "Differential Equations for Engineering" Textmap. This is a textbook targeted for a one semester firs... Article typeSection or PageAuthorJiří LeblAutonumber Section Headingstitle with colon delimitersLicenseCC BY-SALicense Version4.0Show Page TOCno Tags Euler’s formula 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 ☰ 2.1: Second order linear ODEs 2.3: Higher order linear ODEs Complete your gift to make an impact
17585	https://artofproblemsolving.com/wiki/index.php/Line?srsltid=AfmBOorhLDosA6eD7_BRsRJRMZNQ3ah6gc5P9txM8mhgfx5vT8-NpHR_	Art of Problem Solving Line - 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 Line Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Line A line in the euclidean sense is defined as the shortest distance between two points. It is defined to be in 1 direction only, i.e. infinitely thin but also infinitely long. In the Cartesian coordinate system, it is usually described as an equation in x and y of the form , where m is the slope of the line and b is the y-intercept. Any two points define a line, and given specific one can solve for the line's equation. Introductory Example Problem 2006 AMC 10B Problem 12 See also Linear 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.
17586	https://www.youtube.com/watch?v=nj9ryI9f2mw	AKPotW: Proof that -(-a) = a [in the Reals] Center of Math 45100 subscribers 151 likes Description 19681 views Posted: 8 Jun 2017 Full problem and solution transcript here: 11 comments Transcript: Hello and welcome to the worldwide center of mathematics advanced knowledge problem of the week. Um this week we're just proving that uh negative negative a is equal to a um for a and the real numbers. Um so we know the real numbers are a field. So we can prove this supposedly just using the axioms of a field. Um and actually one other thing that's not exactly an axiom but so we know this to be true. um we want to prove it and the way we're going to do so is kind of intuitive but it's also pretty satisfying to do. So the way we're going to do this is by um well we know that for any number um there exists its additive inverse so that we get n plus the additive inverse of n is equal to zero which is just n uh if we're in the real numbers. Um so we have two things we we're looking for this negative a and this a. So let why don't we just substitute these uh numbers as our n. Um so we have these two things. We might already see it but um just so that we can uh look at this and get more information from it just going to use a fact that addition commutes and rewrite this. So now this means that for negative a we have two uh additive inverses which are a and negative negative a. And since uh additive inverses are unique, that means that a has to equalative a. So uh that's your proof. I'm sure you've used this a lot and maybe never thought about it. So that's why I thought this would be a good advanced knowledge problem of the week. Um this is u some we used fields here. We do have a a book on fields available on our store. Stick around for a link to our website. Um, also if you enjoyed this problem of the week, then stick around for a uh playlist or to subscribe to our YouTube channel. Thank you for watching and have a good day.
17587	https://www.youtube.com/watch?v=7lJ77I5A0As	🧮 How to CALCULATE FOLD CHANGE AND PERCENTAGE DIFFERENCE Adwoa 2680 subscribers 217 likes Description 34822 views Posted: 30 Jan 2021 Subscribe for a fun approach to learning lab techniques: A fold change is simply a ratio. It measures the number of times that something has changed. For example, if you are comparing an experimental group and a control group, you would divide the result for the experimental group by that of the control group. If you are measuring a signal over time, you would divide the signal at each point with the signal at the beginning. Or you may divide the signal from the latest time-point to that of the previous. COMMUNICATING/INTERPRETING FOLD CHANGE A twofold increase shows that an amount doubled. Determining the fold change allows us to answer questions about increases and decreases in an experiment. PERCENTAGE CHANGE A percentage difference on the other hand looks at the difference in the new value compared to the original value. It is calculated as: new value - initial value / initial value. Other ways to analyse your data: Average/Mean Since experiments are often conducted with multiple observations (replicates), the raw data can be presented as a singular value that represents what is called the mean of the data. This mean (also known as an average) is calculated by adding up on the values and dividing by the number of values. Standard deviation. Since multiple observations are made ( as mentioned above), it is necessary to give the audience a sense of the variation in the raw data that was used to calculate the average/mean. The standard deviation communicates the variability in the underlying data. It is calculated by subtracting each value from the mean/average and then squaring it. You then take the average of the squared differences. Finally, the square root of the squared mean/average is then taken. Excel has a function/formula you can use. Standard error of the mean. The standard deviation is divided by the square root of the total number of observations/values. Normalisation. It's often handy to present data in reference to a particular time-point so that comparisons are easier to make. This scaling of data is called normalisation and we'll look at how to calculate it in this video. NB: In other disciplines normalisation is used to scale all values between 0-1. This again makes comparisons obvious. Graphing the data. This is what is meant by data visualisation. It converts the numerical values (averages/mean, standard deviations, standard errors and fold changes etc) into a pictorial/graphical format. This makes seeing the story in the data easier for an audience. Finally, remember that noticing differences in data cannot be said to be significant, until statistical tests are done. Where averages/means of two variables are calculated, a student t-test (t test) is used to verify that the observed difference have a probability value of less than 0.05 that it occurred by chance. Where multiple means are being compared statistically, an Analysis of Variance (ANOVA) is the correct test to use. Where the comparisons are not based on means/averages, different statistical tests are used. For example, to examine whether one variable influences another statistical, a correlation test such as linear regression is performed. . . //MUSIC Music: Shine by Joakim Karud . . . . . . . . 36 comments Transcript: Intro if you've ever wondered what a fold change is referring to when it's used in medical research context in the context of data analysis then this video is for you. A fold change has nothing to do with folding.. yeah i know surprising fold change is simply a ratio it measures the number What is full change of times that something has changed for example if you are comparing an experimental group in a control group you will divide the results for the experimental group by that of the control group if you're measuring a signal over time you may want to look at the signal in reference to the very first time point or you may want to look at the signal from the latest time point to the one just before it the previous time point as an example let's say the intensity that i got for protein expression for one time point was this value then i can look at and this was the experimental group then i can look at the control group and the control group has this value so to look at i may want to see what's the fold change what's the ratio here so i can just go ahead and do the experimental group and divide that by the control group Percentage Difference sometimes it may be more meaningful to look at the percentage difference so actually the difference is what you are concentrating on so in this case you will take away the value the new value you take away the initial value and then divided by the initial value and often it's also meaningful to present it as a percentage so you multiply it per 100 and that just makes it easier to get meaning Antibody Response now if for example we're looking at an antibody response following infection we may easily report it as a fold change so we will get the original expression the original quantity which we may have measured via Western blot or an ELISA: Original Expression enzyme-linked immunosorbent essay so then we can just simply divide into the original amount that we got it's just a ratio of the results then easily express that to make it meaningful at a glance as a percentage and that's just by multiplying it by a hundred how about d3 we would simply just divide that value into the original so we'll divide the signal at the new time point or the day by the signal at the beginning like the initial infection so what we see that we have a nine fold change it's 9.8 so almost a ten-fold change from the first the day one post infection to this uh second we have almost a ten-fold change then by day three between day three and the and the day one there's a uh 1.74 fold change and so on and so forth for the remainder of the data and then you can just look at that in terms present that as percentage just to make it easier for your audience for the person looking at your data to just immediately get a sense of what that means so if we just multiply that by a hundred Percent Difference we may also want to present the data in terms of the difference in the data rather than the absolute fold change and in that case it's our new value minus the original divide by the original present that as percentage just to make it easier for your audience when i leave the number just as the difference without multiplying by a hundred you can see you know point two two point two two is a small value it's not immediately clear to someone what you know they don't get a good sense of the number because we don't normally deal with that kind of scale so just by multiplying by 100 we'll get a sense we'll... it will be like 22 and yeah we're comfortable with that so it's good to scale it to a percentage if we want the rest of the cells to be calculated without us going through this process of doing it ourselves we simply constrain the cell by just placing a dollar sign in front of it so we place the dollar sign in front of the row as well as the column and then by dragging it down excel noise to do exactly the same thing for the res the remainder of the cells and then we can go ahead and plot it and plot in excel is super easy you just have to select the two columns that you want to plot and go to insert and you can select the type of chart that you want so for this kind of time series or related data it is best to use a line graph or a scatter graph we can use a line graph in this case because it is related and we can go ahead and and it will automatically have a line that tracks it for us and just by making some simple adjustments we can make the graph look really attractive now i will not go into too much detail because this is just on fold change and on the percentage difference calculations if you would like to see the graphs done in excel then please let me know but i will leave it here for now and if you are interested in joining the study group go ahead and hit the subscribe button and i will catch you in the next video God bless you and take care.
17588	https://genchem.chem.umass.edu/summer/chem112/112_Experiment_5.htm	Chem 112, Exp 5: Determining Ka's Using pH Titration Curves Experiment 5 Polyprotic Acids Determining Ka's Using pH Titration Curves Introduction: pH Titration Curves 'Idealized': To date the equivalence point of an acid base reaction has been determined using an indicator. In this experiment we are going to monitor the changes in pH that occurs during the titration of a weak polyprotic acid with a strong base. At the equivalence point one should expect to see a dramatic change in pH as the solution goes from acidic to strongly basic. Depicted on the left is an idealized pH titration curve for a weak diprotic acid. The first thing that you should notice is that there are two regions where we see a significant pH change. These, if you wish, correspond to two separate titrations. Titration 1 is the reaction of the first proton with the base (in this case sodium hydroxide). H 2 X(aq) + NaOH(aq) = NaHX(aq) + H 2 O(l) The second titration corresponding to the reaction of the second proton with sodium hydroxide NaHX(aq) + NaOH(aq) = Na 2 X(aq) + H 2 O(l) So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a combination of a number of titrations depending on the number of acidic protons on the polyprotic acid. The overall reaction is the sum of the two titration's H 2 X(aq) + 2 NaOH = Na 2 X(aq) + 2 H 2 O(l) In determining the quantity of the acid or the molarity of the acid, we are normally just interested in the final equivalence point. In a pH titration plot, this is determined by finding the point of inflection on the final area where we see a significant rise in pH (This can be approximated by determining the midpoint.) However, this plot contains some other interesting features. First off, if we look at the area corresponding to the first titration, it should come as no surprise that its equivalence point corresponds to the addition of exactly 1/2 the volume of NaOH required to reach the final equivalence point. The real neat point comes at the 1/2 way point of each titration. Let us focus on the Titration 1. At the 1/2 way point, the concentration of H 2 X(aq) remaining in the solution is equal to 1/2 the initial concentration of H 2 X! The concentration of NaHX(aq) produced is also numerically equal to 1/2 the initial concentration of H 2 X! So what, you may ask. Let's focus for a moment on the acid equilibrium associated with the acid that we are dealing with in titration 1. H 2 X(aq) + H 2 O(l) ® HX- + H 3 O+ Ka = [H 3 O+][HX-]/[H 2 X] or written in another way [H 3 O+] = Ka{[H 2 X]/[HX-]} using the concentrations that we know for H 2 X and HX- (=NaHX) at the 1/2 way point we get [H 3 O+] = Ka{1/2[H 2 X]initial/[1/2H 2 X]initial} [H 3 O+] = Ka From the graph we can determine the pH at this point and since pH=-log10[H 3 O+], we can determine [H 3 O+] at this point and thus obtain the Ka for this equilibrium. Neat! Since this is a polyprotic acid, this corresponds to Ka1. Guess what you can determine from the pH at the midpoint of the second titration. This information can be used to help identify the acid in question since Ka for a large number of polyprotic acids are known. The first acid that you will be following today is citric acid which is an acid that falls into the idealized category. You should see three areas where the pH undergoes significant changes and should be able to determine the three Ka values for citric acid and compare the result to the three known values given below. H 3 C 6 H 5 O 7(aq) + H 2 O(l)<=>H 2 C 6 H 7 O 7- + H 3 O+ K a1 = 7.4x10-3 @ 25 o C H 2 C 6 H 5 O 7- + H 2 O(l) <=>HC 6 H 6 O 7 2- + H 3 O+ K a2 = 1.7x10-5 @ 25 o C HC 6 H 5 O 7 2- + H 2 O(l) <=>C 6 H 5 O 7 3- + H 3 O+ K a3 = 4.0x10-7 @ 25 o C pH Titration Curves 'The Real World': In reality, many polyprotic acids only show one discernable equivalence point! The vast majority of the time, this corresponds to final equilibrium. If this is the case then, all the other equivalence points can be determined by knowing what type of polyprotic acid one is dealing with, i.e., diprotic or tri protic. For a triprotic acid, the other two equivalence points should correspond to 1/3 and 2/3 of the volume of the base required to reach the final one and thus one can still determine the Ka values. Note, however, I did say the vast majority of the time. How one knows how to determine whether the observed equivalence point equals the removal of the final proton I leave for you to explore! [Hint: what is the pH at the 1/2 equivalence point of a titration of a polyprotic acid equal to] The second acid that you will be looking at in this lab is phosphoric acid, a triprotic acid whose Ka values are given below. H 3 PO 4(aq) + H 2 O(l) <=>H 2 PO 4- + H 3 O+ K a1 = 7.5x10-3 @ 25 o C H 2 PO 4- + H 2 O(l) <=>HPO 4 2- + H 3 O+ K a2= 6.2x10-8 @ 25 o C HPO 4 2- + H 2 O(l) <=>PO 4 3- + H 3 O+ K a3= 3.6x10-13 @ 25 o C Experimental Procedure: Your TA will also demonstrate the best set up for this experiment. Using a graduated cylinder, place ~ 20mL of the ~0.02M citric acid into a small beaker. If necessary add distilled water such that the tip of the pH probe is covered. Fill your buret with the ~0.02M NaOH solution. Record the exact molarity of this solution. Record the initial buret reading. Remember that this corresponds to 0.00mL of NaOH added. Record the initial pH of the Citric acid. Carefully add the NaOH recording the volume of NaOH required to effect a pH change of 0.2. Continue this process until the pH reaches 12. Plot a graph of 'pH' versus 'Volume of NaOH" added and from this graph determine: The Ka values for citric acid. The exact concentration of the citric acid. Repeat steps one through five using the ~0.02M phosphoric acid and determine the Ka values for phosphoric acid and the exact molarity of the phosphoric acid solution. Write-up: In the discussion portion of your write up, be sure to address the correlation between the Ka values that you obtained and those given to you in this procedure. Address any unusual problems that you encountered.
17589	https://en.wikipedia.org/wiki/Perfect_digital_invariant	Jump to content Perfect digital invariant Add links From Wikipedia, the free encyclopedia Number that is the sum of its own digits, each raised to a given power In number theory, a perfect digital invariant (PDI) is a number in a given number base () that is the sum of its own digits each raised to a given power (). Definition [edit] Let be a natural number. The perfect digital invariant function (also known as a happy function, from happy numbers) for base and power is defined as: where is the number of digits in the number in base , and is the value of each digit of the number. A natural number is a perfect digital invariant if it is a fixed point for , which occurs if . and are trivial perfect digital invariants for all and , all other perfect digital invariants are nontrivial perfect digital invariants. For example, the number 4150 in base is a perfect digital invariant with , because . A natural number is a sociable digital invariant if it is a periodic point for , where for a positive integer (here is the th iterate of ), and forms a cycle of period . A perfect digital invariant is a sociable digital invariant with , and a amicable digital invariant is a sociable digital invariant with . All natural numbers are preperiodic points for , regardless of the base. This is because if , , so any will satisfy until . There are a finite number of natural numbers less than , so the number is guaranteed to reach a periodic point or a fixed point less than , making it a preperiodic point. Numbers in base lead to fixed or periodic points of numbers . Proof If , then the bound can be reduced. Let be the number for which the sum of squares of digits is largest among the numbers less than . : because Let be the number for which the sum of squares of digits is largest among the numbers less than . : because Let be the number for which the sum of squares of digits is largest among the numbers less than . Let be the number for which the sum of squares of digits is largest among the numbers less than . . Thus, numbers in base lead to cycles or fixed points of numbers . The number of iterations needed for to reach a fixed point is the perfect digital invariant function's persistence of , and undefined if it never reaches a fixed point. is the digit sum. The only perfect digital invariants are the single-digit numbers in base , and there are no periodic points with prime period greater than 1. reduces to , as for any power , and . For every natural number , if , and , then for every natural number , if , then , where is Euler's totient function. Proof Let be a natural number with digits, where , and , where is a natural number greater than 1. According to the divisibility rules of base , if , then if , then the digit sum If a digit , then . According to Euler's theorem, if , . Thus, if the digit sum , then . Therefore, for any natural number , if , and , then for every natural number , if , then . No upper bound can be determined for the size of perfect digital invariants in a given base and arbitrary power, and it is not currently known whether or not the number of perfect digital invariants for an arbitrary base is finite or infinite. F2,b [edit] By definition, any three-digit perfect digital invariant for with natural number digits , , has to satisfy the cubic Diophantine equation . has to be equal to 0 or 1 for any , because the maximum value can take is . As a result, there are actually two related quadratic Diophantine equations to solve: : when , and : when . The two-digit natural number is a perfect digital invariant in base This can be proven by taking the first case, where , and solving for . This means that for some values of and , is not a perfect digital invariant in any base, as is not a divisor of . Moreover, , because if or , then , which contradicts the earlier statement that . There are no three-digit perfect digital invariants for , which can be proven by taking the second case, where , and letting and . Then the Diophantine equation for the three-digit perfect digital invariant becomes for all values of . Thus, there are no solutions to the Diophantine equation, and there are no three-digit perfect digital invariants for . F3,b [edit] There are just four numbers, after unity, which are the sums of the cubes of their digits: These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to the mathematician. (sequence A046197 in the OEIS) — G. H. Hardy, A Mathematician's Apology By definition, any four-digit perfect digital invariant for with natural number digits , , , has to satisfy the quartic Diophantine equation . has to be equal to 0, 1, 2 for any , because the maximum value can take is . As a result, there are actually three related cubic Diophantine equations to solve : when : when : when We take the first case, where . b = 3k + 1 [edit] Let be a positive integer and the number base . Then: is a perfect digital invariant for for all . Proof Let the digits of be , , and . Then Thus is a perfect digital invariant for for all . is a perfect digital invariant for for all . Proof Let the digits of be , , and . Then Thus is a perfect digital invariant for for all . is a perfect digital invariant for for all . Proof Let the digits of be , , and . Then Thus is a perfect digital invariant for for all . Perfect digital invariants \| \| \| \| \| \| \| 1 \| 4 \| 130 \| 131 \| 203 \| \| 2 \| 7 \| 250 \| 251 \| 305 \| \| 3 \| 10 \| 370 \| 371 \| 407 \| \| 4 \| 13 \| 490 \| 491 \| 509 \| \| 5 \| 16 \| 5B0 \| 5B1 \| 60B \| \| 6 \| 19 \| 6D0 \| 6D1 \| 70D \| \| 7 \| 22 \| 7F0 \| 7F1 \| 80F \| \| 8 \| 25 \| 8H0 \| 8H1 \| 90H \| \| 9 \| 28 \| 9J0 \| 9J1 \| A0J \| b = 3k + 2 [edit] Let be a positive integer and the number base . Then: is a perfect digital invariant for for all . Proof Let the digits of be , , and . Then Thus is a perfect digital invariant for for all . Perfect digital invariants \| \| \| \| \| 1 \| 5 \| 103 \| \| 2 \| 8 \| 205 \| \| 3 \| 11 \| 307 \| \| 4 \| 14 \| 409 \| \| 5 \| 17 \| 50B \| \| 6 \| 20 \| 60D \| \| 7 \| 23 \| 70F \| \| 8 \| 26 \| 80H \| \| 9 \| 29 \| 90J \| b = 6k + 4 [edit] Let be a positive integer and the number base . Then: is a perfect digital invariant for for all . Proof Let the digits of be , , and . Then Thus is a perfect digital invariant for for all . Perfect digital invariants \| \| \| \| \| 0 \| 4 \| 021 \| \| 1 \| 10 \| 153 \| \| 2 \| 16 \| 285 \| \| 3 \| 22 \| 3B7 \| \| 4 \| 28 \| 4E9 \| Fp,b [edit] All numbers are represented in base . \| \| \| Nontrivial perfect digital invariants \| Cycles \| --- --- \| \| 2 \| 3 \| 12, 22 \| 2 → 11 → 2 \| \| 4 \| \| \| \| 5 \| 23, 33 \| 4 → 31 → 20 → 4 \| \| 6 \| \| 5 → 41 → 25 → 45 → 105 → 42 → 32 → 21 → 5 \| \| 7 \| 13, 34, 44, 63 \| 2 → 4 → 22 → 11 → 2 16 → 52 → 41 → 23 → 16 \| \| 8 \| 24, 64 \| 4 → 20 → 4 5 → 31 → 12 → 5 15 → 32 → 15 \| \| 9 \| 45, 55 \| 58 → 108 → 72 → 58 75 → 82 → 75 \| \| 10 \| \| 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 \| \| 11 \| 56, 66 \| 5 → 23 → 12 → 5 68 → 91 → 75 → 68 \| \| 12 \| 25, A5 \| 5 → 21 → 5 8 → 54 → 35 → 2A → 88 → A8 → 118 → 56 → 51 → 22 → 8 18 → 55 → 42 → 18 68 → 84 → 68 \| \| 13 \| 14, 36, 67, 77, A6, C4 \| 28 → 53 → 28 79 → A0 → 79 98 → B2 → 98 \| \| 14 \| \| 1B → 8A → BA → 11B → 8B → D3 → CA → 136 → 34 → 1B 29 → 61 → 29 \| \| 15 \| 78, 88 \| 2 → 4 → 11 → 2 8 → 44 → 22 → 8 15 → 1B → 82 → 48 → 55 → 35 → 24 → 15 2B → 85 → 5E → EB → 162 → 2B 4E → E2 → D5 → CE → 17A → A0 → 6A → 91 → 57 → 4E 9A → C1 → 9A D6 → DA → 12E → D6 \| \| 16 \| \| D → A9 → B5 → 92 → 55 → 32 → D \| \| 3 \| 3 \| 122 \| 2 → 22 → 121 → 101 → 2 \| \| 4 \| 20, 21, 130, 131, 203, 223, 313, 332 \| \| \| 5 \| 103, 433 \| 14 → 230 → 120 → 14 \| \| 6 \| 243, 514, 1055 \| 13 → 44 → 332 → 142 → 201 → 13 \| \| 7 \| 12, 22, 250, 251, 305, 505 \| 2 → 11 → 2 13 → 40 → 121 → 13 23 → 50 → 236 → 506 → 665 → 1424 → 254 → 401 → 122 → 23 51 → 240 → 132 → 51 160 → 430 → 160 161 → 431 → 161 466 → 1306 → 466 516 → 666 → 1614 → 552 → 516 \| \| 8 \| 134, 205, 463, 660, 661 \| 662 → 670 → 1057 → 725 → 734 → 662 \| \| 9 \| 30, 31, 150, 151, 570, 571, 1388 \| 38 → 658 → 1147 → 504 → 230 → 38 152 → 158 → 778 → 1571 → 572 → 578 → 1308 → 660 → 530 → 178 → 1151 → 152 638 → 1028 → 638 818 → 1358 → 818 \| \| 10 \| 153, 370, 371, 407 \| 55 → 250 → 133 → 55 136 → 244 → 136 160 → 217 → 352 → 160 919 → 1459 → 919 \| \| 11 \| 32, 105, 307, 708, 966, A06, A64 \| 3 → 25 → 111 → 3 9 → 603 → 201 → 9 A → 82A → 1162 → 196 → 790 → 895 → 1032 → 33 → 4A → 888 → 1177 → 576 → 5723 → A3 → 8793 → 1210 → A 25A → 940 → 661 → 364 → 25A 366 → 388 → 876 → 894 → A87 → 1437 → 366 49A → 1390 → 629 → 797 → 1077 → 575 → 49A \| \| 12 \| 577, 668, A83, 11AA \| \| \| 13 \| 490, 491, 509, B85 \| 13 → 22 → 13 \| \| 14 \| 136, 409 \| \| \| 15 \| C3A, D87 \| \| \| 16 \| 23, 40, 41, 156, 173, 208, 248, 285, 4A5, 580, 581, 60B, 64B, 8C0, 8C1, 99A, AA9, AC3, CA8, E69, EA0, EA1 \| \| \| 4 \| 3 \| \| 121 → 200 → 121 122 → 1020 → 122 \| \| 4 \| 1103, 3303 \| 3 → 1101 → 3 \| \| 5 \| 2124, 2403, 3134 \| 1234 → 2404 → 4103 → 2323 → 1234 2324 → 2434 → 4414 → 11034 → 2324 3444 → 11344 → 4340 → 4333 → 3444 \| \| 6 \| \| \| \| 7 \| \| \| \| 8 \| 20, 21, 400, 401, 420, 421 \| \| \| 9 \| 432, 2466 \| \| \| 5 \| 3 \| 1020, 1021, 2102, 10121 \| \| \| 4 \| 200 \| 3 → 3303 → 23121 → 10311 → 3312 → 20013 → 10110 → 3 3311 → 13220 → 10310 → 3311 \| Extension to negative integers [edit] Perfect digital invariants can be extended to the negative integers by use of a signed-digit representation to represent each integer. Balanced ternary [edit] In balanced ternary, the digits are 1, −1 and 0. This results in the following: With odd powers , reduces down to digit sum iteration, as , and . With even powers , indicates whether the number is even or odd, as the sum of each digit will indicate divisibility by 2 if and only if the sum of digits ends in 0. As and , for every pair of digits 1 or −1, their sum is 0 and the sum of their squares is 2. Relation to happy numbers [edit] Main article: Happy number A happy number for a given base and a given power is a preperiodic point for the perfect digital invariant function such that the -th iteration of is equal to the trivial perfect digital invariant , and an unhappy number is one such that there exists no such . Programming example [edit] The example below implements the perfect digital invariant function described in the definition above to search for perfect digital invariants and cycles in Python. This can be used to find happy numbers. ``` def pdif(x: int, p: int, b: int) -> int: """Perfect digital invariant function.""" total = 0 while x > 0: total = total + pow(x % b, p) x = x // b return total def pdif_cycle(x: int, p: int, b: int) -> list[int]: seen = [] while x not in seen: seen.append(x) x = pdif(x, p, b) cycle = [] while x not in cycle: cycle.append(x) x = pdif(x, p, b) return cycle ``` See also [edit] Arithmetic dynamics Dudeney number Factorion Happy number Kaprekar's constant Kaprekar number Meertens number Narcissistic number Perfect digit-to-digit invariant Sum-product number References [edit] ^ Jump up to: a b Perfect and PluPerfect Digital Invariants Archived 2007-10-10 at the Wayback Machine by Scott Moore ^ PDIs by Harvey Heinz External links [edit] Digital Invariants \| v t e Classes of natural numbers \| \| --- \| \| \| Powers and related numbers \| \| --- \| \| Achilles Power of 2 Power of 3 Power of 10 Square Cube Fourth power Fifth power Sixth power Seventh power Eighth power Perfect power Powerful Prime power \| \| \| \| \| \| Of the form a × 2b ± 1 \| \| --- \| \| Cullen Double Mersenne Fermat Mersenne Proth Thabit Woodall \| \| \| \| \| \| Other polynomial numbers \| \| --- \| \| Hilbert Idoneal Leyland Loeschian Lucky numbers of Euler \| \| \| \| \| \| Recursively defined numbers \| \| --- \| \| Fibonacci Jacobsthal Leonardo Lucas Narayana Padovan Pell Perrin \| \| \| \| \| \| Possessing a specific set of other numbers \| \| --- \| \| Amenable Congruent Knödel Riesel Sierpiński \| \| \| \| \| \| Expressible via specific sums \| \| --- \| \| Nonhypotenuse Polite Practical Primary pseudoperfect Ulam Wolstenholme \| \| \| \| \| \| Figurate numbers \| \| --- \| \| \| \| \| \| \| \| \| --- --- --- \| \| 2-dimensional \| \| \| \| --- \| \| centered \| Centered triangular Centered square Centered pentagonal Centered hexagonal Centered heptagonal Centered octagonal Centered nonagonal Centered decagonal Star \| \| non-centered \| Triangular Square Square triangular Pentagonal Hexagonal Heptagonal Octagonal Nonagonal Decagonal Dodecagonal \| \| \| 3-dimensional \| \| \| \| --- \| \| centered \| Centered tetrahedral Centered cube Centered octahedral Centered dodecahedral Centered icosahedral \| \| non-centered \| Tetrahedral Cubic Octahedral Dodecahedral Icosahedral Stella octangula \| \| pyramidal \| Square pyramidal \| \| \| 4-dimensional \| \| \| \| --- \| \| non-centered \| Pentatope Squared triangular Tesseractic \| \| \| \| \| \| \| \| Combinatorial numbers \| \| --- \| \| Bell Cake Catalan Dedekind Delannoy Euler Eulerian Fuss–Catalan Lah Lazy caterer's sequence Lobb Motzkin Narayana Ordered Bell Schröder Schröder–Hipparchus Stirling first Stirling second Telephone number Wedderburn–Etherington \| \| \| \| \| \| Primes \| \| --- \| \| Wieferich Wall–Sun–Sun Wolstenholme prime Wilson \| \| \| \| \| \| \| \| --- \| \| Carmichael number Catalan pseudoprime Elliptic pseudoprime Euler pseudoprime Euler–Jacobi pseudoprime Fermat pseudoprime Frobenius pseudoprime Lucas pseudoprime Lucas–Carmichael number Perrin pseudoprime Somer–Lucas pseudoprime Strong pseudoprime \| \| \| \| \| \| Arithmetic functions and dynamics \| \| --- \| \| \| \| \| --- \| \| Divisor functions \| Abundant Almost perfect Arithmetic Betrothed Colossally abundant Deficient Descartes Hemiperfect Highly abundant Highly composite Hyperperfect Multiply perfect Perfect Practical Primitive abundant Quasiperfect Refactorable Semiperfect Sublime Superabundant Superior highly composite Superperfect \| \| Prime omega functions \| Almost prime Semiprime \| \| Euler's totient function \| Highly cototient Highly totient Noncototient Nontotient Perfect totient Sparsely totient \| \| Aliquot sequences \| Amicable Perfect Sociable Untouchable \| \| Primorial \| Euclid Fortunate \| \| \| \| \| \| \| \| \| --- \| \| Blum Cyclic Erdős–Nicolas Erdős–Woods Friendly Giuga Harmonic divisor Jordan–Pólya Lucas–Carmichael Pronic Regular Rough Smooth Sphenic Størmer Super-Poulet \| \| \| \| \| \| Numeral system-dependent numbers \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Arithmetic functions and dynamics \| Persistence + Additive + Multiplicative \| \| \| --- \| \| Digit sum \| Digit sum Digital root Self Sum-product \| \| Digit product \| Multiplicative digital root Sum-product \| \| Coding-related \| \| \| Other \| Dudeney Factorion Kaprekar Kaprekar's constant Keith Lychrel Narcissistic Perfect digit-to-digit invariant Perfect digital invariant + Happy \| \| \| P-adic numbers-related \| Automorphic + Trimorphic \| \| Digit-composition related \| Palindromic Pandigital Repdigit Repunit Self-descriptive Smarandache–Wellin Undulating \| \| Digit-permutation related \| Cyclic Digit-reassembly Parasitic Primeval Transposable \| \| Divisor-related \| Equidigital Extravagant Frugal Harshad Polydivisible Smith Vampire \| \| Other \| \| \| \| \| \| \| \| Binary numbers \| \| --- \| \| Evil Odious Pernicious \| \| \| \| \| \| Generated via a sieve \| \| --- \| \| Lucky Prime \| \| \| \| \| \| Sorting related \| \| --- \| \| Pancake number Sorting number \| \| \| \| \| \| Natural language related \| \| --- \| \| Aronson's sequence Ban \| \| \| \| \| \| Graphemics related \| \| --- \| \| \| \| \| \| \| Mathematics portal \| \| Retrieved from " Categories: Arithmetic dynamics Base-dependent integer sequences Diophantine equations Hidden categories: Webarchive template wayback links Articles with short description Short description is different from Wikidata Articles with example Python (programming language) code
17590	https://pmc.ncbi.nlm.nih.gov/articles/PMC6560656/	Performance of the proposed ACR–EULAR classification criteria for systemic lupus erythematosus (SLE) in a cohort of patients with SLE with neuropsychiatric symptoms - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Service Alert: Planned Maintenance on July 26th Most services will be unavailable for 12+ hours starting 6 AM EDT. Learn more about the maintenance. 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Learn more: PMC Disclaimer \| PMC Copyright Notice RMD Open . 2019 Jun 5;5(1):e000895. doi: 10.1136/rmdopen-2019-000895 Search in PMC Search in PubMed View in NLM Catalog Add to search Performance of the proposed ACR–EULAR classification criteria for systemic lupus erythematosus (SLE) in a cohort of patients with SLE with neuropsychiatric symptoms Maka Gegenava Maka Gegenava 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands Find articles by Maka Gegenava 1,#, Hannelore Jacqueline Lucia Beaart Hannelore Jacqueline Lucia Beaart 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands Find articles by Hannelore Jacqueline Lucia Beaart 1,#, Rory Caitlin Monahan Rory Caitlin Monahan 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands Find articles by Rory Caitlin Monahan 1, Elisabeth Brilman Elisabeth Brilman 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands Find articles by Elisabeth Brilman 1, Liesbeth J J Beaart-van de Voorde Liesbeth J J Beaart-van de Voorde 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands Find articles by Liesbeth J J Beaart-van de Voorde 1, Cesar Magro-Checa Cesar Magro-Checa 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands 2 Department of Rheumatology, Zuyderland Medical Center, Heerlen, the Netherlands Find articles by Cesar Magro-Checa 1,2, Tom W J Huizinga Tom W J Huizinga 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands Find articles by Tom W J Huizinga 1,✉, Gerda M Steup-Beekman Gerda M Steup-Beekman 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands 3 Department of Rheumatology, Haaglanden Medical Center, The Hague, the Netherlands Find articles by Gerda M Steup-Beekman 1,3 Author information Article notes Copyright and License information 1 Department of Rheumatology, Leiden University Medical Center, Leiden, the Netherlands 2 Department of Rheumatology, Zuyderland Medical Center, Heerlen, the Netherlands 3 Department of Rheumatology, Haaglanden Medical Center, The Hague, the Netherlands ✉ Correspondence to Dr Tom W J Huizinga; t.w.j.huizinga@lumc.nl ✉ Corresponding author. Contributed equally. Series information Short report Received 2019 Jan 5; Revised 2019 Apr 9; Accepted 2019 May 16; Collection date 2019. Keywords: SLE, NPSLE, classification criteria © Author(s) (or their employer(s)) 2019. Re-use permitted under CC BY-NC. No commercial re-use. See rights and permissions. Published by BMJ. This is an open access article distributed in accordance with the Creative Commons Attribution Non Commercial (CC BY-NC 4.0) license, which permits others to distribute, remix, adapt, build upon this work non-commercially, and license their derivative works on different terms, provided the original work is properly cited, appropriate credit is given, any changes made indicated, and the use is non-commercial. See: PMC Copyright notice PMCID: PMC6560656 PMID: 31245049 Key messages. What is already known about this subject? Recently, new American College of Rheumatology–EULAR classification criteria for systemic lupus erythematosus (SLE) have been proposed. What does this study add? In our cohort of patients with (suspected) SLE with neuropsychiatric (NP) symptoms, sensitivity of the proposed criteria was high, but specificity was suboptimal. Sensitivity further improved by including patients with antinuclear antibody–negative lupus nephritis (LN) and specifying the neurological domain. How might this impact on clinical practice? By further improving sensitivity as suggested, more patients with NPSLE and LN will be able to participate in future clinical SLE trials. Introduction Systemic lupus erythematosus (SLE) is a heterogenic connective tissue disease with a broad spectrum of clinical and laboratory manifestations. Due to this heterogeneity, SLE remains challenging to diagnose in clinical practice. In order to create a more homogeneous patient group, the American College of Rheumatology (ACR) developed classification criteria for research purposes in 1972.1 These criteria were revised in 1982 and 1997.2 3 In 2012, the Systemic Lupus International Collaborating Clinics (SLICC) developed and validated new SLE classification criteria.4 The SLICC 2012 criteria performed better than the revised 1997 ACR criteria in terms of sensitivity (97% vs 83%), but were less specific (84% vs 96%).4 Recently, new ACR–EULAR criteria have been proposed in order to improve specificity, while keeping the optimal sensitivity of the SLICC 2012 criteria. Several new elements were added, including the presence of antinuclear antibodies (ANA) as entry criterium, weighted scores for each criterium and domain scores.5–7 It is currently unknown how these criteria perform in patients with SLE with neuropsychiatric (NP) symptoms, one of the least understood manifestations of SLE. Therefore, we aimed to evaluate the performance of the proposed ACR–EULAR criteria in our cohort of patients with SLE presenting with NP symptoms. Methods A retrospective cohort study was performed using electronic medical records of patients referred to the NPSLE clinic in the Leiden University Medical Center (LUMC). Information regarding the proposed ACR–EULAR criteria, the SLICC 2012 criteria and the 1997 ACR criteria was collected. Items of the different classification criteria that were attributed to other causes, for example, medication, were not counted. All patients underwent standardised multidisciplinary assessment, as described previously.8 In case of definitive NPSLE diagnosis, the applicable 1999 ACR NPSLE syndromes were assigned. In the LUMC, ANA is detected using immunofluorescence on immobilised HEp-2 cells (Biomedical Diagnostics). ANA is considered positive at a titre of ≥1:40. Sensitivity, specificity and accuracy were calculated for the proposed ACR–EULAR criteria, the SLICC 2012 criteria and the 1997 ACR criteria, using the clinical diagnosis as golden standard. Statistical analysis was performed using SPSS for Windows V.23.0. Also, 95% CIs were calculated using the Clopper-Pearson CIs. Results A total of 360 patients were included, of which 294 (82%) had the clinical diagnosis of SLE. Mean age was 43 years and the majority of patients were women (86%). Baseline characteristics are presented in table 1. Table 1. Prevalence of different symptoms in patients with and without clinically diagnosed SLE SLE (n=294)No SLE (n=66) Age, years (mean±SD)43±14 46±15 Gender (female, %)256 (87.1)55 (83.3) ACR 1997 criteria (n, %) Malar rash 129 (43.9)9 (13.6) Discoid rash 52 (17.7)7 (10.6) Photosensitivity 155 (52.7)22 (33.3) Oral ulcers 126 (42.9)19 (28.8) Non-erosive arthritis 183 (62.2)11 (16.7) Pleuritis or pericarditis 76 (25.9)2 (3.0) Renal disorder 88 (29.9)3 (4.5) Neurological disorder 37 (12.6)6 (9.1) Haematological disorder 145 (49.3)10 (15.2) Immunological disorder 221 (75.2)20 (30.3) Positive ANA 283 (96.3)50 (75.8) SLICC 2012 criteria (n, %) Acute or subacute cutaneous lupus 148 (50.3)12 (18.2) Chronic cutaneous lupus 52 (17.7)7 (10.6) Oral ulcers 126 (42.9)19 (28.8) Non-scarring alopecia 44 (15.0)2 (3.0) Arthritis 183 (62.2)11 (16.7) Serositis 76 (25.9)2 (1.0) Renal 88 (29.9)3 (4.5) Neurological Acute confusional state/delirium 9 (2.7)0 (0.0) Psychosis 12 (4.1)2 (3.0) Seizure 28 (9.5)2 (3.0) Mononeuritis 5 (1.7)1 (1.5) Myelitis 9 (3.1)2 (3.0) Neuropathy 10 (3.4)4 (6.1) Leucopenia 140 (47.6)15 (22.7) Thrombocytopenia 58 (19.7)4 (6.1) Immunological criteria ANA 283 (96.3)50 (75.8) Anti-Sm 36 (12.2)2 (3.0) Antiphospholipid antibody (IgG, IgM, LAC, anti-β2 glycoprotein)143 (48.6)17 (25.8) Low complement (C3, C4)168 (57.1)14 (21.2) Direct Coombs test 21 (7.1)0 (0.0) Proposed ACR–EULAR criteria Constitutional domain Fever 71 (24.2)11 (16.7) Mucocutaneous domain Non-scarring alopecia 44 (15.0)2 (3.0) Oral ulcers 126 (42.9)19 (28.8) Subacute cutaneous/discoid lupus 94 (32.0)17 (25.8) Acute cutaneous lupus 148 (50.3)12 (18.2) Musculoskeletal domain Arthritis 183 (62.2)11 (16.7) Serositis domain Pleural or pericardial effusion 58 (19.8)1 (1.5) Acute pericarditis 34 (11.6)1 (1.5) Haematological domain Leucopenia 118 (40.3)10 (15.2) Thrombocytopenia 58 (19.7)4 (6.1) Autoimmune haemolysis 43 (14.6)1 (1.5) Renal domain Proteinuria 66 (22.4)1 (1.5) Lupus nephritis class II/V 31 (10.5)2 (3.0) Lupus nephritis class III/IV 37 (12.6)0 (0.0) Complement protein domain Low C3 or low C4 168 (57.3)14 (21.5) Low C3 and low C4 101 (34.5)2 (3.1) Highly specific antibodies domain Anti-dsDNA antibody 169 (57.7)15 (22.7) Anti-Smith antibody 36 (12.2)2 (3.0) Antiphospholipid antibodies domain IgG, IgM, β2 glycoproteins, lupus anticoagulant 143 (48.6)17 (25.8) Neurological domain Acute confusional state/delirium 9 (2.7)0 (0.0) Psychosis 12 (4.1)2 (3.0) Seizure 28 (9.5)2 (3.0) 1999 ACR neuropsychiatric syndromes Aseptic meningitis 0 (0.0) Cerebrovascular disease 30 (10.2)– Demyelinating syndrome 0 (0.0)– Headache 12 (4.1)– Movement disorder (chorea)3 (1.0)– Myelopathy 6 (2.0)– Seizure disorders 11 (3.7)– Acute confusional state 6 (2.0)– Anxiety disorder 2 (0.7)– Cognitive dysfunction 31 (10.5)– Mood disorder 9 (3.1)– Psychosis 7 (2.4)– Acute inflammatory demyelinating polyradiculoneuropathy 0 (0.0)– Autonomic disorder 0 (0.0)– Mononeuropathy 0 (0.0)– Myasthenia gravis 0 (0.0)– Neuropathy, cranial 0 (0.0)– Plexopathy 0 (0.0)– Polyneuropathy 3 (1.0)– Open in a new tab The 1999 ACR NPSLE syndromes were assigned after the presence of NPSLE was confirmed during multidisciplinary assessment. ACR, American College of Rheumatology; ANA, antinuclear antibody; NP, neuropsychiatric; SLE, systemic lupus erythematosus; SLICC, Systemic Lupus International Collaborating Clinics. Of the 66 patients without the clinical diagnosis of SLE, 20 patients had SLE-like disease, 12 patients undifferentiated connective tissue disease, 11 primary Sjögren’s syndrome and 8 patients had mixed connective tissue disease. The diagnoses of the remaining 15 patients were chilblain LE, chronic discoid lupus, subacute cutaneous lupus, antiphospholipid syndrome, dermatomyositis, Calcinosis-Raynaud phenomenon-esophageal involvement-sclerodactyly-telangiectasia (CREST) syndrome, juvenile idiopathic arthritis, Behçet-like disease and somatoform disorder. Of the 294 patients with the clinical diagnosis SLE, 257 (87%) fulfilled the proposed ACR–EULAR criteria, 249 (85%) patients fulfilled the 2012 SLICC criteria and 261 (89%) patients fulfilled the 1997 ACR criteria. The sensitivity, specificity and accuracy of the different criteria are presented in table 2. The proposed ACR–EULAR criteria showed a sensitivity of 87% (95% CI 83% to 91%), specificity of 74% (95% CI 62% to 84%) and an accuracy of 85% (95% CI 81% to 89%). The 2012 SLICC criteria had a sensitivity of 85% (95% CI 80% to 89%), a specificity of 76% (95% CI 64% to 85%) and an accuracy of 83% (95% CI 79% to 87%). The 1997 ACR criteria had a sensitivity of 89% (95% CI 85% to 92%), a specificity of 89% (95% CI 80% to 96%) and an accuracy of 89% (95% CI 85% to 92%). Table 2. Sensitivity, specificity and accuracy of the ACR 1997, proposed ACR–EULAR criteria and SLICC 2012 criteria ACR 1997 Proposed ACR–EULAR criteria SLICC 2012 Sensitivity (95% CI)89% (85% to 92%)87% (83% to 91%)85% (80% to 89%) Specificity (95% CI)89% (80% to 96%)74% (62% to 84%)76% (64% to 85%) Accuracy (95% CI)89% (85% to 92%)85% (81% to 89%)83% (79% to 87%) Open in a new tab ACR, American College of Rheumatology; SLICC, Systemic Lupus International Collaborating Clinics. Discussion We investigated the performance of the proposed ACR–EULAR criteria for SLE in our cohort of patients with (suspected) SLE and NP symptoms and demonstrated that sensitivity was high, but specificity was suboptimal. Our finding contrasts the results of Aringer et al.6, who found that the proposed criteria had a sensitivity close to the SLICC 2012 criteria, while specificity was similar to the ACR 1997 criteria. There are several possible explanations for this difference. The first explanation relates to the design of our cohort, which is a selected population of patients with (suspected) SLE and NP symptoms. Patients are generally referred to our clinic when diagnostic difficulties arise, which means that patients without the clinical diagnosis of SLE often have SLE-mimicking syndromes. Specificity might therefore be lower than expected. This is demonstrated for example by the high prevalence of anti-dsDNA positivity in patients without the clinical diagnosis of SLE (23%), leading to a low sensitivity (57%) and specificity (72%) of this test in our cohort. We expect that this is due to a strong referral bias: patients with autoimmune phenomena and a positive anti-dsDNA are more likely to be referred to our NPSLE clinic in case of NP symptoms than patients with a negative anti-dsDNA. Furthermore, due to the retrospective design of this study, some criteria were difficult to retrieve. This could have led to an underestimation of, for example, the constitutional domain, thereby leading to a lower sensitivity of the proposed criteria in our cohort than truly present. The second explanation relates to limitations we encountered using the proposed ACR–EULAR criteria. First of all, ANA has been proposed as entry criterium. However, in our cohort, we found nine patients with the clinical diagnosis of SLE and a sufficient amount of points for the proposed criteria, but with negative ANA. All but one of these patients also had negative anti-dsDNA. Seven of these patients had biopsy-proven lupus nephritis (LN), of which five had LN class IV, one LN class V and one LN class III. In addition, two patients had LN for which they received immunosuppressive treatment (class unknown). When ANA was not used as entry criterium, sensitivity increased to 90% (95% CI 87% to 94%), while specificity remained similar. As a consequence, using the proposed criteria would exclude patients with ANA-negative LN from (future) clinical studies. In addition, it is known from previous studies that patients with early SLE can have negative ANA as well.9 Therefore, we think that using ANA positivity as an entry criterium should be reconsidered, especially in the case of biopsy-proven LN. Second, in the proposed ACR–EULAR criteria, the definition of the NP domain is limited. It does not specify a time correlation between NP symptoms and the (suspected) diagnosis of SLE. In our cohort, 52 of the 360 patients (one patient had two syndromes) had a positive NP domain, after NP symptoms attributed to other causes were excluded. We recalculated sensitivity and specificity after excluding patients with NP symptoms >1 year prior to (suspected) SLE, as previously proposed by Bortoluzzi et al.10 This led to the exclusion of 14 patients (26.9%). Although this did not influence the sensitivity or specificity in our cohort, we feel that accurate attribution of NP symptoms is important. This is also demonstrated in our cohort in a different way, as nine patients had definitive NPSLE (and the clinical diagnosis of SLE), but did not meet the proposed criteria and did not fulfil the NP domain. The following NPSLE syndromes were present in these patients: chorea (n=1), myelopathy (n=2), cerebral vasculitis (n=2), mood disorder (n=1), cognitive dysfunction (n=1) and cerebrovascular disease (n=3). Adding these syndromes to the NP domain would lead to a maximum increase of sensitivity to 90%, without changing specificity. If the NP domain is adjusted and ANA is not used as entry criterium in case of LN, sensitivity of the proposed ACR–EULAR criteria maximally increases to 93%. In conclusion, we demonstrate that in our cohort of referred patients with (suspected) SLE with NP symptoms, sensitivity of the proposed ACR–EULAR criteria for SLE is high, but specificity remains suboptimal. Including patients with ANA-negative LN and specifying the NP domain might further improve the sensitivity of the proposed criteria. Acknowledgments We would like to thank all our contributors. Footnotes MG and HJLB contributed equally. Twitter: Sensitivity of the proposed ACR-EULAR criteria is high, but specificity remains suboptimal In patients with SLE presenting with neuropsychiatric symptoms. Contributors: All authors contributed to study conception, drafting and approving the manuscript. Funding: The authors have not declared a specific grant for this research from any funding agency in the public, commercial or not-for-profit sector. Competing interests: None declared. Patient consent for publication: Not required. Ethics approval: The Ethics Committee of Leiden University Medical Center. Provenance and peer review: Not commissioned; externally peer reviewed. Data sharing statement: No additional data are available. References 1.Fries JF, Siegel RC. Testing the 'preliminary criteria for classification of SLE'. Ann Rheum Dis 1973;32:171–7. 10.1136/ard.32.2.171 [DOI] [PMC free article] [PubMed] [Google Scholar] 2.Tan EM, Cohen AS, Fries JF, et al. The 1982 revised criteria for the classification of systemic lupus erythematosus. Arthritis Rheum 1982;25:1271–7. 10.1002/art.1780251101 [DOI] [PubMed] [Google Scholar] 3.Hochberg MC. Updating the American College of Rheumatology revised criteria for the classification of systemic lupus erythematosus. Arthritis Rheum 1997;40 10.1002/art.1780400928 [DOI] [PubMed] [Google Scholar] 4.Petri M, Orbai A-M, Alarcón GS, et al. Derivation and validation of the Systemic Lupus International Collaborating Clinics classification criteria for systemic lupus erythematosus. Arthritis Rheum 2012;64:2677–86. 10.1002/art.34473 [DOI] [PMC free article] [PubMed] [Google Scholar] 5.Tedeschi SK, Johnson SR, Boumpas D, et al. Developing and refining new candidate criteria for systemic lupus erythematosus classification: an international collaboration. Arthritis Care Res 2018;70:571–81. 10.1002/acr.23317 [DOI] [PMC free article] [PubMed] [Google Scholar] 6.Aringer M, Costenbader KH, Brinks R. Validation of new systemic lupus erythematosus classification criteria. Annals of the rheumatic diseases 2018;77(Suppl 2):60. [Google Scholar] 7.Johnson SR, Khanna D, Daikh D, et al. Use of consensus methodology to determine candidate items for systemic lupus erythematosus classification criteria. J Rheumatol 2018. 10.3899/jrheum.180478 [DOI] [PubMed] [Google Scholar] 8.Zirkzee EJM, Steup-Beekman GM, van der Mast RC, et al. Prospective study of clinical phenotypes in neuropsychiatric systemic lupus erythematosus; multidisciplinary approach to diagnosis and therapy. J Rheumatol 2012;39:2118–26. 10.3899/jrheum.120545 [DOI] [PubMed] [Google Scholar] 9.Mosca M, Costenbader KH, Johnson SR, et al. Brief report: how do patients with newly diagnosed systemic lupus erythematosus present? A multicenter cohort of early systemic lupus erythematosus to inform the development of new classification criteria. Arthritis Rheumatol 2019;71 10.1002/art.40674 [DOI] [PubMed] [Google Scholar] 10.Bortoluzzi A, Scirè CA, Bombardieri S, et al. Development and validation of a new algorithm for attribution of neuropsychiatric events in systemic lupus erythematosus. Rheumatology 2015;54:891–8. 10.1093/rheumatology/keu384 [DOI] [PubMed] [Google Scholar] Articles from RMD Open are provided here courtesy of BMJ Publishing Group ACTIONS View on publisher site PDF (247.2 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Introduction Methods Results Discussion Acknowledgments Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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Your students become detectives as they solve 12 math problems to uncover WHO took the class pet, WHERE they took it from, WHICH getaway method helped them flee, and WHAT silly weapon they used. Each correct answer unlocks a clue that helps them eliminate suspects, locations, and methods until only t 6 th - 8 th Arithmetic, Basic Operations, Numbers CCSS 6.RP.A.3c $5.00 Original Price $5.00 Add to cart Wish List Part Whole Percent Word Problem Worksheets 6th Grade Created by Simone's Math Resources Part Whole Percent Word Problem Worksheets 6th Grade Included:6 worksheets with answer keys. More PercentsPercent Problems Task Cards Percent Word Problems Scavenger Hunt Bingo Common Core State Standards6.RP.3c Use ratio and rate reasoning to solve real-world and mathematical problems, Find a percent of a quantity as a rate per 100 (e.g., 30% of a 6 th Math, Other (Math) Also included in:Part, Whole, or Percent Word Problems Bundle $3.00 Original Price $3.00 Rated 4.85 out of 5, based on 112 reviews 4.9(112) Add to cart Wish List Percent, Part, and Whole Word Problem Pixel Art \| 4 Levels of Word Problems Created by Misspelled Word Problems Engage Your Students with Math Pixel Art!Bring math to life with this fun, interactive Pixel Art Activity! Perfect for 6th-10th graders, this digital activity allows students to answer math questions to reveal hidden pixel art images. With four levels of pixel art to complete, students must solve problems correctly for each pixel to populate, creating a fun, visual reward for their hard work. What’s Included:Four Unique Pixel Art Levels: Each level presents a new challenge, keeping students moti 6 th - 10 th Math, Math Test Prep, Other (Math) Also included in:Middle School Math Review Centers PIXEL ART Google Activities BUNDLE $2.00 Original Price $2.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Percent Part and Whole Thanksgiving Word Problems Mystery Reveal Created by Loving Math 143 This Digital Self-Checking Math Activity allows middle school students to practice working with percentage word problems including finding the part, the whole, and the percent in a fun and interactive way. Students will apply their understanding of percentages and math word problems for 12 problems! Each correct answer will reveal part of a fun Mystery Picture for Thanksgiving! This is a perfect activity to check for understanding of ratios and proportions during distance learning and stude 6 th - 8 th Math, Other (Math) CCSS 6.RP.A.3c , 7.RP.A.3 $3.00 Original Price $3.00 Rated 5 out of 5, based on 2 reviews 5.0(2) Add to cart Wish List Percent, Part and Whole Word Problems Worksheets Created by Box Number 101 Percent, Part and Whole Word Problems Worksheets Help students build confidence in solving real-life math problems with these Percent, Part, and Whole Word Problems Worksheets. This resource focuses on teaching students how to identify and solve word problems involving percentages, parts, and wholes. Whether students are finding a missing percent, part, or whole, these worksheets are designed to strengthen critical thinking and problem-solving skills in an engaging and structured way. What’s I 5 th - 7 th Basic Operations, Decimals, Fractions Also included in:Finding Percent, Part and Whole Practice Worksheets and Anchor Chart $2.50 Original Price $2.50 Add to cart Wish List Percent of a Number \| Percent Part Whole \| Word Problems and Worksheets Bundle Created by ElementaryStudies Percent Word Problems and Worksheets bundle pack. Includes the following 2 resources: 1) Percent of a Number Word Problems \| Part, Whole, and PercentPercent Word Problems Worksheets. Students will solve the problems to find the Percent, Part, or Whole. There are 30 story problems included in this resource. The problems are included in two different formats: Worksheets format with 6 problems per sheetAssignment format with 10 problems per sheet2) Percent of a Number Worksheets - Part Whole Perc 6 th Applied Math, Arithmetic, Basic Operations Bundle (2 products) $5.00 Original Price $5.00 $4.00 Price $4.00 Add to cart Wish List 6.RP.A.3c Sage and Scribe "Part, Whole, Percent Word Problems" Created by Kelly Bryant Mastering Whole, Part, &Percent Word Problems: Sage & Scribe Activity! Tired of students struggling to set up and solve percent word problems? This "no-frills" Sage and Scribe activity is a powerful, cooperative learning tool designed to help students master finding the whole, part, or percent in real-world contexts! As a math teacher with 24 years of experience, I know the value of structured collaboration, and this activity ensures deep processing and active participation from every stud 6 th Other (Math) $0.50 Original Price $0.50 Add to cart Wish List Percent of a Number Word Problems \| Part, Whole, and Percent Created by ElementaryStudies Engage with the "Percent of a Number Word Problems - Part Whole Percent" resource! This comprehensive package is designed to improve students' skills in solving real-world scenarios involving percentages, parts, and wholes. Featuring a total of 30 thought-provoking story problems, this resource aims to challenge students and promote a better understanding of percentages. Choose from two convenient formats that best suit your instructional preferences: Worksheets Format:Each sheet is packed 6 th Algebra, Fractions, Other (Math) CCSS 6.RP.A.3c Also included in:Percent of a Number \| Percent Part Whole \| Word Problems and Worksheets Bundle $3.00 Original Price $3.00 Add to cart Wish List Missing Whole Given Percent and Part Word Problems Matching Center Created by To the Square Inch- Kate Bing Coners Finding the Missing Whole Matching Center \| Part&Percent Word ProblemsLooking for a fun and interactive way to help your students master finding the missing whole when given the part and percent? This Finding the Missing Whole Matching Center is the perfect hands-on activity for 6th and 7th-grade students to practice and reinforce their understanding of percent problems in real-world scenarios! What’s Included?✅ 20 matching cards featuring real-world percent problems where students solve 5 th - 7 th Fractions, Math, Numbers CCSS 6.RP.A.3c Also included in:Math Matching Game Bundle $3.00 Original Price $3.00 Add to cart Wish List Percent Word Problems: Given Part and Whole&Percent of a Whole Created by The Mythical People from Math Problems Word problems involved percent with 2 types of problems Type 1: Given a part and whole, find percent Type 2: Given a percent of whole, find the part Each problems are provided with break down of smaller steps and a table that summarize the information. 6 th - 10 th, Adult Education Math CCSS 6.RP.A.3c , 7.RP.A.3 , MP1 +1 FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Find the Percent (Given Part and Whole): Math Mystery Picture Created by Mata Math and Science About this resource:In this activity, students will read a real-world situation to create a fraction, then convert it to a percent. There are 15 total word problems What's included?Printable Math Mystery PictureGoogle Sheets™ versionInteractive Notebook Cover SheetShow work templates (small and large boxes)Answer keyStandards alignment:TEKS - 6.5BCCSS - 6.RP.A.3cTry It FREE: Download my FREE Add Integers: Math Mystery PictureWorks Great As:- Distance Learning - Sub Plans - Centers or Stations - 5 th - 9 th Applied Math, Math, Other (Math) CCSS 6.RP.A.3c Also included in:Math Mystery Picture Mega Bundle \| Middle School Review Activities $3.50 Original Price $3.50 Rated 4.96 out of 5, based on 46 reviews 5.0(46) Add to cart Wish List Percentages Word Problems Part Whole Percent of a Number Worksheets Activity Created by Games 4 Gains Make solving word problems with percentages fun with this Percents Word Problems Color-by-Number Worksheet and Task Cards Activity! This part, whole, and percent of a number activity includes 24 task cards that require students to solve real-world percent problems and a color-by-number worksheet. The cards contain both one-step and multi-step problems and increase gradually in difficulty. The coloring aspect of this percents activity motivates even the most reluctant learners to work through the 6 th - 7 th Math Test Prep, Other (Math) CCSS 6.RP.A.3 , 6.RP.A.3c $3.75 Original Price $3.75 Rated 4.88 out of 5, based on 60 reviews 4.9(60) Add to cart Wish List Percent Proportion Word Problems Coloring Activity Created by Live Long and Learn This product was designed as a self-checking activity for students to practice finding the missing percent, missing parts and missing wholes in real life situations. I prefer to use this as an insert into my interactive notebooks. I have created a page that will allow students to still see the problems and the coloring page at the same time. An answer key is included. As well as a final version of the colored hearts. 6 th - 8 th Math, Other (Math) CCSS 7.EE.B.3 $3.00 Original Price $3.00 Rated 4.96 out of 5, based on 26 reviews 5.0(26) Add to cart Wish List Percents: Intro, Finding Part/Whole Plus Solving Percent Word Problem Note Pages Created by 4 the Love of Math Tired of using traditional methods to teach this topic? With engaging doodles and step-by-step explanations, these introduction to finding percent note pages are the perfect tool for helping your students' master percent. The first page defines percent and introduces changing ratios to percent and changing percents to decimals, while the second page delves into finding parts and wholes using information given and solving percent word problems. With two versions of each page included (one with e 6 th - 8 th Algebra, Math, Numbers CCSS 6.RP.A.3c Also included in:Ratios and Proportional Relationships No Prep Note Bundle $2.00 Original Price $2.00 Rated 4.88 out of 5, based on 25 reviews 4.9(25) Add to cart Wish List Percent Proportion (Part, Whole, Percent) Word Problems Winter Digital Activity Created by Playful Pythagoras Engage your students this winter as they build a snowman in this holiday self-checking digital activity. Students will practice finding the part, whole, and percent while solving percent proportion word problems and receive immediate feedback. How does it work?If correct, a snowball appears, and the answer box will turn green. If they are incorrect, their answer turns red to let them know they should go back and revise their answer. But wait, there’s more! Not only do students get immediate fee 6 th - 7 th Math, Other (Math) CCSS 6.RP.A.3 , 6.RP.A.3c $3.50 Original Price $3.50 Rated 4.75 out of 5, based on 4 reviews 4.8(4) Add to cart Wish List Find Percent, Part, &Whole Proportion Stations Activity \| Math Centers Created by Make Sense of Math These percent proportion stations are perfect for your middle school math students to set up and solve proportions to find the percent, part, and whole. Students will do a variety of activities that include: setting up, solving, and creating percent, part, whole problems using proportions. The following math stations are included:Percent Word Problems: Given word problems to find the percent, part or whole, students will solve the problem by setting up and solving proportions.Percent Practice: 6 th - 7 th Math, Numbers, Other (Math) CCSS 6.RP.A.3c Also included in:Middle School Math Stations Bundle \| Math Centers $4.50 Original Price $4.50 Rated 4.88 out of 5, based on 8 reviews 4.9(8) Add to cart Wish List 6.RP.3 Task Cards ✦ Unit Rate &Percent Word Problems ✦ Google Classroom™ Created by Math Lessons by NUMBEROCK This year, go green and let Google's powerful engine do the grading for you while breathing new life into your teaching of common core standards 6.RP.3b & 6.RP.3c (Unit Rate & Percents). We cordially invite you to take a look at the PREVIEW above to get a closer look at this product by clicking on the green button above this description. ============================6.RP.3b & 6.RP.3c LEARNING GOALSI can apply the concept of unit rate to solve real-world problems involving unit pricing. I can a 6 th Math, Math Test Prep, Other (Math) CCSS 6.RP.A.3a Also included in:6th Grade Math Review Digital Task Cards \| Word Problems for Google Classroom™ $3.00 Original Price $3.00 Rated 4.92 out of 5, based on 12 reviews 4.9(12) Add to cart Wish List Percent of a Number Word Problems Digital and Printable Scavenger Hunt Activity Created by 2ndary Math This Scavenger Hunt is a great way to engage your 7th Grade students through this self-checking activity on calculating the percent of a number through word problems. Students can solve these percent of a number in real-world applications algebraically or using the percent proportion. This activity is great for Math centers, test prep, or early finishers. Skills Covered: Calculating the percent- given the part and the wholeCalculating the part- given the percent and the wholeCalculating the who 7 th Math CCSS 7.RP.A.3 Also included in:Percent of a Number and Percent Application Activities Bundle $4.00 Original Price $4.00 Rated 5 out of 5, based on 2 reviews 5.0(2) Add to cart Wish List Percent Word Problems, Part, Whole, Percent of a Number Activity Truth or Dare Created by Cognitive Cardio Math Solving percent problems and finding part, whole, and percent of a number is so engaging with this digital math game. Digital Truth or Dare is a self-checking, easy-prep way to review concepts - as a whole class, in small groups, or independently. The game begins with a Truth grid and Dare grid that leads students to select questions and self-check answers as they navigate through 36 question cards and 36 answer cards in Google Slides. As long as you have access to Slides, you can use this mat 6 th - 7 th Math, Math Test Prep, Other (Math) CCSS 6.RP.A.3c , 7.RP.A.3 Also included in:Fun Friday Math, Middle School Math Resources 5th, 6th Grade Truth or Dare Games $3.00 Original Price $3.00 Rated 4.78 out of 5, based on 29 reviews 4.8(29) Add to cart Wish List Finding Percent, Part and Whole Practice Worksheets and Anchor Chart Created by Box Number 101 Finding Percent, Part and Whole Practice Worksheets and Anchor ChartHelp your students build a strong understanding of percent, part, and whole with this complete math resource designed for clarity, practice, and confidence. The combination of practice worksheets and a visual anchor chart supports different learning styles, making it easier for learners to connect concepts and apply them in problem-solving. From converting between fractions, decimals, and percents to solving real-world per 4 th - 7 th Basic Operations, Decimals, Fractions Bundle (5 products) $12.10 Original Price $12.10 $8.47 Price $8.47 Add to cart Wish List Percent Proportion: Identifying the Part&Whole Created by Goodbye Monotonous Middle School Math This activity is designed to target student difficulty with properly setting up a percent proportion when solving word problems. It is designed to model proper metacognition for students as they think through given information in a problem. Given 6 real world problems, students describe in words what the part (is) &whole (of) represent in each situation. This information is then used to help them set up a percent proportion that can be used to solve for the unknown value. 6 th - 7 th Basic Operations, Math, Other (Math) $2.75 Original Price $2.75 Rated 5 out of 5, based on 23 reviews 5.0(23) Add to cart Wish List Percent of a Number Worksheet for 6th Grade Math Practice and Review - Percents Created by Teaching With The Dollhouse Collector Challenge your 6th grade math students to solve equations and real world problems involving percent! They will practice calculations of percent, part, and whole and solve a multi-part word problem involving a real world situation. What's included?1 worksheet with 12 problemsanswer key digital Easel activityUse this for: math centersindependent practicehomeworkfast finishersClick on the green star under my name to follow me and be the first to know when new items are posted! Your feedback is al 5 th - 6 th Math, Numbers CCSS 6.RP.A.3c Also included in:Ratios, Rates, and Percent Task Cards and Word Wall Cards for 6th Grade Math FREE Rated 4.83 out of 5, based on 12 reviews 4.8(12) View 2 Files Wish List Multi-Step Percent Problems Worksheets \| Percent Word Problems Created by Make Sense of Math Calculating multi-step percent problems such as: simple interest, tax, markups, markdowns, gratuities, and commissions is an important skill for your 7th grade and pre algebra students. These no prep worksheets are perfect to reinforce this concept. These work great at homework or in-class work. THIS IS THE PAPER VERSION - CLICK HERE FOR THE GOOGLE VERSIONEach worksheet includes 4 sections: Review, Fluency, Application, and Analysis. Check out the preview for a clear understanding of how these 6 th - 8 th Algebra, Math, Numbers CCSS 7.RP.A.3 Also included in:7th Grade Math Curriculum : Guided Notes, Activities, Tests & More - Common Core $3.00 Original Price $3.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Percent of a Number Word Problems \| Part, Whole, Percent \| Benchmark Percentages Created by Level Their Learning In this engaging Real Life Food Truck activity, students will solve percent of a number word problems using benchmark percentages (10%, 20%, 25%, 75%, 100%). Students will solve for the part, the whole, and the percent in these percent of a number word problems. 12 different percent of a number word problems for them to solve! Each of these percent of a number word problems also has a bonus question! This gives your learner 24 learning opportunities for practicing percent of a number word proble 6 th - 8 th Math CCSS 6.RP.A.3 , 6.RP.A.3c Also included in:Percent of a Number Word Problems \| Part, Whole, Percent \| Benchmark Percentages $2.50 Original Price $2.50 Rated 5 out of 5, based on 2 reviews 5.0(2) Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 690+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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17592	https://mathoverflow.net/questions/198085/stirling-numbers-of-the-second-kind-with-maximum-part-size	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 Stirling numbers of the second kind with maximum part size Ask Question Asked Modified 3 years, 11 months ago Viewed 1k times 5 $\begingroup$ The stirling number of the second kind $S(n,k)$ counts the number of partitions of the set $[n]$ into $k$ non-empty parts. I found a definition for the numbers called the $r$-associated stirling numbers of the second kind in wikipedia. These count the number of partitions of $[n]$ into $k$ non-empty parts such that all of the parts have at least $r$ elements. They seem to be "not too hard to calculate" via the recursion: $S_r(n+1,k)=kS_r(n,k)+\binom{n}{r-1}S_r(n-r+1,k-1)$. I am interested in a very similar construction, for which I have been unable to find any references. I would like to count $F_r(n,k)$ defined as the number of partitions of $[n]$ into $k$ non-empty parts so that each of them has size $r$ or less. Most of all I am interested in $\sum\limits_{k=0}^nF_r(n,k)$, basically the number of partitions of $[n]$ into parts of size $r$ or less. Thank you very much in advance Regards. I am aware that the problem of finding compositions of $n$ into numbers less than or equal to $k$ has been studied. But I do not think one of them can be used to compute the other. reference-request co.combinatorics enumerative-combinatorics Share Improve this question edited Feb 21, 2015 at 4:57 GorkaGorka asked Feb 21, 2015 at 3:38 GorkaGorka 1,90522 gold badges1919 silver badges3333 bronze badges $\endgroup$ 1 $\begingroup$ Did you manage to find an answer? I asked a similar question on maths stack exchange here. math.stackexchange.com/questions/2302204/… There is a recurrence there but I haven't found a way to solve it. $$F_r(n+1,k)=kF_r(n,k)+F_r(n,k1){n\choose r}F_r(nm,k1) \qquad n,k¥1$$ $\endgroup$ videlity – videlity 2017-06-20 02:04:28 +00:00 Commented Jun 20, 2017 at 2:04 Add a comment \| 2 Answers 2 Reset to default 5 $\begingroup$ This is a routine application of the exponential formula. If $S$ is any subset of the positive integers and $f_S(n)$ is the number of partitions of $[n]$ into parts all belonging to $S$, then $$ \sum_{n\geq 0} f_S(n) \frac{x^n}{n!}= \exp \sum_{i\in S}\frac{x^i}{i!}. $$ Thus in your case we get $\exp \sum_{i=1}^r \frac{x^i}{i!}$. We also have the recurrence $$ f_S(n+1) =\sum_{i\in S}{n\choose i-1}f_S(n-i+1). $$ Share Improve this answer answered Feb 22, 2015 at 2:14 Richard StanleyRichard Stanley 53.5k1515 gold badges163163 silver badges298298 bronze badges $\endgroup$ 1 $\begingroup$ Thank you very much. Do you know if there is a simpler way to calculate $\sum\limits_{k=0}^nF_r(n,k)$? Other than writing up the recurrence for the individual $F_r(n,k)$ and adding? $\endgroup$ Gorka – Gorka 2015-02-22 17:08:04 +00:00 Commented Feb 22, 2015 at 17:08 Add a comment \| 2 $\begingroup$ This is the so-called restricted Stirling number of the second kind. Reference: Incomplete poly-Bernoulli numbers associated with incomplete Stirling numbers Share Improve this answer answered Oct 16, 2021 at 1:39 Yijun YuanYijun Yuan 55922 silver badges1212 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions reference-request co.combinatorics enumerative-combinatorics See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Related 5 Acyclic orientations of complete graphs in terms of Stirling numbers? 5 Alternating sums of alternate Stirling numbers 5 Signless Stirling Numbers of 1st Kind and Probabilistic Descent Who first noticed that Stirling numbers of the second kind count partitions? congruence for Stirling numbers of the first kind 3 Simplifying a Taylor polynomial that involves Stirling numbers of the second kind 5 Bijective proof for an identity concerning Stirling numbers of second kind Question feed
17593	https://opencw.aprende.org/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/video-lectures/lecture-14/	Lecture 14: Accelerated Charges, Poynting Vector, and Power \| Video Lectures \| Physics III: Vibrations and Waves \| Physics \| MIT OpenCourseWare Subscribe to the OCW Newsletter Help\|Contact Us Courses Find courses by: Topic MIT Course Number Department Collections Audio/Video Lectures Online Textbooks New Courses Most Visited Courses OCW Scholar Courses This Course at MIT Supplemental Resources Cross-Disciplinary Topic Lists Energy Entrepreneurship Environment Introductory Programming Life Sciences Transportation Translated Courses 繁體字 / Traditional Chinese 简体字 / Simplified Chinese Español / Spanish Português / Portuguese ภาษาเขียน / Thai فارسی / Persian Türkçe / Turkish (비디오)한국 / Korean More... About About MIT OpenCourseWare Site Stats OCW Stories Media Coverage Newsletter Press Releases OCW's Next Decade Donate Make a Donation Why Donate? Become a Course Champion Our Supporters Other Ways to Contribute Shop OCW Become a Corporate Sponsor Featured Sites Highlights for High School OCW Educator MITx Courses on edX Teaching Excellence at MIT Open Education Consortium Advanced Search Home » Courses » Physics » Physics III: Vibrations and Waves » Video Lectures » Lecture 14: Accelerated Charges, Poynting Vector, and Power Lecture 14: Accelerated Charges, Poynting Vector, and Power Course Home Syllabus Calendar Lecture Notes Assignments Exams Video Lectures Download Course Materials {'English - US': '/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/video-lectures/lecture-14/ocw-8.03-lec-mit-28oct2004-0014-220k.srt'} About JW Player 6.8.4616... 00:00 00:00 00:00 Off English - US About this Video Playlist Related Resources Transcript Download this Video Topics covered: Accelerated Charges - Poynting Vector - Power - Rayleigh Scattering - Polarization Instructor/speaker: Prof. Walter Lewin Course Introduction Lecture 1: Periodic Oscilla... Lecture 2: Beats, Damped Fr... Lecture 3:Forced Oscillatio... Lecture 4: Forced Oscillati... Lecture 5: Coupled Oscillators Lecture 6: Driven Coupled O... Lecture 7: Coupled Oscillat... Lecture 8: Traveling Waves,... Lecture 9: Musical Instrume... Lecture 10: Exam 1 Review Lecture 11: Fourier Analysi... Lecture 12: Dispersion, Pha... Lecture 13: Electromagnetic... Now Playing Lecture 14: Accelerated Cha... Lecture 15: Doppler Effect,... Lecture 16: Interactions of... Lecture 17: Wave Guides, Re... Lecture 18: Boundary Condit... Lecture 19: Exam 2 Review Lecture 20: Huygens' Princi... Lecture 21: Diffraction, Gr... Lecture 22: Rainbows, Haloe... Lecture 23: Farewell Lecture Related Resources Syllabus Assignments Download this transcript - PDF (English - US) WALTER LEWIN: There is energy in an electric field, and there's energy in a magnetic field. You remember that from 8.02. And the energy density-- mind the units, in terms of joules per cubic meter-- for the electric field, we write for that a U equals 1/2 epsilon 0 E squared. There's no such thing as a free lunch. You have to do work to create an electric field. You have to assemble charges and bring them together. That means work. That creates an electric field. And the same is true for a magnetic field. And you have a solenoid, and you create the magnetic field inside the solenoid, that costs energy. And the energy density-- mind the word density, it is per cubic meter-- for a magnetic field is B squared divided by 2 mu 0. Now traveling EM wave, traveling electromagnetic wave, so this is now a traveling wave. We know that the magnitude of B is the magnitude of E, at any moment in time, divided by c. So I can write this as E squared divided by 2 mu 0 times c squared. But c squared is 1 over epsilon 0, mu 0. So this is also 1/2 epsilon 0 E squared. And what you see now is so wonderful, so beautifully symmetric in electromagnetic waves. One is the same as the other. One cannot exist without the other. And look at the energy density in the electric field of a traveling wave is exactly the same as the energy density in the magnetic field. And so the total energy density is the sum of the two, is epsilon 0 times E squared. But you can also write for that, epsilon 0 times E times Bc, if you prefer that. This, of course, is only true in vacuum, when the speed of propagation is c. Now, there is a traveling wave, and this traveling wave moves. And so it carries energy with it. And now the question is, how much energy flows through an area, which is, say, one square meter area, perpendicular to the direction of propagation? So suppose I have a box here. And this side is 1 square meter. And I want to know how much radiation comes out of there in one second. So the radiation is flowing in this direction with speed c. So in one second-- this box is quite a large box, 3 times 10 to the 8 meters. And all that energy, which is in their, will flow through this one square meter, in the time of one second. And so the dimensions that we're talking about now is joules per second per square meter, which is also watts per square meter. And that now is, of course, the total energy density times c, the speed of light. For which you can write epsilon 0 times E times B times c squared, if you like that. But you also write for that EB divided by mu 0, because c squared is 1 over epsilon 0 mu. And this should remind you of something that is in your far distant past, which is what we earlier have called, in 8.02, the Poynting vector. And the Poynting vector S was E cross B divided by mu 0. And the units were watts per square meter, which is exactly what this is. And the reason why the cross disappears here, is that with electromagnetic traveling waves, E is always perpendicular to B. So that takes care of the cross. Now both E and B are time variable. And so the Poynting vector will, obviously, also be time variable. E is going to be proportional-- you can write down E as some E0 times cosine omega t, and B is some B0 times cosine omega t. So in the Poynting vector, you get the cosine square of omega t. But since we never interested in the Poynting vector on a time scale smaller than the period of oscillations-- we want to know the average over many oscillations-- what matters there is the average value of cosine squared omega t. One from the E and the other one from the B. And that is one half. And so we can conclude then, that the average value of the Poynting vector, time averaged, is 1/2 that value, 1/2 EB divided by mu 0. For which you can also write, 1/2-- if you want to kill completely. Oh, by the way, this is 1/2 E0 B0. It's important that you have to E0, because the cosine square average is 1/2. So here are the amplitudes. So you can also write for that 1/2 E0 squared. And so you write down for BE divided by c. And then you get, downstairs, mu 0 divided by c. And the reason why I write it in this form is that it tells you that, if you know what the strength of the E field is, that alone tells you then what the Poynting vector is. Because B is coupled to E, through Maxwell's equations. That is where B is E over C. And so all that matters then, if you want to calculate what the Poynting vector is, is E0. Of course B 0 alone would also be fine. So let's take an example. Suppose we have an electromagnetic wave, whereby E0 were 100 volts per meter, then I can calculate, now-- and it isn't traveling electromagnetic wave-- what the average value of the Poynting vector is. So that would become 1/2 times 100 squared divided by mu 0 times c. And if you do your homework on that, you will find that it is 13 watts per square meter. Now if you exposed yourself to 13 watts per square meter, visible light and infrared light, you take all your clothes off, and you expose yourself to that, your body will absorb that. It will not go through your body. X-rays may go through your body, gamma rays, certainly. But infrared radiation and visible light get absorbed by your body. And so the question now is, will that harm you? And then the answer is, no, you would hardly notice 13 watts per square meter. Your body, itself, radiates about 100 joules per second, because of the body heat that you have. And so the 13 watts per square meter that you absorb-- let's say your cross sectional area is about one square meter, just to round it off. So that means 13 joules per second would be absorbed by your body. And that will not affect you in any way. But let's now take a situation that we have E0 equals 10 times higher, which is 1,000 volts per meter. Now the Poynting vector goes up by the square of E. So now you're going to get that the mean for Poynting vector is 1.3 kilowatts per square meter. And that will fry you. If you walk naked in a field of infrared or optical, whereby the absorption, 1.3 kilowatts per square meter, that is very dangerous. You'll get skin cancer and worse. Now, why do I mention that? And why do I focus on that 1.3 kilowatts per square meter? Because that is exactly what the sun does. Here is the sun. And the sun is a rather powerful light bulb. About 3.9 times 10 to the 26 watts. And this is where you are on Earth. And the distance to Earth is 150 million kilometers. So I can calculate now how many joules per second go through one square meter here. And this one square meter is held perpendicular to the direction to the sun. So that value for S is of course the radiation that leaves here. That is my 3.9 times 10 to the 26. And now I have to divide it by the entire surface area of this sphere, which goes in all directions, that has this radius. And that surface area is 4 pi times that radius square. So that is 150 times 10 to the 9th-- because I have to go to MKS units-- squared. And that is now the number of watts per square meter that we receive at Earth from the sun. And that number is a very famous number. That is 1.4 kilowatts per square meter. And that's why I picked the 1.3 to show you that, if you walk around on the beach, and you don't take care of yourself, if you do that too long, that is very dangerous, because your body absorbs that. This number is called the solar constant. It has major implications, of course, for people who want to harvest solar energy. The maximum that you can ever harvest is, for every square meter that you dedicate to solar energy, you only get 1,400 joules per second. You can never get more. The electric power capacity of the United States is 700,000 megawatts, for which you need 700 power plants. A full size power plant is about 1,000 megawatt power plant. You need 700 of these power plants to have the capacity for the United States. We are energy hungry. We consume more than 1/4 of all the energy in the world. So if you want to replace this by solar energy, that's a major problem. Because you can only get 1.4 kilojoules per second for every square meter. You can calculate how many hundreds of square miles in the desert you would have to commit, with solar cells, which are extremely expensive, in order to get electricity. And the efficiency, of course, is never 100%. And also, when the sun is low in the horizon, then you don't have this one square meter perpendicular to the direction of the sun. So all of that has to be taken into account. Solar energy is not very important in our lives, unfortunately. And this is the number that is the ultimate limit. Now comes the question, is there such a thing as a electric field of 1,000 volts per meter in the solar radiation? Could we actually measure the electric field just from the radiation of the sun? And the answer is, no. And the reason is that the radiation is not really in the form or our idealized plane waves. But more important than anything else is that there is no such thing as just one wave from the sun that has the amplitude of 1,000 volts per meter. In fact, the radiation reaches us in small packages, broken up in pieces, so to speak. However, since the energy flow is 1.4 kilowatts per square meter, it is perfectly OK with me that you refer to this number as the Poynting vector. I have no problem with that. But it is a little bit naive to associate with that an electric field that can be measured, that has a amplitude then of 1,000 volts per meter. And so this now is the right time to take a close look at how electromagnetic waves are produced. In a nut shell, it comes down to this. You can create electromagnetic waves if you accelerate charges. Charges that are stationary or moving at constant velocity, are surrounded by a radial field, pointing away, or pointing inwards, depending upon whether the charge is positive or negative. And there's no kink anywhere in these fields. So whether it has a constant velocity, or whether it stands still, they are radially electric field lines. The moment, however, that you accelerate it, as you will see today, you introduce a kink in those field lines. And that kink is responsible for that electromagnetic radiation. It manifests itself as electromagnetic radiation. I will follow a classic derivation that is verbatim given that way in Bekefi and Barrett. I, therefore, advise you strongly, for the next 13 minutes, not to take any notes, but try to follow my arguments. That will help you way more than that you try to also take notes. Because it's really verbatim from Bekefi and Barrett. It is a classical derivation. With many simplifying assumptions, but it gives a very nice result, which has great practical applications. Suppose I have here a charge, q, which is located at position, a. So we have a charge, q. It is at a, at O, and it is at rest. And I'm going to accelerate that in this direction. So I'm going to accelerate it with an acceleration, which is in this direction. I will not put the vector in there now. I will do that later. Otherwise the figure becomes too complicated. And I do that for delta t seconds. Only very brief. And then it ends up at location O prime, which is here. So now it has a velocity, in this direction. And that velocity, u, in this direction, is now a times delta t, that's 8.01. And so it's cruising now with constant velocity. And we just let it cruise all the way. It's now again a charge with constant velocity. And I look at it, where it is t seconds later. And so at time, t, we'll find it in O double prime. And it's still cruising. And we just let it cruise. We're not going to interfere with it anymore. And here is then O double prime. The whole exercise, from O to O double prime, took so many seconds. That means there is a sphere around point O. And that sphere has a radius, which is c times t plus delta t. Outside that sphere, the world has no knowledge that this charge was accelerated. Because that message has to travel with the speed of light. So I'm going to draw a circle-- but in reality it is a sphere, in all directions-- which has a radius, c, t plus delta t. Delta t, by the way, is way much smaller than t. And outside that sphere, there is no knowledge, the world doesn't have any clue about the fact that this object was being accelerated. So I will mark again, to make sure that you can make a connection. So this one has its center at O and the radius is c times t plus delta t. And if you ask me, what is the electric field, right there, that is radially pointing outwards, if q is positive? So it would be radially pointing outward. This world does not know yet that there was any acceleration. And if you ask me, what is the electric field here? The field line is like so. And so the electric field, it is a positive charge, is pointing radially outwards and has no knowledge that there was any acceleration. And so I call this is my position vector, r. But you can take vector r in any direction that you want. Now let's look at the world that does know that there was a change. When the object was at O prime, the acceleration stopped, and so it started cruising with a constant velocity. And so from that moment on, the field lines are, again, nicely, radially outwards. And so by the time that it reaches point O double prime, I can point, I can draw these field lines, radially outwards. And I think of this original field line as a stick that was connected to the charge in that direction. I could have picked another direction. And so that field line is now here, outwards. And the world that knows about this is the world which has a sphere around O prime, with a radius c times t. And so I'm going to draw another circle, which in reality is a sphere, about point O prime. And this sphere here has now a radius ct. And so this, to remind you what it is, it has the origin at O prime. And the radius is ct. And so everything inside that sphere recognizes that the field is radially outward, from all point O double prime. And everything outside here, still thinks that the electric field is like this. But this field line was this one, that was the stick that I attached to it. And so there must be a connection between here and there. And that connection is only in that very thin shell, which has a thickness c delta t. Let me first make a drawing of a triangle. That's going to be important. Which is this triangle. This length here has a length, u perpendicular times t. You can easily see why that is the case. Because this distance here is ut. Now, you may say, well, it wasn't going with velocity u, here. I grant you that. But keep in mind, in whatever follows, that u is way, way smaller than c. Delta t is way, way smaller than the t. And so r, which is c times t, is way, way larger than ut. So when you see a little distortion in this picture, I cannot of course, make delta t way, way smaller than t. Then you wouldn't see any decent picture. Let me first go to u perpendicular t. So there is here a velocity vector, u. And the component perpendicular to r is a component in this direction. It cruised for t seconds. So this the length is u perpendicular t, to a very good approximation. Forgetting about this little teeny weeny, little distance. So we know what this is. So u perpendicular is the component of the vector, u, perpendicular to my position vector, r. And this little section here must, of course, have thickness c delta t. That is the difference, in this radius and this radius, c delta t. So look at this green triangle. Now comes the key thing. This electric field line must connect with that electric field line. It was one and the same. It's been broken now. And the connection must go through this thin layer. I will draw here the connection. And then here, you have to round it off a little. It cannot be sharp. And this runs off a little. So this point here represents the beginning of the acceleration. And this represents the end of the acceleration. And so then the field line goes [WOOSH] [WOOSH] [WOOSH]. And now comes my task. And that is to calculate the electric field inside that shell. Because that and only that is responsible for the electromagnetic radiation. And so the field line goes like this. So there must be a component of the electric field, which is in this direction. And there must be a component, which is in this direction. I just decompose it. And I call this one E parallel. My notation, parallel, always means parallel to r. And my notation, perpendicular, always means perpendicular to r. And so this then would be called E perpendicular. And this line here is a triangle. This triangle, that you see here, is congruent with this green one. That's why I gave you the dimension of the green one. And so now the question is, what is this component, E perpendicular? You could just feel in your stomach that that is the one that is responsible, that is the electric field in the travelling wave that moves outwards. Because this whole shell moves outwards with a speed, c. And we notice, this is perpendicular to the direction that I have chosen of propagation. You're watching here. And this shell comes over you. And it's always the E field, that is perpendicular to your line of sight, that is the E field in a traveling wave. So our task is now to calculate the E vector perpendicular here. Well, if you look at the fact that this triangle is congruent with this one, then you see that E perpendicular divided by E parallel must be u perpendicular t divided by c delta t. But u perpendicular t-- we know that u is a t. So u perpendicular is a perpendicular times t. a perpendicular is now the component of the acceleration perpendicular to r. I don't want to put it in here, because it becomes too cluttered. I will make a new drawing, shortly. I'll put the a in there. Acceleration is in this direction. And so a perpendicular is like this. And so I can replace u perpendicular t, I can replace it by a perpendicular. Sorry, this is a perpendicular delta t. You guys should have screamed, because the acceleration only lasts for delta t seconds. Yea? Only delta t seconds. So now I get a perpendicular times delta t times t divided by c delta t. And so that is a perpendicular times t divided by c. If now I can calculate what E parallel is, I'm done. Because then I know what E perpendicular is. I put E parallel here, and I am in business. Before I do that, I want to eliminate t. And I'm going to write for this t, r divided by c. So it is also a perpendicular times r divided by c squared. And so I can write down now that E perpendicular is this times E parallel. How do we find E parallel? 8.02, Gauss's law. I make a pill box. And the pill box is going to be like this. I'll make a drawing of it. And this surface is here, in the world that doesn't know yet what happened. And this surface here is in the world which is in turmoil, which is in the transition. So I'm going to make that pill box for you. And so I draw here, again. This is a line that goes somewhere to this point here. It's not this one. It's somewhere there. And here is my pill box. That's my pill box. And so in that pill box, I have in the outside world, the outside world here, I have that E field. So I will put that in as an E. It's in the world that doesn't know yet what happened. That's radially going through point O. Inside the box, I have this E parallel, coming in like this. And then going straight through these sides of the box is this one, E perpendicular. So this comes in, E perpendicular and that E perpendicular goes out. Now, since this is in vacuum, there is no charge density inside this box, so the divergence of E must be 0. And that means that the E vector here must be exactly the same as the E vector here. Because the contribution to these two is 0, so this must also be 0. There's no charge inside the box. But I do know what this E field is. That is 8.02. If someone tells you that I have a charge here, sitting there, what is the electric field at a distance, r? That is Coulomb's law. That's not Coulomb's law. It's a Gauss's law. But in that case, you will find very easy that this E vector is q divided by 4 pi epsilon 0 r squared. It falls off as 1 over r squared. You people should remember that from 8.02. And so now we have accomplished-- the 13 minutes are almost up. E perpendicular now is therefore a perpendicular times r divided by c squared times q divided by 4 pi epsilon 0 times r squared. And you lose one r. And so this is the classic derivation. Already were known in the late 19th century. Which is now the electric vector. The strength of the vector, which clearly is responsible for the electromagnetic traveling wave, because it's perpendicular to my direction, r, that I have chosen. It is inversely proportional to r. I will come back to that. That is a natural consequence of the conservation of energy. And this field is very different from static electric fields, which fall off as one over r squared. This is a traveling wave, which has an E field that falls off as 1 over r, not as 1 over r squared. Now, before I write down the final form, we also have to take into account the fact that there is a time delay. So I am an observer. At time t, the acceleration took place at an earlier time, because the radiation has to reach me. And so t prime is then t minus r divided by c. So there is always a time delay. So that is a reason why we write down this equation differently. So I write it now then in all its glory, the way that you would find it in most books. So this is now the E vector. Due to an accelerated charge, you may choose the direction, r. And it is a function of time. And that now becomes minus. I will get back to the minus sign. Then I get a perpendicular at time t prime, whereas this at time t. And that is the connection between the two. Proportional with the charge, divided by c squared 4 pi epsilon 0 and just one r. You have the privilege to convert or not convert, to write down for me or for yourself, what the associated B field is. It's very easy of course. The strength of the B field is c times smaller. And you must make sure that E cross B is in the direction of propagation. So I leave you with that. And then we get that the Poynting vector, as a function of rt, is then E cross B divided by mu 0. And there you have the entire set that tells me what the electric field is due to an acceleration. What the magnetic field is due to an acceleration. And now, what the Poynting vector is, how many joules per second per square meter, flow out. a perpendicular is different for different directions. If I call this angle theta, then a perpendicular is 0, if I am looking down here. Because then a is towards me, there is no perpendicular component. If I am in this direction, then a perpendicular is the same as a. So you see that the strength of the electric field is a strong function of that angle theta. So what you're going to see is there are spherical waves going out from this accelerated charge. But the electric field strength, in that spherical wave going on in all directions, is the strongest in this plane here, perpendicular to a, and is 0 in that direction. And so I would like to make a somewhat different, somewhat simplified picture for you, to stress the connection, and to also address the minus sign. So if this is the direction of the acceleration, a. So I put the a vector in here. Then this is a perpendicular. Let's first choose the direction of r. This is r. You can choose it any way you want to. So then this would be a perpendicular. And if the charge is positive, which I have chosen here. Because look, the field lines go away. Then the E vector perpendicular is in this direction, is like so. And that is the reason for that minus sign. Namely, if q is positive, then E is in the opposite direction of a perpendicular. If q is negative, then it is in the same direction. And then you get that this angle theta is very important. The E vector is proportional to sine of theta, because a perpendicular is, of course, a times the sine of theta. And so the E vector and B vector have sines of theta in them. And the Poynting vector have sine squared theta in them. Because this has a sine of theta, and this has a sine of theta. So the Poynting vector, S, is proportional to the sine squared of theta. And so the spherical waves going in all directions, the Poynting vector is very different for the different directions. The amount of energy per second that flows out, per square meter, in the plane perpendicular to the acceleration is very high. And it is 0 in this direction. If you looked here, you would see nothing. If you looked here, you would see a maximum. And anything in between, you would see less than the maximum, but more than 0. So this is really nothing like a plane wave, infinitely at all. If anything, it's more spherical, then that it is a plane wave. However, if you're very far away from the origin, you can probably, locally, approximate it by a plane wave solution. So now I want to summarize for you, basically, what I want you to remember, and which is all that I remember. And that is how do you find the E vector, if you know the acceleration, and you know where you are in space? So we accelerate q. And we do that in the direction a. And you are at some position in space, r. Now you observe an electromagnetic wave, but you observe it with a time delay. Which I'll not further expand on. But now comes the key thing. That the E field is in the plane of r and a. And you can confirm that with this picture there. So whatever r you choose, you may choose any r you want to. Also outside the blackboard of course, this is just for simplicity. The E vector must be in that plane. But the E vector must be perpendicular to r. Maxwell's equations demand that E vector is perpendicular to the direction of propagation for a traveling wave. The magnitude of the E vector is proportional to the magnitude of the a perpendicular. And the magnitude of E is proportional to 1 over r. And is q is positive, then E is in the direction minus a perpendicular. If q is negative, then it is in the direction of a perpendicular. The 1 over r is an obvious thing, as a consequence of the conservation of energy, because the Poynting vector is proportional to E squared. Because it's proportional E cross B. So the Poynting vector is proportional E squared. So imagine now that you have a sphere with radius, r. And you want to know all the energy that flows through that sphere. So you have to integrate over the entire sphere. But if you double the radius, the surface area of that sphere goes up by factor of 4. So the only way that energy can be conserved, at the same energy flows through a small sphere then flows through a large sphere, is that the electric field goes down by factor of 2. So that the Poynting vector goes down by factor of 4. So the surface area increases by factor of 4, but the Poynting vector goes down by a factor of 4, because E goes down by a factor of 2. We have here an 80 megahertz transmitter. We've seen that before, wavelength 3.75 meters. And I demonstrated this before, to show you that the radiation is linearly polarized. And I've demonstrated it by having this receiving antenna and by rotating it like this, when the light went off, and rotating it like this, when the light goes on. Today, that's not my objective. Today, I'm interested in the sine squared theta term, for which I need a little bit of assistance from someone. So I want someone to hold this here. And you will see that the light will be on. And then I will rotate that, so that the angle of theta changes. If I have it here, I have maximum. Because, you see, the charge is accelerated in this direction. So this angle is theta 90 degrees. That's optimum. That is theta 0. It's 90 degrees. That's optimum. But the moment that you rotate that 90 degrees, or that you rotate this 90 degrees, then it's 0. And that's exactly what I want to demonstrate. So I need assistance. Can you help me? Make sure you hold it here. If you don't, you get electrocuted. So you have a choice. Hold it above your head. Now you see the light is on. So now, this is the direction of acceleration. And I'm going to point the direction of acceleration towards him. If the acceleration is like this, there's no radiation going in his direction, and that light will go out. Move very slowly. Move carefully. It's already dimming. It's already dimming. And now it's [INAUDIBLE]. And now, I'll get it back. There it goes on. And there it goes off. If you come a little closer, then the light is a little brighter. Hold it a little further down, a little down. That's fine. I'll do it once more. So I change the angle theta, and now it goes out. And now it's in. Great. You're lucky you survived this one. Some students don't. So this was very qualitatively, to show you the effect of the sine squared theta relationship. Now, in order to calculate the total power that is produced by-- the total power in the electromagnetic wave that is produced by the oscillating charge, you will have to integrate this Poynting vector over one complete sphere. Which is not a total trivial integral, because the Poynting vector changes with sine square theta. So I will leave you with that exercise. But I will give you the result, which is not so surprising. At least, it is not surprising that the quantities that you will see. Can most of you see this part of the blackboard? Still that I put it better there. Oh, I can erase some here. Let me erase some here. So this is now the integration of the Poynting vector over a complete sphere that I put around the charge. And the result then is that the power, which is no longer watts per square meter per second, which is simply now in watts. How many energy flows per second through a sphere. And that now becomes q squared. That's no surprise, because q is in a, in the E vector. And q is in the B vector. so you get a q squared. You got an a squared here. No surprise, because the a is in the B. And the a is in the E. And then you get this divided by 6 pi epsilon 0. And I think it is c to the power 2. Did you see my hesitation? I finally arrived at c squared. But it really is c to the power 3. So the downstairs is not c squared, but it is c to the power 3. Sorry for that. And so this is the famous Larmor result. It takes energy to accelerate a single charge in vacuum. And I'm not talking here about kinetic energy, in terms of 1/2 mv squared. But I'm talking here about the creation of electromagnetic fields due to the motion of the charge. Now comes the question, how do we accelerate charges? And I thought about that for a few days, how I was going to tell you that. And the answer that I have to give you is a little bit embarrassing. And that is, we accelerate charges by exposing them to electromagnetic radiation. And then you say, well, isn't this Catch-22. Because you have to accelerate them to create the electromagnetic radiation. How do you accelerate them? Well, you have to expose them to electromagnetic radiation. For whatever that's worth, of course, there is electromagnetic radiation that comes out of all kinds of sources. And so I will show you now that, if we take an electromagnetic wave, we'll take a linearly polarized traveling wave, and we will expose an electron to that radiation. And then we will see what the electron will do with that incoming, traveling wave, a plane wave. So here is an electron. It has charge, q, which is now negative. It has mass, m. And from the left, comes linearly polarized radiation. That's what I have chosen. E0 cosine omega t. So it just comes in. It's a plane wave. Infinite in size, it comes in like that. And this electron begins to shake. And the force that the electron will experience, F is q times E0 cosine omega t. That is the definition of electric field, as it is connected to force. I do this for bound electrons. So these electrons are bound in an atom or in a molecule. So they're bound electrons. So when you try to move them away from equilibrium, they don't like that. They experience a restoring force, because of the Coulomb interaction in the atom. And they want to go back. So there is a restoring force that acts, to some degree, like a spring. And so there is also then a resonance frequency associated with that restoring force. And so that resonance frequency, I will simply call, omega 0 squared, which is then K over m. And K is then, the equivalent of the spring constant, which now acts in the atom. I will ignore, purposely, for the reason of simplicity, any form of damping. And so I'm going to write down now Newton's Second Law. And Newton's Second Law becomes now x double dot plus omega 0 squared times x equals this driving force. Which is q times E 0 divided by m times the cosine omega t. And the reason why there is an m here, because I have already removed the m here. And I have removed the m there. And so this is an equation that we have seen before, when we were dealing with force oscillations. And it has a very simple solution that x is A times the cosine of omega t. This is omega of the driver. And so the object will follow, in steady state solution. And this A is the amplitude. And the amplitude I even remember, if you ask me three months from now, I will not remember anymore. This is always upstairs, remember. And then downstairs, you get all that stuff, omega 0 squared minus omega squared plus omega squared gamma squared, but gamma is 0. So you have no damping. And so you get downstairs here, omega 0 squared minus omega squared. So that is the amplitude of the oscillating electron. But now we want to know how much radiation it's going to produce. So now we have to calculate what x double dot is, because x double dot is the acceleration of the electron. And then we want to see how much radiation comes out. And so the x double dot becomes minus omega squared times x. Just taking the second derivative. You've seen that before. And so that becomes then, minus q E0 divided by m (omega 0 squared minus omega squared). And then I got upstairs that omega squared. And that is the acceleration of this electron. But now, I know how much power goes into that field, because the power that goes into that field-- where do I have? Did I erase my-- the Larmor? I wanted it for that reason. So the total power that is radiated is proportional to A squared. I hope you remember that. I erased the wrong part of the blackboard. And so the A squared is what I really want to know. And so the power that is radiated by that shaking electron is proportional-- forget all the constants-- omega to the 4th divided by omega 0 squared minus omega squared, squared. Forget all the rest. All the rest is constant. And now, if omega is way below resonance, this is proportional to omega to the 4th. And that's a very famous result. Which is known in the literature as Rayleigh scattering. And the reason why we call it scattering, Rayleigh scattering, that you must understand that the radiation that comes in has this frequency. But the radiation that comes out has exactly the same frequency. Look. That is the oscillation of the electron. So there's no change in the frequency. Radiation comes in a certain color, red light. And then there is an electron which starts to shake, gets very nervous, because of the E field. And it radiates, again, red light. The color does not change. But it scatters it. It goes off in different directions. And that's why we call this Rayleigh scattering. And if you take oxygen and nitrogen, in our atmosphere, then the resonance frequency is way above the frequency of visible light. The resonant frequency for oxygen and nitrogen is in the far UV. And so we meet that condition that a visible light strikes oxygen and nitrogen molecules, and for that matter, many other molecules, this relationship holds. And what that tells you then is that blue light has a way higher probability to be scattered by moving electrons then red light, because the wavelength or the frequency, I should say, of blue light is substantially larger than the frequency of red light. And so let me try to stay on the center board then and show you what the difference is in frequency. So omega blue divided by omega red is approximately 1.5. Taking either side of the spectrum, about 6,500 angstroms is red and 4,500 angstroms is blue. And that means that 1.5 to the power of 4 is about 5. And so the blue light has a 5 times higher probability to be scattered then the red light. And this holds, as long as the particles, off which the radiation scatters, are smaller than the wavelength of light, typically a few tenths of a micron. So then we have Rayleigh scattering, then we have the dependence on wavelengths to the power of 4. If the particles grow in size, this dependence weakens. And then the difference between the red and the blue becomes less. And by the time you reach 5 microns, particles that have a size of 5 microns and 10 microns, then the probability is the same for all colors. And that is what I would like to demonstrate to you, very shortly. There is an extra bonus. And the extra bonus is-- this is not so intuitive, but I will convince you of that-- that radiation that was scattered over 90 degrees becomes linearly polarized, 100 percent. Unpolarized radiation comes in, here are scatterers. Light goes off in all directions. But what goes off at 90 degrees is 100% linearly polarized. I will first convince you of that, and then I will do a demonstration to-- that shows that, indeed, it is linearly polarized. That's why I wanted you to bring your polarizers. You see radiation coming from this side. And here is this electron that starts to shake. And if you have very small particles, 1/10 and 2/10 of a micron, there are electrons in those particles which begin to shake. And so this electron starts to shake up and down. And the electric field must be in the plane of r and a. Look very carefully. That this electric field is indeed in the plane or r and a. And this electric field is in the plane of r and a. And this electric field is in the plane or r and a. And so this is the direction of oscillation of the E field. In the direction of theta 0, there is no radiation going, because the acceleration is like this. And if you go off at a different angle, then the electric vector oscillates perpendicular to its direction of propagation. Why now is radiation 100 percent polarized if it scatters over 90 degrees? Well, let's have a beam of unpolarized radiation that comes to you. Linearly polarized light comes to you, [WOOSH] [WOOSH] [WOOSH]. There it is. There's the next wave, and the next wave, and the next wave, and the next wave, and the next wave. We call that unpolarized light. So one comes in like this. The next one comes in like this. One comes in like this. We call that unpolarized light. You are looking here, in the blackboard, at this radiation. And there are scatterers here. There's dust here, very fine dust. And so some light comes to you. The angle is 90 degrees. Because it came like this, and now it does this. The whole board is 90 degrees. Someone who is here, also 90 degrees. What is the direction of the E vector when this light reaches my eye? It's in the plane of a and r, that means it must be in the blackboard. Remember, the E vector is in the plane of acceleration. This acceleration is like this. That's the plane of the backboard. This acceleration is like this. And r is also in the blackboard. So the E factor must be in the plane of the blackboard. But it must also be perpendicular to r. So it must also light like this. And here it must also light like this. And if you look here, it must light like this. So whenever you look at scattered light under 90 degrees, if it is Rayleigh scattering, you will see that the radiation is 100% linearly polarized. An amazing thing. And that is my goal, to demonstrate this to you now. It was a very dramatic experiment in which I put my health on the line. I'm going to smoke a cigarette. And I advise you not too. And in cigarette smoke are extremely fine dust particles, a few tenths of a micron in size, ideal for Rayleigh scattering. And then we have light coming from below. And that light is unpolarized light, light bulbs there. Here we will have dust. And this very fine dust scatters the light in your direction. What will be the color of that light? Blue. Because the white light contains red, green, yellow, blue, everything. But the blue light has a higher probability to be scattered. So my smoke, if the particles are small enough, will look blue. If the particles are not small enough, then it won't be blue. It will be blue, believe me. But not only that. The radiation comes up like this, and every thing, every person in the audience is in ideal position, you're all at 90 degrees relative to the scattering, because that's the way I'm made the arrangements. Light comes up like this. And then it goes like this. So all of you are very close to 90 degrees. So get your polarizers out. And you should see that the E vector is in this direction. For you there, like this. For you there, like this. And for you, like this. So we're going to kill two birds with one stone. It will be blue. And it will be linearly polarized. So you ready for this? Now, look at the smoke. Do you agree that it's bluish? Who agrees, just say, yeah. AUDIENCE: Yeah. WALTER LEWIN: Who doesn't agree, just say no. Thank you. Now, use your polarizers. Can you see that when you rotate you polarizers that it's polarized. It's polarized in this direction. Who can see that, say yes. AUDIENCE: Yes. WALTER LEWIN: Who cannot see it, say no. Thank you. Now comes the double header. If I keep that rotten smoke in my lungs for a while, the vapor in my lungs will precipitate onto these small particles, which initially are smaller than a few tenths of a micron, so you get Rayleigh scattering. But now, when I puff out the smoke, you will have teeny-weeny, little water drops, because of the water vapor in my lungs. And now the particles are too large for Rayleigh scattering, and so the scattered light will now be white. In other words, there's no preference anymore for blue over red. And so, what I will do is, I will hold the smoke in my lungs for while. Just before I puff it out, I will show you, again, this smoke, so that you can have a reference of the color, so that you really believe this is blue. And then when I puff it out, you will see there is a distinct difference in color. It becomes white. And so you've seen three things then. Number one, that light that scatters on very small particles, a few tenths of a micron, prefers the blue. Blue has a higher probability. 90 degree scattering, 100% linearly polarized. But if the particles grow beyond a certain size, there is no longer any preference for the blue. Big difference. Who saw the difference? Say yeah, otherwise I'll do it again. AUDIENCE: Yeah. WALTER LEWIN: Who didn't see the difference in color? Ah, thank goodness. OK, we'll have a break, so I can recover, and you can recover. Five minutes. Four minutes. So the sky is blue, because light scatters on very fine dust and even on density fluctuations in the atmosphere, due to the thermal motion of the molecules. And the clouds are not blue, because the clouds have very small water drops, like I have in my lungs. And so that's why the clouds are white. Let's take a look at you're standing on Earth, and let the sun be in this direction, midday. And here is the atmosphere, thickness of about-- depends on how you measure it, how you define it-- 60, 70, 80 kilometers. And you look in this direction, and there's white light coming in. But the probability that blue is scattered in your direction is larger than red, and so the sky looks blue. If you look in this direction, wow, this angle is 90 degrees. Not only is the sky blue there, but the light, from the sky at a 90 degree angle away from the sun, is 100% linearly polarized. And it is linearly polarized in this direction, perpendicular to the blackboard. And you should be able to figure that out for yourself now. And so you have your linear polarizers. And you can go out and look at the sky. There is always a great circle in the sky, which is 90 degrees away from the sun. And anywhere on that great circle, the light from the blue sky is 100% linearly polarized. It's very easy to show with your linear polarizers. Now, when the sun is high in the sky, the amount of light that is scattered in the atmosphere is only 1%. Thank goodness, otherwise there would be rather dark on Earth, right? But if the angle of elevation above the horizon is 5 degrees, then already the amount of scattered light goes up to about 10%. And the lower the sun is on the horizon, the longer it has to travel, the more dust it has to see in the Earth's atmosphere. And so the more the blue will be scattered out. The green will be scattered out. The yellow will be scattered out. And the only color that hangs in there, that survives is the red. That's the reason why sunsets and sunrises are red. So here now is the sun near the horizon. A huge amount of dust on the way to you. And there is a nice cloud here in the sky. And the light that finally gets filtered through the atmosphere is red. And so this side of the cloud is red. And you look at the cloud, and you say there's a red cloud. Because all the light is red. And so the sun is red when it sets, when it rises. The moon is red. The planets are red. The stars are red. But if a planet rises, and it is red, it doesn't make your clouds red. So you don't notice that. But it's the same effect. And you can see that. I often watch planets just coming above the horizon, and they are just as red as the sun is, when the sun sets. So needless to say, that the dirtier the atmosphere, the more pollution we have, the more beautiful sunsets are. Because you get more crud into the atmosphere. After volcanic eruptions, it's well known that people see spectacular sunsets, spectacular sunrises. I now want to show you three slides, which in a rather dramatic way, make the point that I was trying to make. And we're going to make it very dark for that. So Marco will come with the first slide, which is a picture of the Pleiades. The Pleiades is a small group of stars in our galaxy. And these are these bright stars, very hot stars. They produce white light. But there's dust around them. And the dust manifests itself by this blue light. Very fine dust, you can even conclude that it must be very fine, otherwise it wouldn't be blue, the light. And it scatters in news directions, so it takes off in a different direction. It comes to you, and it is blue. That's Rayleigh scattering that tells you that there is very fine dust there. The next picture is a man walking on the moon. And this man is walking like this. And as he does that, there is some dust from the surface of the moon, which he throws up, comes up. In addition, as I learned from my friend Jeffrey Hoffman, who was an astronaut, Jeffrey told me that there is also water vapor that is released from this package. His body is being cooled, because it is very hot in the sunlight on the moon. And his body is being cooled with water, with an ice pack. And that water is released into the nothing. Because I wouldn't even say atmosphere, because it's vacuum on the moon. And there must also be water vapor here. And then there is the dust. But the net result is that this person creates, around him, his own blue sky. You are looking at the light that's from the sun that comes in this direction. And it is scattered to you. And if the particles are fine enough, then you see them blue. So this is dramatic example of Rayleigh scattering. The next one is a picture of a piece of aerogel. I don't know whether any of you have ever had aerogel in your hands. But aerogel has a density which is only four times the density of air. And it is 99.8% porous. And there are extremely small silica particles, which typically have only a size of 1 to 2 nanometers. It's way smaller than the wavelengths of light. And now look at this. This tells you why the sky is blue. But it also tells you why sunsets are red. Because light comes from above. So the light that is scattered to your direction looks distinctly bluish. But the light that makes it through there, the blue has been removed, and maybe the green has been removed. So whatever comes through is red. And so you can even see this in the laboratory. With just a piece of this very amazing material, aerogel, you see the effect of Rayleigh scattering. Who has seen the total lunar eclipse last night? Did you notice that when the moon was completely in the shadow of the Earth, that the moon was red? Who noticed that? Who knows the explanation for that? AUDIENCE: [INAUDIBLE] WALTER LEWIN: You know. And I believe you, that you know. Tell me. AUDIENCE: The light is going through the Earth's atmosphere, and because it has to go through so much atmosphere, it turns red. WALTER LEWIN: A plus. What you see is light from the Earth. The Earth is four times larger in the sky then the sun when you are on the moon. Because the moon is four times smaller than the Earth. The moon can just cover the sun, seen from the Earth. But seen from the moon, the Earth is four times bigger than the sun. So here is the sun during a total eclipse. So you are now on the moon. You are a moon walker. You have this beautiful halo around you, blue. And you're looking at the Earth. And here is the Earth, dark. And the sun is behind it. And here you have this extremely thin layer, only 1% of the diameter of the Earth. It's a very thin layer of atmosphere. But the sunlight that hits that atmosphere, that gets scattered. And so it can make it to you. It can change direction. But it has to go through an extremely thick layer. It's like having the sun at the horizon. And so the light that makes it through it is red. So imagine, if you stand on the moon, and you see that the Earth, four times larger than the size of the sun, there's this gorgeous red ring around it. And it is that light, from the earth, that goes to the moon, makes the moon red. And then reflects to you. And that's why the moon is red brown. Remarkable phenomenon. Rayleigh scattering in the Earth's atmosphere illuminates the moon. Now comes the demonstration that I think is the best of all demonstrations, the mother of all demonstrations. I'm going to show you-- it is one demonstration, kill three birds with one stone. I'm going to create the blue sky. I'm going to create a red sunset. And I'm going to show you that the sky at 90 degree angles to the sun is linearly polarized. All of that in one demonstration. I have here a bucket of sodium thiosulfate. It there are chemists in my audience, they can tell you and me what that is, sodium thiosulfate. I will put in the sodium thiosulfate some sulfuric acid. Then a chemical reaction will precipitate, small particles of sulfur, in the beginning, very small, smaller than a micron. The light that comes from this direction will be scattered off these very small dust particles, which are sulfur. And they come in your direction. Now very little is coming in your direction, because this is very clear. But now when the sulfur precipitate, it comes in your direction. What color will it have? Blue. The blue is preferred. For those of you who are sitting here, the scattering angle is 90 degrees. Get your polarizers out. The chance of a lifetime. It will be polarized like this. Those of you there, who didn't pay as much as they do, partially polarized, not 0, but not 100%. But there is ideal, 100% polarized in this direction. This is the plane perpendicular to the direction of the radiation, and so if I look here it will be polarized light here. If I look here, it will be polarized light there. If you look there, it will be polarized light there. So you can actually turn the light on and off. It's blue. You can turn it on, and you could turn it off. You, turn it on a little bit and turn it off a little bit. But now, as time goes on, the sulfur will become more and more and more. And so more and more of the blue light will scatter. What do you think the sun will do? The sun will turn red. And as we approach 7 o'clock in the evening, it will be really red. Because that means the atmosphere is thicker and thicker and thicker. Let me first put this stuff in, in daylight. And then I will make it dark. I will get also my polarizer out. And then we will observe this extremely romantic way of ending this lecture. The sun is just normal. It is probably, maybe, 3 o'clock in the afternoon. And the sulfur, ah, the sulfur is already beginning. Look, I already see blue light. And boy is it polarized, look. For those of you, I use my polarizer for you. See the difference? See the difference? Ah, look at the sun. Oh, I think we are approaching 5:00 PM already. The sun is already turning a little bit, in the direction of the red. Oh man, I always feel butterflies in my stomach and ants in my pants. So romantic. Walking on the beach, pinky-pinky, it's wonderful. You want to see the polarization again? And you can use your own linear polarizer. Let's now enjoy this wonderful sunset. The sun gets redder. More and more light, sky is nice and blue. Sun gets redder. Ah, it's a cloud coming in front of the sun. Do you see that? That happens sometimes. Or was that just my imagination? I would guess it's now 6:30, roughly. What time does the sun set? Ah, you must be kidding. Maybe in Australia, but not here. I think we are very close to sunset now. Yeah, yeah I knew it. I knew it. See you Tuesday. 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17594	https://mathematica.stackexchange.com/questions/168140/implement-commutativity-and-associativity-for-custom-operation	physics - Implement commutativity and associativity for custom operation - 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. 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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 Implement commutativity and associativity for custom operation Ask Question Asked 7 years, 6 months ago Modified7 years, 6 months ago Viewed 152 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Just to give context: I want to implement quantum mechanical ladder operators and for this I want Mathematica to realize when it can just commute my ladder operators with numbers or variables because they only act on Kets and Bras. So what I'm trying to implement is something like this: Mathematica should realize that mathematica Times[CenterDot[ladder_op,variable],Ket[__Integer]] is the same as mathematica Times[variable,CenterDot[ladder_op,Ket[__Integer]]] where CenterDot defines the operation of ladder operators on Kets. Basically, I want to be able to write ladder_op [CenterDot] variable [Times] Ket[__Integer] or even ladder_op [Times] variable [CenterDot] Ket[__Integer] The only way I could think of was to do this: mathematica CenterDot[ladder_op_, HoldPattern[Times[variable__, Ket[y__]]]] / := Times[variable, CenterDot[ladder_op, Ket[y]]] but then I need to write the above expression with parenthesis ladder_op [CenterDot] ( variable [Times] Ket[__Integer] ) and I would like to be able to neglect these parenthesis. So basically I just want to get as close as possible to how I would write this down on paper. Can anybody help? physics algebra Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications asked Mar 16, 2018 at 18:06 uitty400uitty400 23 3 3 bronze badges 5 This is neither commutativity nor associativity.Patrick Stevens –Patrick Stevens 2018-03-16 18:16:24 +00:00 Commented Mar 16, 2018 at 18:16 You are right, what would you call it?uitty400 –uitty400 2018-03-16 18:18:12 +00:00 Commented Mar 16, 2018 at 18:18 This seems to a rehash of this closed question. Why should we not close this one, too?m_goldberg –m_goldberg 2018-03-16 18:26:58 +00:00 Commented Mar 16, 2018 at 18:26 You might find useful information in this documentation articlem_goldberg –m_goldberg 2018-03-16 18:29:30 +00:00 Commented Mar 16, 2018 at 18:29 Its a different problem I encountered. Its not a rehash.uitty400 –uitty400 2018-03-16 18:33:12 +00:00 Commented Mar 16, 2018 at 18:33 Add a comment\| 0 Sorted by: Reset to default Know someone who can answer? Share a link to this question via email, Twitter, or Facebook. Your Answer Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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17595	https://courses.lumenlearning.com/suny-mcc-chemistryformajors-2/chapter/lewis-acids-and-bases-2/	15.2 Complex Ions \| General College Chemistry II Skip to main content General College Chemistry II Chapter 15: Aqueous Equilibria Search for: 15.2 Complex Ions Learning Objectives By the end of this module, you will be able to: Write equations for the formation of complex ions Perform equilibrium calculations involving formation constants Many slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the complex ionAg(NH 3)2+Ag(NH 3)2+. The Lewis structure of the Ag(NH 3)2+Ag(NH 3)2+ ion is: The equations for the dissolution of AgCl in a solution of NH 3 are: AgCl(s)⟶Ag+(a q)+Cl−(a q)Ag+(a q)+2 NH 3(a q)⟶Ag(NH 3)2+(a q)Net:AgCl(s)+2 NH 3(a q)⟶Ag(NH 3)2+(a q)+Cl−(a q)AgCl(s)⟶Ag+(a q)+Cl−(a q)Ag+(a q)+2 NH 3(a q)⟶Ag(NH 3)2+(a q)Net:AgCl(s)+2 NH 3(a q)⟶Ag(NH 3)2+(a q)+Cl−(a q) Aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion Al(OH)4−Al(OH)4−. The Lewis structure of the Al(OH)4−Al(OH)4− ion is: The equations for the dissolution are: Al(OH)3(s)⟶Al 3+(a q)+3 OH−(a q)Al 3+(a q)+4 OH−(a q)⟶Al(OH)4−(a q)Net:Al(OH)3(s)+OH−(a q)⟶Al(OH)4−(a q)Al(OH)3(s)⟶Al 3+(a q)+3 OH−(a q)Al 3+(a q)+4 OH−(a q)⟶Al(OH)4−(a q)Net:Al(OH)3(s)+OH−(a q)⟶Al(OH)4−(a q) Mercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S 2– ion: HgS(s)⟶Hg 2+(a q)+S 2−(a q)Hg 2+(a q)+2 S 2−(a q)⟶HgS 2 2−(a q)Net:HgS(s)+S 2−(a q)⟶HgS 2 2−(a q)HgS(s)⟶Hg 2+(a q)+S 2−(a q)Hg 2+(a q)+2 S 2−(a q)⟶HgS 2 2−(a q)Net:HgS(s)+S 2−(a q)⟶HgS 2 2−(a q) A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called ligands. These ligands can be neutral molecules like H 2 O or NH 3, or ions such as CN– or OH–. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands aggregate themselves around the central atom, creating a new ion with a charge equal to the sum of the charges and, most often, a transitional metal ion. This more complex arrangement is why the resulting ion is called a complex ion. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters. The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a formation constant (K f) (sometimes called a stability constant). For example, the complex ion Cu(CN)2−Cu(CN)2− is shown here: It forms by the reaction: Cu+(a q)+2 CN−(a q)⇌Cu(CN)2−(a q)Cu+(a q)+2 CN−(a q)⇌Cu(CN)2−(a q) At equilibrium: K f=Q=[Cu(CN)2−][Cu+][CN−]2 K f=Q=[Cu(CN)2−][Cu+][CN−]2 The inverse of the formation constant is the dissociation constant (K d), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. Formation Constants for Complex Ionsand Table 1 are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of K sp values, the stoichiometry of the compound must be considered. \| Table 1. Common Complex Ions by Decreasing Formulation Constants \| \| Substance \| K f at 25 °C \| \| [Cd(CN)4]2−[Cd(CN)4]2− \| 1.3 ×× 10 7 \| \| Ag(NH 3)2+Ag(NH 3)2+ \| 1.7 ×× 10 7 \| \| [AlF 6]3-[AlF 6]3- \| 7 ×× 10 19 \| As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag+ ([Ag+] = 1.3 ×× 10–5 M): AgCl(s)⇌Ag+(a q)+Cl−(a q)AgCl(s)⇌Ag+(a q)+Cl−(a q) However, if NH 3 is present in the water, the complex ion, Ag(NH 3)2+Ag(NH 3)2+, can form according to the equation: Ag+(a q)+2 NH 3(a q)⇌Ag(NH 3)2+(a q)Ag+(a q)+2 NH 3(a q)⇌Ag(NH 3)2+(a q) with K f=[Ag(NH 3)2+][Ag+][NH 3]2=1.6×10 7 K f=[Ag(NH 3)2+][Ag+][NH 3]2=1.6×10 7 The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH 3 to form Ag(NH 3)2+Ag(NH 3)2+. As a consequence, the concentration of silver ions, [Ag+], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag+][Cl–], falls below the solubility product of AgCl: Q=[Ag+][Cl−]<K sp Q=[Ag+][Cl−]<K sp More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves. Example 1:Dissociation of a Complex Ion Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to Ag(NH 3)2+Ag(NH 3)2+. Show Answer We use the familiar path to solve this problem: Determine the direction of change. The complex ion Ag(NH 3)2+Ag(NH 3)2+ is in equilibrium with its components, as represented by the equation:Ag+(a q)+2 NH 3(a q)⇌Ag(NH 3)2+(a q)Ag+(a q)+2 NH 3(a q)⇌Ag(NH 3)2+(a q)We write the equilibrium as a formation reaction becauseFormation Constants for Complex Ionslists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [K f = 1.6 ×× 10 7, and Q=0.10 0×0 Q=0.10 0×0, it is infinitely large], so the reaction shifts to the left to reach equilibrium. Determine x and equilibrium concentrations. We let the change in concentration of Ag+ be x. Dissociation of 1 mol of Ag(NH 3)2+Ag(NH 3)2+ gives 1 mol of Ag+ and 2 mol of NH 3, so the change in [NH 3] is 2 x and that of Ag(NH 3)2+Ag(NH 3)2+ is –x. In summary: Solve for x and the equilibrium concentrations. At equilibrium:K f=[Ag(NH 3)2+][Ag+][NH 3]2 K f=[Ag(NH 3)2+][Ag+][NH 3]2 1.6×10 7=0.10−x(x)(2 x)2 1.6×10 7=0.10−x(x)(2 x)2 Both Q and K f are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10:1.6×10 7=0.10(x)(2 x)2 1.6×10 7=0.10(x)(2 x)2 x 3=0.10 4(1.6×10 7)=1.6×10−9 x 3=0.10 4(1.6×10 7)=1.6×10−9 x=3√1.6×10−9=1.2×10−3 x=1.6×10−9 3=1.2×10−3 Because only 1.2% of the Ag(NH 3)2+Ag(NH 3)2+ dissociates into Ag+ and NH 3, The assumption that x is small is justified.Now we determine the equilibrium concentrations: [Ag+]=0+x=1.2×10−3 M[Ag+]=0+x=1.2×10−3 M [NH 3]=0+2 x=2.4×10−3 M[NH 3]=0+2 x=2.4×10−3 M [Ag(NH 3)2+]=0.10−x=0.10−0.0012=0.099[Ag(NH 3)2+]=0.10−x=0.10−0.0012=0.099 The concentration of free silver ion in the solution is 0.0012 M. 4. Check the work. The value of Q calculated using the equilibrium concentrations is equal to K f within the error associated with the significant figures in the calculation. Check Your Learning Calculate the silver ion concentration, [Ag+], of a solution prepared by dissolving 1.00 g of AgNO 3 and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because Q<K f, assume the reaction goes to completion then calculate the [Ag+] produced by dissociation of the complex.) Show Answer 3×10−22 M 3×10−22 M Key Takeaways Complex ions are examples of Lewis acid-base adducts. In a complex ion, we have a central atom, often consisting of a transition metal cation, which acts as a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions form by sharing electron pairs to form coordinate covalent bonds. The equilibrium reaction that occurs when forming a complex ion has an equilibrium constant associated with it called a formation constant, K f. This is often referred to as a stability constant, as it represents the stability of the complex ion. Formation of complex ions in solution can have a profound effect on the solubility of a transition metal compound. Exercises Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water? Explain why the addition of NH 3 or HNO 3 to a saturated solution of Ag 2 CO 3 in contact with solid Ag 2 CO 3 increases the solubility of the solid. Calculate the cadmium ion concentration, [Cd 2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO 3)2 with 1.150 L of 0.100 NH 3(aq). Explain why addition of NH 3 or HNO 3 to a saturated solution of Cu(OH)2 in contact with solid Cu(OH)2 increases the solubility of the solid. Sometimes equilibria for complex ions are described in terms of dissociation constants, K d. For the complex ion AlF 6 3-AlF 6 3- the dissociation reaction is:AlF 3−6⇌Al 3++6 F−AlF 6 3−⇌Al 3++6 F−and K d=[Al 3+][F−]6[AlF 3−6]=2×10−24 K d=[Al 3+][F−]6[AlF 6 3−]=2×10−24 Calculate the value of the formation constant, K f, for AlF 3−6 AlF 6 3−. Using the value of the formation constant for the complex ion Co(NH 3)6 2+Co(NH 3)6 2+, calculate the dissociation constant. Using the dissociation constant, K d = 7.8 ×× 10–18, calculate the equilibrium concentrations of Cd 2+ and CN– in a 0.250-M solution of Cd(CN)2−4 Cd(CN)4 2−. Using the dissociation constant, K d = 3.4 ×× 10–15, calculate the equilibrium concentrations of Zn 2+ and OH– in a 0.0465-M solution of Zn(OH)2−4 Zn(OH)4 2−. Using the dissociation constant, K d = 2.2 ×× 10–34, calculate the equilibrium concentrations of Co 3+ and NH 3 in a 0.500-M solution of Co(NH 3)3+6 Co(NH 3)6 3+. Using the dissociation constant, K d = 1 ×× 10–44, calculate the equilibrium concentrations of Fe 3+ and CN– in a 0.333 M solution of Fe(CN)3−6 Fe(CN)6 3−. Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 ×× 10–2 mol of silver cyanide, AgCN. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 ×× 10–3 mol of silver bromide. A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na 2 S 2 O 3.5H 2 O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as Ag(S 2 O 3)2 3-Ag(S 2 O 3)2 3- (K f = 4.7 ×× 10 13)? Calculate [HgCl 4 2−][HgCl 4 2−] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl 2 solution. In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO 3 is added before precipitation begins. [The reaction of Ag+ with CN– goes to completion, producing the Ag(CN)2−Ag(CN)2− complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of Ag(CN)2−Ag(CN)2−. How many grams of NaCN were in the original sample? What are the concentrations of Ag+, CN–, and Ag(CN)2−Ag(CN)2− in a saturated solution of AgCN? Show Selected Answers 1.When the amount of solid is so small that a saturated solution is not produced. 3.Cadmium ions associate with ammonia molecules in solution to form the complex ion [Cd(NH 3)4]2+[Cd(NH 3)4]2+, which is defined by the following equilibrium: Cd 2+(a q)+4 NH 3(a q)⟶[Cd(NH 3)4]2+(a q)K f=4.0×10 6 Cd 2+(a q)+4 NH 3(a q)⟶[Cd(NH 3)4]2+(a q)K f=4.0×10 6 The formation of the complex ion requires 4 mol of NH 3 for each mol of Cd 2+. First, calculate the initial amounts of Cd 2+ and of NH 3 available for association: [Cd 2+]=(0.100 L)(0.0100 mol L−1)0.250 L=4.00×10−3 M[Cd 2+]=(0.100 L)(0.0100 mol L−1)0.250 L=4.00×10−3 M [NH 3]=(0.150 L)(0.100 mol L−1)0.250 L=6.00×10−2 M[NH 3]=(0.150 L)(0.100 mol L−1)0.250 L=6.00×10−2 M For the reaction, 4.00 ×× 10–3 mol/L of Cd 2+ would require 4(4.00 ×× 10–3 mol/L) of NH 3 or a 1.6 ×× 10–2–M solution. Due to the large value of K f and the substantial excess of NH 3, it can be assumed that the reaction goes to completion with only a small amount of the complex dissociating to form the ions. After reaction, concentrations of the species in the solution are [NH 3] = 6.00 ×× 10–2 mol/L – 1.6 ×× 10–2 mol L–1 = 4.4 ×× 10–2 M Let x be the change in concentration of [Cd 2+]: \| \| [Cd(NH 3)4 2+] \| [Cd 2+] \| [NH 3] \| --- --- \| \| Initial concentration (M) \| 4.00× 10−3 \| 0 \| 4.4× 10−2 \| \| Equilibrium (M) \| 4.00× 10−3 − x \| x \| 4.4× 10−2 +4 x \| K f=4.0×10 6=[Cd(NH 3)4 2+][Cd 2+][NH 3]4 K f=4.0×10 6=[Cd(NH 3)4 2+][Cd 2+][NH 3]4 4.00×10 6=(4.00×10−3−x)(x)(4.4×10−2+4 x)4 4.00×10 6=(4.00×10−3−x)(x)(4.4×10−2+4 x)4 As x is expected to be about the same size as the number from which it is subtracted, the entire expression must be expanded and solved, in this case, by successive approximations where substitution of values for x into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have: 4.0 ×× 10 6 x (4.4 ×× 10–2 + 4 x)4 = 4.00 ×× 10–3 – x 4.0 ×× 10 6 x (3.75 ×× 10–6 + 1.36 ×× 10–3 x + 0.186 x 2 + 11.264 x 3 +256 x 4) = 4.00 ×× 10–3 16 x + 5440 x 2 + 7.44 ×× 10 5 x 3 + 4.51 ×× 10 7 x 4 + 1.024 ×× 10 9 x 5 = 4.00 ×× 10–3 Substitution of different values x will give a number to be compared with 4.00 ×× 10–3. Using 2.50 ×× 10–4 for x gives 4.35 ×× 10–3 compared with 4.00 ×× 10–3. Using 2.40 ×× 10–4 gives 4.16 ×× 10–3 compared with 4.00 ×× 10–3. Using 2.30 ×× 10–4 gives 3.98 ×× 10–3 compared with 4.00 ×× 10–3. Thus 2.30 ×× 10–4 is close enough to the true value of x to make the difference equal to zero. If the approximation to drop 4 x is compared with 4.4 ×× 10–2, the value of x obtained is 2.35 ×× 10–4 M. 5.For the formation reaction: Al 3+(a q)+6 F−(a q)⇌AlF 6 3-(a q)K f=[AlF 6 3-][Al 3+][F−]6=1 K d=1 2×10−24=5×10 23 Al 3+(a q)+6 F−(a q)⇌AlF 6 3-(a q)K f=[AlF 6 3-][Al 3+][F−]6=1 K d=1 2×10−24=5×10 23 7. \| \| [Cd(CN)4 2−] \| [CN−] \| [Cd 2+] \| --- --- \| \| Initial concentration (M) \| 0.250 \| 0 \| 0 \| \| Equilibrium (M) \| 0.250−x \| 4 x \| x \| K d=[Cd 2+][CN−][Cd(CN)4 2−]=7.8×10−18=x(4 x)4 0.250−x K d=[Cd 2+][CN−][Cd(CN)4 2−]=7.8×10−18=x(4 x)4 0.250−x Assume that x is small when compared with 0.250 M. 256 x 5 = 0.250 ×× 7.8 ×× 10–18 x 5 = 7.617 ×× 10–21 x = [Cd 2+] = 9.5 ×× 10–5 M 4 x = [CN–] = 3.8 ×× 10–4 M 9. \| \| [Co(NH 3)6 3+] \| [Co 3+] \| [NH 3] \| --- --- \| \| Initial concentration (M) \| 0.500 \| 0 \| 0 \| \| Equilibrium (M) \| 0.500−x \| x \| 6 x \| K d=[Co 2+][NH 3]6[Co(NH 3)6 3+]=x(6 x)6 0.500−x=2.2×10−34 K d=[Co 2+][NH 3]6[Co(NH 3)6 3+]=x(6 x)6 0.500−x=2.2×10−34 Assume that x is small when compared with 0.500 M. 4.67 ×× 104 x 7 = 0.500 ×× 2.2 ×× 10–34 x 7 = 2.358 ×× 10–39 x = [Co 3+] = 3.0 ×× 10–6 M 6 x = [NH 3] = 1.8 ×× 10–5 M 11.Because K sp is small and K f is large, most of the Ag+ is used to form Ag(CN)2−Ag(CN)2−; that is: [Ag+]<[Ag(CN)−2][Ag(CN)−2]≈2.0×10−1 M[Ag+]<[Ag(CN)2−][Ag(CN)2−]≈2.0×10−1 M The CN– from the dissolution and the added CN– exist as CN– and Ag(CN)2−Ag(CN)2−. Let x be the change in concentration upon addition of CN–. Its initial concentration is approximately 0. [CN–] + 2 [Ag(CN)2−][Ag(CN)2−] = 2 ×× 10–1 + x Because K sp is small and K f is large, most of the CN– is used to form [Ag(CN)2−][Ag(CN)2−]; that is: [CN−]<2[Ag(CN)2−]2[Ag(CN)2−]≈2.0×10−1+x[CN−]<2[Ag(CN)2−]2[Ag(CN)2−]≈2.0×10−1+x 2(2.0 ×× 10–1) – 2.0 ×× 10–1 = x 2.0 ×× 10–1 M×× L = mol CN– added The solution has a volume of 100 mL. 2 ×× 10–1 mol/L ×× 0.100 L = 2 ×× 10–2 mol mass KCN = 2.0 ×× 10–2 mol KCN ×× 65.120 g/mol = 1.3 g 13.The reaction is governed by two equilibria, both of which must be satisfied: AgBr(s)⇌Ag+(a q)+Br−(a q);K sp=3.3×10−13 Ag+(a q)+2 S 2 O 2−3(a q)⇌Ag(S 2 O 3)3−2(a q);K f=4.7×10 13 AgBr(s)⇌Ag+(a q)+Br−(a q);K sp=3.3×10−13 Ag+(a q)+2 S 2 O 3 2−(a q)⇌Ag(S 2 O 3)2 3−(a q);K f=4.7×10 13 The overall equilibrium is obtained by adding the two equations and multiplying their K s: [Ag(S 2 O 3)3-][Br−][S 2 O 3 2−]2=15.51[Ag(S 2 O 3)3-][Br−][S 2 O 3 2−]2=15.51 If all Ag is to be dissolved, the concentration of the complex is the molar concentration of AgBr. formula mass (AgBr) = 187.772 g/mol moles present=0.27 g AgBr 187.772 g mol−1=1.438×10−3 mol moles present=0.27 g AgBr 187.772 g mol−1=1.438×10−3 mol Let x be the change in concentration of S 2 O 3 2−:S 2 O 3 2−: \| \| [Ag+] \| [S 2 O 3 2−] \| --- \| Initial concentration (M) \| 0 \| 0 \| \| Equilibrium (M) \| 1 2 x 1 2 x \| x \| (1.438×10−3)(1.438×10−3)x 2=15.51(1.438×10−3)(1.438×10−3)x 2=15.51 x 2 = 1.333 ×× 10–7 x=3.65×10−4 M=[S 2 O 3 2−]x=3.65×10−4 M=[S 2 O 3 2−] The formula mass of Na 2 S 2 O 3•5H 2 O is 248.13 g/mol. The total [S 2 O 3 2−][S 2 O 3 2−] needed is: 2(1.438 ×× 10–3) + 3.65 ×× 10–4 = 3.241 ×× 10–3 mol g(hypo) = 3.241 ×× 10–3 mol ×× 248.13 g/mol = 0.80 g The equilibrium is:Ag+(a q)+2 CN−(a q)⇌Ag(CN)2−(a q)K f=1×10 20 Ag+(a q)+2 CN−(a q)⇌Ag(CN)2−(a q)K f=1×10 20 The number of moles of AgNO 3 added is: 0.02872 L ×× 0.0100 mol/L = 2.87 ×× 10–4 mol This compound reacts with CN– to form Ag(CN)2−Ag(CN)2−, so there are 2.87 ×× 10–4 mol Ag(CN)2−Ag(CN)2−. This amount requires 2 ×× 2.87 ×× 10–4 mol, or 5.74 ×× 10–4 mol, of CN–. The titration is stopped just as precipitation of AgCN begins: AgCN 2−(a q)+Ag+(a q)⇌2 AgCN(s)AgCN 2−(a q)+Ag+(a q)⇌2 AgCN(s) so only the first equilibrium is applicable. The value of K f is very large. mol CN–<[Ag(CN)2−][Ag(CN)2−] mol NaCN = 2 mol [ Ag(CN)2−Ag(CN)2− ] = 5.74 ×× 10–4 mol mass(NaCN)=5.74×10−4 mol×49.007 g 1 mol=0.0281 g mass(NaCN)=5.74×10−4 mol×49.007 g 1 mol=0.0281 g Glossary complex ion:ion consisting of a transition metal central atom and surrounding molecules or ions called ligands dissociation constant:(K d) equilibrium constant for the decomposition of a complex ion into its components in solution formation constant:(K f) (also, stability constant) equilibrium constant for the formation of a complex ion from its components in solution ligand:molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases Candela Citations CC licensed content, Shared previously Chemistry. Provided by: OpenStax College. Located at: License: CC BY: Attribution. License Terms: Download for free at Licenses and Attributions CC licensed content, Shared previously Chemistry. Provided by: OpenStax College. 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17596	https://pmc.ncbi.nlm.nih.gov/articles/PMC5033096/	Diffuse periventricular calcification and brain atrophy: A case of neonatal central nervous system cytomegalovirus infection - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Neuroradiol J . 2016 Aug 16;29(5):314–316. doi: 10.1177/1971400916665372 Search in PMC Search in PubMed View in NLM Catalog Add to search Diffuse periventricular calcification and brain atrophy: A case of neonatal central nervous system cytomegalovirus infection Thomas R Sanchez Thomas R Sanchez 1 Division of Pediatric Radiology, University of California, USA Find articles by Thomas R Sanchez 1,✉, Mitchell D Datlow Mitchell D Datlow 1 Division of Pediatric Radiology, University of California, USA Find articles by Mitchell D Datlow 1, Anna E Nidecker Anna E Nidecker 1 Division of Pediatric Radiology, University of California, USA Find articles by Anna E Nidecker 1 Author information Article notes Copyright and License information 1 Division of Pediatric Radiology, University of California, USA ✉ Thomas R Sanchez, Division of Pediatric Radiology, University of California, Davis Medical Center Children’s Hospital, 4860 Y St., Suite 3100 ACC, Sacramento, CA 95817, USA. Email: trsanchez@ucdavis.edu Issue date 2016 Oct. © The Author(s) 2016 PMC Copyright notice PMCID: PMC5033096 PMID: 27531861 Abstract TORCH refers to the most common congenitally acquired infections: toxoplasma, rubella, cytomegalovirus, and herpes simplex virus. Neonatal cytomegalovirus infection remains a common cause of congenital infection worldwide with effects ranging from hearing impairment to significant neurological morbidity. We report a case of a term neonate with ventriculomegaly on prenatal ultrasound who presented with low birth weight, small head circumference, hepatosplenomegaly, and purpuric rash on physical exam. Central nervous system cytomegalovirus infection typically shows periventricular calcifications and associated deep white matter damage and ventriculomegaly. Ultrasound, computed tomography, and magnetic resonance imaging have different roles in the diagnosis of congenital central nervous system cytomegalovirus infection. Many imaging features of congenital cytomegalovirus are distinctive, and can spur a diagnostic work-up as well as help provide a prognosis. Keywords: TORCH infection, congenital CNS infection, cytomegalovirus Introduction TORCH refers to the most common congenitally acquired infections: toxoplasma, rubella, cytomegalovirus, and herpes simplex virus (HSV). Congenital rubella was the leading cause of intracranial congenital infections but is now rarely seen due to a successful vaccination campaign over the last several decades. Cytomegalovirus remains endemic without an effective vaccine, making it the most common cause of congenital central nervous system (CNS) infection in developed countries at present, occurring in one out of 100 live births.1 In the USA alone, about 40,000 infants are diagnosed with cytomegalovirus each year.2 Most of these cytomegalovirus infections are asymptomatic, with 10% presenting with clinical manifestations at birth, and an additional 10–15% developing symptoms in the first year of life.3 However, the effects can be devastating. Systemic disease manifests as hepatosplenomegaly, thrombocytopenia, and jaundice,2 while CNS abnormalities include periventricular calcifications, ventriculomegaly, and various developmental anomalies.4 Case report We report a case of a term neonate who on prenatal ultrasound had intrauterine growth retardation, ventriculomegaly, and oligohydramnios. At birth there was note of microcephaly, hepatosplenomegaly, jaundice, thrombocytopenia, and diffuse purpuric rash. Initial cranial ultrasound at day one of life (Figure 1) showed non-specific periventricular echogenicities and mild ventriculomegaly. Subsequent computed tomography (CT) scan of the head performed in the first week of life demonstrated central brain atrophy and extensive periventricular calcifications (Figure 2). Cytomegalovirus infection was confirmed on urine polymerase chain reaction (PCR), blood and nasopharyngeal mucosal cultures. Cultures were negative for rubella and HSV. Ophthalmological exam showed no evidence of chorioretinitis to support toxoplasmosis. The patient was then promptly started on ganciclovir. Figure 1. Open in a new tab Head ultrasound at birth. Echogenic foci line the bilateral periventricular regions with associated mild dilatation of the lateral ventricles. Figure 2. Open in a new tab Axial non-contrast head computed tomography shows extensive periventricular calcifications, which are concerning for a neonatal central nervous system TORCH (toxoplasma, rubella, cytomegalovirus, and herpes simplex virus) infection. Due to progressive neurological deterioration, significant developmental delay and persistent seizures, a follow-up head CT was acquired at one year of age (Figure 3), which showed worsening ex-vacuo dilatation of the ventricles indicating significant destruction of the periventricular white matter and central brain atrophy. At present, the patient is five years old and has significant neurological impairment with cerebral palsy, spastic quadriplegia, and cortical blindness. Figure 3. Open in a new tab Follow-up axial non-contrast head computed tomography after one year. There is progressive volume loss with ex-vacuo dilatation of the ventricles and prominent extra-axial fluid. Discussion Cytomegalovirus is the most common cause of congenital TORCH infections in developed countries. Congenital transmission is hematogenous through the placenta, and the earlier the infection during gestation, the more severe the consequences. Infection can also be acquired from the vaginal canal during birth.5 Approximately 10% of infected infants are symptomatic at birth, and an additional 10–15% can present with symptoms in the first year of life, including hearing loss and failure to thrive. The disease manifests systemically as hepatosplenomegaly, thrombocytopenia with petechial rash, jaundice, and elevated transaminases. In the 70% of symptomatic infants who demonstrate clinically evident CNS involvement at birth, the effects can be devastating and prognosis is guarded.6 These infants can have microcephaly, seizures, developmental delay, cerebral palsy, visual disturbances, and/or hearing loss, which is often progressive. Intracranial calcifications are seen in up to 50% of these affected infants, and is a significant predictor of poor prognosis.2,7 High-resolution ultrasound can demonstrate typical ventriculomegaly and echogenic periventricular changes (Figure 1), but the extent and location of parenchymal calcifications are best demonstrated on CT (Figure 2). Although any part of the brain can be involved, the preferential damage to the developing germinal matrix by cytomegalovirus classically results in a predominantly periventricular distribution of calcification.8 The eventual consequence of this deep white matter injury is central atrophy and ventriculomegaly (Figure 3), which are more pronounced in patients who are infected earlier in gestation. Early second trimester cytomegalovirus infection can also result in various developmental anomalies including cerebellar hypoplasia and lissencephaly, while later second trimester infection can result in polymicrogyra.3 These developmental abnormalities are better demonstrated on magnetic resonance imaging (MRI) and can be helpful for determining prognosis and for counseling parents. In contradistinction, symptomatic congenital toxoplasmosis is about a tenth as common as cytomegalovirus. The pattern of parenchymal calcification in toxoplasmosis tends to be more random, involving basal ganglia, periventricular region, and cortex, and these can resolve over time with antibiotic treatment.3 In addition, there are no associated sulcation or migration anomalies.9 The other herpes virus in the TORCH group, HSV, is typically not truly congenital, but is transmitted via the birth canal during delivery. This type of neonatal HSV manifests as multilobar meningoencephalitis and is neurologically devastating.3 True congenital HSV is uncommon, comprising about 5% of neonatal herpes. It causes microcephaly, skin lesions, chorioretinits, and hydrancephaly, and is often fatal.10 The majority of patients who survive the devastating neurological sequelae of neonatal cytomegalovirus will have permanent disability. Even those who are infected but asymptomatic at birth still have a significant risk of developing sensorineural impairment, which is often progressive.2 Children who are asymptomatic at birth should be followed closely for failure to thrive and hearing impairment. In conclusion, neonatal cytomegalovirus infection remains a common cause of congenital infection worldwide, with effects ranging from hearing impairment to significant neurological morbidity and early mortality. Periventricular calcifications, deep white matter damage, and ventriculomegaly, as well as associated sulcation and migration anomalies are the predominant findings on neuroimaging. Confirmation can be obtained from urine PCR, blood and mucosal nasopharyngeal culture. Funding This work received no grant from any funding agency in the public, commercial, or not-for-profit sectors. Conflict of interest The authors declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article. References 1.Malm G, Engman ML. Congenital cytomegalovirus infections. Semin Fetal Neonatal Med 2007; 12(3): 154–159. [DOI] [PubMed] [Google Scholar] 2.Bale JF., Jr Cytomegalovirus infections. Semin Pediatr Neurol 2012; 19(3): 101–106. [DOI] [PubMed] [Google Scholar] 3.Barkovich AJ. Pediatric Neuroimaging, 4th ed Philadelphia, PA: Lippincott Williams and Wilkins, 2005. [Google Scholar] 4.Lanari M, Capretti MG, Lazzarotto T, et al. Neuroimaging in CMV congenital infected neonates: how and when. Early Hum Dev 2012; 88(Suppl 2): S3–S5. [DOI] [PubMed] [Google Scholar] 5.Shaw DW, Cohen WA. Viral infections of the CNS in children: imaging features. AJR Am J Roentgenol 1993; 160(1): 125–133. [DOI] [PubMed] [Google Scholar] 6.Osborn A, Tong K. Handbook of Neuroradiology: Brain and Skull, 2nd ed St Louis, MO: Mosby, 1996. [Google Scholar] 7.Boppana SB, Fowler KB, Vaid Y, et al. Neuroradiographic findings in the newborn period and long-term outcome in children with symptomatic congenital cytomegalovirus infection. Pediatrics 1997; 99(3): 409–414. [DOI] [PubMed] [Google Scholar] 8.Hunter JV, Morriss MC. Neuroimaging of central nervous system infections. Semin Pediatr Infect Dis 2003; 14(2): 140–164. [DOI] [PubMed] [Google Scholar] 9.Altman NR. Intracranial infection in children. Top Magn Reson Imaging 1993; 5(3): 143–160. [PubMed] [Google Scholar] 10.Baskin HJ, Hedlund G. Neuroimaging of herpesvirus infections in children. Pediatr Radiol 2007; 37: 947–963. 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17597	https://chemed.chem.purdue.edu/genchem/probsolv/stoichiometry/empirical2/ef2.4.html	Empirical Formula 2 Empirical Formula: The simplest ratio of the atoms present in a molecule. Problem: Find the empirical formula of a compound that is 48.38% carbon, 8.12% hydrogen, and 53.5% oxygen by mass. Strategy: As with most stoichiometry problems, it is necessary to work in moles. The ratio of the moles of each element will provide the ratio of the atoms of each element. Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages). Remember that percentages are a ratio multiplied by 100. You must convert percentages back to their decimal value before working with them. (.4838) (100 g) = 48.38 g C (.0812 ) (100 g) = 8.12 g H (.5350) (100 g) = 53.38 g O 2. Convert the mass of each element to moles of each element using the atomic masses.(48.38 g C) (1 mol/ 12.10 g C) =4.028 mol C (8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H (53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O 3. Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles. There are fewer moles of oxygen than any other element, so we will assume one mole of oxygen to establish the ratios.(3.336 mol O/ 3.336) = 1 mol O (4.028 mol C/ 3.336) = 1.2 mol C (8.056 mol H/ 3.336) = 2.4 mol H 4. Use the mole ratio to write the empirical formula. The mole ratio did not turn out to be whole numbers. Since the we cannot have partial atoms in the empirical formula, a multiplication factor must be applied to get whole numbers. In this case, 5 is the factor we need. (1 mol O) (5) = 5 mol O (1.2 mol C) (5) = 6 mol C (2.4 mol H) (5) = 12 mol H Now, the ratios are whole numbers, and we can write the empirical formula: C6H12O5 \| \| \| --- \| \| Gen Chem Topic Review \| General Chemistry Help Homepage \|
17598	https://quizlet.com/201496350/chapter-16-flash-cards/	Chapter 16 Flashcards \| Quizlet hello quizlet Study tools Subjects Create Log in Science Chemistry Chapter 16 5.0 (1 review) Save What two ions are central to the Arrhenius definitions of acids and bases? Click the card to flip 👆 H+,OH- (Hydrogen Ion, Hydroxide) Click the card to flip 👆 1 / 12 1 / 12 Flashcards Learn Test Blocks Match Created by Ariel_RoseteTeacher Students also studied Flashcard sets Study guides Chem 1 8 terms JS8686 Preview Comprehensive Chemistry: Atomic Structure, States of Matter, and Laws 307 terms HailahS Preview Six strong acidsss 6 terms jlybeckb Preview Polyatomic ions 13 terms Maelove011 Preview Nuclear Fission and Nuclear Fusion 38 terms doggydogc Preview Chemistry Lab: Lab quiz 2: Nuclear Radiation 7 terms dani_123villa Preview Group 5 Epilepsy Cell Signaling and Disease Interactive Quiz for the Background Information 10 terms TB12izGOAT Preview Lab Quiz Wastewater and Water Treatment 27 terms briakroberts Preview Chemistry Reactions 14 terms Hannah_Fry28 Preview WVIT 300 Midterm Flashcards 58 terms katelynnhoang Preview Exam 1 87 terms Pukki23 Preview Polyatomic Ions 38 terms Melanie_Frick Preview practice exam 3 + practice questions chapter 19 and 20 11 terms enleclaire Preview Isotopic Symbols 6 terms Richie_13 Preview chem 35 terms mallorie_luitwieler Preview ch 10 chem kaplan 56 terms jamasian122 Preview Acids and bases 13 terms jaymemorrison3 Preview Ordering! 25 terms crocker_naomilee Preview Chemistry Polyatomics 7 terms quizlette29351414 Preview Chemistry 114: Chapter 4 44 terms evatheresabyrne Preview KMT 27 terms iborgula Preview Chemistry Ionic Prefixes 64 terms Austin_Brown3785 Preview NS Exam 2 ch 6 pt. 1 27 terms JadeAlexis1234 Preview Veterinary Physiology - Acid Base Balance (Exam 3) 57 terms danielle_shead Preview Biochem Lab Final 40 terms kcloew22 Preview Textiles-Exam 2 77 terms Allison_Dailey80 Preview polyatomic ions 49 terms tucrystal Preview Chemistry 20 terms emerson_hines Preview Terms in this set (12) What two ions are central to the Arrhenius definitions of acids and bases? H+,OH- (Hydrogen Ion, Hydroxide) What is the conjugate base of HSO3−? SO3{2-} What is the conjugate acid of HPO32− ? H2PO3- What are Monoprotic Acids? A monoprotic acid is an acid that donates only one proton or hydrogen atom per molecule to an aqueous solution. This is in contrast to acids capable of donating more than one proton or hydrogen, which are called polyprotic acids. Given that HClO4 is a strong acid, how would you classify the basicity of ClO−4? Negligible Basicity. Because Perchloric Acid is a strong acid, it's Conjugate Base, ClO4- is Weak, and is negligible because it's not that basic. Which solution has the higher pH, a 0.001 M solution of NaOH or a 0.001 M solution of Ba(OH)2? .001 M of Ba(OH)2. What are the 6 Common Strong Acids? Hydrochloric Acid (HCl) Hydrobromic Acid (HBr) Hydroiodic Acid (HI) Nitric Acid (HNO3) Perchloric Acid(HClO4) Sulfuric Acid (H2SO4) Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH. Rank the solutions from the highest to lowest pH: N2H4, HOCl, Ba(OH)2, HCl, NaOH Ba(OH)2 >NaOH > N2H4 > HOCl> HCl Ba(OH)2 is a strong base and will dissociate completely at concentrations of 0.01M or less. NaOH is also a strong base, but it will dissociate completely at a concentration of 1.0M not at 0.01M, making Ba(OH)2 the stronger base out of the two strong bases in the list. N2H4 is a weak base, and HOCl and HCl are both acids. What is the pH of N2H4? Kb= 1.310^-6 10.56 What is the pH of HOCl? Ka= 3.510^-8 4.23 Identify the following salts as neutral acidic or basic: LiNO3, SrBr2, LiF, KCN, NH4ClO4, NH4Br, NH4CN LiNO3,- neutral (salt of a strong acid and a strong base) SrBr2, -neutral (salt of a strong acid and a strong base) NH4ClO4,-acidic (salt of a strong acid and a weak base) KCN, - basic (salt of a weak acid and a strong base) NH4Br = acid NH4CN- basic, LiF = basic See an expert-written answer! We have an expert-written solution to this problem! Will NO3− ions affect the pH of a solution? What about CO3{2-} ions? NO3- ions will not affect the pH. It is the weak conjugate base of Strong Acid HNO3. Carbonate ions are obtained from weak acids H2CO3 and HCO3-, so the conjugate base CO3{2-} wil l be strong and accepts protons from water. 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17599	https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem?srsltid=AfmBOorov3sUEClM1RX-xlXOeVz2LSDcQzNSr7yQaDOUttgKZrC1-9HA	Art of Problem Solving Power of a Point Theorem - 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 Power of a Point Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Power of a Point Theorem The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two linesintersect a circle and each other. Contents [hide] 1 Statement 1.1 Case 1 (Inside the Circle): 1.2 Case 2 (Outside the Circle): 1.2.1 Classic Configuration 1.2.2 Tangent Line 1.3 Case 3 (On the Border/Useless Case): 1.4 Alternate Formulation 1.5 Hint for Proof 2 Notes 3 Proof 3.1 Case 1 (Inside the Circle) 3.2 Case 2 (Outside the Circle) 3.3 Case 3 (On the Circle Border) 4 Problems 4.1 Introductory 4.2 Intermediate 4.3 Olympiad 5 See Also 5.1 External Links Statement There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application. Case 1 (Inside the Circle): If two chords and intersect at a point within a circle, then Case 2 (Outside the Circle): Classic Configuration Given lines and originate from two unique points on the circumference of a circle ( and ), intersect each other at point , outside the circle, and re-intersect the circle at points and respectively, then Tangent Line Given Lines and with tangent to the related circle at , lies outside the circle, and Line intersects the circle between and at , Case 3 (On the Border/Useless Case): If two chords, and , have on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is so no matter what, the constant product is . Alternate Formulation This alternate formulation is much more compact, convenient, and general. Consider a circle and a point in the plane where is not on the circle. Now draw a line through that intersects the circle in two places. The power of a point theorem says that the product of the length from to the first point of intersection and the length from to the second point of intersection is constant for any choice of a line through that intersects the circle. This constant is called the power of point . For example, in the figure below Hint for Proof Draw extra lines to create similar triangles (Draw on all three figures. Draw another line as well.) Notice how this definition still works if and coincide (as is the case with ). Consider also when is inside the circle. The definition still holds in this case. Notes One important result of this theorem is that both tangents from any point outside of a circle to that circle are equal in length. The theorem generalizes to higher dimensions, as follows. Let be a point, and let be an -sphere. Let two arbitrary lines passing through intersect at , respectively. Then Proof. We have already proven the theorem for a -sphere (a circle), so it only remains to prove the theorem for more dimensions. Consider the plane containing both of the lines passing through . The intersection of and must be a circle. If we consider the lines and with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds. Proof Case 1 (Inside the Circle) Join and . In (Angles subtended by the same segment are equal) (Vertically opposite angles) (Corresponding sides of similar triangles are in the same ratio) Case 2 (Outside the Circle) Join and (Why?) Now, In (shown above) (common angle) (Corresponding sides of similar triangles are in the same ratio) Case 3 (On the Circle Border) Length of a point is zero so no proof needed:) Problems Introductory Find the value of in the following diagram: Solution Find the value of in the following diagram: Solution (ARML) In a circle, chords and intersect at . If and , find the ratio . Solution (ARML) Chords and of a given circle are perpendicular to each other and intersect at a right angle at point . Given that , , and , find . Solution Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is (Source) Intermediate Two tangents from an external point are drawn to a circle and intersect it at and . A third tangent meets the circle at , and the tangents and at points and , respectively (this means that T is on the minor arc ). If , find the perimeter of . (Source) Square of side length has a circle inscribed in it. Let be the midpoint of . Find the length of that portion of the segment that lies outside of the circle. (Source) is a chord of a circle such that and Let be the center of the circle. Join and extend to cut the circle at Given find the radius of the circle. (Source) Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find (Source) Let be a triangle inscribed in circle . Let the tangents to at and intersect at point , and let intersect at . If , , and , can be written as the form , where and are relatively prime integers. Find . (Source) Olympiad Given circles and intersecting at points and , let be a line through the center of intersecting at points and and let be a line through the center of intersecting at points and . Prove that if and lie on a circle then the center of this circle lies on line . (Source) Let be a point interior to triangle (with ). The lines , and meet again its circumcircle at , , respectively . The tangent line at to meets the line at . Show that from follows . (Source) See Also Geometry Planar figures External Links Handout on AoPS Forums This article is a stub. Help us out by expanding it. Retrieved from " Categories: Geometry Theorems 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.

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