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14700	https://www.tiktok.com/@learnbyosmosis/video/7221643501815991594	Primera hemostasia: Formación del tapón plaquetario \| TikTok TikTok Log in TikTok Search For You Explore Following Upload LIVE Profile More Log in Company Program Terms & Policies © 2025 TikTok 2720 86 20 00:05 / 00:27 learnbyosmosis Osmosis from Elsevier · 2023-4-13 Follow more You know that one Beyoncé lyric? "Okay ladies, let's get in (platelet plug) formation!" Today's #ClinicalCuts is all about primary hemostasis! Platelet plug formation, also called primary hemostasis, is the first of two steps needed for hemostasis. Hemostasis is how the body prevents blood loss a blood vessel is injured and broken. Without hemostasis even a minor injury would be life-threatening – imagine dying from a nosebleed! During primary hemostasis, platelets clump up together and form a plug around the site of injury. The clumping up of platelets can be further divided into five steps: endothelial injury, exposure, adhesion, activation, and aggregation. Find the full video to learn more about platelet plug formation on Osmosis, link in our bio! You can also tap on our profile to find our TikTok about coagulation (secondary hemostasis)! #learnbyosmosis#clinicalcuts#hemostasis#primaryhemostasis#plateletplugformation#blood#hematology#arteries#veins#hemodynamics#physiology#anatomy#biology#medicine#doctorsoftiktok#healthtok#medicalschool#medstudent#foryoupage#fyp original sound - Osmosis from Elsevier 0 comment Log in to comment You may like Log in Log in Introducing keyboard shortcuts! Go to previous video Go to next video Like video Mute / unmute video Play / Pause Skip forward Skip backward Full screen Unmute
14701	https://www.youtube.com/watch?v=ibeyn2QGjCM	Solving square-root equations: no solution \| Mathematics III \| High School Math \| Khan Academy Khan Academy 9090000 subscribers 131 likes Description 129420 views Posted: 8 Mar 2016 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Sal solves the equation √(3x-7)+√(2x-1)=0, only to find out that the single solution is extraneous, which means the equation has no solution. Watch the next lesson: Missed the previous lesson? High School Math on Khan Academy: Did you realize that the word "algebra" comes from Arabic (just like "algorithm" and "al jazeera" and "Aladdin")? And what is so great about algebra anyway? This tutorial doesn't explore algebra so much as it introduces the history and ideas that underpin it. About Khan Academy: Khan Academy is a nonprofit with a mission to provide a free, world-class education for anyone, anywhere. We believe learners of all ages should have unlimited access to free educational content they can master at their own pace. We use intelligent software, deep data analytics and intuitive user interfaces to help students and teachers around the world. Our resources cover preschool through early college education, including math, biology, chemistry, physics, economics, finance, history, grammar and more. We offer free personalized SAT test prep in partnership with the test developer, the College Board. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide every year. For more information, visit www.khanacademy.org, join us on Facebook or follow us on Twitter at @khanacademy. And remember, you can learn anything. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s High School Math channel: Subscribe to Khan Academy: 12 comments Transcript: let's say that we have the radical equation the square Ro T of 3x - 7 plus the square < TK of 2x - 1 is equal to Zer I encour you to pause the video and see if you could solve for x before we work through it together all right so one thing we could do is we could try to isolate each of the radicals on either side of the equation so let's subtract two let's subtract this one from both sides so I can get it onto the right hand side so or a version of it on the right hand side so I'm subtracting it from the left hand side and from the right and from the right hand side and so this is going to get us that is going to get us on the left hand side I just have square root these cancel out so I'm just left with the square < TK of 3x - 7 is going to be equal to this the netive of the < TK of 2x -1 so now we can square both sides and we always have to be careful when we're doing that because whether we're squaring the positive or the negative square root here we're going to get the same value so the solution we might get might be the version when we when we're solving for the positive square root not when we take the negative of it so we have to test our Solutions at the end to make sure that they're actually valid for our original equation but if you square both sides on the left hand side we are going to get three 3x - 7 and on the right hand side a negative square is just a positive and the square < TK of 2x - 1^ 2 is going to be 2x- 1 and I'll see we can subtract 2x from both sides to get all of our x's on one side so I'm trying to get rid of this and we can add seven to both sides because I'm trying to get rid of the negative s so add seven to both sides and we are going to get we are going to get 3x - 2x is X is = to -1 + 7 x is = to 6 now let's verify that this actually works so if we look at our original equation the square < TK of 3 6 - 7 - 7 needs to plus plus the < TK of 2 6 - 1 needs to be equal to 0er so does this actually work out 3 6 - 7 so this is going to be the square < TK of 11 plus the square < TK of 11 needs to be equal to 0 which clearly is not going to be equal to Zer this is two square roots of 11 which does not equal zero so this does not work and you might say wait how did this happened I I did all of this nice neat Algebra I didn't make any mistakes but I got something that doesn't work well this right here is an extraneous solution why is it an extraneous solution because it's actually the solution to the equation it's a solution to the equation the < TK of 3x - 7us the < TK of 2x - one is equal to zero and you might say well if it's the solution to that if it's the solution to this thing right over here how did I get the answer while I'm trying to do algebraic steps there well the key is if when we added when we took this on the right hand side and squared it well it all boiled down to this regardless of what starting point you started with if you did the exact same thing you would have gotten to that same this point right over here so the solution to this ended up being the solution to this starting point versus the one that we originally started with so this one interestingly has no Solutions and it would actually be fun to think about why it has no Solutions we've we've shown to a certain degree the only solution you got by taking reasonable algebraic steps is an extraneous one it's a solution to a different equation that gets that has a common intermediate step but it's also fun to think about why why this right over here is impossible
14702	https://englishforeveryone.org/PDFs/Three-Step%20Method.pdf	englishforeveryone.org T Th hr re ee e--S St te ep p M Me et th ho od d f fo or r S So ol lv vi in ng g A An na al lo og gi ie es s P Pr ro ob bl le em ms s The best way to solve analogies problems is to attack them using this universal method. Each of the following steps outlined in our Five-Step Method will work with any analogy you may encounter, regardless of the relationship shared between the word pair. S St te ep p 1 1: : Build a strong bridge sentence relating the words in the question pair. The bridge should be as short and clear as possible. F Fu un nc ct ti io on n B Br ri id dg ge e: : 1 1) ) bag : buy 2 2) ) baby : cry 3 3) ) cloud : rain 4 4) ) fork : eat 5 5) ) car : stop In the question to the left, the question pair is SHOVEL : DIG. First, think about a necessary relationship shared between this pair: A s sh ho ov ve el l i is s u us se ed d t to o d di ig g. . Remember, no proper analogies question will put words in the question pair that do not share a d di ir re ec ct t a an nd d n ne ec ce es ss sa ar ry y relationship. For example, you will never see the words M MA AN N : : S ST TR RO ON NG G used in the question pair, because they do not share a necessary relationship. A man may or may not be strong. S St te ep p 2 2: : Now, use this bridge with each answer choice, inserting them in place of the words in the question pair. 1 1) ) A A b ba ag g i is s u us se ed d t to o b bu uy y. . Perhaps, but this is not necessarily true. You may use a shopping cart to buy something. Or, you may use money to buy something. So this doesn’t work. 2 2) ) A A b ba ab by y i is s u us se ed d t to o c cr ry y. . This doesn’t work; a baby isn’t used to do anything. 3 3) ) A A c cl lo ou ud d i is s u us se ed d t to o r ra ai in n. . A cloud produces rain, but a cloud isn’t used to rain. Again, a cloud isn’t necessarily used to do anything. 4 4) ) A A f fo or rk k i is s u us se ed d t to o e ea at t. . Here we have a strong relationship. A fork is definitely used to eat. This seems to be the correct answer, but let’s try the last one just to make sure. 5 5) ) A A c ca ar r i is s u us se ed d t to o s st to op p. . One of the operations of a car is to stop. However, a car is not necessarily used to stop. A car is used to travel, or to transport. So, this doesn’t work. Therefore, answer choice #4 must be the correct answer. S St te ep p 3 3: : If after completing steps 1 and 2 you still have not found an answer pair that works, then it may be necessary to adjust the bridge sentence. Let’s look back at our original example: F Fu un nc ct ti io on n B Br ri id dg ge e: : 1 1) ) bag : buy 2 2) ) baby : cry 3 3) ) cloud : rain 4 4) ) fork : eat 5 5) ) car : stop It may be that our original bridge was not strong enough. Imagine if our bridge was: A s sh ho ov ve el l m ma ay y b be e u us se ed d t to o d di ig g. . Now, it is possible to come up with two correct answers: 1 1) ) A A b ba ag g m ma ay y b be e u us se ed d t to o b bu uy y. . 4 4) ) A A f fo or rk k m ma ay y b be e u us se ed d t to o e ea at t. . Our original bridge wasn’t strong enough. Therefore, we need to strengthen our bridge by making it more specific, and repeat steps 1 and 2.
14703	https://unacademy.com/content/jee/study-material/mathematics/properties-of-arithmetic-progression/	© 2023 Sorting Hat Technologies Pvt Ltd Properties Of Arithmetic Progression Arithmetic progression states that the difference between each term should be constant/same. This article discussed the properties of AP with proof. What is arithmetic progression (AP)? AP is said to be a sequence of series. Moreover, if we add or substrate a constant number in each term, we will get the same difference. For instance, the series 11,13,14,16,18 is an arithmetic progression, as the difference between terms is the same, i.e. 2. However, a series like 11,13,15,18,21 is not an arithmetic progression; it will vary if we make the difference between these terms. For instance, 11 – 13 = 2 13 – 15 = 2 15 – 18 = 3 18 – 21 = 3 So, the differences between the terms are not constant. As a result, it is not an arithmetic progression. ‘Arithmetic’ means there are three numbers. For instance, if we have c, B, and d, then B is called the arithmetic mean of the numbers c and d. Questions related to average arithmetic mean are generally asked in the exams. Moreover, notes on average arithmetic mean play an important role in preparation for the exams. Properties of Arithmetic Progression Property 1: If we add or subtract a constant from each term of AP, then we will get the resulting sequence as an AP only. Moreover, the difference between the terms should be the same or common. Only then it can be called an arithmetic progression. Proof: Let a1, a2, a3, and so on be an AP; the difference is common and is denoted by d. And let the fixed constant be m, which is added to each of the terms of AP. Then the sequence will come as: a1 + m, a2+ m, a3 + m… Let bn = an + m, n = 1,2…, then the new sequence will come as b1, b2, b3,… We have, bn + 1 – bn = (an+1 + m) – (an + m) an + 1 – an = d for all n belongs to N Property 2 If we multiply or divide each term of AP by a non-zero constant m, then AP will be the resulting sequence only. Moreover, md or d/m is a common difference, whereas the given AP’s common difference is d. Proof: Let a1, a2, a3,… be an AP, with d as a common difference, and let the non-zero constant be m. Let b1, b2, b3,… be the sequence got by multiplying each term of given AP by m, then. b1 = a1 m, b2 = a2 m ,…, bn = an m Now, bn + 1 – bn = an + 1 m – an m (an + 1 – an)m = dm for all n belong to N Thus, the new sequence is an arithmetic progression, with dm as a common difference. Property 3 In a finite arithmetic progression, the sum of the terms equidistant from beginning to end is always the same. Moreover, it is equal to the sum of the first and last term i.e. ak + an – (k – 1) = a1 + an for all k = 1, 2, 3,…, n-1. Property 4 If the nth term is a linear expression in n, i.e. an = Cn + D, where C and D are considered as constants, then the sequence is arithmetic progression. Proof: Let an be the last term Then, Tn = a + (n – 1)d an = a + nd – d dn + (a-d) an = Cn+D Where C=d and D= a – d Property 5 A sequence is said to be an arithmetic progression if the sum of its first n terms is of the form Cn² + Dn, where C and D are constants independent of n. The common difference in such a case is 2A, i.e. 2 of the coefficient of n². Property 6 If the terms of an arithmetic progression are chosen at regular intervals, they form an AP. Property 7 If an, an + 1, and an + 2 are three consecutive terms of an arithmetic progression, then 2an + 1 =( an + an + 2) So, these were the seven properties of arithmetic progression. Properties of arithmetic Conclusion In this article on the properties of arithmetic progression, first, we have discussed the meaning of arithmetic progression and arithmetic mean. Also, we have seen several properties of arithmetic progression with proofs to make the concept clear for the students. Appropriate formulas and properties of an arithmetic progression are essential to solve a question, as a minor error can make the whole question wrong. So, it is imperative to remember every property and concept of arithmetic progression to score decent marks in the exams, as questions related to arithmetic progression are not difficult but require a good IQ. Frequently asked questions Get answers to the most common queries related to the JEE Examination Preparation. What is the use of arithmetic progression? What is the Tn in arithmetic progression? What are the properties of arithmetic progression? Discuss merits and demerits of arithmetic mean? How are arithmetic sequences used in real life? Ans. As we have already discussed in arithmetic progression, the difference between each consecutive number should be the same. If we talk about the use of the arithmetic progression, it helps us see the patterns of our daily routine. Moreover, it helps make predictions. Ans. As we have already discussed in arithmetic progression, the difference between each consecutive number should be the same. If we talk about the use of the arithmetic progression, it helps us see the patterns of our daily routine. Moreover, it helps make predictions. Ans. There is a formula to find the nth term in a series of arithmetic progression. That formula is Tn = a + (n-1)d, where TN represents the nth term in the series. In addition, the first term is represented by a, and d indicates the common difference between each consecutive term. Ans. There is a formula to find the nth term in a series of arithmetic progression. That formula is Tn = a + (n-1)d, where TN represents the nth term in the series. In addition, the first term is represented by a, and d indicates the common difference between each consecutive term. Ans. Properties of an arithmetic progression are as follows: Commutative property Identity element property Associative property Inverse element Distributive property Ans. Properties of an arithmetic progression are as follows: Ans. There are several merits and demerits of the arithmetic mean. Here are the merits: It can be easily calculated and comprehended. Mathematical formulas are applied, so the outcomes are the same. The demerits are as follows: It isn’t easy to locate on the graph. A minor mistake can make the whole thing wrong. Sometimes, we have to answer irrelevant questions that don’t make any sense Ans. There are several merits and demerits of the arithmetic mean. Here are the merits: The demerits are as follows: Ans. There are a couple of things we do in our daily lives and in which arithmetic sequences are used. For instance: Piling up chairs Helpful in pyramids Filling something, like glass or any container Arranging tables or sitting around them Fencing These are examples of things we do in our day-to-day life. Here, arithmetic progression helps to see that the differences between the terms are constant. Ans. There are a couple of things we do in our daily lives and in which arithmetic sequences are used. For instance: These are examples of things we do in our day-to-day life. Here, arithmetic progression helps to see that the differences between the terms are constant. Unacademy is India’s largest online learning platform. Download our apps to start learning Starting your preparation? 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14704	https://www.osti.gov/servlets/purl/1361588	LLNL-JRNL-711699 A comprehensive iso-octane combustion model with improved thermochemistry and chemical kinetics N. Atef, G. Kukkadapu, S. Y. Mohamed, M. Al-Rashidi, C. Banyon, M. Mehl, A. Heufer, E. R. Nasir, A. Alfazazi, A. K. Das, C. K. Westbrook, W. J. Pitz, T. Lu, A. Farooq, C. J. Sung, H. J. Curran, S. M. Sarathy November 29, 2016 Combustion and Flame Disclaimer This document was prepared as an account of work sponsored by an agency of the United States government. Neither the United States government nor Lawrence Livermore National Security, LLC, nor any of their employees makes any warranty, expressed or implied, or assumes any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States government or Lawrence Livermore National Security, LLC. The views and opinions of authors expressed herein do not necessarily state or reflect those of the United States government or Lawrence Livermore National Security, LLC, and shall not be used for advertising or product endorsement purposes. 1 A comprehensive iso-octane combustion model with improved thermochemistry and chemical kinetics Nour Atef a , Goutham Kukkadapu b , Samah Y. Mohamed a, Mariam Al Rashidi a, Colin Banyon c,Marco Mehl d , Karl Alexander Heufer c, Ehson F. Nasir a, A. Alfazazi a, Apurba K. Das b , Charles K. Westbrook d , William J. Pitz d , Tianfeng Lu e, Aamir Farooq a, Chih-Jen Sung b , Henry J. Curran c, S. Mani Sarathy a a King Abdullah University of Science and Technology (KAUST), Clean Combustion Research Center (CCRC), Thuwal 23955-6900, Saudi Arabia b Department of Mechanical Engineering, University of Connecticut, Storrs, CT, United States c Combustion Chemistry Centre, Ryan Institute, School of Chemistry, National University of Ireland, Galway, Ireland d Lawrence Livermore National Laboratory, Livermore, CA, United States e Department of Mechanical Engineering, University of Connecticut, Storrs, CT 06269-3139, USA Corresponding Authors: Nour Atef Nour.atef@kaust.edu.sa S. Mani Sarathy Mani.sarathy@kaust.edu.sa Al Kindi Building 5 Room 4336 KAUST Thuwal, Saudi Arabia 23955 Abstract iso-Octane (2,2,4-trimethylpentane) is a primary reference fuel and an important component of gasoline fuels. Moreover, it is a key component used in surrogates to study the ignition and burning characteristics of gasoline fuels. This paper presents an updated chemical kinetic model for iso-octane combustion. Specifically, the thermodynamic data and reaction kinetics of iso-octane have been re-assessed based on new thermodynamic group values and recently evaluated 2rate coefficients from the literature. The adopted rate coefficients were either experimentally measured or determined by analogy to theoretically calculated values. Furthermore, new alternative isomerization pathways for peroxy-alkyl hydroperoxide ( ȮOQOOH) radicals were added to the reaction mechanism. The updated kinetic model was compared against new ignition delay data measured in rapid compression machines (RCM) and a high-pressure shock tube. These experiments were conducted at pressures of 20 and 40 atm, at equivalence ratios of 0.4 and 1.0, and at temperatures in the range of 632–1060 K. The updated model was further compared against shock tube ignition delay times, jet-stirred reactor oxidation speciation data, premixed laminar flame speeds, counterflow diffusion flame ignition , and shock tube pyrolysis speciation data available in the literature. Finally, the updated model was used to investigate the importance of alternative isomerization pathways in the low temperature oxidation of highly branched alkanes. When compared to available models in the literature, the present model represents the current state-of-the-art in fundamental thermochemistry and reaction kinetics of iso-octane; and thus provides the best prediction of wide ranging experimental data and fundamental insights into iso-octane combustion chemistry. Keywords : iso-Octane ,combustion kinetics, thermodynamics, gauche, alternative isomerisation 1- Introduction A better understanding of fuel oxidation chemistry and kinetics is crucial for improving the design of modern internal combustion engines (ICE) . For example, end-gas autoignition (i.e., knock) in spark ignition (SI) engines and combustion phasing in homogeneous charge compression ignition engines (HCCI) and partially premixed combustion (PPC) engines are driven by chemical kinetic processes [2-4]. A thoroughly tested kinetic model enables the accurate simulation of engine experiments and gives insight into the effect of fuel composition on HCCI engine combustion phasing . Iso-octane (2,2,4-trimethylpentane) is a gasoline primary reference fuel (PRF) that is highly knock resistant, with an assigned octane number of 100. Since PRF mixtures may be used as surrogates of gasoline, their kinetics and ignition properties have been extensively investigated over the past few decades, both experimentally and theoretically. Most pratical gasoline fuels have octane numbers above 85, and therefore iso-3octane is the principal component in PRF surrogates and strongly governs their combustion chemistry. Various fundamental combustion experiments have been conducted for iso-octane, including ignition delay time and pyrolysis measurements in shock tubes [6-12] and rapid compression machines (RCM) [13-20], premixed laminar flame speed measurements [21-29], and ignition temperature measurements in counterflow flames [30, 31]. However, the majority of experiments were conducted at relatively high temperatures because iso-octane is generally less reactive at lower temperatures (i.e., compared to normal alkanes such as n-heptane). The measurement of intermediate species for iso-octane low temperature oxidation is limited to the measurement of species’ concentration profiles in a jet-stirred reactor (JSR) [32-34] using gas chromatography and mass spectrometry. A comprehensive chemical kinetic model for iso-octane was presented by Curran et al. . The model was developed based on the low- and high-temperature reaction pathways and rate rules proposed earlier for the Lawrence Livermore National Laboratory (LLNL) n-heptane model ; however, some modifications were made to better predict experimental iso-octane reactivity. The model was compared against various experimental data sets and showed good agreement. Later, Mehl et al. [1, 37] further developed the low temperature reaction mechanism for iso-octane oxidation to better predict low temperature heat release (LTHR) in HCCI engines. Although both versions of the model are capable of matching a wide range of experimental data sets available in the literature, there have been some experimental data sets that are difficult to match, in particular, shock tube ignition data and HCCI LTHR profiles of PRF mixtures containing a large amount of iso-octane (i.e., PRF 85–100) [37-40]. The aforementioned studies [9, 37-40] indicate that the currently available iso-octane kinetic models are not sufficiently reactive at low and intermediate temperatures and lean conditions encountered in HCCI engines. Experimental data of iso-octane ignition at 600-800 K and lean conditions are needed to further improve the model. Since the publication of the latest comprehensive iso-octane model , several computational studies have been reported for iso-octane thermochemistry and kinetics. Snitsiriwat and Bozzelli computed the standard heat of formation, entropy, and specific heat capacities of iso-octane and its radicals at the CBS-QB3//B3LYP/6-31G(d,p) level of theory. Later, Auzmendi-Murua 4and Bozzelli performed energy calculations on the potential energy surface of the secondary iso-octyl radical + O 2 system at the CBS-QB3//B3LYP/6-31G(d,p) level of theory, and provided corresponding thermochemical and kinetic data. Recently, the thermochemistry of species and kinetics of reactions lying on the potential energy surface of tertiary iso-octyl + O 2 system were computationally studied at the CBS-QB3//B3LYP/6-31G(d,p) level of theory . Finally, the pressure-dependent kinetics of iso-octane unimolecular decomposition, radical β-scission, and radical isomerization were computed using the CASPT2-(2e,2o)/6-31+G(d,p)//CAS(2e,2o)/6-31+G(d,p) level of theory and Rice −Ramsperger −Kassel −Marcus/Master Equation (RRKM/ME) analysis . In addition, several computational studies [45-48] have been recently reported regarding the low-temperature oxidation kinetics of normal and branched alkanes. The calculated rate coefficients have been implemented in the oxidation mechanisms of three pentane isomers (i.e., n-pentane, iso-pentane and neo-pentane) , n-hexane , and 2-methylhexane . The use of these coefficients and the addition of alternative peroxy-alkyl hydroperoxide (ȮOQOOH) isomerization pathways were found to improve the agreement between simulated and experimental ignition delay data for the fuels considered. In the present work, a comprehensive approach is taken to update the iso-octane detailed chemical kinetic model developed by Curran et al. and Mehl et al. . Specifically, we updated thermochemistry group values and reaction rate coefficients based on new experimental or theoretical studies. We also incorporated alternative isomerization pathways for peroxy-alkylhydroperoxide species ( ȮOQOOH) and third O 2 addition pathways suggested by Wang et al. [52, 53]. The updated model was compared against new RCM ignition delay data from three different facilities at King Abdullah University of Science and Technology (KAUST), National University of Ireland Galway (NUIG), and University of Connecticut (UCONN). The conditions covered by these facilities were low and intermediate temperatures (632–1060K), equivalence ratios of 1 and 0.4, and nominal pressures of 20 and 40 atm. In addition, new high-pressure shock tube ignition delay data at 20 atm from KAUST were obtained. The updated model was also tested against data of ignition delay times [6-9], JSR oxidation speciation data , premixed laminar flame speeds [21, 23-26, 28, 29], counterflow flame ignition data , and 5shock tube pyrolysis speciation data available in the literature. Chemical kinetic mechanism reduction was utilized to generate a skeletal mechanism that was used in flame simulations. 2- Chemical kinetic modeling The updated comprehensive chemical kinetic model for iso-octane was developed by jointly researchers at KAUST, LLNL, NUIG, and UCONN. The thermochemical data for all species in the iso-octane sub-mechanism were updated using new group values and detailed analysis of intra-molecular interactions. Our updated thermochemistry was compared to literature data available for a limited number of species that have been subject to high-level theoretical calculations. The chemical kinetic reaction mechanism was also updated with new reaction classes and rate rules. Further details about the thermochemistry and reaction mechanism are provided in the following sections. 2-1 Thermochemistry Chemical kinetic models require thermochemical data, not only for calculating the mixture temperature, but also for calculating the rates of reversible reactions. In this work, we re-evaluated the thermochemical data of all species involved in the iso-octane oxidation sub-mechanism using Benson’s group additivity method , as implemented in THERM software . The new group values suggested by Burke et al. and Simmie and Somers were used, and gauche interactions were accounted for based on the work of Sabbe et al. . Iso-octane is a highly branched molecule that has three full alkane gauche (AG) interactions according to the Newman projection shown in Fig. 1a and 1b. Each gauche interaction is estimated to add 0.8 kcal to the heat of formation of the molecule . In the case of iso-octyl radicals, the contribution of the gauche interaction to the heat of formation depends on the nature of the radical site. In their work, Sabbe et al. distinguished between two different types of gauche interactions for radicals: i) interaction between two sp3 -hybridized carbon groups, one of which is connected to the radical site (RG1), and ii) interaction between an sp2 -hybridized and an sp3 -hybridized carbon groups (RG2). According to the calculations performed by , RG1 and RG2 interactions are 40% and 76% lower in energy than a full gauche interaction, respectively. For implementation in the THERM program, the total gauche interactions for each radical were 6rounded off to a total number of whole gauche interactions. Therefore, only two gauche interactions are considered for primary iso-octyl radicals (A Ċ8 H17 and D Ċ8 H17 ) and 3 gauche interactions for secondary and tertiary iso-octyl radicals (B Ċ8 H17 and C Ċ8 H17 ) (refer to the Supplementary material for species nomenclature). For cyclic ethers, gauche interactions are only considered if one or both of the interacting groups are not within the cycle, as shown in Fig. 1e and 1f. This methodology was tested by comparing standard enthalpy of formation and entropies from THERM and those from computational chemistry as discussed below. Figure 1. Newman projections showing gauche interactions for the iso-octane molecule along (a) secondary-quaternary bond and (b) secondary-tertiary bond; radical interactions (c) RG2, (d) RG1, and (e,f) cyclic ethers; and (g) structure of iso-octane with site labels. 7Based on the methodology proposed by for olefinic species, gauche interactions between an sp3 -hybridized carbon center and an sp2 -hybridized one, such as the interaction between C1 and C6 in 2,2,4 trimethyl-pent-3-ene (Fig. 2a), were not included because they are already accounted for in other group values. For example, the C/C3/CD group used for the quaternary carbon center in 2,2,4 trimethyl-pent-3-ene includes the interaction between groups 1 and 6. Figure 2. 2,2,4 trimethyl-pent-3-ene (a) and iso-Octane (b) Another destabilizing interaction in the iso-octane molecule is the 1,5 hydrogen interaction resulting from the repulsion between the two methyl groups lying on the same side of the molecule (e.g., C1 and C7 in the carbon skeletal Fig. 2b). This adds 1.5 kcal mol –1 to the heat of formation as reported by Benson et al. . This destabilizing effect is lower in the case of radicals, as reported by Sabbe et al. , especially for the tertiary iso-octyl radical (-0.43 kcal mol –1 ). When calculating the thermodynamic properties of radicals, standard THERM hydrogen interaction groups were used; the number of standard groups included was chosen to most closely match the specific radical hydrogen interactions. Thus, the 1,5 hydrogen interaction in our group additivity thermochemistry estimation for the tertiary iso-octyl radical (-0.43 kcal mol -1 ) is neglected. In addition to gauche and 1,5 hydrogen interactions, we added an optical isomer group (OI) for every chiral center and hydroperoxy OOH functional group (pseudo-chiral center). This adds R×ln 2 to the entropy of the species, which affects the equilibrium constant: ܭ௘௤ ൌ ௞ ೑௞್ ൌ expሺି ௱ு ோ் ൅ ௱ௌ ோ )where K eq is the equilibrium constant, k f and k b are rates of forward and backward reactions respectively, ΔH and ΔS are the change in the enthalpy and entropy of chemical reaction. 8Computational standard heat of formation values of iso-octane , iso-octyl radicals , and species involved in the potential energy surfaces of secondary and tertiary iso-octyl radical + O 2 are available in the literature. These values, calculated at the CBS-QB3 level of theory, were compared to those calculated using group additivity, as shown in Table 1. The agreement between the two sets of values is within the uncertainty limits of the computational values (± 1.5–2.0 kcal mol –1 ), except for species with an OOH group attached to the tertiary site and for four membered ring cyclic ethers. The differences observed for these species are likely due to uncertainties in the magnitudes of the gauche and 1,5 hydrogen interactions. To correct for such systematic discrepancies, we applied a +2 kcal mol –1 correction to the heat of formation of all species with a tertiary OOH group and a –4 kcal mol –1 correction to the heat of formation of all four membered ring cyclic ethers. These corrections did not affect simulated ignition delay times or jet stirred reactor species profiles (comparison shown in Fig. S1 of the Supplementary material). High-level quantum chemical calculations with uncertainties of less than 1 kcal mol –1 are needed to develop a rigorous set of group values and counting schemes that allow for more accurate estimation of thermochemical properties of complex species. A comparison between calculated and estimated entropy values of various species is presented as Supplementary material (Table S2) along with a species dictionary and the input groups employed in THERM (Table S1). Table 1. Comparison of calculated [41-43] and THERM estimated heats of formation of different species. Species that exhibit large discrepancies are shown in bold. All values in kcal mol −1 . Calculated [41-43] Estimated using THERM Difference Corrected Difference after correction Applied Correction Fuel and fuel radicals IC8H18 −54.40 ± 1.60 −53.14 1.26 AĊ8H 17 −5.00 ± 1.69 −4.89 0.11 BĊ8H 17 −9.03 ± 1.84 −7.17 1.86 CĊ8H 17 −12.30 ± 2.02 −10.22 0.06 DĊ8H 17 −5.18 ± 1.69 −4.89 0.29 ROOH,QOOH, RO2 BC 8H 17OOH −75.47 −75.77 −0.30 BC 8H 17 Ȯ2−42.75 −42.60 0.15 BĊ8H 16OOH-A −26.30 −27.52 −1.22 BĊ8H 16OOH-C −31.00 −31.35 −0.35 9BĊ8H 16OOH-D −26.88 −27.52 −0.64 CC 8H 17OOH −77.85 −80.10 −2.25 −78.10 −0.25 (+2) CC 8H 17 Ȯ2−45.80 −46.93 −1.13 CĊ8H 16OOH-B −31.90 −34.13 −2.23 −32.13 −0.23 (+2) CĊ8H 16OOH-D −28.50 −31.85 −3.35 −29.85 −1.35 (+2) CĊ8H 16OOH-A −29.70 -31.85 −2.15 −29.85 −0.15 (+2) Cyclic ethers IC8ETERCD −55.12 −54.17 0.95 IC8ETERAC −76.82 −76.74 0.08 IC8ETERAB −57.42 −53.44 3.98 −57.44 −0.02 (−4) IC8ETERBC −58.72 −58.20 0.52 IC8ETERBD −58.32 −53.44 4.88 −57.44 0.88 (−4) Alkoxy radicals BC 8H 17 Ȯ−38.30 −40.46 −2.16 CC 8H 17 Ȯ−45.50 −42.93 2.57 2-2 Chemical kinetic mechanism The updated iso-octane oxidation sub-mechanism was built upon that of H2 and C1 –C 7hydrocarbons in a hierarchical manner. AramcoMech 2.0 was used for the C 0 –C 4 base chemistry, whereas the chemistry of C 5 –C 7 species was taken from LLNL’s gasoline surrogate mechanism , with the addition of the mechanism by Metcalfe et al. for the di-isobutylene isomers that are produced from iso-octane’s alkylperoxy radicals (RO Ȯ) by concerted H Ȯ2radical elimination reactions and C-H β-scission of the iso-octyl radicals. The original high- and low-temperature chemistry of iso-octane was taken from , and the rate coefficients for nine reaction classes associated with low-temperature kinetics were updated. The updated classes are shown in Fig. 3. These include H-atom abstraction by ȮH radicals, alkyl radical addition to O 2 ,concerted elimination of H Ȯ2 radicals from alkylperoxy radicals, alkylperoxy/hydroperoxy alkyl isomerization, formation of cyclic ethers, β-scission reactions of hydroperoxy alkyl radicals, QOOH radical addition to O2 , and isomerization of ȮOQOOH species. Alternative isomerizations, concerted eliminations, hydrogen-exchange and third O 2 addition pathways for peroxy-alkyl hydroperoxide ( ȮOQOOH) radicals were also added to the mechanism. Since the kinetics of these reactions has never been studied before, their rate coefficients were estimated by analogy. The complete updated mechanism consists of 9220 reactions and 2768 species, with 481 reactions and 179 species specific to the iso-octane sub-mechanism. 10 For all pressure-dependent low temperature oxidation reactions, only the high pressure limit rates were included following the recommendations of Villano et al. . They conducted simulations for n-butyl radical + O 2 using two mechanisms; one including pressure dependent rate coefficients and another using only the high pressure limit rates. The authors concluded that the high pressure limit for RO 2 and QOOH reactions are sufficient under conditions relevant to practical combustors (i.e., pressures higher than a few atmospheres). They also showed that RO 2radicals are collisionally stabilized at elevated pressure conditions, which results in well-skipping R+O 2 reactions to form alkenes+HO 2 and cyclic ethers+OH being less important. Simulations of low temperature oxidation below one atmosphere may require pressure-dependent rate coefficients, but such conditions less relevant to practical engines. Figure 3. Reaction classes for which rate coefficients were updated. The newly added pathways are shown in red. 11 2-2-1 Hydrogen abstraction reactions In this work, the rate coefficients of all H-atom abstraction reactions were updated based on the rate rules suggested by Sarathy et al. , except for H-atom abstraction by ȮH radicals, for which experimentally-derived rate rules were recently made available [64, 65]. Sivaramakrishnan and co-workers provided rate rules that were inferred from shock tube kinetic measurements of H-atom abstraction by ȮH radicals from five saturated hydrocarbons comprising different types of abstraction sites (primary, secondary and tertiary). The next-nearest-neighbor (NNN) effects, discussed by Cohen , were accounted for, and abstraction sites were further distinguished based on the chemical nature of the neighboring atoms. The compounds that were chosen for the study led to rate rules for P 1 , P 2 , P 3 , S 00 , S 01 , S 11 , T 000 and T002 abstraction sites, where P, S and T refer to primary, secondary, and tertiary sites, and the subscript denotes the number of C atoms bonded to the NNN C atom of the C-H bond of interest . For example, P 3 is a primary carbon bonded to a carbon connected to three other carbons (i.e., the neighbor is a quarternary site); and T 002 is a tertiary carbon bonded to two primary carbons and one tertiary carbon. Badra et al. conducted a similar study wherein they measured the total rate of H-atom abstraction by ȮH radicals from different molecules, including iso-octane. Based on their measurements, and the rate rules reported by , Badra et al. derived rate rules for H-atom abstraction by ȮH radicals from S 20 , S 21 , S 22 , S 30 , S 31 , S 32 , S 33 , T 100 and T 101 sites. The classification of abstraction sites in iso-octane based on NNN effects is shown in Fig.4. Figure 4. Carbon sites in iso-octane labeled according to the NNN methodology Badra et al proposed iso-octane rate coefficients for H-abstraction by ȮH radicals by adopting P 2 and P 3 rates from Sivaramakrishnan et al. and S 32 and T 100 from their own work . These rate expressions yield a total abstraction rate constant profile that agrees with the experimental measurements for iso-octane . The rates from agree better than that which was previously employed in the iso-octane kinetic model by Mehl et al. , as demonstrated in Fig. 5. However, slight deviations between measurements and fittings were P3S32 T100 P2P3P3P2 12 observed at low temperatures, as shown in Fig. 5. Therefore, we re-calculated the rate coefficients of H-atom abstraction by ȮH radicals from S 32 and T 100 to better fit the experimental total rate constant profile reported by . The re-calculated rate expressions are given below: S32 , k(cm 3 mol −1 s −1 )= 18.069 T ଷ.ହ expሺ ହଽଵ.ଷସସ ୘ )T100 , k(cm 3 mol −1 s −1 )= 2.4092 ൈ 10 ଽ T expሺ ଶ.ହଵ଺ ୘ ) Figure 5. Experimental and calculated total rate constant profiles for H-atom abstraction by ȮH radicals from iso-octane. 2-2-2 Addition of alkyl and hydroperoxy alkyl radicals to O 2 The rate rules proposed by Miyoshi were used for the addition of iso-octyl radicals to O 2 to form alkylperoxy radicals (RO Ȯ). These rate rules are derived from energy calculations using variational transition state theory (VTST) and (RRKM)/master equation. The same rules were used for the addition of hydroperoxy-alkyl radicals (QOOH) to O2 to form peroxy-alkylhydroperoxide radicals ( ȮOQOOH); however, the rates were divided by 2, based on the findings of Goldsmith et al. . Their calculations showed that the rate for second addition to O2 (QOOH + O 2 ) is half that of the first addition to O 2 (R + O2 ), and in accordance with other modeling studies [50, 51, 69]. 2-2-3 Concerted elimination from RO Ȯ / ȮOQOOH to form olefin + H Ȯ2 The third updated class is the concerted elimination to form olefins and H Ȯ2 radicals from alkylperoxy radicals (RO Ȯ). This class of formally chain propagation reactions produces 1.E+12 1.E+13 0.6 1.1 1.6 2.1 2.6 3.1 3.6 k (cm 3/mol.K) 1000/T (1/K) Measured Original model Fitting New fitting 13 relatively stable olefins and relatively less reactive H Ȯ2 radicals, effectively inhibiting reactivity and slowing ignition. Villano et al. calculated the rate coefficients of these reactions for C 2to C 6 alkylperoxy radicals using CBS-QB3 electronic structure calculations combined with B3LYP/6-31G(d,p) optimized geometries , and provided rate rules depending on the degree of substitution in the olefins produced. Based on their reported values, the formation of highly substituted olefins from alkylperoxy species is two to three times faster than that of less substituted olefins, which contributes to the lower reactivity of branched alkanes. The reaction kinetics involved in the potential energy surfaces of secondary and tertiary iso-octyl radicals + O2 , including concerted H Ȯ2 radical elimination reactions, were studied computationally by and , respectively, both using optimized geometries from B3LYP/6-31G(d,p) and CBS-QB3 for the energy calculation. In both studies, it was shown that the calculated rate constants for different classes are in good agreement with those of analogous reactions available in the literature [45-48], except for the concerted elimination of H Ȯ2 radicals from the tertiary site. Figure 6. The rates of (upper left) CC 8H17 Ȯ2 <=> IC 8H16 + H Ȯ2 and (upper right) CC 8 H17 Ȯ2 <=>JC 8H16 +H Ȯ2 from Snitsiriwat et al. vs Villano et al. ; and (bottom) 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 0.9 1.1 1.3 1.5 k ( mol/cm 3 .s) 1000/T (1/K) 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 0.9 1.1 1.3 1.5 k ( mol/cm 3.s) 1000/T (1/K) 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 0.9 1.1 1.3 1.5 k (mol/cm 3 .s) 1000/T (1/K) 14 BC 8H17 Ȯ2<=>IC 8H16 +H Ȯ2 from Auzemendi-Murua et al. vs Villano et al. . See Supplementary material for species dictionary. Figure 6 compares rate constants of concerted H Ȯ2 radical elimination for secondary and tertiary peroxy radicals calculated using Villano et al.’s rate rules to those reported for iso-octyl peroxy by Auzmendi-Murua and Bozzelli and Snitsiriwat and Bozzelli , respectively. As shown in Fig. 6, the rate constants obtained from for the H Ȯ2 radical elimination from tertiary iso-octyl peroxy radicals are almost two orders of magnitude higher than those reported by for tertiary peroxy, regardless of temperature. Therefore, for consistency, Villano et al.’s rate rules of concerted HO 2 -elimination were used for primary and secondary peroxy radicals, whereas Snitsiriwat and Bozzelli’s rate coefficients were used for the tertiary iso-octyl peroxy radical. By analogy, the rates calculated by Villano et al. were also used for the concerted elimination from ȮOQOOH. 2-2-4 Isomerization of RO Ȯ and ȮOQOOH species and ketohydroperoxide (KHP) decomposition Isomerization of RO Ȯ species involves the intra-molecular migration of a hydrogen atom from an alkyl group to the peroxy group via cyclic transition states, leading to the formation of hydroperoxy-alkyl radicals (QOOH). The rate coefficients of these reactions were taken from the computational study of Villano et al. , which provides rate rules depending on both the size of the transition state ring and the chemical nature of the alkyl group. The model includes isomerizations via 5,6,7 and 8 membered-ring transition states. The size of the transition and the nature of the migrating hydrogen atom (primary , secondary and tertiary) was considered when assigning the rates, as recommended by . A comparison of isomerization rates for various transition state ring sizes and migrating hydrogen atoms is provided in Supplementary Fig.S3. For ȮOQOOH isomerization, Sharma et al. conducted a computational study at the CBS-QB3 and B3LYP/CBSB7 levels of theory with accurate treatment of hindered rotors. Their rate rules for isomerization of ȮOQOOH to ketohydroperoxides (KHP) were adopted in our study. However, since Sharma et al. do not provide a rate rule for intra-molecular H-atom migration to a peroxy group at a tertiary site, the rate coefficients of this type of reaction were estimated based on the knowledge that the rate of abstraction gets progressively slower from tertiary to secondary and then primary carbons. The subsequent decomposition of KHP via O–O scission was assigned a rate constant with a frequency (A-) factor of 1.5×10 16 s –1 and an 15 activation energy of 42.3 kcal mol –1 . In addition, the H-exchange reactions in ȮOQOOH radicals from the hydroperoxy site to the peroxy (i.e., ȮOQOOH=HOOQO Ȯ) were added to the mechanism. The rate rules from Miyoshi were adopted for this class. 2-2-5 Cyclic ether formation, and β-scission of QOOH species Hydroperoxy-alkyl radicals (QOOH) undergo addition to molecular oxygen, cyclic ether formation, and β-scission reactions. The reactions involving QOOH radical addition to molecular oxygen have already been discussed above. The β-scission reactions are divided into two classes; β-QOOH = olefin + H Ȯ2 and β-QOOH = olefin + carbonyl + OH. The formation of H Ȯ2 and olefin is favored due to the weakness of the C–OOH bond compared to a C–C or C–H bond. The rates of this reaction class were updated based on the computational rate rules published by Villano et al. . They showed that the kinetics of this class is insensitive to carbon substitution(s) at the hydroperoxy or the radical site, and thus, they only provided one rate rule for this reaction class. The other class involving the formation of olefin, carbonyl and OH is less energetically favorable than the other pathways (addition to O 2 and cyclic ethers formation) due to the higher bond dissociation energy of C–C bonds compared to O–O and C–O bonds. This class was updated using the rate rules recommended by Bugler and co-workers . Another propagation reaction pathway for QOOH radicals is the formation of cyclic ethers + ȮH. The kinetics of such reactions were computationally studied by Villano et al. and Miyoshi at the CBS-QB3 level of theory. Villano et al. found that the activation energies of these reactions vary depending on the thermochemistry of the specific reaction. Therefore, they reported rate rules with activation energies written as a function of standard heat of reaction ( ΔHrxn ). Meanwhile, Miyoshi distinguished between different cyclic ether formation reactions based on the degree of branching. As the number of substituents on the formed ring increases, the exothermicity of the reaction increases, implying that the rate of the reaction depends on the nature of the radical site and hydroperoxy site in the QOOH radical. In their work, Villano et al. compared their calculated rate constants to those from Miyoshi’s work and both sets of rate rules gave good agreement. 16 However, when applied to iso-octane, the rate rules from and did not give similar rate constants for cyclic ether formation reactions. Figure 7 compares the rate constants of cyclic ether formation from BĊ8 H16 OOH-A (7a), CĊ8 H16 OOH-B (7b), CĊ8 H16 OOH-D (7c) and CĊ8 H16 OOH-A (7d) using the rate rules from Miyoshi and Villano et al. (see Supplementary material for species glossary). When using Villano et al.’s ΔHrxn -dependent rate rules, three values of ΔHrxn were tested in the comparison: i) computational values for ΔHrxn from Bozzelli’s group [42, 43] (solid black lines in Fig. 7), ii) THERM-calculated values with the corrections mentioned previously (dashed black line in Fig. 7), and iii) THERM-calculated values without the corrections (dotted black line in Fig. 7). Figure 7 clearly shows that the rate constants of cyclic ether formation reactions are highly sensitive to the thermochemistry of the reaction. The best agreement is observed between Miyoshi’s rate constants and those obtained using Villano et al.’s rate rules with ΔHrxn calculated using computational values form Bozzelli’s group [42, 43]. Since computational thermochemical properties are not available for all iso-octane QOOH and cyclic ether species, Miyoshi’s rate rules for cyclic ether formation reactions were chosen in our model. The effect of using different sources for the rate constants of cyclic ether formation on iso-octane ignition delay profiles is shown in Fig. S2 of the Supplementary material. 17 Figure 7. Rates of cyclic ether formation from (a) B Ċ8 H16 OOH-A, (b)C Ċ8 H16 OOH-B, (c) CĊ8 H16 OOH-D, and (d) C Ċ8 H16 OOH-A calculated using Miyoshi’s rate rules (grey line), Villano’s rate rules with computationally calculated heats of formation from Bozzelli and co-workers [42,43] (solid black line), heat of formation from THERM after adding the corrections to the groups (dashed black line) and without adding these corrections (dotted black line). 2-3 Alternative isomerization pathways In previously developed mechanisms [35, 36], the only reaction pathway available for peroxy-alkylhydroperoxide radicals ( ȮOQOOH) was the formation of ketohydroperoxides (KHP) via intra-molecular H-atom migration from the C–H site connected to the peroxy group (i.e. COOH); this is followed by cleavage of the O–OH bond to form KHP and an ȮH radical. The H-atom at the COOH site is abstracted to form the ketohydroperoxide due to the weak C-H bond. However, recent studies [48-51, 69, 70] show that the intra-molecular migration of H-atoms from other carbon sites in ȮOQOOH radicals can be competitive. In their study on 2,5-dimethylhexane oxidation in a jet stirred reactor, Wang et al. detected species with four and five oxygen 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 1.E+10 0.6 1.1 1.6 k (1/s) 1000/T (1/ K) (a) 1.E+06 1.E+07 1.E+08 1.E+09 1.E+10 1.E+11 0.6 1.1 1.6 k (1/s) 1000/T (1/ K) (c) 1.E+07 1.E+08 1.E+09 1.E+10 1.E+11 0.6 1.1 1.6 k (1/s) 1000/T (1/ K) (b) 1.E+06 1.E+07 1.E+08 1.E+09 0.6 1.1 1.6 k (1/s) 1000/T (1/K) (d) 18 atoms that were not accounted for in conventional combustion mechanisms. Based on their findings, they proposed new oxidation pathways that are instigated by alternative isomerization of ȮOQOOH species to produce alkyl-dihydroperoxides (P(OOH) 2 ), followed by a third O 2addition, or β-scission, or the formation of hydroperoxy cyclic ethers, as shown in Fig. 3. Later, Wang and Sarathy showed the importance of the third O 2 addition pathway in simulating combustion in HCCI engines, and recommended the addition of this pathway to mechanisms of large alkanes (> C 6). Therefore, alternative isomerization and subsequent reactions of P(OOH) 2species were added to our updated iso-octane model. Figure 8. Conventional and alternative isomerization for one of the ȮOQOOH species Since the added reactions have only been recently postulated, no information exists in the literature regarding their kinetics. Therefore, we used analogies and estimations to assign rate coefficients of these reactions. For alternative isomerization and β-scission of P(OOH) 2 species, we used Villano et al.’s rate coefficients for RO Ȯ/QOOH isomerization and β-scission of QOOH, respectively. Meanwhile, Miyoshi’s rate coefficients of cyclic ether formation were used for the formation of hydroperoxy cyclic ethers. Finally, Miyoshi’s coefficients for alkyl radical addition to O 2 were used for the third O 2 addition (P(OOH) 2 + O 2 ), with an A-factor divided by two, similar to the second addition to O 2 reactions. 19 The subsequent reaction pathway of ȮOP(OOH) 2 species is isomerization to keto-dihydroperoxides (KDHP), for which Sharma et al.’s KHP formation rate coefficients were used. Similar to KHP, the KDHP species decompose via O–OH bond scission. However, unlike the decomposition of KHP, for which the activation energy is 42.3 kcal mol –1 , the activation energy of KDHP decomposition was chosen as 39 kcal mol –1 following the recommendations of Sarathy et al. . This value was assigned to better match experimental ignition delay profiles of iso-octane. 2-4 Performance of updated model The addition of all of the aforementioned reaction pathways and rate constant updates led to a model that is too fast at low and intermediate temperatures, compared to experimental data , with an example comparison shown in Fig. 9. Figure 9. Simulated IDT using the updated and untuned model compared against experimental data from literature circles and open squares at 40 atm, stoichiometric iso-octane/air mixture Further modifications to the rate rules, based on ȮH radical sensitivity analysis performed using CHEMKIN-PRO , were required to achieve better agreement between model-simulated and experimental ignition delay time profiles. Figure 10 shows that the QOOH radical plus O 2reactions exhibit high and positive sensitivity coefficients, meaning that they promote the production of ȮH radicals and increase reactivity. Meanwhile, the competing pathways of cyclic ether formation from QOOH radicals have negative sensitivity coefficients and decrease 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 0.7 0.9 1.1 1.3 1.5 IDT (μs) 1000/T (1/K) φ=1 40 atm 20 reactivity. The formation of cyclic ethers is a chain propagation pathway while the O 2 addition is a chain branching pathway, which illustrates their opposite effect on reactivity. The rate constants of these sensitive reactions were modified within their uncertainty limits, and model agreement with experimental data was tested after each modification. The best agreement was obtained when the rate constants of cyclic ether and hydroperoxide cyclic ether formation reactions for iso-octane moieties were doubled. The tuned reaction classes are shown in Table 2. Table 2. Modifications to the rate parameters to fit the experimental data for iso-octane. Rate Reference Modification QOOH = cyclic ether + ȮH A-factor × 2 P(OOH) 2 = hydroperoxy cyclic ether + ȮHKDHP = ȮH + β-scission products A-factor × 2 Ea=39 kcal Figure 10. Sensitivity analysis to ȮH radicals at 750 K,40 atm, φ =1 at just before ignition. Rates of cyclic ethers formation and second O 2 addition are shown in dash-outlined and solid-outlined boxes, respectively. -10 -8 -6 -4 -2 0246 IC8H18+OH<=>CC8H17+H2O AC8H17O2<=>AC8H16OOH ‐CAC8H15(OOH)BD<=>D(OOH)ETRAB+OH BC8H17O2<=>IC8H16+HO2 DC8H17O2<=>JC8H16+HO2 AC8H16OOH ‐B=>IC8ETERAB+OH AC8H16OOH ‐C=>IC8ETERAC+OH DC8H16OOH ‐B=>IC8ETERBD+OH BC8H16OOH ‐D=>IC8ETERBD+OH AC8H15(OOH)AB<=>A(OOH)ETRAB+OH DC8H16OOH ‐BO2=>AC8H15(OOH)BD BC8H16OOH ‐A=>IC8ETERAB+OH BC8H16OOH ‐D+O2<=>BC8H16OOH ‐DO2 DC8H16OOH ‐B+O2<=>DC8H16OOH ‐BO2 BC8H16OOH ‐A+O2<=>BC8H16OOH ‐AO2 IC8H18+OH<=>BC8H17+H2O AC8H16OOH ‐B+O2<=>AC8H16OOH ‐BO2 IC8H18+OH<=>DC8H17+H2O IC8H18+OH<=>AC8H17+H2O Sensitivity Coefficient 21 3- Experimental 3-1 UCONN RCM Experimental ignition delay time measurements at low and intermediate temperatures were conducted using rapid compression machines (RCM) at UCONN, NUIG and at KAUST. The UCONN RCM consists of a creviced piston that is driven pneumatically, and brought to rest hydraulically towards the end of compression. Compression occurs in a single stroke and the compression time is about 30–40 ms. The compression ratio can be varied by changing the stroke length and clearance volume independently. Pressure time history from the start of compression to post-ignition is measured using a Kistler 6125C pressure transducer along with a 5010B charge amplifier. Compressed pressures as high as 70 atm and pre-heated temperatures up to about 420 K can be attained in this RCM. Further details regarding this machine can be found in [72, 73] Homogeneous fuel air mixtures were prepared in a stainless steel mixing chamber. Fuel was injected into the mixing chamber on a gravimetric measure using a syringe, whereas oxygen and diluent gases were filled in on a barometric measure. Ultra high purity (>99.99%) gases supplied from Airgas were used to prepare the mixtures, except for iso-octane, which was supplied by Sigma-Aldrich at high purity (>99.8%) by Sigma-Aldrich. A magnetic stirrer at the bottom of the mixing chamber was used to ensure homogeneity of the prepared gas mixtures. The fuel-oxidizer mixtures were allowed to mix in the chamber for about 2.5 hours. The homogeneous fuel-air mixtures were compressed rapidly to the desired pressure and temperature. The compressed temperature was estimated using the adiabatic-core hypothesis. Subsequent to compression and after an induction period, a rapid rise in pressure was observed due to auto-ignition. Ignition delay time is defined as the time interval between the end of compression and the maximum rate of pressure rise. First- ( 1 ) and total ignition delays ( ) are defined similarly. This definition of ignition delay is illustrated in Fig. S4. A minimum of 5 concordant experimental runs were carried out for each data point reported. A representative experimental pressure trace close to the mean is chosen for reporting the ignition delay. The typical scatter in ignition delay time is ~10% of the reported value and the temperature uncertainty is ±5 K. Figure S5 demonstrates the typical repeatability. Inert runs wherein oxygen 22 was replaced by nitrogen were conducted at every set of temperature, pressure and equivalence ratio condition in order to infer the heat loss characteristics during the compression stroke and the post-compression period. Experiments were conducted in the low-to-intermediate temperature range with temperature at end of compression ( TC) varying between 640 K and 960 K at equivalence ratios of 0.4 and 1 in air. High compressed pressures of nominal pC near 20 and 40 atm are investigated herein. 3-2 KAUST RCM The rapid compression machine (RCM) facility at KAUST is based on a twin-opposed piston design similar to the one used at NUIG . The combustion chamber bore is 5.08 cm and, based on the final piston spacing, the volumetric compression ratio can be adjusted up to 16.8. The characteristic time required for the final 50% of the pressure rise during the compression phase is about 3 ms. The piston heads are creviced to minimize boundary layer mixing with the core gas. The gas mixtures are prepared in a 20 L stainless steel magnetically stirred mixing vessel. The temperature of the vessel and the combustion chamber can be increased using heating jackets. The compressed gas temperature at the end-of-compression (EOC) is computed by applying the following isentropic relation on the measured pressure trace: න γγ െ 1dTTT CT I = ln ൬p ஼pI ൰ where, the subscripts C and I refer to the EOC (end of compression) and initial conditions, respectively. The diluent was varied from pure nitrogen to a mixture of nitrogen and argon to reach temperatures up to 950 K and a mixture of nitrogen and carbon dioxide to reach temperatures down to 635 K. Six axisymmetric ports are present on the central axial location of the combustion chamber for optical access, gas exchange and the pressure transducer (Kistler 6045A). The ignition delay time for this work was determined from the maximum pressure rise in the measured pressure trace. The uncertainty in the ignition delay measurement was approximately +/- 10%. 23 3-3 NUIG RCM Low- to intermediate-temperature (650-1000 K) ignition delay times for stoichiometric mixtures of iso-octane in a 21:79 oxygen to diluent bath gas at nominal pressures of 15 near 15 and 20 atm were measured in two RCM facilities at NUIG; low compression piston head (LCPH) and high compression piston head (HCPH). The details of the experimental platform and the methodology used to measure fuel ignition delay times within the device have been extensively documented in [51, 75-78] therefore only a brief description of the apparatus is given here. Briefly, the device volumetrically compresses a test gas to a high temperature and pressure condition, and allows an isochoric reaction to proceed after compression within the high-energy gas. The compressed gas pressure is varied by adjusting the initial gas pressure in the reaction chamber, while the compressed gas temperature is varied by altering the initial temperature of the reaction chamber as well as by tailoring the composition (specific heat) of the non-reactive diluent gases. The transient pressure history of the test gas is monitored during the experiment by a Kistler 6045A pressure transducer that is mounted in the sidewall of the reaction chamber. Ignition delay times are measured from the test gas pressure history, and defined as the time difference between the local maximum in pressure that occurs near the end of volumetric compression and the maximum rate of pressure rise due to ignition. The widely used adiabatic core model [72, 79] is employed to evaluate the compressed gas temperatures from pressure histories. Furthermore, the adiabatic core model is also used to transiently prescribe the energy supplied to or lost from the test gases due to volumetric compression and heat losses, respectively, in order to adequately compare experimental data sets with mechanism simulations. These transport processes are isolated from chemical ones by compressing an analogous non-reactive gas for each experimental condition (where the reactive gas oxygen fraction is substituted by nitrogen), and quantified by deriving the specific volume history of the adiabatic core gas from the measured pressure history of the non-reactive experiment that is then used as an input boundary condition for a simulated variable volume 0-D reactor. The uncertainty in the ignition delay measurement is approximately +/- 15%. The fuel burned in this study was provided by Sigma-Aldrich (99.8% pure), while nitrogen (99.95%), oxygen (99.5%) and argon (99.5%) gases were supplied by BOC Ireland. 24 3-4 KAUST HPST The high-pressure shock tube (HPST) facility has been previously described in detail in , so only a brief overview is given here. The facility consists of a stainless steel, electropolished tube divided by aluminum diaphragms into a 6.6 m driven section and a modular driver section that can be extended up to 6.6 m. The inner diameter of the tube is 10 cm. The measurements were conducted at the reflected shock conditions via a pressure transducer (Kistler 6045A) mounted at a distance of 1 cm from the shock tube endwall. The reflected shock conditions are determined from the incident shock wave velocity measurements using five axially spaced pressure transducers (PCB113B26) over the last 3.2 m length of the driven section. The reactant mixture was prepared in a 40 L stainless steel mixing vessel equipped with a magnetic stirrer. The vessel and shock tube were heated to 80 0C for these experiments using heating jackets to prevent fuel condensation. As with the RCM, the ignition delay time is determined by the time of maximum pressure rise. The uncertainty in the ignition delay measurement is +/-20 %. A non-ideal pressure rise ( dp/dt ) of approximately +3% /ms was observed behind the reflected shock wave in these experiments. This non-ideality is accounted for in numerical shock tube simulations using a representative volume profile. Example pressure traces for shock tube ignition delay measurements are illustrated in Fig. S6 of the Supplementary material. Several pressure traces at lower temperatures (i.e., longer ignition delay times) exhibited significant pre-ignition pressure rise that may be caused by inhomogeneous ignition [81, 82]. 4- Results and Discussion 4-1 Model validation against new experimental data The chemical kinetic model was compared against experimental data at a range of conditions, including newly acquired data and data available in the literature, as shown in Table 3. The experimentally measured ignition delay time profiles were simulated using the closed homogeneous batch reactor model in CHEMKIN-PRO software along with the updated iso-octane model. In order to account for heat losses, RCM experiments were simulated using experimentally determined variable volume profiles. These volume profiles are available as supplementary material. 25 Table 3. Summary of experimental conditions employed for validating iso-octane model. Technique Pressure (atm) Equivalence Ratio φ Temperature range (K) % Fuel Diluent Reference ST ignition 50 0.5,1 700-1200 100% air [9, 10] ST ignition 10-50 0.25,0.5, 1 870-1250 100% air ST ignition 14,17,50 0.5, 1 855-1170 100% air ST ignition 1.4 0.5,1,2 1250-1810 0.25%,0.5%,1% Ar ST ignition 20 1 893-1060 100% air KAUST JSR oxidation 10 0.3,0.5,1,1.5 750-1100 0.1% N2 Laminar flame speed 1 0.6-1.7 298,355 100% air [21, 24-26, 28, 29] Laminar flame speed 10 0.8-1.1 323,373,423,473 100% air RCM ignition 15,20 1635-787 100% 100% N 2NUIG 765-965 100% 25%N 2/75%Ar RCM ignition 20,40 0.4,1 632-958 100% 100% N 2 UCONN RCM ignition 20 1632,655 100% 80%CO 2/20%N 2KAUST 666-813 100% 100% N 2851-948 100% 20%N 2/80%Ar ST pyrolysis 25,50 900-1700 137, 149 ppm Ar This study Figure 11 shows a comparison of the simulated ignition delay time (IDT) profiles to those measured experimentally using the RCM facility at UCONN. Simulations were conducted using the updated iso-octane model developed in this study, as well as the original model developed by Mehl et al. . As shown in Fig. 11, the updated model matches the experimental data better than the original one, particularly at φ = 0.4. Matching the reactivity under fuel-lean conditions is crucial for HCCI engine simulations because those engines are typically operated at equivalence 26 ratios of less than 0.5. The updated model under predicts the measured ignition delay times by nearly a factor of two at the lowest temperature conditions at φ = 0.4, 40 bar and φ = 1, 20 and 40 bar. At φ = 0.4, 20 bar, the model slightly over predicts the measured ignition delay time. Despite the improved agreement between the updated model and experimental data at φ = 0.4 and pC = 20 atm, the updated model shows higher reactivity at φ = 1 and pC = 20 atm for temperatures higher than 750 K. This discrepancy warranted further investigation. In order to ensure the validity of RCM experiments conducted at UCONN, particularly at conditions where significant discrepancies were observed between experimental and simulated data, the same experiments ( φ = 1 and p C = 20 atm) were run using RCM facilities at NUIG and at KAUST. However, as mentioned earlier, experiments at UCONN were conducted in air (O 2 /N 2 ), whereas Argon was added to Nitrogen bath gas at NUIG and at KAUST to achieve high temperatures at EOC (end of compression). Figure 12a shows that for temperatures greater than 850 K, NUIG and KAUST RCM ignition delay data do not match those recorded at UCONN. This discrepancy is a attributed to the diluent effect, which was thoroughly discussed by Würmel et al. ; they argued that substituting nitrogen by argon increases the ignition delay time in RCM experiments due to the cooling effect of argon in the post compression period. As shown in Fig. 12, for temperatures greater than 850 K the ignition delay times measured by NUIG (diluent gas is 25% N 2 and 75% Ar) and KAUST (diluent gas is 20% N 2 and 80% Ar) are longer than those measured by UCONN (diluent gas is only N 2 ). The diluent effect is not observed at 630 K and 655 K where 80% CO 2 /20% N 2 were used as diluent at KAUST compared to 100% N 2 at NUIG and UCONN. This is largely due to the lower heat loss due to the lower thermal diffusivities (with correspondingly lower thermal conductivities and higher heat capacities) of N 2and CO 2 .The diluent effect was not observed in simulated ignition delay times when comparing those under NUIG and UCONN conditions, as shown in Fig. 12a. For example, simulated ignition delay times using the inert volume profiles and bath gas conditions from UCONN and NUIG show similar ignition delay times at all temperatures. Simulations at the KAUST conditions (i.e., volume profiles and inert composition) show longer ignition delay times compared to the UCONN simulations due to the use of Ar as the diluent, but the differences are not as prominent as those observed experimentally. A comparison of the inert pressure and temperature traces at 27 850 K from UCONN (O 2 /N 2 ) , NUIG (O 2 / N 2/ Ar) and KAUST (O 2 / N 2/ Ar) is shown in Fig. S7 in the Supplementary Material. It is interesting to note that the pressure and temperature profiles after TDC are similar for both facilities, which indicates similar heat-loss characteristics. Figure 11 . Iso-octane/air ignition delay times measured at the UCONN RCM facility and the corresponding simulations. Left panels are 20 atm while right panel are 40 atm. Upper panels are 0.4 equivalence ratio and lower panels are stoichiometric. Solid and dashed lines represent simulations using the updated model and original model , respectively. The insets show first stage IDT profiles. Wagnon and Wooldridge studied the effect of buffer gas composition on autoignition of stoichiometric mixtures of three different fuels; iso-octane, n-heptane and n-butanol. Their results emphasized the effect of the bath gas composition on low temperature combustion 847.5 898 948.5 330 300 1.02 1.07 1.12 1.17 Temperature (K) IDT (ms) 1000/T (1/K) 625 675 725 110 100 1.38 1.48 1.58 Temperature (K) IDT (ms) 1000/T (1/K) 110 100 1.3 1.4 1.5 1.6 φ =0.4 20 atm φ =0.4 40 atm 625 725 825 110 100 1.1 1.3 1.5 Temperature (K) IDT (ms) 1000/T (1/K) 110 100 1.3 1.5 φ =1 20 atm 625 645 665 110 100 1.45 1.5 1.55 1.6 Temperature (K) IDT (ms) 1000/T (1/K) φ =1 40 atm 28 chemistry through H Ȯ2 radical and H 2 O2 recombination and decomposition reactions, which are pressure dependent. They showed that different bath gases have different collision efficiencies, which affects the second stage ignition delay time. They compared experimental data for n-heptane ignition using different diluents (Ar and N 2 ) and showed that ignition is faster in the presence of Ar at high temperatures (>750 K); however their simulations were not able to capture the same difference. Figure 12a shows a typical trend for the reactivity of alkanes. At low temperature, reactivity increases with increasing temperature and then starts to be inversely proportional to temperature. This region is known as the negative temperature coefficient regime (NTC), which is characterized by the competition between chain branching reactions, mainly isomerization of ȮOQOOH compounds (production of keto hydroperoxides), and chain propagation reactions, such as the formation of cyclic ethers and olefins from QOOH species (QOOH = cyclic ether + ȮH and QOOH = olefin + H Ȯ2 ). After the NTC region, reactivity is more controlled by high-temperature chemistry and HȮ2 radical chemistry, and again increases with increasing temperature. Figure 12. Iso-octane/oxygen/diluent ignition delay times at φ = 1 and 20 atm from (a) various RCM facilities and (b) KAUST ST and RCM. Solid lines are variable volume simulations while dashed lines are constant volume simulations in (b). Open and closed circles represent RCM experiments with only N 2 and N 2 +Ar bath gases, respectively. Closed squares represent RCM 625 700 775 850 925 1000 110 100 11.2 1.4 1.6 Temperature (K) IDT(ms) 1000/T(1/K) UCONN NUI Galway KAUST 625 775 925 1075 1225 0.1 110 100 0.8 11.2 1.4 1.6 Temperature (K) IDT(ms) 1000/T(1/K) (b) KAUST affected by preignition (a) 29 experiments with N 2 +CO 2 diluent while open squares in (b) represent KAUST shock tube experiments. Dashed circle identified shock tube experiments that exhibited significant pre-ignition heat release. See digital paper for color version of this figure To further understand discrepancies between simulated and experimental measurements, ignition delay times measured in the KAUST HPST are presented in Fig. 12b and compared with the KAUST RCM data and simulations. The HPST simulations include 3% per millisecond ( dp/dt )pressure rise rate. An interesting observation in Fig. 12b is a factor of 3 discrepancy in IDT between the HPST and RCM experiments at an intermediate temperature TC of ~ 870 K. This phenomenon was noticed in previous studies [38, 84, 85] of other high-octane fuels, but the reason for this difference is not yet identified. It may be attributed to the invalidity of the homogeneous core assumption of the RCM due to mass loss in the facility crevices or heat loss through walls. Inhomogeneity and boundary layer phenomenon in the shock tube may also contribute to the discrepancy. The updated kinetic model captures the IDTs measured by the HPST at temperature above 900 K, but the model is more reactive when compared to the RCM data these temperatures. It is worth restating that HPST experiments at low temperatures (933 K, 928 K and 894 K see dashed circle in Fig. 12b) exhibited non-ideal pre-ignition pressure rise, as shown in Fig. S5 in the Supplementary material. These conditions exhibited significant pre-ignition pressure rise (above 3%/ms) due to inhomogeneous ignition phenomenon [81, 82]. Various shock tube groups, including KAUST, are exploring such inhomogeneous ignition phenomena in detail to understand their causes, their effect on ignition measurements and potential remedies. Such discussions and illustrations are beyond the scope of the current work. It should be noted that experimental measurements exhibiting pre-ignition pressure rise cannot be accurately modeled using 0-D simulations, and therefore are not suitable targets for kinetic model validation/optimization. The performance of the updated model is also compared to a recently optimized model by Cai and Pitsch in which the rate rules assigned by Curran et al. were calibrated to match the ignition delay time for iso-octane in , which limits the agreement of the model to a broader set of target data. The simulation results from both models compared to the new RCM experiments from UCONN are shown in Fig. S8 in the Supplementary material. The present model reproduces the RCM experimental data better than the optimized model of Cai and Pitsch 30 . The optimized model could be improved by including the new data as targets for the model optimization. Figure 13 shows the ignition delay time for stoichiometric iso-octane mixtures using different diluents at 15 atm and 20 atm in the NUIG RCM. Decreasing the pressure decreases the reactivity due to lower rate of production of ȮH radicals from H 2 O2 (+M) = ȮH + ȮH (+M), a pressure dependent reaction. In addition, high pressure promotes low temperature chemistry, specifically alky radicals addition to O 2 . Overall, the model shows good agreement with the experimental data at low and intermediate temperatures; however, the model is more reactive than the experiments at higher temperatures. Figure 13. Iso-octane/air ignition delay times at φ =1 and 15 atm and 20 atm. Closed circles and straight lines represent experiments and simulations respectively with only N 2 as diluent in LCPH RCM. Triangles and dotted lines represent experiments and simulations in HCPH RCM with only N 2 as diluent. Open circles and dashed lines represent experiments and simulations with 20% N 2 and & 80% Ar diluent. 4-2 Model validation against experimental data in literature 4-2-1 Shock tube ignition delay data The updated chemical kinetic model was used to simulate shock tube experiments from the literature. The simulations were carried out using the constant volume batch reactor code in 625 700 775 850 925 1000 110 100 1000 11.2 1.4 1.6 Temperature (K) IDT (ms) 1000/T (1/K) 15 bar 20 bar 31 CHEMKIN-PRO assuming homogeneous and adiabatic conditions behind the reflected shock wave. Hartmann et al. and Fieweger et al. measured ignition delay times of iso-octane/air mixtures at 40 atm and φ = 0.5 and 1.0 in a high pressure shock tube. Figure 14 shows the experimental and simulated ignition delay time profiles over the temperature range 700–1200 K. The reactivity at this temperature range is sensitive to the equivalence ratio of the mixture because chain branching is dominated by KHP production. The leaner the mixture, the lower the fuel concentration and the less KHP produced, which results in a less reactive system . Figure 14 also shows that the present model offers improved agreement with experimental data when compared with the original model by Mehl et al. . Figure 14. Iso-octane/air ignition delay times at 40 atm and equivalence ratios of 0.5 and 1.0. Experiments are from (circles) and (open squares). Simulations (lines) with the updated and original model are represented by solid and dashed lines, respectively. Shen and co-workers carried out ignition delay time experiments of iso-octane/air mixtures in a high-pressure shock tube at different equivalence ratios (0.25, 0.5 and 1.0) and at pressures of 10, 12, 25, 45 and 50 atm. Their experimental data, and those simulated by our updated model and the original model at the same conditions, are presented in Fig. 15. Both models match the experimental data relatively well at all investigated conditions, with better predictions for the updated model at high pressures and low temperatures. 700 800 900 1000 1100 1200 00110 100 0.8 11.2 1.4 Temperatur (K) IDT (ms) 1000/T (1/K) φ=1 φ=0.5 32 Figure 15. Iso-octane/air ignition delay times at various pressures and (a) φ = 1.0 , (b) φ = 0.5, and (c) φ = 0.25. Experiments are from (symbols). Simulations (lines) with the updated and original model are represented by solid and dashed lines, respectively. See digital paper for color version of this figure. Shock tube iso-octane oxidation experiments were also conducted by Davidson and coworkers for fuel/air mixtures at pressures between 14 atm and 59 atm, temperatures between 855 K and 1270 K, and equivalence ratios of 0.5 and 1. Figure 16 compares the experimental IDT profiles from to those simulated using the updated and original iso-octane models. The updated model is more reactive than the original one at high pressures and low temperatures, which gives better agreement with experimental data at these conditions, particularly at φ =1.0. 870 970 1070 1170 1270 5.E ‐02 5.E ‐01 5.E+00 0.75 0.85 0.95 1.05 1.15 Temperature (K) IDT (ms) 1000/T (1/K) 12 atm 45 atm 1429 1449 1469 1489 3.E ‐02 3.E ‐01 3.E+00 0.7 0.8 0.9 11.1 Temperature (K) IDT (ms) 1000/T (1/K) 12 atm 45 atm (c) 870 970 1070 1170 1.E ‐02 1.E ‐01 1.E+00 1.E+01 0.8 0.9 11.1 Temperature (K) IDT (ms) 1000/T (1/K) 10 atm 25 atm 50 atm (a) (b) 33 Figure 16. Iso-octane/air ignition delay times at two pressure and (a) φ = 1.0 and (b) φ =0.5. Experiments are from (symbols). Simulations (lines) with the updated and original model are represented by solid and dashed lines, respectively. Oehlschlaeger et al. used a shock tube to measure IDT for iso-octane/O 2 /Ar mixtures at 1.4 atm and 0.5 < φ < 2 using different fuel concentrations (0.25%, 0.5% and 1%). Figure 17 compares experimental data from to those simulated using the updated and original models. Both models show good agreement with the experimental data. However, for 1% fuel in air mixtures, the updated model is more reactive than the original one, especially at lower temperatures, and shows better agreement with experimental data. 1000 1050 1100 1150 1200 1250 8.E ‐02 8.E ‐01 8.E+00 0.8 0.85 0.9 0.95 1 Temperature (K) IDT (ms) 1000/T (/K) (b) 14 atm 50 atm 833 933 1033 1133 1233 1.0E ‐01 1.0E+00 1.0E+01 0.8 0.9 11.1 1.2 Temperature (K) IDT (ms) 1000/T (/K) (a) 17 atm 50 atm 34 Figure 17. Iso-octane/O 2 /Ar ignition delay times at 1.4 atm, various equivalence ratios, and fuel concentrations of (a) 1%, (b) 0.25%, and (c) 0. 5%. Experiment are from (symbols). Simulations with updated and original models are represented by solid and dashed lines, respectively. 4-2-2 Jet Stirred Reactor data Dagaut et al. carried out iso-octane oxidation experiments in a JSR for 0.1% fuel/oxidizer mixtures at φ = 0.3, 0.5, 1.0 and 1.5. The experiments were performed at 10 atm and with a mean residence time of 1 s. Gas chromatography and mass spectrometry were used to identify and quantify intermediates formed during iso-octane oxidation. The updated iso-octane model was compared against species concentration profiles measured in using the perfectly stirred reactor code in CHEMKIN-PRO with a transient solver to obtain a steady-state solution. 35 Figures [18-21] show the experimental and simulation profiles for different species at the four experimentally investigated equivalence ratios. Qualitatively, the updated model matches the experimental concentration profiles of all species detected species which include iso-octane, CO, CO 2 , O 2 , CH 4 , H 2 , CH 2 O, C 2 H4 , C 3 H6 , iso-butene, 2,4-dimethyl-1-pentene, 2,4-dimethyl-2-pentene and 4,4-dimethyl-1-pentene. The profiles of iso-octane, carbon monoxide (CO), and carbon dioxide (CO 2 ) are well captured by the model at different equivalence ratios. Based on the shape of the iso-octane and CO profiles, neither the simulations nor experiments showed NTC behavior at the conditions studied. At all investigated conditions, CO concentrations are lower than those of CO 2 ; however, the concentration profiles of these species cross over at lean conditions as a result of oxidation of CO to form CO 2 at high temperatures. The leaner the mixture, the lower is the crossover temperature. In the presented data, the crossover temperature is 1000 K and 1050 K for φ = 0.3 and 0.5, respectively. The simulated concentration profiles of H2 , CH 4 and CH 2 O agree well with those measured experimentally at φ = 0.5, 1.0 and 1.5. For φ = 0.3, H 2 concentrations are under predicted by 51% whereas CH 2 O concentrations are over predicted by 36% at peak values. CH 4 concentrations are under-predicted below 900 K and over-predicted above 900 K. The concentration profiles of C 2 H4 and C 3 H6 are quantitatively well predicted by the model at fuel-rich and stoichiometric conditions; however, the peak concentrations are under-predicted by 50–61% and 41–44%, respectively, at fuel-lean conditions. The predicted profiles as a function are temperature for 2,4-dimethyl-1-pentene (XC 7H14 ), 2,4-dimethyl-2-pentene (YC 7 H14 ) and 4,4-dimethyl-1-pentene (PC 7 H14 ), the β-scission products of iso-octyl radicals, are generally broader that the experimental measurements. However, the maximum concentrations of YC 7 H14 and PC 7 H14 are under predicted with maximum deviation at φ = 0.3 of 52% and 73%, respectively. Also Figs. S9-S12 in the Suppllementary material show the simulated species profiles using the Mehl et al. model. Generally, the updated model shows a better agreement for all species. 36 Figure 18. JSR species concentration profiles (points) and simulations (lines) using updated model (straight lines) and original model (dashed lines) for 0.1 % iso-octane at φ =1.5. Experiments are from . See digital paper for color version of this figure. Figure 19. JSR species concentration profiles (points) and simulations using updated model (straight lines) and original model (dashed lines) for 0.1 % iso-octane at φ =1.0. Experiments are from . See digital paper for color version of this figure. 0123456750 850 950 1050 mole fraction (x10 3 )Temperature (K) IC8H18 CO CO2 0246810 12 14 750 850 950 1050 mole fraction (x10 4)Temperature (K) CH4 H2 CH2O 0123456750 850 950 1050 mole fraction (x10 4)Temperature (K) C2H6 C2H4 C3H6 IC4H8 0246810 12 14 16 18 750 950 mole fraction (x10 6)Temperature (K) YC7H14 XC7H14 PC7H14 0123456750 850 950 1050 mole fraction (x10 3)Temperature (K) IC8H18 CO CO2 01234567750 850 950 1050 mole fraction (x10 4)Temperature (K) CH4 H2 CH2O 012345750 850 950 1050 mole fraction (x10 4)Temperature (K) C2H6 C2H4 C3H6 IC4H8 0246810 12 14 750 850 950 1050 mole fraction (x10 6)Temperature (K) YC7H14 XC7H14 PC7H14 37 Figure 20. JSR species concentration profiles (points) and simulations using updated model (straight lines) and original model (dashed lines) for 0.1 % iso-octane at φ =0.5. Experiments are from . See digital paper for color version of this figure. Figure 21. JSR species concentration profiles (points) and simulations using updated model (straight lines) and original model (dashed lines) for 0.1 % iso-octane at φ =0.3. Experiments are from . See digital paper for color version of this figure. 0123456600 800 1000 mole fraction (x10 3)Temperature (K) IC8H18 CO CO2 01122334750 850 950 1,050 mole fraction (x10 5)Temperature (K) CH4 H2 CH2O 0510 15 20 25 30 35 750 850 950 1050 mole fraction (x10 5)Temperature (K) C2H6 C2H4 C3H6 IC4H8 0246810 12 750 850 950 1050 mole fraction (x10 6)Temperature (K) YC7H14 XC7H14 012345678750 850 950 1050 mole fraction (x10 3)Temerature (K) IC8H18 CO CO2 0510 15 20 25 750 850 950 1050 mole fraction (x10 5)Temerature (K) CH4 H2 CH2O 0112233750 850 950 1050 mole fraction (x10 4)Temperature (K) C2H6 C2H4 C3H6 IC4H8 0246810 12 750 850 950 1,050 mole fraction (x10 6)Temperature (K) YC7H14 XC7H14 PC7H14 38 4-2-3 Laminar Flame Speed In addition to IDT measurements, the updated model was tested against laminar flame speed data available in the literature carried out at atmospheric pressure [21, 24-26, 28, 29] and high pressure . Flame speeds depend strongly on the base chemistry; however, different fuels produce different concentrations of C 1 -C 4 species and this impacts the flame speed. In their analysis, Ji et al. showed that a high concentration of iso-butene is produced during the oxidation of iso-octane, mainly from the β-scission of iso-octyl radicals, which affects its flame speed. To minimize computational time and cost, the proposed updated mechanism was reduced and only high temperature reaction pathways were retained. The model was reduced using the method of direct relation graph with expert knowledge (DRG-X) . This approach assumes that some species are weakly coupled to others, so they do not play a significant role in combustion processes, and as such, they can be removed whilst retaining the chemical fidelity of the remaining species. Also, contrary to the original DRG approach that is restricted by a uniform error tolerance for all the species, the DRG-X method allows a separate controlled reduction error for heat release rate and species of interest. Therefore, the DRG-X method can result in a smaller reduced mechanism with similar chemical fidelity compared to the original DRG method. In this work, the detailed mechanism is reduced by specifying the error tolerance for heat release as 0.01, for Ḣ atoms and ȮH radicals as 0.1, for iso-octane as 0.5, and for other species as 0.4. The target temperatures and pressures for the reduction are 1000-2300 K and 1-10 atm respectively. A skeletal mechanism with 189 species was generated from the detailed mechanism. The transport parameters for some species in the reduced model were estimated by analogy, except for the Lennard-Jones collision parameters that were estimated using critical temperature, critical pressure, and acentric factor properties based on Tee et al.’s method. The reduced mechanism and the transport file are provided as Supplementary Material. The simulations were carried out on CloudFlame [89-91], a cyber-infrastructure that presents a web front-end for Cantera . The premixed laminar flame speed module was used with a domain size of 5 cm and a initial base grid of 20 points; the final solution exhibited grid-convergence with approximately 260 grid points. Figure 22 shows the experimental and simulation results for flame speeds measured at 1 atm and at 298 K and 353 K. Simulations show 39 good agreement with experimental data for fuel-lean conditions at both temperatures; however the model under predicts flame speeds at fuel-rich conditions at 298 K. In general, the model matches well with the experimental data at 355 K [24,29] across the entire range of equivalence ratios when considering the typical experimental uncertainty of ± 2 cm s –1 .In addition, the model was used to simulate high pressure flame speed experiments carried out at 10 atm and temperatures of 323 K, 373 K, 423 K and 473 K with a high-speed shadowgraph system . Figure 23 shows that the updated model agrees within the experimental uncertainty to the experimental data of . Figure 22. Iso-octane/air premixed laminar flame speeds at 1 atm and (a) 298 K and (b) 355 K. Experimental data (symbols) are given for open squares , closed triangles , crosses , open triangles , open circles , and closed squares . Solid lines represent simulations using the updated model. 0510 15 20 25 30 35 40 0.4 0.9 1.4 Laminar Flame Speed (cm/s) Equivalence Ratio (a) 10 15 20 25 30 35 40 45 50 0.6 0.8 11.2 1.4 1.6 Laminar Flame Speed (cm/s) Equivalence Ratio (b) 40 Figure 23. Iso-octane/aire premixed laminar flame speeds at 10 atm initial pressure and different initial temperatures. Experiments (symbols) are from . Solid lines represent simulations using the updated model.See digital paper for color version of this figure. 4-2-4 Counterflow Diffusion Flame Numerical simulations were also carried out to compare the skeletal model against ignition temperature measurements by at different strain rates and iso-octane mass fractions. The experiment were carried out in counterflow diffusion flames. The fuel stream was comprised pre-vaporized isooctane in nitrogen diluent, while the oxidizer stream was air. Counterflow flame ignition simulations were performed using the OPPDIF solver in CHEMKIN-PRO . A temperature profile was first established with cold mixtures at the oxidizer and fuel inlets, and then the temperature of the oxidizer stream was gradually raised until ignition. The composition of the fuel and the oxidizer streams and the temperature of the fuel stream, T1, were maintained constant while carrying out this procedure [93, 94]. The calculations were carried out with thermal diffusion, mixture-averaged transport and convergence parameters of GRAD = 0.25 and CURVATURE = 0.25. Overall, the model captured the trend in the experiment, with a maximimum 10 K difference observed in the ignition temperatures, as shown in Fig. 24. 10 15 20 25 30 35 40 45 0.75 0.8 0.85 0.9 0.95 11.05 1.1 1.15 Lamiar Flame Speed ( cm/s) Equivalence Ratio 473 K 423 K 373 K 323 K 41 Figure 24. Iso-octane/air counterflow diffusion flame ignition temperature. Experiments (symbols) are from . Solid lines represent simulations using the updated model. 4-2-5Pyrolysis in shock tube The updated model was also tested against shock tube pyrolysis experiments from Malewicki et al. . Generally, pyrolysis mechanisms consist of unimolecular decomposition of fuel, β-scission, isomerization of alky radicals reactions and abstraction by CH 3 . The first 3 reaction classes were not updated in this study even though these classes have been recently studied. Wang et al. calculated the rate coefficients of radical isomerization reactions at the CBS-QB3 level of theory. Their calculated values are similar to those used in the original model with a maximum difference in rate constants of 13% at 1200 K. The rates are compared in Fig. S13 in the Supplementary material. Meanwhile, the kinetics of unimolecular decomposition of iso-octane has been computationally studied by Ning et al. . Their calculated rate constants are much higher than those of Tsang et al. that are used in the original mechanism. Both sets of values have been tested against the experimental pyrolysis concentration profile of iso-octane measured by . Malewicki et al. conducted pyrolysis experiments for diluted iso-octane mixtures (137 ppm and 149 ppm) in argon at 25 atm and 50 atm, and at different reaction times over the temperature range of 900–1700 K in a high pressure shock tube. They measured the concentrations of stable species using gas chromatography. Figure 25 compares the experimental concentration profile of 1200 1210 1220 1230 1240 1250 1260 1270 200 300 400 500 600 Ignition Tempearture (K) Strain Rate (1/s) 1240 1245 1250 1255 1260 1265 1270 1275 1280 0.2 0.3 0.4 0.5 Ignition Tempearture (K) Fuel mass fraction 42 iso-octane to those simulated using unimolecular decomposition rate constants from Tsang et al. and from Ning et al. . The figure clearly shows that the Tsang et al.’s values are needed in order for the model to match the experimental profile. Therefore, Tsang et al.’s values were retained in the updated mechanism. Figure 25. Experimental (symbols) and simulated (lines) concentration profiles of iso-octane pyrolysis at 25 atm (grey) and 50 atm (black). Solid and dashed line represents simulations using unimolecular decomposition rates from and , respectively. Figure 26 shows the concentration profiles of the major measured species at 25 atm and 50 atm. The model matches the experimental concentration profiles of all detected species, except for those of acetylene (C 2 H2 ) and methane (CH 4 ), which are over-predicted and under-predicted, respectively. 020 40 60 80 100 120 140 160 900 1000 1100 1200 1300 Concentration (ppm) Temperature (K) 43 Figure 26. Concentration profiles of propene (C 3 H6 ), iso-butene+1-butene (C 4 H8 ), propyne (C 3 H4 -P), ethane (C 2 H6 ), methane (CH 4 ) and acetylene (C 2 H2) produced from iso-octane pyrolysis at 25 atm (grey) and 50 atm (black). Symbols and lines represent experiments and simulations, respectively. 4-3 Sensitivity Analyses In order to elucidate the reactions responsible for iso-octane ignition under different conditions, brute force sensitivity analyses were carried out using CHEMKIN-PRO for stoichiometric iso-octane/air mixtures at 20 atm and at 650 K, 750 K and 850 K. 020 40 60 80 100 120 800 1000 1200 1400 1600 1800 Concentration (ppm) Temperature (K) C3H6 020 40 60 80 100 120 800 1000 1200 1400 1600 Concentartion (ppm) Temperature (K) C4H8 050 100 150 200 250 300 800 1000 1200 1400 1600 1800 Concentration (ppm) Temperature (K) C2H2 010 20 30 40 50 60 70 80 900 1100 1300 1500 1700 Concentration (ppm) Temperature (K) C2H6 010 20 30 40 50 60 1000 1200 1400 1600 1800 Concentration (ppm) Temperature (K) C3H4 ‐P 050 100 150 200 850 1050 1250 1450 1650 1850 Concentration (ppm) Temperature (K) CH4 44 First, ȮH and H Ȯ2 radicals sensitivity analyses were performed at each condition to identify reactions that should be considered for brute force sensitivity analysis. The sensitivity coefficient of every reaction, at each temperature condition, was then calculated using the following equation: ൌ ߪ logሺ߬ ߬ଶ ଴.ହ ሻlogሺ 20.5 ሻ where τ2 and τ0.5 are the ignition delay times computed with rate constant multiplied and divided by a factor of two, respectively. A positive sensitivity coefficient ( σ) indicates that the specified reaction increases the simulated ignition delay time, and thus, decreases reactivity, and vice versa. Plots of ȮH and H Ȯ2 radicals sensitivity analyses are available as Supplementary Material, Figs. S14–S19. The figures demonstrates that, both analyses show the same important reactions. Figure 27 shows the results of the brute force sensitivity analyses at 650, 750 and 850 K. At all the considered temperatures, the ignition delay time is most sensitive to rates of H-atom abstraction by ȮH radicals. However, abstractions from primary (A Ċ8 H17 and D Ċ8 H17 ) and secondary (B Ċ8 H17 ) sites increase reactivity, whereas abstractions from the tertiary site (C Ċ8 H17 )decreases reactivity due to the lack of low-temperature chain-branching paths from this site and the production of relatively unreactive olefins, as will be discussed in the next section. The sensitivity analysis also shows that the rate of H-atom abstraction from the secondary site by HȮ2 radicals is more important than ȮH radicals only at 850K, where the high activation barrier of H-atom abstraction by H Ȯ2 radicals is overcome. The isomerization of AC 8 H17 Ȯ2 to A Ċ8 H16 OOH-C has negative effect on sensitivity because these reactions compete with the formation of A Ċ8 H16 OOH-B with which addition to O 2 is very promoting (Fig. 27). Also the formation of H 2 O2 molecules from H Ȯ2 radical self-reaction exhibits a positive sensitivity, as this is a chain termination step where H Ȯ2 radicals form two stable species. The formation of cyclic ethers from hydroperoxy alkyl radicals (QOOH) exhibits a positive sensitivity coefficient at all temperatures while its competing pathway; addition to O 2 (second O 245 addition) shows a negative sensitivity coefficient . The effect of both classes is more significant at 750 K (NTC region). Moreover, the rates of production of hydroperoxy cyclic ethers (e.g. D(OOH)ETRBD, Fig. 27) and 3 rd addition to O 2 (e.g. C8(OOH)BD-DO 2 ) showed a positive and negative sensitivity for the low temperature case (650 K), respectively. The appearance of these rates in the sensitivity analysis manifests the importance of these added pathways. Figure 27. Brute force sensitivity analyses for ignition delay of stoichiometric iso-octane/air mixtures at 20 atm and different temperatures. (See Supplementary Material for species dictionary) 4-4 Rate of production analysis Rate of production (ROP) analysis for stoichiometric iso-octane/air mixture at 20 atm, 750 K, and 20% fuel conversion is presented in Fig. 28. The ROP demonstrates that the major radical produced from iso-octane is A Ċ8 H17 due to the presence of nine equivalent hydrogen atoms at this site. Also the figure shows that the predominant pathway for alkyl radicals at 750 K is addition to molecular oxygen. Once the alkylperoxy radical is formed, it either undergoes ‐1‐0.5 00.5 IC8H18+OH<=>DC8H17+H2O IC8H18+OH<=>AC8H17+H2O IC8H18+HO2<=>BC8H17+H2O2 IC8H18+OH<=>BC8H17+H2O AC8H16OOH ‐B+O2<=>AC8H16OOH ‐BO2 BC8H16OOH ‐A+O2<=>BC8H16OOH ‐AO2 DC8H15(OOH)BD+O2<=>C8(OOH)BD ‐DO2 DC8H15(OOH)BD<=>D(OOH)ETRBD+OH DC8H16OOH ‐B=>IC8ETERBD+OH AC8H16OOH ‐B=>IC8ETERAB+OH AC8H17O2<=>AC8H16OOH ‐C2HO2<=>H2O2+O2 IC8H18+OH<=>CC8H17+H2O σ 850 750 650 46 concerted elimination to form an olefin + H Ȯ2 radical or it isomerizes to hydroperoxy alkyl (QOOH) radicals. For example, 98 % of the formed tertiary peroxy radical (C Ċ8 H17 ) follows the concerted elimination pathway, which agrees with the findings of Snitsiriwat and Bozzelli , who argued that the main product of the tertiary iso-octyl radical + O 2 system is iso-octene + HȮ2 . The concerted elimination chain termination pathway of the tertiary peroxy radical is favored by the absence of 6-membered ring isomerization pathways. Hence the production of tertiary iso-octyl radicals inhibits reactivity. On the other hand, DC 8 H17 Ȯ2 , a primary iso-octyl peroxy radical, produces two different hydroperoxy alkyl radicals through 6-membered ring transition states, which are more favorable than concerted elimination due to the lower ring strain. The other primary iso-octyl peroxy radical, AC 8 H17 Ȯ2 , does not undergo concerted elimination due to the absence of beta hydrogens, and rather undergoes isomerization reactions involving six-membered ring transition states that lead to low temperature chain-branching and promote reactivity. None of the hydroperoxy alkyl radicals (QOOH) shown in Fig. 28 are β-QOOH radicals, therefore, the primary consumption pathways are: formation of cyclic ethers and addition to O 2 .In general, the pathway forming cyclic ethers shows a higher flux than addition to O 2 due to the low thermodynamic stability of the ȮOQOOH radicals. Once formed at the relatively high temperature investigated, they favor back dissociation to O 2 and QOOH. RO 2 isomerizations involving seven-membered ring transition states often lead to cyclic ether formation rather than low-temperature chain branching because these isomerizations have lower rate constants than those involving six-membered rings that lead to chain branching. This behavior is evident for the seven-membered ring RO 2 isomerization leading to A Ċ8 H16 OOH-C radicals (98% of these radicals form the cyclic ether (IC 8 ETERAC)). Figure 28 also shows the ROP of different species from the original model , which highlights the mechanisitic difference between both models. Generally, the rate of consumption of IC 8 H18 is lower using the original model when compared to the updated one, as previously demonstrated by the JSR profiles (Figs. S8-S11). The original model shows a higher flux going to C Ċ8 H17 compared to B Ċ8 H17, which contradicts the updated model. The updated model has different branching ratios due to the implementation of updated iso-octane+OH abstraction rates that account for NNN effects. Moreover, the original model shows that addition to O 2 is favored by the alkyl radicals except for C Ċ8 H17 where β-scission is more favored. This is attributed to the 47 absence of consumption pathways for the peroxy-hydroperoxy alkyls ( ȮOQOOH) with a tertiary hydroperoxy group, which prevents the initial tertiary radical from contributing towards radical chain branching. Figure 28. ROP analysis for stoichiometric iso-octane/air mixture at 20 atm-750 K and 20% fuel consumption corresponding to 788 K for the updated model in bold and 791 K for original model in italic . 4-5 Effect of Alternative Isomerization and 3 rd O 2 Addition Pathways on Reactivity Including the alternative isomerization pathway in the mechanism required the addition of 206 reactions and 100 new species to the model. Figure 29 demonstrates that the addition of this pathway increased the reactivity of the mechanism at low temperature and NTC. This agrees with the findings of Silke et al. and Mohamed et al. for n-heptane and 2-methylhexane models, respectively. These findings encourage more thorough investigation of other low temperature oxidation pathways, which may be relevant to combustion kinetic models. 48 Peroxy-hydroperoxy alkyls ( ȮOQOOH) with a tertiary hydroperoxy group do not form KHPs due to absence of -hydrogen atoms. In the original mechanism, these species were accumulated in the system, but in the current model they form alkyl-dihydroperoxides (P(OOH) 2 ). Figure 29. Ignition delay simulation at 20 atm and φ = 1.0 presented to demonstrate the effect of alternative isomerization pathways. To better understand the importance of the added pathways, ROP analyses were performed for the 3 P(OOH) 2 radical with the highest concentrations at a pressure of 20 atm, equivalence ratio of 1 , and temperatures of 650 K and 750 K , where 650 K shows behavior in the low temperature region and 750 K in the middle of the NTC region. The analyses were conducted at 20% of fuel consumption for each case and results are shown in Fig.30. Figure 30 shows that the formation of alkyl-dihydroperoxides from ȮOQOOH radicals may be more important than the formation of KHPs, depending on the size of the transition state rings. Alternative isomerization pathways with 6-membered ring transition states are the most favored due to lower ring strain energy compared to other isomerization reactions. However, in cases where the migrating hydrogen atom is at a tertiary site, the 7- membered ring transition state isomerization is more favorable than the 6-membered ring one that migrates an H-atom from the primary site, based on the reaction rate rules assigned in Section 2-3. For example, in Fig. 30-a , at 750 K, 51% of BC 8 H16 OOH-A Ȯ2 forms CC 8 H15 (OOH)AB via a 7-membered ring transition state compared to 28 % for the formation of AC 8 H15 (OOH)AB via a 6-membered ring transition state. 0110 100 1000 0.7 0.9 1.1 1.3 1.5 1.7 IDT (ms) 1000/T (1/K) With alternative pathway No alternative pathway 49 Figure 30 also shows that the reactivity of ȮOQOOH species at low and intermediate temperatures are very similar. However, temperature-dependent variations in ROP are observed for P(OOH) 2 species. The third O 2 addition reactions are more favorable at low temperatures. For example, 44% of AC 8 H15 (OOH)BD (Fig. 30(a)) undergoes 3 rd O2 addition at 650 K compared to 13.3% at 750 K. Similarly, 24.6% and 59.4% of DC 8 H15 (OOH)BD (Fig. 30-a) and AC 8 H15 (OOH)AB (Fig. 30-b) species undergo 3 rd O2 addition at 650 K compared to 4.55% and 13.1% at 750 K, respectively. It is notable that for the added alternative isomerization pathways both hydroperoxy cyclic ether (HPCE) formation and 3 rd O2 addition followed by formation of ketodihydroperoxides (KDHP) are chain branching pathways. Although H Ȯ2 radical elimination reactions from P(OOH) 2 species have been added to the model, they do not affect reactivity as shown in Fig. 30. For iso-octane, hydroperoxy olefins may only be produced from P(OOH) 2 species via 5-membered ring transition states that are energetically un-favored. Figure 30 also shows that the H-transfer from the hydroperoxy group attached to a primary site to the peroxy group attached to a secondary site via H-exchange pathway is more favorable than the opposite direction. Figure 30. Flux analysis for alkyl-dihyroperoxides with the OOH on the (a) B-D sites and (b) A-B sites for stoichiometric iso-octane/air mixture at 20 atm at 20% fuel consumption. Values in 50 italic and bold represent ROP percentages at 650 K (Low temperature region) and 750 K (NTC region), respectively. 5- Conclusions This paper presents a comprehensive chemical kinetic model for iso-octane, an important gasoline primary reference fuel surrogate. The iso-octane thermochemistry and mechanism were thoroughly updated using newly evaluated group values, reaction pathways, and rate rules. An alternative pathway for the isomerization of peroxy-alkyl hydroperoxides ( ȮOQOOH) was included in the updated model along with 3 rd O2 addition reactions. The new model was compared against new rapid compression machine and high pressure shock tube experiments. The new experiments covered a range of temperatures and pressures that were not previously available in literature. Additional comparisons were presented against ignition delay, laminar flame speed, counterflow diffusion flame ignition, and speciation measurements available in the literature. The present model is compared against experimental data acquired across atemperature range of 630-1800K, pressure range of 10-50 atm, and equivalence ratio range of 0.25-2. The present model shows improved agreement at lower equivalence ratios when compared to other mechanisms available in literature. Thus, the present model is appropriate for simulating engine combustions modes operating at lean conditions, such as various low temperature combustion engines. Sensitivity analysis showed that the model is highly sensitive to H-atom abstraction from fuel molecules by ȮH radical. Flux analysis showed that the main pathway available for tertiary iso-octyl peroxy radical is concerted elimination, and this decreases the reactivity. The importance of the additional reaction pathways was also investigated. The addition of alternative isomerization increases the model reactivity, as all subsequent reaction pathways result in chain branching. The 3 rd O2 addition reaction pathway was shown to be more important at lower temperatures. These findings on the significance of new reaction pathways highlight the need for theoretical and experimental investigations of low temperature oxidation mechanisms. Indeed, the kinetic model presented here only represents the current state of knowledge on alkane auto-oxidation, and reaction pathways that are currently unexplored are not included in the model. 51 The rigorous and comprehensive kinetic model for iso-octane developed herein indicates that more fundamental research is needed to improve predictive capabilities. The thermochemistry of low temperature species derived from highly branched alkanes (e.g., iso-octane) need to be determined using computational chemistry at a high level of theory. This will facilitate the extrapolation of group values and correction parameters to account for various intra-molecular interactions (e.g., gauche and 1,5-interactions). Rate rules for highly branched alkanes also need to be determined using rigorous computational methods. Specifically, the rates for cyclic ether formation, alternative isomerization reactions, and 3 rd addition to O 2 reactions require attention. Finally, kinetic model optimization with uncertainty quantification can be used to improve the iso-octane model’s predictions of combustion across a broad range of conditions. Acknowledgement The authors are grateful of insightful scientific discussions with Dr. Zhandong Wang (KAUST), Dr. Kuiwen Zhang (NUIG), Dr. John Bugler (NUIG), and Dr. Jihad Badra (Saudi Aramco). The presented work was supported by Saudi Aramco under the FUELCOM program and by the King Abdullah University of Science and Technology (KAUST) with competitive research funding given to the Clean Combustion Research Center (CCRC). The work at UCONN was supported by the National Science Foundation under Grant No. CBET-1402231. The work at LLNL was supported by the U.S. Department of Energy, Vehicle Technologies Office, program managers Gurpreet Singh and Leo Breton and was performed under the auspices of the U.S. Department of Energy by Lawrence Livermore National Laboratories under contract DE-AC52-07NA27344 References M. Mehl, W.J. Pitz, C.K. Westbrook, H.J. Curran, Kinetic modeling of gasoline surrogate components and mixtures under engine conditions, Proceedings of the Combustion Institute 33 (2011) 193-200. G.T. Kalghatgi, L. Hildingsson, A.J. Harrison, B. 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14705	https://www.ck12.org/flexi/cbse-math/prime-factorization/what-is-the-prime-factorisation-of-375/	What is the prime factorisation of 375? - Steps \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Prime Factorization Question What is the prime factorisation of 375? Flexi Says: Prime factorisation of 375 is 3 x 5 x 5 x 5. Any number can be split up into factors. If we keep splitting the number into factors, ultimately, we reach a stage when all the factors are prime factors. Breaking down a number into a product of all prime numbers is called prime factorization. The prime factorization of 375 can be found by dividing it by its prime factors until only primes are left. Steps to Prime Factorise a Number using the Division Method To find the prime factorisation of a number using the division method; we use the following steps: Step 1: Divide the given number by its smallest prime factor. Step 2: Divide the quotient obtained in step 1 by its smallest prime factor. Step 3: Continue until the quotient is a 1. Step 4: Write the given number as the product of all the primes that are the divisors of the division. On factorising 375 by the division method, we get 375 = 3 x 5 x 5 x 5 Click here to learn more about prime factorization! Analogy / Example Try Asking: What is the prime factorization of 36 in exponential form?What are the prime factors of 19440?What is the prime factorisation of 6838? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
14706	https://emedicine.medscape.com/article/985766-overview	close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Pediatrics: General Medicine Bone Mineralization and Related Disorders Updated: Jun 14, 2024 Author: Horacio B Plotkin, MD, FAAP; Chief Editor: Jatinder Bhatia, MBBS, FAAP more...;) 13 Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Bone Mineralization and Related Disorders Sections Bone Mineralization and Related Disorders Practice Essentials Vitamin D Metabolism Pathophysiology Rickets Renal Causes Tumor-Induced Osteomalacia Other Causes Show All Media Gallery;) References;) Practice Essentials Several diseases can result in disorders of bone mineralization, which can be defined as the process by which osteoid becomes calcified. This process depends on adequate levels of ionized calcium and phosphate in the extracellular fluid. Vitamin D influences these levels after its dihydroxylation into calcitriol. Disorders of bone mineralization Diseases that can cause disorders of bone mineralization in children include rickets, renal diseases (renal osteodystrophy, Fanconi syndrome), tumor-induced osteomalacia, hypophosphatasia, McCune-Albright syndrome, and osteogenesis imperfecta with mineralization defect (syndrome resembling osteogenesis imperfecta [SROI]). These conditions may result in failure of osteoid calcification (rickets) in children because of a disruption in the pathway of either vitamin D or phosphate metabolism. Rickets, once thought defeated, is reappearing and remains a major health problem in many developing and developed countries. See the image below. Radiograph in a 4-year-old girl with rickets depicts bowing of the legs caused by loading. View Media Gallery) Types of rickets include the following: Nutritional rickets Congenital rickets Rickets of prematurity Vitamin D resistance (type I and type II) Neoplastic rickets Hypophosphatemic rickets Drug-induced rickets Clinical and laboratory findings Clinical results and laboratory examination findings vary with each disorder. Low phosphate and high alkaline phosphatase levels characterize most of the disorders. Exceptions are noted in the discussion of each disorder. Next: Vitamin D Metabolism Vitamin D Metabolism The primary absorption site for vitamin D is the jejunum. The 2 main sources of vitamin D in humans are vitamin D3 (cholecalciferol), produced by the skin after ultraviolet (UV) radiation (290-320nm)–dependent conversion of 7-dehydrocholesterol, and dietary intake of either vitamin D2 (ergocalciferol) or vitamin D3. Both forms of vitamin D have identical biologic actions. The initial step in the metabolic activation process is the introduction of a hydroxyl group at the side chain at C-25 by the hepatic enzyme, CYP 27 (a vitamin D-25-hydroxylase). The products of this reaction are 25-(OH)D2 and 25-(OH)D3, respectively. Further hydroxylation of these metabolites occurs in the mitochondria of kidney tissue, catalyzed by renal 25-hydroxyvitamin D-1α-hydroxylase to produce 1α,25-(OH)2 D2 (activated vitamin D2 or 1,25[OH]2 D2), the primary biologically active form of vitamin D2, and 1α,25-(OH)2 D3 (calcitriol or 1,25[OH]2 D3), the biologically active form of vitamin D3. Of note, the kidney generates at least 30 other vitamin D metabolites, but their biologic significance is not clear. The pathophysiology of rickets is not completely understood, nor is the role of the many vitamin D metabolites. Calcitriol levels may be normal in patients with rickets, suggesting that it is not the only active form of the vitamin. Causes of rickets related to phosphate deficiency are discussed in the article Hypophosphatemic Rickets. Previous Next: Vitamin D Metabolism Pathophysiology Calcification of osteoid depends on adequate levels of ionized calcium and phosphate in the extracellular fluid. Vitamin D influences these levels after its dihydroxylation into calcitriol (at the 25 position in the liver and the 1 position in the kidney). If the enzyme that controls either of these steps is deficient because of a mutation, vitamin D function is less than normal. In addition, a renal tubular defect that reduces reabsorption may alter phosphate metabolism. Finally, a genetic absence of the receptor for calcitriol results in deficient calcification. X-linked hypophosphatemic rickets and autosomal recessive hypophosphatemic rickets are the result of mutations in PHEX (a phosphate-regulating gene with homologies to endopeptidases on the X chromosome) and dentin matrix protein 1 (DMP1), respectively. Degradation of matrix extracellular phosphoglycoprotein (MEPE) and DMP-1 and release of acidic serine-rich and aspartate-rich MEPE-associated motif (ASARM) peptides are chiefly responsible for the hypophosphatemic rickets mineralization defect and changes in osteoblast-osteoclast differentiation. In patients with oncogenic osteomalacia, intact and C-terminal fibroblast growth factor-23 (FGF-23) levels are elevated, and the tumors responsible for this disease show increased expression of FGF-23 messenger ribonucleic acid (mRNA). Previous Next: Vitamin D Metabolism Rickets Nutritional rickets A recommended daily allowance (RDA) for vitamin D has not been defined. Because no strong data support an RDA, recommendations for vitamin D intake actually refer to "adequate intake." Dietary rickets can be a consequence of inadequate intake of calcium, vitamin D, phosphate, or a combination of these. Infants fed exclusively with mother's milk can develop nutritional rickets because of the low content of vitamin D in breast milk (4-100 IU/L). In premature infants, insufficient amounts of calcium and phosphorus may cause nutritional rickets. Furthermore, reserves of vitamin D in the neonate highly depend on the mother's vitamin D status. Infants with low or no sun exposure may develop rickets, particularly if they have dark skin, because of decreased vitamin D production by the skin after exposure to UV light. Maternal hypovitaminosis D may cause congenital rickets in infants. In infants, clinical features of hypocalcemia and hyperphosphatemia include seizures, apnea, and tetany. In children, clinical features of rickets include the following (see the images below): Delayed motor milestones Hypotonia Enlargement of wrists Progressive bowing of long bones Rachitic rosary Harrison sulcus Violin case deformity of the chest Late closure of anterior fontanelle Parietal and frontal bossing Craniotabes Craniosynostosis Delay in teeth eruption Enamel hypoplasia Decreased bone mineral density Myopathy with normal deep tendon reflexes Propensity for infections - As a consequence of impaired phagocytosis and neutrophil motility Radiograph in a 4-year-old girl with rickets depicts bowing of the legs caused by loading. View Media Gallery) Findings in patients with rickets. View Media Gallery) Fractures occur in older infants and toddlers with overt rickets and can be seen using radiography. Of note, the fractures do not resemble nonaccidental trauma fractures. Radiologic features include widening of the epiphysial plate, cupping, and deformities in the shaft of long bones. Of note, radiographs of the costochondral junction are not useful in the diagnosis of rickets. The healing process is characterized by broadened bands of increased density. Different treatment modalities are available for nutritional rickets. Oral doses of 5,000-15,000 IU/day of vitamin D for 4 weeks are generally safe and effective. If compliance cannot be assured, 100,000-500,000 IU can be given orally or intramuscularly every 6 months or 600,000 IU may be given in a single intramuscular dose. Calcium intake must be optimized at the same time. Calcium, phosphorus, and parathyroid hormone concentrations should normalize within 1-3 weeks. Radiologic lesions and clinical symptoms improve rapidly with treatment, although alkaline phosphatase levels may remain elevated for several months after radiologic resolution. A study by Dabas et al compared the efficacy of daily versus weekly oral vitamin D3 therapy in the radiologic healing of nutritional rickets. Children who received daily supplementation had greater increases in their radiologic scores from baseline than those who received weekly therapy. Vitamin D–dependent rickets (type I) Also known as vitamin D–pseudodeficiency rickets (PDDR), this disorder results from a genetic deficiency in the enzyme that converts calcidiol to calcitriol in the kidney. Inheritance is autosomal recessive, and the gene is located in band 12q13.3. Clinical and laboratory examination findings are similar to those associated with nutritional rickets, with low levels of 1,25(OH)2 vitamin D. Levels of 1,25(OH)2 vitamin D may be normal but inadequately low for the levels of calcium, phosphorus, and parathyroid hormone. These patients develop rickets despite receiving vitamin D at the recommended preventive doses. Medical treatment consists of oral calcitriol (0.5-1.5 mcg/day). These patients may also respond to pharmacologic doses of vitamin D (5,000-10,000 U/day). Receptor defect rickets (type II vitamin D–dependent rickets) Receptor defect rickets (hereditary 1,25-dihydroxyvitamin D–resistant rickets [HVDRR]) results from a recessively inherited abnormality in the calcitriol receptor, causing an end-organ resistance to the vitamin. The clinical picture, which is evident early in life, consists of rickets with very severe hypocalcemia and alopecia, although a variant without alopecia has been reported. Patients without alopecia appear to respond better to treatment with vitamin D metabolites. Serum levels of 1,25(OH)2 vitamin D3 are typically elevated. HVDRR can be lethal in the perinatal period. Because calciferol receptors are in many tissues, other, more subtle dysfunctions may occur. Patients are hypocalcemic and usually normophosphatemic. Several mutant forms of receptor defect rickets are recognized, with a wide range of severity and response to calcitriol therapy. Some patients are totally resistant to therapy. Some others have benefited from intravenous calcium (400-1400 mg/m2/day) followed by oral therapy with high doses of calcium (with secondary risk of nephrocalcinosis, hypercalciuria, nephrolithiasis, and cardiac arrhythmias). Patients with mutations in the ligand-binding domain (LBD) region of the receptor are more likely to respond to high-dose vitamin D treatment than are patients with mutations in the deoxyribonucleic acid (DNA)–binding domain (DBD) region of the receptor. Defective 25-hydroxylase Two cases of 25-hydroxylase deficiency have been reported, one involving a family in the United States and the other involving a family in Germany. Inheritance is likely autosomal recessive. The clinical picture resembles that observed in nutritional rickets, with a later age of onset. Treatment with calcidiol in physiologic amounts is sufficient for this condition. Calcidiol is a natural metabolite of vitamin D. Calcidiol is hydroxylated once at the 25 position and is the circulating form for vitamin D in plasma. Familial hypophosphatemia Several different familial and acquired conditions may lead to hypophosphatemia in children. In familial hypophosphatemia, the kidneys fail to reabsorb sufficient phosphate, leading to low levels of serum phosphate. This is usually evident only after age 6-10 months. Prior to this occurrence, the glomerular filtration rate is low, which sustains an adequate phosphate level. Once renal maturity is reached, phosphate levels are usually less than 3.5 mg/dL and are often less than 2.5 mg/dL. Levels of 1,25(OH)2 vitamin D are actually normal in these patients, owing to an abnormal response to hypophosphatemia, in which levels of 1,25(OH)2 vitamin D should increase. Mutations in PHEX and DMP1 result in X-linked hypophosphatemic rickets and autosomal recessive hypophosphatemic rickets, respectively. (Most families of patients with familial hypophosphatemia exhibit X-linked dominant inheritance.) PHEX, a phosphate-regulating gene, codes for a protease, which is an enzyme that catalyzes the hydrolysis of a protein. Degradation of MEPE and DMP-1 and release of ASARM peptides are chiefly responsible for the hypophosphatemic rickets mineralization defect and changes in osteoblast-osteoclast differentiation. FGF-23 has been implicated in the renal phosphate wasting in tumor-induced osteomalacia and autosomal dominant hypophosphatemic rickets. Mutations in the gene that codes for the main renal sodium-phosphate cotransporter (NPT2a) have been reported in some patients with familial renal calcium stones and hypophosphatemia owing to a decrease in renal phosphate reabsorption. These patients have hypercalciuria and elevated levels of 1,25(OH)2 vitamin D3. Hereditary hypophosphatemic rickets with hypercalciuria (HHRH) is a metabolic disorder caused by homozygous loss-of-function mutations in the SLC34A3 gene, which encodes the renal type IIc sodium-phosphate cotransporter (NaPi-IIc). The typical presentation is severe rickets, hypophosphatemia, and hypercalciuria. Autosomal recessive and autosomal dominant inheritance have each been found and have been associated with the same clinical phenotype. In approximately one third of patients, the disease appears to occur as a consequence of a new mutation. Clinical findings are similar to those of nutritional rickets but without proximal myopathy. These patients usually have high bone density. As hypophosphatemia is usually clinically evident at a later age, infantile skull defects are not apparent. Because calcium levels remain normal, neither tetany nor secondary hyperparathyroidism are present. Treatment Optimal therapy consists of oral phosphate to provide 1-3 g of elemental phosphate per day in 5 divided doses plus oral calcitriol (0.5-1.5 mcg/day). Calcitriol prevents increases in parathyroid hormone caused by phosphate therapy. The phosphate mixture contains mineral salts of phosphoric acid. Raising the concentration of plasma phosphate facilitates calcification of osteoid. Of note, phosphate half-life in serum is short, which usually causes low phosphate levels in fasting serum samples, despite proper therapy. Efficacy is reflected by proper linear growth. Minor changes in calcitriol dose may produce hypercalcemia and renal damage. The calcium-creatinine (mg/mg) ratio in urine must be closely monitored at first and then every 3-6 months. An elevated phosphate intake may produce secondary hyperparathyroidism. Therefore, only experienced practitioners should treat these patients. Drug-induced rickets Different medications may affect bone in different ways. Chronic anticonvulsant therapy (particularly with phenobarbital and phenytoin) may cause rickets, regardless of appropriate vitamin D intake. The main mechanism is related to induction of hepatic cytochrome P-450 hydroxylation, generating inactive metabolites. Levels of 25-hydroxyvitamin D3 were reported to be low in children on long-term anticonvulsant therapy. Fractures were associated with the use of anticonvulsants in patients with cerebral palsy. A down-regulation of 25-hydroxylation by phenobarbital may explain, at least in part, the increased risk of osteomalacia, bone loss, and fractures associated with long-term phenobarbital therapy. Conversely, calcitriol levels in plasma are reportedly not low in patients taking medication for seizures. The dose of vitamin D required to prevent this type of rickets is unclear. Supplementation may not be needed. Approximately 800-1000 IU/day, plus good calcium intake, may be sufficient. Previous Next: Vitamin D Metabolism Renal Causes Fanconi syndrome Fanconi syndrome is a disorder of proximal renal tubular transport. Phosphate, amino acid, glucose, bicarbonate, and uric acid wasting characterize this disorder. Dysfunctions in tubular phosphate reabsorption via the sodium-phosphate cotransporter, endocytotic reabsorption of the vitamin D–vitamin D–binding protein complex mediated by megalin and cubilin, and acid-base regulation are the most important factors that cause bone mineralization defects in these patients. Lowe disease and Dent disease are familial forms of Fanconi. Two different genes have been identified as being involved in the development of Dent disease. CLCN5 is affected in Dent disease type 1 and OCRL1 is affected in Dent disease type 2. Other genes may also be involved, because mutations in CLCN5 and OCRL1 are not found in some patients. In Fanconi syndrome, which includes cystinosis and tyrosinemia, renal phosphate wasting may occur, along with aminoaciduria and glycosuria. Fanconi syndrome can have a genetic cause (as in Lowe and Dent disease), or it may be acquired from various toxins, including heavy metals (eg, mercury, lead) and drugs. The clinical picture varies with age and cause and includes severe hypophosphatemic rickets, failure to thrive, and metabolic acidosis. A potential drug-induced Fanconi syndrome has been noticed in children treated with ifosfamide, a derivative of cyclophosphamide. The syndrome presents with radiologic changes compatible with rickets. Most patients respond to a combination of managing the underlying cause when possible and vitamin D therapy. These patients do not necessarily appear to require treatment with calcitriol. Renal tubular acidosis, through phosphate wasting, may also cause rickets. Renal osteodystrophy In end-stage renal disease, renal 1-hydroxylase is diminished or lost, and excretion of phosphate is defective. This leads to low levels of 1,25(OH) 2 vitamin D, hypocalcemia, and failure of osteoid calcification. Osteodystrophy (ie, renal rickets) is the only type of rickets with a high serum phosphate level. It can be adynamic (a reduction in osteoblastic activity) or hyperdynamic (increased bone turnover). Calcium receptors (CaRs) have been discovered in bone, kidney, and intestine and also in organs not directly related to calcium regulation. Mutations that cause loss of function in the CaRs result in familial benign hypocalciuric hypercalcemia and neonatal severe hyperparathyroidism. Familial benign hypocalciuric hypercalcemia is usually associated with heterozygous inactivating mutations of the CAR gene, whereas neonatal severe hyperparathyroidism is usually due to homozygous inactivation of the CAR gene. Familial benign hypocalciuric hypercalcemia is generally asymptomatic and is characterized by mild to moderate, lifelong hypercalcemia; relative hypocalciuria; and normal intact parathyroid hormone. Individuals with neonatal severe hyperparathyroidism frequently develop life-threatening hypercalcemia. Treatment of these patients includes phosphate binders, a low phosphate intake, and calcitriol and other vitamin D analogs. Previous Next: Vitamin D Metabolism Tumor-Induced Osteomalacia Tumor-induced osteomalacia (TIO) is a paraneoplastic syndrome with hypophosphatemia secondary to decreased renal phosphate reabsorption, normal or low serum 1,25-dihydroxyvitamin D concentration, osteomalacia, and myopathy. Several mesenchymal tumors of bone or connective tissue (including nonossifying fibromas, fibroangioma, and giant cell tumors) secrete a phosphaturic substance (parathyroidlike protein) that results in rickets. The age of onset has been late childhood, adolescence, or young adulthood. The clinical characteristics are similar to those associated with familial hypophosphatemia. FGF-23 causes renal phosphate wasting in tumor-induced osteomalacia. Treatment is surgical removal of the tumor (if it can be located) with excellent results. If resection of the tumor is not possible, medical therapy with phosphate and active vitamin D is indicated. Burosumab is a monoclonal antibody against the phosphaturic hormone FGF-23. It is indicated for X-linked hypophosphatemia in adult and pediatric patients aged 6 months or older. It is also approved for FGF-23–related hypophosphatemia in tumor-induced osteomalacia associated with phosphaturic mesenchymal tumors that cannot be curatively resected or localized in adults and pediatric patients aged 2 years or older. Previous Next: Vitamin D Metabolism Other Causes Hypophosphatasia This autosomal recessive condition, which results in low activity of the tissue-nonspecific isoenzyme of alkaline phosphatase (TNSALP), causes rickets without disturbance of calcium and phosphate metabolism. Levels of TNSALP substrates, namely pyridoxal-5'-phosphate (PLP), inorganic pyrophosphate (PPi), and phosphoethanolamine (PEA) in serum and urine, are increased. Clinical severity widely varies, ranging from death in utero to pathologic fractures first presenting only in adulthood. Six clinical forms of hypophosphatasia have been distinguished, although form assignment may be challenging in some cases. This classification is based on the age at which skeletal lesions are discovered: perinatal (lethal), infantile, childhood, and adult. Two particular forms include odontohypophosphatasia (only biochemical and dental manifestations are present, with no clinical changes in long bones) and pseudohypophosphatasia. The forms of hypophosphatasia that appear in childhood or adulthood are typically less severe than those that appear in infancy and during the perinatal period. The effects of bone marrow transplant in hypophosphatasia are transient, and bone lesions may recur 6 months after the transplant. Nonsteroidal anti-inflammatory drugs (NSAIDs) have been used in patients with childhood hypophosphatasia with some clinical improvement. The US Food and Drug Administration has approved asfotase alfa as the first permitted treatment for perinatal, infantile and juvenile-onset hypophosphatasia. A study by Whyte et al found that asfotase alfa enzyme replacement therapy is effective and safe for treating children with hypophosphatasia. McCune-Albright syndrome Patients with McCune-Albright syndrome may have hypophosphatemia secondary to urinary phosphate leak, which may cause osteomalacia. Fasting phosphate levels should always be monitored in these patients, and phosphate supplements prescribed when indicated. Syndrome resembling osteogenesis imperfecta Syndrome resembling osteogenesis imperfecta (SROI) with mineralization defect is clinically indistinguishable from moderate to severe osteogenesis imperfecta. (This rare form, in fact, has been termed type VI osteogenesis imperfecta.) It can only be diagnosed with bone biopsy, in which a mineralization defect that affects the bone matrix and sparing growth cartilage are evident. These patients have neither dentinogenesis imperfecta nor Wormian bones. Despite the histologic mineralization defect, no radiologic signs of growth plate involvement are seen. The pattern of inheritance is not clear, but a case of 2 siblings from healthy consanguineous parents has been described, suggesting gonadal mosaicism or a somatic recessive trait. No mutations of COL1A1 and COL1A2 genes have been found in these patients, and collagen structure appears to be normal. This form shares several characteristics with fibrogenesis imperfecta ossium. A mild, rare form of this condition may occur (3 patients in a series of 128 bone biopsies performed to assess bone fragility). These patients do not appear to respond well to treatment with intravenous bisphosphonates. Previous Next: Vitamin D Metabolism References Michigami T. Skeletal mineralization: mechanisms and diseases. Ann Pediatr Endocrinol Metab. 2019 Dec. 24 (4):213-9. [QxMD MEDLINE Link]. Martin A, David V, Laurence JS, et al. Degradation of MEPE, DMP1, and release of SIBLING ASARM-peptides (minhibins): ASARM-peptide(s) are directly responsible for defective mineralization in HYP. Endocrinology. 2008 Apr. 149(4):1757-72. [QxMD MEDLINE Link]. Vieth R, Fraser D. Vitamin D insufficiency: no recommended dietary allowance exists for this nutrient. CMAJ. 2002 Jun 11. 166(12):1541-2. [QxMD MEDLINE Link]. Chapman T, Sugar N, Done S, Marasigan J, Wambold N, Feldman K. Fractures in infants and toddlers with rickets. Pediatr Radiol. 2009 Dec 9. Epub:[QxMD MEDLINE Link]. Dabas A, Dabas V, Dabla PK, et al. Daily v. weekly oral vitamin D(3) therapy for nutritional rickets in Indian children: a randomised controlled open-label trial. Br J Nutr. 2022 May 13. 1-8. [QxMD MEDLINE Link]. Tencza AL, Ichikawa S, Dang A, et al. Hypophosphatemic rickets with hypercalciuria due to mutation in SLC34A3/type IIc sodium-phosphate cotransporter: presentation as hypercalciuria and nephrolithiasis. J Clin Endocrinol Metab. 2009 Nov. 94(11):4433-8. [QxMD MEDLINE Link]. [Full Text]. Henderson RC, Lin PP, Greene WB. Bone-mineral density in children and adolescents who have spastic cerebral palsy. J Bone Joint Surg Am. 1995 Nov. 77(11):1671-81. [QxMD MEDLINE Link]. Hosseinpour F, Ellfolk M, Norlin M, Wikvall K. Phenobarbital suppresses vitamin D3 25-hydroxylase expression: a potential new mechanism for drug-induced osteomalacia. Biochem Biophys Res Commun. 2007 Jun 8. 357(3):603-7. [QxMD MEDLINE Link]. Prie D, Huart V, Bakouh N, et al. Nephrolithiasis and osteoporosis associated with hypophosphatemia caused by mutations in the type 2a sodium-phosphate cotransporter. N Engl J Med. 2002 Sep 26. 347(13):983-91. [QxMD MEDLINE Link]. Florenzano P, Hartley IR, Jimenez M, Roszko K, Gafni RI, Collins MT. Tumor-induced osteomalacia. Calcif Tissue Int. 2020 Jun 5. [QxMD MEDLINE Link]. Crysvita (burosumab) [package insert]. Novato, CA: Ultragenyx Pharmaceutical Inc. June 2020. Available at [Full Text]. Reis FS, Lazaretti-Castro M. Hypophosphatasia: from birth to adulthood. Arch Endocrinol Metab. 2023 May 25. 67 (5):e000626. [QxMD MEDLINE Link]. [Full Text]. FDA approves new treatment for rare metabolic disorder. U.S Food & Drug Administration. Available at October 23, 2015; Accessed: March 28, 2017. Whyte MP, Madson KL, Phillips D, Reeves AL, McAlister WH, Yakimoski A, et al. Asfotase alfa therapy for children with hypophosphatasia. JCI Insight. 2016 Jun 16. 1 (9):e85971. [QxMD MEDLINE Link]. Plotkin H. Syndromes with congenital brittle bones. BMC Pediatr. 2004 Aug 31. 4:16. [QxMD MEDLINE Link]. ADHR Consortium. Autosomal dominant hypophosphataemic rickets is associated with mutations in FGF23. Nat Genet. 2000 Nov. 26(3):345-8. [QxMD MEDLINE Link]. Anatoliotaki M, Tsilimigaki A, Tsekoura T, et al. Congenital rickets due to maternal vitamin D deficiency in a sunny island of Greece. Acta Paediatr. 2003. 92(3):389-91. [QxMD MEDLINE Link]. Chesney RW, Zimmerman J, Hamstra A, et al. Vitamin D metabolite concentrations in vitamin D deficiency. Are calcitriol levels normal. Am J Dis Child. 1981 Nov. 135(11):1025-8. [QxMD MEDLINE Link]. Intakes SCotSEoDR. Dietary Reference Intakes: Calcium, Phosphorus, Magnesium, Vitamin D, and Fluoride. Washington, DC: National Academy Press; 1997. Juppner H. Novel regulators of phosphate homeostasis and bone metabolism. Ther Apher Dial. 2007. 11:S3-S22. [QxMD MEDLINE Link]. Plotkin H, Lifshitz F. Rickets and osteoporosis. Lifshitz F, ed. Pediatric Endocrinology. 5th ed. 2006. Media Gallery Radiograph in a 4-year-old girl with rickets depicts bowing of the legs caused by loading. Findings in patients with rickets. of 2 Tables Back to List Contributor Information and Disclosures Author Horacio B Plotkin, MD, FAAP Chief Medical Officer, Retrophin, Inc; Adjunct Associate Professor of Pediatrics and Orthopedic Surgery, University of Nebraska College of Medicine Horacio B Plotkin, MD, FAAP is a member of the following medical societies: American Academy of Pediatrics Disclosure: Received salary from PPD, Inc for: PPD. Coauthor(s) Laurence Finberg, MD Clinical Professor, Department of Pediatrics, University of California, San Francisco, School of Medicine and Stanford University School of Medicine Laurence Finberg, MD is a member of the following medical societies: American Medical Association Disclosure: Nothing to disclose. Chief Editor Jatinder Bhatia, MBBS, FAAP Professor of Pediatrics, Medical College of Georgia, Georgia Regents University; Chief, Division of Neonatology, Director, Fellowship Program in Neonatal-Perinatal Medicine, Director, Transport/ECMO/Nutrition, Vice Chair, Clinical Research, Department of Pediatrics, Children's Hospital of Georgia Jatinder Bhatia, MBBS, FAAP is a member of the following medical societies: Academy of Nutrition and Dietetics, American Academy of Pediatrics, American Association for the Advancement of Science, American Pediatric Society, American Society for Nutrition, American Society for Parenteral and Enteral Nutrition, Society for Pediatric Research, Southern Society for Pediatric Research Disclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: NestleServe(d) as a speaker or a member of a speakers bureau for: NestleReceived income in an amount equal to or greater than $250 from: Nestle. Acknowledgements Maria Rebello Mascarenhas, MBBS Associate Professor of Pediatrics, University of Pennsylvania School of Medicine; Section Chief Nutrition, Division of Gastroenterology and Nutrition, Director, Nutrition Support Service, Children's Hospital of Philadelphia Maria Rebello Mascarenhas, MBBS is a member of the following medical societies: American Gastroenterological Association, American Society for Parenteral and Enteral Nutrition, and North American Society for Pediatric Gastroenterology and Nutrition Disclosure: Nothing to disclose. Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Nothing to disclose. encoded search term (Bone Mineralization and Related Disorders) and Bone Mineralization and Related Disorders What to Read Next on Medscape
14707	https://stackoverflow.com/questions/3953623/is-there-a-simple-algorithm-for-calculating-the-maximum-inscribed-circle-into-a	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Is there a simple algorithm for calculating the maximum inscribed circle into a convex polygon? Ask Question Asked Modified 2 years, 10 months ago Viewed 18k times This question shows research effort; it is useful and clear 16 Save this question. Show activity on this post. I found some solutions, but they're too messy. algorithm linear-algebra linear-programming convex-optimization convex-polygon Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited May 15, 2017 at 2:04 Robin 84522 gold badges99 silver badges2727 bronze badges asked Oct 17, 2010 at 14:31 m88m88 1,55933 gold badges1212 silver badges1818 bronze badges 0 Add a comment \| 4 Answers 4 Reset to default This answer is useful 10 Save this answer. Show activity on this post. Yes. The Chebyshev center, x, of a set C is the center of the largest ball that lies inside C. [Boyd, p. 416] When C is a convex set, then this problem is a convex optimization problem. Better yet, when C is a polyhedron, then this problem becomes a linear program. Suppose the m-sided polyhedron C is defined by a set of linear inequalities: ai^T x <= bi, for i in {1, 2, ..., m}. Then the problem becomes ``` maximize R such that ai^T x + R\|\|a\|\| <= bi, i in {1, 2, ..., m} R >= 0 ``` where the variables of minimization are R and x, and \|\|a\|\| is the Euclidean norm of a. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 27, 2013 at 9:44 Tobias Kienzler 27.7k2222 gold badges138138 silver badges230230 bronze badges answered Oct 31, 2010 at 3:32 Steve TjoaSteve Tjoa 61.3k1818 gold badges9292 silver badges103103 bronze badges 4 There is an iterative solution on that: link.springer.com/article/10.1007%2FBF01204183 – math Commented Jan 23, 2014 at 10:49 A simple O(n log n) algorithm is given in stackoverflow.com/questions/27872964/…. But the answer would better fit to this question here. – S. Huber Commented Oct 22, 2017 at 17:54 1 I assume you used this (en.wikipedia.org/wiki/Distance_from_a_point_to_a_line) formula to calculate the distance from point to a line. I am wondering, how did you get rid of the absolute value? I am stuck with \|ai^T x - bi\| / \|\|a\|\| >= r and have no idea how to continue. – karlosss Commented Nov 3, 2019 at 13:04 An example with CVXR: <web.cvxr.com/cvx/examples/cvxbook/Ch04_cvx_opt_probs/html/…> – Stéphane Laurent Commented May 17, 2023 at 10:59 Add a comment \| This answer is useful 7 Save this answer. Show activity on this post. Perhaps these "too messy" solutions are what you actually looking for, and there are no simplier ones? I can suggest a simple, but potentially imprecise solution, which uses numerical analysis. Assume you have a resilient ball, and you inflate it, starting from radius zero. If its center is not in the center you're looking for, then it will move, because the walls would "push" it in the proper direction, until it reaches the point, from where he can't move anywhere else. I guess, for a convex polygon, the ball will eventually move to the point where it has maximum radius. You can write a program that emulates the process of circle inflation. Start with an arbitrary point, and "inflate" the circle until it reaches a wall. If you keep inflating it, it will move in one of the directions that don't make it any closer to the walls it already encounters. You can determine the possible ways where it could move by drawing the lines that are parallel to the walls through the center you're currently at. In this example, the ball would move in one of the directions marked with green: (source: coldattic.info) Then, move your ball slightly in one of these directions (a good choice might be moving along the bisection of the angle), and repeat the step. If the new radius would be less than the one you have, retreat and decrease the pace you move it. When you'll have to make your pace less than a value of, say, 1 inch, then you've found the centre with precision of 1 in. (If you're going to draw it on a screen, precision of 0.5 pixel would be good enough, I guess). If an imprecise solution is enough for you, this is simple enough, I guess. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Aug 7, 2019 at 5:38 Glorfindel 22.8k1313 gold badges9494 silver badges122122 bronze badges answered Oct 17, 2010 at 14:53 P ShvedP Shved 99.8k1919 gold badges128128 silver badges168168 bronze badges 1 This worked well for me. It is easy to implement. I even used a variation of this to find the maximal inscribed circle in a (not necessarily convex) blob with holes. – JDiMatteo Commented Mar 13, 2017 at 22:01 Add a comment \| This answer is useful 6 Save this answer. Show activity on this post. Summary: It is not trivial. So it is very unlikely that it will not get messy. But there are some lecture slides which you may find useful. Source: Your problem is not trivial, and there is no C# code that does this straight out of the box. You will have to write your own. I found the problem intriguing, and did some research, so here are a few clues that may help. First, here's an answer in "plain English" from mathforum.org: Link The answer references Voronoi Diagrams as a methodology for making the process more efficient. In researching Voronoi diagrams, in conjunction with the "maximum empty circle" problem (same problem, different name), I came across this informative paper: It was written by Martin Held, a Computational Geometry professor at the University of Salzberg in Austria. Further investigation of Dr. Held's writings yielded a couple of good articles: Further research into Vornoi Diagrams yielded the following site: This site has lots of information, code in various languages, and links to other resources. Finally, here is the URL to the Mathematics and Computational Sciences Division of the National Institute of Standards and Technology (U.S.), a wealth of information and links regarding mathematics of all sorts: -- HTH, Kevin Spencer Microsoft MVP Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 6, 2022 at 18:13 Glorfindel 22.8k1313 gold badges9494 silver badges122122 bronze badges answered Oct 17, 2010 at 14:40 Jungle HunterJungle Hunter 7,2961111 gold badges4646 silver badges6868 bronze badges 5 1 This bulk of this reply text from Kevin Spencer refers to the problem involving concave or convex polygons. The problem might be much easier in the convex case. – Josephine Commented Oct 18, 2010 at 14:41 @Josephine: Might be. I haven't gone through in detail, but might be. Perhaps you have something on your mind? – Jungle Hunter Commented Oct 18, 2010 at 17:31 After commenting here, I also posted an answer. My ideas are there. – Josephine Commented Oct 19, 2010 at 12:00 The problem is definitely easier when the polygon is convex. The problem becomes a linear program. I posted an answer, too. – Steve Tjoa Commented Oct 31, 2010 at 3:43 Links are broken. – Emad Commented Aug 18, 2021 at 13:04 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. The largest inscribed circle (I'm assuming it's unique) will intersect some of the faces tangentially, and may fail to intersect others. Let's call a face "relevant" if the largest inscribed circle intersects it, and "irrelevant" otherwise. If your convex polygon is in fact a triangle, then the problem can be solved by calculating the triangle's incenter, by intersecting angle bisectors. This may seem a trivial case, but even when your convex polygon is complicated, the inscribed circle will always be tangent to at least three faces (proof? seems geometrically obvious), and so its center can be calculated as the incenter of three relevant faces (extended outwards to make a triangle which circumscribes the original polygon). Here we assume that no two such faces are parallel. If two are parallel, we have to interpret the "angle bisector" of two parallel lines to mean that third parallel line between them. This immediately suggests a rather terrible algorithm: Consider all n-choose-3 subsets of faces, find the incenters of all triangles as above, and test each circle for containment in the original polygon. Maximize among those that are legal. But this is cubic in n and we can do much better. But it's possible instead to identify faces that are irrelevant upfront: If a face is tangent to some inscribed circle, then there is a region of points bounded by that face and by the two angle bisectors at its endpoints, wherein the circle's center must lie. If even the circle whose center lies at the farthest tip of that triangular region is "legal" (entirely contained in the polygon), then the face itself is irrelevant, and can be removed. The two faces touching it should be extended beyond it so that they meet. By iteratively removing faces which are irrelevant in this sense, you should be able to reduce the polygon to a triangle, or perhaps a trapezoid, at which point the problem will be easily solved, and its solution will still lie within the original polygon. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Oct 18, 2010 at 15:53 JosephineJosephine 1,44911 gold badge99 silver badges1111 bronze badges 6 What about a rectangle? What about a modified rectangle which has it's smaller sides slightly convexed out as two sides of a triangle, each? ...but even when your convex polygon is complicated, the inscribed circle will always be tangent to at least three faces (proof? seems geometrically obvious) doesn't seem to hold. – Jungle Hunter Commented Oct 19, 2010 at 13:47 2 In both cases you describe, "the" largest inscribed circle is not unique, but among all largest inscribed circles, at least one intersects three sides. – Josephine Commented Oct 19, 2010 at 19:34 even if it was only for such cases, you need to somehow know if the largest inscribed circle is not unique. – Jungle Hunter Commented Oct 20, 2010 at 13:17 2 the polygon is convex, so the largest inscribed circle is unique. – m88 Commented Nov 5, 2010 at 21:11 2 @m88: That's not true. For example, any rectangle having different width and height is (a) convex and (b) has an infinite number of maximum-size inscribed circles. – j_random_hacker Commented Jun 22, 2015 at 22:55 \| Show 1 more comment Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm linear-algebra linear-programming convex-optimization convex-polygon See similar questions with these tags. 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14708	https://en.wikipedia.org/wiki/White_Americans	Jump to content Search Contents 1 History 2 Self Reported Ancestry 3 Genetics 4 Historical and present definitions 4.1 US census definition 4.2 Social definition 5 Demographic information 5.1 Geographic distribution 5.2 Income and educational attainment 5.3 Proportion in each county 5.3.1 White Americans of one race or alone from 2000 to 2020 5.3.1.1 Non-Hispanic population 6 Politics 7 Health 8 Culture 8.1 Albion's Seed: Four British Folkways in America 9 Admixture 9.1 Admixture in non-Hispanic whites 9.2 Admixture in Hispanic whites 10 See also 11 Notes 12 References 13 External links White Americans العربية Azərbaycanca تۆرکجه 閩南語 / Bn-lm-gí Български Буряад Čeština Dansk Deutsch Ελληνικά Español فارسی Français 客家語 / Hak-k-ngî 한국어 Hrvatski Bahasa Indonesia Italiano עברית مصرى Português Romnă Русский Shqip Simple English Српски / srpski Srpskohrvatski / српскохрватски தமிழ் Türkçe Українська Tiếng Việt 中文 Edit links Article Talk Read View source View history Tools Actions Read View source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia White people of the United States "White America" redirects here. For Non-Hispanic whites, see Non-Hispanic whites. For the white supremacist organization, see White America, Inc. For the song by Eminem, see White America (song). Ethnic group White Americans \| Proportion of White Americans in each county as of the 2020 US census \| \| Total population \| \| Alone (one race) 204,277,273 (2020 census) 61.63% of the total US populationIn combination (multiracial) 31,134,234 (2020 census) 9.39% of the total US populationAlone or in combination 235,411,507 (2020 census) 71.02% of the total US population \| \| Regions with significant populations \| \| All areas of the United States \| \| California \| 16,296,122 \| \| Texas \| 14,609,365 \| \| Florida \| 12,422,961 \| \| New York \| 11,143,349 \| \| Pennsylvania \| 9,750,687 \| \| Languages \| \| Majority: EnglishMinority: German · Spanish · Irish · Italian · Polish · French · Arabic · Scots · Norwegian · Russian · Dutch · Swedish · Portuguese \| \| Religion \| \| Protestant 48% Catholic 19% Mormon 2% Jewish 3% Other 3% Irreligious 24% (among non-Hispanic whites only) \| \| Related ethnic groups \| \| European AmericansNorth African AmericansMiddle Eastern Americans \| White Americans (sometimes also called Caucasian Americans although such usage has been criticized) are Americans who identify as white people. In a more official sense, the United States Census Bureau, which collects demographic data on Americans, defines "white" as "[a] person having origins in any of the original peoples of Europe, the Middle East, or North Africa". Individuals within this group tend to have light skin tones and various hair colors, mainly brown or blonde and to a lesser extent black or red due to their primarily English and German origins, although Irish, Italian and White Hispanic origin are also prominent. White Americans have historically constituted the majority population in the United States, though their share has been gradually declining in recent decades. As of the latest American Community Survey conducted by the U.S. Census Bureau in 2024, an estimated 59.8% of the U.S. population—approximately 203.3 million people—identify as White alone, while Non-Hispanic Whites account for 56.3% of the population, or roughly 191.4 million people. Overall, 72.1% of Americans identify as White either alone or in combination with one or more other racial groups. European Americans are by far the largest panethnic group of white Americans and have constituted the majority population of the United States since the nation's founding. Middle Eastern Americans constitute a much smaller demographic of white Americans, making up around 1.1% of the US population in 2020. According to the 2020 census, 61.6% of Americans, or 204,277,273 people, identified as White alone. This represented a national decrease from a 72.4% white alone share of the US population in the 2010 census. The share of Americans identifying as White alone or in combination (including multiracial white people) was 71.0% in 2020, a smaller decline from 74.8% of the population in 2010. As opposed to the declines seen in the white alone population, the number of people identifying as part white (in combination with other races) saw a large increase, growing from 2.4% of the population in 2010, to 9.4% in 2020. While the large decline in the white alone population observed between 2010 and 2020 has been partly attributed to natural trends, researchers have found that most of the sharp growth in the multiracial population, and commensurate decline in the white alone population, were due to changes in the methodology used by the Census Bureau, leading to a significant number of people who previously identified as white alone in 2010, mostly those identifying as White Hispanics, being reclassified as multiracial in 2020. In 2010, around 53% of Hispanics in the country identified as white alone, while in 2020, this number had declined to only 20.3% of Hispanics. The US Census Bureau uses a particular definition of "white" that differs from some colloquial uses of the term. The Bureau defines "White" people to be those "having origins in any of the original peoples of Europe, the Middle East or North Africa". Within official census definitions, people of all racial categories may be further divided into those who identify as "not Hispanic or Latino" and those who do identify as "Hispanic or Latino". The term "non-Hispanic white", rather than just "white", may be the census group corresponding most closely to those persons who identify as and are perceived to be white in common usage; similarly not all Hispanic/Latino people identify as "white", "black", or any other listed racial category. In 2015, the Census Bureau announced their intention to make Hispanic/Latino and Middle Eastern/North African racial categories similar to "white" or "black", with respondents able to choose one, two, or more racial categories; this change was canceled during the Trump administration. Other persons who are classified as "white" by the US census but may or may not identify as or be perceived as "white" include Arab Americans and Jewish Americans of European or MENA descent. In the United States, the term White people generally denotes a person of European ancestry, but has been legally extended to people of West Asian and North African (Middle Eastern, West Asian, and North African) ancestry. However, in 2024, the Office of Management and Budget announced that the race categories used by the federal government would be updated, and that Middle Eastern and North African Americans will no longer be classified as white in the upcoming 2030 Census. History The Western concept of Whiteness originated in Iberia in the 15th century in the aftermath of the Reconquista. The Spanish described Europeans as White and pure (literally, having clean blood) in contrast to the darker Arabs, Jews, Gypsies, and other racial minorities who contaminated the blood of their homeland. As the pioneer of Western colonization, Spain spread its racial terminology across Europe and the world, with other European societies incorporating Spanish terms such as mulatto, negro, indio ("Indian"), and so on, into their own languages. By the 17th century, most European societies were using the term "White race" or "White people". The Spanish caste system, accompanied by the enslavement and dispossession of indigenous peoples and sub-Saharan Africans in the Spanish Empire, was directly transplanted to its colonies, as were similar caste systems in other European colonies. In the context of American history, the term "white" was first used in early colonial Virginia as a way to justify racism against African Americans. The white ruling class in the colonies deemed black people inferior and suitable for slavery, and sought to draw a clear line between European settlers and white indentured servants, who were protected as citizens under colonial law, and chattel slaves of African descent, who possessed no legal rights. The population of what would become the United States has been enumerated by racial categories since the first permanent British settlers arrived at Jamestown in 1607. A colonial census from 1620, just after the first enslaved Africans were transported to Virginia, recorded the colony's population as including 2,282 "White" individuals and 20 "Negroes". Estimates indicate that whites made up the majority of non-Indigenous inhabitants of the colonies in every year from 1620 to 1780. Following the independence of the United States, every census since 1790 has enumerated the population by perceived race or "color." Although the categories used by the census have varied over time, the classification of a "white" race has been included on every census conducted in the US. The white population, numbering 3,172,006 in 1790, was primarily composed of English descendants, with a smaller minority of Germans, Irish, and Scots. Over the following decades, the white population would grow steadily as the country expanded westward, reaching a population of 19,553,068 in 1850, or 80.7% of the total population. Immigration waves during the 19th and early 20th centuries further grew the white population, which reached 110,286,740 in 1930. The white population also grew considerably in diversity during that time period, driven by large amounts of immigration from Ireland, Italy, Eastern Europe, and Southern Europe, transforming a historically "White Anglo-Saxon Protestant" society into a melting pot of different European ethnic groups. The white population peaked as a percent of the population in 1940, at just under 90% of the total population, and has been gradually declining in share since then, reaching a historic low of 61.6% in 2020, as increased immigration from Asia and Latin America has gradually displaced Europe as the primary source of immigrants to the US. Self Reported Ancestry Ethnic groups according to the 2020 US Census British (8.30%) German (4.67%) Irish (3.29%) Italian (2.00%) Polish (0.81%) Scandinavian (0.76%) West Asian / North African (0.86%) French (0.54%) Dutch (0.27%) Russian (0.30%) Other or multiple white ethnicities (White alone) (40.0%) Multiracial White (9.39%) Non-white (28.8%) The most commonly reported ancestries of White Americans include English (12.5%), German (7.6%), Irish (5.3%), Italian (3.2%), and Polish (1.3%). It is difficult to track full or partial ancestry from Spain in White Hispanics, Mestizos, or Mulattoes since people of direct Spanish descent are also classified as Hispanic, and though the census does track Hispanics' national origin, it does not classify it by race. In 2020, 1,896,300 people claimed ancestry from Spain, 0.6% of the total population. However, genetic studies have found that the vast majority of Hispanics in the US have varying amounts of European ancestry, with the largest component being Spanish or Iberian. The English Americans' demography is also considered a serious under-count, as the stock tend to self-report and identify as simply "Americans" (7%), due to the length of time they have inhabited the United States, particularly if their family arrived prior to the American Revolution. The following table lists all self-reported European and Middle Eastern ancestries with over 50,000 members, according to 2022 estimates from the American Community Survey: \| Ancestry \| Number in 2022 (Alone) \| Number as of 2022 (Alone or in any combination) \| % Total \| --- --- \| \| German \| 13,241,923 \| 41,137,168 \| 12.3% \| \| English \| 12,331,696 \| 31,380,620 \| 9.4% \| \| Irish \| 8,649,243 \| 30,655,612 \| 9.2% \| \| American (Mostly old-stock British Americans) \| 14,929,899 \| 17,786,214 \| 5.3% \| \| Italian \| 5,766,634 \| 16,009,774 \| 4.8% \| \| Polish \| 2,658,632 \| 8,249,491 \| 2.5% \| \| French (Not including French Canadian) \| 1,360,631 \| 6,310,548 \| 1.9% \| \| Scottish \| 1,555,579 \| 5,352,344 \| 1.6% \| \| Broadly "European" (No country specified) \| 3,718,055 \| 4,819,541 \| 1.4% \| \| Swedish \| 740,478 \| 3,936,772 \| 1.2% \| \| Norwegian \| 1,224,373 \| 3,317,462 \| 1.0% \| \| Dutch \| 858,809 \| 3,019,465 \| 0.9% \| \| Scotch-Irish \| 940,337 \| 2,524,746 \| 0.8% \| \| Arab (Including Lebanese (583,719), Egyptian (334,574), Syrian (203,282), Palestinian (171,969), Iraqi (164,851), Moroccan (140,196), and all other Arab ancestries) \| 1,502,360 \| 2,237,982 \| 0.7% \| \| Russian \| 747,866 \| 2,099,079 \| 0.6% \| \| Spanish (Including responses of "Spaniard," "Spanish," and "Spanish American." Many Hispanos of New Mexico identify as Spanish/Spaniard) \| — \| 1,926,228 \| 0.6% \| \| French Canadian \| 694,089 \| 1,626,456 \| 0.5% \| \| Welsh \| 293,551 \| 1,521,565 \| 0.5% \| \| Portuguese \| 543,531 \| 1,350,442 \| 0.4% \| \| Hungarian \| 390,561 \| 1,247,165 \| 0.4% \| \| Greek \| 486,878 \| 1,200,706 \| 0.4% \| \| Broadly "British" (Not further specified) \| 503,077 \| 1,196,265 \| 0.4% \| \| Czech \| 340,768 \| 1,188,711 \| 0.4% \| \| Ukrainian \| 565,431 \| 1,164,728 \| 0.3% \| \| Danish \| 268,019 \| 1,127,518 \| 0.3% \| \| Broadly "Eastern European" (Not further specified) \| 566,715 \| 951,384 \| 0.3% \| \| Broadly "Scandinavian" (Not further specified) \| 372,673 \| 935,153 \| 0.3% \| \| Swiss \| 196,120 \| 847,247 \| 0.3% \| \| Finnish \| 189,603 \| 606,028 \| 0.2% \| \| Slovak \| 186,902 \| 602,949 \| 0.2% \| \| Lithuanian \| 167,355 \| 598,508 \| 0.2% \| \| Austrian \| 123,987 \| 584,517 \| 0.2% \| \| Canadian \| 249,309 \| 542,459 \| 0.2% \| \| Iranian \| 392,051 \| 519,658 \| 0.2% \| \| Armenian \| 282,012 \| 458,841 \| 0.1% \| \| Romanian \| 251,069 \| 450,751 \| 0.1% \| \| Broadly "Northern European" (No country specified) \| 273,675 \| 434,292 \| 0.1% \| \| Croatian \| 128,623 \| 389,272 \| 0.1% \| \| Belgian \| 96,361 \| 316,493 \| 0.1% \| \| Turkish \| 168,354 \| 239,667 \| 0.07% \| \| Pennsylvania German \| 155,563 \| 228,634 \| 0.07% \| \| "Czechoslovakian" (Not further specified) \| 79,992 \| 227,217 \| 0.07% \| \| Albanian \| 182,625 \| 223,984 \| 0.07% \| \| "Yugoslavian" (Not further specified) \| 129,759 \| 198,687 \| 0.06% \| \| Serbian \| 96,388 \| 191,538 \| 0.06% \| \| Afghan \| 169,255 \| 189,493 \| 0.06% \| \| Slovene \| 48,809 \| 153,589 \| 0.05% \| \| Israeli \| 80,336 \| 144,202 \| 0.04% \| \| Broadly "Slavic" (No country specified) \| 57,491 \| 140,956 \| 0.04% \| \| Bulgarian \| 75,386 \| 106,896 \| 0.03% \| \| Assyrian \| 64,349 \| 93,542 \| 0.03% \| \| Latvian \| 33,742 \| 91,859 \| 0.03% \| \| Cajun \| 59,046 \| 91,706 \| 0.03% \| \| Australian \| 37,180 \| 88,999 \| 0.03% \| \| Macedonian \| 39,586 \| 65,107 \| 0.02% \| \| Basque \| 24,219 \| 62,731 \| 0.02% \| \| Icelandic \| 18,978 \| 53,415 \| 0.02% \| Genetics A 2015 genetic study published in the American Journal of Human Genetics analyzed the genetic ancestry of 148,789 European Americans. The study concluded that British/Irish ancestry is the most common European ancestry among white Americans, with this component ranging between 20% (Wisconsin, Minnesota, North Dakota) and 55% (Mississippi, Tennessee, Arkansas). These states strongly correlated with those where the largest number of people identified with "American" ancestry on the census. Many white Americans also have ancestry from multiple countries. According to the 2022 American Community Survey, 76,678,228 Americans identified with multiple European, Middle Eastern, or North African ancestry groups, with the large majority of these identifying with various European groups. Historical and present definitions Main article: Definitions of whiteness in the United States Further information: One-drop rule Definitions of who is "White" have changed throughout the history of the United States. US census definition The term "white American" can encompass many different ethnic groups. Although the United States census purports to reflect a social definition of race, the social dimensions of race are more complex than Census criteria. The 2000 US census states that racial categories "generally reflect a social definition of race recognized in this country. They do not conform to any biological, anthropological or genetic criteria." The Census question on race lists the categories White or European American, Black or African American, American Indian and Alaska Native, Native Hawaiian or Other Pacific Islander, Asian, plus "Some other race", with the respondent having the ability to mark more than one racial or ethnic category. The Census Bureau defines White people as follows: "White" refers to a person having origins in any of the original peoples of Europe, the Middle East or North Africa. It includes people who indicated their race(s) as "White" or reported entries such as German, Italian, Lebanese, Arab, Moroccan, or Caucasian. In US census documents, the designation White overlaps, as do all other official racial categories, with the term Hispanic or Latino, which was introduced in the 1980 census as a category of ethnicity, separate and independent of race, despite treating as ethnic groups nationalities from the Americas as ethnically and racially diverse as the United States. Hispanic and Latino Americans as a whole make up a racially diverse group and are the largest minority in the country. Beginning in 1930, Mexican was added as a distinct race on the US census with the explanation that "practically all Mexican laborers are of a racial mixture difficult to classify". The Mexican racial category was removed in 1940, with new direction that "Mexicans are to be regarded as white unless definitely of Indian or other nonwhite race"; this was continued in 1950. 1970 saw the creation of the Spanish Origin category, which superseded previous classifications for Mexicans, Central and South Americans and is now represented by the Hispanic or Latino "ethnic" category. Hispanic or Latino was again to be raised to racial status for the 2020 census (along with Middle Eastern and North African), but this was canceled by President Donald J. Trump. The characterization of Middle Eastern and North African Americans as white has been a matter of controversy. In the early 20th century, there were a number of cases where people of Arab descent were denied entry into the United States or deported, because they were characterized as nonwhite. In the early 21st century, MENA (Middle Eastern, North African) Americans began lobbying for the creation of their own racial group and were successful; in 2015, the US Census Bureau announced that it had responded to their requests and would add a "Middle Eastern and North African" racial category to the 2020 census. The Trump administration nullified this change after coming to power in 2016. However, in 2024, the Office of Management and Budget under the Biden administration reinstated the proposed changes, announcing that the race categories used by the federal government would be updated, and that Middle Eastern and North African Americans will no longer be classified as white in the upcoming 2030 Census, and Hispanic and Latino will also be treated similar to a racial, rather than ethnic, category. The Census Bureau defines the planned definition of White people as follows: "Individuals with origins in any of the original peoples of Europe, including, for example, English, German, Irish, Italian, Polish, and Scottish." President Abraham Lincoln was descended from Samuel Lincoln and was of English and Welsh ancestry. Gloria Vanderbilt, noted artist and designer, was of Dutch descent. In cases where individuals do not self-identify, the US census parameters for race give each national origin a racial value. On some government documents, such as the 2007 SEER program's Coding and Staging Manual, people who reported Muslim (or a sect of Islam such as Shia or Sunni), Jewish, Zoroastrian, Caucasian, Middle Eastern, North African, Mexican, Central American or South American ethnicity as their race in the "Some other race" section, without noting a country of origin or Native American tribal affiliation, were automatically tallied as White. The 1990 US census Public Use Microdata Sample (PUMS) listed "Caucasian" or "Aryan" among other terms as subgroups of "white" in their ancestry code listing, but 2005 and proceeding years of PUMS codes do not. Social definition Social perceptions of whiteness have evolved over the course of American history. For example, Benjamin Franklin commented that the Saxons of Germany and the English "make the principal Body of White People on the Face of the Earth". Historically, many groups of European descent, such as the Irish, Italians, Greeks, Poles, and Spaniards, were not readily integrated into mainstream American society. The perception of Finns as a Mongoloid rather than a European peoples led to prejudice and discrimination against Finns and debates surrounding Finnish whiteness. In the contemporary United States, any one of European descent is typically considered to be white. David Roediger argues that the construction of the white race in the United States was an effort to mentally distance slave owners from slaves. The process of officially being defined as white by law often came about in court disputes over pursuit of citizenship, which made the social integration of other racial groups, such as African-Americans, difficult for decades (Jim Crow laws). Demographic information See also: Americans, White people, and White demographic decline \| White alone 1790–2020 \| \| Year \| Population \| % ofthe US \| % change(raw) \| % change (share) \| \| 1790 \| 3,172,006 \| 80.7 \| \| \| \| 1800 \| 4,306,446 \| 81.1 \| 35.8% \| 0.4 \| \| 1810 \| 5,862,073 \| 81.0 \| 36.1% \| -0.1 \| \| 1820 \| 7,866,797 \| 81.6 \| 34.2% \| 0.6 \| \| 1830 \| 10,532,060 \| 81.9 \| 33.9% \| 0.3 \| \| 1840 \| 14,189,705 \| 83.2 \| 34.7% \| 1.3 \| \| 1850 \| 19,553,068 \| 84.3 \| 37.8% \| 0.9 \| \| 1860 \| 26,922,537 \| 85.6 \| 37.7% \| 1.3 \| \| 1870 \| 33,589,377 \| 87.1 \| 24.8% \| 1.5 \| \| 1880 \| 43,402,970 \| 86.5 \| 29.2% \| -0.6 \| \| 1890 \| 55,101,258 \| 87.5 \| 26.9% \| 1.0 \| \| 1900 \| 66,809,196 \| 87.9 \| 21.2% \| 0.4 \| \| 1910 \| 81,731,957 \| 88.9 \| 22.3% \| 1.0 \| \| 1920 \| 94,820,915 \| 89.7 \| 16.0% \| 1.2 \| \| 1930 \| 110,286,740 \| 89.8 (highest) \| 16.3% \| 0.1 \| \| 1940 \| 118,214,870 \| 89.8 \| 7.2% \| 0.0 \| \| 1950 \| 134,942,028 \| 89.5 \| 14.1% \| -0.3 \| \| 1960 \| 158,831,732 \| 88.6 \| 17.7% \| -0.9 \| \| 1970 \| 178,119,221 \| 87.6 \| 12.1% \| -1.1 \| \| 1980 \| 188,371,622 \| 83.1 \| 5.8% \| -4.4 \| \| 1990 \| 199,686,070 \| 80.3 \| 6.0% \| -2.8 \| \| 2000 \| 211,460,626 \| 75.1 \| 5.9% \| -4.8 \| \| 2010 \| 223,553,265 \| 72.4 \| 5.7% \| -2.7 \| \| 2020 \| 204,277,273 \| 61.6 \| 8.4% \| -9.8 \| \| Source: United States census bureau. \| The fifty states, the District of Columbia, and Puerto Rico as of the 2020 United States census White Americans constitute the majority of the 332 million people living in the United States, with 71% of the population in the 2020 United States census, including 61.6% who identified as "white alone". This represented a 10.6 percentage point national white demographic decline, from a 72.4% share of the US's self-identified white alone population in 2010.[note 1] The white birth rate is below the replacement level. The largest ethnic groups (by ancestry) among White Americans were English or British, followed by Germans and Irish. In the 1980 census 49,598,035 Americans cited that they were of English ancestry, making them 26% of the country and the largest group at the time, and in fact larger than the population of England itself. Slightly more than half of these people would cite that they were of "American" ancestry on subsequent censuses and virtually everywhere that "American" ancestry predominates on the 2000 census corresponds to places where "English" predominated on the 1980 census. White American groups according to the census \| Years \| Non-Hispanic Whites \| White Hispanics \| Total \| \| # \| % \| # \| % \| \| 2020 \| 191,697,647 \| 57.84% \| 12,579,626 \| 3.80% \| 204,277,273 \| Geographic distribution White Americans alone (including White Hispanics) are the majority racial group in most of the United States. As of 2022, they are not the majority in Hawaii, California, Texas, New Mexico, Nevada, and Maryland, making up just under half of the population in the last four states. If White Hispanics are excluded, they are also a minority in Georgia. They are also a minority in many American Indian reservations, parts of the South, especially areas part of the "black belt", the District of Columbia, all US territories, and in many urban areas throughout the country. However, when including multiracial Americans, those who identify as part or fully White make up the majority of the population in every state except for Hawaii, along with Puerto Rico. Overall the highest concentration of those referred to as "non-Hispanic whites" by the Census Bureau are found in the Midwest, New England, the northern Rocky Mountain states, Kentucky, West Virginia, and East Tennessee. The lowest concentration of whites is found in southern, mid-Atlantic, and southwestern states. White Population in all 50 states and D.C. (2020 Census) \| State or district \| Total Population \| White alone population \| % White Alone \| White alone or in any combination population \| % White Alone or in Combination \| \| Alabama \| 5,024,279 \| 3,220,452 \| 64.1% \| 3,458,850 \| 68.8% \| \| Alaska \| 733,391 \| 435,392 \| 59.4% \| 516,525 \| 70.4% \| \| Arizona \| 7,151,502 \| 4,322,337 \| 60.4% \| 5,271,038 \| 73.7% \| \| Arkansas \| 3,011,524 \| 2,114,512 \| 70.2% \| 2,317,826 \| 77.0% \| \| California \| 39,538,223 \| 16,296,122 \| 41.2% \| 21,597,610 \| 54.6% \| \| Colorado \| 5,773,714 \| 4,082,927 \| 70.7% \| 4,757,752 \| 82.4% \| \| Connecticut \| 3,605,944 \| 2,395,128 \| 66.4% \| 2,692,022 \| 74.7% \| \| Delaware \| 989,948 \| 597,763 \| 60.4% \| 665,198 \| 67.2% \| \| District of Columbia \| 689,545 \| 273,194 \| 39.6% \| 319,816 \| 46.4% \| \| Florida \| 21,538,187 \| 12,422,961 \| 57.7% \| 15,758,296 \| 73.2% \| \| Georgia \| 10,711,908 \| 5,555,483 \| 51.9% \| 6,212,741 \| 58.0% \| \| Hawaii \| 1,455,271 \| 333,261 \| 22.9% \| 609,215 \| 41.9% \| \| Idaho \| 1,839,106 \| 1,510,360 \| 82.1% \| 1,659,230 \| 90.2% \| \| Illinois \| 12,812,508 \| 7,868,227 \| 61.4% \| 8,934,277 \| 69.7% \| \| Indiana \| 6,785,528 \| 5,241,795 \| 77.2% \| 5,653,387 \| 83.3% \| \| Iowa \| 3,190,369 \| 2,694,521 \| 84.5% \| 2,865,585 \| 89.8% \| \| Kansas \| 2,937,880 \| 2,222,462 \| 75.6% \| 2,490,266 \| 84.8% \| \| Kentucky \| 4,505,836 \| 3,711,254 \| 82.4% \| 3,942,244 \| 87.5% \| \| Louisiana \| 4,657,757 \| 2,657,652 \| 57.1% \| 2,903,192 \| 62.3% \| \| Maine \| 1,362,359 \| 1,237,041 \| 90.8% \| 1,299,963 \| 95.4% \| \| Maryland \| 6,177,224 \| 3,007,874 \| 48.7% \| 3,421,858 \| 55.4% \| \| Massachusetts \| 7,029,917 \| 4,896,037 \| 69.6% \| 5,399,122 \| 76.8% \| \| Michigan \| 10,077,331 \| 7,444,974 \| 73.9% \| 8,044,575 \| 79.8% \| \| Minnesota \| 5,706,494 \| 4,423,146 \| 77.5% \| 4,748,348 \| 83.2% \| \| Mississippi \| 2,961,279 \| 1,658,893 \| 56.0% \| 1,759,356 \| 59.4% \| \| Missouri \| 6,154,913 \| 4,740,335 \| 77.0% \| 5,132,279 \| 83.4% \| \| Montana \| 1,084,225 \| 916,524 \| 84.5% \| 985,660 \| 90.9% \| \| Nebraska \| 1,961,504 \| 1,538,052 \| 78.4% \| 1,674,853 \| 85.4% \| \| Nevada \| 3,104,614 \| 1,588,463 \| 51.2% \| 1,981,814 \| 63.8% \| \| New Hampshire \| 1,377,529 \| 1,216,203 \| 88.3% \| 1,290,770 \| 93.7% \| \| New Jersey \| 9,288,994 \| 5,112,280 \| 55.0% \| 5,897,538 \| 63.5% \| \| New Mexico \| 2,117,522 \| 1,078,937 \| 51.0% \| 1,485,973 \| 70.2% \| \| New York \| 20,201,249 \| 11,143,349 \| 55.2% \| 12,534,037 \| 62.0% \| \| North Carolina \| 10,439,388 \| 6,488,459 \| 62.2% \| 7,128,036 \| 68.3% \| \| North Dakota \| 779,094 \| 645,938 \| 82.9% \| 685,762 \| 88.0% \| \| Ohio \| 11,799,448 \| 9,080,688 \| 77.0% \| 9,717,936 \| 82.4% \| \| Oklahoma \| 3,959,353 \| 2,514,885 \| 63.5% \| 2,991,001 \| 75.5% \| \| Oregon \| 4,237,256 \| 3,169,096 \| 74.8% \| 3,593,558 \| 84.8% \| \| Pennsylvania \| 13,002,700 \| 9,750,687 \| 75.0% \| 10,451,170 \| 80.4% \| \| Rhode Island \| 1,097,379 \| 782,920 \| 71.3% \| 860,658 \| 78.4% \| \| South Carolina \| 5,118,425 \| 3,243,442 \| 63.4% \| 3,516,966 \| 68.7% \| \| South Dakota \| 886,667 \| 715,336 \| 80.7% \| 759,608 \| 85.7% \| \| Tennessee \| 6,910,840 \| 4,990,938 \| 72.2% \| 5,379,080 \| 77.8% \| \| Texas \| 29,145,505 \| 14,609,365 \| 50.1% \| 19,528,528 \| 67.0% \| \| Utah \| 3,271,616 \| 2,573,413 \| 78.7% \| 2,839,674 \| 86.8% \| \| Vermont \| 643,077 \| 577,751 \| 89.8% \| 613,912 \| 95.5% \| \| Virginia \| 8,631,393 \| 5,208,856 \| 60.3% \| 5,848,488 \| 67.8% \| \| Washington \| 7,705,281 \| 5,130,920 \| 66.6% \| 5,912,348 \| 76.7% \| \| West Virginia \| 1,793,716 \| 1,610,749 \| 89.8% \| 1,692,816 \| 94.4% \| \| Wisconsin \| 5,893,718 \| 4,737,545 \| 80.4% \| 5,080,160 \| 86.2% \| \| Wyoming \| 576,851 \| 488,374 \| 84.7% \| 530,590 \| 92.0% \| \| United States \| 331,449,281 \| 204,277,273 \| 61.6% \| 235,411,507 \| 71.0% \| White population in all 50 states (plus D.C. and Puerto Rico), as of 2022 \| State or territory \| Population (2022 est.) \| White alone (Non Hispanic) \| White alone \| White alone or in combination \| \| Alabama \| 5,074,296 \| 64.1% \| 65.1% \| 69.8% \| \| Alaska \| 733,583 \| 57.4% \| 59.6% \| 72.6% \| \| Arizona \| 7,359,197 \| 51.8% \| 57.8% \| 76.5% \| \| Arkansas \| 3,045,637 \| 67.5% \| 69.1% \| 79.3% \| \| California \| 39,029,344 \| 33.7% \| 38.9% \| 56.6% \| \| Colorado \| 5,839,926 \| 65.0% \| 70.3% \| 84.3% \| \| Connecticut \| 3,626,205 \| 62.0% \| 65.0% \| 75.2% \| \| Delaware \| 1,018,396 \| 58.9% \| 59.9% \| 68.2% \| \| District of Columbia \| 671,803 \| 36.7% \| 38.4% \| 47.3% \| \| Florida \| 22,244,824 \| 50.8% \| 55.9% \| 73.9% \| \| Georgia \| 10,912,876 \| 49.6% \| 51.3% \| 58.7% \| \| Hawaii \| 1,440,196 \| 20.7% \| 22.2% \| 43.8% \| \| Idaho \| 1,939,033 \| 79.0% \| 81.9% \| 91.5% \| \| Illinois \| 12,582,032 \| 58.5% \| 61.1% \| 71.3% \| \| Indiana \| 6,833,037 \| 76.0% \| 77.5% \| 84.1% \| \| Iowa \| 3,200,517 \| 82.8% \| 84.4% \| 90.6% \| \| Kansas \| 2,937,150 \| 73.1% \| 76.3% \| 85.8% \| \| Kentucky \| 4,512,310 \| 82.2% \| 83.1% \| 88.8% \| \| Louisiana \| 4,590,241 \| 56.7% \| 57.6% \| 63.9% \| \| Maine \| 1,385,340 \| 90.2% \| 90.8% \| 95.9% \| \| Maryland \| 6,164,660 \| 47.1% \| 48.4% \| 55.4% \| \| Massachusetts \| 6,981,974 \| 67.0% \| 68.8% \| 77.8% \| \| Michigan \| 10,034,118 \| 72.6% \| 74.0% \| 80.7% \| \| Minnesota \| 5,717,184 \| 76.2% \| 77.2% \| 83.5% \| \| Mississippi \| 2,940,057 \| 55.3% \| 55.7% \| 59.8% \| \| Missouri \| 6,177,957 \| 76.6% \| 77.6% \| 84.6% \| \| Montana \| 1,122,867 \| 83.5% \| 85.1% \| 91.7% \| \| Nebraska \| 1,967,923 \| 75.8% \| 78.4% \| 86.8% \| \| Nevada \| 3,177,772 \| 44.4% \| 49.1% \| 65.2% \| \| New Hampshire \| 1,395,231 \| 86.6% \| 87.5% \| 93.9% \| \| New Jersey \| 9,261,699 \| 51.5% \| 54.1% \| 64.8% \| \| New Mexico \| 2,113,344 \| 34.8% \| 46.4% \| 70.8% \| \| New York \| 19,677,152 \| 52.9% \| 55.2% \| 63.4% \| \| North Carolina \| 10,698,973 \| 60.7% \| 62.2% \| 69.4% \| \| North Dakota \| 779,261 \| 82.0% \| 83.2% \| 88.2% \| \| Ohio \| 11,756,058 \| 76.1% \| 77.1% \| 83.1% \| \| Oklahoma \| 4,019,800 \| 62.6% \| 65.2% \| 78.6% \| \| Oregon \| 4,240,137 \| 71.6% \| 74.5% \| 85.8% \| \| Pennsylvania \| 12,972,008 \| 73.1% \| 74.4% \| 80.9% \| \| Puerto Rico \| 3,221,789 \| 0.6% \| 26.3% \| 60.7% \| \| Rhode Island \| 1,093,734 \| 68.2% \| 70.5% \| 80.1% \| \| South Carolina \| 5,282,634 \| 62.5% \| 63.6% \| 69.5% \| \| South Dakota \| 909,824 \| 79.9% \| 80.8% \| 86.7% \| \| Tennessee \| 7,051,339 \| 71.9% \| 73.0% \| 79.5% \| \| Texas \| 30,029,572 \| 38.9% \| 47.6% \| 70.6% \| \| Utah \| 3,380,800 \| 75.6% \| 79.2% \| 87.7% \| \| Vermont \| 647,064 \| 90.2% \| 90.9% \| 96.2% \| \| Virginia \| 8,683,619 \| 58.7% \| 60.2% \| 68.6% \| \| Washington \| 7,785,786 \| 63.5% \| 65.9% \| 77.7% \| \| West Virginia \| 1,775,156 \| 89.8% \| 90.3% \| 94.9% \| \| Wisconsin \| 5,892,539 \| 79.0% \| 80.4% \| 88.0% \| \| Wyoming \| 581,381 \| 81.4% \| 84.6% \| 92.6% \| Although all large geographical areas are dominated by White Americans, much larger differences can be seen between specific parts of large cities, as well as regions within certain states, especially in the South, where many rural regions are predominantly African American, or the Southwest, where large rural areas, such as the Colorado Plateau, are predominantly populated by Native Americans. States with the highest percentages of White Americans, either White Alone or in combination with another race as of 2020:[failed verification] Vermont 95.6% Maine 95.4% West Virginia 94.4% New Hampshire 93.7% Wyoming 92.0% Montana 90.9% Idaho 90.2% Iowa 89.8% North Dakota 88.0% Kentucky 87.5% States with the highest percentages of non-Latino/Hispanic whites, alone or in combination, as of 2020:[failed verification] Maine 92.0% Vermont 91.3% New Hampshire 91.3% West Virginia 90.4% Wyoming 90.7% Idaho 90.7% Utah 88.7% Iowa 88.7% Montana 86.7% Nebraska 86.0% Income and educational attainment Main article: Affluence in the United States § Race Further information: Personal income in the United States and Household income in the United States personal and household income in the United States Census in 2005 \| \| \| This section needs to be updated. Please help update this article to reflect recent events or newly available information. (August 2020) \| White Americans have the second highest median household income and personal income levels in the nation, by cultural background, only behind Asian Americans. The median income per household member was also the highest, since White Americans had the smallest households of any racial demographic in the nation. In 2006, the median individual income of a White American age 25 or older was $33,030, with those who were full-time employed, and of age 25 to 64, earning $34,432. Since 42% of all households had two income earners, the median household income was considerably higher than the median personal income, which was $48,554 in 2005. Jewish Americans rank first in household income, personal income, and educational attainment among White Americans. In 2005, White households had a median household income of $48,977, which is 10% above the national median of $44,389. Among Cuban Americans, with 86% classified as White, those born in the US have a higher median income and educational attainment level than most other Whites. The poverty rates for White Americans are the second-lowest of any racial group, with 11% of non-Hispanic white individuals living below the poverty line, 3% lower than the national average. However, due to Whites' majority status, 48% of Americans living in poverty are non-Hispanic white. White Americans' educational attainment is the second-highest in the country, after Asian Americans'. Overall, nearly one-third of White Americans had a Bachelor's degree, with the educational attainment for Whites being higher for those born outside the United States: 38% of foreign born, and 30% of native born Whites had a college degree. Both figures are above the national average of 27%. Gender income inequality was the greatest among Whites, with White men outearning White women by 48%. Census Bureau data for 2005 reveals that the median income of White females was lower than that of males of all races. In 2005, the median income for White American females was only slightly higher than that of African American females. White Americans are more likely to live in suburbs and small cities than their black counterparts. Proportion in each county White American (Alone) population distribution over time 1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1940 1950 1960 1970 1980 1990 2000 2010 2020 White Americans of one race or alone from 2000 to 2020 White American (of one race or alone) population as of 2000, 2010 and 2020 censuses \| State \| 2000 \| 2010 \| 2020 \| Growth \| \| Pop. 2000 \| % 2000 \| Pop. 2010 \| % 2010 \| Pop. 2020 \| % 2020 \| % growth between 2000 and 2010 \| \| Alabama \| 3,162,808 \| 71.1% \| 3,275,394 \| 68.5% \| 3,220,452 \| 64.1% \| +3.6% \| \| Alaska \| 434,534 \| 69.3% \| 473,576 \| 66.7% \| 435,392 \| 59.4% \| +9.0% \| \| Arizona \| 3,873,611 \| 75.5% \| 4,667,121 \| 73.0% \| 4,322,337 \| 60.4% \| +20.5% \| \| Arkansas \| 2,138,598 \| 80.0% \| 2,245,229 \| 77.0% \| 2,114,512 \| 70.2% \| +5.0% \| \| California \| 20,170,059 \| 59.5% \| 21,453,934 \| 57.6% \| 16,296,122 \| 41.2% \| +6.4% \| \| Colorado \| 3,560,005 \| 82.8% \| 4,089,202 \| 81.3% \| 4,082,927 \| 70.7% \| +14.9% \| \| Connecticut \| 2,780,355 \| 81.6% \| 2,772,410 \| 77.6% \| 2,395,128 \| 66.4% \| -0.3% \| \| Delaware \| 584,773 \| 74.6% \| 618,617 \| 68.9% \| 597,763 \| 60.4% \| +5.8% \| \| District of Columbia \| 176,101 \| 30.8% \| 231,471 \| 38.5% \| 273,194 \| 39.4% \| +31.4% \| \| Florida \| 12,465,029 \| 78.0% \| 14,109,162 \| 75.0% \| 12,422,961 \| 57.7% \| +13.2% \| \| Georgia \| 5,327,281 \| 65.1% \| 5,787,440 \| 59.7% \| 5,555,483 \| 51.9% \| +8.6% \| \| Hawaii \| 294,102 \| 24.3% \| 336,599 \| 24.7% \| 333,261 \| 22.9% \| +14.4% \| \| Idaho \| 1,177,304 \| 91.0% \| 1,396,487 \| 89.1% \| 1,510,360 \| 82.1% \| +18.6% \| \| Illinois \| 9,125,471 \| 73.5% \| 9,177,877 \| 71.5% \| 7,868,227 \| 61.4% \| +0.6% \| \| Indiana \| 5,320,022 \| 87.5% \| 5,467,906 \| 84.3% \| 5,241,791 \| 77.2% \| +2.8% \| \| Iowa \| 2,748,640 \| 93.9% \| 2,781,561 \| 91.3% \| 2,694,521 \| 84.5% \| +1.2% \| \| Kansas \| 2,313,944 \| 86.1% \| 2,391,044 \| 83.8% \| 2,222,462 \| 75.6% \| +3.3% \| \| Kentucky \| 3,640,889 \| 90.1% \| 3,809,537 \| 87.8% \| 3,711,254 \| 82.4% \| +4.6% \| \| Louisiana \| 2,856,161 \| 63.9% \| 2,836,192 \| 62.6% \| 2,675,652 \| 57.1% \| -0.7% \| \| Maine \| 1,236,014 \| 96.9% \| 1,264,971 \| 95.2% \| 1,237,041 \| 90.8% \| +2.3% \| \| Maryland \| 3,391,308 \| 64.0% \| 3,359,284 \| 58.2% \| 3,007,874 \| 48.7% \| -0.9% \| \| Massachusetts \| 5,367,286 \| 84.5% \| 5,265,236 \| 80.4% \| 4,896,037 \| 69.6% \| -1.9% \| \| Michigan \| 7,966,053 \| 80.2% \| 7,803,120 \| 78.9% \| 7,444,974 \| 73.9% \| -2.0% \| \| Minnesota \| 4,400,282 \| 89.4% \| 4,524,062 \| 85.3% \| 4,423,146 \| 77.5% \| +2.8% \| \| Mississippi \| 1,746,099 \| 61.4% \| 1,754,684 \| 59.1% \| 1,658,893 \| 56% \| +0.5% \| \| Missouri \| 4,748,083 \| 84.9% \| 4,958,770 \| 82.8% \| 4,740,335 \| 77% \| +4.4% \| \| Montana \| 817,229 \| 90.6% \| 884,961 \| 89.4% \| 916,524 \| 84.5% \| +8.3% \| \| Nebraska \| 1,533,261 \| 89.6% \| 1,572,838 \| 86.1% \| 1,538,052 \| 78.4% \| +2.6% \| \| Nevada \| 1,501,886 \| 75.2% \| 1,786,688 \| 66.2% \| 1,588,463 \| 51.2% \| +19.0% \| \| New Hampshire \| 1,186,851 \| 96.0% \| 1,236,050 \| 92.3% \| 1,216,203 \| 88.3% \| +4.1% \| \| New Jersey \| 6,104,705 \| 72.6% \| 6,029,248 \| 68.6% \| 5,112,280 \| 55% \| -1.2% \| \| New Mexico \| 1,214,253 \| 66.8% \| 1,407,876 \| 68.4% \| 1,078,927 \| 51% \| +15.9% \| \| New York \| 12,893,689 \| 67.9% \| 12,740,974 \| 65.7% \| 11,143,349 \| 55.2% \| -1.2% \| \| North Carolina \| 5,804,656 \| 72.1% \| 6,528,950 \| 68.5% \| 6,448,459 \| 62.2% \| +12.5% \| \| North Dakota \| 593,181 \| 92.4% \| 605,449 \| 90.0% \| 645,938 \| 82.9% \| +2.1% \| \| Ohio \| 9,645,453 \| 85.0% \| 9,539,437 \| 82.7% \| 9,080,688 \| 77% \| -1.1% \| \| Oklahoma \| 2,628,434 \| 76.2% \| 2,706,845 \| 72.2% \| 2,514,884 \| 63.5% \| +3.0% \| \| Oregon \| 2,961,623 \| 86.6% \| 3,204,614 \| 83.6% \| 3,169,096 \| 74.8% \| +8.2% \| \| Pennsylvania \| 10,484,203 \| 85.4% \| 10,406,288 \| 81.9% \| 9,750,687 \| 75% \| -0.7% \| \| Rhode Island \| 891,191 \| 85.0% \| 856,869 \| 81.4% \| 782,920 \| 71.3% \| -3.8% \| \| South Carolina \| 2,695,560 \| 67.2% \| 3,060,000 \| 66.2% \| 3,243,442 \| 63.4% \| +13.5% \| \| South Dakota \| 669,404 \| 88.7% \| 699,392 \| 85.9% \| 715,336 \| 80.7% \| +4.5% \| \| Tennessee \| 4,563,310 \| 80.2% \| 4,921,948 \| 77.6% \| 4,990,938 \| 72.2% \| +7.9% \| \| Texas \| 14,799,505 \| 71.0% \| 17,701,552 \| 70.4% \| 14,609,365 \| 50.1% \| +19.6% \| \| Utah \| 1,992,975 \| 89.2% \| 2,379,560 \| 86.1% \| 2,573,413 \| 78.7% \| +19.4% \| \| Vermont \| 589,208 \| 96.8% \| 596,292 \| 95.3% \| 577,751 \| 89.8% \| +1.2% \| \| Virginia \| 5,120,110 \| 72.3% \| 5,486,852 \| 68.6% \| 5,208,856 \| 60.3% \| +7.2% \| \| Washington \| 4,821,823 \| 81.8% \| 5,196,362 \| 77.3% \| 5,130,920 \| 66.6% \| +7.8% \| \| West Virginia \| 1,718,777 \| 95.0% \| 1,739,988 \| 93.9% \| 1,610,749 \| 89.8% \| +1.2% \| \| Wisconsin \| 4,769,857 \| 88.9% \| 4,902,067 \| 86.2% \| 4,737,545 \| 80.4% \| +2.8% \| \| Wyoming \| 454,670 \| 92.1% \| 511,279 \| 90.7% \| 488,374 \| 84.7% \| +12.4% \| \| United States of America \| 211,460,626 \| 75.1% \| 223,553,265 \| 72.4% \| 204,277,273 \| 61.6% \| +5.7% \| White population by state (includes Hispanics who identify as white) \| State \| Pop. 2016 \| % 2016 \| Pop. 2017 \| % 2017 \| percentagegrowth \| numericgrowth \| \| Alabama \| 3,371,066 \| 69.35% \| 3,374,131 \| 69.22% \| -0.13% \| +3,065 \| \| Alaska \| 490,864 \| 66.20% \| 486,724 \| 65.79% \| -0.41% \| -4,140 \| \| Arizona \| 5,753,506 \| 83.28% \| 5,827,866 \| 83.06% \| -0.22% \| +74,360 \| \| Arkansas \| 2,372,843 \| 79.41% \| 2,381,662 \| 79.27% \| -0.14% \| +3,740 \| \| California \| 28,560,032 \| 72.68% \| 28,611,160 \| 72.37% \| -0.31% \| +51,128 \| \| Colorado \| 4,837,197 \| 87.47% \| 4,894,372 \| 87.29% \| -0.18% \| +57,175 \| \| Connecticut \| 2,891,943 \| 80.60% \| 2,879,759 \| 80.26% \| -0.34% \| -12,184 \| \| Delaware \| 667,076 \| 70.02% \| 670,512 \| 69.70% \| -0.32% \| +3,436 \| \| District of Columbia \| 305,232 \| 44.60% \| 313,234 \| 45.14% \| +0.54% \| +8,002 \| \| Florida \| 16,022,497 \| 77.56% \| 16,247,613 \| 77.43% \| -0.13% \| +225,116 \| \| Georgia \| 6,310,426 \| 61.18% \| 6,341,768 \| 60.81% \| -0.37% \| +31,342 \| \| Hawaii \| 370,362 \| 25.92% \| 366,546 \| 25.67% \| -0.25% \| -3,816 \| \| Idaho \| 1,567,868 \| 93.32% \| 1,599,814 \| 93.18% \| -0.2% \| +31,946 \| \| Illinois \| 9,909,184 \| 77.20% \| 9,864,942 \| 77.06% \| -0.14% \| -44,242 \| \| Indiana \| 5,679,252 \| 85.61% \| 5,690,929 \| 85.36% \| -0.25% \| +11,677 \| \| Iowa \| 2,860,136 \| 91.35% \| 2,864,664 \| 91.06% \| -0.29% \| +4,528 \| \| Kansas \| 2,519,340 \| 86.64% \| 2,519,176 \| 86.47% \| -0.17% \| -164 \| \| Kentucky \| 3,901,878 \| 87.96% \| 3,908,964 \| 87.76% \| -0.20% \| +7,086 \| \| Louisiana \| 2,958,471 \| 63.13% \| 2,951,003 \| 63.00% \| -0.13% \| -7,468 \| \| Maine \| 1,261,247 \| 94.81% \| 1,264,744 \| 94.67% \| -0.14% \| +3,497 \| \| Maryland \| 3,572,673 \| 59.30% \| 3,568,679 \| 58.96% \| -0.34% \| -3,994 \| \| Massachusetts \| 5,575,622 \| 81.71% \| 5,576,725 \| 81.29% \| -0.42% \| +1,103 \| \| Michigan \| 7,906,913 \| 79.60% \| 7,914,418 \| 79.44% \| -0.16% \| +7,505 \| \| Minnesota \| 4,687,397 \| 84.84% \| 4,708,215 \| 84.43% \| -0.41% \| +20,818 \| \| Mississippi \| 1,771,276 \| 59.33% \| 1,766,950 \| 59.21% \| -0.12% \| -4,326 \| \| Missouri \| 5,069,869 \| 83.23% \| 5,080,444 \| 83.10% \| -0.13% \| +10,575 \| \| Montana \| 926,475 \| 89.20% \| 935,792 \| 89.08% \| -0.12% \| +9,317 \| \| Nebraska \| 1,693,622 \| 88.78% \| 1,700,881 \| 88.58% \| -0.20% \| +7,259 \| \| Nevada \| 2,208,915 \| 75.15% \| 2,235,657 \| 74.57% \| -0.58% \| +26,742 \| \| New Hampshire \| 1,251,836 \| 93.77% \| 1,256,807 \| 93.59% \| -0.18% \| +4,971 \| \| New Jersey \| 6,499,057 \| 72.38% \| 6,489,409 \| 72.06% \| -0.32% \| -9,648 \| \| New Mexico \| 1,716,662 \| 82.31% \| 1,715,623 \| 82.16% \| -0.15% \| -1,039 \| \| New York \| 13,856,651 \| 69.85% \| 13,807,127 \| 69.56% \| -0.29% \| -49,524 \| \| North Carolina \| 7,212,423 \| 71.01% \| 7,276,995 \| 70.83% \| -0.18% \| +64,572 \| \| North Dakota \| 663,424 \| 87.81% \| 661,217 \| 87.53% \| -0.28% \| -2,207 \| \| Ohio \| 9,578,424 \| 82.41% \| 9,579,207 \| 82.16% \| -0.25% \| +783 \| \| Oklahoma \| 2,923,751 \| 74.56% \| 2,921,390 \| 74.32% \| -0.24% \| -2,361 \| \| Oregon \| 3,569,538 \| 87.29% \| 3,607,515 \| 87.08% \| -0.21% \| +37,977 \| \| Pennsylvania \| 10,525,562 \| 82.31% \| 10,507,780 \| 82.06% \| -0.25% \| -17,782 \| \| Rhode Island \| 892,287 \| 84.37% \| 890,883 \| 84.07% \| -0.30% \| -1,404 \| \| South Carolina \| 3,393,346 \| 68.2% \| 3,440,141 \| 68.47% \| +0.27% \| +46,795 \| \| South Dakota \| 733,199 \| 85.10% \| 738,554 \| 84.92% \| -0.18% \| +5,355 \| \| Tennessee \| 5,231,987 \| 78.68% \| 5,276,748 \| 78.57% \| -0.11% \| +44,761 \| \| Texas \| 22,166,782 \| 79.44% \| 22,404,118 \| 79.15% \| -0.29% \| +237,336 \| \| Utah \| 2,774,606 \| 91.14% \| 2,820,387 \| 90.93% \| -0.21% \| +45,781 \| \| Vermont \| 589,836 \| 94.62% \| 589,163 \| 94.47% \| -0.15% \| -673 \| \| Virginia \| 5,891,174 \| 70.01% \| 5,904,472 \| 69.71% \| -0.30% \| +13,298 \| \| Washington \| 5,820,007 \| 79.93% \| 5,887,060 \| 79.49% \| -0.44% \| +67,053 \| \| West Virginia \| 1,712,647 \| 93.66% \| 1,699,266 \| 93.58% \| -0.08% \| -13,381 \| \| Wisconsin \| 5,049,698 \| 87.47% \| 5,060,891 \| 87.32% \| -0.15% \| +11,193 \| \| Wyoming \| 543,224 \| 92.87% \| 537,396 \| 92.76% \| -0.11% \| -5,828 \| \| United States \| 248,619,303 \| 76.87% \| 249,619,493 \| 76.64% \| -0.23% \| +1,000,190 \| Non-Hispanic population Main article: List of U.S. states by non-Hispanic white population Non-Hispanic white population by state \| State \| Pop. 2016 \| % 2016 \| Pop. 2017 \| % 2017 \| percentagegrowth \| numericgrowth \| \| Alabama \| 3,198,381 \| 65.80% \| 3,196,852 \| 65.58% \| -0.22% \| -1,529 \| \| Alaska \| 454,651 \| 61.31% \| 449,776 \| 60.80% \| -0.51% \| -4,875 \| \| Arizona \| 3,819,881 \| 55.29% \| 3,849,130 \| 54.86% \| -0.43% \| +29,249 \| \| Arkansas \| 2,175,521 \| 72.80% \| 2,177,809 \| 72.49% \| -0.31% \| +2,288 \| \| California \| 14,797,971 \| 37.66% \| 14,696,754 \| 37.17% \| -0.49% \| -101,217 \| \| Colorado \| 3,791,612 \| 68.56% \| 3,827,750 \| 68.26% \| -0.30% \| +36,135 \| \| Connecticut \| 2,428,332 \| 67.68% \| 2,404,792 \| 67.02% \| -0.66% \| -23,540 \| \| Delaware \| 597,728 \| 62.74% \| 599,260 \| 62.30% \| -0.44% \| +1,532 \| \| District of Columbia \| 249,141 \| 36.40% \| 255,387 \| 36.80% \| +0.40% \| +6,246 \| \| Florida \| 11,273,388 \| 54.57% \| 11,343,977 \| 54.06% \| -0.51% \| +70,589 \| \| Georgia \| 5,499,055 \| 53.32% \| 5,507,334 \| 52.81% \| -0.51% \| +8,279 \| \| Hawaii \| 317,026 \| 22.19% \| 312,492 \| 21.89% \| -0.30% \| -4,534 \| \| Idaho \| 1,382,934 \| 82.32% \| 1,408,294 \| 82.02% \| -0.30% \| +25,360 \| \| Illinois \| 7,915,013 \| 61.65% \| 7,849,887 \| 61.32% \| -0.33% \| -65,126 \| \| Indiana \| 5,280,029 \| 79.59% \| 5,280,420 \| 79.20% \| -0.39% \| +391 \| \| Iowa \| 2,696,686 \| 86.13% \| 2,695,962 \| 85.70% \| -0.43% \| -724 \| \| Kansas \| 2,215,920 \| 76.21% \| 2,209,748 \| 75.86% \| -0.35% \| -6,172 \| \| Kentucky \| 3,767,092 \| 84.92% \| 3,768,891 \| 84.61% \| -0.31% \| +1,799 \| \| Louisiana \| 2,760,416 \| 58.91% \| 2,747,730 \| 58.66% \| -0.25% \| -12,686 \| \| Maine \| 1,243,741 \| 93.50% \| 1,246,478 \| 93.30% \| -0.20% \| +2,737 \| \| Maryland \| 3,098,543 \| 51.43% \| 3,077,907 \| 50.86% \| -0.57% \| -20,636 \| \| Massachusetts \| 4,972,010 \| 72.86% \| 4,953,695 \| 72.21% \| -0.65% \| -18,315 \| \| Michigan \| 7,489,609 \| 75.40% \| 7,488,326 \| 75.17% \| -0.23% \| -1,283 \| \| Minnesota \| 4,442,684 \| 80.41% \| 4,455,605 \| 79.89% \| -0.52% \| +12,921 \| \| Mississippi \| 1,697,562 \| 56.86% \| 1,691,566 \| 56.69% \| -0.17% \| -5,996 \| \| Missouri \| 4,855,156 \| 79.71% \| 4,859,227 \| 79.48% \| -0.23% \| +4,071 \| \| Montana \| 897,790 \| 86.44% \| 905,811 \| 86.23% \| -0.21% \| +8,021 \| \| Nebraska \| 1,515,494 \| 79.44% \| 1,516,962 \| 79.00% \| -0.44% \| +1,468 \| \| Nevada \| 1,465,888 \| 49.87% \| 1,470,855 \| 49.06% \| -0.81% \| +4,967 \| \| New Hampshire \| 1,212,377 \| 90.81% \| 1,215,447 \| 90.52% \| -0.29% \| +3,070 \| \| New Jersey \| 5,002,866 \| 55.72% \| 4,962,470 \| 55.10% \| -0.62% \| -40,396 \| \| New Mexico \| 789,869 \| 38.31% \| 783,064 \| 37.50% \| -0.81% \| -6,805 \| \| New York \| 11,047,456 \| 55.69% \| 10,972,959 \| 55.28% \| -0.41% \| -74,497 \| \| North Carolina \| 6,447,852 \| 63.48% \| 6,486,100 \| 63.13% \| -0.35% \| +38,248 \| \| North Dakota \| 641,945 \| 84.96% \| 639,029 \| 84.59% \| -0.37% \| -2,916 \| \| Ohio \| 9,229,932 \| 79.41% \| 9,219,577 \| 79.08% \| -0.33% \| -10,355 \| \| Oklahoma \| 2,592,571 \| 66.12% \| 2,581,568 \| 65.67% \| -0.45% \| -11,003 \| \| Oregon \| 3,115,656 \| 76.25% \| 3,139,685 \| 75.79% \| -0.46% \| +24,029 \| \| Pennsylvania \| 9,841,619 \| 76.96% \| 9,796,510 \| 76.50% \| -0.44% \| -45,109 \| \| Rhode Island \| 773,405 \| 73.13% \| 768,229 \| 72.50% \| -0.63% \| -5,176 \| \| South Carolina \| 3,165,176 \| 63.82% \| 3,203,045 \| 63.75% \| -0.07% \| +37,869 \| \| South Dakota \| 710,509 \| 82.47% \| 714,881 \| 82.20% \| -0.27% \| +4,372 \| \| Tennessee \| 4,931,609 \| 74.17% \| 4,963,780 \| 73.91% \| -0.26% \| +32,171 \| \| Texas \| 11,862,697 \| 42.51% \| 11,886,381 \| 42.00% \| -0.51% \| +23,684 \| \| Utah \| 2,400,885 \| 78.86% \| 2,434,785 \| 78.49% \| -0.37% \| +33,900 \| \| Vermont \| 580,238 \| 93.08% \| 579,149 \| 92.86% \| -0.22% \| -1,089 \| \| Virginia \| 5,247,231 \| 62.36% \| 5,241,262 \| 61.88% \| -0.48% \| -5,969 \| \| Washington \| 5,049,817 \| 69.36% \| 5,091,370 \| 68.75% \| -0.61% \| +41,553 \| \| West Virginia \| 1,688,472 \| 92.33% \| 1,674,557 \| 92.22% \| -0.11% \| -13,915 \| \| Wisconsin \| 4,710,928 \| 81.60% \| 4,713,993 \| 81.34% \| -0.26% \| +3,065 \| \| Wyoming \| 492,235 \| 84.16% \| 486,565 \| 83.99% \| -0.17% \| -5,670 \| \| United States \| 197,834,599 \| 61.17% \| 197,803,083 \| 60.73% \| -0.44% \| -31,516 \| Politics A majority of White Americans have voted for the Republican Party since the 1968 United States presidential election, with the 1964 United States presidential election being the last election when the Democratic Party won a majority of White voters.[citation needed] In 2012, 88% of Romney voters were non-Hispanic white while 56% of Obama voters were non-Hispanic white. In the 2008 presidential election, John McCain won 55% of non-Hispanic white votes. In the 2010 House election, Republicans won 60% of the non-Hispanic white votes. Some academics and commentators have argued that Donald Trump's presidential election victory in 2016 is an example of "White backlash". \| Year \| Candidate ofthe plurality \| Political party \| % ofNon-Hispanic Whitevote[citation needed] \| Result \| --- --- \| 1980 \| Ronald Reagan \| Republican Party \| 56% \| Won \| \| 1984 \| Ronald Reagan \| Republican \| 66% \| Won \| \| 1988 \| George H. W. Bush \| Republican \| 59% \| Won \| \| 1992 \| George H. W. Bush \| Republican \| 40% \| Lost \| \| 1996 \| Bob Dole \| Republican \| 46% \| Lost \| \| 2000 \| George W. Bush \| Republican \| 55% \| Won \| \| 2004 \| George W. Bush \| Republican \| 58% \| Won \| \| 2008 \| John McCain \| Republican \| 55% \| Lost \| \| 2012 \| Mitt Romney \| Republican \| 59% \| Lost \| \| 2016 \| Donald Trump \| Republican \| 57% \| Won \| \| 2020 \| Donald Trump \| Republican \| 58% \| Lost \| \| 2024 \| Donald Trump \| Republican \| 57% \| Won \| Health Main article: Health status of White Americans Culture From their earliest presence in North America, White Americans have contributed literature, art, cinema, religion, agricultural skills, foods, science and technology, fashion and clothing styles, music, language, legal system, political system, and social and technological innovation to American culture. White American culture derived its earliest influences from English, German, Scottish, Welsh, and Irish settlers and is quantitatively the largest proportion of American culture. The overall American culture reflects White American culture. The culture has been developing since long before the United States formed a separate country. Much of White American culture shows influences from British culture. Colonial ties to Great Britain spread the English language, legal system and other cultural attributes. Albion's Seed: Four British Folkways in America In his 1989 book Albion's Seed: Four British Folkways in America, David Hackett Fischer explores the details of the folkways of four groups of settlers from the British Isles that moved to the American colonies during the 17th and 18th centuries from distinct regions of Britain and Ireland. His thesis is that the culture of each group persisted (albeit in modified form), providing the basis for the modern United States. According to Fischer, the foundation of America's four regional cultures was formed from four mass migrations from four regions of the British Isles by four distinct ethno-cultural groups. New England's formative period occurred between 1629 and 1640 when Puritans, mostly from East Anglia, settled there, thus forming the basis for the New England regional culture. The next mass migration was of southern English Cavaliers and their working class British Isles servants to the Chesapeake Bay region between 1640 and 1675. This spawned the creation of the American Southern culture. Then, between 1675 and 1725, thousands of Irish, Cornish, English and Welsh Quakers plus many Germans sympathetic to Quaker ideas, led by William Penn, settled the Delaware Valley. This resulted in the formation of the General American culture, although, according to Fischer, this is really a "regional culture", even if it does today encompass most of the US from the mid-Atlantic states to the Pacific Coast. Finally, a huge number of settlers from the borderlands between England and Scotland, sometimes by way of northern Ireland, migrated to Appalachia between 1717 and 1775. This resulted in the formation of the Upland South regional culture, which has since expanded to the west to West Texas and parts of the American Southwest. In his book, Fischer brings up several points. He states that the US is not a country with one "general" culture and several "regional" cultures, as is commonly thought. Rather, there are only four regional cultures as described above, and understanding this helps one to more clearly understand American history as well as contemporary American life. Fischer asserts that it is not only important to understand where different groups came from, but when. All population groups have, at different times, their own unique set of beliefs, fears, hopes and prejudices. When different groups moved to America and brought certain beliefs and values with them, these ideas became, according to Fischer, more or less frozen in time, even if they eventually changed in their original place of origin. Admixture See also: Race and genetics Admixture in non-Hispanic whites White Americans have a mean of 98.6% European, 0.19% sub-Saharan African, and 0.18% Native American ancestry according to a study specifically focusing on customers who took a 23andMe DNA test. However, non-European ancestry in White Americans is highly variable; for example, Black ancestry (2% or greater) is found in over five percent of European Americans in Louisiana and South Carolina, and Native American ancestry (2% or greater) is found in over three percent of European Americans in Louisiana and North Dakota. African ancestry is most common in the South and least common in the Midwest; Native American ancestry is more common in Western states than Eastern states. Older studies have also been performed. DNA analysis on White Americans by geneticist Mark D. Shriver showed an average of 0.7% sub-Saharan African admixture and 3.2% Native American admixture. The same author, in another study, claimed that about 30% of all White Americans, approximately 66 million people, have a median of 2.3% of Black African admixture. Shriver discovered his ancestry is 10 percent African, and Shriver's partner in DNA Print Genomics, J.T. Frudacas, contradicted him two years later stating "Five percent of European Americans exhibit some detectable level of African ancestry." In a 2007 study, Gonçalves et al. reported sub-Saharan and Amerindian mtDNA lineages at a frequency of 3.1% (respectively 0.9% and 2.2%) in a sample of 1387 American Caucasians as compared to 62% in white Brazilians (respectively 29% and 33%), 98% for white Colombians (respectively 8% and 90%) and 96% for white Costa Ricans (respectively 7% and 89%). A 2003 study on Y-chromosomes and mtDNA found African admixture in European Americans to be "below the limits of detection". Admixture in Hispanic whites In contrast to non-Hispanic or Latino whites, whose average European ancestry is 98.6%, genetic research has found much higher amounts of non-European ancestry in White Hispanics. A study from 2014 found that the average European admixture among self-identified White Hispanic and Latino Americans is 73%, while the average European admixture for Hispanic Americans overall (regardless of their self-identified race) was 65.1%. However, this study only obtained its genetic data from people who took a paid ancestry test from 23andMe, and as such may not be fully representative of the general Hispanic population in the US. Another study from 2019 focusing on Native American ancestry in the general US population, which did not differentiate Hispanics by self-identified race, found an average of around 55% European ancestry among all Hispanic Americans, compared to over 98% for self identified non-Hispanic whites, and around 20% in African-Americans. See also Europe portal Middle East portal United States portal Health status of White Americans White demographic decline#United States American ancestry – People in the United States who self-identify their ancestral origin or descent as "American" Anglo Emigration from Europe European Americans Hyphenated American Middle Eastern Americans Non-Hispanic or Latino whites – White Americans who are not HispanicPages displaying short descriptions of redirect targets Race and ethnicity in the United States Racism in the United States Stereotypes of white Americans White Anglo-Saxon Protestant White ethnic White Latino Americans White Puerto Ricans White Southerners White Americans in California White Americans in Maryland History of White Americans in Baltimore White Americans in Texas White supremacy White nationalism White people in Hawaii White trash – American English slur for poor white people, especially in the American South Poor White – United States social caste and ethnic group Mountain white – White settlers of the rural Appalachian Mountains in the United States White Americans in Louisiana Notes ^ Of the foreign-born population from Europe (4,817 thousand), in 2010, 62% were naturalized. References ^ a b c d e f g h "Race and Ethnicity in the United States". United States Census Bureau. August 12, 2021. Retrieved August 17, 2021. ^ "Religious tradition by race/ethnicity (2014)". The Pew Forum on Religion & Public Life. Retrieved April 5, 2019. ^ Herbst, Philip (June 15, 1997). The color of words: an encyclopaedic dictionary of ethnic bias in the United States. Intercultural Press. ISBN 978-1-877864-97-1. Though discredited as an anthropological term and not recommended in most editorial guidelines, it is still heard and used, for example, as a category on forms asking for ethnic identification. It is also still used for police blotters (the abbreviated Cauc may be heard among police) and appears elsewhere as a euphemism. Its synonym, Caucasoid, also once used in anthropology but now dated and considered pejorative, is disappearing. ^ Mukhopadhyay, Carol C. (June 30, 2008). "Getting Rid of the Word 'Caucasian'". In Mica Pollock (ed.). Everyday Antiracism: Getting Real About Race in School. New Press. pp. 14–. ISBN 978-1-59558-567-7. Yet there is one striking exception in our modern racial vocabulary: the term 'Caucasian'. Despite being a remnant of a discredited theory of racial classification, the term has persisted into the twenty-first century, within as well as outside of the educational community. It is high time we got rid of the word Caucasian. Some might protest that it is 'only a label'. But language is one of the most systematic, subtle, and significant vehicles for transmitting racial ideology. Terms that describe imagined groups, such as Caucasian, encapsulate those beliefs. Every time we use them and uncritically expose students to them, we are reinforcing rather than dismantling the old racialized worldview. Using the word Caucasian invokes scientific racism, the false idea that races are naturally occurring, biologically ranked subdivisions of the human species and that Caucasians are the superior race. Beyond this, the label Caucasian can even convey messages about which groups have culture and are entitled to recognition as Americans. ^ Dewanjuly, Shaila (July 6, 2013). "Has 'Caucasian' Lost Its Meaning?". The New York Times. Retrieved March 16, 2018. AS a racial classification, the term Caucasian has many flaws, dating as it does from a time when the study of race was based on skull measurements and travel diaries ... Its equivalents from that era are obsolete – nobody refers to Asians as 'Mongolian' or blacks as 'Negroid'. ... There is no legal reason to use it. It rarely appears in federal statutes, and the Census Bureau has never put a checkbox by the word Caucasian. (White is an option.) ... The Supreme Court, which can be more colloquial, has used the term in only 64 cases, including a pair from the 1920s that reveal its limitations ... In 1889, the editors of the original Oxford English Dictionary noted that the term Caucasian had been 'practically discarded'. But they spoke too soon. Blumenbach's authority had given the word a pseudoscientific sheen that preserved its appeal. Even now, the word gives discussions of race a weird technocratic gravitas, as when the police insist that you step out of your 'vehicle' instead of your car ... Susan Glisson, who as the executive director of the William Winter Institute for Racial Reconciliation in Oxford, Miss., regularly witnesses Southerners sorting through their racial vocabulary, said she rarely hears 'Caucasian'. 'Most of the folks who work in this field know that it's a completely ridiculous term to assign to whites,' she said. 'I think it's a term of last resort for people who are really uncomfortable talking about race. 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PMID 31545791.{{cite journal}}: CS1 maint: article number as page number (link) External links White Population 2000 from the US census \| v t e Demographics of the United States \| \| Demographic history \| \| \| Socioeconomic status \| \| Affluence Gender inequality Educational attainment Emigration Home-ownership Household income Housing segregation Immigration Income inequality Language LGBT Middle classes Personal income Poverty Racial inequality + Race and health + Racial achievement gap + Racial wage gap Social class Standard of living Unemployment by state Wealth Ethnocultural politics \| \| \| \| Religions \| \| Baha'is Buddhists Christians + Catholics + Coptics + Mormons + Protestants Hindus Jains Jews Muslims + Ahmadiyyas + Five Percenters + Moorish Scientists + Nation of Islam + Value Creators Native American Neopagans Non-religious Rastafaris Scientologists Sikhs Zoroastrians \| \| \| \| v t e Ancestry and ethnicity in the United States \| \| \| General/miscellaneous ethno-racial classifications \| \| \| \| \| --- \| \| General groups \| African Americans Asian Americans Native Americans + Alaska Natives Pacific Islanders + Native Hawaiians White/European Americans + Non-Hispanic Whites - 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14709	https://www.chegg.com/homework-help/questions-and-answers/given-cos-60-1-2-sin-60-3-2-determine-following-values-exact-form-cos-120-b-sin-120-q48403398	Solved Given that cos (60°) = 1/2 and sin(60°) = √3/2 , \| 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 Math Algebra Algebra questions and answers Given that cos (60°) = 1/2 and sin(60°) = √3/2 , determine the following values in exact form. (a) cos 120°(b) sin 120° 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: Given that cos (60°) = 1/2 and sin(60°) = √3/2 , determine the following values in exact form. (a) cos 120°(b) sin 120° Given that cos (60°) = 1/2 and sin(60°) = √3/2 , determine the following values in exact form. (a) cos 120° (b) sin 120° Here’s the best way to solve it.Solution Share Share Share done loading Copy link View the full answer Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. 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14710	https://assess.ly/en/number-series-practice/alternating-pattern/1/	Number Series Explanation 3 - Alternating Pattern Number Sequences Explanation - Alternating Pattern With these types of patterns you will often have no idea what is happening with the numbers in the series. It is important to neatly write down the most logical arches. When a number gets bigger, you have to check for multiplication or addition in that step. If a number becomes smaller, you check whether it is a subtraction or division. For example: from number 1 to number 2, both -2 and / 2 are possible. From number 2 to number 3 both 3 and +4 are possible. From number 3 to number 4 realistically only -2 is possible. From number 3 to number 4 both +8 and 3 are possible. There are many more different possibilities of alternating patterns, but they are all based on the same principle. Especially look for this pattern if the differences between the numbers are changing every step. We will first work out 3 examples and show you how to draw the arches. Then you can do 3 practice exercises yourself. Assessments
14711	https://en.wikipedia.org/wiki/Sector_mass_spectrometer	Published Time: 2005-07-11T18:35:41Z Sector mass spectrometer - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Theory 2 Classic geometriesToggle Classic geometries subsection 2.1 Bainbridge–Jordan 2.2 Mattauch–Herzog 2.3 Nier–Johnson 2.4 Hinterberger–Konig 2.5 Takeshita 2.6 Matsuda 3 See also 4 References 5 Further reading Sector mass spectrometer [x] 8 languages Čeština Deutsch 日本語 Norsk bokmål Português Русский Türkçe 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Class of mass spectrometer This article is about mass spectrometers. For sector compasses, see sector (instrument). A five sector mass spectrometer A sector instrument is a general term for a class of mass spectrometer that uses a static electric (E) or magnetic (B) sector or some combination of the two (separately in space) as a mass analyzer. Popular combinations of these sectors have been the EB, BE (of so-called reverse geometry), three-sector BEB and four-sector EBEB (electric-magnetic-electric-magnetic) instruments. Most modern sector instruments are double-focusing instruments (first developed by Francis William Aston, Arthur Jeffrey Dempster, Kenneth Bainbridge and Josef Mattauch in 1936) in that they focus the ion beams both in direction and velocity. Theory [edit] The behavior of ions in a homogeneous, linear, static electric or magnetic field (separately) as is found in a sector instrument is simple. The physics are described by a single equation called the Lorentz force law. This equation is the fundamental equation of all mass spectrometric techniques and applies in non-linear, non-homogeneous cases too and is an important equation in the field of electrodynamics in general. F=q(E+v×B),{\displaystyle \mathbf {F} =q(\mathbf {E} +\mathbf {v} \times \mathbf {B} ),} where E is the electric field strength, B is the magnetic field induction, q is the charge of the particle, v is its current velocity (expressed as a vector), and × is the cross product. So the force on an ion in a linear homogenous electric field (an electric sector) is: F=q E{\displaystyle F=qE\,}, in the direction of the electric field, with positive ions and opposite that with negative ions. Electric sector from a Finnigan MAT mass spectrometer (vacuum chamber housing removed) The force is only dependent on the charge and electric field strength. The lighter ions will be deflected more and heavier ions less due to the difference in inertia and the ions will physically separate from each other in space into distinct beams of ions as they exit the electric sector. And the force on an ion in a linear homogenous magnetic field (a magnetic sector) is: F=q v B{\displaystyle F=qvB\,}, perpendicular to both the magnetic field and the velocity vector of the ion itself, in the direction determined by the right-hand rule of cross products and the sign of the charge. The force in the magnetic sector is complicated by the velocity dependence but with the right conditions (uniform velocity for example) ions of different masses will separate physically in space into different beams as with the electric sector. Classic geometries [edit] These are some of the classic geometries from mass spectrographs which are often used to distinguish different types of sector arrangements, although most current instruments do not fit precisely into any of these categories as the designs have evolved further. Bainbridge–Jordan [edit] The sector instrument geometry consists of a 127.30° (π 2){\displaystyle \left({\frac {\pi }{\sqrt {2}}}\right)} electric sector without an initial drift length followed by a 60° magnetic sector with the same direction of curvature. Sometimes called a "Bainbridge mass spectrometer," this configuration is often used to determine isotopicmasses. A beam of positive particles is produced from the isotope under study. The beam is subject to the combined action of perpendicular electric and magnetic fields. Since the forces due to these two fields are equal and opposite when the particles have a velocity given by v=E/B{\displaystyle v=E/B\,} they do not experience a resultant force; they pass freely through a slit, and are then subject to another magnetic field, transversing a semi-circular path and striking a photographic plate. The mass of the isotope is determined through subsequent calculation. Mattauch–Herzog [edit] The Mattauch–Herzog geometry consists of a 31.82° (π/4 2{\displaystyle \pi /4{\sqrt {2}}} radians) electric sector, a drift length which is followed by a 90° magnetic sector of opposite curvature direction. The entry of the ions sorted primarily by charge into the magnetic field produces an energy focussing effect and much higher transmission than a standard energy filter. This geometry is often used in applications with a high energy spread in the ions produced where sensitivity is nonetheless required, such as spark source mass spectrometry (SSMS) and secondary ion mass spectrometry (SIMS). The advantage of this geometry over the Nier–Johnson geometry is that the ions of different masses are all focused onto the same flat plane. This allows the use of a photographic plate or other flat detector array. Nier–Johnson [edit] The Nier–Johnson geometry consists of a 90° electric sector, a long intermediate drift length and a 60° magnetic sector of the same curvature direction. Hinterberger–Konig [edit] The Hinterberger–Konig geometry consists of a 42.43° electric sector, a long intermediate drift length and a 130° magnetic sector of the same curvature direction. Takeshita [edit] The Takeshita geometry consists of a 54.43° electric sector, and short drift length, a second electric sector of the same curvature direction followed by another drift length before a 180° magnetic sector of opposite curvature direction. Matsuda [edit] The Matsuda geometry consists of an 85° electric sector, a quadrupole lens and a 72.5° magnetic sector of the same curvature direction. This geometry is used in the SHRIMP and Panorama (gas source, high-resolution, multicollector to measure isotopologues in geochemistry). See also [edit] Mass-analyzed ion kinetic energy spectrometry Charge remote fragmentation Kenneth Bainbridge Alfred O. C. Nier References [edit] ^IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "electric sector". doi:10.1351/goldbook.E01938 ^Arthur Jeffrey Dempster (American physicist) at the Encyclopædia Britannica ^Burgoyne, Thomas W.; Gary M. Hieftje (1996). "An introduction to ion optics for the mass spectrograph". Mass Spectrometry Reviews. 15 (4): 241–259. Bibcode:1996MSRv...15..241B. CiteSeerX10.1.1.625.841. doi:10.1002/(SICI)1098-2787(1996)15:4<241::AID-MAS2>3.0.CO;2-I. PMID27082712. Archived from the original(abstract) on 2012-12-10. ^Klemm, Alfred (1946). "Zur Theorie der für alle Massen doppelfokussierenden Massenspektrographen" [The theory of a mass-spectrograph with double focus independent of mass]. Zeitschrift für Naturforschung A. 1 (3): 137–141. Bibcode:1946ZNatA...1..137K. doi:10.1515/zna-1946-0306. S2CID94043005. ^Schilling GD; Andrade FJ; Barnes JH; Sperline RP; Denton MB; Barinaga CJ; Koppenaal DW; Hieftje GM (2006). "Characterization of a second-generation focal-plane camera coupled to an inductively coupled plasma Mattauch–Herzog geometry mass spectrograph". Anal. Chem. 78 (13): 4319–25. doi:10.1021/ac052026k. PMID16808438. ^De Laeter; J. & Kurz; M. D. (2006). "Alfred Nier and the sector field mass spectrometer". Journal of Mass Spectrometry. 41 (7): 847–854. Bibcode:2006JMSp...41..847D. doi:10.1002/jms.1057. PMID16810642. ^IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "Nier-Johnson geometry". doi:10.1351/goldbook.N04141 ^US 4553029, Matsuda, Hisashi, "Mass spectrometer", published 1985-11-12, assigned to Jeol Ltd. Further reading [edit] Thomson, J. J.: Rays of Positive Electricity and their Application to Chemical Analyses; Longmans Green: London, 1913 \| hide v t e Mass spectrometry \| \| Mass m/z Mass spectrum MS software Acronyms \| \| Ion source \| APCI APLI APPI CI DAPPI DART DESI DIOS EESI EI ESI FAB FD GD IA ICP LAESI MALDI MALDESI MIP PTR SESI SIMS SS SSI SELDI TI TS \| \| Mass analyzer \| Sector Wien filter Time-of-flight Quadrupole mass filter Quadrupole ion trap Penning trap FT-ICR Orbitrap \| \| Detector \| Electron multiplier Microchannel plate detector Daly detector Faraday cup Langmuir–Taylor detector \| \| MS combination \| MS/MS QqQ AMS Hybrid MS GC/MS LC/MS IMS/MS CE-MS \| \| Fragmentation \| BIRD CID ECD EDD ETD HCD IRMPD NETD SID \| \| Category Commons WikiProject \| Retrieved from " Categories: Mass spectrometry Measuring instruments Hidden categories: Articles with Encyclopædia Britannica links Articles with short description Short description is different from Wikidata This page was last edited on 22 April 2025, at 18:52(UTC). 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14712	https://uomustansiriyah.edu.iq/media/lectures/3/3_2025_01_15!11_06_19_PM.pdf	1 ORAL PATHOLOGY Non-odontogenic cysts Cysts are epithelium-lined pathological cavities, usually filled with fluid, semi-solid material, or cellular debris. Clinically, they present as a soft or fluctuant swelling. Cysts of the oral and maxillofacial region are divided into, non-odontogenic, odontogenic, pseudocysts, and neck cysts. Pseudocysts differ from true cysts in that they lack an epithelial lining. Cysts of the maxilla, mandible, and perioral regions vary markedly in histogenesis, incidence, behavior, and treatment. Also are called fissural cysts or occlusion cysts, because they arise from embryonic epithelium that becomes entrapped during embryogenesis. 1-Nasolabial Cyst (Nasoalveolar cyst) is a rare developmental soft tissue cyst that develops in the upper lip in the canine region. Etiology: Unclear. Although there are two major theories:- One theory considers this cyst to be a fissural cyst arising from epithelial remnants entrapped along the line of fusion of the maxillary, medial nasal, and lateral nasal processes. A second theory suggests that these cysts develop from the misplaced epithelium of the nasolacrimal duct. Clinical and radiographical features: -  It appears as a soft-tissue swelling in the mucobuccal fold of the maxilla, lateral to the midline.  the patient may complain of nasal obstruction, discomfort, or difficulties in wearing dentures.  The cyst is more common in women, usually between 40 and 50 years of age.  Because this cyst arises in soft tissues, in most cases there are no radiographic changes, but resorption of the underlying bone may occur. Histopathological examination shows the lining of the cyst pseudostratifiedied columnar epithelium, often showing goblet cells, and cilia. The cyst wall is composed of fibrous connective tissue with adjacent skeletal muscle. Differential diagnosis:- Soft-tissue abscess, tooth abscess, mucocele, radicular cyst, salivary gland neoplasms, and mesenchymal neoplasms. Treatment: Surgical excision. 2- Nasopalatine duct cyst(Incisive canal cyst) : It is the most common non-odontogenic cyst of the oral cavity. Etiology: It arises from epithelial rests in the incisive foramen. Clinical and radiographical features: It appears as a slow-growing soft swelling of the palatine papilla, covered with normal mucosa. The cyst, after mechanical irritation, may be inflamed and become painful due to local infection. Radiographically: usually demonstrate well-circumscribed RL in or near the midline of the maxilla, between and apical to the central incisor teeth. 2  Note: It may be difficult to distinguish a small nasopalatine duct cyst from a large incisive foramen. It is generally accepted that a diameter of (6 mm) is the upper limit of normal size for incisive foramen. Therefore, a radiolucency that is (6 mm) or smaller in this area is usually considered a normal foramen unless other clinical signs and symptoms are present. Histologically The clinical diagnosis should be confirmed by histopathological examination that showed epithelial lining composed of the followings: a. Stratified squamous epithelium. b. Pseudostratified columnar epithelium. c. Simple columnar epithelium. d. Simple cuboidal epithelium. Differential diagnosis: - Tooth and periodontal abscesses, mechanical trauma of the palatine papilla, fibroma, lipoma. Note: to distinguish between the nasopalatine cyst & P.A. cyst: 1- In N.P.cyst, the tooth is vital, but the t in the case of P.A. cyst the associated tooth is nonvital. 2- Because the N.P.cyst is not related (attached) to the apex of the root, so by changing the direction of the X-ray beam we see if the lesion remains attached to the apex of the root, it means it’s radicular cyst, if not it means N.P. cyst. Treatment: Surgical removal. 3-Globulomaxillary cyst Theory of origin: Globulomaxillary cyst were once considered fissural cysts, located between the globular and maxillary processes. The former theory of origin is related to epithelial entrapment within a line of embryologic closure with subsequent cystic change. Embryologic evidence now shows that the premaxilla and maxillary processes do not fuse in this manner, and thus there can be no fusion-related mechanism to account for a distinct globulomaxillary cyst in this location that’s why a current theory holds that most of the cysts that develop in the globulomaxillary area, are of odontogenic origin. Radiolucencies in this location, when reviewed microscopically, have been shown to represent radicular cysts, periapical granulomas, lateral periodontal cysts, OKCs, central giant cell granulomas, calcifying odontogenic cysts odontogenic myxomas. Thus today the term globulomaxillary can be justified only in an anatomic sense, with a definitive diagnosis of lesions located in this area made by combined clinical and microscopic examination. Radiologically: a globulomaxillary lesion appears as a well-defined radiolucency, often producing divergence of the roots of the maxillary lateral incisor and canine teeth. Radicular cysts and periapical granuloma can be ruled out with pulp vitality testing. Histologically: - lining epithelium is stratified squamous and sometimes pseudostratified ciliated columnar respiratory epithelium. Thin C.T. wall which is free from inflammation. Treatment and prognosis are determined by the definitive microscopic diagnosis. 4-Oral Lymphoepithelial Cyst Definition: Lymphoepithelial cyst is an uncommon developmental lesion of the oral mucosa. Etiology: Probably caused by cystic degeneration of glandular or surface epithelium entrapped in lymphoid tissue during embryogenesis. Clinical features:  It presents as an asymptomatic, mobile, well-defined nodule, usually firm on palpation and elevated, with a yellowish or whitish color.  The size ranges from 0.5 cm to 2 cm in diameter 3  It occurs intraorally and usually located in the posterior part of tongue or in the floor of the mouth and sometime near the soft palate and the pharynx and in the tonsilar area (lymphoid tissue). Histologically: Lymphoepithelial cysts are similar to the branchial cleft cysts that develop in the lateral neck. Histopathological examination showed epithelial lining of stratified squamous that may or may not be keratinized. The wall of the cyst typically contains lymphoid tissue often demonstrating germinal center formation. Differential diagnosis: lymphoid tissue aggregation, dermoid cyst, mucocele, lipoma, fibroma and other benign tumors. Treatment: Surgical removal. 5-Thyroglossal Duct Cyst Definition: Thyroglossal duct cyst is a rare developmental lesion that may form along the thyroglossal tract. Etiology: Remnants of thyroglossal duct epithelium. Clinical features:  The cyst is usually located under the hyoid bone but can be located anywhere from the suprasternal notch to the foramen cecum of the dorsal tongue.  Intraorally, it appears as a painless, fluctuant swelling usually 1–3 cm in diameter, located in the midline of the dorsum of the tongue close to the foramen caecum. Occasionally, a fistula may form following infection.  The cyst is most often diagnosed in patients less than 20 years of age. Histopathological examination showed a lining epithelium of stratified squamocolumnar, small intestinal epithelium, or a mixture of them. The C.T. tissue wall may contain normal thyroid tissue. Differential diagnosis: Median rhomboid glossitis, benign and malignant tumors. Treatment: Surgical removal. 6-Median mandibular cysts:- Like globulomaxillary cysts, were once considered fissural cysts, in which a fissural origin was based on the theory of epithelial entrapment in the midline of the mandible during the "fusion" of each half of the mandibular arch. There is now embryologic evidence of an isthmus of mesenchyme between the mandibular processes that is gradually eliminated as growth continues, and therefore no evidence of epithelial fusion. So recently it is thought to be of odontogenic origin. Clinical and radiographical features: Swelling In the midline of the mandible In x-ray it appears a well-circumscribed RL between the two lower central incisors in the midline. Histopathology ling epithelium is mainly stratified squamous. C.T.’s wall is free from inflammation. Treatment: surgical removal 7-Median palatal cyst (palatine cyst): It a is rare fissural cyst that develops from epithelium entrapped along the embryonic line of fusion of lateral palatal shelves. This cyst may be difficult to distinguish a from nasopalatine duct cyst. Clinical and radiographical features:  This cyst is present as firm or fluctuant swelling in the midline of the hard palate posterior to the palatine papilla. Most of these cysts are asymptomatic, but sometimes pain may be present.  X-ray: occlusal radiograph show well-circumscribed radiolucency(RL) in the midline of the hard palate. A midline RL without clinical evidence of expansion is probably a nasopalatine duct cyst. 4 Histopathology: - Cyst is usually lined by stratified squamous epithelium. Area of ciliated pseudostratified columnar epithelium may be present in some cases. Chronic inflammation may be present in the cyst wall. Treatment: surgical removal 8-Dermoid & Epidmoid cyst These represents a simple form of cystic teratoma derived from skin epithelium entrapped during embryonic development. Clinically:  Most of these cysts occur in the head & neck region, primarily in the skin around the eyes & the anterior upper neck, extending superiorly into the floor of the mouth.  Mostly occur in young adults, present as painless swelling exhibiting a doughy consistency on palpation,& may cause elevation of the tongue & can interfere with eating & speaking. Histopathology: The cyst lined by a layer of orthokeratinized squamous epithelium, surrounding by C.T. capsule. In dermoid cyst in addition to these, the lesion exhibiting variable numbers of dermal appendages including hair follicles, sebaceous glands. Treatment: surgical excision. Dermoid cyst: These cysts probably form as a result of some abnormality of development of the branchial arches or pharyngeal pouches. It is generally classified as a benigin form of Teratoma. Clinical features Dermoid cysts develop between the hyoid and jaw or may form immediately beneath the tongue. They are sometimes filled with desquamated keratin giving them a semi- solid consistency If the cyst develops above the geniohyoid muscle a sublingual swelling may displace the tongue upward and create difficulty in eating, speaking or even breathing. Cysts that occur below the geniohyoid muscle often produce a submental swelling with a double chin appearance. Dermoid cyst is more deeply placed than a ranula; the latter is obviously superficial, having a thin wall and a bluish appearance. A dermoid cyst causes no symptoms until large enough to interfere with speech or eating. Pathology: The lining of epidermoid cysts is keratinising stratified squamous epithelium alone. Less often, cysts also have dermal appendages (sebaceous gland, hair follicle or sweat gland) in the wall and are then referred to as dermoid cysts. These cysts should be removed surgically.
14713	http://www.journal-1.eu/2018/Grozdev-Okumura-Dekov-Intouch-Triangle.pdf	International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 c ⃝IJCDM Volume 3, 2018, pp.55-61 Received 1 February 2018. Published on-line 14 April 2018 web: c ⃝The Author(s) This article is published with open access1. Problems for Students about Intouch Triangle Sava Grozdeva, Hiroshi Okumurab and Deko Dekovc 2 a VUZF University of Finance, Business and Entrepreneurship, Gusla Street 1, 1618 Soﬁa, Bulgaria e-mail: sava.grozdev@gmail.com b Maebashi Gunma, 371-0123, Japan e-mail: hokmr@protonmail.com cZahari Knjazheski 81, 6000 Stara Zagora, Bulgaria e-mail: ddekov@ddekov.eu web: Abstract. We present problems for students about triangles similar (but not ho-mothetic) with the Intouch triangle. The problems are discovered by the computer program “Discoverer” created by the authors. Keywords. Euclidean geometry; triangle geometry; computer discovered math-ematics; “Discoverer”. Mathematics Subject Classiﬁcation (2010). 51-04, 68T01, 68T99. The Intouch triangle of a triangle ABC, also called the Contact triangle, is the triangle formed by the points of tangency of the incircle of triangle ABC with triangle ABC. The Intouch triangle is also the Cevian triangle of triangle ABC with respect to the Gergonne point. See also Contact triangle in . We present problems for triangles similar (but not homothetic) with the Intouch triangle. The problems are discovered by the computer program “Discoverer” , , created by the authors. We encourage the students and teachers to solve the problems. We denote the side lengths of triangle ABC by a = BC, b = CA and c = AB. Given triangles PaPbPc and QaQbQc. The triangles are similar if and only if all the corresponding sides have lengths in the same ratio: PbPc QbQc = PcPa QcQa = PaPb QaQb 1This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. 2Corresponding author 55 56 Problems for Students about Intouch Triangle We denote by k = PbPc QbQc the ratio of similarity of PaPbPc to QaQbQc. Reference for Problem 1: Fuhrmann triangle in . Problem 1. The Intouch triangle PaPbPc is similar with triangle QaQbQc = the Fuhrmann triangle. The ratio of similarity is k = (b + c −a)(c + a −b)(a + b −c) 2 √ abc √ E where E = a3 + b3 + c3 + 3abc −a2b −a2c −ab2 −ac2 −bc2 −cb2. A B C Pa Pb Pc Qa Qb Qc Figure 1. Figure 1 illustrates Problem 1. Reference for Problem 2: Pedal triangle in , Inversion in . Problem 2. The Intouch triangle PaPbPc is similar with triangle QaQbQc = the Pedal Triangle of the Inverse of the Incenter in the Circumcircle. The ratio of similarity is k = √ E √ abc where E is as in Problem 1 A B C Pa Pb Pc Qa Qb Qc I O Q c Figure 2. Figure 2 illustrates Problem 2. In ﬁgure 2: Sava Grozdev, Hiroshi Okumura and Deko Dekov 57 • PaPbPc is the Intouch triangle, • I is the Incenter, • O is the circumcenter, • c is the circumcircle, • Q is the inverse point of the Incenter with respect to circcumcircle, • QaQbQc is the Pedal triangle of point Q. Reference for Problem 3: Circumcevian triangle in , Inversion in . Problem 3. The Intouch triangle PaPbPc is similar with triangle QaQbQc = the Circumcevian Triangle of the Inverse of the Incenter in the Circumcircle. The ratio of similarity is k = (b + c −a)(c + a −b)(a + b −c) 2abc . A B C Pa Pb Pc Qa Qb Qc I O c Figure 3. Figure 3 illustrates Problem 3. In ﬁgure 3: • PaPbPc is the Intouch triangle, • I is the Incenter, • O is the circumcenter, • c is the circumcircle, • QaQbQc is the Circumcevian Triangle of the Inverse of the Incenter in the Circumcircle. Problem 4. The Intouch triangle PaPbPc is similar with triangle QaQbQc = the Triangle of Reﬂections of the Inverse of the Incenter in the Circumcircle in the Sidelines of Triangle ABC. The ratio of similarity is k = √ E √ 2abc , where E is as in Theorem 1. Figure 4 illustrates Problem 4. In ﬁgure 4: • PaPbPc is the Intouch triangle, • I is the Incenter, • O is the circumcenter, 58 Problems for Students about Intouch Triangle A B C Pa Pb Pc Qa Qb Qc I O c Figure 4. • c is the circumcircle, • QaQbQc is Triangle of Reﬂections of the Inverse of the Incenter in the Circumcircle in the Sidelines of Triangle ABC. Let RaRbRc be the Circumcevian triangle of a point P. Denote by Qa the mid-point of segment ARa, by Qb the midpoint of segment BRb,and by Qc the mid-point of segment CRc. Then QaQbQc is the Half-Circumcevian triangle of point P. Problem 5. The Intouch triangle PaPbPc is similar with triangle QaQbQc = the Half-Circumcevian Triangle of the Incenter. The ratio of similarity is k = (b + c −a)(c + a −b)(a + b −c) √ abcE , where E is as in Theorem 1. A B C Pa Pb Pc Qa Qb Qc Ra Rb Rc I O c Figure 5. Figure 5 illustrates Problem 5. In ﬁgure 5: • PaPbPc is the Intouch triangle, • I is the Incenter, • O is the circumcenter, Sava Grozdev, Hiroshi Okumura and Deko Dekov 59 • c is the circumcircle, • RaRbRc is the Circumcevian triangle of the Incenter, • Qa the midpoint of segment ARa, • Qb the midpoint of segment BRb, • Qc the midpoint of segment CRc, • QaQbQc is the Half-Circumcevian Triangle of the Incenter. Reference for Problem 6: Nine-Point Center in . Problem 6. Denote by I the Incenter of triangle ABC. Denote by Qa the Nine-Point Center of triangle IBC, by Qb the Nine-Point Center of triangle AIC, and by Qc the Nine-Point Center of triangle ABI. The Intouch triangle PaPbPc is similar with triangle QaQbQc = the Triangle of the Nine-Point Centers of the Triangulation Triangles of the Incenter. The ratio of similarity is k = (b + c −a)(c + a −b)(a + b −c) √ abcE , where E is as in Theorem 1. A B C Pa Pb Pc Qa Qb Qc I Figure 6. Figure 6 illustrates Problem 6. In ﬁgure 6: • PaPbPc is the Intouch triangle, • I is the Incenter, • Qa is the Nine-Point Center of triangle IBC, • Qb is the Nine-Point Center of triangle AIC, • Qc is the Nine-Point Center of triangle ABI, • QaQbQc is the Triangle of the Nine-Point Centers of the Triangulation Triangles of the Incenter. Solution to Problem 6. We use barycentric coordinates . The Intouch triangle PaPbPc is the Cevian triangle (, Section 8), of the Gergonne point (, Section 7) , hence it has barycentric coordinates Pa = (0, (b −c + a)(b + c −a), (c −a + b)(c + a −b)), Pb = ((a −b + c)(a + b −c), 0, (c −a + b)(c + a −b)), Pc = ((a −b + c)(a + b −c), (b −c + a)(b + c −a), 0). By using the distance formula (9) in , we ﬁnd the lengths of segments form the Incenter I = (a, b, c) to vertices A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1) as 60 Problems for Students about Intouch Triangle follows: PA = s bc(b + c −a) (a + b + c) , PB = s ac(a + c −b) (a + b + c) , PC = s ab(a + b −c) (a + b + c) . Hence, the side lengths of triangle T1 = IBC are a1 = a, b1 = PC, c1 = PB, the side lengths of triangle T2 = AIC are a2 = PC, b2 = b, c2 = PA, and the side lengths of triangle T3 = ABI are a3 = PB, b3 = PA, c3 = c. Now by using the change of coordinates formula (10) in we obtain the barycen-tric coordinates wrt triangle ABC of Qa = Nine-Point Center (, Section 7) of triangle IBC. Similarly, we ﬁnd the barycentric coordinates of Qb and Qc. The barycentric coordinates are as follows: Qa = (a(b + c), b2 + 2bc −a2 −ab + c2, 2bc + c2 −a2 −ac + b2), Qb = (−ab −b2 + 2ac + c2 + a2, b(a + c), 2ac + c2 + a2 −b2 −bc), Qc = (2ab + b2 −ac −c2 + a2, a2 + 2ab −bc −c2 + b2, c(a + b)). Now by using the distance formula (9) in we can calculate the lengths of sides of triangles PaPbPc and QaQbQc, Denote E = a3 + b3 + c3 + 3abc −a2b −a2c −ab2 −ac2 −bc2 −cb2. . We obtain PbPc = (b + c −a) p (c + a −b)(a + b −c) 2 √ bc , PcPa = (c + a −b) p (a + b −c)(b + c −a) 2√ca , PaPb = (a + b −c) p (b + c −a)(c + a −b) 2 √ ab , QbQc = √ aE 2 p (c + a −b)(a + b −c) , QcQa = √ bE 2 p (a + b −c)(b + c −a) , QaQb = √ cE 2 p (b + c −a)(c + a −b) . Hence PbPc QbQc = PcPa QcQa = PaPb QaQb = (b + c −a)(c + a −b)(a + b −c) √ abcE . Sava Grozdev, Hiroshi Okumura and Deko Dekov 61 Acknowledgement The authors are grateful to Professor Ren´ e Grothmann for his wonderful computer program C.a.R. See also The authors are also grateful to Professor Troy Henderson for his wonderful computer program MetaPost Previewer for creation of eps graphics References S. Grozdev and D. Dekov, A Survey of Mathematics Discovered by Computers, Inter-national Journal of Computer Discovered Mathematics, 2015, vol.0, no.0, 3-20. http: //www.journal-1.eu/2015/01/Grozdev-Dekov-A-Survey-pp.3-20.pdf. S. Grozdev, H. Okumura and D. Dekov, A Survey of Mathematics Discovered by Computers. Part 2, Mathematics and Informatics, 2017, vol.60, no.6, 543-550. papers/Grozdev-Okumura-Dekov-A-Survey-2017.pdf. S. Grozdev and D. Dekov, Barycentric Coordinates: Formula Sheet, International Journal of Computer Discovered Mathematics, vol.1, 2016, no 2, 75-82. eu/2016-2/Grozdev-Dekov-Barycentric-Coordinates-pp.75-82.pdf. E. W. Weisstein, MathWorld - A Wolfram Web Resource, com/.
14714	https://www.oncotarget.com/article/17309/text/	Differentiation of pancreatic neuroendocrine carcinoma from pancreatic ductal adenocarcinoma using magnetic resonance imaging: The value of contrast-enhanced... \| Oncotarget Online ISSN: 1949-2553 Search: Oncotarget Journal Content Home Editorial Board Submission Current Volume Archive Scientific Integrity Editorial Policies Publication Ethics Statements Special Collections Interviews with Outstanding Authors Oncotarget In The News Sponsored Conferences Search Contact Information Oncotarget (a primarily oncology-focused, peer-reviewed, open access journal) aims to maximize research impact through insightful peer-review; eliminate borders between specialties by linking different fields of oncology, cancer research and biomedical sciences; and foster application of basic and clinical science. Its scope is unique. 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Chuangen Guo, Xiao Chen, Zhongqiu Wang, Wenbo Xiao, Qidong Wang, Ke Sun and Xiaoling Zhuge_ Abstract Chuangen Guo 1,, Xiao Chen 2,3,, Zhongqiu Wang 2, Wenbo Xiao 1, Qidong Wang 1, Ke Sun 4 and Xiaoling Zhuge 5 1 Department of Radiology, The First Affiliated Hospital, College of Medicine Zhejiang University, Hangzhou 310003, China 2 Department of Radiology, The Affiliated Hospital of Nanjing University of Chinese Medicine, Nanjing 2100029, China 3 Division of Nephrology, Zhongshan Hospital Fudan University, Shanghai 200032, China 4 Department of Pathology, The First Affiliated Hospital, College of Medicine Zhejiang University, Hangzhou 310003, China 5 Department of Laboratory Medicine, The First Affiliated Hospital, College of Medicine Zhejiang University, Hangzhou 310003, China These authors have contributed equally to this work Correspondence to: Xiaoling Zhuge, email: xlzhgzy@qq.com Chuangen Guo, email: zyyfy001@sina.cn Keywords: pancreatic ductal adenocarcinoma, pancreatic neuroendocrine carcinoma, magnetic resonance imaging, diffusion-weighted imaging Received: December 28, 2016 Accepted: April 05, 2017 Published: April 21, 2017 ABSTRACT Pancreatic neuroendocrine carcinoma (PNEC) is often misdiagnosed as pancreatic ductal adenocarcinoma (PDAC). This retrospective study differentiated PNEC from PDAC using magnetic resonance imaging (MRI), including contrast-enhanced (CE) and diffusion-weighted imaging (DWI). Clinical data and MRI findings, including the T1/T2 signal, tumor boundary, size, enhancement degree, and apparent diffusion coefficient (ADC), were compared between 37 PDACs and 13 PNECs. Boundaries were more poorly defined in PDAC than PNEC (97.3% vs. 61.5%, p<0.01). Hyper-/isointensity was more common in PNEC than PDAC at the arterial (38.5% vs. 0.0), portal (46.2% vs. 2.7%) and delayed phases (46.2% vs. 5.4%) (all p<0.01). Lymph node metastasis (97.3% vs. 61.5%, p<0.01) and local invasion/distant metastasis (86.5% vs. 46.2%, p<0.01) were more common in PDAC than PNEC. Enhancement degree via CE-MRI was higher in PNEC than PDAC at the arterial and portal phases (p<0.01). PNEC ADC values were lower than those of normal pancreatic parenchyma (p<0.01) and PDAC (p<0.01). Arterial and portal phase signal intensity ratios and ADC values showed the largest areas under the receiver operating characteristic curve and good sensitivities (92.1%–97.2%) and specificities (76.9%–92.3%) for differentiating PNEC from PDAC. Thus the enhancement degree at the arterial and portal phases and the ADC values may be useful for differentiating PNEC from PDAC using MRI. INTRODUCTION Pancreatic ductal adenocarcinoma (PDAC) is the most common malignant tumor of the pancreas. It is highly aggressive and rapidly fatal, with a five-year survival rate <5% . While resection can be curative, resection rates remain low at 10–15% due to local invasion or distant metastases. The PDAC characteristic vascularization pattern can be visualized using computed tomography (CT) or magnetic resonance imaging (MRI) [3, 4], and PDACs are often hypovascularized as compared to adjacent normal tissue . Pancreatic neuroendocrine neoplasms (PNENs) account for 1–2% of pancreatic tumor cases . The WHO 2010 classification separates PNENs into grade 1 (G1), grade 2 (G2), and neuroendocrine carcinoma (NEC, G3) based on histological differentiation, including mitoses and Ki-67 proliferation index. Pancreatic neuroendocrine carcinomas (PNECs) are very rare, accounting for only 2–3% of PNENs [8, 9]. PNENs are hypervascular lesions with marked enhancement via CT or MRI [10–13]. Jang, et al. showed differences in tumor margin, enhancement pattern, bile duct dilatation, pancreatic duct dilatation, and pancreatic atrophy between PNENs and PDAC. Several studies also showed that contrast-enhanced ultrasonography (3D US and harmonic endoscopic US) is useful for differential diagnosis of PDAC and PNEN [14, 15]. Kim, et al. observed differences in transfer coefficient (K(trans)), rate constant (K(ep)), and initial area under the concentration curve over 60 sec (iAUC) between PDAC and PNEN using dynamic contrast-enhanced MRI. However, recent studies indicated that G1 and G2 PNEN and PNEC enhancement patterns differ [11, 12, 17]. Most PNECs exhibit arterial and portal phase hypoenhancement, indicating that PNECs and PDACs have similar enhancement degrees. Kimura, et al. reported a PNEC case, misdiagnosed as PDAC, that exhibited low vascularity on enhanced CT. Lewis, et al. also demonstrated that PNEC MRIs resemble those of PDACs, including T1 and T2 signals, and hypoenhancement. Therefore, we speculate that qualitative imaging is not effective for differentiating PNEC from PDAC. PDAC and PNEC treatment strategies and prognoses differ. For PNEC, surgery is indicated if curative resection is possible, even in those cases with limited metastases, for example to liver [19, 20]. In addition, targeted therapy with sunitinib or everolimus and somatostatin analogues (octrecotide) , or radionuclide-labeled somatostatin (DOTATATE) may be valuable for PNEC, along with cytotoxic chemotherapy (e.g., cisplatin with etoposide). PNEC are generally less aggressive and have better outcomes than PDAC. The PNEC five-year survival rate is approximately 27.7% , which is higher than that of PDAC (<5%). Pretreatment differentiation of PNEC from PDAC is important in determining therapeutic strategies. To the best of our knowledge, no study has explored differences in imaging features between PDAC and PNEC. MRI, particularly in diffusion-weight imaging (DWI), has been used to differentiate pancreatitis and pancreatic cancer [25–28], and MRI has a similar or better performance in PDAC evaluation [29, 30]. The present study assessed the value of MRI, including DWI and dynamic contrast-enhanced imaging, for differentiating PNEC and PDAC. RESULTS Patient and tumor characteristics Thirty-seven PDAC and 13 PNEC patients were analyzed (Figure 1, Table 1) in this retrospective study. Thirty-one PDAC patients underwent surgery and six underwent biopsy. Twelve PNEC patients underwent surgery and one underwent biopsy. We compared demographic data and clinical symptoms between PDAC and PNEC patients. No differences were found for age, gender, or clinical symptoms between those two groups. However, yellow urine or icterus was more common in PDAC compared with PNEC patients (27.0% vs. 7.7%, p>0.05). PNEC tended to occur in men compared with PDAC (76.9% vs. 62.2%, p>0.05). Carbohydrate antigen 19-9 (CA19-9) and CA125 levels were higher in PDAC than in PENC patients. Abnormal CA19-9 level was more common in PDAC than PNEC (81.8% vs 30.8%, p<0.05). In our series, 89.2% of PDAC patients were correctly diagnosed via MRI, while eight (61.5%) PNEC patients were misdiagnosed as PDAC via MRI. Figure 1: Flow diagram of the study patients with pancreatic ductal adenocarcinoma (PDAC) and pancreatic neuroendocrine carcinoma (PNEC). Table 1: Clinical data of patients \| Variables \| PDAC(n=37) \| PNEC(n=13) \| P \| --- --- \| \| Age (years) \| 61.2±8.5(43-79) \| 53.9±13.1(32-71) \| 0.06 \| \| Gender \| \| \| >0.05 \| \| Male \| 23(62.2%) \| 10(76.9%) \| \| \| Female \| 14(37.8%) \| 3(23.1%) \| \| \| Clinical symptoms \| \| \| \| \| Abdominal Pain \| 19(51.4%) \| 10(76.9%) \| >0.05 \| \| Confusion of consciousness or dizziness \| 0 \| 1(7.7%) \| >0.05 \| \| Diarrhea or Abdominal bloating \| 8(21.6%) \| 2(15.4%) \| >0.05 \| \| Yellow urine or icterus \| 10(27.0%) \| 1(7.7%) \| >0.05 \| \| Marasmus \| 4(10.8%) \| 0 \| >0.05 \| \| Others \| 2(5.4%) \| 1(7.7%) \| >0.05 \| \| Asymptomatic \| 5(13.5%) \| 3(23.1%) \| >0.05 \| \| Surgery \| 31(84.6%) \| 12(92.3%) \| >0.05 \| \| Biopsy \| 6(15.4%) \| 1(7.7%) \| >0.05 \| \| CA19-9 (U/ml) \| 211.4(5-12000) \| 12.5(2.9-638) \| <0.001 \| \| >37 \| 30(81.8%) \| 4(30.8%) \| <0.001 \| \| CA125(U/ml) \| 21.9(6.6-188.5) \| 8.9(4.7-43.4) \| <0.05 \| \| CEA(ng/ml) \| 3.9(0.8-55.7) \| 2.3(0.9-159) \| >0.05 \| \| Imaging diagnosis \| \| \| <0.05 \| \| Pancreatic cancer \| 33(89.2%) \| 8(61.5%) \| \| \| PNEC \| 0 \| 1(7.7%) \| \| \| Pancreatic cystadenoma or cystadenocarcinoma \| 2(5.4%) \| 0 \| \| \| Others \| 2(5.4%) \| 4(30.7%) \| \| Data are shown as median. CA19-9: Carbohydrate antigen 19-9; CA125: Carbohydrate antigen 125; CEA: carcino-embryonic antigen; PNEC: pancreatic neuroendocrine carcinoma MRI findings: qualitative analysis Qualitative MRI data were summarized in Table 2. PDAC occurred more commonly in pancreatic head-neck compared with PNEC (70.3% vs. 46.2%), but the difference was not significant. A well defined boundary was more prevalent in PNEC compared with PDAC (38.5% vs. 2.7%, p<0.01). Differences in T1 signal intensity and DWI signal were not significant between PDAC and PNEC. However, isointensity in T2 weighted images was more common in PNEC compared with PDAC (23.1% vs. 2.7%, p=0.05). Moreover, hyper- and isointensity were more common in PNEC than PDAC at arterial phase (38.5% vs. 0.0, p<0.01), portal phase (46.2% vs. 2.7%, p<0.01), and delayed phase (46.2% vs. 5.4%, p< 0.01). Lymph node metastases and local invasion/distant metastases were more common in PDAC than PNEC (97.3% vs. 61.5%, 86.5 vs. 46.2%, respectively, p<0.01). Intra/extrahepatic bile duct dilatations were also more common in PDAC than in PNEC (54.1% vs. 23.1%), but this difference was not significant. Representative unenhanced T1- and T2-weighted PDAC and PNEC images, and gadolinium-enhanced images are shown in Figures 2 and 3. PDAC showed hypointensity in T1 weighted images, hyperintensity in T2 weighted images, and marked hypointensity compared with normal pancreas in contrast-enhanced images. Figure 2C showed bile duct dilatation in PDAC. PNEC also showed hypointensity in T1 weighted images (Figure 3A), hyperintensity in T2 weighted images (Figure 3B), and isointensity in contrast-enhanced T1 weighted images (Figure 3C–3E). Additionally, PNEC boundaries (Figure 3) were relatively well defined compared with those of PDAC (Figure 2). Table 2: The summary of MRI findings \| MR findings \| PDAC(n=37) \| PNEC(n=13) \| p \| --- --- \| \| Location \| \| \| >0.05 \| \| Pancreatic Head-neck \| 26(70.3%) \| 6(46.2%) \| \| \| Pancreatic Body-tail \| 11(29.7%) \| 7(53.9%) \| \| \| Boundary \| \| \| <0.01 \| \| Well-circumscribed \| 1(2.7%) \| 5((38.5%) \| \| \| Ill-defined \| 36(97.3%) \| 8(61.5%) \| \| \| MRI signal of tumor \| \| \| \| \| TIWI \| \| \| >0.05 \| \| Isointense \| 4(10.8%) \| 3(23.1%) \| \| \| Hypointense \| 29(78.4%) \| 10(76.9%) \| \| \| Iso-/Hypo intensity \| 4(10.8%) \| 0 \| \| \| T2WI \| \| \| 0.05 \| \| Isointense \| 1(2.7%) \| 3(23.1%) \| \| \| Hyperintense \| 36(97.3%) \| 10(76.9%) \| \| \| DWI \| \| \| 0.15 \| \| Isointense \| 2(5.4%) \| 0 \| \| \| Moderate Hyperintense \| 8(21.6%) \| 1(7.7%) \| \| \| Marked Hyperintense \| 27(73.0%) \| 12(92.3%) \| \| \| Enhancement degree \| \| \| \| \| Arterial phases \| \| \| <0.01 \| \| Hyper- intense \| 0 \| 1(7.7%) \| \| \| Iso-intense \| 0 \| 4(30.8%) \| \| \| Hypointense \| 37(100%) \| 8(61.5%) \| \| \| Portal phases \| \| \| <0.01 \| \| Hyper-intense \| 0 \| 1(7.7%) \| \| \| Iso-intense \| 1(2.7%) \| 5(38.5%) \| \| \| Hypo-intense \| 36(97.3%) \| 7(53.8%) \| \| \| Delayed phases \| \| \| \| \| Hyperintense \| 0 \| 1(7.7%) \| <0.01 \| \| Iso-intense \| 2(5.4%) \| 5(38.5%) \| \| \| Hypointense \| 35(94.6) \| 7(53.8%) \| \| \| Pancreatic duct dilatation \| 26(70.5%) \| 8(61.5%) \| >0.05 \| \| Intra-, extrahepatic bile duct dilatation \| 20(54.1%) \| 3(23.1%) \| >0.05 \| \| Pancreatic atrophy \| 8(21.6%) \| 3(23.1%) \| >0.05 \| \| Lymphnodes invasion \| 36(97.3%) \| 8(61.5%) \| <0.01 \| \| Local invasion or Metastases \| 32(86.5%) \| 6(46.2%) \| <0.01 \| \| Size (cm) \| 3.3±1.5(1.2-7.1) \| 5.2±4.4(2.2-19) \| 0.03 \| compared with hyper- and isointensity DWI: diffusion weighted imaging Figure 2: A 57 year old female patient with pathologically proven pancreatic ductal adenocarcinoma. On fat-suppressed LAVA T1- (A) and T2- (B) weighted imaging, the tumor showed hypointensity and hyperintensity with an ill-defined boundary, respectively. MRI cholangiopancreatography (MRCP) showed the common bile duct and intra-, extrahepatic bile ducts were markedly dilated (C). Gadolinium enhanced images in arterial (D), portal (E) and delayed phase (F) both showed the tumor were hypointensity compared with the pancreatic parenchyma. Figure 3: A 54 year old male patient with pathologically proven pancreatic neuroendocrine carcinoma. The tumor showed hypointensity and hyperintensity with relatively well-defined boundary on fat-suppressed LAVA T1- (A) and T2- (B) weighted imaging, respectively. The tumor showed hypo- to isointensity at arterial phase (C), isointensity at portal (D) and delayed phases (E) compared with the pancreatic parenchyma in contrast-enhanced images. Diffusion-weighted images showed the tumor was marked hyperintensity (F). MRI findings: quantitative analyses We quantitatively analyzed tumor sizes, signal intensities in T1 weighted images, and apparent diffusion coefficient (ADC) values. Mean PNEC tumor size was greater than that of PDAC (5.2 cm vs. 3.3 cm, p=0.03) (Table 2). Figure 4 shows the signal intensity ratio in unenhanced (pre-contrast image) and contrast-enhanced T1 weighted images measured by two readers. PDAC signal intensity ratios were lower than those of PNEC at arterial and portal phases (p≤0.01) (Figure 4). Figure 4: The signal intensity ratio compared with parenchyma in pancreatic ductal adenocarcinoma (PDAC) and pancreatic neuroendocrine carcinoma (PNEC) calculated by two readers. Signal intensity ratios in PDAC were lower than PNEC at arterial and portal phases. p≤ 0.01, compared with PDAC. The bias between the two readers were -1.2%(-9.6%, 7.2%), -2.2% (-14.0%, 9.5%) and 1.4% (-6.8%, 9.5%) for signal intensity ratio at arterial, portal and delayed phases, respectively. Representative DWI and ADC value maps for PDAC and PNEC are shown in Figure 5. PDAC DWI signal intensity was lower than that of PNEC. Mean ADC values for pancreatic parenchyma, PDAC, and PNEC were 1.38, 1.04, and 0.87×10-3 mm 2/s (pooled data), respectively (Figure 6). In both readers, mean PDAC and PNEC ADC values were lower than in normal pancreas parenchyma (p<0.05 or 0.01). PNEC ADC values were also lower compared with those of PDAC (p< 0.01). Figure 5: Diffusion-weighted images and ADC maps in pancreatic ductal adenocarcinoma (PDAC) and pancreatic neuroendocrine carcinoma (PNEC). PNEC showed higher DWI signal and lower ADC value than PDAC. Figure 6: Apparent diffusion coefficients (ADC) values in pancreatic ductal adenocarcinoma (PDAC) and pancreatic neuroendocrine carcinoma (PNEC) measured by two readers. The ADC values of PDAC and PNEC were both lower than the normal parenchyma. In addition, the ADC value of PNEC was also lower than PDAC. The bias between the two readers was -0.3%(-10.7%, 10%) for ADC value. In addition, we also evaluated the measurement agreement between the readers. The correlations between the two readers were 0.87-0.93 for ADC values and signal intensity ratio. The bias between the two readers were -0.3%(-10.7%, 10%) for ADC value, -1.2%(-9.6%, 7.2%) for signal intensity ratio at arterial phase, -2.2%(-14.0%, 9.5%) at portal phase and 1.4% (-6.8%, 9.5%) at delayed phase. Imaging feature diagnostic performances The sensitivity and specificity of the different imaging features for PNEC identification (vs. PDAC) ranged from 52.6%–97.2% and 38.5%–100%, respectively (Table 3). The area under the curve (AUC) ranged from 0.667–0.954 (Table 3). Signal intensity ratio at arterial and portal phases, and ADC value had the largest AUC, indicating that these features can potentially differentiate PNEC from PDAC (Figure 7). Cutoff values were 0.768 for signal intensity ratio at arterial phase with 97.2% sensitivity and 92.3% specificity, 0.823 for signal intensity ratio at portal phase with 97.2% sensitivity and 76.9% specificity, and 1.0×10-3 mm 2/s for ADC values with 92.1% sensitivity and 91.7% specificity. Table 3: Diagnostic performances of clinical and imaging features \| Variables \| AUC \| Sensitivity(95%CI)(%) \| Specificity(95%CI)(%) \| --- --- \| \| AER \| 0.954 \| 97.2(85.5-99.9) \| 92.3(64.0-99.8) \| \| PER \| 0.865 \| 97.2(85.5-99.9) \| 76.9(46.2-95.0) \| \| DER \| 0.769 \| 55.6(38.1-72.1) \| 100(75.3-100) \| \| ADC \| 0.910 \| 92.1(78.6-98.3) \| 91.7(61.5-99.8) \| \| Sizes \| 0.694 \| 70.1(52.5-83.9) \| 65.3(31.6-86.7) \| \| Bile duct dilatation \| 0.680 \| 52.6(35.8-69.0) \| 83.3(51.6-97.9) \| \| Invasion or Metastases \| 0.769 \| 87.2(72.6-95.7) \| 66.7(34.9-90.1) \| \| Boundary \| 0.667 \| 94.9(82.7-99.4) \| 38.5(13.9-68.4) \| \| CA19-9 \| 0.840 \| 78.1(60.0-90.7) \| 78.6(54.4-93.9) \| The data from two readers were pooled together. AER: signal intensity ratio at arterial phase; PER: signal intensity ratio at portal phase; DER: signal intensity ratio at delayed phase; ADC: apparent diffusion coefficients; AUC: area under curve; CA19-9: Carbohydrate antigen 19-9; CI: confidential interval Figure 7: ROC curve of the signal intensity ratio at arterial (AER), portal (PER) and delayed phases (DER), and mean apparent diffusion coefficients (ADC) value for differentiating neuroendocrine carcinoma (PNEC) from pancreatic ductal adenocarcinoma (PDAC). The data from two readers were pooled together. The area under the curve is 0.954, 0.865, 0.769 and 0.910, respectively. DISCUSSION PNEC can mimic PDAC in CT or MR images due in part to a characteristic hypoenhanced pattern [8, 18]. It is valuable to accurately diagnose patient tumor type (PNEC vs. PDAC) before surgery, because treatment strategies and prognoses differ between the two types. The present study compared PNEC MR imaging features with those of PDAC. We found that ill-defined tumor boundaries, hypointensity in contrast-enhanced images, lymph node metastases, and local invasion/distant metastases were more common in PDAC than PNEC. Quantitative data further indicated that signal intensity ratios at arterial and portal phases, and ADC values have potential for differentiating those two tumors. PDAC is a fast-progressing malignancy with surrounding tissue invasion and metastases to distant organs. Tumors usually exhibit ill-defined margins . Lymph nodes metastases and local invasion or distant metastases are also common in PDAC . In our series, 97% of PDAC cases had ill-defined margins and 97% exhibited lymph node metastases, which was consistent with previous studies. PNECs also frequently exhibited ill-defined boundaries (61%), and often metastasized to lymph nodes (61%) or invaded the surrounding tissues. However, these features were still more common in PDAC than PNEC. To some degree, these qualitative features may be useful for differentiating PNEC from PDAC. Contrast-enhanced CT and MRI are helpful for differential diagnosis of benign and malignant lesions based on characteristic vascularization patterns [13, 32, 33]. Several studies showed differences between pancreatic carcinoma and mass-forming focal pancreatitis using contrast-enhanced approaches [31, 33]. PDACs are frequently hypovascular, and enhancement degree is lower than the surrounding pancreatic parenchyma, in particular at arterial and portal phases. However, most PNECs also showed hypoenhancement at arterial and portal phases. Thus, it is challenging to discriminate PNEC from PDAC using regularly qualitative features in contrast-enhanced images. We speculated that quantitative analysis of imaging features would provide more useful diagnostic information. Our data showed that PNEC enhancement degree is higher than that of PDAC at arterial and portal phases. The signal intensity ratios at arterial and portal phases showed good sensitivity and specificity in differentiating PNEC from PDAC. Our results confirmed that quantitative assessment provides reliable information in differentiating PNEC and PDAC. DWI has also been widely used to differentiate benign and malignant pancreatic lesions [27, 33], and may be applicable for grading PNENs [12, 34] and differentiating PNENs and PDAC [35, 36]. Lee, et al. and Wang, et al. found no differences in ADC values between PNENs and PDAC [35, 36]. However, G1 and G2 PNETs, which may have higher ADC values than PNECs [12, 35], were included in these studies. Therefore, we speculate that it is important to consider PNEN grade when evaluating ADC values. As PNECs frequently exhibited high mitotic ability (>20 mitoses per 10 high powered fields (HPFs)) and proliferation (Ki-67 index >20%) in our study, PNECs should therefore exhibit high tissue cellularity and restricted water mobility. Our data supported this speculation. PNEC ADC values were lower than those of PDAC. Moreover, ADC showed very good sensitivity and specificity in differentiating PNEC from PDAC. Our study had several limitations. First, retrospective studies may exhibit selection and verification bias. Some PNEC patients were excluded due to lack of MRI examinations, and the PNEC study population could not reflect the entire PNEC spectrum. Second, only CE-MR and routine DWI were studied, and some novel approaches, such as intravoxel incoherent motion (IVIM) DWI and texture analysis, would provide more information [37, 38]. Finally, larger number of b values would be more sensible rather to use only two b values . In conclusion, we compared PDAC and PNEC MR imaging features, and found that ill-defined margins, lymph node metastases, and local invasion are more common in PDAC. In addition, PNEC enhancement degree at arterial and portal phases are higher than those of PNEC. PNEC usually exhibits lower ADC values than PDAC. Enhancement degree at arterial and portal phases and ADC value may be useful for differentiating PNEC from PDAC. MATERIALS AND METHODS Patient selection This study included patients treated for PDAC and PNEC at our institution between September 2015 and July 2016, and September 2012 and July 2016, respectively. A total of 69 patients with surgically or biopsy diagnosed PDAC were identified. Twenty-two patients were excluded due to lack of MRI examination. Ten patients were excluded because of only a single-phase scan (n=5) or lack of DWI (n=5). We identified 19 patients with surgically or biopsy diagnosed PNEC. Six were excluded due to metastatic NEC (n=2), or lack of MRI examination (n=3) or DWI (n=1). In total, 37 PDAC and 13 PNEC patients were included in this study (Figure 1). The pathological diagnosis of PNEC was based on the WHO 2010 classification for NENs: NEC G3, >20 mitoses per 10 HPF, Ki-67 index >20%. In addition, demographic and clinical data were retrieved from medical records. This retrospective study was approved by our institutional review board with waiver of patient formal consent. MRI examinations All MR scans were performed on a 3.0 superconducting system (Signa HDx 3.0-T, GE Medical Systems, USA) using eight-channel phased-array torso coils. Patients fasted for 8 h prior to MR examination. The protocol included 3D T1-weighted fat-suppressed liver acquisition with volume acceleration-extended volume (LAVA-XV, GE) [TR/TE: 3.1/1.5 ms; imaging duration, 1–2 min]; fast spin-echo T2-weighted fat-suppressed sequence (TE/TR: 4000–8000/80–90 ms; imaging duration, 2–3min) with 3–5 mm slice thickness, 1–2 mm interslice gap, 384×256 matrix and 300–400 mm field-of-view, and the axial DW sequence using the respiratory-triggered single shot echo-planar sequence [TR/TE: 6000–8000/60–70 ms; imaging duration, 2–3 min] with b values of 0 and 1000 s/mm 2 before contrast administration. Based on the DWI signal at two different b values, tissue ADC values were obtained. MRI cholangiopancreatography (MRCP) was performed using heavily T2-weighted fast acquisition spin echo sequence (TR/TE: 2500–6000 /500–800 ms, imaging duration, 2.5 min). Gadopentetate dimeglumine (Magnevist, Bayer HealthCare Pharmaceuticals, Berlin, Germany) was injected at a dose of 0.1 mmol/kg of body weight (2.5 ml/s) following by a 20-ml saline flush, and then axial and coronal T1 images were obtained at 25–35 s (arterial phases), 60–70 s (portal phases) and 200–240 s (delayed phase). MR imaging analysis All MR images were retrospectively reviewed by two abdominal radiologists with more than eight years of experience in abdominal MRI examination on a picture archiving communication system (PACS) workstation. The radiologists were blinded to the final histopathological results and MR diagnosis. The following imaging information was reviewed: tumor position (head-neck or body-tail), tumor margin [well-defined: smooth or lobulating margin with few spiculations or infiltrations (<20%); ill-defined: the perimeter of the tumor showed spiculation or infiltration (>20%)] (11), size, presence of cystic components (solid, cystic components <25%; or mixed cystic-solid), T1 and T2 signal, signal on DWI, enhancement degree at arterial, portal, and delayed phases (hypo-, iso-, or hyperintense compared with normal pancreas). The presence of intrahepatic/extrahepatic bile duct dilatation, pancreatic duct dilatation, pancreatic parenchymal atrophy, lymph node metastases, and local invasion/distant metastases were also reviewed. Pancreatic duct dilation was defined as the main pancreatic duct diameter ≥4 mm. Intrahepatic and extrahepatic bile duct dilatation were confirmed if the duct diameter was >5 mm and >8 mm, respectively. Areas that were hypointense on precontrast T1-weighted images, markedly hyperintense on T2-weighted images, and with no enhancement were identified as cystic components. Quantitative analyses were performed using ADC values and tumor signal intensities on unenhanced and enhanced T1 weighted images by two abdominal radiologists. The signal intensity ratio of tumor to pancreas [signal intensity ratio=signal intensities of tumors/normal pancreatic parenchyma] was calculated. Cystic components were avoided during signal intensity analysis. On DWI and T1 weighted images, regions of interest (ROI) were centered on the solid tumor portions while avoiding necrotic or cystic components, and the most peripheral portions that might result in partial volume effects of adjacent extra-lesional tissues. Tumor and pancreatic parenchyma ADC values were measured. For the normal pancreas, signal intensities and ADC values were noted at a similar ROI avoiding the main duct. ADC values and signal intensities were measured at least three times by each radiologist. The means and the agreement between the two readers were analyzed. Pathology analysis Tumor tissue specimens were fixed with 10% formalin, embedded in paraffin, and sliced for hematoxylin-eosin (H&E) staining. PDAC was typically characterized by moderately to poorly differentiated glandular structures. PNEC was diagnosed based on the 2010 WHO classification for neuroendocrine neoplasms by counting the number of mitoses per 10 HPFs and assessing Ki-67 proliferation index (percentage of positive cells in areas of highest nuclear labeling). NEC G3 was regarded as >20 mitoses per 10 HPF, Ki-67 index >20%. Statistical analysis Data were analyzed using SPSS 16.0 (SPSS Inc, Chicago, IL). Quantitative data were presented as means (standard deviation) or median, and were analyzed via independent samples t test, Mann-Whitney U test, or one-way analysis of variance (ANOVA). Categorical variables were represented as the number of cases (percentage) and were analyzed using Chi-square or Fisher’s exact tests when necessary. The data from the two readers were pooled together for diagnostic performance analysis. The diagnostic values, sensitivities, and specificities of ADC and signal intensity ratio for differentiating PDAC from PENC were assessed by the receiver operating characteristic (ROC) curve, and inter-readers agreement were determined by Bland-Altman plot using Medcalc software (Mariakerke, Belgium). P<0.05 was considered statistically significant. ACKNOWLEDGMENTS This study was funded by Zhejiang Medical Science and Technology Project (2017KY331). CONFLICTS OF INTEREST The authors declare no conflicts of interest. REFERENCES Jemal A, Siegel R, Ward E, Hao Y, Xu J, Murray T, Thun MJ. Cancer statistics, 2008. CA Cancer J Clin. 2008; 58:71–96. Schima W, Ba-Ssalamah A, Kölblinger C, Kulinna-Cosentini C, Puespoek A, Götzinger P. 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14715	https://pressbooks.online.ucf.edu/osuniversityphysics2/chapter/electric-generators-and-back-emf/	Skip to content Chapter 13. Electromagnetic Induction 13.6 Electric Generators and Back Emf Learning Objectives By the end of this section, you will be able to: Explain how an electric generator works Determine the induced emf in a loop at any time interval, rotating at a constant rate in a magnetic field Show that rotating coils have an induced emf; in motors this is called back emf because it opposes the emf input to the motor A variety of important phenomena and devices can be understood with Faraday’s law. In this section, we examine two of these. Electric Generators Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Motional Emf. We now explore generators in more detail. Consider the following example. Example Calculating the Emf Induced in a Generator Coil The generator coil shown in Figure 13.27 is rotated through one-fourth of a revolution (from to) in 15.0 ms. The 200-turn circular coil has a 5.00-cm radius and is in a uniform 0.80-T magnetic field. What is the emf induced? Strategy Faraday’s law of induction is used to find the emf induced: We recognize this situation as the same one in Example 13.6. According to the diagram, the projection of the surface normal vector to the magnetic field is initially and this is inserted by the definition of the dot product. The magnitude of the magnetic field and area of the loop are fixed over time, which makes the integration simplify quickly. The induced emf is written out using Faraday’s law: Solution Show Answer We are given that , , and The area of the loop is Entering this value gives Significance This is a practical average value, similar to the 120 V used in household power. The emf calculated in Example 13.9 is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width w and height l in a uniform magnetic field, as illustrated in Figure 13.28. Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be , where the velocity v is perpendicular to the magnetic field B. Here the velocity is at an angle with B, so that its component perpendicular to B is v sin (see Figure 13.28). Thus, in this case, the emf induced on each side is , and they are in the same direction. The total emf around the loop is then This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity . The angle is related to angular velocity by so that Now, linear velocity v is related to angular velocity by Here, so that and Noting that the area of the loop is and allowing for N loops, we find that This is the emf induced in a generator coil of N turns and area A rotating at a constant angular velocity in a uniform magnetic field B. This can also be expressed as where is the peak emf, since the maximum value of . Note that the frequency of the oscillation is and the period is Figure 13.29 shows a graph of emf as a function of time, and it now seems reasonable that ac voltage is sinusoidal. The fact that the peak emf is makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater ), the greater the emf. This is noticeable on bicycle generators—at least the cheaper varieties. Figure 13.30 shows a scheme by which a generator can be made to produce pulsed dc. More elaborate arrangements of multiple coils and split rings can produce smoother dc, although electronic rather than mechanical means are usually used to make ripple-free dc. In real life, electric generators look a lot different from the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 13.31 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. The generation of electrical energy from mechanical energy is the basic principle of all power that is sent through our electrical grids to our homes. Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In the next section, we further explore the action of a motor as a generator. Back Emf Generators convert mechanical energy into electrical energy, whereas motors convert electrical energy into mechanical energy. Thus, it is not surprising that motors and generators have the same general construction. A motor works by sending a current through a loop of wire located in a magnetic field. As a result, the magnetic field exerts torque on the loop. This rotates a shaft, thereby extracting mechanical work out of the electrical current sent in initially. (Refer to Force and Torque on a Current Loop for a discussion on motors that will help you understand more about them before proceeding.) When the coil of a motor is turned, magnetic flux changes through the coil, and an emf (consistent with Faraday’s law) is induced. The motor thus acts as a generator whenever its coil rotates. This happens whether the shaft is turned by an external input, like a belt drive, or by the action of the motor itself. That is, when a motor is doing work and its shaft is turning, an emf is generated. Lenz’s law tells us the emf opposes any change, so that the input emf that powers the motor is opposed by the motor’s self-generated emf, called the back emf of the motor (Figure 13.32). The generator output of a motor is the difference between the supply voltage and the back emf. The back emf is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster, the back emf grows, always opposing the driving emf, and reduces both the voltage across the coil and the amount of current it draws. This effect is noticeable in many common situations. When a vacuum cleaner, refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feeder lines by the large current drawn by the motor. When a motor first comes on, it draws more current than when it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motor slows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can overheat it (via resistive power in the coil, ), perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it increases its angular velocity until the back emf is nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction. Eddy currents in iron cores of motors can cause troublesome energy losses. These are usually minimized by constructing the cores out of thin, electrically insulated sheets of iron. The magnetic properties of the core are hardly affected by the lamination of the insulating sheet, while the resistive heating is reduced considerably. Consider, for example, the motor coils represented in Figure 13.32. The coils have an equivalent resistance of and are driven by an emf of 48.0 V. Shortly after being turned on, they draw a current and thus dissipate of energy as heat transfer. Under normal operating conditions for this motor, suppose the back emf is 40.0 V. Then at operating speed, the total voltage across the coils is 8.0 V (48.0 V minus the 40.0 V back emf), and the current drawn is Under normal load, then, the power dissipated is This does not cause a problem for this motor, whereas the former 5.76 kW would burn out the coils if sustained. Example A Series-Wound Motor in Operation The total resistance of a series-wound dc motor is (Figure 13.33). When connected to a 120-V source (), the motor draws 10 A while running at constant angular velocity. (a) What is the back emf induced in the rotating coil, (b) What is the mechanical power output of the motor? (c) How much power is dissipated in the resistance of the coils? (d) What is the power output of the 120-V source? (e) Suppose the load on the motor increases, causing it to slow down to the point where it draws 20 A. Answer parts (a) through (d) for this situation. Strategy The back emf is calculated based on the difference between the supplied voltage and the loss from the current through the resistance. The power from each device is calculated from one of the power formulas based on the given information. Solution Show Answer The back emf is Since the potential across the armature is 100 V when the current through it is 10 A, the power output of the motor is A 10-A current flows through coils whose combined resistance is , so the power dissipated in the coils is Since 10 A is drawn from the 120-V source, its power output is Repeating the same calculations with , we find The motor is turning more slowly in this case, so its power output and the power of the source are larger. Significance Notice that we have an energy balance in part (d): Summary An electric generator rotates a coil in a magnetic field, inducing an emf given as a function of time by where A is the area of an N-turn coil rotated at a constant angular velocity in a uniform magnetic field The peak emf of a generator is . Any rotating coil produces an induced emf. In motors, this is called back emf because it opposes the emf input to the motor. Problems Design a current loop that, when rotated in a uniform magnetic field of strength 0.10 T, will produce an emf where and Show Solution three turns with an area of 1 m2 A flat, square coil of 20 turns that has sides of length 15.0 cm is rotating in a magnetic field of strength 0.050 T. If the maximum emf produced in the coil is 30.0 mV, what is the angular velocity of the coil? A 50-turn rectangular coil with dimensions rotates in a uniform magnetic field of magnitude 0.75 T at 3600 rev/min. (a) Determine the emf induced in the coil as a function of time. (b) If the coil is connected to a resistor, what is the power as a function of time required to keep the coil turning at 3600 rpm? (c) Answer part (b) if the coil is connected to a 2000- resistor. Show Solution a. b. c. The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V. (a) What is the frequency of the output voltage? (b) What is the strength of the magnetic field in which the coil is turning? A flip coil is a relatively simple device used to measure a magnetic field. It consists of a circular coil of N turns wound with fine conducting wire. The coil is attached to a ballistic galvanometer, a device that measures the total charge that passes through it. The coil is placed in a magnetic field such that its face is perpendicular to the field. It is then flipped through and the total charge Q that flows through the galvanometer is measured. (a) If the total resistance of the coil and galvanometer is R, what is the relationship between B and Q? Because the coil is very small, you can assume that is uniform over it. (b) How can you determine whether or not the magnetic field is perpendicular to the face of the coil? Show Solution a. B is proportional to Q; b. If the coin turns easily, the magnetic field is perpendicular. If the coin is at an equilibrium position, it is parallel. The flip coil of the preceding problem has a radius of 3.0 cm and is wound with 40 turns of copper wire. The total resistance of the coil and ballistic galvanometer is When the coil is flipped through in a magnetic field a change of 0.090 C flows through the ballistic galvanometer. (a) Assuming that and the face of the coil are initially perpendicular, what is the magnetic field? (b) If the coil is flipped through what is the reading of the galvanometer? A 120-V, series-wound motor has a field resistance of 80 and an armature resistance of 10 . When it is operating at full speed, a back emf of 75 V is generated. (a) What is the initial current drawn by the motor? When the motor is operating at full speed, where are (b) the current drawn by the motor, (c) the power output of the source, (d) the power output of the motor, and (e) the power dissipated in the two resistances? Show Solution a. 1.33 A; b. 0.50 A; c. 60 W; d. 37.5 W; e. 22.5W A small series-wound dc motor is operated from a 12-V car battery. Under a normal load, the motor draws 4.0 A, and when the armature is clamped so that it cannot turn, the motor draws 24 A. What is the back emf when the motor is operating normally? Glossary back emf : emf generated by a running motor, because it consists of a coil turning in a magnetic field; it opposes the voltage powering the motor peak emf : maximum emf produced by a generator electric generator : device for converting mechanical work into electric energy; it induces an emf by rotating a coil in a magnetic field Licenses and Attributions Electric Generators and Back Emf. Authored by: OpenStax College. Located at: License: CC BY: Attribution. License Terms: Download for free at
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14717	https://www3.stat.sinica.edu.tw/statistica/oldpdf/A6n410.pdf	Statistica Sinica 6(1996), 957-974 DISTRIBUTION THEORY OF RUNS AND PATTERNS ASSOCIATED WITH A SEQUENCE OF MULTI-STATE TRIALS James C. Fu National Donghwa University Abstract: Runs and patterns in a sequence of Bernoulli trials and multi-state trials have broadly been used for various purposes and in many areas of statistics and applied probability. Recently, Fu and Koutras (1994) developed a new method based on ﬁnite Markov chain imbedding technique to study the exact distributions for the number of speciﬁed runs and patterns in a sequence of Bernoulli trials. In this manuscript, a “forward and backward principle” for the method of ﬁnite Markov chain imbedding is introduced to study the exact and joint distributions for the numbers of runs and patterns in a sequence of multi-state trials. Waiting time distribution for a pattern is also obtained. Key words and phrases: Multi-state trials, distributions of runs and patterns, Markov chain, transition probability matrix. 1. Introduction Run statistics and patterns in a sequence of Bernoulli trials have been suc-cessfully used in various areas, such as hypothesis testing (run-test, Wald and Wolfowitz (1940) and Walsh (1965)), statistical quality control (Mosteller (1941) and Wolfowitz (1943)), DNA sequencing, psychology, and ecology (Schwager (1983)). The distribution theory of runs seems to have started at the end of the nineteenth century (Stevens (1939)). There were a considerable amount of research works on the distribution theory of runs around 1940, for example, Wishart and Hirsheld (1936), Cochran (1938), Mood (1940), Wald and Wolfowitz (1940) and Wolfowitz (1943). Most of the research during this period was focused on the study of the conditional distributions of runs given the total number of successes. There was very little research on the exact and limiting distributions of runs during the period between 1950 and 1970. However, there were some very interesting approximate formulae for the distributions of runs and patterns de-veloped during this period (Walsh (1965) and Gibbons (1985)). During the late 1980’s and early 1990’s , this area became very active again. Recent publications in this area are mainly focused on studying the exact and limiting distributions for certain patterns and runs in a sequence of independent identically distributed (i.i.d.) Bernoulli trials. 958 JAMES C. FU The number Nn,k of non-overlapping consecutive k successes (in the sense of Feller’s counting (1968, Ch. XIII)) in a sequence of Bernoulli trials is probably one of the most studied run statistic. In the context of distribution theory, the distribution of Nn,k is called a binomial distribution of order k. The run statistic Nn,k also plays an even more important role in ﬁnding the distributions for other runs. Its distribution has been applied successfully in various areas, especially to the reliabilities of some engineering systems. For example the probability of Nn,k = 0 is the reliability of a consecutive-k-out-of-n : F system and the tail probability of Nn,k < m is the reliability of a m-consecutive-k-out-of-n : F system. It is heavily studied by many researchers in the area of reliability theory. Traditionally, combinatorial methods were used to ﬁnd the exact distribu-tions for the numbers of runs and patterns. Even in the case of i.i.d. Bernoulli trials, the formulae for the distributions of runs and patterns are often complex and tedious. It is usually diﬃcult to extend those i.i.d. results by combinatorial methods to a sequence of independent but non-identical Bernoulli trials, or to extend to a more complex case such as a sequence of m-step Markov dependent multi-state trials. Recently, Fu and Koutras (1994) took a Markov chain ap-proach to study the exact distributions for the numbers of runs and patterns in a sequence of Bernoulli trials. Their ideas were largely borrowed from the sequence of papers by Fu (1986) and Chao and Fu (1989, 1991) for studying the reliabilities of linearly connected engineering systems. The exact distribution of a speciﬁed run statistic is studied under the framework of a ﬁnite Markov chain, and is expressed in terms of its transition probability matrices. In that paper, they have obtained the exact distributions of ﬁve often used run statistics in a sequence of i.i.d. Bernoulli trials (viz., the number En,k of success runs of size exactly k, the number Gn,k of success runs of size greater than or equal to k, the number Nn,k of non-overlapping consecutive k successes, the number Mn,k of overlapping consecutive k successes, and the length Ln of the longest success run). Until recently, there were not many general results on the exact and limit-ing distributions for runs and patterns associated with a sequence of multi-state trials. This was probably due to the complexity of combinatorial analysis in the case of multi-state trials. Schwager (1983) gave a very interesting paper that studied the probability of a simple run (run containing one or two symbols) in a sequence of n multi-state trials each of which has s (s ≥2) possible outcomes (symbols) by using the recursive method of renewal equations. He argued that the recursive approach is superior to the generating function approach (see Feller (1968), Ch. XIII) to treat the runs in the latter can only be used for the multiple runs (viz., runs containing more than two symbols) and the generalized multi-ple runs (viz., a collection of runs) for the i.i.d. multi-state trials. For ﬁnding DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 959 the exact and joint distributions of runs under independent but non-identical multi-state trials or Markov dependent multi-state trials, both methods become tedious and intractable. There have been no general theorems to deal with those dependent multi-state trials. There were also considerable applications of waiting time (sooner or later) distribution of a speciﬁed run to various areas of statistics, especially biostatis-tics and quality control. Most of these results were focused on i.i.d. Bernoulli trials (see Hahn and Gage (1983), Benevento (1984), Ebneshahrashoob and So-bel (1990), and Balasubramanian, Viveros and Balakrishnan (1993)). There were very few general results regarding the waiting time distribution for the case of multi-state trials and there was no result in Markov dependent multi-state trials. This article mainly introduces a “forward and backward principle” for the method of ﬁnite Markov chain imbedding developed by Fu and Koutras (1994) to study the exact and joint distributions for the runs and patterns in four diﬀerent cases; (1) i.i.d., (2) independent but non-identical, (3) homogeneous Markov dependent, and (4) non-homogeneous Markov dependent sequences of multi-state trials. It also demonstrates that this method could be used to obtain the waiting time distribution of a speciﬁed run. The rest of this manuscript is organized in the following ways. In Section 2, the method of Markov chain imbedding for ﬁnding the exact distribution of runs is introduced. In Section 3, the “forward and backward principle” for constructing a Markov chain associated with a speciﬁed run in a sequence of multi-state trials is systematically developed. Section 4 studies mainly the marginal and joint distributions for success runs and failure runs in a sequence of Bernoulli trails. Section 5 gives a general result regarding the waiting time distribution of the mth occurrence of a speciﬁed pattern. In Section 6, several numerical examples are given to illustrate the main results. In Section 7, several technical remarks, discussions and possible extensions are given. 2. Finite Markov Chain Imbedding Let Z1, . . . , Zn be a sequence of n i.i.d. multi-state trials, each of which has s (s ≥2) states (or symbols), labeled b1, . . . , bs and their corresponding probabilities p1, . . . , ps of occurring. Throughout this section, we denote the random variable Xn(Xn = φ(Z1, . . . , Zn)) by the number of runs (or patterns) in a sequence of n multi-state trials. For a given positive integer n, let Γn = {0, 1, . . . , n} be a ﬁnite index set and Ω= {a1, . . . , am} be a state space having m states. Deﬁnition 2.1. The random variable Xn = φ(Z1, . . . , Zn), the number of a speciﬁed runs occurring in a sequence of n multi-state trials Z1, . . . , Zn, is ﬁnite Markov chain imbeddable if 960 JAMES C. FU (i) there exists a ﬁnite Markov chain {Yt, t ∈Γn} deﬁned on a ﬁnite state space Ωwith transition matrices Mt, t ∈Γn, and (ii) there exists a partition {Cx, x = 0, 1, . . . , l} on the state space Ω(where Cx and l may depend on n), such that, for every x = 0, 1, . . . , l, P(Xn = x) = P(Yn ∈Cx). (2.1) It is well known that the probabilistic behavior of a Markov chain is uniquely characterized by its own transition matrices. The following specially constructed formula (2.2) for computing the probability of Yn ∈Cx introduced by Fu and Koutras (1994) is a matrix version of the Chapman-Kolmogorov Theorem. It has been successfully used to obtain the exact distribution (also moments and generating function) for the number of runs (or patterns) in Bernoulli trials. It will play an even more important role for developing the distribution of the number of runs in a sequence of multi-state trials. For this reason, we repeat the formula here (see Fu and Koutras (1994) for details). Let {Yt, Ω, Mt : t ∈Γn} be a Markov chain with initial probabilities π0 = (p1, . . . , pm), where pj = P(Y0 = aj), for j = 1, . . . , m, U(ai) = (0, . . . , 0, 1, 0, . . . , 0) be a unit vector with 1 at ith place and 0 elsewhere. For given n, if Xn is ﬁnite Markov chain imbeddable, then P(Xn = x) = πo n t=1 Mt U′(Cx), (2.2) where U(Cx) = ai∈Cx U(ai). Similarly, the moments and the generating func-tion of the random variable Xn can be obtained by replacing the vector U′(Cx) in (2.2) with vectors V ′ r and W ′(s) respectively, where Vr = l x=0 xrU(Cx) and W(s) = l x=0 sxU(Cx). Remark. The random variable Xn is ﬁnite Markov chain imbeddable for every n that does not imply the sequence {Xn} is a Markov chain. 3. Forward and Backward Principle The Markov chain imbedding technique for ﬁnding the distribution of a spec-iﬁed run introduced by Fu and Koutras (1994) for Bernoulli trials involves basi-cally three essential steps: (i) the construction of a proper state space Ωbased on the structure of the speciﬁed run, (ii) the construction of a ﬁnite Markov chain and its transition probability matrices, and (iii) the construction of a partition {Cx} on the state space Ωwhich is one to one corresponding to the random variables Xn in the sense that P(Xn = x) = P(Yn ∈Cx) for all x. Even in the case of a sequence of i.i.d. Bernoulli trials, the task of ﬁnding the exact distribu-tion of a speciﬁed pattern (run) by using the Markov chain imbedding technique DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 961 may not always be trivial. There is no good deﬁnition about what is a pattern (or run). Wolfowitz (1943) wrote “ · · · we shall not give a general deﬁnition here, because new advances and application of new criteria to new problems will probably soon render most deﬁnitions obsolete”. In this manuscript we consider mainly two types of often used patterns: Deﬁnition 3.1. Λ is a simple pattern if Λ is composed of a speciﬁed sequence of k symbols, i.e. Λ = bi1, . . . , bik (the length of the pattern k is ﬁxed, and the symbols in the pattern are allowed to be repeated). Deﬁne Λ1 ∪Λ2 to be either the pattern Λ1 occurred or the pattern Λ2 oc-curred. Deﬁnition 3.2. Λ is a compound pattern if it is a union of m distinct simple patterns, i.e. Λ = Λ1 ∪. . . ∪Λm (m is ﬁxed). To ﬁnd the exact distribution for a speciﬁed pattern not only requires a deep understanding of the structure of the speciﬁed pattern but also its counting procedure throughout the sequence of n multi-state trials. In order to facilitate our study of the general ﬁnite Markov chain imbedding technique for simple and compound patterns in multi-state trials without heavy mathematics, let us consider a sequence of n i.i.d. three-state trials Z1, . . . , Zn, each of which has three possible outcomes (symbols), labeled b1, b2, and b3 with corresponding probabilities p1, p2, and p3 of occurring. Suppose we are interested in ﬁnding the exact distribution of the random variable Xn, the number of non-overlapping speciﬁed simple patterns Λ = b1b1b2 in a sequence of n three-state trials. In the following we introduce the forward and backward principle for the ﬁnite Markov chain imbedding technique to ﬁnd the exact distribution of a given simple pattern Λ. (i) Decompose the pattern Λ = b1b1b2 forward into three sub-patterns labeled 0, 1 = b1, and 2 = b1b1, where the label “0” stands for neither sub-pattern “1” nor sub-pattern “2”. We shall refer to these three sub-patterns’ 0, 1, and 2 as ending blocks. (ii) Let ω = (z1, . . . , zn) be a realization of a sequence of n three-state trials where zi is the outcome of the ith trials. We deﬁne the Markov chain {Yt : t = 1, . . . , n} operating on ω as Yt(ω) = (u, v) for every t = 1, . . . , n, where u = the total number of non-overlapping patterns Λ that occurred in the ﬁrst t trials (counting forward from the ﬁrst trials to the tth trials), v = the sub-pattern (ending block), counting backward from t. The following diagram illustrates the deﬁnitions of u and v in a sequence of t trials: 962 JAMES C. FU 1 2 3 4 5 t′ t u of non-overlapping Λ patterns -Counting Forward sub-pattern v Counting Backward no Λ pattern For example, let us consider a realization ω = (b3b1b1b2b1b1b3) of a sequence of seven three-state trials. Applying the forward and backward principle, then the realization of the imbedded Markov chain Yt on ω is given by {Y1(ω) = (0, 0), Y2(ω) = (0, 1), Y3(ω) = (0, 2), Y4(ω) = (1, 0), Y5(ω) = (1, 1), Y6(ω) = (1, 2) and Y7(ω) = (1, 0)}. Note that for every given ω, the realization of the imbedded Markov chain Yt(ω) = (u, v) is uniquely by determined by (i) and (ii) under non-overlapping counting. Based on (i) and (ii), the state space Ωassociated with the imbedded Markov chain is deﬁned by Ω= {(u, v) : u = 0, 1, . . . , l, and v = 0, 1, . . . , k −1} (3.1) with m = (l + 1)k states, where k is the length of the simple pattern Λ and l = [n/k] is the maximum number of patterns Λ possible in the sequence of n multi-state trials. (iii) For t = 1, . . . , n, the transition probabilities of the transition matrix Mt is determined by the following two equations : for u = 0, 1, . . . , l −1, P(Yt = (u + 1, 0)\|Yt−1 = (u, k −1)) = pj, (3.2) where pj is the probability of bj the last symbol of the speciﬁed pattern Λ, for u = 0, 1, . . . , l and v, v′ = 0, 1, . . . , k −1, P(Yt = (u, v′)\|Yt−1 = (u, v)) = v→v′ pi, (3.3) where v→v′ sums over all pi corresponding to bi of which the ending block v is changed to the ending block v′, and zero otherwise. (iv) The partition {Cx = [(x, v) : (x, v) ∈Ω, v = 0, 1, . . . , k −1], for x = 0, 1, . . . , l} on Ωis one-to-one corresponding to the random variable Xn in the sense that P(Xn = x) = P(Yn ∈Cx) for every x = 0, 1, . . . , l. It follows from our construction that the transition probabilities given by (3.2) and (3.3) depend only on Yt−1 = (u, v); hence the sequence {Yt} is a Markov chain. The four steps (i), (ii), (iii), and (iv) of construction mentioned DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 963 above establish that the random variable Xn, the number of simple patterns Λ in a sequence of n independent multi-state trials, is ﬁnite Markov chain imbeddable. For our example of n = 7 and Λ = b1b1b2, it follows from the forward and backward principle that the imbedded Markov chain {Yt : t = 1, . . . , 7} deﬁned on the state space Ω= {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)} has transition probability matrices Mt =                  p2 + p3 p1 0 \| 0 0 0 \| 0 0 0 p2 + p3 0 p1 \| 0 0 0 \| 0 0 0 p3 0 p1 \| p2 0 0 \| 0 0 0 −−− − − \| −−− − − \| −−− − − 0 0 0 \| p2 + p3 p1 0 \| 0 0 0 0 0 0 \| p2 + p3 0 p1 \| 0 0 0 0 0 0 \| p3 0 p1 \| p2 0 0 −−−− − − \| −−− − − \| −−− − − 0 0 0 \| 0 0 0 \| p2 + p3 p1 0 0 0 0 \| 0 0 0 \| p2 + p3 0 p1 0 0 0 \| 0 0 0 \| 0 0 1                  (3.4) for t = 1, 2, . . . , 7. The partition associated with the random variable X7 is generated by the subsets Cx = {(x, v) : v = 0, 1, 2}, x = 0, 1, 2. The distribution of X7 could be easily computed by using the equation (2.2). The numerical result is given in Section 6. To demonstrate the usefulness of this approach for the compound pattern, let us consider an example of n = 5 and the speciﬁed compound pattern Λ = Λ1∪Λ2 being a union of two simple patterns Λ1 = b1b2 and Λ2 = b3b1. We are interested in the distribution of the random variable X5, the number of patterns either Λ1 or Λ2 occurring in a sequence of ﬁve i.i.d. three-state trials. Note that the compound pattern Λ1 ∪Λ2 can be decomposed into 0, 1 = b1, and 2 = b3 three ending blocks, where the ending block “0” stands for neither the ending block “1” nor the ending block “2”. The imbedded Markov chain associated with the compound pattern has a state space Ω= {(u, v) : u = 0, 1, 2 and v = 0, 1, 2} and transition probability matrices Mt =                  p2 p1 p3 \| 0 0 0 \| 0 0 0 0 p1 p3 \| p2 0 0 \| 0 0 0 p2 0 p3 \| p1 0 0 \| 0 0 0 − − − \| − − − \| − − − 0 0 0 \| p2 p1 p3 \| 0 0 0 0 0 0 \| 0 p1 p3 \| p2 0 0 0 0 0 \| p2 0 p3 \| p1 0 0 − − − \| − − − \| − − − 0 0 0 \| 0 0 0 \| p2 p1 p3 0 0 0 \| 0 0 0 \| 0 1 0 0 0 0 \| 0 0 0 \| 0 0 1                  (3.5) 964 JAMES C. FU for t = 1, . . . , 5. The numerical results are given in Section 6. With some modiﬁcations of the forward and backward principle, the method could also be extended to a sequence of homogeneous (or non-homogeneous) Markov dependent multi-state trials. We provide the following example. Let {Zi; i = 1, . . . , n} be a three-state {a, b, c} homogeneous Markov chain with a 3 × 3 transition probability matrix (pij), where i, j = a, b, and c. Let Λ = aa be the pattern of interest and the random variable Xn(Λ) be the number of patterns Λ in a sequence of n Markov dependent trials. Deﬁne the state space Ω= {(x, y) : x = 0, 1, . . . , [n/2], and y = a∗, a, b, c} and partition {Cx = [(x, y) : y = a∗, a, b, c], for x = 0, 1, . . . , [n/2]}. For every t and ω = (z1, . . . , zn), deﬁne Yt(ω) = (x, y), where x = number of a full pattern Λ in the sequence z1, . . . , zt, y =              a∗, if a full pattern Λ occurs at the tth trial in non-overlapping, a, if a half pattern occurs at the tth trial in non-overlapping, b or c, if the tth trial Zt = b or c respectively. We give the three key cases of transition probabilities of Yt+1 = (u, v) given Yt = (x, y) below; for x = 0, 1, . . . , ([n/2] −1), (i) P(Yt+1 = (x + 1, a∗)\|Yt = (x, a)) = paa, (ii) P(Yt+1 = (x, a)\|Yt = (x, a∗)) = paa, and (iii) P(Yt+1 = (x, v)\|Yt = (x, y)) = pyv, for y = b, c, and v = a, b, and c, and leave the remaining cases to the readers. The exact distribution of Xn(Λ) can be obtained by equation (2.2) with ease. For n s-state (s ≥2) trials, the total number of possible realizations {ω = (z1, . . . , zn)} is sn which tends to inﬁnity exponentially fast. When n the number of trials is relatively large, say n ≥50, the total number of possible realizations becomes extremely large and the computer cannot handle the case. For the case of a simple pattern, the forward and backward principle classiﬁes a realization ω according to the number of patterns occurring in the realization and its ending block. It reduces the number of states to a linear function of n (at most m = (l+ 1)k states). This reduction is signiﬁcant and makes the computation manageable. 4. Exact and Joint Distributions of Runs in Bernoulli Trials In this section, the “forward and backward principle” and the Markov chain imbedding technique are used to study the marginal and joint distributions of the DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 965 random variable R(n, S), the number of success runs, and the random variable R(n, F), the number of failure runs, under three diﬀerent types of Bernoulli trials; (i) i.i.d., (ii) independent but not identically distributed, and (iii) one-step homogeneous or non-homogeneous Markov chain dependency. 4.1. Exact distributions Let Z1, . . . , Zn be n i.i.d. Bernoulli trial and for each trials with probability p(0 < p < 1) for S (Success) and probability q (q = 1 −p) for F (Failure). (i) In the sequential counting procedure, the pattern “success run” can be decomposed into two ending blocks (or sub-patterns) F and S respectively (it means that a sequence ends with either F or S). (ii) For 1 ≤t ≤n, deﬁne the Markov chain Yt(ω) = (u, v), with the under-standing that the sequence ω has u success runs with an ending block v (v = F or S) in the ﬁrst t trials. Therefore the state space Ωcan be deﬁned as Ω= {(0, F)} ∩{(u, v) : u = 1, . . . , [(n + 1)/2] and v = S, F}, (4.1) where the state (0, F) indicates that either all the outcomes are failures or when the system is at the initial time, t = 0. The partition of the state space Ω associated with the random variable R(n, S) is given by {C0, C1, . . . , Cl; l = [(n + 1)/2]}, (4.2) where C0 = {(0, F)} and Cu = {(u, S) and (u, F)}, for u = 1, . . . , l. (iii) The transition probability matrix associated with the Markov chain is (0, F) (1, S) (1, F) · · · · · · (l, S) (l, F) M(S) = (0, F) (1, S) (1, F) . . . . . . (l, S) (l, F)               q p 0 · · · 0 0 p q 0 · · · 0 0 ... q p 0 0 . . . ... ... ... ... . . . 0 0 q p 0 0 · · · 0 p q 0 0 0 · · · 0 0 1               , (4.3) where the states (u, v) are arranged lexicographically. It is a (2l + 1) × (2l + 1) stochastic matrix. Given π0 = (1, 0, . . . , 0), it follows from our construction and equation (2.2) that the distribution of success runs is speciﬁed by the following equation, for r = 0, 1, . . . , l, P(R(n, S) = r) = π0Mn(S)U′(Cr). (4.4) 966 JAMES C. FU For independent but non-identically distributed Bernoulli trials with pi and qi being success and failure probabilities respectively, the exact distribution of a success run can be represented by, P(R(n, S) = r) = π0 n t=1 Mn(S) U′(Cr) for r = 0, 1, . . . , l, (4.5) where the transition probability matrix Mt(S) has the same form as M(S) except that p is replaced by pt and q is replaced by qt. Further the result can also be easily extended to a sequence of one-step homogeneous Markov dependent Bernoulli trials {Zt : t = 1, . . . , n} having a transition probability matrix A = pSS pSF pF S pF F . (4.6) Given the initial condition that P(Z1 = S) = p and P(Z1 = F) = q, the exact distribution of success runs can also be represented by the equation (4.5) with M1 = M(S) and Mt =               pF F pF S pSS pSF 0 pF F pF S ... ... ... ... 0 pSS pSF 1               for t = 2, 3, . . . , n. (4.7) The above results (4.4), (4.5) and (4.7) can be easily extended to the distri-bution of a failure run R(n, F) by interchanging the S with F. Note that all the results derived from this method are simple and tractable whether they are in a sequence of i.i.d., or non-identical, or one-step homogeneous Markov dependent Bernoulli trials. They diﬀer only in the transition matrices which are rather minor. 4.2. Joint distribution Let R(n) be the total number of success runs and failure runs in a sequence of n Bernoulli trials. It follows that for every n R(n) = R(n, S) + R(n, F). (4.8) DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 967 Note that the random variables R(n, S) and R(n, F) are highly related in the following ways. If there are x of success runs, then there can only be x + 1, x, or x+1 of failure runs. If R(n) = 2x is an even number, then R(n, S) = R(n, F) = x, and if R(n) = 2x−1 is an odd number, then either R(n, S)+1 = R(n, F) = x or in the situation R(n, S) = R(n, F) + 1 = x. Hence, for any sequence of Bernoulli trials it can only be classiﬁed into one for the following four types of states: (i) (x, x −1) being a sequence which has x success runs and (x −1) failure runs with an ending block S. (ii) (x, x + 1) being a sequence which has x success runs and (x + 1) failure runs with an ending block F. (iii) (x, x, F) being a sequence which has x success runs and x failure runs with an ending block F. (iv) (x, x, S) being a sequence which has x success runs and x failure runs with an ending block S. For ﬁnding the distribution of R(n), we could ﬁrst ﬁnd the joint distribution of R(n, S) and R(n, F), then project the joint distribution of (R(n, S), R(n, F)) on the partition generated by R(n, S) + R(n, F). Let {Yt} be a Markov chain deﬁned on the state space speciﬁed by (i) to (iv), Ω= {(1, 0), (0, 1), (1, 1, S), (1, 1, F), . . . , (l, 1, S), (l, 1, F)}. (4.9) For each ω = (SSFS . . . FF), we deﬁne Yt(ω) = (the number of success runs, the number of failure runs, ending sub-pattern) as an element of Ω. In the case of independent but non-identical Bernoulli trials, the deﬁnition of Yt, t = 1, . . . , n yields the initial distribution π1 = (p1, q1, 0, . . . , 0) of Y1 and the transition prob-ability matrices for t = 2, 3, . . . , n Mt = (1, 0) (0, 1) (1, 1, S) (1, 1, F) · · · (l, l −1) (l −1, l) (l, l, S) (l, l, F)                      pt 0 0 qt qt pt 0 0 0 pt 0 0 qt qt pt · · · · · · · · · · · · pt 0 0 qt qt pt 0 0 1 0 1                      . (4.10) The exact distribution of the total number of runs can be represented by P(R(n) = r) = (p1, q1, 0, . . . , 0) n t=2 Mt U′(Cr), for r = 1, . . . , l, (4.11) 968 JAMES C. FU where Cr = {(r/2, r/2, S), (r/2, r/2, F)}, if r is an even number, and Cr = {((r + 1)/2, (r −1)/2), ((r −1)/2, (r + 1)/2)}, if r is an odd number. The marginal distributions of R(n, S) and R(n, F) can also be obtained by projecting the joint distribution to the partitions generated by random variables R(n, S) and R(n, F) respectively. Again, with some simple modiﬁcations, the result (4.11) also holds for the both homogeneous and non-homogeneous Markov dependent Bernoulli trials. 5. Waiting Time Distribution Consider a simple pattern Λ = bi1, . . . , bik composed of a sequence of k speciﬁed symbols, bi1, . . . , and bik. The pattern Λ can always be decomposed forward as k + 1 ending blocks 0,1 = bi1, . . . , k −1 = bi1, . . . , bik. We deﬁne the state space Ωm,k associated with the pattern Λ as Ωm,k = {(x, y) : x = 0, 1, . . . , m −1 and y = 0, 1, . . . , k −1} ∪{α}, (5.1) where the α is an absorbing state. Construct a Markov chain {Yt : t = 1, . . . , n} on the state space Ωm,k with Yt(ω) =            (x, y), if there are x Λ patterns with an ending block y in the ﬁrst t trials, α, if there are m or more than m Λ patterns in the ﬁrst t trials, and the transition probability matrix Mt(m) = (0, 0) (0, 1) . . . (m −1, k −1) −−−−−−− α           \| \| p(x,y),(u,v)(t) \| p(x,y),α(t) \| −− −−−−− − − −−−− 0 · · · 0 \| 1           , (5.2) where α is an absorbing state and Mt(m) is a (mk + 1) × (mk + 1) matrix whose transition probabilities p(x,y),(u,v)(t) and p(x,y),α(t) are determined by the equation (3.2) and (3.3). Let Xn(Λ) be the number of patterns occurring in a sequence of n multi-state trials and W(m, Λ) be the waiting time ( the number of trials required) for the mth (m = 1, 2, . . .) pattern to have occurred. For example, W(1, Λ) = 5 means the ﬁrst time the pattern Λ occurred at the 5th trial. DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 969 Given Λ, m and n(mk ≤n), it follows from the deﬁnitions that the two random variables Xn(Λ) and W(m, Λ) are one-to-one related in the following way: W(m, Λ) ≤n if and only if Xn(Λ) ≥m. (5.3) Since the exact distribution of Xn(Λ) can be obtained by using the method of Markov chain imbedding, hence the relationship (5.3) yields the distribution for the waiting time W(m, Λ): P(W(m, Λ) = n) = π0 n−1 t=1 Mt(m) (I −Mn(m))U′(m) (5.4) for n ≥m × k, where π0 = (1, 0, . . . , 0) is a 1 × (km + 1) vector, Mt(m) is given by (5.2), k is the length of the pattern Λ, and U′(m) = m−1 x=0 U′(Cx). By properly deﬁning the transition matrix, this general theorem covers many well-known results about the waiting time distributions given by Hahn and Gage (1983), Benevento (1984), Ebneshahrashoob and Sobel (1990), Balassubrama-nian, Viveros and Balakrishnan (1993), and Fu and Koutras (1994). For ex-ample, let Λ = SS . . . SS be a pattern of consecutive k successes in a sequence of i.i.d. Bernoulli trials having probabilities p and q for success and failure re-spectively. The simple pattern Λ = SS . . . SS of consecutive k successes can be decomposed into k ending blocks 0 = F, 1 = S, 2 = SS, . . . , k −1 consecutive successes. The state space Ω= {(0, y) : y = 0, 1, . . . , k −1} ∪{α} contains k + 1 states having α as absorbing state. For the i.i.d. Bernoulli trials, the transition matrices deﬁned by (5.2) become, M = (0, 0) (0, 1) · · (0, k −1) α           q p · · · 0 0 q 0 p · · 0 0 · · · · · · · · · · · · · · q 0 0 · · 0 p 0 0 0 · · 0 1           for all t = 1, . . . , n. (5.5) It follows immediately from (5.4) that the distribution of the waiting time for the ﬁrst occurrence of the pattern Λ = SS . . . SS, is given by P(W(1, Λ) = n) = π0Mn−kU′pk, (5.6) where π0 = (1, 0, . . . , 0), U = (1, 0, . . . , 0), and M is given by (5.5). By the same token, the above result (5.4) is also true for the compound pattern. 970 JAMES C. FU 6. Numerical Examples Several numerical examples are given here to illustrate our method and re-sults developed in previous sections. Example 1. Consider 10 i.i.d. Bernoulli trials each with p = 0.4 and q = 1 −p = 0.6. By (2.2) and transition matrix (4.10), the following Table 1 gives the joint distribution and marginal distributions of success runs and failure runs respectively. Table 1. Joint and marginal distributions of runs R(10, S) and R(10, F) R(10, F) 1 2 3 4 5 R(10, S) R(10, S) 0 .006047 .006047 1 .000105 .023557 .073503 .097165 2 .020726 .169656 .175211 .365593 3 .079272 .225210 .096082 .400564 4 .054057 .059985 .009157 .123199 5 .005839 .000796 .006635 R(10, F) .000105 .050330 .322431 .454478 .161907 .009953 1.0000 Projecting the joint distribution of R(10, S) and R(10, F) on the partition generated by R(10) = R(10, S) + R(10, F), this yields the distribution of the total number of success and failure runs shown by the following Table 2. Table 2. Distribution of the total number R(10) of runs in ten trials. R(10) 1 2 3 4 5 6 7 8 9 10 Prob. .006152 .023557 .094229 .169656 .254483 .225210 .150139 .059985 .001505 .000796 Example 2. Consider a sequence of i.i.d. trials each of them having three states b1, b2, and b3 and corresponding probabilities of occurring p1 = p2 = 0.3 and p3 = 0.4. Suppose we are interested in ﬁnding the exact distribution for the number of a speciﬁed compound pattern Λ = b1b2 ∪b3b1 in n trials. Applying the forward and backward principle, this yields a Markov chain Yt deﬁned on the state space Ω= {(u, v) : u = 0, 1, . . . , l, l = [n/2] and v = 0, 1, 2}, having transition matrices Mt =             A B 0 0 0 0 0 A B 0 0 0 · · · · · · · · · · · · · · · · · · 0 0 0 0 A B 0 0 0 0 0 A∗             , for t = 1, . . . , n, DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 971 where A =    0.3 0.3 0.4 0 0.3 0.4 0.3 0 0.4   , B =    0 0 0 0.3 0 0 0.3 0 0   , and A∗=    0.3 0.3 0.4 0 1 0 0 0 1   , and the partition {Cx = [(x, v) : v = 0, 1, 2]; x = 0, 1, . . . , l.}. The following Table 3 provides numerical results for the distributions of the number of patterns Λ for n = 7, 11, and 21. Table 3. Exact distributions for the number of patterns Λ in n = 7, 11 and 21 three-state trials. n x p(x) n x p(x) 7 0 .241887 21 0 .0092905 1 .468508 1 .0600785 2 .257324 2 .1680750 3 .032281 3 .2654140 4 .2587810 11 0 .095315 5 .1600370 1 .307973 6 .0620535 2 .365549 7 .0143712 3 .189825 8 .0017987 4 .039238 9 .0000999 5 .002100 10 .155×107 The above numerical results were done with Mathematica on an IBM 386 PC. For each case, the CPU time is rather minimum usually it takes no more than a few seconds even in the case of n = 21. 7. Discussions The results about the exact distributions of number of patterns obtained by combinatorial, generating function, and recursive equation methods for i.i.d. multi-state trials are usually diﬃcult to extend to Markov chain dependent trials. The recursive method of the renewal equation (see Schwager (1983)) could not be used for searching the joint distribution of runs. On the contrary, from examples in the previous sections, the results for i.i.d. trials obtained by using the Markov chain imbedding technique can be easily extended to trials with one-step (or m-step) Markov chain dependency. The basic idea of handling the one step (or m-step) Markov chain dependence is to have an additional coordinate to indicate the outcome of the last trial (or last m trials). Suppose Xn(Λ), the number of patterns Λ in a sequence of n multi-state trials, could be imbedded into a ﬁnite Markov chain {Yt : t = 1, . . . , n} deﬁned 972 JAMES C. FU on the state space Ω= {(x, y) : x = 0, . . . , ln and y = 0, . . . , mx −1} = {Cx : x = 0, . . . , ln}, (7.1) where ln = [n/k], k is the length of the pattern and mx and card(Cx), and having transition matrices of the following form Mt =                At(0) Bt(0) At(1) Bt(1) 0 · · · · · · · · 0 · Bt(ln−1) At(ln)                (7.2) for t = 1, . . . , n. There are many runs and patterns of interest whose transition matrices do have the above form. For instance, all the transition matrices asso-ciated with the run statistics Nn,k, Mn,k, Gn,k, (see Fu and Koutras (1994)) and patterns Λ = b1b1b2 and Λ = b1b2 ∪b3b1 (see Section 3) all have the form (7.2). Denote vectors, αt(x) = (P(Yt = (x, 0)), . . . , P(Yt = (x, m −1))) for t = 1, . . . , n. The probability of Xn(Λ) = x can be represented as P(Xn(Λ) = x) = αn(x)1′, for all x = 0, 1, . . . , ln, (7.3) where 1′ = (1, . . . , 1)′. Since Mt = Kt + Ht, where Kt is a diagonal matrix with components At(x), for x = 0, 1, . . . , ln and Ht is an upper diagonal matrix with components Bt(x), for x = 0, 1, . . . , ln−1, it follows from the fact π0(t j=1 Mj) = π0(t−1 j=1 Mj)Mt for every t = 1, . . . , n that the following recursive equations hold      αt(0) = αt−1(0)At(0) αt(x) = αt−1(x −1)Bt(x −1) + αt−1(x)At(x), x = 1, . . . , ln, (7.4) for every t = 1, . . . , n. The above recursive formula (7.4) provides an eﬃcient way to compute the probabilities P(Xn(Λ) = x) = αn(x)1′, for all x = 0, 1, . . . , ln, especially when the dimension of the transition matrix Mt is so large that the formula π0(n t=1 Mt)U′(Cx) takes longer time to do the computation. From back-ward multiplication, the ﬁnite Markov chain imbedding technique always provides a recursive equation like (7.4) automatically for the distribution of Xn(Λ) which could not be easily obtained through the combinatorial or renewal equation meth-ods. The recursive equation (7.4) provides a very useful tool to study the limiting DISTRIBUTION THEORY OF RUNS AND PATTERNS OF MULTI-STATE TRIALS 973 distribution for the random variable Xn(Λ). In spite of their potential interests, we will not pursuit these here. The Markov chain imbedding method can also be modiﬁed to study the distributions of certain runs and patterns on permutations of n symbols or per-mutations of several symbols alike. The extensions of the method for these cases are often non-trivial. Let {πn = (3, 2, n, . . . , 5)} be all permutation of n inte-gers 1,2,. . . , n. Deﬁne the index functions of successions for i = 1, . . . , n −1, Ii(πn) = 1 if πn(i) + 1 = πn(i + 1), and 0 otherwise, where πn(i) is the number of ith coordinate of the permutation πn and the number of successions in a random permutation, Xn(πn) = n i=1 Ii(πn). By extending the Markov chain imbedding technique to random permutation, Fu (1995) obtained the exact distribution for the random variable Xn(πn), the number of successes, and also showed that the limiting distribution is a Poisson distribution with parameter λ = 1. For the compound pattern Λ = ∪s−1 i=1Λi, where Λi = bibi+1, i = 1, . . . , s−1, if s = n and pi = 1/n for all i = 1, . . . , n we believe that the random variable Xn(Λ) also has a Poisson distribution as n →∞for the same reason that the number of successions in a random permutation of n integers has a Poisson distribution as n →∞. In practice, this means that when the number of states s in each trial is very large, then the distribution of the number of patterns Λ can be approximated by a Poisson distribution. The ﬁnite Markov chain imbedding may not be unique. Often there are several diﬀerent ways to imbed a random variable. To ﬁnd the best imbedded Markov chain it requires experience and understanding the structure of the count-ing process associated with the random variable. Sometimes in order to obtain the distribution of a pattern Λ1, it may be more eﬃcient and simpler to ﬁnd the joint distribution of pattern Λ1 and Λ2 ﬁrst, then project the joint distribution on the partition {Cx} associated with the pattern Λ1. This can be seen from the procedure for ﬁnding the distribution of the total number of runs given in the Section 4. With today’s high speed computers, the exact distributions (also means and variances) of runs and patterns can be obtained by the method of ﬁnite Markov chain imbedding with ease. All our numerical results in Section 6 were carried out on an IBM 386 PC with less then one minute in CPU time in each case. For very large n, if the probability of pattern Λ occurring is very small, say having an order of λ/n, then the distribution of the pattern Λ occurring can be approximated by a Poisson distribution with parameter λ. This fact had been broadly used in studying reliabilities of certain engineering systems when the failure probabilities of components are very small, for instance the reliability of consecutive-k-out-of-n : F system. 974 JAMES C. FU Acknowledgement The author would like to thank the referee for his useful suggestions and comments. This work was supported in part by the Natural Sciences and Engi-neering Research Council of Canada under Grant NSERC A-9216, and National Science Council of Republic of China under Grant 85-2121-M-259-003. References Balasubramanian, K., Viveros, R. and Balakrishnan, N. (1993). Sooner and later waiting time problems for Markovian Bernoulli trials. Statist. Probab. Lett. 18, 153-161. Benevento, R. V. (1984). The occurrence of sequence patterns in ergodic Markov chains. Stochastic Process. Appl. 17, 369-373. Chao, M. T. and Fu, J. C. (1989). A limit theorem of certain repairable systems. Ann. Inst. Statist. Math. 41, 809-818. Chao, M. T. and Fu, J. C. (1991). The reliability of large series systems under Markov structure. Adv. Appl. Probability 23, 894-908. Cochran, W. G. (1938). An extension of Gold’s method for examining the apparent persistence of one type of weather. Roy. Meteorol. Soc. Quart. Journal 64, 631-634. Ebneshahrashoob, M. and Sobel, M. (1990). Sooner and later waiting time problems for Bernoulli trials : frequency and run quotas. Statist. Probab. Lett. 9, 5-11. Feller, W. (1968). An Introduction to Probability Theory and Its Applications, Vol.I, 3rd ed., John Wiley, New York. Fu, J. C. (1986). Reliability of consecutive-k-out-of-n : F systems with (k−1)-step Markov dependence. IEEE Trans. Reliability R35, 602-606. Fu, J. C. (1995). Exact and limiting distributions of the number of successions in a random permutation. Ann. Inst. Statist. Math. 47, 435-446. Fu, J. C. and Koutras, M. V. (1994). Distribution theory of runs : A Markov chain approach. J. Amer. Statist. Assoc. 89, 1050-1058. Gibbons, J. D. (1985). Non-parametric Statistical Inference. M. Dekker, New York. Hahn, G. J. and Gage, J. B. (1983). Evaluation of a start-up demonstration test. J. Qual. Technol. 15, 103-106. Koutras, M. V. and Papastavridis, S. G. (1993). On the number of runs and related statistics. Statist. Sinica 3, 277-294. Mood, A. M. (1940). The distribution theory of runs. Ann. Math. Statist. 11, 367-392. Mosteller, F. (1941). Note on an application of runs to quality control charts, Ann. Math. Statist. 12, 228-232. Schwager, S. J. (1983). Run probabilities in sequences of Markov-dependent trials. J. Amer. Statist. Assoc. 78, 168-175. Stevens, W. L. (1939). Distributions of groups in a sequence of alternatives. Ann. of Eugenics 9, 10-17. Wald, A. and Wolfowitz, J. (1940). On a test whether two samples are from the same population. Ann. Math. Statist. 11, 147-162. Walsh, J. E. (1965). Handbook of Non-Parametric Statistics. D. Van Nostrand Company, Inc. New York. Wishart, J. and Hirshfeld, H. O. (1936). A theorem concerning the distribution of joins between line segments. London Math. Soc. J. 11, 227-235. Wolfowitz, J. (1943). On the theory of runs with some applications to quality control, Ann. Math. Statist. 14, 280-288. Institute of Applied Mathematics, National Donghwa University, Hualien, Taiwan. (Received August 1994; accepted January 1996)
14718	https://pmc.ncbi.nlm.nih.gov/articles/PMC4689754/	SEOM clinical guidelines for the treatment of Hodgkin’s lymphoma - 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 Clin Transl Oncol . 2015 Oct 26;17:1005–1013. doi: 10.1007/s12094-015-1429-1 Search in PMC Search in PubMed View in NLM Catalog Add to search SEOM clinical guidelines for the treatment of Hodgkin’s lymphoma A Rueda Domínguez A Rueda Domínguez 1 Área de Oncología y Hematología, Hospital Costa del Sol, Autovía A-7, km 187, 29603 Marbella, Málaga Spain Find articles by A Rueda Domínguez 1,✉, J Alfaro Lizaso J Alfaro Lizaso 2 Servicio de Oncología Médica, Instituto Oncológico de Guipúzcoa, San Sebastian, Spain Find articles by J Alfaro Lizaso 2, L de la Cruz Merino L de la Cruz Merino 3 Servicio de Oncología Médica, Complejo Hospitalario Regional Virgen Macarena, Seville, Spain Find articles by L de la Cruz Merino 3, J Gumá i Padró J Gumá i Padró 4 Servicio de Oncología Médica, Hospital Universitari de Sant Joan de Reus, Reus, Spain Find articles by J Gumá i Padró 4, C Quero Blanco C Quero Blanco 5 Servicio de Oncología Médica, Complejo Hospitalario Regional y Virgen de la Victoria, Málaga, Spain Find articles by C Quero Blanco 5, J Gómez Codina J Gómez Codina 6 Servicio de Oncología Médica, Hospital Universitari i Politècnic la Fe, Valencia, Spain Find articles by J Gómez Codina 6, M Llanos Muñoz M Llanos Muñoz 7 Servicio de Oncología Médica, Hospital Universitario de Canarias (H.U.C), San Cristóbal De La Laguna, Tenerife Spain Find articles by M Llanos Muñoz 7, N Martinez Banaclocha N Martinez Banaclocha 8 Servicio de Oncología Médica, Hospital General Universitario de Elche y Vega Baja, Elche, Spain Find articles by N Martinez Banaclocha 8, D Rodriguez Abreu D Rodriguez Abreu 9 Servicio de Oncología Médica, Hospital Universitario Insular de Gran Canaria, Las Palmas De Gran Canarias, Spain Find articles by D Rodriguez Abreu 9, M Provencio Pulla M Provencio Pulla 10 Servicio de Oncología Médica, Hospital Universitario Puerta de Hierro Majadahonda, Madrid, Spain Find articles by M Provencio Pulla 10 Author information Article notes Copyright and License information 1 Área de Oncología y Hematología, Hospital Costa del Sol, Autovía A-7, km 187, 29603 Marbella, Málaga Spain 2 Servicio de Oncología Médica, Instituto Oncológico de Guipúzcoa, San Sebastian, Spain 3 Servicio de Oncología Médica, Complejo Hospitalario Regional Virgen Macarena, Seville, Spain 4 Servicio de Oncología Médica, Hospital Universitari de Sant Joan de Reus, Reus, Spain 5 Servicio de Oncología Médica, Complejo Hospitalario Regional y Virgen de la Victoria, Málaga, Spain 6 Servicio de Oncología Médica, Hospital Universitari i Politècnic la Fe, Valencia, Spain 7 Servicio de Oncología Médica, Hospital Universitario de Canarias (H.U.C), San Cristóbal De La Laguna, Tenerife Spain 8 Servicio de Oncología Médica, Hospital General Universitario de Elche y Vega Baja, Elche, Spain 9 Servicio de Oncología Médica, Hospital Universitario Insular de Gran Canaria, Las Palmas De Gran Canarias, Spain 10 Servicio de Oncología Médica, Hospital Universitario Puerta de Hierro Majadahonda, Madrid, Spain ✉ Corresponding author. Received 2015 Oct 7; Accepted 2015 Oct 9; Issue date 2015. © The Author(s) 2015 Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. PMC Copyright notice PMCID: PMC4689754 PMID: 26497354 Abstract Hodgkin lymphoma (HL) is an uncommon B cell lymphoid malignancy representing approximately 10–15% of all lymphomas. HL is composed of two distinct disease entities; the more commonly diagnosed classical HL and the rare nodular lymphocyte-predominant HL. An accurate assessment of the stage of disease and prognostic factors that identify patients at low or high risk for recurrence are used to optimize therapy. Patients with early stage disease are treated with combined modality strategies using abbreviated courses of combination chemotherapy followed by involved-field radiation therapy, while those with advanced stage disease receive a longer course of chemotherapy often without radiation therapy. High-dose chemotherapy (HDCT) followed by an autologous stem cell transplant (ASCT) is the standard of care for most patients who relapse following initial therapy. Brentuximab vedotin should be considered for patients who fail HDCT with ASCT. Keywords: Oncohematology malignancies, Hodgkin lymphoma, Hodgkin lymphoma therapy Methods-methodology To identify the main topics published in medical literature, a search in "PubMed" and "isiknowledge" (that includes both full papers and abstracts) has been performed. Sentences used were "Hodgkin Lymphoma," "Hodgkin disease" "Hodgkin Lymphoma staging” “Hodgkin Lymphoma treatment," and "Hodgkin Lymphoma new therapies." Main recent reviews on the topics: ESMO clinical guides, NCCN guides, Annual Clinical Updates in Hematology Malignances of the American Journal of Hematology and Italian guideline for HL have been consulted. Introduction Hodgkin Lymphoma (HL, formerly called Hodgkin Disease) is a malignant disease with an incidence of 3.7 (male) and 2.6 (female) cases/100000 (adjusted world estimates rates) in Spain . HL shows an age-related bimodal incidence. The first peak occurs in young adults aged 20–40 years and a much smaller peak occurs after the age of 55 years. Over the last 4 decades, advances in radiation therapy and the addition of combination chemotherapy have significantly increased the cure rate of patients with HL. Currently, more than 80% of all newly diagnosed patients younger than 60 years are likely to be cured of their disease. However, most patients with HL die due to acute or late complications, principally treatment induced second solid tumors and cardiovascular disease. This fact must be taken into account when choosing the optimal first-line treatment for an individualized patient. Diagnosis At the time of diagnosis, the majority of patients with HL present with supradiaphragmatic lymphadenopathy. Patients commonly present with cervical, anterior mediastinal, supraclavicular, and axillary lymph node involvement, while the inguinal areas are less frequently involved. Approximately one-third of patients present with systemic symptoms that include fever, night sweats, and weight loss; some patients also present with chronic pruritus. Although the disease most commonly involves contiguous lymph node groups, HL may also affect extranodal tissues by direct invasion or by hematogenous spread. The most commonly involved extranodal sites are the lungs, bone, liver, and bone marrow. A fine-needle aspirate is inadequate for initial diagnosis. An incisional or excisional biopsy is preferred to provide adequate tissue for different studies (morphology, immunohistochemistry, and molecular biology) but a core-needle biopsy can be considered when excisional biopsy is not possible . HL is a malignancy in that the tumor cells constitute the minority of the cellular population and an inadequate biopsy may fail to include malignant cells in the specimen. To confirm the diagnosis, it is necessary to identify the malignant Reed–Sternberg (RS) cell, which is of follicular center B-cell origin, within the appropriate cellular environment of normal reactive lymphocytes, eosinophils, and histiocytes. Two histological categories have been defined by the WHO classification : the classical variant and the nodular lymphocyte predominant variant. Classical Hodgkin Lymphoma includes four subtypes: nodular sclerosis, mixed cellularity, lymphocyte-rich and lymphocyte-depleted, and represents about 95% of all HL cases. Most of cases have expression of CD30 and CD15 but no CD45. Nodular sclerosis Hodgkin’s lymphoma (NSHL) is the most common subtype of HL and represents about 60% of cases. Morfologic feature has a partially nodular pattern with fibrous bands separating the nodules in most cases; diffuses areas are common, as is necrosis. The characteristic cell is the lacunar-type RS cell. Mixed-cellularity Hodgkin’s lymphoma (MCHL) comprises 20–25% of HL cases. The infiltrate is usually diffuse. RS cells are of the classic type. Lymphocyte-rich classic Hodgkin’s lymphoma (LRCHL) represents 1% of HL cases. It has a background infiltrate that consists predominantly of small lymphocytes similar than nodular lymphocyte predominant variant but RS cells are classic or lacunar type. Lymphocyte-depleted Hodgkin’s lymphoma (LDHL) represents fewer than 1% of the cases. The infiltrate is diffuse and often hypocellular. It is the most common in individuals positive for human immunodeficiency virus (HIV). Nodular lymphocyte predominant Hodgkin’s lymphoma is a very rare neoplasm with indolent course and relatively good prognosis. It has a nodular growth pattern and may have diffuse areas. The characteristic neoplastic cell is “pop corn” cell or L&H cell. The background is constituted predominantly by lymphocytes. In contrast to classical HL, the atypical cells are CD45+ and express B-cell-associated antigens (CD20 and CD79a+). Staging, prognosis, and response criteria An accurate assessment of the stage of disease in patients with HL is critical for the selection of the appropriate therapy. The staging system for patients with HL is based on whether the involved lymph nodes are on one or both sides of the diaphragm, the number of involved sites, whether the sites of involvement are bulky, whether there is contiguous extranodal involvement or disseminated extranodal disease, and whether typical systemic symptoms (B symptoms) are present. Fluorodeoxyglucose positron emission tomography (FDGPET) scanning has emerged as an important tool in the staging of patients with HL in that it significantly adds to the staging information obtained using other standard radiographic methods . The Cotswolds modifications of the Ann-Arbor recommendations are the current staging system used for patients with Hodgkin’s lymphoma (Table1). The recommended staging evaluation should be the following: Clinical evaluation: Age, sex, B Symptoms (fevers to more than 38.3°C, drenching night sweats or unexplained weight loss more than 10% of body mass over 6 months), history of malignancy. Fatigue, pruritus, and alcohol-induced pain in patients with HL should also be noted. Physical examination includes measurement of accessible nodal groups and the size of the spleen and liver in cm in the midclavicular line. Laboratory tests: CBC with differential and platelet count, erythrocyte sedimentation rate (ESR), biochemical tests of liver, bone and renal function, LDH, albumin and calcium concentration. HBV, HCV and HIV tests. Pregnancy test for women of childbearing age. Chest X-ray. CT scan of the neck, chest, abdomen, and pelvis with contrast. PET–CT. In Lugano 2011 the consensus was that PET–CT should be recommended for routine staging as the gold standard . Bone marrow biopsy from at least one site for patients with clinical stage III–IV or stage II disease with anemia or another blood count depression. However, if PET–CT is performed, a bone marrow biopsy is no longer required for the routine evaluation of patients with HL . Measures to preserve fertility should be offered to all HL patients before treatment attending to age, patient's wishes and risk of infertility due to therapy. Table 1. Cotswolds staging classification for Hodgkin’s lymphoma (The Ann Arbor staging system with Cotswolds modifications) Stage I: Involvement of a single lymph node region (e.g., cervical, axillary, inguinal, mediastinal) or lymphoid structure such as the spleen, thymus, or Waldeyer’s ring Stage II: Involvement of two or more lymph node regions or lymph node structures on the same side of the diaphragm. Hilar nodes should be considered to be “lateralized” and when involved on both sides, constitute stage II disease. For the purpose of defining the number of anatomic regions, all nodal disease within the mediastinum is considered to be a single lymph node region and hiliar involvement constitutes an additional site of involvement. The number of anatomic regions should be indicated by a subscript (e.g., II-3) Stage III: Involvement of lymph node regions or lymphoid structures on both sides of the diaphragm. This may be subdivided stage III-1 or III-2: stage III-1 is used for patients with involvement of the spleen or splenic hilar, celiac or portal nodes; and stage III-2 is used for patients with involvement of the paraaortic, iliac, inguinal, or mesenteric nodes Stage IV: Diffuse or disseminated involvement of one or more extranodal organs or tissue beyond that designated E, with or without associated lymph node involvement All cases are subclassified to indicate the absence (A) or presence (B) of the systemic symptoms of significant unexplained fever, night sweats, or unexplained weight loss exceeding 10% of body weight during the 6 months prior to diagnosis The designation “E” refers to extranodal contiguous extension (i.e., proximal or contiguous extranodal disease) that can be encompassed within an irradiation field appropriate for nodal disease of the same anatomic extent. More extensive extranodal disease is designated stage IV The subscript “X” is used if bulky disease is present. This is defined as a mediastinal mass with a maximum width that is equal to or greater than one-third of the internal transverse diameter of the thorax at the level of T5/6 interspace or >10 cm maximum dimension of a nodal mass. No subscripts are used in the absence of bulk Patients can be clinically or pathologically staged. Splenectomy, liver biopsy, lymph node biopsy, and bone marrow biopsy are mandatory for the establishment of pathological stage. The pathologic stage at a given site is denoted by a subscript (e.g., M=bone marrow, H=liver, L=lung, O=bone, P=pleura, and D=skin) Open in a new tab The predominant factors that determine the initial choice of therapy for HL patients are the histology of the disease (classical HL or nodular lymphocyte-predominant HL), the anatomical stage of disease (limited or advanced disease), and the presence of poor prognostic features. Among patients with early disease (stage I or II), there is subsequent stratification into favorable and unfavorable prognosis disease based upon the presence or absence of certain clinical features. The two most commonly used definitions of favorable disease are those proposed by the European Organization for the Research and Treatment of Cancer (EORTC) and the German Hodgkin Study Group (GHSG) . The GHSG defines the limited stage favorable prognostic group as patients with no more than two nodal sites; no extranodal extension; no mediastinal mass measuring one-third the maximum thoracic diameter or greater; and ESR less than 50 mm/h (less than 30 mm/h if B symptoms present). Patients with at least one of these factors are considered as unfavorable prognostic early disease. GHSG definition is preferred today because treatment recommendations for these patients are based on the results of GHSG trials. In contrast, in patients with advanced HL (stage III or IV), disease bulk and other traditional prognostic variables have been found to be less predictive of outcome. A different prognostic scoring system was, therefore, developed for these patients by the International Prognostic Factor Project on advanced HL . This study identified seven variables (age >45 years, presence of stage IV disease, male sex, white blood count >15,000 cells/mL, lymphocyte count <600 cells/mL, albumin <4.0 g/dL, hemoglobin <10.5 g/dL) that predicted patient outcome in a multivariate analysis. Patients with five or more factors were found to have a 5-year freedom from progression of 42% while patients with no negative prognostic factors had an 84% likelihood of being free from progression at 5 years. Response evaluation by contrast-enhanced CT should be carried out after completion of chemotherapy/before RT in early stages and after four cycles of chemotherapy as well as before RT in advanced stages. Final evaluation should be carried out after completion of treatment. Physical examination, laboratory analyses, and contrast-enhanced CT are mandatory. In addition, PET should be carried out at final response evaluation . In patients with early stage, an interim PET made after 2 or 3 cycles of chemotherapy can be useful to avoid RT in selected patients (Table2). Table 2. Recommendation for the management of classical Hodgkin’s lymphoma Recommendation 1: A bone marrow biopsy is not required if a PET/CT is performed during routine staging of HL (II, A) Recommendation 2: Two cycles of ABVD followed by IFRT (20 Gy) is the preferred treatment for favorable early-stage HL (IA). However, for patients with high risk of secondary solid neoplasm, RT could be avoided if a PET CR is achieved after 3–4 ABVD cycles (IB) Recommendation 3: Four cycles of ABVD followed by IFRT (30 Gy) is the preferred treatment for unfavorable early-stage HL (IA). However, for patients with high risks of secondary solid neoplasm and no bulky disease, RT could be avoided if a PET CR is achieved after 6 ABVD cycles (IIB) Recommendation 4: Six to eight cycles of ABVD is the preferred treatment for advanced-stage HL (IA). Only patients in PR after chemotherapy should received complementary RT (IA) Recommendation 5: Salvage chemotherapy followed by high-dose therapy and autologous stem cell transplant is the best option for most patients with relapsing and refractory disease (IB). Brentuximab Vedotin is the preferred option for patients relapsing after ASCT (IIB) Recommendation 6: Anamnesis and physical examination at 4–6 months intervals for the first 5 years and yearly thereafter is the mainstay of follow-up (IIB). Blood and imaging test are optional and should be individualized (IIB) Open in a new tab Recommendation 1:A bone marrow biopsy is not required if a PET/CT is performed during routine staging of HL (II, A). Treatment of classical Hodgkin’s lymphoma Patients with HL have an excellent outcome with current management approaches. Treatment requires a careful balance between optimum disease control and the risk of long-term treatment-related side effects. Outcome of this population is so successful that even the overall mortality rate from causes other than HL may exceed those seen from Hodgkin’s lymphoma after 10–30 years. The current standard of care for HL is to have different treatment strategies for HL patients with early stage disease with favorable prognostic features, those with early stage disease but who have poor prognostic features, or those with advanced disease. Favorable prognosis early-stage Hodgkin’s lymphoma (Fig.1) Fig.1. Open in a new tab Treatment algorithm for favorable prognosis early (stage I–II) classical Hodgkin’s lymphoma. Source Modified from reference For decades, extended field radiation therapy (EFRT) has been an essential part of treatment in early-stage HL. However, combined modality treatment (chemotherapy plus less extensive radiotherapy) is the actual standard. In most of randomized clinical trials, combined modality treatment has resulted in higher rates of freedom from recurrence without differences in overall survival. This lack of overall survival benefit may be related to the effectiveness of salvage chemotherapy after failure of radiation therapy. In a report from the German Hodgkin Study Group (GHSG), HD7 trial, two cycles of ABVD followed by extended-field radiation therapy (EFRT 30 Gy plus 10 Gy to the involved field) was more effective than EFRT alone . Several studies have investigated the reduction of number of cycles of chemotherapy and radiation field size. The EORTC H8F trial compared three cycles of MOPP/ABV hybrid plus involved-field radiation therapy (IFRT) to EFRT. This trial was the first to demonstrate a significant 10-year overall survival benefit in favor of combined modality treatment when compared with radiotherapy alone . In HD10 trial (GHSG), patients were randomized to receive four versus two cycles of ABVD and 30 Gy versus 20 Gy of IFRT. Results after 7.5 years of follow-up showed no differences in survival rates among treatment arms . Chemotherapy alone has also been investigated as a treatment option for patients with early-stage HL. A systematic review of randomized trials showed similar CR rates with a detriment in tumor control and OS in some of them, but this is controversial because both the types of chemotherapy as the volume of radiation therapy utilized was not optimal . Two randomized trials have examined the role of FDG-PET in identifying an early favorable HL patient population in which radiation could be omitted without compromising PFS. Both of these trials used noninferiority designs. The RAPID trial randomized patients who had a negative PET scan after 3 cycles of ABVD to receive additional IFRT or no further therapy. The 3-year PFS rate was superior for the combined treatment arm (93.8 vs. 90.7%). This study did not demonstrate non-inferiority of the two approaches in PFS . Nevertheless, patients who are PET negative after chemotherapy have a very good outcome with or without consolidation radiotherapy. In EORTC/LYSA/FILH10 trial, involved node radiotherapy (INRT) was omitted in patients with a PET-negative scan after 2 cycles of ABVD. A planned interim analysis for futility led to the closure of the experimental no-radiation arm based on an increased number of progression events when radiation was omitted . Recommendation 2:Two cycles of ABVD followed by IFRT (20 Gy) is the preferred treatment for favorable early-stage HL (IA). However, for patients with high risks of secondary solid neoplasm, RT could be avoided if a PET CR is achieved after 3–4 ABVD cycles (IB). Unfavorable prognosis early-stage Hodgkin’s lymphoma (Fig.2) Fig.2. Open in a new tab Treatment algorithm for unfavorable prognosis early (stage I–II) classical Hodgkin’s lymphoma. Source Modified from reference Different chemotherapeutic regimens have been evaluated in combined modality treatment without identifying any differences in overall survival. ABVD is more effective than MOPP (freedom from progression, FFP) with less hematologic and late gonadal toxicity (H6U trial) but increase in pulmonary toxicity. Less toxic chemotherapy regimens (EVE, EBVP, EBVM) have failed to demonstrate better results. To address the issue of the number of cycles of chemotherapy necessary in combined modality treatment, the three-arm EORTC H8U trial compared four versus six cycles of hybrid MOPP/ABV regimen in addition to involved versus extended field radiation without differences in survival . Preliminary results of H9U trial showed similar results in terms of survival between six cycles of ABVD, four cycles of ABVD or four cycles of BEACOPP followed by IFRT 30 Gy in all arms, but increased toxicity was seen with BEACOPP . To determine the radiation dose needs to be applied and looking for an improvement in results with more intensive chemotherapy, the GHSG HD11 trial randomly assigned in a 2×2 factorial design to either ABVD or BEACOPPbaseline followed by 20 or 30 Gy of IFRT. BEACOPPbaseline did not significantly improve outcome and four cycles of ABVD should be followed by 30 Gy of IFRT . In HD14 trial, patients were randomly assigned to either four cycles of ABVD or an intensified treatment consisting of two cycles of escalated BEACOPP followed by two cycles of ABVD (2+2). Chemotherapy was followed by 30 Gy IFRT in both arms. Intensified chemotherapy achieved a small significant improvement in freedom from treatment failure, mainly in patients with bulky disease, without differences in overall survival. More gonadal and severe acute hematological toxicities were seen with intensive treatment . Recommendation 3:Four cycles of ABVD followed by IFRT (30 Gy) is the preferred treatment for unfavorable early-stage HL (IA). However, for patients with high risks of secondary solid neoplasm and no bulky disease, RT could be avoided if a PET CR is achieved after 6 ABVD cycles (IIB). Advanced Hodgkin’s lymphoma (Fig.3) Fig.3. Open in a new tab Treatment algorithm for advanced stage disease in classical Hodgkin’s lymphoma. Source Modified from reference Approximately three-quarters of patients with Hodgkin’s lymphoma in advanced stages (stages III and IV) can be cured with chemotherapy. The chemotherapy scheme most widely used is the combination of doxorubicin, bleomicin, vinblastin, and dacarbazine (ABVD). However, results in terms of response and progression free survival are only slightly better than the classic MOPP (meclorethamine, vincristine, procarbazine, and prednisone). The low frequency of long-term toxicities for ABVD, especially second neoplasms and sterility, leads to the abandonment of MOPP and the generalized adoption of ABVD in this setting. Alternating schedules of MOPP and ABVD, and the so-called hybrid scheme (MOPP-ABV) were also compared with ABVD alone, and none of these was associated with a higher overall survival, and the toxicity profile favored ABVD . The number of cycles of ABVD usually given to treat a patient with advanced Hodgkin’s disease is between six and eight. If there is an early metabolic complete response by Positron Emission Tomography (PET) (e.g., after two or three cycles), six cycles of ABVD are probably sufficient. For slower responders, a total of eight cycles may be needed. However, there is no general agreement on the value of early PET in deciding the total number of chemotherapy cycles to be given . Other chemotherapy combinations have been compared with ABVD with the aim of improving survival in advanced Hodgkin’s lymphoma. Of these, the most relevant are Stanford V and BEACOPP. The Stanford V and BEACOPP original scheme incorporated radiotherapy at the end of chemotherapy in pretreatment-affected areas larger than five cm. Three randomized trials failed to demonstrate the superiority of Stanford V over ABVD. In spite of this, Stanford V was inferior to ABVD when the amount of radiotherapy given was less than that planned in the original phase II trial [18–20]. BEACOPP, and especially its escalated variant (with higher dose of etoposide and cyclophosphamide), demonstrated an improvement in progression free survival, but not in overall survival compared with ABVD, if relapses are treated properly with high-dose chemotherapy. Moreover, BEACOPP was associated with an increased acute and late toxicity (myelotoxicity, secondary leukemia and solid tumors, and sterility) . In patients with advanced stage consolidation radiotherapy can be omitted if a complete response is achieved with chemotherapy [22, 23]. Radiotherapy is recommended if only a partial response is achieved, especially if it is corroborated by PET [23, 24]. Recommendation 4:Six to eight cycles of ABVD is the preferred treatment for advanced-stage HL (IA). Only patients in PR after chemotherapy should received complementary RT (IA). Therapy of relapsed/resistant disease (Fig.4) Fig.4. Open in a new tab Treatment algorithm for relapsed and resistant disease. Source Modified from reference Approximately 10–15% of patients in early stage and 20–40% of patients with advanced stage experience relapse after first-line treatment, generally within the first 12 months. The choice of the best salvage approach should rely on the evaluation of prognostic factors and clinical characteristics of patients. Salvage therapy can achieve durable responses in one-half of these patients. The length of remission after first-line therapy is the most important prognostic factor and has a significant effect on the success of subsequent salvage treatment. Patients with progressive disease during first line therapy or in the first 3 months after remission are considered to have primary resistant disease and have a cure rate less than 30%. Early relapse is defined as relapse that occurs within 12 months from remission and late relapse if it occurs beyond this term . At relapse, a new histologic analysis should be performed because of the increased risk of second tumors (NHL or solid tumors) or benign diseases (sarcoidosis and others). Rebiopsy is probably unnecessary in early recurrences with incomplete remissions, especially in symptomatic patients. Salvage chemotherapy followed by high-dose therapy and autologous stem cell transplant (ASCT) in young patients with relapsing and refractory disease has significantly better results over conventional chemotherapy in terms of disease free survival and is considered standard of care [26–28]. Conventional-dose chemotherapy as salvage treatment is twofold: to achieve a maximum tumor reduction and to mobilize progenitor cells into peripheral blood for subsequent autologous rescue. Conventional-dose chemotherapy alone has no curative potential in patients with refractory and early-relapsing disease. However, it may be considered the treatment of choice (often followed with radiotherapy) in patients with a late relapse (>12 months after completion of initial therapy), asymptomatic presentation and low burden disease . There are not randomized trials comparing the effectiveness of different conventional salvage chemotherapy regimens and clinical practice varies widely. Regimens most commonly used in this setting are ICE, GDP, GVD, GEM-P, DHAP, ESHAP, and mini-BEAM . With respect to RT, it may have a role when failure occurs in limited nodal sites and prior RT has not been delivered. In addition, RT to residual nodal disease is advisable in patients with residual disease after salvage therapy with or without ASCT. On the contrary, there exists controversy in relation to the eventual benefit of consolidation RT to sites of previous bulky disease . Brentuximab vedotin is an immunotoxin that comprised a CD30 antibody linked to the antitubulin agent monomethyl auristatin E (MMAE). FDA and EMA granted approval to Brentuximab for the treatment of patients with HL after failure of ASCT or after failure of at least two prior multiagent chemotherapy regimens in patients who are not candidates for ASCT. This approval was based on the results of a phase II open-label trial conducted on 102 patients with relapsed or refractory HL after previous ASCT, which were treated with Brentuximab vedotin 1.8 mg/kg every 3 weeks for up to 16 cycles . 75% of patients achieved an objective response and 34% of patients achieved CR. After a median follow-up of 33 months, 25% of the patients with an objective response to brentuximab vedotin (18 out of 103, 16 of them complete responses) were still in remission without the start of new therapy, other than a consolidative allogeneic stem cell transplant (allo-SCT) that was performed in six of 18 patients. The proportion of patients with a best response of CR who remain in remission without a consolidative allo-SCT was 43% (12/28) . Therefore, consolidative allo-SCT for patients in CR after Brentuximab vedotin should be considered investigational. Classic Hodgkin’s lymphomas include small numbers of malignant Reed–Sternberg cells within an extensive inflammatory infiltrate, and thus it seems an interesting disease to explore activity of new immunotherapies. In this sense, the genes encoding the PD-1 ligands, PDL1 and PDL2, are key targets of chromosome 9p24.1 amplification, a recurrent genetic abnormality in the nodular sclerosis type of Hodgkin’s lymphoma. JAK-STAT activity induces PD-1 ligand transcription and overexpression of the PD-1 ligands on Reed–Sternberg cells in patients with Hodgkin’s lymphoma. Nivolumab and pembrolizumab are two IgG4 monoclonal antibodies against PD1 that have demonstrated an unexpected activity in two phase 1 studies that included heavily pretreated patients (previous ASCT and Brentuximab in 78%), with mild toxicities [33, 34]. Other phase 2 trials are ongoing to elucidate clinical effect of these immunotherapies in HL, especially after ASCT. Allo-SCT is an option in selected patients relapsing after an autologous transplant. Reduced-intensity conditioning (RIC) allo-SCT can induce long-term progression-free survival (PFS), and even curation in a small subset of patients. However, its use is associated with high rates of progression and non-relapse mortality . Recommendation 5:Salvage chemotherapy followed by high-dose therapy and autologous stem cell transplant is the best option for most patients with relapsing and refractory disease (IB). Brentuximab Vedotin is the preferred option for patients relapsing after ASCT (IIB). Treatment of lymphocyte-predominant Hodgkin’s lymphoma Lymphocyte-Predominant Hodgkin’s lymphoma (LPHL) is characterized by an indolent course. Usually it involves peripheral lymph nodes with sparing of the mediastinum, retroperitoneum, and the spleen. Early-stage LPHL has a better prognosis than classical HL. Involved-field radiation therapy (IFRT) 30–36 Gy is recommended for all patients with stage IA or IIA disease. For the rare patients with stage I to II who have B symptoms, combined modality therapy with chemotherapy and IFRT is recommended . Rarely (20% of cases), patients present as III or IV stage disease, with a concomitantly worse prognosis. Outcome in these cases is similar than classical HL and treatment should be the same. Late relapses are frequent, regardless of first-line treatment. Biopsy should be performed because high risk of transformation to non-Hodgkin’s lymphoma or classical HL. Limited relapses can receive “involved field” irradiation again. The monoclonal antibody rituximab has been tested in LPHL with high response rates. Follow-up Follow-up in Hodgkin’s lymphoma is focused on detecting disease relapse and late treatment toxicities. Anamnesis and physical examination at four- to six-month intervals for the first 5 years and yearly thereafter is the mainstay of follow-up. It seems clear that imaging tests in follow-up does not translate into an improvement in survival . However, it is a common practice to perform a CT scan every 6 months for the first 2 years and yearly until the fifth year. A blood test with CBC, ESR, and LDH is usually on a similar schedule to imaging, and some authors recommend performing it yearly for life. If neck radiation therapy is carried out, we suggest including thyroidal function in the blood test. If mediastinal radiation therapy is part of the primary treatment, especially in females younger than twenty, a yearly bilateral breast MRI is recommended from the eighth year post-therapy, in order to screen for breast cancer. The risk of developing lung cancer for heavy smokers who have received mediastinal radiation therapy is well known. Nevertheless it is less clear if a low-dose chest CT scan might be useful in secondary prevention. Recommendation 6:Anamnesis and physical examination at 4- to 6-month intervals for the first 5 years and yearly thereafter is the mainstay of follow-up (IIB). Blood and imaging test are optional and should be individualized (IIB). Compliance with ethical standards Conflict of interest The authors declare that they have no conflict of interest. References 1.Galcerán J, Ameijide A, Carulla M, Mateos A, Quirós JR, Alemán A, et al. Estimaciones de la incidencia y la supervivencia del cáncer en España y su situación en Europa. Red Española de Registros de Cáncer (REDECAN). 2014. 2.Cheson BD, Fisher R, Barrington SF, Cavalli F, Schwartz LH, Zucca E, et al. Recommendations for initial evaluation, staging, and response assessment of Hodgkin an Non-Hodgkin Lymphoma: the Lugano Classification. J Clin Oncol. 2014;32:3059–3067. doi: 10.1200/JCO.2013.54.8800. 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[DOI] [PubMed] [Google Scholar] Articles from Clinical & Translational Oncology are provided here courtesy of Springer ACTIONS View on publisher site PDF (916.8 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Methods-methodology Introduction Diagnosis Staging, prognosis, and response criteria Treatment of classical Hodgkin’s lymphoma Treatment of lymphocyte-predominant Hodgkin’s lymphoma Follow-up Compliance with ethical standards 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
14719	https://math.stackexchange.com/questions/298306/when-does-a-system-of-equations-have-no-solution	Skip to main content When does a system of equations have no solution? Ask Question Asked Modified 6 years, 6 months ago Viewed 66k times This question shows research effort; it is useful and clear 8 Save this question. Show activity on this post. I have performed Gaussian elimination on this matrix to reduce it to ⎡⎣⎢⎢⎢−300−1−5302103a+2183b+65⎤⎦⎥⎥⎥ I thought that setting a equal to −2 and having b not equal to −65 would be the answer to this problem, but it apparently isn't. Could someone please explain why? linear-algebra matrices Share CC BY-SA 3.0 Follow this question to receive notifications edited Feb 8, 2013 at 21:18 Asaf Karagila♦ 407k4848 gold badges646646 silver badges1.1k1.1k bronze badges asked Feb 8, 2013 at 21:06 user1709173user1709173 10522 gold badges44 silver badges88 bronze badges 3 Yes, you did, thank you. user1709173 – user1709173 2013-02-08 21:14:52 +00:00 Commented Feb 8, 2013 at 21:14 How about using Cramer's rule? Lazar Ljubenović – Lazar Ljubenović 2013-02-08 21:17:37 +00:00 Commented Feb 8, 2013 at 21:17 Tags go to the tags, not the title. Asaf Karagila – Asaf Karagila ♦ 2013-02-08 21:19:00 +00:00 Commented Feb 8, 2013 at 21:19 Add a comment \| 3 Answers 3 Reset to default This answer is useful 13 Save this answer. Show activity on this post. there is no solution when the matrix is inconsistent. This means you will have a zero row in your reduced matrix corresponding to a non-zero entry of the desired solution eg. ⎡⎣⎢⎢−300−1−53021030183any non-zero⎤⎦⎥⎥ this is because the third row would imply 0∗x+0∗y+0∗z=0=c≠0 which is obviously false Share CC BY-SA 3.0 Follow this answer to receive notifications answered Feb 8, 2013 at 21:22 BenBen 1,45011 gold badge1414 silver badges2020 bronze badges Add a comment \| This answer is useful 5 Save this answer. Show activity on this post. Given a system of linear equations represented by the matrix equation: Ax⃗ =b⃗ , there is no unique set of solutions for detA=0. Therefore, in your case: ⎡⎣⎢−300−1−5302103a+2⎤⎦⎥⎡⎣⎢xyz⎤⎦⎥=⎡⎣⎢⎢183b+65⎤⎦⎥⎥ So we are interested in the case when: ∣∣∣∣∣−300−1−5302103a+2∣∣∣∣∣=0⟹(5a+10)=0⟹a=−105=−2 Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 8, 2013 at 21:21 answered Feb 8, 2013 at 21:12 Thomas RussellThomas Russell 10.6k66 gold badges4141 silver badges6767 bronze badges 2 Edit: As a follow-up, could I ask when the matrix will have no solution? Will it be when the bottom row is entirely 0s? user1709173 – user1709173 2013-02-08 21:14:12 +00:00 Commented Feb 8, 2013 at 21:14 So the asnwer to the question is? Julien – Julien 2013-02-08 21:28:01 +00:00 Commented Feb 8, 2013 at 21:28 Add a comment \| This answer is useful 5 Save this answer. Show activity on this post. Answering the title of the post: a linear system Ax=y,x∈Rn,y∈Rm admits solutions if and only if the matrix (A\|y), obtained by juxtaposing the column y and the matrix A, has the same rank of A itself. This happens if and only if y is a linear combination of the columns of A. Geometrically, this means that y lies in the image of the linear map x↦Ax. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Feb 20, 2019 at 20:00 answered Feb 8, 2013 at 21:15 pppqqqpppqqq 2,23011 gold badge2020 silver badges2525 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra matrices See similar questions with these tags. Featured on Meta stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 1 When does Ax +b = 0 not have a solution? Related 1 For which values of a,b does the system of equations not have any solutions? 7 Linear Algebra solution when determinant is zero 1 Relationship among b1, b2 and b3 to have a solution 0 Show that the system of equations Ax=b is not consistent for all b in R3 8 If and when a linear system has exactly three solutions 0 System of Linear Equations, when does it have infnitely many solutions? 3 For what values of k does this system of equations have a unique / infinite / no solutions? 2 Finding values that make a system of equations inconsistent 3 How to solve this system of linear equations using Gaussian elimination? 1 Help with Gaussian elimination for a 3×3 matrix Hot Network Questions Can chess engines solve the bishop and knight mate without the use of an endings tablebase? Why does truth seem to lack compelling power? Why can we rarely convince anyone to change their mind? 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14720	https://math.stackexchange.com/questions/921476/find-the-domain-of-the-function-fx-fracx4x2-9	Find the domain of the function $f(x) =\frac{x+4}{x^2-9}$ - 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 Find the domain of the function f(x)=x+4 x 2−9 f(x)=x+4 x 2−9 Ask Question Asked 11 years ago Modified11 years ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I need to find the domain of the function f(x)=x+4 x 2−9.f(x)=x+4 x 2−9. My answer was: (−∞,−3)∪(3,∞)(−∞,−3)∪(3,∞). The book's answer was: (−∞,−3)∪(−3,−3)∪(3,+∞)(−∞,−3)∪(−3,−3)∪(3,+∞) It's question 25 btw. Could the book have possibly made a mistake? functions rational-functions Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Sep 7, 2014 at 12:52 amWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges asked Sep 6, 2014 at 15:31 ChachaChacha 11 2 2 bronze badges 2 Related : math.stackexchange.com/questions/472169/…lab bhattacharjee –lab bhattacharjee 2014-09-06 15:33:08 +00:00 Commented Sep 6, 2014 at 15:33 1 Perhaps you could think of it as "all real numbers except 3 3 and −3−3" and then compare with your answer and the book's answer.recmath –recmath 2014-09-06 15:34:16 +00:00 Commented Sep 6, 2014 at 15:34 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The book is right. It is not defined only for exactly two points, ±3±3. I assume you meant (−3,3)(−3,3) for the middle interval. It certainly is defined well for say x=0,1 x=0,1. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 6, 2014 at 15:33 NemoNemo 1,340 12 12 silver badges 14 14 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Your domain excludes all valuesbetween and including−3−3 and 3 3. The only points to exclude are x=−3,x=3 x=−3,x=3, and that's just what the book does although the middle interval should read (−3,3)(−3,3) (and not (−3,−3)(−3,−3), as you've typed it): Domain: x∈(−∞,−3)∪(−3,3)∪(3,+∞)x∈(−∞,−3)∪(−3,3)∪(3,+∞) Test for yourself: pick a point in the interval (−3,3)(−3,3) and test it. You'll see the function is indeed defined there. Another, perhaps simpler way to write the domain is as follows: x∈R−{−3,3}=R∖{−3,3},x∈R−{−3,3}=R∖{−3,3}, essentially, "the domain consists of all real numbers except −3−3 and 3 3. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Sep 6, 2014 at 15:46 answered Sep 6, 2014 at 15:40 amWhyamWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. x 2−9≠0⟺x≠3,−3⟺x∈R−{3,3}⟺x 2−9≠0⟺x≠3,−3⟺x∈R−{3,3}⟺ ⟺x∈(−∞,−3)∪(−3,3)∪(3,∞)⟺x∈(−∞,−3)∪(−3,3)∪(3,∞) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 6, 2014 at 15:44 Adi DaniAdi Dani 17k 5 5 gold badges 33 33 silver badges 53 53 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Informally, the domain of the function is the set of all input values of the function. We can input any real number into this function, exceptx=±3 x=±3 (in which case we're dividing by 0 0, which is not allowed), so the domain is all real numbers that are not equal to ±3±3. There are loads of different ways of writing dom(f)dom(f). e.g. R∖{−3,3}R∖{−3,3} (−∞,−3)∪(−3,3)∪(3,∞)(−∞,−3)∪(−3,3)∪(3,∞) (the book's answer). Your answer is wrong because you've forgotten about the numbers between −3−3 and 3 3 (exclusive). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Sep 6, 2014 at 15:46 answered Sep 6, 2014 at 15:41 beep-boopbeep-boop 11.8k 12 12 gold badges 49 49 silver badges 79 79 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 functions rational-functions See similar questions with these tags. 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14721	https://math.stackexchange.com/questions/993944/solving-lnx-e-x	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 Solving $\ln(x) = e^{-x}$ Ask Question Asked Modified 9 years, 9 months ago Viewed 3k times 3 $\begingroup$ I'm trying to solve $\ln(x) = e^{-x}$ but I can't really get how to do it :( (Removing a statement that was incorrect, as explained by the comments below) Additionally, while I started to solve it I ended up with something really weird and I can't really understand what is the wrong passage: Start with: $$ \ln(x) = e^{-x} $$ My understanding is that if $\ln(x) = a$ then that is equivalent to say that $e^{a} = x$ (assuming $x>0$), so, in the previous case: $$ e^{e^{-x}} = x $$ And applying the exponent property that: $a^{m^{n}} = a^{mn}$ $$ e^{-ex} = x $$ But at this point it's really weird so I decided to take logarithm of both sides: $$ \ln(e^{-ex}) = \ln(x)\ -ex\ln(e) = \ln(x)\ -ex = ln(x)\ $$ But that clashes with $ln(x) = e^{-x}$ because clearly $e^{-x}$ is not the same as $-ex$. So could you please tell me which passage I'm doing wrong and how would you approach the initial equation? Thanks a lot! logarithms exponentiation Share asked Oct 27, 2014 at 20:03 complexguestcomplexguest 40344 silver badges99 bronze badges $\endgroup$ 3 1 $\begingroup$ What? In general, $a^{m^n}\ne (a^m)^n=a^{mn}.$ $\endgroup$ mfl – mfl 2014-10-27 20:05:15 +00:00 Commented Oct 27, 2014 at 20:05 2 $\begingroup$ "And applying the exponent property that: $a^{m^{n}} = a^{mn}$" <- No, that's wrong. $(a^m)^n = a^{mn}$, not $a^{m^n}$, which is $a^{(m^n)}$. $\endgroup$ Daniel Fischer – Daniel Fischer 2014-10-27 20:05:40 +00:00 Commented Oct 27, 2014 at 20:05 $\begingroup$ Argh sorry you are right :\| $\endgroup$ complexguest – complexguest 2014-10-27 20:06:48 +00:00 Commented Oct 27, 2014 at 20:06 Add a comment \| 3 Answers 3 Reset to default 3 $\begingroup$ You won't find a closed formula for this equation. (Nor for the simpler $e^{-x}=x$, equivalent to $\ln(x)=-x$, though these can be expressed in terms of the Lambert W function, as $W(1)$.) You must resort to numerical root finding. In this particular case, you will notice that the fixed-point method will work: start from any real value $x_0$ and iterate $$x_{n+1}=e^{e^{-x_n}}.$$ This sequence always converges and at convergence, $x=e^{e^{-x}}$. 0 2.71828182846 1.06821393681 1.41004511415 1.27651316255 1.32181853101 1.30558334634 1.31129394458 1.30927185914 1.30998618969 1.30973363268 1.30982289996 1.30979134482 1.30980249885 1.30979855610 1.30979994979 1.30979945714 1.30979963128 1.30979956973 1.30979959149 ... Share edited Oct 27, 2014 at 21:08 answered Oct 27, 2014 at 20:26 user65203user65203 $\endgroup$ Add a comment \| 1 $\begingroup$ Newton's Method says that we should iterate $$ x_{n+1}=\frac{x_ne^{-x_n}e^{e^{-x_n}}+e^{e^{-x_n}}}{e^{-x_n}e^{e^{-x_n}}+1} $$ This is because the right hand side is $$ x-\frac{f(x)}{f'(x)} $$ where $f(x)=e^{e^{-x}}-x$. For example: \begin{align} x_0&=0\ x_1&=0.731058578630004879251159241822\ x_2&=1.229791323145443105484685957301\ x_3&=1.308681010785501028518648717394\ x_4&=1.309799378172371994132312272864\ x_5&=1.309799585804143328882655453650\ x_6&=1.309799585804150477669233701960\ x_7&=1.309799585804150477669233701968\ x_8&=1.309799585804150477669233701968 \end{align} Share edited Dec 20, 2015 at 17:31 answered Oct 27, 2014 at 20:56 robjohn♦robjohn 355k3838 gold badges499499 silver badges892892 bronze badges $\endgroup$ 1 $\begingroup$ I wanted the table on the left. Did it seem that off-balance? $\endgroup$ robjohn – robjohn ♦ 2014-10-27 21:12:00 +00:00 Commented Oct 27, 2014 at 21:12 Add a comment \| -1 $\begingroup$ by a numerical method i got this here $1.309799585804150477669233701968172506010868896430480435558475367426214513358226234915421428122420845$ you can use the Newton method Share edited Oct 27, 2014 at 20:12 answered Oct 27, 2014 at 20:07 Dr. Sonnhard GraubnerDr. Sonnhard Graubner 97.5k44 gold badges4242 silver badges8080 bronze badges $\endgroup$ 2 $\begingroup$ how do you get to that number? indeed it is the right result $\endgroup$ complexguest – complexguest 2014-10-27 20:09:03 +00:00 Commented Oct 27, 2014 at 20:09 $\begingroup$ @complexguest: I don't know how the author did it, but this is the answer that Mathematica returns for FindRoot[Exp[Exp[-x]]-x,{x,1},WorkingPrecision->100] $\endgroup$ robjohn – robjohn ♦ 2014-10-27 21:20:51 +00:00 Commented Oct 27, 2014 at 21:20 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 logarithms exponentiation 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 3 How to solve $e^x \ln(x) = a$? 0 General solution of $a^{-x} = \log_a(x)$ Related Why is $x^{\log_x n}=n$? 5 Solving an equation in which the variable appears as both an exponent and a base 0 $1500=P \times { (1 + 0.02) }^{ 24 }$, what is the value of $P$? 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14722	https://www.clippercontrols.com/application%20notes/ultrasonic-sound-speeds-in-solids-and-liquids	Ultrasonic Sound Speeds in Solids and Liquids Skip to searchSkip to main content (844) 880-2469 About Us Join Our Crew What We Do Industries Products Manufacturers Field Services Resources Clipper Academy Nautical Fun Contact Us Our Product Categories Analyzers Control Displays Flow Gas, Flame, & Leaks Level Moisture Power Pressure Samplers Shelters Signal Interface Temperature Vibration Vision Systems Explore All Products Ultrasonic Sound Speeds Quick Navigation: Table 1: Ultrasonic Sound Speeds in Solids Table 2: Ultrasonic Sound Speeds in Liquids Table 3: Ultrasonic Sound Speeds in Water at Selected Temperatures Ultrasonic Sound Speeds in Solids The values in this table are critical for configuring and interpreting ultrasonic measurements in solid materials. Whether you're dealing with carbon steel in pipeline systems, stainless steel in cleanroom infrastructure, or specialized alloys like Inconel in high-temperature environments, knowing the accurate sound velocity helps ensure correct flow measurement and thickness gauging. Ultrasonic flowmeters and thickness gauges rely on these nominal sound speeds to calculate transit time and determine material boundaries. Products such as the Panametrics PT900 portable flowmeter or fixed systems like the DigitalFlow XMT1000 utilize these solid material profiles when measuring flow in lined pipes, pipe walls, or tank structures. If your application involves clamp-on ultrasonic sensors or non-intrusive flow measurement across metal or plastic surfaces, referencing these material-specific sound velocities will help ensure precision in both calibration and long-term monitoring. These values are nominal. Solids may be inhomogeneous and anisotropic. Actual values depend on exact composition, temperature, and to a lesser extent, on pressure or stress.\| \| Shear Wave \| Shear Wave \| Long Wave \| Long Wave \| --- --- \| Material \| m/s \| ft/s \| mm/µs \| in./µs \| \| Steel, 1% Carbon, hardened \| 3150 \| 10335 \| 5.88 \| 0.2315 \| \| Carbon Steel \| 3230 \| 10598 \| 5.89 \| 0.2319 \| \| Mild Steel \| 3235 \| 10614 \| 5.89 \| 0.2319 \| \| Steel, 1% Carbon \| 3220 \| 10565 \| \| \| \| 302 Stainless Steel \| 3120 \| 10236 \| 5.69 \| 0.224 \| \| 303 Stainless Steel \| 3120 \| 10236 \| 5.64 \| 0.222 \| \| 304 Stainless Steel \| 3141 \| 10306 \| 5.92 \| 0.233 \| \| 304L Stainless Steel \| 3070 \| 10073 \| 5.79 \| 0.228 \| \| 316 Stainless Steel \| 3272 \| 10735 \| 5.72 \| 0.225 \| \| 347 Stainless Steel \| 3095 \| 10512 \| 5.72 \| 0.225 \| \| Aluminum \| 3100 \| 10171 \| 6.32 \| 0.2488 \| \| Aluminum (rolled) \| 3040 \| 9974 \| \| \| \| Copper \| 2260 \| 7415 \| 4.66 \| 0.1835 \| \| Copper (annealed) \| 2325 \| 7628 \| \| \| \| Copper (rolled) \| 2270 \| 7448 \| \| \| \| CuNi (70%Cu 30%Ni) \| 2540 \| 8334 \| 5.03 \| 0.198 \| \| CuNi (90%Cu 10%Ni) \| 2060 \| 6759 \| 4.01 \| 0.1579 \| \| Brass (Naval) \| 2120 \| 6923 \| 4.43 \| 0.1744 \| \| Gold (hard-drawn) \| 1200 \| 3937 \| 3.24 \| 0.1276 \| \| Inconel \| 3020 \| 9909 \| 5.82 \| 0.2291 \| \| Iron (electrolytic) \| 3240 \| 10630 \| 5.9 \| 0.2323 \| \| Iron (Armco) \| 3240 \| 10630 \| 5.9 \| 0.2323 \| \| Ductile Iron \| 3000 \| 9843 \| \| \| \| Cast Iron \| 2500 \| 8203 \| 4.55 \| 0.1791 \| \| Monel \| 2720 \| 8924 \| 5.35 \| 0.2106 \| \| Nickel \| 2960 \| 9712 \| 5.63 \| 0.2217 \| \| Tin, rolled \| 1670 \| 5479 \| 3.32 \| 0.1307 \| \| Titanium \| 3125 \| 10253 \| 6.1 \| 0.2402 \| \| Tungsten, annealed \| 2890 \| 9482 \| 5.18 \| 0.2039 \| \| Tungsten, drawn \| 2640 \| 8661 \| \| \| \| Tungsten, carbide \| 3980 \| 13058 \| \| \| \| Zinc, rolled \| 2440 \| 8005 \| 4.17 \| 0.1642 \| \| Glass, Pyrex \| 3280 \| 10761 \| 5.61 \| 0.2209 \| \| Glass, heavy silicate flint \| 2380 \| 7808 \| \| \| \| Glass, light borate crown \| 2840 \| 9318 \| 5.26 \| 0.2071 \| \| Nylon \| 1150 \| 3772 \| 2.4 \| 0.0945 \| \| Nylon, 6-6 \| 1070 \| 3510 \| \| \| \| Polyethylene (HD) \| 2.31 \| 0.0909 \| \| \| \| Polyethylene (LD) \| 540 \| 1772 \| 1.94 \| 0.0764 \| \| PVC, CPVC \| 1060 \| 3477 \| 2.4 \| 0.0945 \| \| Acrylic \| 1430 \| 4690 \| 2.73 \| 0.1075 \| \| Asbestos Cement \| 2.2 \| 0.0866 \| \| \| \| Tar Epoxy \| 2 \| 0.0787 \| \| \| \| Mortar \| 2.5 \| 0.0984 \| \| \| \| Rubber \| 1.9 \| 0.0748 \| \| \| Ultrasonic Sound Speeds in Liquids For ultrasonic measurements in liquid media, accurate knowledge of the sound speed is essential to ensure proper calibration and real-time compensation for temperature, pressure, and composition variations. This is particularly important for non-intrusive flow measurement and liquid identification applications. The table below includes common process liquids such as water, alcohols, oils, and chemical solvents. These values are often used in conjunction with Panametrics ultrasonic flowmeters and level sensors which automatically adjust for expected sound speed based on media type and conditions. Whether you're monitoring clean water, hydrocarbons, or more complex chemical mixtures, our ultrasonic instruments—like the AquaTrans AT600 or DigitalFlow GC868—use this baseline data to deliver dependable flow and velocity measurements. For specialty applications, such as custody transfer, wastewater treatment, or chemical dosing, our technical team can assist with configuring the appropriate media-specific sound profiles for your sensors and transmitters. All values are nominal and may vary with temperature, pressure, and sample purity. All data given at 25°C (77°F) unless otherwise noted. \| Substance \| Chemical Formula \| Specific Gravity \| Sound Speed (m/s) \| Sound Speed (ft/s) \| ∆v/°C (m/s/°C) \| Kinematic Viscosity x10-6 (m²/s) \| Kinematic Viscosity x10-6 (ft²/s) \| --- --- --- --- \| \| Acetic anhydride (22) \| (CH3CO)2O \| 1.082 (20°C) \| 1180 \| 3871.4 \| 2.5 \| 0.769 \| 8.274 \| \| Acetic acid, anhydride (22) \| (CH3CO)2O \| (20°C) 1.082 \| 1180 \| 3871.4 \| 2.5 \| 0.769 \| 8.274 \| \| Acetic acid, nitrile \| C2H3N \| 0.783 \| 1290 \| 4232.3 \| 4.1 \| 0.441 \| 4.745 \| \| Acetic acid, ethyl ester (33) \| C4H8O2 \| 0.901 \| 1085 \| 3559.7 \| 4.4 \| 0.467 \| 5.025 \| \| Acetic acid, methyl ester \| C3H6O2 \| 0.934 \| 1211 \| 3973.1 \| 0.407 \| 4.379 \| \| \| Acetone \| C3H6O \| 0.791 \| 1174 \| 3851.7 \| 4.5 \| 0.399 \| 4.293 \| \| Acetonitrile \| C2H3N \| 0.783 \| 1290 \| 4232.3 \| 4.1 \| 0.441 \| 4.745 \| \| Acetonylacetone \| C6H10O2 \| 0.729 \| 1399 \| 4589.9 \| 3.6 \| \| \| \| Acetylen dichloride \| C2H2Cl2 \| 1.26 \| 1015 \| 3330.1 \| 3.8 \| 0.4 \| 4.304 \| \| Acetylene tetrabromide (47) \| C2H2Br4 \| 2.966 \| 1027 \| 3369.4 \| \| \| \| \| Acetylene tetrachloride (47) \| C2H2Cl4 \| 1.595 \| 1147 \| 3763.1 \| (15°C) 1.156 \| (59°F) 12.438 \| \| \| Alcohol \| C2H6O \| 0.789 \| 1207 \| 3960 \| 4 \| 1.396 \| 15.02 \| \| Alkazene-13 \| C15H24 \| 0.86 \| 1317 \| 4320.9 \| 3.9 \| \| \| \| Alkazene-25 \| C10H12Cl2 \| 1.2 \| 1307 \| 4288.1 \| 3.4 \| \| \| \| 2-Aminoethanol \| C2H7NO \| 1.018 \| 1724 \| 5656.2 \| 3.4 \| \| \| \| 2-Aminotolidine (46) \| C7H9N \| 0.999 (20°C) \| 1618 \| 5308.4 \| (20°C) 4.394 \| 47.279 (68°F) \| \| \| 4-Aminotolidine (46) \| C7H9N \| 0.966 (45°C) \| 1480 \| 4855.6 \| (50°C) 1.863 \| 20.045 (122°F) \| \| \| Ammonia (35) \| NH3 \| 0.771 \| 1,729 (-33°C) \| 5,672.6 (-27°F) \| 6.68 \| (-0.292 33°C) \| (-3.141 27°F) \| \| Amorphous Polyolefin \| 0.98 \| 962.6 (190°C) \| 3158.2 (374°F) \| 26600 \| 286000 \| \| \| \| t-Amyl alcohol \| C5H12O \| 0.81 \| 1204 \| 3950.1 \| 4.374 \| 47.064 \| \| \| Aminobenzene (41) \| C6H5NO2 \| 1.022 \| 1639 \| 5377.3 \| 4 \| 3.63 \| 39.058 \| \| Aniline (41) \| C6H5NO2 \| 1.022 \| 1639 \| 5377.3 \| 4 \| 3.63 \| 39.058 \| \| Argon (45) \| Ar \| 1.400 (-188°C) \| 853 (-188°C) \| 2798.6 (-306°F) \| \| \| \| \| Azine \| C6H5N \| 0.982 \| 1415 \| 4642.4 \| 4.1 \| 0.992 (20°C) \| 10.673 (68°F) \| \| Benzene (29, 40, 41) \| C6H6 \| 0.879 \| 1306 \| 4284.8 \| 4.65 \| 0.711 \| 7.65 \| \| Benzol (29, 40, 41) \| C6H6 \| 0.879 \| 1306 \| 4284.8 \| 4.65 \| 0.711 \| 7.65 \| \| Bromine (21) \| Br2 \| 2.928 \| 889 \| 2916.7 \| 3 \| 0.323 \| 3.475 \| \| Bromo-benzene (46) \| C6H5Br \| 1.522 \| 1,170 (20°C) \| 3,838.6 (68°F) \| 0.693 \| 7.456 \| \| \| 1-Bromo-butane (46) \| C4H9Br \| 1.276 (20°C) \| 1,019 (20°C) \| 3,343.2 (68°F) \| 0.49 (15°C) \| 5.272 (59°F) \| \| \| Bromo-ethane (46) \| C2H5Br \| 1.460 (20°C) \| 900 (20°C) \| 2,952.8 (68°F) \| 0.275 \| 2.959 \| \| \| Bromoform (46, 47) \| CHBr3 \| (20°C) 2.89 \| 918 \| 3011.8 \| 3.1 \| 0.654 \| 7.037 \| \| n-Butane (2) \| C4H10 \| 0.601 (0°C) \| 1,085 (-5°C) \| 3,559.7 (23°F) \| 5.8 \| \| \| \| 2-Butanol \| C4H10O \| 0.81 \| 1240 \| 4068.2 \| 3.3 \| 3.239 \| 34.851 \| \| sec-Butylalcohol \| C4H10O \| 0.81 \| 1240 \| 4068.2 \| 3.3 \| 3.239 \| 34.851 \| \| n-Butyl bromide (46) \| C4H9Br \| 1.276 (20°C) \| 1,019 (20°C) \| 3,343.2 (68°F) \| 0.49 (15°C) \| 5.272 (59°F) \| \| \| n-Butyl chloride (22, 46) \| C4H9Cl \| 0.887 \| 1140 \| 3740.2 \| 4.57 \| 0.529 (15°C) \| 5.692 (59°F) \| \| tert Butyl chloride \| C4H9Cl \| 0.84 \| 984 \| 3228.3 \| 4.2 \| 0.646 \| 6.95 \| \| Butyl oleate \| C22H42O2 \| 1404 \| 4606.3 \| 3 \| \| \| \| \| 2, 3 Butylene glycol \| C4H10O2 \| 1.019 \| 1484 \| 4868.8 \| 1.51 \| \| \| \| Cadmium (7) \| Cd \| 2,237.7 (400°C) \| 7,341.5 (752°F) \| 1.355cp (440°C) \| 14.579 (824°F) \| \| \| \| Carbinol (40, 41) \| CH4O \| (20°C) 0.791 \| 1076 \| 3530.2 \| 2.92 \| 0.695 \| 7.478 \| \| Carbitol \| C6H14O3 \| 0.988 \| 1458 \| 4783.5 \| \| \| \| \| Carbon dioxide (26) \| CO2 \| 1.101 (-37°C) \| 839 (-37°C) \| 2,752.6 (-35°F) \| 7.71 \| (-0.137 37°C) \| (-1.474 35°F) \| \| Carbon disulphide \| CS2 \| 1.261 (22°C) \| 1149 \| 3769.7 \| 0.278 \| 2.991 \| \| \| Carbon tetrachloride (33, 35, 47) \| CCl4 \| 1.595 (20°C) \| 926 \| 3038.1 \| 2.48 \| 0.607 \| 6.531 \| \| Carbon tetrafluoride (14) (Freon 14) \| CF4 \| 150°C) 1.75 (- \| (-875.2 150°C) \| (-2,871.5 238°F) \| 6.61 \| \| \| \| Cetane (23) \| C16H34 \| 0.773 (20°C) \| 1338 \| 4389.8 \| 3.71 \| 4.32 \| 46.483 \| \| Chloro-benezene \| C6H5Cl \| 1.106 \| 1273 \| 4176.5 \| 3.6 \| 0.722 \| 7.768 \| \| 1-Chloro-butane (22, 46) \| C4H9Cl \| 0.887 \| 1140 \| 3740.2 \| 4.57 \| 0.529 (15°C) \| 5.692 (59°F) \| \| Chloro-diFluoromethane (3)(Freon 22) \| CHClF2 \| 1.491 (-69°C) \| 893.9 (-50°C) \| 2,932.7 (-58°F) \| 4.79 \| \| \| \| Chloroform (47) \| CHCl3 \| 1.489 \| 979 \| 3211.9 \| 3.4 \| 0.55 \| 5.918 \| \| 1-Chloropropane (47) \| C3H7Cl \| 0.892 \| 1058 \| 3471.1 \| 0.378 \| 4.067 \| \| \| Chlorotrifluoromethane (5) \| CClF3 \| 724 (-82°C) \| 2,375.3 (-116°F) \| 5.26 \| \| \| \| \| Cinnamaldehyde \| C9H8O \| 1.112 \| 1554 \| 5098.4 \| 3.2 \| \| \| \| Cinnamic aldehyde \| C9H8O \| 1.112 \| 1554 \| 5098.4 \| 3.2 \| \| \| \| Colamine \| C2H7NO \| 1.018 \| 1724 \| 5656.2 \| 3.4 \| \| \| \| o-Cresol (46) \| C7H8O \| 1.047 (20°C) \| 1,541 (20°C) \| 5,055.8 (68°F) \| 4.29 (40°C) \| 46.16 (104°F) \| \| \| m-Cresol (46) \| C7H8O \| 1.034 (20°C) \| 1,500 (20°C) \| 4,921.3 (68°F) \| 5.979 (40°C) \| 64.334 (104°F) \| \| \| Cyanomethane \| C2H3N \| 0.783 \| 1290 \| 4232.3 \| 4.1 \| 0.441 \| 4.745 \| \| Cyclohexane (15) \| C6H12 \| 0.779 (20°C) \| 1248 \| 4094.5 \| 5.41 \| (17°C) 1.31 \| 14.095 (63°F) \| \| Cyclohexanol \| C6H12O \| 0.962 \| 1454 \| 4770.3 \| 3.6 \| 0.071 (17°C) \| 0.764 (63°F) \| \| Cyclohexanone \| C6H10O \| 0.948 \| 1423 \| 4668.6 \| 4 \| \| \| \| Decane (46) \| C10H22 \| 0.73 \| 1252 \| 4107.6 \| 1.26 (20°C) \| 13.55 (68°F) \| \| \| 1-Decene (27) \| C10H20 \| 0.746 \| 1235 \| 4051.8 \| 4 \| \| \| \| n-Decylene (27) \| C10H20 \| 0.746 \| 1235 \| 4051.8 \| 4 \| \| \| \| Diacetyl \| C4H6O2 \| 0.99 \| 1236 \| 4055.1 \| 4.6 \| \| \| \| Diamylamine \| C10H23N \| 1256 \| 4120.7 \| 3.9 \| \| \| \| \| 1,2 Dibromoethane (47) \| C2H4Br2 \| 2.18 \| 995 \| 3264.4 \| 0.79 (20°C) \| 8.5 (68°F) \| \| \| trans-1,2- Dibromoethene (47) \| C2H2Br2 \| 2.231 \| 935 \| 3067.6 \| \| \| \| \| Dibutyl phthalate \| C8H22O4 \| 1408 \| 4619.4 \| \| \| \| \| \| Dichloro-t-butyl alcohol \| C4H8Cl2O \| 1304 \| 4278.2 \| 3.8 \| \| \| \| \| 2,3 Dichlorodioxane \| C2H6Cl2O2 \| 1391 \| 4563.6 \| 3.7 \| \| \| \| \| Dichlorodifluoromethane(3) (Freon 12) \| CCl2F2 \| 1.516 (40°C) \| 774.1 \| 2539.7 \| 4.24 \| \| \| \| 1,2 Dichloro ethane (47) \| C2H4Cl2 \| 1.253 \| 1193 \| 3914 \| 0.61 \| 6.563 \| \| \| cis1,2-Dichloro-ethene (3, 47) \| C2H2Cl2 \| 1.284 \| 1061 \| 3481 \| \| \| \| \| trans1,2-Dichloro-ethene (3, 47) \| C2H2Cl2 \| 1.257 \| 1010 \| 3313.6 \| \| \| \| \| Dichloro-fluoromethane (3)(Freon 21) \| CHCl2F \| 1.426 (0°C) \| 891 (0°C) \| 2,923.2 (32°F) \| 3.97 \| \| \| \| 1-2-Dichlorohexafluorocyclobutane (47) \| C4Cl2F6 \| 1.654 \| 669 \| 2194.9 \| \| \| \| \| 1-3-Dichloroisobutane \| C4H8Cl2 \| 1.14 \| 1220 \| 4002.6 \| 3.4 \| \| \| \| Dichloro methane (3) \| CH2Cl2 \| 1.327 \| 1070 \| 3510.5 \| 3.94 \| 0.31 \| 3.335 \| \| 1,1-Dichloro-1,2,2,2 tetra fluoroethane \| CClF2-CClF2 \| 1.455 \| 665.3 (-10°C) \| 2,182.7 (14°F) \| 3.73 \| \| \| \| Diethyl ether \| C4H10O \| 0.713 \| 985 \| 3231.6 \| 4.87 \| 0.311 \| 3.346 \| \| Diethylene glycol \| C4H10O3 \| 1.116 \| 1586 \| 5203.4 \| 2.4 \| \| \| \| Diethylene glycol, monoethyl ether \| C6H14O3 \| 0.988 \| 1458 \| 4783.5 \| \| \| \| \| Diethylenimide oxide \| C4H9NO \| 1 \| 1442 \| 4731 \| 3.8 \| \| \| \| 1,2-bis(DiFluoramino) butane (43) \| C4H8(NF2)2 \| 1.216 \| 1000 \| 3280.8 \| \| \| \| \| 1,2-bis(DiFluor amino)-2-methylpropa ne (43) \| C4H9(NF2)2 \| 1.213 \| 900 \| 2952.8 \| \| \| \| \| 1,2-bis(DiFluor amino) propane (43) \| C3H6(NF2)2 \| 1.265 \| 960 \| 3149.6 \| \| \| \| \| 2,2-bis(DiFluor amino propane (43) \| C3H6(NF2)2 \| 1.254 \| 890 \| 2920 \| \| \| \| \| 2,2-Dihydroxydiethyl ether \| C4H10O3 \| 1.116 \| 1586 \| 5203.4 \| 2.4 \| \| \| \| Dihydroxy ethane \| C2H6O2 \| 1.113 \| 1658 \| 5439.6 \| 2.1 \| \| \| \| 1,3-Dimethylbenzene (46) \| C8H10 \| (15°C) 0.868 \| (20°C) 1,343 \| 4,406.2 (68°F) \| (15°C) 0.749 \| (59°F) 8.059 \| \| \| 1,2-Dimethylbenzene (29, 46) \| C8H10 \| (20°C) 0.897 \| 1331.5 \| 4368.4 \| 4.1 \| (20°C) 0.903 \| (68°F) 9.716 \| \| 1,4-Dimethylbenzene (46) \| C8H10 \| (20°C) 1,334 \| 4,376.6 (68°F) \| 0.662 \| 7.123 \| \| \| \| 2,2-Dimethylbutane (29, 33) \| C6H14 \| (20°C) 0.649 \| 1079 \| 3540 \| \| \| \| \| Dimethyl ketone \| C3H6O \| 0.791 \| 1174 \| 3851.7 \| 4.5 \| 0.399 \| 4.293 \| \| Dimethyl pentane (47) \| C7H16 \| 0.674 \| 1063 \| 3487.5 \| \| \| \| \| Dimethyl phthalate \| C8H10O4 \| 1.2 \| 1463 \| 4799.9 \| \| \| \| \| Diiodo-methane \| CH2I2 \| 3.235 \| 980 \| 3215.2 \| \| \| \| \| Dioxane \| C4H8O2 \| 1.033 \| 1376 \| 4514.4 \| \| \| \| \| Dodecane (23) \| C12H26 \| 0.749 \| 1279 \| 4196.2 \| 3.85 \| 1.8 \| 19.368 \| \| 1,2-Ethanediol \| C2H6O2 \| 1.113 \| 1658 \| 5439.6 \| 2.1 \| \| \| \| Ethanenitrile \| C2H3N \| 0.783 \| 1290 \| 4232.3 \| 0.441 \| 4.745 \| \| \| Ethanoic anhydride (22) \| (CH3CO)2O \| 1.082 \| 1180 \| 3871.4 \| 0.769 \| 8.274 \| \| \| Ethanol \| C2H6O \| 0.789 \| 1207 \| 3960 \| 4 \| 1.39 \| 14.956 \| \| Ethanol amide \| C2H7NO \| 1.018 \| 1724 \| 5656.2 \| 3.4 \| \| \| \| Ethoxyethane \| C4H10O \| 0.713 \| 985 \| 3231.6 \| 4.87 \| 0.311 \| 3.346 \| \| Ethyl acetate (33) \| C4H8O2 \| 0.901 \| 1085 \| 3559.7 \| 4.4 \| 0.489 \| 5.263 \| \| Ethyl alcohol \| C2H6O \| 0.789 \| 1207 \| 3960 \| 4 \| 1.396 \| 15.02 \| \| Ethyl benzene (46) \| C8H10 \| 0.867 (20°C) \| 1,338 (20°C) \| 4,389.8 (68°F) \| 0.797 (17°C) \| 8.575 (63°F) \| \| \| Ethyl Bromide (46) \| C2H5Br \| 1.461 (20°C) \| 900 (20°C) \| 2,952.8 (68°F) \| 0.275 (20°C) \| 2.959 (68°F) \| \| \| Ethyliodide (46) \| C2H5I \| 1.950 (20°C) \| 876 (20°C) \| 2874 (68°F) \| 0.29 \| 3.12 \| \| \| Ether \| C4H10O \| 0.713 \| 985 \| 3231.6 \| 4.87 \| 0.311 \| 3.346 \| \| Ethyl ether \| C4H10O \| 0.713 \| 985 \| 3231.6 \| 4.87 \| 0.311 \| 3.346 \| \| Ethylene bromide (47) \| C2H4Br2 \| 2.18 \| 995 \| 3264.4 \| 0.79 \| 8.5 \| \| \| Ethylene chloride (47) \| C2H4Cl2 \| 1.253 \| 1193 \| 3914 \| 0.61 \| 6.563 \| \| \| Ethylene glycol \| C2H6O2 \| 1.113 \| 1658 \| 5439.6 \| 2.1 \| 17.208 (20°C) \| 185.158 (68°F) \| \| d-Fenochone \| C10H16O \| 0.947 \| 1320 \| 4330.7 \| 0.22 \| 2.367 \| \| \| d-2- Fenechanone \| C10H16O \| 0.947 \| 1320 \| 4330.7 \| 0.22 \| 2.367 \| \| \| Fluorine \| F \| 0.545 (-143°C) \| 403 (-143°C) \| 1322.2 (-225°F) \| 11.31 \| \| \| \| Fluoro-benzene (46) \| C6H5F \| 1.024 (20°C) \| 1189 \| 3900.9 \| (20°C) 0.584 \| (68°F) 6.283 \| \| \| Formaldehyde, methyl ester \| C2H4O2 \| 0.974 \| 1127 \| 3697.5 \| 4.02 \| \| \| \| Formamide \| CH3NO \| 1.134 (20°C) \| 1622 \| 5321.5 \| 2.2 \| 2.91 \| 31.311 \| \| Formic acid, amide \| CH3NO \| 1.134 (20°C) \| 1622 \| 5321.5 \| 2.91 \| 31.311 \| \| \| Freon R12 \| 774.2 \| 2540 \| \| \| \| \| \| \| Furfural \| C5H4O2 \| 1.157 \| 1444 \| 4737.5 \| 3.7 \| \| \| \| Furfuryl alcohol \| C5H6O2 \| 1.135 \| 1450 \| 4757.2 \| 3.4 \| \| \| \| Fural \| C5H4O2 \| 1.157 \| 1444 \| 4737.5 \| 3.7 \| \| \| \| 2-Furaldehyde \| C5H4O2 \| 1.157 \| 1444 \| 4737.5 \| 3.7 \| \| \| \| 2-Furancarbox aldehyde \| C5H4O2 \| 1.157 \| 1444 \| 4737.5 \| 3.7 \| \| \| \| 2-FurylMethanol \| C5H6O2 \| 1.135 \| 1450 \| 4757.2 \| 3.4 \| \| \| \| Gallium \| Ga \| 6.095 \| 2,870 (30°C) \| 9416 (86°F) \| \| \| \| \| Glycerin \| C3H8O3 \| 1.26 \| 1904 \| 6246.7 \| 2.2 \| 757.1 \| 8081.836 \| \| Glycerol \| C3H8O3 \| 1.26 \| 1904 \| 6246.7 \| 2.2 \| 757.1 \| 8081.836 \| \| Glycol \| C2H6O2 \| 1.113 \| 1658 \| 5439.6 \| 2.1 \| \| \| \| 50% Glycol / 50% H2O \| 1578 \| 5177 \| \| \| \| \| \| \| Helium (45) \| He4 \| 0.125 (-269°C ) \| 183 (-269°C) \| 600.4 (-452°F) \| 0.025 \| 0.269 \| \| \| Heptane (22, 23) \| C7H16 \| 0.684 (20°C) \| 1131 \| 3710.6 \| 4.25 \| (20°C) 0.598 \| (68°F) 6.434 \| \| n-Heptane (29, 33) \| C7H16 \| (20°C) 0.684 \| 1180 \| 3871.3 \| 4 \| \| \| \| HexachloroCyclopentadiene (47) \| C5Cl6 \| 1.718 \| 1150 \| 3773 \| \| \| \| \| Hexadecane (23) \| C16H34 \| 0.773 (20°C) \| 1338 \| 4389.8 \| 3.71 \| (20°C) 4.32 \| 46.483 (68°F) \| \| Hexalin \| C6H12O \| 0.962 \| 1454 \| 4770.3 \| 3.6 \| 70.69 (17°C) \| 760.882 (63°F) \| \| Hexane (16, 22, 23) \| C6H14 \| 0.659 \| 1112 \| 3648.3 \| 2.71 \| 0.446 \| 4.798 \| \| n-Hexane (29, 33) \| C6H14 \| (20°C) 0.649 \| 1079 \| 3540 \| 4.53 \| \| \| \| 2,5- Hexanedione \| C6H10O2 \| 0.729 \| 1399 \| 4589.9 \| 3.6 \| \| \| \| n-Hexanol \| C6H14O \| 0.819 \| 1300 \| 4265.1 \| 3.8 \| \| \| \| Hexahydro benzene (15) \| C6H12 \| 0.779 \| 1248 \| 4094.5 \| 5.41 \| (17°C) 1.31 \| 14.095 (63°F) \| \| Hexahydro phenol \| C6H12O \| 0.962 \| 1454 \| 4770.3 \| 3.6 \| \| \| \| Hexamethylene (15) \| C6H12 \| 0.779 \| 1248 \| 4094.5 \| 5.41 \| 1.31 (17°C) \| 14.095 (63°F) \| \| Hydrogen (45) \| H2 \| 0.071 (-256°C ) \| 1,187 (-256°C) \| 3,894.4 (-429°F) \| 0.003 (-256°C) \| 0.032 (-429°F) \| \| \| 2-Hydroxytoluene (46) \| C7H8O \| (20°C) 1.047 \| (20°C) 1,541 \| 5,055.8 (68°F) \| (40°C) 4.29 \| (104°F) 46.16 \| \| \| 3-Hydroxytoluene (46) \| C7H8O \| 1.034 (20°C) \| 1,500 (20°C) \| 4,921.3 (68°F) \| 5.979 (40°C) \| 64.334 (104°F) \| \| \| Iodo-benzene (46) \| C6H5I \| 1.823 \| 1,114 (20°C) \| 3,654.9 (68°F) \| 0.954 \| \| \| \| Iodo-ethane (46) \| C2H5I \| 1.950 (20°C) \| 876 (20°C) \| 2,874 (68°F) \| 0.29 \| 3.12 \| \| \| Iodo-methane \| CH3I \| 2.28 (20°C) \| 978 \| 3208.7 \| 0.211 \| 2.27 \| \| \| Isobutyl acetate (22) \| C6H12O \| 1,180 (27°C) \| 3,871.4 (81°F) \| 4.85 \| \| \| \| \| Isobutanol \| C4H10O \| 0.81 (20°C) \| 1212 \| 3976.4 \| \| \| \| \| Iso-Butane \| 1219.8 \| 4002 \| \| \| \| \| \| \| Isopentane (36) \| C5H12 \| 0.62 (20°C) \| 980 \| 3215.2 \| 4.8 \| 0.34 \| 3.658 \| \| Isopropanol (46) \| C3H8O \| 0.785 (20°C) \| 1,170 (20°C) \| 3,838.6 (68°F) \| 2.718 \| 29.245 \| \| \| Isopropyl alcohol (46) \| C3H8O \| 0.785 (20°C) \| 1,170 (20°C) \| 3,838.6 (68°F) \| 2.718 \| 29.245 \| \| \| Kerosene \| 0.81 \| 1324 \| 4343.8 \| 3.6 \| \| \| \| \| Ketohexa methylene \| C6H10O \| 0.948 \| 1423 \| 4668.6 \| 4 \| \| \| \| Lithium fluoride (42) \| LiF \| 2,485 (900°C) \| 8,152.9 (1652°F) \| 1.29 \| \| \| \| \| Mercury (45) \| Hg \| 13.594 \| 1,449 (24°C) \| 4,753.9 (75°F) \| 0.114 \| 1.226 \| \| \| Mesityloxide \| C6H16O \| 0.85 \| 1310 \| 4297.9 \| \| \| \| \| Methane (25, 28, 38, 39) \| CH4 \| (-0.162 89°C) \| (-89°C) 405 \| (- 1,328.7 128°F) \| 17.5 \| \| \| \| Methanol (40, 41) \| CH4O \| 0.791 (20°C) \| 1076 \| 3530.2 \| 2.92 \| 0.695 \| 7.478 \| \| Methyl acetate \| C3H6O2 \| 0.934 \| 1211 \| 3973.1 \| 0.407 \| 4.379 \| \| \| o-Methylaniline (46) \| C7H9N \| 0.999 (20°C) \| 1618 \| 5308.4 \| (20°C) 4.394 \| 47.279 (68°F) \| \| \| 4-Methylaniline (46) \| C7H9N \| 0.966 (45°C) \| 1480 \| 4855.6 \| (50°C) 1.863 \| 20.095 (122°F) \| \| \| Methyl alcohol (40, 44) \| CH4O \| 0.791 (20°C) \| 1076 \| 3530.2 \| 2.92 \| 0.695 \| 7.478 \| \| Methyl benzene (16, 52) \| C7H8 \| 0.867 \| 1,328 (20°C) \| 4,357 (68°F) \| 4.27 \| 0.644 \| 7.144 \| \| 2-Methyl-butane (36) \| C5H12 \| 0.62 (20°C) \| 980 \| 3215.2 \| 0.34 \| 3.658 \| \| \| Methyl carbinol \| C2H6O \| 0.789 \| 1207 \| 3960 \| 4 \| 1.396 \| \| \| Methylchloroform (47) \| C2H3Cl3 \| 1.33 \| 985 \| 3231.6 \| (20°C) 0.902 \| (68°F) 9.705 \| \| \| Methyl-cyanide \| C2H3N \| 0.783 \| 1290 \| 4232.3 \| 0.441 \| 4.745 \| \| \| 3-Methyl cyclohexanol \| C7H14O \| 0.92 \| 1400 \| 4593.2 \| \| \| \| \| Methylene chloride (3) \| CH2Cl2 \| 1.327 \| 1070 \| 3510.5 \| 3.94 \| 0.31 \| 3.335 \| \| Methylene iodide \| CH2I2 \| 3.235 \| 980 \| 3215.2 \| \| \| \| \| Methyl formate (22) \| C2H4O2 \| 0.974 (20°C) \| 1127 \| 3697.5 \| 4.02 \| \| \| \| Methyl iodide \| CH3I \| 2.28 (20°C) \| 978 \| 3208.7 \| 0.211 \| 2.27 \| \| \| a-Methyl naphthalene \| C11H10 \| 1.09 \| 1510 \| 4954.1 \| 3.7 \| \| \| \| 2-Methylphenol (46) \| C7H8O \| 1.047 (20°C) \| 1,541 (20°C) \| 5,055.8 (68°F) \| 4.29 (40°C) \| 46.16 (104°F) \| \| \| 3-Methylphenol (46) \| C7H8O \| 1.034 (20°C) \| 1,500 (20°C) \| 4,921.3 (68°F) \| 5.979 (40°C) \| 64.334 (104°F) \| \| \| Milk, homogenized \| 1548 \| 5080 \| \| \| \| \| \| \| Morpholine \| C4H9NO \| 1 \| 1442 \| 4731 \| 3.8 \| \| \| \| Naphtha \| 0.76 \| 1225 \| 4019 \| \| \| \| \| \| Natural Gas (37) \| 0.316 (-103°C) \| 753 (-103°C) \| 2,470.5 (-153°F) \| \| \| \| \| \| Neon (45) \| Ne \| 1.207 (-246°C ) \| 595 (- 246°C) \| 1,952.1 (-411°F) \| \| \| \| \| Nitrobenzene (46) \| C6H5NO2 \| 1.204 (20°C) \| 1,415 (20°C) \| 4,642.4 (68°F) \| 1.514 \| 16.29 \| \| \| Nitrogen (45) \| N2 \| 0.808 (-199°C) \| 962 (-199°C) \| 3,156.2 (-326°F) \| 0.217 (-199°C) \| 2.334 (-326°F) \| \| \| Nitromethane (43) \| CH3NO2 \| 1.135 \| 1300 \| 4265.1 \| 4 \| 0.549 \| 5.907 \| \| Nonane (23) \| C9H2O \| 0.718 (20°C) \| 1207 \| 3960 \| 4.04 \| (20°C) 0.99 \| (68°F) 10.652 \| \| 1-Nonene (27) \| C9H18 \| 0.736 (20°C) \| 1207 \| 3960 \| 4 \| \| \| \| Octane (23) \| C8H18 \| 0.703 \| 1172 \| 3845.1 \| 4.14 \| 0.73 \| 7.857 \| \| n-Octane (29) \| C8H18 \| 0.704 (20°C) \| 1212.5 \| 3978 \| 3.5 \| 0.737 \| .930) \| \| 1-Octene (27) \| C8H16 \| 0.723 (20°C) \| 1175.5 \| 3856.6 \| 4.1 \| \| \| \| Oil of Camphor Sassafrassy \| 1390 \| 4560.4 \| 3.8 \| \| \| \| \| \| Oil, Car (SAE 20a.30) \| 1.74 \| 870 \| 2854.3 \| 190 \| 2,045.09 3 \| \| \| \| Oil, Castor \| C11H10O10 \| 0.969 \| 1477 \| 4845.8 \| 3.6 \| 0.67 \| 7.209 \| \| Oil, Diesel \| 0.8 \| 1250 \| 4101 \| \| \| \| \| \| Oil, Fuel AA gravity \| 0.99 \| 1485 \| 4872 \| 3.7 \| \| \| \| \| Oil (Lubricating X200) \| 1530 \| 5019.9 \| \| \| \| \| \| \| Oil (Olive) \| 0.912 \| 1431 \| 4694.9 \| 2.75 \| 100 \| 1076.365 \| \| \| Oil (Peanut) \| 0.936 \| 1458 \| 4783.5 \| \| \| \| \| \| Oil (Sperm) \| 0.88 \| 1440 \| 4724.4 \| \| \| \| \| \| Oil, 6 \| 1,509 (22°C) \| 4,951 (72°F) \| \| \| \| \| \| \| 2,2-Oxydi ethanol \| C4H10O3 \| 1.116 \| 1586 \| 5203.4 \| 2.4 \| \| \| \| Oxygen (45) \| O2 \| 1.155 (-186°C) \| 952 (-186°C) \| 3,123.4 (-303°F) \| 0.173 \| 1.861 \| \| \| Pentachloroethane (47) \| C2HCl5 \| 1.687 \| 1082 \| 3549.9 \| \| \| \| \| Pentalin (47) \| C2HCl5 \| 1.687 \| 1082 \| 3549.9 \| \| \| \| \| Pentane (36) \| C5H12 \| 0.626 (20°C) \| 1020 \| 3346.5 \| 0.363 \| 3.905 \| \| \| n-Pentane (47) \| C5H12 \| 0.557 \| 1006 \| 3300.5 \| 0.41 \| 4.413 \| \| \| Perchlorocyclopentadie ne (47) \| C5Cl6 \| 1.718 \| 1150 \| 3773 \| \| \| \| \| Perchloroethylene (47) \| C2Cl4 \| 1.632 \| 1036 \| 3399 \| \| \| \| \| Perfluoro-1- Hepten (47) \| C7F14 \| 1.67 \| 583 \| 1912.7 \| \| \| \| \| Perfluoron-Hexane (47) \| C6F14 \| 1.672 \| 508 \| 1666.7 \| \| \| \| \| Phene L(29, 40, 41) \| C6H6 \| 0.879 \| 1306 \| 4284.8 \| 4.65 \| 0.711 \| 7.65 \| \| b-Phenyl acrolein \| C9H8O \| 1.112 \| 1554 \| 5098.4 \| 3.2 \| \| \| \| Phenylamine (41) \| C6H5NO2 \| 1.022 \| 1639 \| 5377.3 \| 4 \| 3.63 \| 39.058 \| \| Phenyl bromide (46) \| C6H5Br \| 1.522 \| 1,170 (20°C) \| 3,838.6 (68°F) \| 0.693 \| 7.456 \| \| \| Phenyl chloride \| C6H5Cl \| 1.106 \| 1273 \| 4176.5 \| 3.6 \| 0.722 \| 7.768 \| \| Phenyl iodide (46) \| C6H5I \| 1.823 \| 1,114 (20°C) \| 3,654.9 (68°F) \| 0.954 (15°C) \| 10.265 (59°F) \| \| \| Phenyl methane (16, 52) \| C7H8 \| 0.867 (20°C) \| 1,328 (20°C) \| 4,357 (68°F) \| 4.27 \| 0.644 \| 6.929 \| \| 3-Phenyl propenal \| C9H8O \| 1.112 \| 1554 \| 5098.4 \| 3.2 \| \| \| \| Phthalardione \| C8H4O3L \| 1,125 (152°C) \| 3,691 (306°F) \| \| \| \| \| \| Phthalic acid, anhydride \| C8H4O3 \| 1,125 (152°C) \| 3,691 (306°F) \| \| \| \| \| \| Pthalic anhydride \| C8H4O3 \| 1,125 (152°C) \| 3,691 (306°F) \| \| \| \| \| \| Pimelic ketone \| C6H10O \| 0.948 \| 1423 \| 4668.6 \| 4 \| \| \| \| Plexiglas, Lucite, Acrylic \| 2651 \| 8698 \| \| \| \| \| \| \| Polyterpene Resin \| 0.77 \| 1,099.8 (190°C) \| 3,608.4 (374°F) \| 39000 \| 419500 \| \| \| \| Potassium bromide (42) \| KBr \| 1,169 (900°C) \| 3,835.3 (1652°F) \| 0.71 \| (900°C) .715cp \| (1652°F) 7.693 \| \| \| Potassium fluoride (42) \| KF \| 1,792 (900°C) \| 5,879.3 (1652°F) \| 1.03 \| \| \| \| \| Potassium iodide (42) \| KI \| 985 (900°C) \| 3,231.6 (1652°F) \| 0.64 \| \| \| \| \| Potassium nitrate (48) \| KNO3 \| 1.859 (352°C) \| 1,740.1 (352°C) \| 5,709 (666°F) \| 1.1 \| (327°C) 1.19 \| (621°F) 12.804 \| \| Propane (2, 13) (-45° to -130°C) \| C3H8 \| 0.585 (-45°C) \| 1,003 (-45°C) \| 3,290.6 (-49°F) \| 5.7 \| \| \| \| 1,2,3- Propanetriol \| C3H8O3 \| 1.26 \| 1904 \| 6246.7 \| 2.2 \| 0.000757 \| \| \| 1-Propanol (46) \| C3H8O \| 0.78 (20°C) \| 1,222 (20°C) \| 4,009.2 (68°F) \| \| \| \| \| 2-Propanol (46) \| C3H8O \| 0.785 (20°C) \| 1,170 (20°C) \| 3,838.6 (68°F) \| 2.718 \| 29.245 \| \| \| 2-Propanone \| C3H6O \| 0.791 \| 1174 \| 3851.7 \| 4.5 \| 0.399 \| 4.293 \| \| Propene (17, 18, 35) \| C3H6 \| (- 0.563 13°C) \| (-963 13°C) \| 3,159.4 (9°F) \| 6.32 \| \| \| \| n-Propyl acetate (22) \| C5H10O2 \| 1,280 (2°C) \| 4,199 (36°F) \| 4.63 \| \| \| \| \| n-Propyl-alcohol \| C3H8O \| 0.78 (20°C) \| 1,222 (20°C) \| 4,009.2 (68°F) \| 2.549 \| 27.427 \| \| \| Propylchloride (47) \| C3H7Cl \| 0.892 \| 1058 \| 3471.1 \| 0.378 \| 4.067 \| \| \| Propylene (17, 18, 35) \| C3H6 \| (- 0.563 13°C) \| (-963 13°C) \| (3159.4) (9°F) \| 6.32 \| \| \| \| Pyridine \| C6H5N \| 0.982 \| 1415 \| 4642.4 \| 4.1 \| 0.992 (20°C) \| 10.673 (68°F) \| \| Refrigerant 11 (3, 4) \| CCl3F \| 1.49 \| 828.3 (0°C) \| 2,717.5 (32°F) \| 3.56 \| \| \| \| Refrigerant 12 (3) \| CCl2F2 \| 1.516 (-40°C) \| 774.1 (-40°C) \| 2,539.7 (-40°F) \| 4.24 \| \| \| \| Refrigerant 14 (14) \| CF4 \| 1.75 (- 150°C) \| 875.24 (-150°C) \| 2,871.5 (-238°F) \| 6.61 \| \| \| \| Refrigerant 21 (3) \| CHCl2F \| 1.426 (0°C) \| 891 (0°C) \| 2,923.2 (32°F) \| 3.97 \| \| \| \| Refrigerant 22 (3) \| CHClF2 \| 1.491 (-69°C) \| 893.9 (50°C) \| 2,932.7 (122°F) \| 4.79 \| \| \| \| Refrigerant 113 (3) \| CCl2F-CClF2 \| 1.563 \| 783.7 (0°C) \| 2,571.2 (32°F) \| 3.44 \| \| \| \| Refrigerant 114 (3) \| CClF2-CClF2 \| 1.455 \| 665.3 (-10°C) \| 2,182.7 (14°F) \| 3.73 \| \| \| \| Refrigerant 115 (3) \| C2ClF5 \| 656.4 (-50°C) \| 2,153.5 (-58°F) \| 4.42 \| \| \| \| \| Refrigerant C318 (3) \| C4F8 \| 1.62 (-20°C) \| 574 (-10°C) \| 1,883.2 (14°F) \| 3.88 \| \| \| \| Selenium (8) \| Se \| 1,072 (250°C) \| 3,517.1 (482°F) \| 0.68 \| \| \| \| \| Silicone (30 cp) \| 0.993 \| 990 \| 3248 \| 30 \| 322.8 \| \| \| \| Sodium fluoride (42) \| NaF \| 0.877 \| 2,082 (1000°C) \| 6,830.7 (1832°F) \| 1.32 \| \| \| \| Sodium nitrate (48) \| NaNO3 \| 1.884 (336°C) \| 1,763.3 (336°C) \| 5,785.1 (637°F) \| 0.74 \| (336°C) 1.37 \| (637°F) 14.74 \| \| Sodium nitrite (48) \| NaNO2 \| 1.805 (292°C) \| 1,876.8 (292°C) \| 6,157.5 (558°F) \| \| \| \| \| Solvesso #3 \| 0.877 \| 1370 \| 4494.8 \| 3.7 \| \| \| \| \| Spirit of wine \| C2H6O \| 0.789 \| 1207 \| 3960 \| 4 \| 1.396 \| 15.02 \| \| Sulfur (7, 8, 10) \| S \| 1,177 (250°C) \| 3,861.5 (482°F) \| -1.13 \| \| \| \| \| Sulfuric Acid (1) \| H2SO4 \| 1.841 \| 1257.6 \| 4126 \| 1.43 \| 11.16 \| 120.081 \| \| Tellurium (7) \| Te \| 991 (450°C) \| 3,251.3 (842°F) \| 0.73 \| \| \| \| \| 1,1,2,2-Tetra bromoethane (47) \| C2H2Br4 \| 2.966 \| 1027 \| 3369.4 \| \| \| \| \| 1,1,2,2-Tetra chloro-ethane (67) \| C2H2Cl4 \| 1.595 \| 1147 \| 3763.1 \| 1.156 (15°C) \| 12.438 (59°F) \| \| \| Tetrachloro ethane (46) \| C2H2Cl4 \| (20°C) 1.553 \| (20°C) 1,170 \| 3,838.6 (68°F) \| 1.19 \| 12.804 \| \| \| Tetrachloroethene (47) \| C2Cl4 \| 1.632 \| 1036 \| 3399 \| \| \| \| \| TetrachloroMethane (33, 47) \| CCl4 \| 1.595 (20°C) \| 926 \| 3038.1 \| 0.607 \| 6.531 \| \| \| Tetradecane (46) \| C14H3O \| 0.763 (20°C) \| 1,331 (20°C) \| 4,366.8 (68°F) \| 2.86 (20°C) \| 30.773 (68°F) \| \| \| Tetraethylene glycol \| C8H18O5 \| 1.123 \| 1586 \| 5203.4 \| 3 \| \| \| \| Tetrafluoromethane (14) (Freon 14) \| CF4 \| 1.75 (- 150°C) \| 875.24 (-150°C) \| 2,871.5 (-238°F) \| 6.61 \| \| \| \| Tetrahydro-1,4- isoxazine \| C4H9NO \| 1 \| 1442 \| 4731 \| 3.8 \| \| \| \| Toluene (16, 52) \| C7H8 \| 0.867 (20°C) \| 1,328 (20°C) \| 4,357 (68°F) \| 4.27 \| 0.644 \| 6.929 \| \| o-Toluidine (46) \| C7H9N \| 0.999 (20°C) \| 1618 \| 5308.4 \| (20°C) 4.394 \| 47.279 (68°F) \| \| \| p-Toluidine (46) \| C7H9N \| 0.966 (45°C) \| 1480 \| 4855.6 \| (50°C) 1.863 \| (122°F) 20.053 \| \| \| Toluol \| C7H8 \| 0.866 \| 1308 \| 4291.3 \| 4.2 \| 0.58 \| 6.24 \| \| Tribromomethane (46, 47) \| CHBr3 \| 2.89 (20°C) \| 918 \| 3011.8 \| 0.654 \| 7.037 \| \| \| 1,1,1-Trichloro-ethane (47) \| C2H3Cl3 \| 1.33 \| 985 \| 3231.6 \| (20°C) 0.902 \| (68°F) 9.705 \| \| \| Trichloro-ethene (47) \| C2HCl3 \| 1.464 \| 1028 \| 3372.7 \| \| \| \| \| Trichloro-fluoromethane (3)(Freon 11) \| CCl3F \| 1.49 \| 828.3 (0°C) \| 2,717.5 (32°F) \| 3.56 \| \| \| \| Trichloromethane (47) \| CHCl3 \| 1.489 \| 979 \| 3211.9 \| 3.4 \| 0.55 \| 5.918 \| \| 1,1,2-Trichloro- 1,2,2-Trifluoro-Ethane \| CCl2F-CClF2 \| 1.563 \| 783.7 (0°C) \| 2,571.2 (32°F) \| \| \| \| \| Triethyl-amine (33) \| C6H15N \| 0.726 \| 1123 \| 3684.4 \| 4.47 \| \| \| \| Triethylene glycol \| C6H14O4 \| 1.123 \| 1608 \| 5275.6 \| 3.8 \| \| \| \| 1,1,1-Trifluoro- 2-Chloro-2- Bromo-Ethane \| C2HClBrF3 \| 1.869 \| 693 \| 2273.6 \| \| \| \| \| 1,2,2-Trifluoro trichloro- ethane (Freon 113) \| CCl2F-CClF2 \| 1.563 \| 783.7 (0°C) \| 2,571.2 (32°F) \| 3.44 \| \| \| \| d-1,3,3-Tri methylnor camphor \| C10H16O \| 0.947 \| 1320 \| 4330.7 \| 0.22 \| 2.367 \| \| \| Trinitrotoluene (43) \| C7H5(NO2)3 \| 1.64 \| 1,610 (81°C) \| 5,282.2 (178°F) \| \| \| \| \| Turpentine \| 0.88 \| 1255 \| 4117.5 \| 1.4 \| 15.064 \| \| \| \| Unisis 800 \| 0.87 \| 1346 \| 4416 \| \| \| \| \| \| Water, distilled (49, 50) \| H2O \| 0.996 \| 1498 \| 4914.7 \| -2.4 \| 1 \| 10.76 \| \| Water, heavy \| D2O \| 1400 \| 4593 \| \| \| \| \| \| Water, sea \| 1.025 \| 1531 \| 5023 \| -2.4 \| 1 \| 10.76 \| \| \| Wood Alcohol (40, 41) \| CH4O \| 0.791 (20°C) \| 1076 \| 3530.2 \| 2.92 \| 0.695 \| 7.478 \| \| Xenon (45) \| Xe \| 630 (-109°C) \| 2,067 (-164°F) \| \| \| \| \| \| m-Xylene (46) \| C8H10 \| 0.868 (15°C) \| 1,343 (20°C) \| 4,406.2 (68°F) \| 0.749 (15°C) \| 8.059 (59°F) \| \| \| o-Xylene (29, 46) \| C8H10 \| (20°C) 0.897 \| 1331.5 \| 4368.4 \| 4.1 \| (20°C) 0.903 \| (68°F) 9.716 \| \| p-Xylene (46) \| C8H10 \| 1,334 (20°C) \| 4,376.6 (68°F) \| 0.662 \| 7.123 \| \| \| \| Xylene hexafluoride \| C8H4F6 \| 1.37 \| 879 \| 2883.9 \| 0.613 \| 6.595 \| \| \| Zinc (7) \| Zn \| 3,298 (450°C) \| 10,820.2 (842°F) \| \| \| \| \| Ultrasonic Sound Speeds in Water at Selected Temperatures Sound speeds in water vary with temperature. All values are approximate.\| Water Temperature (°C) \| Water Temperature (°F) \| Sound Speed (m/s) \| Sound Speed (ft/s) \| --- --- \| \| 0.0 \| 32.0 \| 1402.0 \| 4600.0 \| \| 1.0 \| 33.8 \| 1407.0 \| 4616.0 \| \| 2.0 \| 35.6 \| 1412.0 \| 4633.0 \| \| 3.0 \| 37.4 \| 1417.0 \| 4649.0 \| \| 4.0 \| 39.2 \| 1421.0 \| 4662.0 \| \| 5.0 \| 41.0 \| 1426.0 \| 4679.0 \| \| 6.0 \| 42.8 \| 1430.0 \| 4692.0 \| \| 7.0 \| 44.6 \| 1434.0 \| 4705.0 \| \| 8.0 \| 46.4 \| 1439.0 \| 4721.0 \| \| 9.0 \| 48.2 \| 1443.0 \| 4734.0 \| \| 10.0 \| 50.0 \| 1447.0 \| 4748.0 \| \| 11.0 \| 51.8 \| 1451.0 \| 4761.0 \| \| 12.0 \| 53.6 \| 1455.0 \| 4774.0 \| \| 13.0 \| 55.4 \| 1458.0 \| 4784.0 \| \| 14.0 \| 57.2 \| 1462.0 \| 4797.0 \| \| 15.0 \| 59.0 \| 1465.0 \| 4807.0 \| \| 16.0 \| 60.8 \| 1469.0 \| 4820.0 \| \| 17.0 \| 62.6 \| 1472.0 \| 4830.0 \| \| 18.0 \| 64.4 \| 1476.0 \| 4843.0 \| \| 19.0 \| 66.2 \| 1479.0 \| 4853.0 \| \| 20.0 \| 68.0 \| 1482.0 \| 4862.0 \| \| 21.0 \| 69.8 \| 1485.0 \| 4872.0 \| \| 22.0 \| 71.6 \| 1488.0 \| 4882.0 \| \| 23.0 \| 73.4 \| 1491.0 \| 4892.0 \| \| 24.0 \| 75.2 \| 1493.0 \| 4899.0 \| \| 25.0 \| 77.0 \| 1496.0 \| 4908.0 \| \| 26.0 \| 78.8 \| 1499.0 \| 4918.0 \| \| 27.0 \| 80.6 \| 1501.0 \| 4925.0 \| \| 28.0 \| 82.4 \| 1504.0 \| 4935.0 \| \| 29.0 \| 84.2 \| 1506.0 \| 4941.0 \| \| 30.0 \| 86.0 \| 1509.0 \| 4951.0 \| \| 31.0 \| 87.8 \| 1511.0 \| 4958.0 \| \| 32.0 \| 89.6 \| 1513.0 \| 4964.0 \| \| 33.0 \| 91.4 \| 1515.0 \| 4971.0 \| \| 34.0 \| 93.2 \| 1517.0 \| 4977.0 \| \| 35.0 \| 95.0 \| 1519.0 \| 4984.0 \| \| 36.0 \| 96.8 \| 1521.0 \| 4990.0 \| \| 37.0 \| 98.6 \| 1523.0 \| 4997.0 \| \| 38.0 \| 100.4 \| 1525.0 \| 5004.0 \| \| 39.0 \| 102.2 \| 1527.0 \| 5010.0 \| \| 40.0 \| 104.0 \| 1528.0 \| 5013.0 \| \| 41.0 \| 105.8 \| 1530.0 \| 5020.0 \| \| 42.0 \| 107.6 \| 1532.0 \| 5026.0 \| \| 43.0 \| 109.4 \| 1534.0 \| 5033.0 \| \| 44.0 \| 111.2 \| 1535.0 \| 5036.0 \| \| 45.0 \| 113.0 \| 1536.0 \| 5040.0 \| \| 46.0 \| 114.8 \| 1538.0 \| 5046.0 \| \| 47.0 \| 116.6 \| 1539.0 \| 5049.0 \| \| 48.0 \| 118.4 \| 1540.0 \| 5053.0 \| \| 49.0 \| 120.2 \| 1541.0 \| 5056.0 \| \| 50.0 \| 122.0 \| 1543.0 \| 5063.0 \| \| 51.0 \| 123.8 \| 1543.0 \| 5063.0 \| \| 52.0 \| 125.6 \| 1544.0 \| 5066.0 \| \| 53.0 \| 127.4 \| 1545.0 \| 5069.0 \| \| 54.0 \| 129.2 \| 1546.0 \| 5072.0 \| \| 55.0 \| 131.0 \| 1547.0 \| 5076.0 \| \| 56.0 \| 132.8 \| 1548.0 \| 5079.0 \| \| 57.0 \| 134.6 \| 1548.0 \| 5079.0 \| \| 58.0 \| 136.4 \| 1549.0 \| 5082.0 \| \| 59.0 \| 138.2 \| 1550.0 \| 5086.0 \| \| 60.0 \| 140.0 \| 1550.0 \| 5086.0 \| \| 61.0 \| 141.8 \| 1551.0 \| 5089.0 \| \| 62.0 \| 143.6 \| 1552.0 \| 5092.0 \| \| 63.0 \| 145.4 \| 1552.0 \| 5092.0 \| \| 64.0 \| 147.2 \| 1553.0 \| 5095.0 \| \| 65.0 \| 149.0 \| 1553.0 \| 5095.0 \| \| 66.0 \| 150.8 \| 1553.0 \| 5095.0 \| \| 67.0 \| 152.6 \| 1554.0 \| 5099.0 \| \| 68.0 \| 154.4 \| 1554.0 \| 5099.0 \| \| 69.0 \| 156.2 \| 1554.0 \| 5099.0 \| \| 70.0 \| 158.0 \| 1554.0 \| 5099.0 \| \| 71.0 \| 159.8 \| 1554.0 \| 5099.0 \| \| 72.0 \| 161.6 \| 1555.0 \| 5102.0 \| \| 73.0 \| 163.4 \| 1555.0 \| 5102.0 \| \| 74.0 \| 165.2 \| 1555.0 \| 5102.0 \| \| 75.0 \| 167.0 \| 1555.0 \| 5102.0 \| \| 76.0 \| 168.8 \| 1555.0 \| 5102.0 \| \| 77.0 \| 170.6 \| 1554.0 \| 5099.0 \| \| 78.0 \| 172.4 \| 1554.0 \| 5099.0 \| \| 79.0 \| 174.2 \| 1554.0 \| 5099.0 \| \| 80.0 \| 176.0 \| 1554.0 \| 5099.0 \| \| 81.0 \| 177.8 \| 1554.0 \| 5099.0 \| \| 82.0 \| 179.6 \| 1553.0 \| 5095.0 \| \| 83.0 \| 181.4 \| 1553.0 \| 5095.0 \| \| 84.0 \| 183.2 \| 1553.0 \| 5095.0 \| \| 85.0 \| 185.0 \| 1552.0 \| 5092.0 \| \| 86.0 \| 186.8 \| 1552.0 \| 5092.0 \| \| 87.0 \| 188.6 \| 1552.0 \| 5092.0 \| \| 88.0 \| 190.4 \| 1551.0 \| 5089.0 \| \| 89.0 \| 192.2 \| 1551.0 \| 5089.0 \| \| 90.0 \| 194.0 \| 1550.0 \| 5086.0 \| \| 91.0 \| 195.8 \| 1549.0 \| 5082.0 \| \| 92.0 \| 197.6 \| 1549.0 \| 5082.0 \| \| 93.0 \| 199.4 \| 1548.0 \| 5079.0 \| \| 94.0 \| 201.2 \| 1547.0 \| 5076.0 \| \| 95.0 \| 203.0 \| 1547.0 \| 5076.0 \| \| 96.0 \| 204.8 \| 1546.0 \| 5072.0 \| \| 97.0 \| 206.6 \| 1545.0 \| 5069.0 \| \| 98.0 \| 208.4 \| 1544.0 \| 5066.0 \| \| 99.0 \| 210.2 \| 1543.0 \| 5063.0 \| \| 100.0 \| 212.0 \| 1543.0 \| 5063.0 \| \| 104.0 \| 220.0 \| 1538.0 \| 5046.0 \| \| 110.0 \| 230.0 \| 1532.0 \| 5026.0 \| \| 116.0 \| 240.0 \| 1524.0 \| 5000.0 \| \| 121.0 \| 250.0 \| 1526.0 \| 5007.0 \| \| 127.0 \| 260.0 \| 1507.0 \| 4944.0 \| \| 132.0 \| 270.0 \| 1497.0 \| 4912.0 \| \| 138.0 \| 280.0 \| 1487.0 \| 4879.0 \| \| 143.0 \| 290.0 \| 1476.0 \| 4843.0 \| \| 149.0 \| 300.0 \| 1465.0 \| 4807.0 \| \| 154.0 \| 310.0 \| 1453.0 \| 4767.0 \| \| 160.0 \| 320.0 \| 1440.0 \| 4725.0 \| \| 166.0 \| 330.0 \| 1426.0 \| 4679.0 \| \| 171.0 \| 340.0 \| 1412.0 \| 4633.0 \| \| 177.0 \| 350.0 \| 1398.0 \| 4587.0 \| \| 182.0 \| 360.0 \| 1383.0 \| 4538.0 \| \| 188.0 \| 370.0 \| 1368.0 \| 4488.0 \| \| 193.0 \| 380.0 \| 1353.0 \| 4439.0 \| \| 199.0 \| 390.0 \| 1337.0 \| 4387.0 \| \| 204.0 \| 400.0 \| 1320.0 \| 4331.0 \| \| 210.0 \| 410.0 \| 1302.0 \| 4272.0 \| \| 216.0 \| 420.0 \| 1283.0 \| 4210.0 \| \| 221.0 \| 430.0 \| 1264.0 \| 4147.0 \| \| 227.0 \| 440.0 \| 1244.0 \| 4082.0 \| \| 232.0 \| 450.0 \| 1220.0 \| 4003.0 \| \| 238.0 \| 460.0 \| 1200.0 \| 3937.0 \| \| 243.0 \| 470.0 \| 1180.0 \| 3872.0 \| \| 249.0 \| 480.0 \| 1160.0 \| 3806.0 \| \| 254.0 \| 490.0 \| 1140.0 \| 3740.0 \| \| 260.0 \| 500.0 \| 1110.0 \| 3642.0 \| Quick Links About Us Join Our Crew What We Do Industries Products Manufacturers Field Services Resources Clipper Academy Nautical Fun Contact Us Toll Free (844) 880-AHOY (2469) Clipper Controls Inc. 660 Bercut Drive Sacramento CA 95811 Let's Connect on LinkedIn! 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14723	https://achievethecore.org/coherence-map/4/18	K.CC Counting And Cardinality K.OA Operations And Algebraic Thinking K.NBT Number And Operations In Base Ten K.MD Measurement And Data K.G 1st Grade 1.OA Operations And Algebraic Thinking 1.NBT Number And Operations In Base Ten 1.MD Measurement And Data 1.G Geometry 2nd Grade 2.OA Operations And Algebraic Thinking 2.NBT Number And Operations In Base Ten 2.MD Measurement And Data 2.G Geometry 3rd Grade 3.OA Operations And Algebraic Thinking 3.NBT Number And Operations In Base Ten 3.NF Number And Operations-Fractions 3.MD Measurement And Data 3.G Geometry 4th Grade 4.OA Operations And Algebraic Thinking 4.NBT Number And Operations In Base Ten 4.NF Number And Operations-Fractions 4.MD Measurement And Data 4.G Geometry 5th Grade 5.OA Operations And Algebraic Thinking 5.NBT Number And Operations In Base Ten 5.NF Number And Operations-Fractions 5.MD Measurement And Data 5.G Geometry 6th Grade 6.RP Ratios And Proportional Relationships 6.NS The Number System 6.EE Expressions And Equations 6.G Geometry 6.SP Statistics And Probability 7th Grade 7.RP Ratios And Proportional Relationships 7.NS The Number System 7.EE Expressions And Equations 7.G Geometry 7.SP Statistics And Probability 8th Grade 8.NS The Number System 8.EE Expressions And Equations 8.F Functions 8.G Geometry 8.SP Statistics And Probability High School HS.N Number and Quantity HS.A Algebra HS.F Functions HS.M Modeling HS.G Geometry HS.S Statistics and Probability Extend Understanding Of Fraction Equivalence And Ordering. Major Cluster 4.NF.A.1 Explain why a fraction) a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models), with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. 4.NF.A.2 Compare two fractions) with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model). Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. Build Fractions From Unit Fractions By Applying And Extending Previous Understandings Of Operations On Whole Numbers. Major Cluster 4.NF.B.3 Understand a fraction) a/b with a > 1 as a sum of fractions 1/b. Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. 4.NF.B.3.a Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. 4.NF.B.3.b Decompose a fraction) into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model). Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8. 4.NF.B.3.c Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction), and/or by using properties of operations) and the relationship between addition and subtraction. 4.NF.B.3.d Solve word problems involving addition and subtraction of fractions) referring to the same whole and having like denominators, e.g., by using visual fraction models) and equations to represent the problem. 4.NF.B.4 Apply and extend previous understandings of multiplication to multiply a fraction) by a whole number). Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. Understand a fraction) a/b as a multiple of 1/b. For example, use a visual fraction model) to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4). Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction) by a whole number). For example, use a visual fraction model) to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n × (a/b) = (n × a)/b.) Solve word problems involving multiplication of a fraction) by a whole number), e.g., by using visual fraction models) and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? Understand Decimal Notation For Fractions, And Compare Decimal Fractions. Major Cluster 4.NF.C.5 Express a fraction) with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100. For example, express 3/10 as 30/100, and add 3/10 + 4/100 = 34/100. Students who can generate equivalent fractions can develop strategies for adding fractions with unlike denominators in general. But addition and subtraction with unlike denominators in general is not a requirement at this grade. Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. 4.NF.C.6 Use decimal notation for fractions) with denominators 10 or 100. For example, rewrite 0.62 as 62/100; describe a length as 0.62 meters; locate 0.62 on a number line diagram). Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. 4.NF.C.7 Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model. Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. Send Feedback
14724	https://blog.mensor.com/blog/what-is-air-data	What is Air Data Calibration? Search Company Products Industries Service / Calibration Downloads Career Contact Company Products Industries Service / Calibration Downloads Career Contact Search this site on Google Search Google Mensor Mensor Calibration Blog What is Air Data Calibration? Share What is Air Data Calibration? Air data calibration is a measurement of the air mass surrounding an airplane. The two physical characteristics measured are pressure and temperature. Air data is acquired through various sensors on the aircraft and is used to calculate altitude, speed, rate of climb or decent, and angle-of-attack or angle-of-sideslip. What pressure data is measured? The pressure measurements in air data calibration consist of static pressure (Ps) and total pressure (Pt). Definitions: Static Pressure (Ps) is the absolute pressure of still air surrounding the aircraft. This is the barometric pressure at the altitude where the aircraft is traveling and is independent of any pressure disturbances caused by the motion of the aircraft. Total Pressure (Pt) is the sum of the local atmospheric pressure (Ps) and the impact pressure (Qc) caused by the aircraft's motion through the air. Impact Pressure (Qc) is the pressure a moving stream of air produces against a surface that brings part of the moving stream to rest. It is the difference between the total pressure (Pt) and the static pressure (Ps). These pressure properties are related by the formula: Qc = Pt - Ps Ps and Pt are acquired by one or more pitot static probes on the aircraft body. Qc is calculated from these values. Ps is used to calculate the altitude of the aircraft. Qc is used to calculate the speed of the aircraft. How is altitude determined? Altitude is derived from a series of equations using the static pressure. Altitude calculations are based on a "standard atmosphere," which assumes a known relationship between pressure, temperature and atmospheric density. Generally, the lower the static pressure, the higher the altitude. Why is temperature measured? Temperature is measured in order to calculate true airspeed (the actual speed of the plane through air) from indicated airspeed and temperature. A temperature probe on the body of the aircraft acquires temperature values. Pressure and temperature data sensing is sometimes combined in the same sensor. Indicated airspeed (IAS), true airspeed (TAS) and Mach number are versions of an aircraft’s speed and have a temperature component incorporated. Definitions: Indicated Airspeed (IAS): The speed of an aircraft relative to the surrounding air. It is uncorrected for any installation or instrument errors. Indicated airspeed equals true airspeed at standard sea level conditions only, and it is a function of impact pressure (Qc). True Airspeed (TAS): Indicated airspeed corrected for nonstandard temperatures that can be determined using Mach number and total temperature. It is the actual aircraft speed through the air mass. Mach Number: The ratio of true airspeed and the speed of sound in the surrounding air. The speed of sound is proportional to the square root of the average temperature. Mach number is calculated using the ratio of impact pressure (Qc) to static pressure (Ps). Reference: Rosemount Aerospace, Air Data Handbook Download the Air Data Infographic Related Reading: A Brief Introduction to Pressure What is Pressure? Definitions and Types Explained What is Bidirectional Pressure? What is Differential Pressure? What is Vacuum Pressure? Effects of Pressure: The Marshmallow Experiment tagLearningCalibrationair data First Name Last Name Email Comment Author ### Larry Mock Marketing Communications Manager LinkedIn Categories Pressure Temperature Applications Learning Products Subscribe for Blog Updates Email Notification Frequency accuracy (12) aerospace (2) air data (5) air data test (1) analog (2) Applications (21) automation (8) automotive (4) aviation (1) barometers (1) barometric pressure (2) branding (2) Calibration (42) company (2) compressibility (3) contamination prevention (2) continuous output (2) continuous readings (1) control stability (2) controllers/calibrators (21) corrections (3) cpc3050 (3) cpc6020 (1) cpc6050 (2) cpc8000 (1) cpc8000-h (4) custom system (3) customer service (2) data logging (1) data stream (1) dead weight testers (2) deadweight testers (11) dry-well (1) end-of-repair (1) healthcare (4) how to (10) hydraulic (5) indicators/gauges (7) instructional (1) Learning (70) low pressure (5) manufacturing (2) measurement specifications (11) metrology (7) nuclear (2) oil and gas (2) overpressure protection (1) pneumatic (6) portable instruments (2) Pressure (53) pressure basics (5) Pressure Controllers (10) pressure media (7) primary standards (2) process tools (3) product return (1) product return form (1) Products (31) service (6) software (1) stability (1) STEP files (1) Temperature (9) temperature calibration (1) temperature calibration equipment (1) temperature calibrator (1) temperature sensor calibration (1) thermometry (6) traceability (2) transducers/sensors (24) uncertainty (3) © 2023 Mensor Privacy policy Legal notice
14725	https://www.youtube.com/watch?v=Y2O02fTACTY	Grade 9 Math - Volume of a cylinder and a cone 1Mjourney 3720 subscribers 12 likes Description 644 views Posted: 25 Jan 2021 Grade 9 Math What is the volume of a cylinder and cone? The formula and explanation is provided in the Grade 9 video! The proof of showing why a cone is 1/3 of a cylinder can be found here: And if you like Calculus, check out this amazing summary of deriving the formula of a cone: Thanks Peter for doing this, awesome job! If this video helps one person, then it has served its purpose! help1inspire1M Entire High School Math Video series: 1mjourney.com (Click on 1Mjourney - Math) Free Math Textbooks Free Graphing Calculator Tool: Free 3D Graphing Tools: Textbook References: Grade 9: Mathematics Dearling, Erdman, Ferneyhough, McCudden, McLaren, Meisel, Speijer, Principles of Mathematics 9, McGraw-Hill Ryerson (2006) Grade 10: Mathematics Canton, Dearling, Erdman, McCudden, McLaren, Meisel, Speijer, Principles of Mathematics 10, McGraw-Hill Ryerson (2007) Small, Kirkpatrick, Trew, D’Agostino, Rodger, Chilvers, Macpherson, Bourassa, Nelson Principles of Mathematics 10 (2009) Grade 11: Functions Kirkpatrick, Alldred, Dmytriw, Godin, Lillo, Pilmer, Trew, Walker; Functions 11, Nelson (2008) Speijer, Meisel, Petro, Stewart, Vukets, Functions 11, McGraw-Hill Ryerson (2009) Grade 12: Data Management and Probability Canton, Erdman, Irvine, Lim, McLaren, Meisel, Miller, Speijer; Mathematics of Data Management, McGraw-Hill Ryerson, (2002) Zimmer, Cooke, Craven, Farahani, Steinke, Kirkpatrick; Mathematics of Data Management, Nelson (2003) Grade 12: Advanced Functions Alldred, Chilvers, Farahani, Farentino, Lillo, Macpherson, Rodger, Trew; Advanced Functions, Nelson (2009) Erdman, Lenjosek, Meisel, Speijer; Advanced Functions 12, McGraw-Hill Ryerson, (2008) Salas, Hille, Etgen, Calculus, One and Several Variables, Wiley 8th ed. (1999) Grade 12: Calculus and Vectors Erdman, Ferguson, Lenjosek, Petro; Calculus and Vectors 12, McGraw-Hill, (2008) Salas, Hille, Etgen, Calculus, One and Several Variables, Wiley 8th ed. (1999) 2 comments Transcript: okay two one welcome back so volume of cylinder and cone so let's just jot these down i'll show you hopefully where these come from and what you can do especially in the younger grades so this is assuming that you're kind of in grade 8 or 9 and you're trying to learn about this volume if you were in the later years i would have shown a little bit different approach to showing you the formulas all right but i will maybe put in the show notes one of my favorite ones which it does use a little bit of calculus so obviously if you're in grade nine you're still might not be there but you can certainly take a look and see how much of it you would understand and where you would get stuck okay but i'll try to keep it as simple as possible which is probably best so volume of a cylinder and a cone so if you're talking about the volume now i just want to remind you okay so for prisms which i discussed i'll put a link up above there you can take a look if you haven't seen that one so for prisms we know that the volume is equal to the base multiplied by the height now because we have a cylinder here which we know the base of and we kind of unknow the height of we can measure those things so we can figure out exactly what the volume of this cylinder is so our base which is this nice region you know right here so that's our base well we know that the base is pi r squared so that is the area so that is the base of the okay so let's say if it was some kind of a prism so in this case it's a cylinder so we know our base so that's right here now we know our height so we know that the volume of a cylinder is pi r squared times h that shouldn't surprise you right because you have the base and then all you're doing is you're going all the way up so if we would fill this up with water or some liquid that would be the volume that we would generate okay and then depending on what the units of r and h are now please make sure that those units are um aligned right so they're consistent so if it's in centimeters then the h is in centimeters if it's all in inches it's all in inches it's not that you have one in meters one in feet or something like that so always make sure that you are having the same unit so cylinder is not bad now what about if you had to figure out a cone so i'm not going to go and try to use calculus here i'll try to use a little bit of intuition and what you could do to practice this out all right to find the volume of a cone so our goal is to find the volume of a cone what it is now from above we know that the volume of the cylinder was equal to pi r squared h now a cone is clearly not a cylinder however what if our cone so our cone right here had the same base so if our base what i'm going to do here is i'm going to take this base all right and let's say that's the base of our cone which is the same as the cylinder now if you filled this cylinder up let's say that you could fill it up with water then we would know what the volume is the volume would have been this right here that would have been our volume of that entire cylinder now if you had a cone that you could put inside of this cylinder with the exact same base you know and you'll be basically placing it all the way inside then what you would run into is you know so i'm going to draw this cone right here you placed it all the way inside the volume of that cylinder is basically the same as whatever water okay if you filled this up with okay so if you fill this up with some liquid so whatever liquid spills out would be the actual volume of your code because whatever has spilled out okay is whatever has been displaced but you putting in the cone inside of that cylinder and now when you do that exercise it actually turns out that what spills out is going to be a proportion okay of the entire cylinder and it turns out that a third of the volume will spill out and that third is the volume of your code so this isn't some kind of a proof by some mathematical means or some others it is basically just using your intuition and thinking about okay what are these volumes all about and in reality that's kind of how it all started right so if you are experimenting you know you're going to start making some hypothesis of like hmm what is the volume of a cone okay and i think the volume of a cone is let's say a quarter of the cylinder and then you would test out your hypothesis and you would turn out and you would see if you kept doing these experiments that you basically would realize that no it's not a quarter it's not a half it's not anything else it is actually a third all right so that's what i wanted to show you so these are the two equations so this one is for the cone and this one is for the cylinder itself and now you can start applying this to many uh other problems which i will do because i'm gonna do a summary test okay towards the end of this little unit that i'm working on all right okay so thanks for watching and see you in a future video bye everybody
14726	https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Spin-orbit_Coupling/Atomic_Term_Symbols	Atomic Term Symbols - Chemistry 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 Spin-orbit Coupling Electronic Spectroscopy { } { Atomic_Term_Symbols : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Molecular_Term_Symbols : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", The_Russell_Saunders_Coupling_Scheme : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { Circular_Dichroism : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electronic_Spectroscopy:Application" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electronic_Spectroscopy-_Interpretation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Electronic_Spectroscopy_Basics : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Fluorescence_and_Phosphorescence : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Jablonski_diagram : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Metal_to_Ligand_and_Ligand_to_Metal_Charge_Transfer_Bands : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Radiative_Decay : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Selection_Rules_for_Electronic_Spectra_of_Transition_Metal_Complexes : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Spin-orbit_Coupling" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Two-photon_absorption" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 30 Jan 2023 07:07:06 GMT Atomic Term Symbols 1775 1775 admin { } Anonymous Anonymous User 2 false false [ "article:topic", "atomic term symbols", "Russell-Saunders coupling", "showtoc:no", "license:ccby", "licenseversion:40", "author@Anda Zheng" ] [ "article:topic", "atomic term symbols", "Russell-Saunders coupling", "showtoc:no", "license:ccby", "licenseversion:40", "author@Anda Zheng" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Bookshelves 3. Physical & Theoretical Chemistry 4. Supplemental Modules (Physical and Theoretical Chemistry) 5. Spectroscopy 6. Electronic Spectroscopy 7. Spin-orbit Coupling 8. Atomic Term Symbols Expand/collapse global location Atomic Term Symbols Last updated Jan 30, 2023 Save as PDF Spin-orbit Coupling Molecular Term Symbols Page ID 1775 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. History 2. Term Symbols 3. Angular momenta of an electron 4. Coupling of the electronic angular momenta 1. LS coupling 2. jj coupling Term symbols for an Electron Configuration Using group theory to determine term symbols Determining the ground state Electronic transitions References Problems Answers In electronic spectroscopy, an atomic term symbol specifies a certain electronic state of an atom (usually a multi-electron one), by briefing the quantum numbers for the angular momenta of that atom. The form of an atomic term symbol implies Russell-Saunders coupling. Transitions between two different atomic states may be represented using their term symbols, to which certain rules apply. History At the beginning, the spectroscopic notation for term symbols was derived from an obsolete system of categorizing spectral lines. In 1885, Johann Balmer, a Swiss mathematician, discovered the Balmer formula for a series of hydrogen emission lines. (1)λ=B⁡(m 2 m 2−4) where B is constant, and m is an integer greater than 2. Later it was extended by Johannes Rydberg and Walter Ritz. Yet this principle could hardly explain the discovery of fine structure, the splitting of spectral lines. In spectroscopy, spectral lines of alkali metals used to be divided into categories: sharp, principal, diffuse and fundamental, based on their fine structures. These categories, or "term series," then became associated with atomic energy levels along with the birth of the old quantum theory. The initials of those categories were employed to mark the atomic orbitals with respect to their azimuthal quantum numbers. The sequence of "s, p, d, f, g, h, i, k..." is known as the spectroscopic notation for atomic orbitals. By introducing spin as a nature of electrons, the fine structure of alkali spectra became further understood. The term "spin" was first used to describe the rotation of electrons. Later, although electrons have been proved unable to rotate, the word "spin" is reserved and used to describe the property of an electron that involves its intrinsic magnetism. LS coupling was first proposed by Henry Russell and Frederick Saunders in 1923. It perfectly explained the fine structures of hydrogen-like atomic spectra. The format of term symbols was developed in the Russell-Saunders coupling scheme. Term Symbols In the Russell-Saunders coupling scheme, term symbols are in the form of 2 S+1 L J, where S represents the total spin angular momentum, L specifies the total orbital angular momentum, and J refers to the total angular momentum. In a term symbol, L is always an upper-case from the sequence "s, p, d, f, g, h, i, k...", wherein the first four letters stand for sharp, principal, diffuse and fundamental, and the rest follow in an alphabetical pattern. Note that the letter j is omitted. Angular momenta of an electron In today's physics, an electron in a spherically symmetric potential field can be described by four quantum numbers all together, which applies to hydrogen-like atoms only. Yet other atoms may undergo trivial approximations in order to fit in this description. Those quantum numbers each present a conserved property, such as the orbital angular momentum. They are sufficient to distinguish a particular electron within one atom. The term "angular momentum" describes the phenomenon that an electron distributes its position around the nucleus. Yet the underlying quantum mechanics is much more complicated than mere mechanical movement. The azimuthal angular momentum: The azimuthal angular momentum (or the orbital angular momentum when describing an electron in an atom) specifies the azimuthal component of the total angular momentum for a particular electron in an atom. The orbital quantum number, l, one of the four quantum numbers of an electron, has been used to represent the azimuthal angular momentum. The value of l is an integer ranging from 0 to n-1, while n is the principal quantum number of the electron. The limits of l came from the solutions of the Schrödinger Equation. The intrinsic angular momentum: The intrinsic angular momentum (or the spin) represents the intrinsic property of elementry particles, and the particles made of them. The inherent magnetic momentum of an electron may be explained by its intrinsic angular momentum. In LS-coupling, the spin of an electron can couple with its azimuthal angular momentum. The spin quantum number of an electron has a value of either ½ or -½, which reflects the nature of the electron. Coupling of the electronic angular momenta Coupling of angular momenta was first introduced to explain the fine structures of atomic spectra. As for LS coupling, S, L, J and M J are the four "good" quantum numbers to describe electronic states in lighter atoms. For heavier atoms, jj coupling is more applicable, where J, M J, M L and M s are "good" quantum numbers. LS coupling LS coupling, also known as Russell-Saunders coupling, assumes that the interaction between an electron's intrinsic angular momentum s and its orbital angular momentum L is small enough to be considered as an perturbation to the electronic Hamiltonian. Such interactions can be derived in a classical way. Let's suppose that the electron goes around the nucleus in a circular orbit, as in Bohr model. Set the electron's velocity to be V e. The electron experiences a magnetic field B due to the relative movement of the nucleus (2)B=1 m e⁢c 2⁢(E×p)=Z⁢e 4⁢π⁢ϵ 0⁢m e⁢c 2⁢r 3⁢L while E is the electric field at the electron due to the nucleus, p the classical monumentum of the electron, and r the distance between the electron and the nucleus. The electron's spin s brings a magnetic dipole moment μ s (3)μ s=−g s⁢e⁡s 2⁢m e⁢(L^⋅s^) where g s is the gyromagnetic ratio of an electron. Since the potential energy of the coulumbic attraction between the electron and nucleus is 6 (4)V⁡(r)=−Z⁢e 2 4⁢π⁢ε 0⁢r the interaction between μs and B is (5)H^s⁢o=g s 2⁢m e 2⁢c 2⁢Z⁢e 2 4⁢π⁢ϵ 0⁢r 2⁢(L⋅s)=g s 2⁢m e 2⁢c 2⁢∂V∂r⁢(L⋅s) After a correction due to centripetal acceleration 10, the interaction has a format of (6)H^s⁢o=1 2⁢μ 2⁢c 2⁢∂V∂r⁢(L⋅s) Thus the coupling energy is (7)⟨ψ n⁢l⁢m⁢\|H^s⁢o\|⁢ψ n⁢l⁢m⟩ In lighter atoms, the coupling energy is low enough be treated as a first-order perturbation to the total electronic Hamiltonian, hence LS coupling is applicable to them. For a single electron, the spin-orbit coupling angular momentum quantum number j has the following possible values j = \|l-s\|, ..., l+s if the total angular momentum J is defined as J = L + s. The azimuthal counterpart of j is m j, which can be a whole number in the range of [-j, j]. The first-order perturbation to the electronic energy can be deduced so 6 (8)ℏ 2⁢[j⁢(j+1)−l⁢(l+1)−s⁢(s+1)]4⁢μ 2⁢c 2⁢∫0∞r 2 d r⁢1 r⁢∂V∂r⁢R n⁢l 2⁡(r) Above is about the spin-orbit coupling of one electron. For many-electron atoms, the idea is similar. The coupling of angular momenta is (9)J=L+S thereby the total angular quantum number J = \|L-S\|, ..., L+S where the total orbital quantum number (10)L=∑i l i and the total spin quantum number (11)S=∑i s i While J is still the total angular momentum, L and S are the total orbital angular momentum and the total spin, respectively. The magnetic momentum due to J is (12)μ J=−g J⁡e⁢J 2⁢m e wherein the Landé g factor is (13)g J=1+J⁢(J+1)+S⁢(S+1)−L⁢(L+1)2⁢J⁢(J+1) supposing the gyromagnetic ratio of an electron is 2. jj coupling For heavier atoms, the coupling between the total angular momenta of different electrons is more significant, causing the fine structures not to be "fine" any more. Therefore the coupling term can no more be considered as a perturbation to the electronic Hamiltonian, so that jj coupling is a better way to quantize the electron energy states and levels. For each electron, the quantum number j = l + s. For the whole atom, the total angular momentum quantum number (14)J=∑i j i Term symbols for an Electron Configuration Term symbols usually represent electronic states in the Russell-Saunders coupling scheme, where a typical atomic term symbol consists of the spin multiplicity, the symmetry label and the total angular momentum of the atom. They have the format of (15)2⁢S+1⁢L J such as 3 D 2, where S = 1, L = 2, and J = 2. Here is a commonly used method to determine term symbols for an electron configuration. It requires a table of possibilities of different "micro states," which happened to be called "Slater's table".6 Each row of the table represents a total magnetic quantum number, while each column does a total spin. Using this table we can pick out the possible electronic states easily since all terms are concentric rectangles on the table. The method of using a table to count possible "microstates" has been developed so long ago and honed by so many scientists and educators that it is hard to accredit a single person. Let's take the electronic configuration of d 3 as an example. In the Slater's table, each cell contains the number of ways to assign the three electrons quantum numbers according to the M S and M L values. These assignments follow Pauli's exclusion law. The figure below shows an example to find out how many ways to assign quantum numbers to d 3 electrons when M L = 3 and M S = -1/2. M S -3/2-1/2 1/2 3/2 M L5 0 1 1 0 4 0 2 2 0 3 1 4 4 1 2 1 6 6 1 1 2 8 8 2 0 2 8 8 2 -1 2 8 8 2 -2 1 6 6 1 -3 1 4 4 1 -4 0 2 2 0 -5 0 1 1 0 Now we start to subtract term symbols from this table. First there is a 2 H state. And now it is subtracted from the table. M S -3/2-1/2 1/2 3/2 M L5 0 0 0 0 4 0 1 1 0 3 1 3 3 1 2 1 5 5 1 1 2 7 7 2 0 2 7 7 2 -1 2 7 7 2 -2 1 5 5 1 -3 1 3 3 1 -4 0 1 1 0 -5 0 0 0 0 And now is a 4 F state. After being subtracted by 4 F, the table becomes M S -3/2-1/2 1/2 3/2 M L5 0 0 0 0 4 0 1 1 0 3 0 2 2 0 2 0 4 4 0 1 1 6 6 1 0 1 6 6 1 -1 1 6 6 1 -2 0 4 4 0 -3 0 2 2 0 -4 0 1 1 0 -5 0 0 0 0 And now 2 G. M S -3/2-1/2 1/2 3/2 M L5 0 0 0 0 4 0 0 0 0 3 0 1 1 0 2 0 3 3 0 1 1 5 5 1 0 1 5 5 1 -1 1 5 5 1 -2 0 3 3 0 -3 0 1 1 0 -4 0 0 0 0 -5 0 0 0 0 Now 2 F. M S -3/2-1/2 1/2 3/2 M L5 0 0 0 0 4 0 0 0 0 3 0 0 0 0 2 0 2 2 0 1 1 4 4 1 0 1 4 4 1 -1 1 4 4 1 -2 0 2 2 0 -3 0 0 0 0 -4 0 0 0 0 -5 0 0 0 0 Here in the table are two 2 D states. M S -3/2-1/2 1/2 3/2 M L5 0 0 0 0 4 0 0 0 0 3 0 0 0 0 2 0 0 0 0 1 1 2 2 1 0 1 2 2 1 -1 1 2 2 1 -2 0 0 0 0 -3 0 0 0 0 -4 0 0 0 0 -5 0 0 0 0 4 P. M S -3/2-1/2 1/2 3/2 M L5 0 0 0 0 4 0 0 0 0 3 0 0 0 0 2 0 0 0 0 1 0 1 1 0 0 0 1 1 0 -1 0 1 1 0 -2 0 0 0 0 -3 0 0 0 0 -4 0 0 0 0 -5 0 0 0 0 And the final deducted state is 2 P. So in total the possible states for a d 3 configuration are 4 F, 4 P, 2 H, 2 G, 2 F, 2 D, 2 D and 2 P. Taken J into consideration, the possible states are: (16)4⁢F 2⁡4⁢F 3⁡4⁢F 4⁡4⁢P 0⁢4⁢P 1⁢4⁢P 2⁢2⁢H 9 2⁡2⁢H 11 2⁡2⁢G 9 2⁡2⁢G 7 2⁡2⁢F 7 2⁡2⁢F 5 2⁡2⁢D 5 2⁢2⁢D 5 2⁢2⁢D 3 2⁢2⁢D 3 2⁢2⁢P 3 2⁢2⁢P 3 2 For lighter atoms before or among the first-row transition metals, this method works well. Using group theory to determine term symbols Another method is to use direct products in group theory to quickly work out possible term symbols for a certain electronic configuration. Basically, both electrons and holes are taken into consideration, which naturally results in the same term symbols for complementary configurations like p 2 vs p 4. Electrons are categorized by spin, therefore divided into two categories, α and β, as are holes: α stands for +1, and βstands for -1, or vice versa. Term symbols of different possible configurations within one category are given. The term symbols for the total electronic configuration are derived from direct products of term symbols for different categories of electrons. For the p 3 configuration, for example, the possible combinations of different categories are e α 3, e β 3, e α 2 e β and e α e β 2. The first two combinations were assigned the partial term of S.6 As e α 2 and e β were given an P symbol, the combination of them gives their direct product P × P = S + [P] + D. The direct product for e α and e β 2 is also P × P = S + [P] + D. Considering the degeneracy, eventually the term symbols for p 3 configuration are 4 S, 2 D and 2 P. There is a specially modified version of this method for atoms with 2 unpaired electrons. The only step gives the direct product of the symmetries of the two orbitals. The degeneracy is still determined by Pauli's exclusion. Determining the ground state In general, states with a greater degeneracy have a lower energy. For one configuration, the level with the largest S, which has the largest spin degeneracy, has the lowest energy. If two levels have the same S value, then the one with the larger L (and also the larger orbital degeneracy) have the lower energy. If the electrons in the subshell are fewer than half-filled, the ground state should have the smallest value of J, otherwise the ground state has the greatest value of J. Electronic transitions Electrons of an atom may undergo certain transitions which may have strong or weak intensities. There are rules about which transitions should be strong and which should be weak. Usually an electronic transition is excited by heat or radiation. Electronic states can be interpreted by solutions of Schrödinger's equation. Those solutions have certain symmetries, which are a factor of whether transitions will be allowed or not. The transition may be triggered by an electric dipole momentum, a magnetic dipole momentum, and so on. These triggers are transition operators. The most common and usually most intense transitions occur in an electric dipolar field, so the selection rules are Δ L= 0, ±1 except L = 0 ‡ L' = 0 Δ S = 0 Δ J = 0; ±1 except J = 0 ‡ J' = 0 where a double dagger means not combinable. For jj coupling, only the third rule applies with an addition rule: Δ j = 0; ±1. References W. Ritz (1908), "On a New Law of Series Spectra", Astrophysical Journal 28(10): 237. doi: 10.1086/141591 §7.12, Jeremy B. Tatum, Stellar Atomsphere, online book. Accessed on line February 6, 2011. Herzberg, Gerhard (1945). Atomic Spectra and Atomic Structure. New York: Dover. pp. 54–5. ISBN 0-486-60115-3. Levine, Ira (2000). Quantum Chemistry (5 ed.). Prentice Hall. pp. 144–145. ISBN 0-13-685512-1 George Kean Sweetnam, The Command of Light - google books. P 182. W. S. Struve (1989), Fundamentals of Molecular Spectroscopy, John Wiley & Sons, Inc. pp. 44-46, 61-62. ISBN 0-471-85424-7 Niels Bohr (1913). "On the Constitution of Atoms and Molecules, Part I". Philosophical Magazine 26: 1–24 M. H. Nayfeh, M. K (1985). Brussel, Electricity and Magnetism, Wiley, New York. ISBN 0-471-87681-X E. Merzbacher (1966), Quantum Mechanics, Wiley, New York. W. H. Furry (1955), "Lorentz transformation and the Thomas precession", American Journal of Physics 23(8); 517-525. doi: 10.1119/1.1934085 Darl H. McDaniel, "Spin factoring as an aid in the determination of spectroscopic terms", Journal of Chemical Education 54(3):147 (1977). doi: 10.1021/ed054p147 Fermi, Enrico (1926). "Sulla quantizzazione del gas perfetto monoatomico" (in Italian). Rend. Lincei 3: 145–9., translated as On the Quantization of the Monoatomic Ideal Gas. 1999-12-14. Problems Potassium has the electronic configuration of [Ar]4s 1, what are the possible term symbols of a neutral K atom? What is the ground state of Cr 2+? Specify the value of J. Why does not the spin selection rule apply for electronic transitions under jj coupling? Why do different term symbols appear as rectangles on the Slater's table? Tanabe-Sugano diagrams tell us the order of energy levels in a complex, usually the order of d electron levels. Look up for the possible term symbols for the d 2 configuration, and determine how many fundamentals (transition from the ground state to an excited state) may occur when Δ = 0 i.e. minor field due to ligands. Answers 2 S. 5 D. Under jj coupling the spin quantum number is not a "good" quantum number any more, which means it cannot describe and differentiate electronic states properly. A certain term symbol represents a certain value of L as well as of S, which implies limited possible values of M L and M s. Following Pauli's exclusion rule, there must be only one possible way to assign the configuration a certain value of M L and M s. No appropriate answer. Atomic Term Symbols is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Anda Zheng. Back to top Spin-orbit Coupling Molecular Term Symbols Was this article helpful? 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14727	https://web.njit.edu/~kn328/files/math645.pdf	Real Analysis: Solved Problems From Royden & Fitzpatrick 4th Edition Kristian Nestor 1 Problem 50 A function f : E →R is Lipschitz if ∃c ≥0 for which \|f(x) −f(y)\| ≤c\|x −y\|, ∀x, y ∈E, but whenever \|x −y\| < δ we have that \|f(x) −f(y)\| ≤c\|x −y\| < cδ = ϵ, ∀x, y ∈E. Therefore, f is uniformly continuous on E. In order to show that there are uniformly continuous functions that are not Lipschitz we just have to find or create such a function. Consider f(x) = √x, 0 ≤x ≤1, which is uniformly continious, since every continuous function on a closed and bounded interval is uniformly continuous, but we can also show that is actually uniformly continuous, as ∀x, y ∈[0, 1], \|f(x) −f(y)\| = \|√x −√y\| = q \|√x −√y\|\|√x −√y\| ≤ q \|√x −√y\|\|√x + √y\| = p \|x −y\| Therefore, whenever \|x −y\| < δ we have that \|f(x) −f(y)\| ≤ p \|x −y\| < √ δ = ϵ. Since, δ does not depend on x, y, f is uniformly continuous. Suppose, that f is Lipschitz then ∃c ≥0 : ∀x, y ∈E \|f(x) −f(y)\| < c\|x −y\| ⇐ ⇒c ≥\|f(x) −f(y)\| \|x −y\| = \|√x −√y\| \|x −y\| = \|√x −√y\|\|√x + √y\| \|x −y\|\|√x + √y\| = 1 \|√x + √y\| But the quantity 1 \|√x+√y\| is unbounded on [0,1] and thus we can make it as large as we want so that there is no c that can exceed it. Hence, f is not Lipschitz. Problem 53 →From Heine-Borel theorem we know that if a set E is closed and bounded every open cover of E has a finite subcover. ←Let H be any collection of open sets such that E ⊂∪H∈HH. We assume that ∃n ∈ N : E ⊂∪n i=1Hi, Hi ∈H, i = 1, 2, ..., n.. Define, H = {x ∈E : x −ϵ < x < x + ϵ} which is an open cover for E. By assumption, there is a finite subcover, Hi = {x ∈E : xi −ϵ < x < xi +ϵ} : E ⊂∪n i=1{x ∈E : xi −ϵ < x < xi +ϵ}. Then, E ⊂(min xi i=1,..,n −ϵ, max xi i=1,..,n +ϵ). Therefore, ∀x ∈E min xi i=1,..,n −ϵ < x < max xi i=1,..,n + ϵ. Hence, E is bounded. Suppose that E is not closed. Therefore, it doesn’t contain all its limit points. We assume, without loss of generality that it doesn’t contain one limit point say x0. Let, H = (−∞, x0 −1 n) ∪(x0 + 1 n, ∞) be an open cover for E. Then, there is a finite subcover of H such that E ⊂∪m n=1(−∞, x0 −1 n) ∪(x0 + 1 n, ∞) = ⇒E ⊂(−∞, x0 −1 m) ∪(x0 + 1 m, ∞). By the density of the reals in between (x0 −1 m, x0) ∃α ∈E which is not covered by the finite subcover. Therefore, there is no finite subcover of E, a contradiction, because we assumed that E is closed. Hence, E is closed and bounded. 2 Problem 1 The idea is to decompose B into a countable union of disjoint sets and at least one of these set has to be A(so we can have m(A)), in order to use the countably additivity over countable disjoint unions property. We can decompose B as B = A ∩(B −A). The two sets A and B −A are obviously disjoint and the union is countable therefore, m(B) = m A ∩{B −A} disj. count. union = m(A) + m(B −A) m(B−A)≥0 ≥ m(A) Problem 4 The counting measure is translation invarient because by shifting the elements on a set, by a constant, doesn’t change the number of elements. Let, {En}∞ n=1 be a countable and disjoint collection of sets. Furthermore, we know that the union of a countable collection of countable sets is countable and so ∪∞ i=1En) is countable. Case 1: If all of En are empty and so their union and trivially c(∪∞ i=1En) = P∞ n=1 c(En) = 0. Case 2: If at least one of the En has infinitely many memebers (c(En) = ∞) and so the union. Then, c(∪∞ i=1En) = ∞and P∞ n=1 c(En) c(.)≥0 = ∞. Therefore, c(∪∞ i=1En) = P∞ n=1 c(En). Case 3: If all of the En are finite and non-empty then c(∪∞ i=1En) = ∞as ∪∞ i=1En is a countably infinte set. Furthermore, P∞ i=1 En = ∞as a countable infinite sum. Therefore, c(∪∞ i=1En) = P∞ n=1 c(En). Case 4: If {En}∞ n=1 are finite and say, without loss of generality, the first m out of them are non-empty, and the rest empty, and let ni = number of elements of Ei. Then, the number of elements of ∪∞ i=1En = ∪m n=1En ∪∞ n=m+1 En are n1 + ... + nm = ⇒c(∪∞ n=1En) = n1 + ... + nm. Furthermore, P∞ n=1 c(En) = Pm i=1 c(En) + P∞ n=m+1 c(En) c(∅)=0 = Pm n=1 c(En) + P∞ n=m+1 0 = n1 + ... + nm. Therefore, c(∪∞ i=1En) = P∞ n=1 c(En). Hence, the counting measure is countably additive and translation invariant. Problem 6 We know that the set of rational numbers Q is countable and Q ∩[0, 1] ⊂Q. Therefore, Q ∩[0, 1] is countable. Furthermore, for any countable sets we know that its outer measure is 0. So, m∗(Q ∩[0, 1]) = 0. Also, we can decompose [0, 1] as [0, 1] = {Qc ∩[0, 1]} ∪{Q ∩[0, 1]}. By the countable sub-additivity of the outer measure me have that m∗([0, 1]) = m∗({Qc ∩[0, 1]} ∪{Q ∩[0, 1]}) ≤m∗({Qc ∩[0, 1]}) + m∗({Q ∩[0, 1]}) (1) But the outer measure of an interval is its length and the outer measure of a countable set is zero. Therefore, (1) takes the form m∗({Qc ∩[0, 1]}) ≥1 (2) 3 Furthermore, Qc ∩[0, 1] ⊂[0, 1] and by the monotonicity of the outer measure Qc ∩[0, 1] ⊂[0, 1] = ⇒m∗(Qc ∩[0, 1]) ≤m∗([0, 1]) = 1 (3) Hence, combining (2) and (3) m∗(Qc ∩[0, 1]) = 1 Problem 11 We know that σ-algebra is closed under compliments and countable unions. Let A be the σ-algebra (a, ∞) ∈A compliment = ⇒ (−∞, a] ∈A count.union = ⇒ ∞ [ n=1 −∞, a −1 n = (−∞, a) ∈A compliment = ⇒ [a, ∞) ∈A (−∞, a), (b, ∞) ∈A count.union = ⇒ (−∞, a) ∪(b, ∞) ∈A compliment = ⇒ [a, b] ∈A (−∞, a], (b, ∞) ∈A count.union = ⇒ (−∞, a] ∪(b, ∞) ∈A compliment = ⇒ (a, b] ∈A (−∞, a), [b, ∞) ∈A count.union = ⇒ (−∞, a) ∪[b, ∞) ∈A compliment = ⇒ [a, b) ∈A (−∞, a], [a, ∞) ∈A count.inters = ⇒ (−∞, a] ∩[a, ∞) = {a} ∈A Hence, if a σ algebra A containts intevrals of the form (a, ∞) then it contains all type of intervals. 4 Problem 18 Let {Ik}∞ k=1 be a countable collection of open intervals that covers E. Then, ∀ϵ > 0 ∞ X k=1 l(Ik) < m∗(E) + ϵ Define O = S∞ k=1 Ik. Then, O is open set containing E. By the definition of the outer measure of O, m∗(O) ≤ ∞ X k=1 l(Ik) < m∗(E) + ϵ, ∀ϵ > 0 = ⇒m∗(O) ≤m∗(E) (4) Let {On}∞ n=1 be a countable collection of such sets. Then we define G = ∩∞ n=1On, which is a Gδ set and so is measurable. Observe, that G is also an open covering for E. By the monotonicity of the outer measure, E ⊂G = ⇒m∗(E) ≤m∗(G) (5) On the other hand G ⊂O, because if x ∈G then x is in every set ∪∞ k=1Ik and so in O. By monotonicity m∗(G) ≤m∗(O) (1) ≤m∗(E) (6) Therefore, by (2) and (3) m∗(E) = m∗(G), where G is a Gδ set that contains E. From the Inner approximation by closed and Fσ sets, E is measurable if ∃F ∈Fσ : F ⊂E : m∗(E −F) = 0. Furhermore, F has a finite outer measure since by monotonicity F ⊂E = ⇒ m∗(F) ≤m∗(E) < ∞. Therefore, by excision property 0 = m∗(F −E) = m∗(F) −m∗(E) = ⇒m∗(E) = m∗(F) 5 Problem 24 We can decompose E1 and E2 as the union of disjoint set and because both of them are measurable we can use the countable additivity proberty of the measure E1 = {E1 −E2} ∪{E1 ∩E2} = ⇒m(E1) = m(E1 −E2) + m(E1 ∩E2) = ⇒m(E1 −E2) = m(E1) −m(E1 ∩E2) (7) E2 = {E2 −E1} ∪{E1 ∩E2} = ⇒m(E2) = m(E2 −E1) + m(E1 ∩E2) = ⇒m(E2 −E1) = m(E2) −m(E1 ∩E2) (8) Furthermore, because E1, E2 are measurable and so is their union and can decompose it as the union of disjoint sets where we can apply also the countable additivity property of the measure E1 ∪E2 = {E1 −E2} ∪{E1 ∩E2} ∪{E2 −E1} m(E1 ∪E2) = m(E1 −E2) + m(E1 ∩E2) + m(E2 −E1) (1),(2) = ⇒m(E1 ∪E2) = m(E1) −m(E1 ∩E2) + m(E2) Note: Each of the decomposed sets belong to the σ-algebra as they can be formed by unions, intersections and compliments of the measurable sets E1, E2 and so they are measurable. Problem 26 We can write the set A ∩S∞ k=1 Ek as S∞ k=1{A ∩Ek}. Therefore, by the sub-additivity property of the outer measure we have m∗ A ∩ ∞ [ k=1 Ek = m∗ ∞ [ k=1 {A ∩Ek} ≤ ∞ X k=1 m∗(A ∩Ek) (9) On the other hand the finite union Sn k=1{A ∩Ek} is a subset of the countable union S∞ k=1{A ∩Ek} and by the monotonicity property of the outer measure n [ k=1 {A ∩Ek} ⊂ ∞ [ k=1 {A ∩Ek} m∗ ∞ [ k=1 {A ∩Ek} ≥m∗ n [ k=1 {A ∩Ek} , for each n = n X k=1 m∗(A ∩Ek), for each n {Ek}∞ k=1 countable disjoint The left hand side of this inequality is independent of n. Therefore, 6 m∗ ∞ [ k=1 {A ∩Ek} ≥ n X k=1 m∗(A ∩Ek) (10) Combining (6) and (7) we have m∗ ∞ [ k=1 {A ∩Ek} = n X k=1 m∗(A ∩Ek) Problem 28 Without loss of generality, let {Ak}∞ k=1 be a collection of disjoint measurable sets, if they were not disjoint we can always construct a disjoint collection. In order to use the continuity of the measure we need somehow to contstuct either an ascending or descending set. Let, Ck = Sk i=1 Ai, which is obviously ascending. Furthermore, the set S∞ k=1 Ck is equal to S∞ k=1 Ak. Therefore, m ∞ [ k=1 Ak = m ∞ [ k=1 Ck = lim k→∞m(Ck) (continuity of measure) = lim k→∞m k [ i=1 Ai = lim k→∞ k X i=1 m(Ai) (finite additivity) = ∞ X i=1 m(Ai) 7 Problem 27 Consider the sequence of measurable functions {fn} such that fn(x) = χ(n,∞)(x), ∀x ∈R. Observe that {fn} p.w. →f where f(x) = 0, ∀x ∈E = R. Then by Egoroff’s theorem ∀ϵ > 0 there is a closed set F contained in R for which {fn} u →f on F and m(R −F) < ϵ From the uniform convergence of fn we have that ∀ϵ > 0, ∃N ∈N : \|fn −f\| < ϵ, ∀n > N. Choose m : m > N sufficiently large and because thats true for every ϵ > 0, choose ϵ : 0 < ϵ < 1. So, χ(m,∞) < ϵ on F χ(m,∞) < ϵ ⇐ ⇒x ∈(−∞, m) Therefore, F must be a closed set subset of {x ∈R : x ∈(−∞, m)}. F ⊂{x ∈R : x ∈(−∞, m)} = ⇒{x ∈R : x ∈(m, ∞)} ⊂R −F By the monotonicity of the measure and the result of Egoroff’s theorem we have m {x ∈R : x ∈(m, ∞)} ≤m(R −F) < ϵ = ⇒∞< ϵ a contradiction, because we choose E to be an unbounded set, with infinite measure. Problem 9 For each c ∈R consider the set {x ∈E : f(x) < c} ⊂E = ⇒m {x ∈E : f(x) < c} ≤m(E) = 0 = ⇒m {x ∈E : f(x) < c} = 0 Every set of measure 0 is measurable. Therefore, {x ∈E : f(x) < c} measurable = ⇒f measurable. Consider, a finite collection of disjoint sets {Ei}n i=1 such that Sn i=1 Ei = E. Then, 0 = m(E) = m n [ i=1 Ei = n X i=1 m(Ei) = ⇒ m(Ei) = 0 ∀i = 1, 2, ..., n Since f is measurable and bounded on E the simple approximation lemma applies. So, there are simple functions ϕ, ψ on E such that ϕ ≤f ≤ψ on E. Let αi, βi be the distinct values that ϕ, ψ take in each Ei, respectively. Then, 8 Z E ϕ = n X i=1 αim(Ei) = 0 and Z E ψ = n X i=1 βim(Ei) = 0 = ⇒sup Z E ϕ : ϕ simple and ϕ ≤f = 0 and inf Z E ψ : ψ simple and f ≤ψ = 0 So, the upper and lower Lebesgue integrals are equal and by definition f is Lebesgue integrable and Z E f = sup Z E ϕ : ϕ simple and ϕ ≤f = inf Z E ψ : ψ simple and f ≤ψ = 0 Problem 10 Since f is measurable and A is a measurable subset of E, fχA is measurable on A. Also, E has a finite measure and so A has. Then, fχA is a bouded (since f is bounded), measurable function on a set of finite measure and so is integrable on A. In addition, E has finite measure. Consider, a finite collection of disjoint sets {Ei}n i=1 such that Sn i=1 Ei = E. From simple approximation lemma we know that there exist simple functions ϕ, ψ such that ϕ ≤f ≤ψ on E and let αi, βi be the distinct values that ϕ, ψ take in each Ei, respectively. Then, ϕχA ≤fχA ≤ψχA on E = ⇒ Z E ϕχA ≤ Z E fχA ≤ Z E ψχA (11) We re-write ϕ and ψ in their canonical representation ϕ = Pn i=1 αiχEi, ψ = Pn i=1 βiχEi. Then, Z E ϕχA = Z E n X i=1 αiχEiχA = Z E n X i=1 αiχEi∩A = n X i=1 αim(Ei ∩A) = Z A ϕ Z A f = sup Z A ϕ : ϕ simple and ϕ ≤f ≤ Z E fχA from (1) (12) Z E ψχA = Z E n X i=1 βiχEiχA = Z E n X i=1 βiχEi∩A = n X i=1 βim(Ei ∩A) = Z A ψ 9 Z A f = inf Z A ψ : ψ simple and f ≤ψ ≥ Z E fχA from (1) (13) From (2) and (3) Z E fχA ≤ Z A f ≤ Z E fχA = ⇒ Z A f = Z E fχA Problem 12 Let E0 = {x ∈E : f(x) ̸= g(x)} and E −E0 = {x ∈E : f(x) = g(x)}, then m(E0) = 0. Since, f = g a.e onE = ⇒g measurable. So, g is a bounded, measurable, on a set of finite measure = ⇒g integrable. For the set of measure zero we have R E0 f = R E0 g = 0. Z E f = Z E−E0 f + Z E0 f f=g on E−E0 = Z E−E0 g + 0 = Z E−E0 g + Z E0 g = Z E g Problem 17 Let E : m(E) = 0 and define {fn} = n be an increasing sequence of measurable functions on E, {fn} p.w. →f = ∞and so the Monotone Convergence Theorem applies Z E f = lim n→∞ Z E fn = lim n→∞ Z E n = lim n→∞nm(E) = lim n→∞0 = 0 (14) 10 Problem 23 f(x) = an, x ∈[n, n + 1) on E = [1, ∞). Then, we can write f(x) = P∞ n=1 anχ[n,n+1)(x) where anχ[n,n+1) is a sequence of non-negative functions as an is a sequence of non-negative real numbers. Then, from the corollary of the monotone convergence theorem we have Z E f = ∞ X n=1 Z E anχ[n,n+1) int of simple function = ∞ X n=1 anm([n, n + 1)) = ∞ X n=1 an Problem 27 From previous homework problem if fn is a sequence of measurable functions then inf{fk : k ≥n} is also measurable. Define gn := inf{fk : k ≥n} and g := limn→∞inf{fk : k ≥n}. Then, gn is an increasing sequence of non-negative measurable functions and gn p.w →g Therefore, the Monotone Convergence Theorem applies Z E g = lim n→∞ Z E gn ≤lim n→∞inf Z E gk : k ≥n (15) In addition, gn = inf{fk : k ≥n} ≤fn = ⇒ Z E gn ≤ Z E fn = ⇒inf Z E gn ≤inf Z E fn (16) Hence, combing (1) and (2) we have Z E lim n→∞inf{fk : k ≥n} = Z E g ≤lim n→∞inf Z E gk : k ≥n ≤lim n→∞inf Z E fk : k ≥n Problem 28 Since, f is integrable so fχC is. Then, by definition Z E fχC := Z E (fχC)+ − Z E (fχC)−= Z E f +χC − Z E f −χC (17) We only need to show that R E f +χC = R C f + and R E f −χC = R C f −. 11 Z E f +χC = sup Z E h : h bounded, measurable, with finite support : h ≤f +χC on E ∗ = sup Z C h : h bounded, measurable, with finite support : h ≤f + on C = Z C f + (18) Similarly, Z E f −χC = sup Z E h : h bounded, measurable, with finite support : h ≤f −χC on E ∗ = sup Z C h : h bounded, measurable, with finite support : h ≤f −on C = Z C f − (19) Therefore, combining (3), (4) and (5) we have Z E fχC := Z E f +χC − Z E f −χC = Z C f + − Z C f −:= Z C f If h ≤f +χC, h ≤f −χC, on E then h ≤f +, h ≤f −on C. 12 Problem 29 Consider the function f(x) = χ[n,n+1)(x) −χ[n+1,n+2)(x) ∞ X n=1 Z n+1 n f(x) = ∞ X n=1 Z n+1 n χn,n+1 −χ[n+1,n+2)(x) = ∞ X n=1 Z n+1 n χ[n,n+1)(x) − Z n+1 n χ[n+1,n+2)(x) = ∞ X n=1 m([n, n + 1]) −m([n + 1, n + 2)) = ∞ X n=1 \|0\| = 0 Hence, the series converges absolutely which implies convergence, too. But the function is not integrable since \|f\| = f + + f −, wheref + = χ[n,n+1)(x) andf −= χ[n+1,n+2)(x) Z [1,+∞) f + = Z [1,+∞) χn,n+1 ∗ = ∞ X n=1 Z [n,n+1] χ[n,n+1) = ∞ X n=1 1 = ∞ ∗simple function is integrable and [1, ∞) = ∪∞ n=1[n, n + 1) Hence, R [1,∞) \|f\| = ∞= ⇒\|f\| not integrable = ⇒f not integrable. So, both of the if and only if statements are not true, as we found a counter-example that disaproves each of one direction, which is enough. Problem 37 We need to show that ∀ϵ > 0, ∃N ∈N : R En f < ϵ ∀n ≥N, which is equivalent of showing that limn→∞ R En f = 0. The countable collection of measurable set En = {x ∈E : \|x\| ≥n} is descening and T∞ n=1 En = ∅= ⇒m T∞ n=1 En = 0. In addition, f is integrable over E, so is finite a.e. and so bounded. From a previous homework problem the integral of a bounded function over a set of measure zero, is zero. So, R T∞ n=1 En f = 0. Therefore, from the continuity of integration lim n→∞ Z En f = Z T∞ n=1 En f = 0 ⇐ ⇒∀ϵ > 0 ∃N ∈N : Z En f < ϵ ∀n ≥N 13 Problem 9 Consider the sequence of measurable functions fn(x) = χn,n+1. Then, fn p.w →f = 0 in E = R. Then in order fn to converge in measure on R to f, we need ∀ϵ > 0 lim n→∞m x ∈R : \|χn,n+1\| > ϵ = 0 Choose ϵ < 1 then, lim n→∞m x ∈R : \|χn,n+1\| > ϵ = lim n→∞m([n, n + 1]) = 1 Therefore, fn fails to converge in measure on R to f. Another, counter example is by choosing gn(x) = χ[n,∞)(x), then, gn p.w →g = 0 in E = R Choose ϵ < 1 then, lim n→∞m x ∈R : \|χ[n,∞)\| > ϵ = lim n→∞m([n, ∞)) = ∞ 14
14728	https://medium.com/woodworkers-of-the-world-unite/sum-of-the-parts-e1e2f131bbe9	Sum of The Parts. Thoughts of creation \| by Paroma Sen \| Woodworkers of the World Unite!!! \| Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Mastodon Woodworkers of the World Unite!!! --------------------------------- · Follow publication A place for duuudes of all sexes, ages, religions, and coffee-preferences to hang out and shoot the sh!t about their latest creations, to brag, lament, query, or quote, it’s all good… Follow publication 1 1 · Top highlight Member-only story Sum of The Parts Thoughts of creation Paroma Sen Follow 2 min read · Feb 6, 2021 835 5 Listen Share Press enter or click to view image in full size Image by author, concept, design & layout by author The whole is greater than the sum of its parts. I’ve always been struck by the depth of this statement. From a human face, to a work of art, to life itself, every single time — the whole is greater than the sum of its parts. The idea behind the phrase is attributed to Aristotle, who used it in a teamwork analogy. But it applies to so much more. Back when I was in school, I illustrated the concept through the picture you see above. I cut up a face from a magazine, and pasted it back, but randomly. Nothing is where it should be. For this picture, the extra is the order in which the parts are placed. For something else, like a human body, the extra can be the life force itself. Powering the body through its lifetime, it disappears when life leaves the body. Where does the life force go? Why can’t we include it in the sum of parts? What is it made of, and how does it transform? As it turns out, many of the things that make life interesting cannot be explained in linear mathematical terms. There’s always something more, defined or undefined, some secret sauce. There are always more questions than answers. I wonder what it is about the human condition that makes us ponder such unanswerables. As I glance at my cat, transfixed by the sight of geese outside the window, I know she is not pondering any of life’s unanswerables. And she is quite happy without them. So perhaps it is best to blow these questions into a balloon, tie a tight knot at the end, and let them escape out the window. Paroma Sen 2021 Create an account to read the full story. The author made this story available to Medium members only. If you’re new to Medium, create a new account to read this story on us. Continue in app Or, continue in mobile web Sign up with Google Sign up with Facebook Sign up with email Already have an account? Sign in 835 835 5 Follow Published in Woodworkers of the World Unite!!! ---------------------------------------------- 1K followers ·Last published 1 day ago A place for duuudes of all sexes, ages, religions, and coffee-preferences to hang out and shoot the sh!t about their latest creations, to brag, lament, query, or quote, it’s all good… Follow Follow Written by Paroma Sen --------------------- 1.3K followers ·687 following “Do not go gentle into that good night, but rage, rage, rage against the dying of the light.” Follow Responses (5) Write a response What are your thoughts? Cancel Respond Amy Marley Feb 10, 2021 So perhaps it is best to blow these questions into a balloon, tie a tight knot at the end, and let them escape out the window. I have been coming to similar conclusions. Why not… hot air be gone… perhaps what we are left with is the definition of pure freeing bliss? Still… but still ??? Ahahah. Love your presence. Thanks for being you. -- 1 reply Reply Joseph Lieungh Feb 10, 2021 My favorite thing to do is sit and ponder. Lovely piece Paroma and Segway into more pondering. Thank you. 😊 -- 1 reply Reply Imad Feb 7, 2021 As it turns out, many of the things that make life interesting cannot be explained in linear mathematical terms. There’s always something more, defined or undefined, some secret sauce. Where mathematics fails, philosophy intercedes with many convoluted arguments that sum up to ‘I don’t know’ but maybe, perhaps…and so many words meandering in the realms of uncertainty. Some things have no answers and like your cat perhaps we need to let the unanswerable be unanswered. Wonderful writing Paroma :) -- 1 reply Reply See all responses More from Paroma Sen and Woodworkers of the World Unite!!! In Scrittura by Paroma Sen Achilles Heel ------------- ### Saturday Prompt: vulnerability Jul 5 10 In Woodworkers of the World Unite!!! by David Rudder Moved by the Wind ----------------- ### Of freedom. 1d ago 7 In Woodworkers of the World Unite!!! by Mia Verita The Power of Words ------------------ ### The magic of the words we wield Mar 5, 2023 20 In Scrittura by Paroma Sen It’s Hard to be Lonely ---------------------- ### 2nd Saturday prompt: traveling solo Apr 12 12 See all from Paroma Sen See all from Woodworkers of the World Unite!!! Recommended from Medium In Lit Up by Edward Swafford Mislaid ------- ### Her abdication Sep 19 15 In Global Literary Theory by Ira Fader Poet’s Choice: Who is “I”? -------------------------- ### Meeting poetry’s “first person singular” Apr 1 10 Alessandra Graziella Mastroianni Di'Stefano Her Mouth Was a Guillotine -------------------------- Sep 6 In Long. Sweet. 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14729	https://link.springer.com/rwe/10.1007/978-1-4020-4425-0_9063	Advertisement Sun Zi 556 Accesses Of all the mathematicians in ancient China, we know the least about the life of Sun Zi. The fact that he was given the honorific designation of Zi (Master) after his surname led some people like Zhu Zunyi (AD 1629–1709) to identify him with Sun Wu, a celebrated tactician in the sixth century BCE. Then Ruan Yuan (AD 1764–1849) assigned him to the late Zhou period of the third century BCE. With regard to the text written by Sun Zi, Dai Zhen, the eighteenth century scholar and mathematician, citing from internal evidence, argued that the text could not have been written earlier than the Han dynasty at the turn of the Christian era. Though this text by Sun Zi was listed in all the bibliographical chapters of the Sui Shu (Standard History of the Sui dynasty), the Jiu Tang Shu (Old Standard History of the Tang Dynasty), and the Xin Tang Shu(New Standard History of the Tang Dynasty), there was no mention of its author. This shows that as early as the middle of the seventh century AD, no one... This is a preview of subscription content, log in via an institution to check access. Access this chapter Subscribe and save Buy Now Tax calculation will be finalised at checkout Purchases are for personal use only Institutional subscriptions References Lam Lay Yong and Ang Tian Se. Fleeting Footsteps: Tracing the Conception of Arithmetic and Algebra in Ancient China. Singapore: World Scientific Publishing Co., 1992. Google Scholar Mikami, Yoshio. The Development of Mathematics in China and Japan. 2nd ed. New York: Chelsea Publishing Company, 1974. Google Scholar Qian, Baozong ed. Suanjing Shi Shu (Ten Mathematical Classics). Beijing: Zhonghua Shuju, 1963. Google Scholar Ruan, Yuan, Chouren Zhuan (Biographies of Mathematicians and Astronomers). Vol. 1. Shanghai: Shangwu Yinshuguan, 1955. Google Scholar Download references Search author on:PubMed Google Scholar Editor information Editors and Affiliations Hampshire College, 893 West Street, 01002, Amherst, MA, USA Helaine Selin (Editor) (Editor) Rights and permissions Reprints and permissions Copyright information © 2008 Springer-Verlag Berlin Heidelberg New York About this entry Cite this entry Se, A.T. (2008). Sun Zi. In: Selin, H. (eds) Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures. Springer, Dordrecht. Download citation DOI: Publisher Name: Springer, Dordrecht Print ISBN: 978-1-4020-4559-2 Online ISBN: 978-1-4020-4425-0 eBook Packages: Humanities, Social Sciences and LawReference Module Humanities and Social SciencesReference Module Humanities Share this entry Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Publish with us Policies and ethics Access this chapter Subscribe and save Buy Now Tax calculation will be finalised at checkout Purchases are for personal use only Institutional subscriptions Search Navigation Discover content Publish with us Products and services Our brands 44.220.174.32 Not affiliated © 2025 Springer Nature
14730	https://calculat.io/en/number/long-division/13--3	Send You can also email us on infocalculat.io Long Division: 13 ÷ 3 Dividend (number to be divided) Divisor (number to divide by) Calculate How to divide 13 by 3 using long division and find the remainder? Visual Long Division This shows the traditional long division layout used in schools, with each step clearly marked. Step-by-Step Explanation Here's how to solve 13 ÷ 3 using long division method step by step: Division Steps \| Step \| Dividend Part \| Quotient Digit \| Multiply \| Subtract \| Remainder \| --- --- --- \| \| 1 \| 13 \| 4 \| 4 × 3 = 12 \| 13 - 12 = 1 \| 1 \| Final remainder: 1 Since 1 is less than 3, we cannot divide further. The final answer is 4 with remainder 1. About Long Division Long division is a method for dividing large numbers that breaks down the division process into smaller, manageable steps. It's taught in elementary school as a systematic way to perform division by hand. The process involves: This method ensures accuracy and helps students understand the relationship between multiplication and division. See Also About "Long Division Calculator" Calculator "Long Division Calculator" Calculator Dividend (number to be divided) Divisor (number to divide by) Calculate FAQ How to divide 13 by 3 using long division and find the remainder? See Also
14731	https://www.keisu.t.u-tokyo.ac.jp/data/2001/METR01-04.pdf	Randomized algorithms to solve parameter-dependent linear matrix inequalities∗ Yasuaki Oishi† and Hidenori Kimura‡ Abstract The randomized algorithm of Polyak and Tempo (2000), which consists of random sampling and subgradient descent, is generalized in order to solve parameter-dependent linear matrix inequalities and its computational complexity is analyzed. This paper ﬁrst examines an algorithm obtained by direct generalization of Polyak and Tempo’s and shows that its expected time to achieve a solution is inﬁnite. Then this paper improves this algorithm so that its expected achievement time becomes ﬁnite. An explicit upper bound of the expected achievement time is given in a special case. A numerical example is provided. Keywords: randomized algorithms, parameter-dependent linear matrix inequalities, computational complexity, curse of dimensionality, linear parameter-varying systems. 1. Introduction Parameter-dependent linear matrix inequalities (LMIs) often appear when we consider analysis and synthesis of linear parameter-varying (LPV) systems, which are useful to model time-varying or nonlinear systems. Parameter-dependent LMIs are discussed in this context by the authors such as Becker, Packard, Philbrick, and Balas (1993), Becker and Packard (1994), and Apkarian, Gahinet, and Becker (1995), and their advanced use taking account of the rate of parameter change is considered by the authors including Watanabe, Uchida, Fujita, and Shimemura (1994), Gahinet, Apkarian, and Chilali (1996), Feron, Apkarian, and Gahinet (1996), Wu, Yang, Packard, and Becker (1996), Yu and Sideris (1997), and Apkarian and Adams (1998). Since a parameter-dependent LMI is equivalent to a set of inﬁnitely many LMIs, it is diﬃcult to solve in general. One conventional approach is to make its parameter dependence aﬃne by approximation (Ohara, Ide, Yamaguchi, & Ohno, 1999) or by introduction of new parameters (Masubuchi & Shimemura, 1999). However, this approach often causes conservatism or increase of the problem size. Another approach is to grid the parameter set and to consider the LMI only at the grid points (Wu, Yang, Packard, & Becker, 1996; Apkarian & Adams, 1998). In this approach, there is a concern that the LMI may not be satisﬁed between the grid points. Moreover, it has a problem called the “curse of ∗Submitted to Automatica as a brief paper on July 17, 2001 †Corresponding author. Department of Mathematical Informatics, Graduate School of Information Science and Tech-nology, The University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan; phone: +81-3-5841-6906; fax: +81-3-5841-6886; email: oishi@simplex.t.u-tokyo.ac.jp ‡Department of Complexity Science and Engineering, Graduate School of Frontier Sciences, The University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan; email: kimura@crux.t.u-tokyo.ac.jp 1 dimensionality,” that is, the number of the grid points increases exponentially fast as the parameter dimension grows. We try in this paper a probabilistic approach to parameter-dependent LMIs. The algorithms to be proposed are obtained by generalizing a recent work of Polyak and Tempo (2000), where they considered design of a robust linear quadratic regulator. Since our approach does not require the number of LMIs to be ﬁnite, it does not suﬀer from the mentioned problems of the conventional approaches. Although there are many works on a probabilistic approach or randomized algorithms (for example, Ray & Stengel, 1993; Khargonekar & Tikku, 1996; Barmish & Lagoa, 1997; Tempo, Bai, & Dabbene, 1997; Vidyasagar, 1999; Chen & Zhou, 2000), most of them utilize random sampling only. The proposed algorithms are more systematic than the existing ones because they use a combination of random sampling and subgradient descent. By utilizing the result of Polyak and Tempo (2000), it is possible to show that the proposed algorithms arrive at a solution after a ﬁnite time with a probability one. However, it is not obvious how long this time is. This paper spends its main eﬀorts in evaluating the expected time to achieve a solution. Particularly, it ﬁrst shows that the expected achievement time is inﬁnite when we use the algorithm obtained by direct generalization of Polyak and Tempo’s one. This implies that this basic algorithm requires long computational time and is not practical. This paper then shows that one can improve this algorithm so that its expected achievement time is ﬁnite. An explicit upper bound of this time is given in a special case. Furthermore, it is shown that the improved algorithm overcomes the curse of dimensionality in the case that the LMI has to be satisﬁed not for all the parameters but for almost all of them. The contents of this paper are to be presented in the conference (Oishi & Kimura, 2001). After ﬁnishing the main body of the present research, the author is notiﬁed that Fujisaki, Dabbene, and Tempo (2001) independently consider generalization of the algorithm of Polyak and Tempo. How-ever, they concentrate on a special type of parameter-dependent LMIs and do not perform so detailed analysis of computational complexity as is found in the present paper. The following notation is used. Let Rn mean the n-dimensional Euclidean space. The symbol ∥· ∥ stands for the 2-norm of a vector and the symbol ∥· ∥F for the Frobenius norm of a matrix. Deﬁne λ(A) as the maximum (i.e., the most positive) eigenvalue of a symmetric matrix A. For two symmetric matrices A and B, the inequalities A < B and A ≤B imply that B −A is positive deﬁnite and is positive semideﬁnite, respectively. For a symmetric matrix A, the symbol A+ means the projection onto the cone of positive semideﬁnite matrices. That is, with an orthogonal matrix U that makes U TAU = Ξ = (ξij) diagonal, the matrix A+ is deﬁned as UΞ+U T, where the (i, j)-element of Ξ+ is ξij if ξij > 0 and is zero otherwise. 2 2. Basic algorithm A problem to be solved is presented ﬁrst. Problem. The parameter set P is the n-dimensional hypercube [−1, 1]n and the variable domain Q is a bounded closed convex subset of Rd having its interior. Moreover, V (q, p) is a symmetric-matrix-valued function of q ∈Q and p ∈P, which is aﬃne with respect to q and is continuously diﬀerentiable with respect to p. Find a q satisfying V (q, p) ≤O for any p ∈P. □ One can formulate analysis and synthesis of LPV systems into this form possibly using information on the rate of parameter change. Details are found in the references cited in the previous section. See Section 5 for an example. Remark. Although the parameter set P is a general hyper rectangle in many LPV systems, it can be transformed to the above hypercube [−1, 1]n without loss of generality. It is not practically diﬃcult to limit the variable domain to a bounded Q by performing some preliminary computation. One may notice that a strict inequality V (q, p) < O is considered in many applications instead of V (q, p) ≤O above. However, this diﬀerence is not essential in numerical computation and the improved algorithm to be introduced gives a solution q that satisﬁes the strict inequality. □ It is assumed throughout this paper that there exists a q satisfying the strict inequality V (q, p) < O for any p ∈P. Let us call the region {q ∈Q : V (q, p) ≤O for any p ∈P} the solution region. The solution region contains by assumption a d-dimensional closed ball whose center q∗attains V (q∗, p) < O for any p ∈P. Fix one such ball and write its radius as r∗. We now present an algorithm to solve the problem above. It is a generalization of the algorithm of Polyak and Tempo (2000), which is for design of a robust quadratic regulator. In order to state the algorithm, some preparation is needed. Choose an initial point q0 in the variable domain Q. We introduce into the parameter set P a probability density function possessing a ﬁnite upper bound µP and a positive lower bound µP . Note that the inequality V (q, p) ≰O is equivalent to ∥V (q, p)+∥F > 0. When ∥V (q, p)+∥F > 0 holds, we deﬁne ∇ ∥V (q, p)+∥F to be the vector whose i-th element is 1 ∥V (q, p)+∥F tr V (q, p)+ ∂V (q, p) ∂qi , where q = [q1 . . . qd]T. It is possible to show that ∥V (q, p)+∥F is a convex function of q and ∇ ∥V (q, p)+∥F is its subgradient. Furthermore, let Π be the projection onto the variable domain Q, that is, Πq := arg min ∈Q ∥ q −q∥for q ∈Rd. Finally, deﬁne r0, r1, . . . to be a sequence of positive numbers that monotonically decreases to zero and satisﬁes ∞ ℓ=0(rℓ)2 = ∞. 3 Algorithm. 0. k := 0. ℓ:= 0. 1. Randomly sample pk from P according to the probability density function on P. 2. If V (qk, pk) ≰O, put qk+1 := Π qk −γk∇ ∥V (qk, pk)+∥F , ℓ:= ℓ+ 1, where γk := ∥V (qk, pk)+∥F + rℓ ∇ ∥V (qk, pk)+∥F ∇ ∥V (qk, pk)+∥F 2 ; Otherwise, just put qk+1 := qk. 3. k := k + 1. Return to Step 1. □ This algorithm produces a sequence of points qk. One can prove the next property on this sequence as is found in Appendix A. Proposition 1. Let ℓ(k) be the value of ℓat the k-th iteration of the preceding algorithm. If rℓ(k) ≤r∗ and V (qk, pk) ≰O, there holds ∥qk+1 −q∗∥2 ≤∥qk −q∗∥2 −[rℓ(k)]2. When the point qk is not in the solution region, the probability that a correction step is made in Step 2 is positive by assumption. This implies that correction of qk is made after a ﬁnite number of iterations with a probability one. Combining this observation with the fact that the squared sum of {rℓ} is inﬁnite, we have the following corollary. See Polyak and Tempo (2000) for more details. Corollary 1. With a probability one, there exists a ﬁnite number k1 such that the point qk produced by the preceding algorithm is contained in the solution region for any k ≥k1. This property still holds in fact even if Q is unbounded. Among the numbers that can be k1, we refer to the smallest one as the achievement time kA. The above corollary claims that there exists a ﬁnite achievement time with a probability one. A problem here is that it is not obvious how large the achievement time kA is. Although Polyak and Tempo (2000) provide an answer to this question, it is not applicable in our generalized setting where the set P is continuous and the number rℓis not constant. 4 3. Inﬁniteness of the expected achievement time Since the achievement time kA is a random number, it is easier to evaluate its expectation EP (kA), which is called the expected achievement time henceforth. The value of EP(kA) depends on the choice of the initial point q0. In order to evaluate it in an averaged case, we suppose that the initial point q0 is chosen according to some probability density function ﬁxed on Q and consider the expectation EQ[EP (kA)]. The claim of this section is that this expectation is inﬁnite. It means that the preceding algorithm often requires a long time to achieve a solution and is not appropriate for a practical use. Improvement of the algorithm is certainly necessary, which is a task of the next section. We add the following assumptions in order to obtain the mentioned result. Suppose that there exists a pair of q ∈Q and p ∈P that satisﬁes V (q, p) ≰O. The probability density function introduced into Q is assumed to have a positive lower bound µQ. It is not diﬃcult to see that the function sup ∈P λ[V (q, p)] is continuous and is convex with re-spect to q. By the assumptions so far, there holds sup ∈P λ[V (q, p)] > 0 at some q ∈Q while sup ∈P λ[V (q∗, p)] < 0 at q∗. Accordingly, on the line segment connecting these two points, there exists one and only one q that attains sup ∈P λ[V (q, p)] = 0. Let us write it as q♯from now on. This q♯ belongs to the interior of Q. It is possible to construct a (d−1)-dimensional manifold in a neighborhood of q♯so that it contains q♯and sup ∈P λ[V (q, p)] = 0 holds anywhere on this manifold. We make here our ﬁnal assumption, that is, there is no p = [p1 . . . pn]T ∈P that satisﬁes det V (q♯, p) = 0, ∂det V (q♯, p) ∂p1 = 0, . . . , ∂det V (q♯, p) ∂pn = 0 simultaneously. This assumption is considered to be mild enough because n + 1 equalities are not simultaneously satisﬁed in an n-dimensional space in general. Theorem 1. Under the assumptions so far, there holds EQ EP (kA) = ∞. The proof of this theorem requires the following lemma, which is proved in Appendix B. Lemma 1. Deﬁne q(α) := q♯+α(q♯−q∗). Under the assumptions so far, there exists a positive C such that Prob{p ∈P : V ( q(α), p) ≰O} ≤Cα holds for any small enough nonnegative α. Proof of Theorem 1. Suppose that the initial point q0 is chosen at q(α), α > 0. Since this q0 is not in the solution region, at least one correction step is required to arrive at a solution. The expected number of steps required to make one correction step is 1 Prob{p ∈P : V ( q(α), p) ≰O}, 5 which is greater than or equal to 1/Cα by Lemma 1. Since the expected achievement time EP (kA) is greater than or equal to this value, we have EP(kA) ≥1/Cα. Consider a (d−1)-dimensional manifold S on which sup ∈P λ[V (q, p)] = 0 and the point q♯is located. If we choose this manifold S small enough, there exists a positive α0 such that the inequality Prob{p ∈P : V (q + α(q♯−q∗), p) ≰O} ≤2Cα holds for any q ∈S and any 0 < α < α0. Since the expectation EQ[EP (kA)] is the integration over the whole Q, the integration over its subset {q + α(q♯−q∗) : q ∈S and 0 < α < α0} is less than or equal to this. We consequently have EQ[EP (kA)] ≥µQD α0 0 EP (kA) dα ≥µQD α0 0 1 2Cα dα = ∞ with some positive number D. □ 4. Improved algorithm As we can see in the proof of Theorem 1, the reason of the inﬁnite expected achievement time is that the probability to make a correction step is close to zero near the solution region. We hence increase this probability hoping to have the ﬁnite expected achievement time. Let a0, a1, . . . be a sequence of positive numbers that monotonically decreases to zero and replace V (qk, pk) in the preceding algorithm by V (qk, pk) + aℓI. In this improved algorithm, a correction step is made if V (qk, pk) + aℓI ≰O, which holds with more probability than V (qk, pk) ≰O. Before investigating the expected achievement time, we show that the achievement time remains ﬁnite with a probability one after the improvement above. Deﬁne a∗:= −sup ∈P λ[V (q∗, p)], which is positive by the deﬁnition of q∗. Let ℓ(k) be the value of ℓat the k-th iteration of the improved algorithm and suppose aℓ(k) < a∗. Then, since the function sup ∈P λ[V (q, p)] is convex, the ball with the center q∗and the radius [1 −aℓ(k)/a∗]r∗is included by the region {q ∈Q : sup ∈P λ[V (q, p) + aℓ(k)I] ≤0}. This enables us to use Proposition 1 with replacing rℓ(k) ≤r∗by rℓ(k) ≤[1 −aℓ(k)/a∗]r∗and V (qk, pk) by V (qk, pk) + aℓ(k)I. Following the reasoning to obtain Corollary 1, we arrive at the result below. Proposition 2. With a probability one, there exists a ﬁnite number k1 such that the point qk produced by the improved algorithm is contained in the solution region for any k ≥k1. Again we refer to the smallest k1 as the achievement time kA. We now consider the expected achievement time EP(kA) in the improved algorithm. As in the previous section, it is assumed that there exists a pair of q ∈Q and p ∈P satisfying V (q, p) ≰O. We do not need here the assumption on the density function in Q or on the point q♯. Let G be a positive number with which e1 ∂V (q, p) ∂p1 + · · · + en ∂V (q, p) ∂pn F < G 6 holds for any q ∈Q, any p ∈P, and any unit vector [e1 . . . en]T. Deﬁne R := sup ∈Q ∥q −q∗∥. The lemma below is important for our purpose. Lemma 2. Let a be any nonnegative number and q♯be any vector satisfying sup ∈P λ[V (q♯, p)] = 0. There then holds Prob{p ∈P : V (q♯, p) + aI ≰O} ≥min 1, µP a √nG n . This lemma is proved in Appendix C. From this lemma the following theorems are derived. Theorem 2. For any initial point q0, the expected achievement time EP(kA) deﬁned for the improved algorithm is ﬁnite. Theorem 3. Choose rℓ:= r0/ √ ℓ+ 1 and aℓ:= a0/ √ ℓ+ 1. Then, there holds EP (kA) < max e(R/r0)2r0 r∗+ a0 a∗+ 1 2 , e(R/r0)2[e(R/r0)2 + 1]n/2(√nG)n µP (a0)n r0 r∗+ a0 a∗+ 1 n+2 . (1) The right-hand side of (1) increases faster than the exponential order of n. This suggests that our algorithm suﬀers from the curse of dimensionality like the gridding algorithm, which was discussed in Section 1. However, the curse of dimensionality is removed if it is suﬃcient to obtain an “approximate solution” that satisﬁes the provided LMI not for all p ∈P but for almost all of them. This is shown next. Theorem 4. Let ϵ and δ be any positive numbers less than unity and let {qk} be a sequence produced by the improved algorithm. With a probability greater than 1 −δ, there exists a nonnegative integer k less than or equal to 2e(R/r0)2 ϵ r0 r∗+ a0 a∗+ 1 2 + 1 2ϵ2 ln 1 δ (2) that satisﬁes Prob{p ∈P : V (qk, p) ≤O} ≥1 −ϵ. Here ln denotes the natural logarithm. Choose ϵ and δ to be small positive numbers. Then, with a high probability, the time to obtain an approximate solution has the upper bound (2), which is independent of the parameter dimension n. The present algorithm is therefore advantageous to the gridding algorithm and other deterministic algorithms when the parameter dimension n is high. Proof of Theorem 2. If q0 is in the solution region, the achievement time is zero. We suppose in the rest of the proof that q0 is not in the solution region. Let ℓ0 be the smallest integer ℓthat attains rℓ≤(1 −aℓ/a∗)r∗. Note that the same inequality remains valid for any ℓ≥ℓ0. Let ℓ1 be the smallest integer satisfying ℓ1−1 ℓ=ℓ0 (rℓ)2 ≥R2. The number of correction steps required to achieve the solution region must be less than or equal to ℓ1 and, thus, is 7 ﬁnite. It is left for us to show that the number of steps required to make one correction step is ﬁnite on the average. Suppose that qk is not in the solution region. Since the corresponding ℓhas to be less than or equal to ℓ1, there holds aℓ≥aℓ1. Furthermore, there exists a point q♯satisfying sup ∈P λ[V (q♯, p)] = 0 on the line segment connecting qk and q∗. There holds by deﬁnition that rℓ1 ≤(1 −aℓ1/a∗)r∗, which implies aℓ1 < a∗and then sup ∈P λ[V (q∗, p) + aℓ1I] < 0. Since λ[V (q, p) + aℓ1I] is a convex function of q, if λ[V (q♯, p) + aℓ1I] > 0, it is inferred that λ[V (qk, p) + aℓ1I] > 0. By these observations, we can see that Prob{p ∈P : V (qk, p) + aℓI ≰O} ≥Prob{p ∈P : V (qk, p) + aℓ1I ≰O} ≥Prob{p ∈P : V (q♯, p) + aℓ1I ≰O} ≥min 1, µP aℓ1 √nG n . (3) The last inequality follows from Lemma 2. The expected number of steps required to make one correction step is the reciprocal of the leftmost expression in (3), which is shown to be ﬁnite. The proof is hence complete. □ Proof of Theorem 3. We evaluate ℓ0 and ℓ1 in the previous proof explicitly. Solve rℓ≤(1 −aℓ/a∗)r∗to have ℓ≥(r0/r∗+ a0/a∗)2 −1, which implies ℓ0 < (r0/r∗+ a0/a∗)2. Since ℓ−1 ℓ=ℓ0(rℓ)2 > ℓ ℓ0(r0)2/(x + 1) dx = (r0)2 ln[(ℓ+ 1)/(ℓ0 + 1)], the inequality ℓ≥e(R/r0)2(ℓ0 + 1) −1 guarantees ℓ−1 ℓ=ℓ0(rℓ)2 > R2. Therefore, we have ℓ1 < e(R/r0)2(ℓ0 + 1) < e(R/r0)2r0 r∗+ a0 a∗ 2 + 1 < e(R/r0)2r0 r∗+ a0 a∗+ 1 2 and 1 aℓ1 = √ℓ1 + 1 a0 < 1 a0 e(R/r0)2 r0 r∗+ a0 a∗ 2 + 1 + 1 < 1 a0 e(R/r0)2 + 1 r0 r∗+ a0 a∗+ 1 2 . The expected achievement time is no greater than the product of ℓ1 and the reciprocal of the rightmost expression in (3). Substitution of the above inequalities gives the desired result. □ Proof of Theorem 4. Let k be the smallest integer greater than (2). Suppose that the inequality Prob{p ∈P : V (qk, p) ≰O} > ϵ (4) holds for any 0 ≤k ≤ k −1. This implies that the point qk does not arrive in the solution region at least until k = k −1, which ensures ℓ< ℓ1, where ℓis the value of ℓat the time k −1 and ℓ1 is the number deﬁned in the proof of Theorem 3. Note that the inequality ℓ< ℓ1 is equivalent to ϵ − ℓ/ k > ϵ −ℓ1/ k. Consider in general that one carries out k Bernoulli trials whose probability of success is s. It is known as Chernoﬀ’s inequality that the probability to have s − ℓ/ k ≥A, where ℓis the number of success and A is any nonnegative number, is less than or equal to exp(−2 kA2) (see, e.g., p. 22, Vidyasagar, 1997). 8 Since we have (4) for any 0 ≤k ≤ k −1 and the number ϵ −ℓ1/ k is nonnegative by assumption, the probability to have ϵ − ℓ/ k > ϵ −ℓ1/ k is less than or equal to exp[−2( kϵ −ℓ1)2/ k]. This quantity is less than δ again by assumption. Now it is seen that the inequality (4) fails to hold at least one 0 ≤k ≤ k −1 with a probability greater than 1 −δ. □ 5. Example The proposed two algorithms, the basic one and the improved one, are applied to an example problem. g ν ℓ θ w L, W Figure 1. A considered plant We consider a plant given in Figure 1, which is a simpliﬁcation of the tower crane investigated by Takagi and Nishimura (1999). A uniform boom of the length L[m] and the weight W[kg] is connected to the ground by a free joint. From the top of the boom, a load of the weight w[kg] is hung down by a rope of the length ℓ[m]. The angles from the vertical line to the boom and to the rope are ν[rad] and θ[rad], respectively. One can add a force g[N] to the top of the boom in the horizontal direction. Especially, the force that makes the equilibrium boom angle equal to some given value ν0[rad] is denoted by g0[N]. Linearize the plant dynamics around the equilibrium point and choose x = [ν −ν0 θ ˙ ν ˙ θ]T as a state vector and u = g −g0 as an input. Then we have an LPV system ˙ x(t) = Ax(t) + bu(t). The matrix A and the column vector b are functions of the parameter p := [ℓ ν0 w]T. Suppose that L = 0.71m and W = 0.205kg here and that the parameter p can take any value in the set P = {p : 0.5 m ≤ℓ≤1 m, (40π/180)rad ≤ν0 ≤(50π/180)rad, 0.1 kg ≤w ≤0.2 kg}. We want to see whether the state feedback u = f Tx, f T = [12.25 −12.21 4.91 1.16], stabilizes this system for any parameter p in P. We suppose here for simplicity that the time derivative of the parameter p is zero. Let us consider a Lyapunov function having the form xTL(p)x with L(p) = L0 +ℓL1 +(cos ν0)L2 +wL3. Then, a suﬃcient condition for stability is the existence of symmetric matrices L0, . . . , L3 that satisfy  [A(p) + b(p)fT]TL(p) + L(p)[A(p) + b(p)f T] −L(p)  < O for any p in P. The problem to ﬁnd L0, . . . , L3 is formulated into the general form described in Section 2. Deﬁne q to be a vector consisting of the elements of L0, . . . , L3 and write the left-hand side of the above inequality as V (q, p). The matrices L0, . . . , L3 are seeked for in the region where each element of them belongs to the interval [−2, 2]. The variable domain Q is deﬁned as such. We choose the probability 9 density functions in P and Q to be those corresponding to the uniform distribution. The sequences {rℓ} and {aℓ} are deﬁned by rℓ= aℓ= 1/ √ ℓ+ 1. 0 1 2 3 4 5 6 0 5 10 15 20 ×105 Achievement time (a) Basic algorithm 0 1 2 3 4 5 6 0 5 10 15 20 ×105 Achievement time (b) Improved algorithm Figure 2. Distribution of the achievement time for 100 trials In Figure 2, (a) and (b) are histograms of the achievement time when we execute the basic and the improved algorithms 100 times for each. While the achievement time of the basic algorithm distributes in a rather wide range, that of the improved algorithm concentrates in the range of k ≤2×105. This shows the eﬀect of the improvement. Strictly speaking, the word “achievement time” is not used here in its original sense. Since it is diﬃcult to judge whether a provided qk satisﬁes V (qk, p) ≤O for all p ∈P, we regard qk to be a solution if this qk satisﬁes the inequality for all of 10000 p’s randomly sampled from P. Hence, this is an achievement time to an “approximate solution” considered in Theorem 4. By the result of Khargonekar and Tikku (1996) and Tempo, Bai, and Dabbene (1997), the fact that V (qk, p) ≤O holds for all of randomly sampled 10000 p’s implies that the inequality Prob{p ∈P : V (qk, p) ≤O} ≥0.999 holds with a probability greater than 0.9999. From this we expect that the achievement time measured here suﬃciently reﬂects the properties of the achievement time in the original sense. The time to execute 2×105 iterations in the improved algorithm is measured on Pentium III 1.0 GHz with 256 MBytes memory. On the average of 100 executions, the time is 106s without the achievement judgment described above. 6. Conclusion Randomized algorithms for parameter-dependent LMIs are proposed as a generalization of the algorithm of Polyak and Tempo (2000) and their computational complexity is analyzed. The following problems are left unsolved. The proposed algorithms do not contain an explicit termi-nation rule. Since it is diﬃcult to judge whether the considered LMI is satisﬁed for all the parameters p in P, it is considered to be practical to terminate the algorithms when the LMI is satisﬁed for all of many p’s randomly sampled from P. If one adds this termination rule to our algorithms, their computa-tional complexity is not suﬃciently characterized by the iteration number k. One has to reexamine their complexity using a more appropriate index. The computational time presented in the previous section 10 is not very long but not very short either. More speedup is preferable. It would be useful if one can explicitly evaluate the upper bounds given in Theorems 3 and 4. However, this is diﬃcult because these upper bounds include unknown numbers r∗and a∗. Explicitly computable upper bounds are desired. Finally, it is assumed throughout this paper that the considered LMI has a solution. It is also important to investigate the case that the LMI has no solution. Acknowledgment Thanks are due to Professors Roberto Tempo and Yasumasa Fujisaki. They provided the authors the full-paper version of Polyak and Tempo (2000). Their comments on an earlier version of this paper motivated improvement of the example in Section 5 and derivation of Theorem 4. A. Proof of Proposition 1 The proof is a complete parallel to that of Theorem 1 of Polyak and Tempo (2000). We write r instead of rℓ(k) for simplicity. Since r ≤r∗, the point q := q∗+ r ∇ ∥V (qk, pk)+∥F ∇ ∥V (qk, pk)+∥F is included in the solution region and thus ∥V ( q, pk)+∥F = 0. From V (qk, pk) ≰O it follows that ∥qk+1 −q∗∥2 ≤ qk −γk∇ ∥V (qk, pk)+∥F −q∗ 2 =∥qk −q∗∥2 + (γk)2 ∇ ∥V (qk, pk)+∥F 2 −2γk ∇ ∥V (qk, pk)+∥F T(qk − q) −2γk ∇ ∥V (qk, pk)+∥F T( q −q∗). (5) Since ∇ ∥V (qk, pk)+∥F is a subgradient, there holds on the second last term of (5) 0 = ∥V ( q, pk)+∥F ≥ ∥V (qk, pk)+∥F + ∇ ∥V (qk, pk)+∥F T( q −qk). On the last term of (5) we have ∇ ∥V (qk, pk)+∥F T( q − q∗) = r ∇ ∥V (qk, pk)+∥F . Substitution of these relations into (5) gives the desired inequality. B. Proof of Lemma 1 The proof requires two lemmas. Let ∇ det V (q♯, p) denote the vector whose i-th element is (∂/∂pi) det V (q♯, p). Lemma 3. There exists a (possibly multiple) p♯∈P that satisﬁes λ[V (q♯, p♯)] = 0. This p♯is on the surface of P and the value of ∥∇ det V (q♯, p♯)∥has a positive lower bound common to all of such p♯. If the parameter dimension n is even, the angle between the two vectors ∇ det V (q♯, p♯) and x −p♯is no greater than π/2 for any x ∈P other than p♯. If n is odd, the same is true with −∇ det V (q♯, p♯) instead of ∇ det V (q♯, p♯). 11 Proof. The existence of p♯is ensured by the deﬁnition of q♯and the compactness of P. Since all the eigenvalues of V (q♯, p) are negative at any p that cannot be p♯, the function det V (q♯, p) takes the minimum value at p♯if n is even and the maximum value otherwise. By assumption, the gradient ∇ det V (q♯, p♯) is not the zero vector, which means that p♯has to be on the surface of P. Because the set {p ∈P : λ[V (q♯, p)] = 0} is closed and bounded, there exists a positive minimum value for ∥∇ det V (q♯, p♯)∥. The last property is a consequence of the Karush-Kuhn-Tucker necessary condition for optimality. □ Lemma 4. There exists a positive number g such that det V ( q(α), p) = 0 implies maxi=1,... ,n \|pi\| ≥ 1 −gα for any small enough nonnegative α and any p ∈P. Proof. Let p♯be a vector considered in Lemma 3. With a guarantee of Lemma 3, one can choose g so that the inequality g∥∇ V (q♯, p♯)∥> (∂/∂α) det V ( q(α), p♯)\|α=0 holds for any p♯. We negate the claim of the lemma with this g and derive contradiction. Assume that there hold det V ( q(αj), xj) = 0 and maxi=1,... ,n \|xj i\| < 1 −gαj for some sequence of positive numbers α1, α2, . . . that monotonically decreases to zero and for some sequence of parameters x1, x2, . . . in P. Here, xj i denotes the i-th element of xj. The sequence {xj} has to have an accumulation point, which has to be one of the considered p♯’s. We can assume that the sequence {xj} converges to this p♯by taking its subsequence if necessary. Expand det V ( q(α), p) around α = 0 and p = p♯to have 0 = det V ( q(αj), xj) = ∂det V ( q( αj), xj) ∂α αj + [∇ det V ( q( αj), xj)]T(xj −p♯), (6) where αj →0 and xj →p♯as j →∞. From the assumption maxi=1,... ,n \|xj i\| < 1 −gαj, it follows that gαj < [∇ det V (q♯, p♯)]T ∥∇ det V (q♯, p♯)∥(xj −p♯) . Take the limit j →∞in this inequality and use (6) to conclude g∥∇ det V (q♯, p♯)∥≤lim j→∞ 1 αj [∇ det V (q♯, p♯)]T(xj −p♯) = lim j→∞ 1 αj [∇ det V ( q( αj), xj)]T(xj −p♯) = ∂det V ( q(α), p♯) ∂α α=0 . This contradicts with the deﬁnition of g. □ We now prove Lemma 1. Suppose that V ( q(α), p) ≰O for some small enough nonnegative α and some p ∈P. Since V ( q(0), p) ≤O, there exists 0 ≤ α < α with which det V ( q( α), p) = 0. Lemma 4 is invoked here to give maxi=1,... ,n \|pi\| ≥1 −g α > 1 −gα. By the above reasoning, the set {p ∈P : 1 ≥maxi=1,... ,n \|pi\| ≥1 −gα} includes the set {p ∈P : V ( q(α), p) ≰O}. The probability measure of the former set is at most µP 2n −(2 −2gα)n , which is less than or equal to µP 2nngα for a small enough nonnegative α. 12 C. Proof of Lemma 2 If a = 0 the considered inequality is trivial. Suppose that 0 < a < 2√nG. Since sup ∈P λ[V (q♯, p)] = 0, there exists p ∈P satisfying λ[V (q♯, p)] = 0. Write it as p♯to have λ[V (q♯, p♯) + aI] = a. From the deﬁnition of G, it follows that G∥p −p′∥> ∥V (q, p) −V (q, p′)∥F ≥ λ[V (q, p) + aI] −λ[V (q, p′) + aI] for any p and p′ in P. This implies that there holds V (q♯, p) + aI ≰O in the set {p : ∥p −p♯∥≤a/G}. Irrespective of the location of p♯, the intersection of this set and P includes the n-dimensional hypercube each edge of which has the length a/√nG. Since this hypercube ensures V (q♯, p) + aI ≰O in its inside and has the volume (a/√nG)n, the desired inequality holds. If a ≥2√nG, again the desired inequality is trivial because the set {p : ∥p −p♯∥≤a/G} includes the whole P. References Apkarian, P., & Adams, R. J. (1998). Advanced gain-scheduling techniques for uncertain systems. IEEE Trans-actions on Control Systems Technology, 6 (1), 21–32. Apkarian, P., Gahinet, P., & Becker, G. (1995). Self-scheduled H∞control of linear parameter-varying systems: a design example. Automatica, 31 (9), 1251–1261. Barmish, B. R., & Lagoa, C. M. (1997). The uniform distribution: a rigorous justiﬁcation for its use in robustness analysis. Mathematics of Control, Signals, and Systems, 10 (3), 203–222. Becker, G., & Packard, A. (1994). Robust performance of linear parametrically varying systems using parametrically-dependent linear feedback. Systems & Control Letters, 23 (3), 205–215. Becker, G., Packard, A., Philbrick, D., & Balas, G. (1993). Control of parametrically-dependent linear systems: a single quadratic Lyapunov approach. Proceedings of the 1993 American Control Conference, San Francisco, CA (pp. 2795–2799). Chen, X. & Zhou, K. (2000). Constrained robustness analysis by randomized algorithms. IEEE Transactions on Automatic Control, 45 (6), 1180–1186. Feron, E., Apkarian, P., & Gahinet, P. (1996). Analysis and synthesis of robust control systems via parameter-dependent Lyapunov functions. IEEE Transactions on Automatic Control, 41 (7), 1041–1046. Fujisaki, Y., Dabbene, F., & Tempo, R. (2001). Probabilistic robust design of LPV control systems. To be pre-sented at the 40th IEEE Conference on Decision and Control, Orlando, FL. Gahinet, P., Apkarian, P., & Chilali, M. (1996). Aﬃne parameter-dependent Lyapunov functions and real para-metric uncertainty. IEEE Transactions on Automatic Control, 41 (3), 436–442. Khargonekar, P., & Tikku, A. (1996). Randomized algorithms for robust control analysis and synthesis have poly-nomial complexity. Proceedings of the 35th IEEE Conference on Decision and Control, Kobe, Japan (pp. 3470– 3475). Masubuchi, I., & Shimemura, E. (1999). On application of the descriptor form to design of gain scheduling systems (in Japanese). Transactions of the Institute of Systems, Control and Information Engineers, 12 (7), 390–394. Ohara, A., Ide, M., Yamaguchi, Y., & Ohno, M. (1999). LPV modeling and gain scheduled control of ALFLEX vehicle ﬂight systems (in Japanese). Transactions of the Institute of Systems, Control and Information Engi-neers, 12 (11), 655–663. 13 Oishi, Y., & Kimura, H. (2001). Randomized algorithms to solve parameter-dependent linear matrix inequalities and their computational complexity. To be presented at the 40th IEEE Conference on Decision and Control, Orlando, FL. Polyak, B. T., & Tempo, R. (2000). Probabilistic robust design with linear quadratic regulators. Proceedings of the 39th IEEE Conference on Decision and Control, Sydney, Australia (pp. 1037–1042). The full-paper version: IRITI-CNR Report 7-00. Ray, L. R., & Stengel, R. F. (1993). A Monte Carlo approach to the analysis of control system robustness. Auto-matica, 29 (1), 229–236. Takagi, K., & Nishimura, H. (1999). Gain-scheduled control of a tower crane considering varying load-rope length. JSME International Journal, Ser. C, 42 (4), 914–921. Tempo, R., Bai, E. W., & Dabbene, F. (1997). Probabilistic robustness analysis: explicit bounds for the mini-mum number of samples. Systems & Control Letters, 30 (5), 237–242. Vidyasagar, M. (1997). A Theory of Learning and Generalization: With Applications to Neural Networks and Control Systems. London, UK: Springer. Vidyasagar, M. (1999). Randomized algorithms for robust controller synthesis using statistical learning theory. In Y. Yamamoto and S. Hara, Learning, Control and Hybrid Systems (pp. 3–24). London, UK: Springer. Watanabe, R., Uchida, K., Fujita, M., & Shimemura, E. (1994). L2 gain and H∞control of linear systems with scheduling parameter. Proceedings of the 33rd IEEE Conference on Decision and Control, Lake Buena Vista, FL (pp. 1412–1414). Wu, F., Yang, X. H., Packard, A., & Becker, G. (1996). Induced L2-norm control for LPV systems with bounded parameter variation rates. International Journal of Robust and Nonlinear Control, 6 (9–10), 983–998. Yu, J., & Sideris, A. (1997). H∞control with parametric Lyapunov functions. Systems & Control Letters, 30 (2–3), 57–69. 14
14732	https://mathoverflow.net/questions/82622/numbers-of-intersection-points-and-lines	Skip to main content Numbers of intersection points and lines Ask Question Asked Modified 8 years, 5 months ago Viewed 4k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. Hello, I don't know if this question has already been posted, I have made a little search with keywords and did not found it, sorry if I missed anything. Is it possible to characterize the set of pairs of integers (l,i) such that one can draw l lines on the euclidean plane with exactly i intersection points? It is quite trivial to see is that given l, an upper bound for i is l(l+1)/2. More generally, given l, any additive decomposition of l of the form ∑kj=1li provides a value for i which is ∑ki=1 ∑kj>ililj if we fix for any i∈[1,k] exactly li parallel lines such that there is no intersection of three or more lines at the same point. It is not difficult to see that there are pairs (l,i) that are not of this form. For instance, if you try all decompositions of 6, you may draw 6 lines with 5, 8, 9, 11, 12, 13, 14 or 15 intersection points with this method, but 7 and 10 are missing (they can be obtained with intersection points of three lines). Here is a link containing some observations ( As far as I know, this is the only place where this problem has been seriously considered, but it is quite old and maybe lacks of results. So any additional comment will be welcomed :) Just a final remark, thanks to some projective properties, this question is the same as finding c circles sharing a common point with exactly i+1 intersection points. Don't know if this can help. co.combinatorics discrete-geometry Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Dec 4, 2011 at 14:09 Edmund Harriss 1,31422 gold badges1616 silver badges3030 bronze badges asked Dec 4, 2011 at 13:58 NekochanNekochan 44933 silver badges1212 bronze badges 6 4 The expert on this topic is Branko Grünbaum, who recently summarized his extensive knowledge in the book, Configurations of Points and Lines: ams.org/bookstore-getitem/item=GSM-103 . – Joseph O'Rourke Commented Dec 4, 2011 at 14:47 So if you count points of infinity, each pair of lines must intersect one. The intersection of k lines has multiplicity k(k−1)/2. Also your upper bound for l lines should probably be l(l−1)/2. So you take l(l−1)/2, partition it into smaller triangular numbers, throw out the ones at infinity, and count the remaining ones. – Will Sawin Commented Dec 4, 2011 at 19:16 @Will: Easier said than done. – Igor Rivin Commented Dec 4, 2011 at 19:49 A related question is to characterize the pairs (ℓ,r) such that one can draw ℓ lines in the plane dividing the plane into exactly r regions. This was discussed at math.stackexchange.com/questions/38350/… and, given Euler's formula, perhaps some of the discussion and references there would be relevant. – Gerry Myerson Commented Dec 4, 2011 at 22:23 Thanks Joseph for the book however it won't be possible for me to read it :/ You're right Will, that's l(l-1)/2, not l(l+1)/2 Thanks Gerry for this discussion, I will take a look at this, maybe this can help – Nekochan Commented Dec 5, 2011 at 12:28 \| Show 1 more comment 1 Answer 1 Reset to default This answer is useful 6 Save this answer. Show activity on this post. I asked Jon Lenchner, an expert on point-line incidences, and he told me the question (in dual form) was posed in Grünbaum's 1971 book Arrangements and Spreads, and fully answered in a paper by Peter Salamon and Paul Erdős: "The solution to a problem of Grünbaum," Canad. Math. Bull. 31: 129-138 (1988). Here are its first two sentences: In the paper below we characterize for large n the possible values of the number of connecting lines determined by a set of Pn points in the plane, where a connecting line is any straight line containing at least two points of Pn. This solves a problem posed by Grünbaum [5,6] which asks for the sequence of all integers m with the property that some configuration of n points determine exactly m lines. ( is Arrangements and Spreads; is Erdős's earlier partial solution.) They obtain exact expressions "for the lower end of the continuum of values leading down from (n2)−4." "The possible values...can be seen to bear a strong resemblance to physical spectra." The lower end of the continuum grows as n3/2 (with constant 1). Here are two figures from the paper: Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 20, 2017 at 9:46 answered Dec 8, 2011 at 12:57 Joseph O'RourkeJoseph O'Rourke 152k3636 gold badges370370 silver badges984984 bronze badges 3 Joseph, can you elaborate on the nature of the duality here? It seems to me there might be something at stake in the difference. In particular, it's easy to get n lines to have zero points of intersection, but a lot harder to get n points to determine less than one line. – Barry Cipra Commented Dec 8, 2011 at 19:25 1 @Barry: I think the distinction only occurs in the Euclidean plane, but in the projective plane, n "parallel" lines meet at one point at infinity, and n collinear points determine just one line. – Joseph O'Rourke Commented Dec 9, 2011 at 0:29 That's very nice, thanks a lot for this reference! – Nekochan Commented Dec 11, 2011 at 13:21 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 co.combinatorics discrete-geometry See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... 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14733	https://www.wyzant.com/resources/answers/932502/identify-the-intervals-on-each-quadratic-function-is-positive	Identify the intervals on each quadratic function is positive \| Wyzant Ask An Expert Log inSign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us All Questions Search for a Question Find an Online Tutor Now Ask a Question for Free Login WYZANT TUTORING Log in Sign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us Subject ZIP Search SearchFind an Online Tutor NowAsk Ask a Question For Free Login Algebra 2 Destiny B. asked • 09/12/23 Identify the intervals on each quadratic function is positive y=x^2+2x-8 Y=x^2-5x-24 Y=-x^2+4x+12 Y=5x^2-3x-8 Follow •1 Add comment More Report 2 Answers By Expert Tutors Best Newest Oldest By: William C.answered • 09/12/23 Tutor 4.9(117) Experienced Tutor Specializing in Chemistry, Math, and Physics About this tutor› About this tutor› −For a quadratic function ax 2 + bx + c where coefficient a is positive (which is the case for each of the four given functions), then the function will be negative in between the zeroes, if they exist, zero at the zeroes (of course!), and positive everywhere else. You can find zeroes by factoring or by using the quadratic formula. y = x 2 + 2x – 8 = (x + 4)(x – 2) so y = 0 when x = −4 or x = 2 as Denise G. showed in her earlier response. So y < 0 on the interval (−4,2) and positive on the intervals (−∞,–4) and (2,∞) y = x 2 − 5x – 24 = (x + 3)(x – 8) so y = 0 when x = −3 or x = 8 So y is negative on the interval (−3,8) and positive on the intervals (−∞,−3) and (8,∞) y = x 2 + 4x + 12 can't be factored and you will find, based on the quadratic formula, that it has no zeroes (at least not any that are in the set of real numbers). You know there are no zeroes when you calculate b 2 − 4ac = (4)2− 4(1)(12) = 16 − 48 < 0. b 2 − 4ac ≥ 0 is required for the quadratic function to have zeroes (one zero if b 2 − 4ac = 0; two zeroes if b 2 − 4ac > 0). since the function y = x 2 + 4x + 12 has no zeroes that means it is never crosses the x-axis. This means that it is positive for all values of x. So y is positive on the interval (−∞,∞) y = 5x 2 − 3x – 8 = (x + 1)(5x – 8) so y = 0 when x = −1 or x = 8/5 So y is negative on the interval (−1,8/5) and positive on the intervals (−∞,−1) and (8/5,∞) 5x 2 − 3x – 8 is a little bit more challenging to factor (because a = 5 instead of 1). But you can always find the same zeroes using the quadratic formula if you get stuck in the factoring step. Upvote • 0Downvote Comment •1 More Report William C. tutor I just realized that I missed minus sign on the first term for 3rd quadratic function, which means I answered the wrong question! The actual function (where the minus is not omitted) can be factored, has two zeroes, and can be solved the same way as the other three. Report 09/12/23 Denise G.answered • 09/12/23 Tutor 5.0(540) Algebra, College Algebra, Prealgebra, Precalculus, GED, ASVAB Tutor About this tutor› About this tutor› y=x 2+2x-8 Factor this to find the zeros y=(x+4)(x-2) Set the factors equal to zero x+4=0 and x-2=0 x=-4 and x=2 are the zeros. These 2 zeros split the number line into 3 parts. Pick a test point in each of the 3 sections and evaluate at these values. I just need to sign to see if it is positive or negative x=-5 (-5+4)(-5-2) = (negative)(negative) = positive x=0 (0+4)(0-2) = (positive)(negative) = negative x=3 (3+4)(3-2) = (positive)(positive) = positive (-∞,-4) U (2,∞) Upvote • 0Downvote Add comment More Report Still looking for help? Get the right answer, fast. Ask a question for free Get a free answer to a quick problem. Most questions answered within 4 hours. OR Find an Online Tutor Now Choose an expert and meet online. 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14734	https://courses.lumenlearning.com/uvu-combinedalgebra/chapter/7-2-transformations-of-the-rational-functions-fx1-x/	7.2: Transformations of the Rational Function \| Intermediate Algebra Skip to main content Intermediate Algebra Chapter 7: Rational Functions Search for: 7.2: Transformations of the Rational Function Learning Objectives For the rational parent function f(x)=1 x f(x)=1 x, Perform vertical and horizontal shifts Perform vertical stretches and compressions Perform reflections across the x x-axis Perform reflections across the y y-axis Determine the transformations performed on the parent function f(x)=1 x f(x)=1 x to get the rational function f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k Vertical Shifts If we shift the graph of the rational function f(x)=1 x f(x)=1 x up 5 units, all of the points on the graph increase their y y-coordinates by 5, but their x x-coordinates remain the same. Therefore, the equation of the function f(x)=1 x f(x)=1 x after it has been shifted up 5 units transforms to f(x)=1 x+5 f(x)=1 x+5. Table 1 shows the changes to specific values of this function, which are replicated on the graph in figure 1. \| x x \| 1 x 1 x \| 1 x+5 1 x+5 \| Figure 1. Shifting the graph of the function f(x)=1 x f(x)=1 x up 5 units. \| --- \| –2–2 \| −1 2−1 2 \| 9 2 9 2 \| \| –1–1 \| –1–1 \| 4 4 \| \| −1 2−1 2 \| –2–2 \| 3 3 \| \| 1 2 1 2 \| 2 2 \| 7 7 \| \| 1 1 \| 1 1 \| 6 6 \| \| 2 2 \| 1 2 1 2 \| 11 2 11 2 \| \| 3 3 \| 1 3 1 3 \| 16 3 16 3 \| \| Table 1. f(x)=1 x f(x)=1 x is transformed to f(x)=1 x+5 f(x)=1 x+5. \| If we shift the graph of the function f(x)=1 x f(x)=1 x down 8 units, all of the points on the graph decrease their y y-coordinates by 8, but their x x-coordinates remain the same. Therefore, the equation of the function f(x)=1 x f(x)=1 x after it has been shifted down 8 units transforms to f(x)=1 x−8 f(x)=1 x−8. Table 2 shows the changes to specific values of this function, which are replicated on the graph in figure 2. \| f(x)f(x) \| 1 x 1 x \| f(x)=1 x−8 f(x)=1 x−8 \| Figure 2. Shifting the graph of the function f(x)=1 x f(x)=1 x down 8 units. \| --- \| –2–2 \| −1 2−1 2 \| −17 2−17 2 \| \| –1–1 \| –1–1 \| –9–9 \| \| −1 2−1 2 \| –2–2 \| –10–10 \| \| 1 2 1 2 \| 2 2 \| –6–6 \| \| 1 1 \| 1 1 \| –7–7 \| \| 2 2 \| 1 2 1 2 \| −15 2−15 2 \| \| 3 3 \| 1 3 1 3 \| −23 3−23 3 \| \| Table 2.f(x)=1 x f(x)=1 x is transformed to f(x)=1 x−8 f(x)=1 x−8. \| Vertical shifts We can represent a vertical shift of the graph of the function f(x)=1 x f(x)=1 x by adding a constant, k k, to the function: f(x)=1 x+k f(x)=1 x+k If k>0 k>0, the graph shifts upwards and if k<0 k<0 the graph shifts downwards. Manipulate the graph in figure 3 to shift the graph vertically. Pay attention to what happens with the function as k k changes value. Also, watch what happens with the asymptotes. Figure 3. Vertical Transformations Example 1 Use figure 3 to graph the function f(x)=1 x−6 f(x)=1 x−6. What is the horizontal asymptote of the function? What is the relationship between the value of k k and the horizontal asymptote? What is the vertical asymptote of the function? Without graphing the function, what is the horizontal asymptote of the function g(x)=1 x+5 g(x)=1 x+5? What is the vertical asymptote? Solution 1. The horizontal asymptote is the line y=−6 y=−6. The value of k k is −6−6 so the horizontal asymptote mimics the value of k k. The vertical asymptote is the line x=0 x=0. Since k=5 k=5 the graph of f(x)=1 x f(x)=1 x gets shifted up by 5 units. This means that the horizontal asymptote is shifted up 5 units to y=5 y=5. The vertical asymptote does not move so is still x=0 x=0. Try It 1 Use figure 3 to graph the function f(x)=1 x+3 f(x)=1 x+3. What is the horizontal asymptote of the function? What is the relationship between the value of k k and the horizontal asymptote? What is the vertical asymptote of the function? Without graphing the function, what is the horizontal asymptote of the function g(x)=1 x−7 g(x)=1 x−7? What is the vertical asymptote? Show Answer 1. The horizontal asymptote is the line y=3 y=3. The value of k k is 3 3 so the horizontal asymptote mimics the value of k k. The vertical asymptote is the line x=0 x=0. The horizontal asymptote is y=−7 y=−7. The vertical asymptote is x=0 x=0. Example 2 The graph of the rational function y=f(x)y=f(x) has a horizontal asymptote at y=9 y=9 and a vertical asymptote at x=0 x=0. Determine a rational functional f(x)f(x) whose graph meets these criteria. Solution Since the vertical asymptote is at x=0 x=0 there is no horizontal shift from the parent function f(x)=1 x f(x)=1 x. Since the horizontal asymptote is at y=9 y=9, there has been a vertical shift of 9 units up from the parent function f(x)=1 x f(x)=1 x. This means that k=9 k=9, and the function is f(x)=1 x+9 f(x)=1 x+9. Try It 2 The graph of the rational function y=f(x)y=f(x) has a horizontal asymptote at y=−12 y=−12 and a vertical asymptote at x=0 x=0. Determine a rational functional f(x)f(x) whose graph meets these criteria. Show Answer f(x)=1 x−12 f(x)=1 x−12 Horizontal Shifts If we shift the graph of the function f(x)=1 x f(x)=1 x right 7 units, all of the points on the graph increase their x x-coordinates by 7, but their y y-coordinates remain the same. The point (1, 1) in the original graph is moved to (8, 1). Any point (x,y)(x,y) on the original graph is moved to (x+7,y)(x+7,y)(figure 4). But what happens to the original function f(x)=1 x f(x)=1 x? An automatic assumption may be that since x x moves to x+7 x+7 that the function will become f(x)=1 x+7 f(x)=1 x+7. But that is NOT the case. Remember that the x x-intercept is moved to (8, 1) and if we substitute x=8 x=8 into the function f(x)=1 x+7 f(x)=1 x+7 we get f(x)=1 15≠1 f(x)=1 15≠1!! The way to get a function value of 1 is for the transformed function to be f(x)=1 x−7 f(x)=1 x−7. Then f(8)=1 8−7=1 f(8)=1 8−7=1. So the function f(x)=1 x f(x)=1 x transforms to f(x)=1 x−7 f(x)=1 x−7 after being shifted 7 units to the right. The reason is that when we move the function 7 units to the right, the x x-value increases by 7 and to keep the corresponding y y-coordinate the same in the transformed function, the x x-coordinate of the transformed function needs to subtract 7 to get back to the original x x that is associated with the original y y-value. Table 3 shows the changes to specific values of this function, and the graph is shown in figure 4. \| x x \| x−7 x−7 \| f(x)=1 x−7 f(x)=1 x−7 \| Figure 4. Shifting the graph of the function f(x)=1 x f(x)=1 x right 7 units. \| --- \| 5 5 \| −2−2 \| −1 2−1 2 \| \| 6 6 \| −1−1 \| −1−1 \| \| 13 2 13 2 \| −1 2−1 2 \| −2−2 \| \| 15 2 15 2 \| 1 2 1 2 \| 2 2 \| \| 8 8 \| 1 1 \| 1 1 \| \| 9 9 \| 2 2 \| 1 2 1 2 \| \| 10 10 \| 3 3 \| 1 3 1 3 \| \| Table 3. Shifting the graph right by 7 units transforms f(x)=1 x f(x)=1 x into f(x)=1 x−7 f(x)=1 x−7 . \| On the other hand, if we shift the graph of the function f(x)=1 x f(x)=1 x left by 11 units, all of the points on the graph decrease their x x-coordinates by 11, but their y y-coordinates remain the same. So any point (x,y)(x,y) on the original graph moves to (x−11,y)(x−11,y). Consequently, to keep the same y y-values we need to increase the x x-value by 11 in the transformed function. The equation of the function after being shifted left 11 units is f(x)=1 x+11 f(x)=1 x+11. Table 4 shows the changes to specific values of this function, and the graph is shown in figure 5. \| x x \| x+11 x+11 \| f(x)=1 x+11 f(x)=1 x+11 \| Figure 5. Shifting the graph of the function f(x)=1 x f(x)=1 x left 11 units. \| --- \| –13–13 \| −2−2 \| −1 2−1 2 \| \| −12−12 \| −1−1 \| −1−1 \| \| −23 2−23 2 \| −1 2−1 2 \| −2−2 \| \| −21 2−21 2 \| 1 2 1 2 \| 2 2 \| \| −10−10 \| 1 1 \| 1 1 \| \| −9−9 \| 2 2 \| 1 2 1 2 \| \| −8−8 \| 3 3 \| 1 3 1 3 \| \| Table 4. Shifting the graph left by 11 units transforms f(x)=1 x f(x)=1 x into f(x)=1 x+11 f(x)=1 x+11. \| horizontal shifts We can represent a horizontal shift of the graph of the function f(x)=1 x f(x)=1 x by subtracting a constant, h h, from the variable x x. f(x)=1 x−h f(x)=1 x−h If h>0 h>0 the graph shifts toward the right and if h<0 h<0 the graph shifts to the left. Manipulate the graph in figure 6 to shift the graph horizontally. Pay attention to what happens with the function as h h changes value. Also, watch what happens with the asymptotes. Figure 6. Horizontal shifts Example 3 Use figure 6 to graph the function f(x)=1 x+6 f(x)=1 x+6. What is the vertical asymptote of the function? What is the relationship between the value of h h and the vertical asymptote? What is the horizontal asymptote of the function? Without graphing the function, what is the vertical asymptote of the function g(x)=1 x+3 g(x)=1 x+3? What is the horizontal asymptote? Solution 1. The vertical asymptote is the line x=−6 x=−6. The value of h h is −6−6 so the vertical asymptote mimics the value of h h. The horizontal asymptote is the line y=0 y=0. Since h=−3 h=−3 the graph of f(x)=1 x f(x)=1 x gets shifted left by 3 units. This means that the vertical asymptote is shifted left 3 units to x=−3 x=−3. The horizontal asymptote does not move so is still y=0 y=0. Try It 3 Use figure 6 to graph the function f(x)=1 x+3 f(x)=1 x+3. What is the vertical asymptote of the function? What is the relationship between the value of h h and the vertical asymptote? What is the horizontal asymptote of the function? Without graphing the function, what is the vertical asymptote of the function g(x)=1 x−7 g(x)=1 x−7? What is the horizontal asymptote? Show Answer 1. The vertical asymptote is the line x=−3 x=−3. The value of h h is −3−3 so the vertical asymptote mimics the value of h h. The horizontal asymptote is the line y=0 y=0. The vertical asymptote is x=7 x=7. The horizontal asymptote is y=0 y=0. Example 4 The graph of the rational function y=f(x)y=f(x) has a horizontal asymptote at y=0 y=0 and a vertical asymptote at x=5 x=5. Determine a rational functional f(x)f(x) whose graph meets these criteria. Solution Since the horizontal asymptote is at y=0 y=0 there is no vertical shift from the parent function f(x)=1 x f(x)=1 x. Since the vertical asymptote is at x=5 x=5, there has been a horizontal shift of 5 units right from the parent function f(x)=1 x f(x)=1 x. This means that h=5 h=5, and the function is f(x)=1 x−5 f(x)=1 x−5. Try It 4 The graph of the rational function y=f(x)y=f(x) has a horizontal asymptote at y=0 y=0 and a vertical asymptote at x=−9 x=−9. Determine a rational functional f(x)f(x) whose graph meets these criteria. Show Answer f(x)=1 x+9 f(x)=1 x+9 Combining Vertical and Horizontal Shifts vertical and horizontal shifts We can represent both a vertical and a horizontal shift of the graph of the function f(x)=1 x f(x)=1 x by the transformed function f(x)=1 x−h+k f(x)=1 x−h+k If h>0 h>0 the graph shifts toward the right and if h<0 h<0 the graph shifts to the left. If k>0 k>0 the graph shifts upwards and if k<0 k<0 the graph shifts downwards. When the parent function f(x)=1 x f(x)=1 x is shifted both vertically and horizontally, there will be non-zero values for both h h and k k. The graph in figure 7 can be manipulated both vertically and horizontally. Pay attention to what happens to the function as the values for h h and k k change. Also notice what happens to the asymptotes. Figure 7. Vertical and horizontal shifts Example 5 The graph of the parent function f(x)=1 x f(x)=1 x is shifted up by 4 units and left by 7 units. Determine the equation of the transformed function. Determine the vertical asymptote. Determine the horizontal asymptote. The point (2,1 2)(2,1 2) lies on the parent function. Where does it end up after the transformation? Solution Shifted up by 4 units means k=4 k=4 and shifted left by 7 units means h=−7 h=−7. Therefore the function is f(x)=1 x−h+k=1 x+7+4 f(x)=1 x−h+k=1 x+7+4. Since the graph is shifted left 7 units, the vertical asymptote is also shifted left 7 units: x=−7 x=−7. Since the graph is shifted up 4 units, the horizontal asymptote is also shifted up 4 units: y=4 y=4. Since the graph is shifted left 7 units, the x x-coordinate is shifted left 7 units to 2 – 7 = –5.Since the graph is shifted up 4 units, the y y-coordinate is shifted up 4 units to 1 2+4=9 2 1 2+4=9 2. Consequently,(2,1 2)(2,1 2) moves to(−5,9 2)(−5,9 2). Try It 5 The graph of the parent function f(x)=1 x f(x)=1 x is shifted down by 3 units and right by 2 units. Determine the equation of the transformed function. Determine the vertical asymptote. Determine the horizontal asymptote. The point (2,1 2)(2,1 2) lies on the parent function. Where does it end up after the transformation? Show Answer f(x)=1 x−2−3 f(x)=1 x−2−3 x=2 x=2 y=−3 y=−3 (4,−5 2)(4,−5 2) Stretching and Compressing If we vertically stretch the graph of the function f(x)=1 x f(x)=1 x by a factor of 7, all of the y y-coordinates of the points on the graph are multiplied by 7, but their x x-coordinates remain the same. The equation of the function after the graph is stretched is f(x)=7×1 x=7 x f(x)=7×1 x=7 x. The reason for multiplying f(x)=1 x f(x)=1 x by 7 is that each y y-coordinate is made 7 times larger, and since y=1 x y=1 x, 1 x 1 x is also made 7 times larger. Table 5 shows this change and the graph is shown in figure 8. \| x x \| 1 x 1 x \| f(x)=7 x f(x)=7 x \| Figure 8. Stretching the graph vertically. \| --- \| \| −2−2 \| −1 2−1 2 \| −7 2−7 2 \| \| −1−1 \| −1−1 \| −7−7 \| \| −1 2−1 2 \| −2−2 \| −14−14 \| \| 1 2 1 2 \| 2 2 \| 14 14 \| \| 1 1 \| 1 1 \| 7 7 \| \| 2 2 \| 1 2 1 2 \| 7 2 7 2 \| \| 3 3 \| 1 3 1 3 \| 7 3 7 3 \| \| Table 5.Stretching the graph vertically 7 times transforms 1 x 1 x into 7 x 7 x. \| On the other hand, if we vertically compress the graph of the function 1 x 1 x into 1 5 1 5 of its original height, all of the y y-coordinates of the points on the graph are divided by 5, but their x x-coordinates remain the same. This means the y y-coordinates are multiplied by 1 5 1 5. The equation of the function after being compressed is f(x)=1 5×1 x=1 5 x f(x)=1 5×1 x=1 5 x. The reason for multiplying f(x)=1 x f(x)=1 x by 1 5 1 5 is that each y y-coordinate becomes 1 5 1 5 of the original value. Table 6 shows this change and the graph is shown in figure 9. \| x x \| 1 x 1 x \| 1 5 x 1 5 x \| Figure 9. Compress the graph into 1 5 1 5 of the original height. \| --- \| \| −2−2 \| −1 2−1 2 \| −1 10−1 10 \| \| −1−1 \| −1−1 \| −1 5−1 5 \| \| −1 2−1 2 \| −2−2 \| −2 5−2 5 \| \| 1 2 1 2 \| 2 2 \| 2 5 2 5 \| \| 1 1 \| 1 1 \| 1 5 1 5 \| \| 2 2 \| 1 2 1 2 \| 1 10 1 10 \| \| 3 3 \| 1 3 1 3 \| 1 15 1 15 \| \| Table 6.Compressing the graph vertically into 1 5 1 5 of the original height transforms f(x)=1 x f(x)=1 x into f(x)=1 5 x f(x)=1 5 x. \| vertical stretching and compressing A vertical stretch or compression of the graph of the function f(x)=1 x f(x)=1 x can be represented by multiplying the function by a constant, a>0 a>0. f(x)=a 1 x f(x)=a 1 x The magnitude of a a indicates the stretch/compression of the graph. If a>1 a>1, the graph is stretched. If [latex]0 Move the red dot in figure 10 to change the value of a a. Notice whether the graph is stretching or compressing depending on the value of a a. Notice also what happens to the function. FIgure 10. Stretching and compressing Example 6 The parent function f(x)=1 x f(x)=1 x is stretched by a factor of 5. Determine the equation of the transformed function.The point (1, 1) lies on the parent function. What happens to this point after the transformation? 2.The parent function f(x)=1 x f(x)=1 x is compressed to 1 2 1 2 its size. Determine the equation of the transformed function.The point (1, 1) lies on the parent function. What happens to this point after the transformation? Solution Stretched by a factor of 5 means a=5 a=5, therefore, the transformed function is f(x)=5(1 x)f(x)=5(1 x). This can also be written as f(x)=5 x f(x)=5 x. When a function is stretched its x x-value stays the same while the y y-value is multiplied by the stretch factor. So, (1, 1) moves to (1, 5). Compressed to 1 2 1 2 its size means a=1 2 a=1 2, therefore, the transformed function is f(x)=1 2(1 x)f(x)=1 2(1 x). This can also be written as f(x)=1 2 x f(x)=1 2 x. When a function is compressed its x x-value stays the same while the y y-value is multiplied by the compression. So, (1, 1) moves to (1,1 5)(1,1 5). Try It 6 The parent function f(x)=1 x f(x)=1 x is stretched by a factor of 4. Determine the equation of the transformed function.The point (1, 1) lies on the parent function. What happens to this point after the transformation? 2.The parent function f(x)=1 x f(x)=1 x is compressed to 1 8 1 8 th its size. Determine the equation of the transformed function.The point (1, 1) lies on the parent function. What happens to this point after the transformation? Show Answer f(x)=4(1 x)=4 x(1,4)f(x)=4(1 x)=4 x(1,4) f(x)=1 8(1 x)=1 8 x(1,1 8)f(x)=1 8(1 x)=1 8 x(1,1 8) Reflections across the x x-axis When the graph of the function f(x)=1 x f(x)=1 x is reflected across the x x-axis, the y y-coordinates of all of the points on the graph change their signs, from positive to negative values or from negative to positive values, while the x x-coordinates remain the same. The equation of the function after f(x)=1 x f(x)=1 x is reflected across the x x-axis is f(x)=−1 x f(x)=−1 x. Table 7 shows the effect of such a reflection on the functions values and the graph is shown in figure 11. \| x x \| 1 x 1 x \| −1 x−1 x \| Figure 11. Reflecting the graph of the function f(x)=1 x f(x)=1 x across the x x-axis. \| --- \| \| –2 \| −1 2−1 2 \| 1 2 1 2 \| \| –1 \| –1 \| 1 \| \| −1 2−1 2 \| –2 \| 2 \| \| 1 2 1 2 \| 2 \| –2 \| \| 1 \| 1 \| –1 \| \| 2 \| 1 2 1 2 \| −1 2−1 2 \| \| 3 \| 1 3 1 3 \| −1 3−1 3 \| \| Table 7.Reflecting the graph of f(x)=1 x f(x)=1 x across the x x-axis transforms f(x)=1 x f(x)=1 x into f(x)=−1 x f(x)=−1 x. \| across the y y-axis When the graph of the function f(x)=1 x f(x)=1 x is reflected across the y y-axis, the x x-coordinates of all of the points on the graph change their signs, from positive to negative values, or from negative to positive values, while the y y-coordinates remain the same. The equation of the function after f(x)=1 x f(x)=1 x is reflected across the y y-axis is f(x)=1−x=−1 x f(x)=1−x=−1 x. This equation is the same as the equation after being reflected across the x x-axis. Therefore, its graph is the same as the graph after being reflected across the x x-axis, even though it got there by a different route (note the differences in figures 11 and 13).Table 8 shows the effect of such a reflection on the functions values and the graph is shown in figure 13. \| x x \| 1 x 1 x \| 1−x 1−x \| Figure 13. Reflecting the graph of the function f(x)=1 x f(x)=1 x across the y y-axis. \| --- \| \| –2 \| −1 2−1 2 \| 1 2 1 2 \| \| –1 \| –1 \| 1 \| \| −1 2−1 2 \| –2 \| 2 \| \| 1 2 1 2 \| 2 \| –2 \| \| 1 \| 1 \| –1 \| \| 2 \| 1 2 1 2 \| −1 2−1 2 \| \| 3 \| 1 3 1 3 \| −1 3−1 3 \| \| Table 8.Reflecting the graph of f(x)=1 x f(x)=1 x across the y y-axis transforms f(x)=1 x f(x)=1 x into f(x)=1−x f(x)=1−x. \| Example 7 Explain why the parent function f(x)=1 x f(x)=1 x is transformed to f(x)=−1 x f(x)=−1 x when it is reflected across the x x-axis. Solution With reflection across the x x-axis, the x x values stay the same and the y y-values change sign. So, y=1 x y=1 x transforms to −y=1 x−y=1 x. If we multiply (or divide) both sides of the equation by –1 we get, y=−1 x y=−1 x. Consequently, the transformed function is f(x)=−1 x f(x)=−1 x. Try It 7 Explain why the parent function f(x)=1 x f(x)=1 x is transformed to f(x)=−1 x f(x)=−1 x when it is reflected across the y y-axis. Show Answer With reflection across the y y-axis, the y y values stay the same and the x x-values change sign. So, y=1 x y=1 x transforms to y=1−x y=1−x, which simplifies to y=−1 x y=−1 x. Consequently, the transformed function is f(x)=−1 x f(x)=−1 x. Combining Transformations Now that we have learned all the transformations for the function f(x)=1 x f(x)=1 x, we should be able to write the transformed function equation given specific transformations, and determine what transformations have been performed on the function f(x)=1 x f(x)=1 x, given an arbitrary transformed function f(x)=a 1 x−h+k f(x)=a 1 x−h+k. Combining Transformations The graph of the function f(x)=1 x f(x)=1 x can be shifted horizontally (h h-value), shifted vertically(k k-value), stretched or compressed (a a-value), reflected across the x x-axis (negative function value} or reflected across the y y-axis (negative x x-value) and can be represented by the function f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k Example 8 The parent function f(x)=1 x f(x)=1 x is transformed. Write the equation of the transformed function. Stretched by a factor of 2; shifted vertically up 3 units; shifted horizontally 5 units right. Shifted horizontally left by 7 units; reflected across the x x-axis. Compressed to 1 3 1 3 rd its height; reflected across the y y-axis; shifted vertically down by 4 units. Solution a=2,k=3,h=5 a=2,k=3,h=5 so function is f(x)=a(1 x−h)+k=2(1 x−5)+3 f(x)=a(1 x−h)+k=2(1 x−5)+3 a=−1,k=0,h=−7 a=−1,k=0,h=−7 so function is f(x)=a(1 x−h)+k=−1 x+7 f(x)=a(1 x−h)+k=−1 x+7 a=1 3,x→−x,k=−4 a=1 3,x→−x,k=−4 so function is f(x)=1 3(1−x)−4=−1 3(1 x)−4 f(x)=1 3(1−x)−4=−1 3(1 x)−4. This can also be written as f(x)=−1 3 x−4 f(x)=−1 3 x−4. Try It 8 The parent function f(x)=1 x f(x)=1 x is transformed. Write the equation of the transformed function. Stretched by a factor of 4; shifted vertically down by 7 units; shifted horizontally 3 units left. Shifted horizontally left by 8 units; reflected across the x x-axis. Compressed to 1 5 1 5 th its height; reflected across the x x-axis; shifted vertically down by 4 units; shifted right by 2 units. Show Answer f(x)=4(1 x+3)−7 f(x)=4(1 x+3)−7 f(x)=−1 x+8 f(x)=−1 x+8 f(x)=−1 5(1 x−2)−4 f(x)=−1 5(1 x−2)−4 Example 9 Determine the transformations made to the parent function f(x)=1 x f(x)=1 x to get the transformed function f(x)=−7 x−4+3 f(x)=−7 x−4+3. Solution f(x)=−7 x−4+3 f(x)=−7 x−4+3 can be written as f(x)=−7(1 x−4)+3 f(x)=−7(1 x−4)+3. Comparing f(x)=−7(1 x−4)+3 f(x)=−7(1 x−4)+3 to f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k, we get a=−7,h=4,k=3 a=−7,h=4,k=3, which translated to a reflection across the x x-axis, a stretch by a factor of 7, a horizontal shift to the right by 4 units, and a vertical shift up by 3 units. Try It 9 Determine the transformations made to the parent function f(x)=1 x f(x)=1 x to get the transformed function f(x)=−5 3 x−7 f(x)=−5 3 x−7. Show Answer Reflection across the x x-axis (negative sign in front) Stretched to 5 3 5 3 rds its height Shifted down by 7 units Formats of Rational Functions You may be wondering why we defined a rational function as f(x)=P(x)Q(x)f(x)=P(x)Q(x) when we write a rational function using transformations in the form f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k. It is simply that rational functions in the form f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k are based on the parent rational function f(x)=1 x f(x)=1 x, and a special case of f(x)=P(x)Q(x)f(x)=P(x)Q(x). We could take the function f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k and write it as a single fraction by using a common denominator: f(x)=a(1 x−h)+k=a(1 x−h)+k⋅x−h x−h=a+k(x−h)x−h=k x+(a−h k)x−h f(x)=a(1 x−h)+k=a(1 x−h)+k⋅x−h x−h=a+k(x−h)x−h=k x+(a−h k)x−h Now the function is written in standard form with P(x)=k x+(a−h k)P(x)=k x+(a−h k) and Q(x)=x−h Q(x)=x−h. Both P(x)P(x) and Q(x)Q(x) have degree 1. Consequently, a rational function in the transformed form f(x)=a(1 x−h)+k f(x)=a(1 x−h)+k is a special case of the standard form f(x)=P(x)Q(x)f(x)=P(x)Q(x) with P(x)P(x) and Q(x)Q(x) having degree 1. We leave the more in depth study of rational functions in the form f(x)=P(x)Q(x)f(x)=P(x)Q(x) to College Algebra. Candela Citations CC licensed content, Original Transformations of the Rational Function. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution All graphs created using desmos graphing calculator. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution Examples and Try Its: hjm245; hjm236; hjm518; hjm208; hjm730; hjm400; hjm614; hjm004; hjm613. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution Formats of Rational Functions. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution Combining Vertical and Horizontal Shifts. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution Licenses and Attributions CC licensed content, Original Transformations of the Rational Function. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution All graphs created using desmos graphing calculator. Authored by: Leo Chang and Hazel McKenna. Provided by: Utah Valley University. Located at: License: CC BY: Attribution Examples and Try Its: hjm245; hjm236; hjm518; hjm208; hjm730; hjm400; hjm614; hjm004; hjm613. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution Formats of Rational Functions. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution Combining Vertical and Horizontal Shifts. Authored by: Hazel McKenna. Provided by: Utah Valley University. 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14735	https://math.stackexchange.com/questions/1647246/how-to-derive-the-equation-of-tangent-to-an-arbitrarily-point-on-a-ellipse	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 derive the equation of tangent to an arbitrarily point on a ellipse? Ask Question Asked Modified 3 years, 3 months ago Viewed 3k times 5 $\begingroup$ Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$ I've tried implicit differentiation $\to \frac{2x\frac{d}{dx}}{a^2}+\frac{2y\frac{d}{dy}}{b^2} = 0$, but not sure where to go from here. Substituting $P$ doesn't seem to help me much, and solving for $y$ from original equation seems to cause me more trouble than help. Please don't give me the solution, rather just give a slight hint or two:) Thanks in advance! Edit: Solving for $\frac{dy}{dx}$ and putting it into point slope gives me: $\frac{d}{dx} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] = \frac{d}{dx}\left[1\right] \to \frac{2x}{a^2} + \frac{y}{b^2}\frac{dy}{dx}= 0\to \frac{dy}{dx}=-\frac{xb^2}{ya^2} \to \frac{dy}{dx}(P)=-\frac{x_0b^2}{y_0a^2}$ Then we get: $y-y_0=\frac{dy}{dx}\left(x-x_0\right)\to y = -\frac{x_0b^2}{y_0a^2}(x-x_0)+y_0\to y = \frac{x_0^2b^2}{y_0a^2}-\frac{xx_0b^2}{y_0a^2} + y_0\to /:b^2,y_0\to \frac{yy_0}{b^2}+\frac{xx_0}{a^2}=\frac{y_0^2}{b^2}$ Which looks close, but not exactly the expression i wanted. Where is the error? calculus conic-sections Share edited Feb 9, 2016 at 9:35 ThomasThomas asked Feb 9, 2016 at 8:32 ThomasThomas 79111 gold badge1010 silver badges1818 bronze badges $\endgroup$ 2 1 $\begingroup$ Solve for $\frac{ dy}{d x}$ from implicit differentiation. Use this as the slope of the tangent line, and $(x_0,y_0)$ as the point on the line for point-slope form of a line. Post your attempt. $\endgroup$ OnceUponACrinoid – OnceUponACrinoid 2016-02-09 08:38:49 +00:00 Commented Feb 9, 2016 at 8:38 1 $\begingroup$ I'm not sure if you understand implicit differentiation correctly but your notation seems off. Your $\frac{d}{dx}$ should be just $dx$ (and the same with $dy$). $\endgroup$ Tunococ – Tunococ 2016-02-09 08:40:56 +00:00 Commented Feb 9, 2016 at 8:40 Add a comment \| 6 Answers 6 Reset to default 6 $\begingroup$ This probably isn't the kind of answer you're looking for, but it's one I like because it doesn't use calculus. The ellipse is the image of the unit circle under the linear mapping $T(x,y) = (ax,by)$. The tangent line to the ellipse is the image under $T$ of the tangent line to the circle. We have $n = T^{-1}(x_0,y_0) = (x_0/a,y_0/b)$, the corresponding point on the unit circle. The equation of the tangent line to the circle at $n$ is $1 = n \cdot (x,y) = xx_0/a + yy_0/b$. To get the equation of the tangent line to the ellipse at $T(n) = (x_0,y_0)$, we substitute $T^{-1}(x,y) = (x/a,y/b)$ into this equation to obtain $$1 = xx_0/a^2 + yy_0/b^2.$$ Share answered Feb 9, 2016 at 8:52 DavidDavid 6,5361818 silver badges4040 bronze badges $\endgroup$ 1 $\begingroup$ Where can i read more of this stuff which doesnt uses calculues and derives such cool things like tagents /normals etc? $\endgroup$ Orion_Pax – Orion_Pax 2022-04-16 20:29:16 +00:00 Commented Apr 16, 2022 at 20:29 Add a comment \| 0 $\begingroup$ Use the fact that the gradient of a differentiable function at a point is orthogonal to the level set of that function passing through that point. Here, $$f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2},$$ and $$\vec\nabla f(x_0,y_0)=\frac{2x_0}{a^2}+\frac{2y_0}{b^2}.$$ Hence, for $(x_0,y_0)\in\mathbb{R}^2$ such that $f(x_0,y_0)=1$, an equation of the tangent line to the ellipse $f(x,y)=1$ at $(x_0,y_0)$ is $$\vec\nabla f(x_0,y_0)\cdot(x-x_0,y-y_0)=0,$$ i.e., $$\frac{2x_0(x-x_0)}{a^2}+\frac{2y_0(y-y_0)}{b^2}=0,$$ i.e., $$\frac{2x_0x}{a^2}+\frac{2y_0y}{b^2}=\frac{2x_0^2}{a^2}+\frac{2y_0^2}{b^2}.$$ The result follows after dividing by $2$ and using the fact that $f(x_0,y_0)=1$. Share answered Feb 9, 2016 at 8:44 gniourf_gniourfgniourf_gniourf 4,3061919 silver badges2323 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Let the parametric equation of the tangent be $$x=x_0+t\cos(\theta),y=y_0+t\sin(\theta),$$ where $\theta$ is unknown. Plug in the equation of the ellipse to get $$\frac{(x_0+t\cos(t))^2}{a^2}+\frac{(y_0+t\sin(t))^2}{b^2}=1\ =\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}+2\left(\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{b^2}\right)t+\left(\frac{\cos^2(\theta)}{a^2}+\frac{\sin^2(\theta)}{b^2}\right)t^2.$$ After simplification, this equation is of the form $\alpha t^2+\beta t=0$. As the line is tangent, the root $t=0$ must be double. This occurs if $\beta=0$, $$\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{b^2}=0.$$ Then for any $t$ $$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=\frac{(x_0+t\cos(\theta))x_0}{a^2}+\frac{(y_0+t\sin(\theta))y_0}{b^2}=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1.$$ Share edited Feb 9, 2016 at 10:24 answered Feb 9, 2016 at 9:56 user65203user65203 $\endgroup$ Add a comment \| 0 $\begingroup$ The answer is given in Perspectives on Projective Geometry by JÃ¼rgen Richter-Gebert (credit to @Jan-MagnusÃkland for this insight): A conic consists of all points p that satisfy an equation $p^TAp = 0.$ The set of all tangents to this conic can be described as ${Ap \| p^TAp = 0}.$ The proof is that since $\forall p,q \in \mathbb R^3,p \cdot q := p^T q$ and $p \cdot q = 0$ implies orthogonality, the equations $q = Ap$ define tangent lines (in line coordinates) to the conic at points $p$. Thus, in general, the tangent lines to any conic (not just ellipses) can be described by $$ \left(\matrix{x & y & 1}\right) A \left(\matrix{ x_0 \ y_0 \ 1 }\right) $$ Share answered Oct 5, 2021 at 18:51 adam.hendryadam.hendry 37211 silver badge1212 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ To show that the given line is tangent to the ellipse, one can show that the system of equations \begin{align} \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \tag{I} \ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} &= \tag{II} 1 \end{align} with $x$, $y$ variables has exactly one solutionnamely $(x_0, y_0)$. We further assume that $(x_0, y_0)$ lies on the ellipse, i.e., $$ \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1. \tag{III} $$ Taking $(\mathrm I) - 2(\mathrm{II}) + (\mathrm{III})$ produces $$ \frac{x^2 - 2xx_0 + x_0^2}{a^2} + \frac{y^2 - 2yy_0 + y_0^2}{b^2} = 0 $$ or $$ \left(\frac{x-x_0}{a}\right)^2 + \left(\frac{y-y_0}{b}\right)^2 = 0. $$ The only way for the sum of two squares to be zero is that $x=x_0$ and $y=y_0$, hence this is indeed the only solution as desired. Share answered Jun 1, 2022 at 12:27 OlinOlin 4955 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ We first differentiate the ellipse equation w.r.t. $x$ to find the slope of the tangent $m$, $$\dfrac{2x_0}{a^2}+\dfrac{2y_0m}{b^2}=0$$ $$m=-\dfrac{x_0}{y_0}\cdot\dfrac{b^2}{a^2}$$ Thus, the tangent is given by $$\color{darkgreen}{y-y_0}=\color{blue}{-}\dfrac{\color{blue}{x_0}}{\color{darkgreen}{y_0}}\cdot\dfrac{\color{darkgreen}{b^2}}{\color{blue}{a^2}}\color{blue}{(x-x_0)}$$ $$\color{darkgreen}{\dfrac{yy_0}{b^2}-\dfrac{y_0^2}{b^2}}=\color{blue}{\dfrac{x_0^2}{a^2}-\dfrac{xx_0}{a^2}}$$ $$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}=\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=1$$ as desired. Hope this helps. :) Share answered Jun 1, 2022 at 12:59 ultralegend5385ultralegend5385 10.3k33 gold badges1919 silver badges4242 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 calculus conic-sections 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 Find equation of ellipse given two tangent lines at given points and a point on ellipse 0 Why can the equation of the tangent can be obtained by replacing $x$ with $x_0$ and similarly for $y$? 2 Show that the tangent to the hyperbola $(x_0 , y_0)$ does not intersect the curve anywhere else. 0 Tangent to an ellipse Question on tangent lines and the center of an ellipse 2 Given an ellipse and a reference point, how to find the two lines that are tangent to the ellipse? 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14736	https://pubs.rsna.org/doi/pdf/10.1148/132.3.683	The Sonographic Appearance of Acute Pyelonephritis \| Radiology Login to your account Username Password Forgot password? New UserInstitutional Login Change Password Old Password New Password Too Short Weak Medium Strong Very Strong Too Long Your password must have 1 characters or more and contain 3 of the following: a lower case character, an upper case character, a special character or a digit Too Short Password Changed Successfully Your password has been changed Create a new account Email Returning user Can't sign in? Forgot your password? Enter your email address below and we will send you the reset instructions Email Cancel If the address matches an existing account you will receive an email with instructions to reset your password. Close Request Username Can't sign in? Forgot your username? Enter your email address below and we will send you your username Email Close If the address matches an existing account you will receive an email with instructions to retrieve your username "skipMainNavigation" closeDrawerMenu openDrawerMenuHome Journals All Journals Radiology RadioGraphics Radiology: Artificial Intelligence Radiology: Cardiothoracic Imaging Radiology: Imaging Cancer Radiology Advances CME CME Catalog Track your CME CME Objectives & Tests Contact Us Subscribe Email Alerts More Search This Journal This Journal All Journals Quick Search in Journals Search Search Search Quick Search anywhere Search Search Search Advanced Search 0 Sign in Menu Latest Articles All Issues Collections COVID-19 Statements and Guidelines Images in Radiology Radiology In Training Generative AI Content Centennial Content EVALI Collection For Authors Instructions for Authors Editorial Policies Scientific Style Guide Submit a Manuscript Author Services Radiology: Behind the Scenes blog Diagnosis Please Multimedia Podcasts Tweetorials Video Summaries Browse by Biomarkers/Quantitative Imaging Breast Imaging Cardiac Radiology Chest Radiology Computed Tomography Education Emergency Radiology Gastrointestinal Radiology General/Multisystem Radiology Genitourinary Radiology Head and Neck Health Policy Informatics Interventional Radiology Leadership & Management Magnetic Resonance Imaging Molecular Imaging Musculoskeletal Radiology Neuroradiology Nuclear Medicine Obstetric/Gynecologic Radiology Oncologic Imaging Other Pediatric Radiology Physics and Basic Science Professionalism Radiation Oncology Research and Statistical Methods Safety and Quality Ultrasound Vascular Radiology Information About Radiology Top 10 from 2024 Citation Metrics Editorial Board Podcasts Margulis Award For Reviewers Editorial Policies Open Access Policy Read and Publish Agreement HomeRadiologyVol. 132, No. 3 PreviousNext Ultrasound The Sonographic Appearance of Acute Pyelonephritis Steven L. Edell 22, John A. Bonavita Steven L. Edell 22, John A. Bonavita Author Affiliations Department of Radiology, Riverside Hospital, Wilmington, DE 19899 Steven L. Edell 22 John A. Bonavita Published Online:Sep 1 1979 More Figures References Related Details PDF Sections Abstract Article History PDF Tools Add to favorites Cite Track Citations Permissions Reprints Share Share on Facebook X Linked In Abstract Acute pyelonephritis and acute ureteral obstruction often present with similar clinical and urographic findings. Ultrasound, however, can easily detect the presence of obstruction as well as demonstrate characteristic findings suggestive of acute pyelonephritis, and thus allows differentiation. In two patients with acute pyelonephritis, the ultrasonic findings consisted of a large swollen kidney with an increased anechoic corticomedullary area, with multiple scattered low-level echoes. Each of the two cases is discussed in detail. Article History Received: Dec 4 1978 Accepted: Mar 30 1979 Revision requested: Mar 30 1979 Revision received: Apr 30 1979 Published in print: Sept 1979 We recommend Radiation appendicitis: demonstration with graded compression US.J B Puylaert, Radiology, 1987 Nondisplaced fractures of the greater tuberosity of the humerus: sonographic detection.R M Patten, Radiology, 1992 Cystic ovarian teratomas: the sonographic appearance of the dermoid plug.S F Quinn, Radiology, 1985 Sonographic appearance of hematoma in liver, spleen, and kidney: a clinical, pathologic, and animal study.E vanSonnenberg, Radiology, 1983 Pancreatic metastases: US evaluation.K Wernecke, Radiology, 1986 Is Bone Marrow Examination Justified in Isolated Childhood Thrombocytopenia?K. N. Abdurrahman, Qatar Med J, 2012 The role of ECMO in acute interstitial lung diseaseNicholas Barrett, Qatar Med J, 2017 Risk factors for severe bronchiolitis among children in the emergency department at Sultan Qaboos University HospitalMaitha S. Al Asmi, Qatar Med J, 2019 An ensemble model for fetal birthweight prediction in third-trimester (Preprint)Jing Gao, Jie Xu, Yujun Yao, et al., JMIR Pediatrics and Parenting, 2025 The Influence of TikTok on Body Satisfaction among Gen-Z: Combination of Quantitative and Qualitative ApproachHanifa Ariana, JMIR Preprints, 2024 Powered by Targeting settings Do not sell my personal information Figures References Related Details None Cited By Revisiting the utility of prenatal ultrasound in the routine workup of first febrile UTI: A systematic review and meta-analysis of the negative predictive value of prenatal ultrasound for identification of urinary tract abnormalities after first febrile urinary tract infection in children Maya R.Overland, KathrynTrandem, Isabel ElaineAllen, Hillary L.Copp 2023Dec1 \| Journal of Pediatric Urology, Vol. 19, No. 6 Point-of-Care Ultrasound: A Multimodal Tool for the Management of Sepsis in the Emergency Department EffiePolyzogopoulou, MariaVelliou, ChristosVerras, IoannisVentoulis, JohnParissis, JosephOsterwalder, BeatriceHoffmann 20 June 2023 \| Medicina, Vol. 59, No. 6 Imaging of Renal Infections and Inflammatory Disease MariaZulfiqar, Cristián VarelaUbilla, RefkyNicola, Christine O.Menias 2020Sep1 \| Radiologic Clinics of North America, Vol. 58, No. 5 Prostate EwaKuligowska, James F.Stinchon 2020Mar5 Imaging the Pediatric Urinary Tract JeffreyTraubici, RuthLim 17 February 2017 A rare case of bilateral multiple renal abscesses ManavBhalla, NamrataBhalla, BirwaBhatt 7 May 2010 \| Ultrasound, Vol. 18, No. 2 Imaging the Pediatric Urinary Tract JeffreyTraubici, RuthLim 2008Jan1 SONOGRAPHIC APPLICATION IN THE DIAGNOSIS OF PYELONEPHRITIS IN CATTLE MARTINAFLOECK 22 December 2006 \| Veterinary Radiology & Ultrasound, Vol. 48, No. 1 Kidney Swelling LENAWALLIN, INGEMARHELIN, MARIKABAJC 1997May1 \| CLINICAL NUCLEAR MEDICINE, Vol. 22, No. 5 Pictorial review: Compute tomography of renal inflammatory disease Lindsey S.Rabushka, Elliot K.Fishman, Stanford M.Goldman 1994Oct1 \| Urology, Vol. 44, No. 4 Renal pelvicalyceal dilation in antepartum pyelonephritis: Ultrasonographic findings DianeTwickler, Bertis B.Little, Andrew J.Satin, Charles E.L.Brown 1991Oct1 \| American Journal of Obstetrics and Gynecology, Vol. 165, No. 4 Inflammatory Renal Disease Stanford M.Goldman 1991Jan1 Sonography of renal inflammatory disease MarkPiccirillo, Christopher M.Rigsby, Arthur T.Rosenfield 1988Dec1 \| Urologic Radiology, Vol. 9, No. 1 Renal parenchymal volume during and after acute pyelonephritis measured by ultrasonography. BJohansson, STroell, UBerg 1988Nov1 \| Archives of Disease in Childhood, Vol. 63, No. 11 Contemporary Imaging of Renal Inflammatory Disease MarkPiccirillo, ChristopherRigsby, Arthur T.Rosenfield 1987Dec1 \| Infectious Disease Clinics of North America, Vol. 1, No. 4 The Role of Imaging Studies in Urinary Tract Infection MarkBenson, Joseph P.Li Puma, Martin I.Resnick 1986Nov1 \| Urologic Clinics of North America, Vol. 13, No. 4 The Role of Sonography and CT in Urosepsis R. BrookeJeffrey, Fred S.Vernacchia 1 May 1986 \| Journal of Diagnostic Medical Sonography, Vol. 2, No. 3 Radiological Diagnosis and Assessment of Inflammatory Renal Diseases Stanford M.Goldman 1986Jan1 Kidney: Native M. LeonSkolnick 1986Jan1 Ultrasound detection of focal bacterial nephritis (lobar nephronia) and its evolution into a renal abscess Roslind I.McCoy, Alfred B.Kurtz, Matthew D.Rifkin, Michael B.Kodroff, Maureen M.Bidula 1985Dec1 \| Urologic Radiology, Vol. 7, No. 1 Ultrasonography of the Upper Genitourinary Tract Beverly G.Coleman 1985Nov1 \| Urologic Clinics of North America, Vol. 12, No. 4 Ultraschalldiagnostik des Harntrakts D.Weitzel, H.Peters 1985Jan1 Renal Ultrasonography 1984: A Practical Overview‡ Edward S.Amis, David S.Hartman 1984Jun1 \| Radiologic Clinics of North America, Vol. 22, No. 2 Harntrakt DieterWeitzel, MatthiasDittrich, HelmutPeters, ErnstDinkel 1984Jan1 Niere WilandFiegler 1984Jan1 Renal Transplant Problems as Assessed by Ultrasound and Nuclear Medicine R. C.Sanders 1983Jan1 Ultrasound evaluation of renal parenchymal disease and hydronephrosis Arthur T.Rosenfield 1982Dec1 \| Urologic radiology, Vol. 4, No. 1 Possibilités et limites de l'échotomographie d'urgence dans l'insuffisance rénale majeure D.Kleinknecht, P.Chauveau, G.Bochereau 1982Dec1 \| La Revue de Médecine Interne, Vol. 3, No. 4 Radiographic findings in acute segmental pyelonephritis Jose R.Sotolongo, HowardSchiff, Mendley A.Wulfsohn 1982Mar1 \| Urology, Vol. 19, No. 3 The Value of Ultrasound in Obstructive Uropathies Ž.Fučkar, Vj.Peterković, M.Aničić, T.Tićac 1 October 1980 \| Urologia Journal, Vol. 47, No. 5 Vol. 132, No. 3 Metrics Downloaded 1,101 times Altmetric Score PDF download back Close Figure Viewer Browse All FiguresReturn to FigureChange zoom level Zoom in Zoom out Previous FigureNext Figure Caption Contact Us About Sign Up for E-mail Alerts 820 Jorie Blvd., Suite 200 Oak Brook, IL 60523-2251 U.S. & Canada: 1-877-776-2636 Outside U.S. & Canada: 1-630-571-7873 Information For Authors For Reviewers For Librarians For Agencies For Advertisers Help Contact Us Publications Staff Login Help Email Alerts Resources Subscribe Permissions Reprints Library Free Online Trial Terms of Use,Policies,Accessibility © 2025 Radiological Society of North America To help offer the best experience possible, RSNA uses cookies on its site. 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14737	https://openstax.org/books/university-physics-volume-1/pages/4-4-uniform-and-nonuniform-circular-motion	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in University Physics Volume 1 4.4 Uniform and Nonuniform Circular Motion University Physics Volume 1 4.4 Uniform and Nonuniform Circular Motion Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Solve for the centripetal acceleration of an object moving on a circular path. Use the equations of circular motion to find the position, velocity, and acceleration of a particle executing circular motion. Explain the differences between centripetal acceleration and tangential acceleration resulting from nonuniform circular motion. Evaluate centripetal and tangential acceleration in nonuniform circular motion, and find the total acceleration vector. Uniform circular motion is a specific type of motion in which an object travels in a circle with a constant speed. For example, any point on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are the second, minute, and hour hands of a watch. It is remarkable that points on these rotating objects are actually accelerating, although the rotation rate is a constant. To see this, we must analyze the motion in terms of vectors. Centripetal Acceleration In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or This is shown in Figure 4.18. As the particle moves counterclockwise in time on the circular path, its position vector moves from to The velocity vector has constant magnitude and is tangent to the path as it changes from to changing its direction only. Since the velocity vector is perpendicular to the position vector the triangles formed by the position vectors and and the velocity vectors and are similar. Furthermore, since and the two triangles are isosceles. From these facts we can make the assertion or Figure 4.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times and (b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector points toward the center of the circle in the limit We can find the magnitude of the acceleration from The direction of the acceleration can also be found by noting that as and therefore approach zero, the vector approaches a direction perpendicular to In the limit is perpendicular to Since is tangent to the circle, the acceleration points toward the center of the circle. Summarizing, a particle moving in a circle at a constant speed has an acceleration with magnitude 4.27 The direction of the acceleration vector is toward the center of the circle (Figure 4.19). This is a radial acceleration and is called the centripetal acceleration, which is why we give it the subscript c. The word centripetal comes from the Latin words centrum (meaning “center”) and petere (meaning “to seek”), and thus takes the meaning “center seeking.” Figure 4.19 The centripetal acceleration vector points toward the center of the circular path of motion and is an acceleration in the radial direction. The velocity vector is also shown and is tangent to the circle. Let’s investigate some examples that illustrate the relative magnitudes of the velocity, radius, and centripetal acceleration. Example 4.10 Creating an Acceleration of 1 g A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. What does the radius of the circle have to be to produce a centripetal acceleration of 1 g on the pilot and jet toward the center of the circular trajectory? Strategy Given the speed of the jet, we can solve for the radius of the circle in the expression for the centripetal acceleration. Solution Set the centripetal acceleration equal to the acceleration of gravity: Solving for the radius, we find Significance To create a greater acceleration than g on the pilot, the jet would either have to decrease the radius of its circular trajectory or increase its speed on its existing trajectory or both. Check Your Understanding 4.5 A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge of the flywheel if it experiences a centripetal acceleration of Centripetal acceleration can have a wide range of values, depending on the speed and radius of curvature of the circular path. Typical centripetal accelerations are given in the following table. \| Object \| Centripetal Acceleration (m/s2 or factors of g) \| --- \| \| Earth around the Sun \| \| \| Moon around the Earth \| \| \| Satellite in geosynchronous orbit \| 0.233 \| \| Outer edge of a CD when playing \| \| \| Jet in a barrel roll \| (2–3 g) \| \| Roller coaster \| (5 g) \| \| Electron orbiting a proton in a simple Bohr model of the atom \| \| Table 4.1 Typical Centripetal Accelerations Equations of Motion for Uniform Circular Motion A particle executing circular motion can be described by its position vector Figure 4.20 shows a particle executing circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle with the x-axis. Vector making an angle with the x-axis is shown with its components along the x- and y-axes. The magnitude of the position vector is and is also the radius of the circle, so that in terms of its components, 4.28 Here, is a constant called the angular frequency of the particle. The angular frequency has units of radians (rad) per second and is simply the number of radians of angular measure through which the particle passes per second. The angle that the position vector has at any particular time is . If T is the period of motion, or the time to complete one revolution ( rad), then Figure 4.20 The position vector for a particle in circular motion with its components along the x- and y-axes. The particle moves counterclockwise. Angle is the angular frequency in radians per second multiplied by t. Velocity and acceleration can be obtained from the position function by differentiation: 4.29 It can be shown from Figure 4.20 that the velocity vector is tangential to the circle at the location of the particle, with magnitude Similarly, the acceleration vector is found by differentiating the velocity: 4.30 From this equation we see that the acceleration vector has magnitude and is directed opposite the position vector, toward the origin, because Example 4.11 Circular Motion of a Proton A proton has speed and is moving in a circle in the xy plane of radius r = 0.175 m. What is its position in the xy plane at time At t = 0, the position of the proton is and it circles counterclockwise. Sketch the trajectory. Solution According to Equation 3.5, Since the period T is the time it takes an object to go once arounce a circle, and the distance around a circle is 2πr, we have: From the given data, the proton has period and angular frequency: The position of the particle at with A = 0.175 m is From this result we see that the proton is located slightly below the x-axis. This is shown in Figure 4.21. Figure 4.21 Position vector of the proton at The trajectory of the proton is shown. The angle through which the proton travels along the circle is 5.712 rad, which a little less than one complete revolution. Significance We picked the initial position of the particle to be on the x-axis. This was completely arbitrary. If a different starting position were given, we would have a different final position at t = 200 ns. Nonuniform Circular Motion Circular motion does not have to be at a constant speed. A particle can travel in a circle and speed up or slow down, showing an acceleration in the direction of the motion. In uniform circular motion, the particle executing circular motion has a constant speed and the circle is at a fixed radius. If the speed of the particle is changing as well, then we introduce an additional acceleration in the direction tangential to the circle. Such accelerations occur at a point on a top that is changing its spin rate, or any accelerating rotor. In Displacement and Velocity Vectors we showed that centripetal acceleration is the time rate of change of the direction of the velocity vector. If the speed of the particle is changing, then it has a tangential acceleration that is the time rate of change of the magnitude of the velocity: The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radially inward toward the center of the circle. Thus, a particle in circular motion with a tangential acceleration has a total acceleration that is the vector sum of the centripetal and tangential accelerations: The acceleration vectors are shown in Figure 4.22. Note that the two acceleration vectors and are perpendicular to each other, with in the radial direction and in the tangential direction. The total acceleration points at an angle between and Figure 4.22 The centripetal acceleration points toward the center of the circle. The tangential acceleration is tangential to the circle at the particle’s position. The total acceleration is the vector sum of the tangential and centripetal accelerations, which are perpendicular. Example 4.12 Total Acceleration during Circular Motion A particle moves in a circle of radius r = 2.0 m. During the time interval from t = 1.5 s to t = 4.0 s its speed varies with time according to What is the total acceleration of the particle at t = 2.0 s? Strategy We are given the speed of the particle and the radius of the circle, so we can calculate centripetal acceleration easily. The direction of the centripetal acceleration is toward the center of the circle. We find the magnitude of the tangential acceleration by taking the derivative with respect to time of using Equation 4.31 and evaluating it at t = 2.0 s. We use this and the magnitude of the centripetal acceleration to find the total acceleration. Solution Centripetal acceleration is directed toward the center of the circle. Tangential acceleration is Total acceleration is and from the tangent to the circle. See Figure 4.23. Figure 4.23 The tangential and centripetal acceleration vectors. The net acceleration is the vector sum of the two accelerations. Significance The directions of centripetal and tangential accelerations can be described more conveniently in terms of a polar coordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is used for motion along curved paths, is discussed in detail later in the book. Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: William Moebs, Samuel J. 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14738	https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19?srsltid=AfmBOoqIqXtXYwYuI6VlkOtn2Z1VQ1fEWfhWMRmhu5-Z-bR4ovAh55ou	Art of Problem Solving 2016 AMC 8 Problems/Problem 19 - 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 2016 AMC 8 Problems/Problem 19 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2016 AMC 8 Problems/Problem 19 Contents 1 Problem 19 2 Solution 1 3 Solution 2 4 Video Solution 5 See Also Problem 19 The sum of consecutive even integers is . What is the largest of these consecutive integers? Solution 1 Let be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to since . Now, . Remembering that this is the 13th integer, we wish to find the 25th, which is . Solution 2 Let be the smallest number. The equation will become, . After you combine like terms, you get which turns into . , so . Then, you add . ~AfterglowBlaziken Video Solution ~savannahsolver See Also 2016 AMC 8 (Problems • Answer Key • Resources) Preceded by Problem 18Followed by Problem 20 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 All AJHSME/AMC 8 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Introductory Algebra Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
14739	https://courses.physics.ucsd.edu/2014/Fall/physics200a/LECTURES/CH03.pdf	Contents Contents i List of Figures iii List of Tables iv 3 Lagrangian Mechanics 1 3.1 Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3.2 Functions and Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3.2.1 Functional Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.3 Examples from the Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.3.1 Example 1 : minimal surface of revolution . . . . . . . . . . . . . . . . . . . . . . . 6 3.3.2 Example 2 : geodesic on a surface of revolution . . . . . . . . . . . . . . . . . . . . 8 3.3.3 Example 3 : brachistochrone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.3.4 Ocean waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.4 More on Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.5 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.6 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.6.1 Invariance of the equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.6.2 Remarks on the order of the equations of motion . . . . . . . . . . . . . . . . . . . 19 3.6.3 Lagrangian for a free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3.7 Conserved Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 i ii CONTENTS 3.7.1 Momentum conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.7.2 Energy conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.8 Choosing Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.9 How to Solve Mechanics Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.10 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.10.1 One-dimensional motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.10.2 Central force in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.10.3 A sliding point mass on a sliding wedge . . . . . . . . . . . . . . . . . . . . . . . . 25 3.10.4 A pendulum attached to a mass on a spring . . . . . . . . . . . . . . . . . . . . . . 26 3.10.5 The double pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.10.6 The thingy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.11 The Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3.12 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.12.1 Continuous Symmetry Implies Conserved Charges . . . . . . . . . . . . . . . . . . 33 3.12.2 Examples of one-parameter families of transformations . . . . . . . . . . . . . . . 34 3.12.3 Conservation of linear and angular momentum . . . . . . . . . . . . . . . . . . . . 36 3.12.4 Invariance of L vs. Invariance of S . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.13 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.13.1 From Lagrangian to Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.13.2 Is H = T + U ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.13.3 Example: A bead on a rotating hoop . . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.13.4 Charged particle in a magnetic ﬁeld . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 3.14 Motion in Rapidly Oscillating Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.14.1 Slow and fast dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.14.2 Example : pendulum with oscillating support . . . . . . . . . . . . . . . . . . . . . 45 3.15 Field Theory: Systems with Several Independent Variables . . . . . . . . . . . . . . . . . . 47 3.15.1 Equations of motion and Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . 47 3.15.2 Gross-Pitaevskii model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.16 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.16.2 Constrained extremization of functions : Lagrange multipliers . . . . . . . . . . . 52 3.16.3 Constraints and variational calculus . . . . . . . . . . . . . . . . . . . . . . . . . . 54 3.16.4 Extremization of functionals : integral constraints . . . . . . . . . . . . . . . . . . . 54 3.16.5 Extremization of functionals : holonomic constraints . . . . . . . . . . . . . . . . . 55 3.16.6 Examples of functional extremization with constraints . . . . . . . . . . . . . . . . 56 3.16.7 Constraints in Lagrangian mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.16.8 Constraints and conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.17 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.17.1 One cylinder rolling off another . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.17.2 Frictionless motion along a curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.17.3 Disk rolling down an inclined plane . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.17.4 Pendulum with nonrigid support . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 3.17.5 Falling ladder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 3.17.6 Point mass inside rolling hoop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 3.18 Appendix : Legendre Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 List of Figures 3.1 The shortest distance between two points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3.2 Extremal path consisting of three line segments . . . . . . . . . . . . . . . . . . . . . . . . . 3 3.3 A path y(x) and its variation y(x) + δy(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3.4 Minimal surface solution, with y(x) = b cosh(x/b) and y(x0) = y0 . . . . . . . . . . . . . . 7 3.5 Shallow water wave propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 iii iv LIST OF FIGURES 3.6 A functional as a continuum limit of a multivariable function . . . . . . . . . . . . . . . . . 13 3.7 A wedge sliding along a horizontal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.8 The spring–pendulum system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.9 The double pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.10 The thingy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3.11 A bead of mass m on a rotating hoop of radius a . . . . . . . . . . . . . . . . . . . . . . . . 41 3.12 An effective potential and its corresponding phase curves . . . . . . . . . . . . . . . . . . . 42 3.13 Dimensionless potential for pendulum with oscillating support . . . . . . . . . . . . . . . 46 3.14 A cylinder of radius a rolls along a half-cylinder of radius R . . . . . . . . . . . . . . . . . 52 3.15 Frictionless motion under gravity along a curved surface . . . . . . . . . . . . . . . . . . . 63 3.16 Finding the local radius of curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 3.17 A hoop rolling down an inclined plane lying on a frictionless surface . . . . . . . . . . . . 65 3.18 A ladder sliding down a wall and across a ﬂoor . . . . . . . . . . . . . . . . . . . . . . . . 68 3.19 Plot of time to fall for the slipping ladder . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.20 A point mass m inside a hoop of mass M . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 3.21 The Legendre transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 List of Tables v vi LIST OF TABLES Chapter 3 Lagrangian Mechanics 3.1 Snell’s Law Warm-up problem: You are standing at point (x1, y1) on the beach and you want to get to a point (x2, y2) in the water, a few meters offshore. The interface between the beach and the water lies at x = 0. What path results in the shortest travel time? It is not a straight line! This is because your speed v1 on the sand is greater than your speed v2 in the water. The optimal path actually consists of two line segments, as shown in ﬁg. 3.1. Let the path pass through the point (0, y) on the interface. Then the time T is a function of y: T(y) = 1 v1 q x2 1 + (y −y1)2 + 1 v2 q x2 2 + (y2 −y)2 . (3.1) To ﬁnd the minimum time, we set dT dy = 0 = 1 v1 y −y1 q x2 1 + (y −y1)2 −1 v2 y2 −y q x2 2 + (y2 −y)2 = sin θ1 v1 −sin θ2 v2 . (3.2) Thus, the optimal path satisﬁes sin θ1 sin θ2 = v1 v2 , (3.3) which is known as Snell’s Law. Snell’s Law is familiar from optics, where the speed of light in a polarizable medium is written v = c/n, where n is the index of refraction. In terms of n, n1 sin θ1 = n2 sin θ2 . (3.4) If there are several interfaces, Snell’s law holds at each one, so that ni sin θi = ni+1 sin θi+1 , (3.5) 1 2 CHAPTER 3. LAGRANGIAN MECHANICS Figure 3.1: The shortest path between (x1, y1) and (x2, y2) is not a straight line, but rather two successive line segments of different slope. at the interface between media i and i + 1. In the limit where the number of slabs goes to inﬁnity but their thickness is inﬁnitesimal, we can regard n and θ as functions of a continuous variable x. One then has sin θ(x) v(x) = y′ v p 1 + y′2 = P , (3.6) where P is a constant. Here wve have used the result sin θ = y′/ p 1 + y′2, which follows from drawing a right triangle with side lengths dx, dy, and p dx2 + dy2. If we differentiate the above equation with respect to x, we eliminate the constant and obtain the second order ODE 1 1 + y′2 y′′ y′ = v′ v . (3.7) This is a differential equation that y(x) must satisfy if the functional T y(x) = Z ds v = x2 Z x1 dx p 1 + y′2 v(x) (3.8) is to be minimized. 3.2 Functions and Functionals A function is a mathematical object which takes a real (or complex) variable, or several such variables, and returns a real (or complex) number. A functional is a mathematical object which takes an entire 3.2. FUNCTIONS AND FUNCTIONALS 3 Figure 3.2: The path of shortest length is composed of three line segments. The relation between the angles at each interface is governed by Snell’s Law. function and returns a number. In the case at hand, we have T y(x) = x2 Z x1 dx L(y, y′, x) , (3.9) where the function L(y, y′, x) is given by L(y, y′, x) = 1 v(x) q 1 + y′2 . (3.10) Here v(x) is a given function characterizing the medium, and y(x) is the path whose time is to be eval-uated. In ordinary calculus, we extremize a function f(x) by demanding that f not change to lowest order when we change x →x + dx: f(x + dx) = f(x) + f ′(x) dx + 1 2 f ′′(x) (dx)2 + . . . . (3.11) We say that x = x∗is an extremum when f ′(x∗) = 0. For a functional, the ﬁrst functional variation is obtained by sending y(x) →y(x) + δy(x), and extracting 4 CHAPTER 3. LAGRANGIAN MECHANICS Figure 3.3: A path y(x) and its variation y(x) + δy(x). the variation in the functional to order δy. Thus, we compute T y(x) + δy(x) = x2 Z x1 dx L(y + δy, y′ + δy′, x) = x2 Z x1 dx L + ∂L ∂y δy + ∂L ∂y′ δy′ + O (δy)2 = T y(x) + x2 Z x1 dx ∂L ∂y δy + ∂L ∂y′ d dx δy = T y(x) + x2 Z x1 dx " ∂L ∂y −d dx ∂L ∂y′ # δy + ∂L ∂y′ δy x2 x1 . (3.12) Now one very important thing about the variation δy(x) is that it must vanish at the endpoints: δy(x1) = δy(x2) = 0. This is because the space of functions under consideration satisfy ﬁxed boundary conditions y(x1) = y1 and y(x2) = y2. Thus, the last term in the above equation vanishes, and we have δT = x2 Z x1 dx " ∂L ∂y −d dx ∂L ∂y′ # δy . (3.13) We say that the ﬁrst functional derivative of T with respect to y(x) is δT δy(x) = " ∂L ∂y −d dx ∂L ∂y′ # x , (3.14) where the subscript indicates that the expression inside the square brackets is to be evaluated at x. The functional T y(x) is extremized when its ﬁrst functional derivative vanishes, which results in a 3.2. FUNCTIONS AND FUNCTIONALS 5 differential equation for y(x), ∂L ∂y −d dx ∂L ∂y′ = 0 , (3.15) known as the Euler-Lagrange equation. L(y, y′, x) independent of y Suppose L(y, y′, x) is independent of y. Then from the Euler-Lagrange equations we have that P ≡∂L ∂y′ (3.16) is a constant. In classical mechanics, this will turn out to be a generalized momentum. For L = 1 v p 1 + y′2, we have P = y′ v p 1 + y′2 . (3.17) Setting dP/dx = 0, we recover the second order ODE of eqn. 3.7. Solving for y′, dy dx = ± v(x) p v2 0 −v2(x) , (3.18) where v0 = 1/P. L(y, y′, x) independent of x When L(y, y′, x) is independent of x, we can again integrate the equation of motion. Consider the quan-tity H = y′ ∂L ∂y′ −L . (3.19) Then dH dx = d dx y′ ∂L ∂y′ −L = y′′ ∂L ∂y′ + y′ d dx ∂L ∂y′ −∂L ∂y′ y′′ −∂L ∂y y′ −∂L ∂x = y′ d dx ∂L ∂y′ −∂L ∂y −∂L ∂x , (3.20) where we have used the Euler-Lagrange equations to write d dx ∂L ∂y′ = ∂L ∂y . So if ∂L/∂x = 0, we have dH/dx = 0, i.e. H is a constant. 6 CHAPTER 3. LAGRANGIAN MECHANICS 3.2.1 Functional Taylor series In general, we may expand a functional F[y + δy] in a functional Taylor series, F[y + δy] = F[y] + Z dx1 K1(x1) δy(x1) + 1 2 ! Z dx1 Z dx2 K2(x1, x2) δy(x1) δy(x2) + 1 3 ! Z dx1 Z dx2 Z dx3 K3(x1, x2, x3) δy(x1) δy(x2) δy(x3) + . . . (3.21) and we write Kn(x1, . . . , xn) ≡ δnF δy(x1) · · · δy(xn) (3.22) for the nth functional derivative. 3.3 Examples from the Calculus of Variations Here we present three useful examples of variational calculus as applied to problems in mathematics and physics. 3.3.1 Example 1 : minimal surface of revolution Consider a surface formed by rotating the function y(x) about the x-axis. The area is then A y(x) = x2 Z x1 dx 2πy s 1 + dy dx 2 , (3.23) and is a functional of the curve y(x). Thus we can deﬁne L(y, y′) = 2πy p 1 + y′2 and make the identiﬁ-cation y(x) ↔q(t). Since L(y, y′, x) is independent of x, we have H = y′ ∂L ∂y′ −L ⇒ dH dx = −∂L ∂x , (3.24) and when L has no explicit x-dependence, H is conserved. One ﬁnds H = 2πy · y′2 p 1 + y′2 −2πy q 1 + y′2 = − 2πy p 1 + y′2 . (3.25) Solving for y′, dy dx = ± s2πy H 2 −1 , (3.26) which may be integrated with the substitution y = H 2π cosh u, yielding y(x) = b cosh x −a b , (3.27) 3.3. EXAMPLES FROM THE CALCULUS OF VARIATIONS 7 Figure 3.4: Minimal surface solution, with y(x) = b cosh(x/b) and y(x0) = y0. Top panel: A/2πy2 0 vs. y0/x0. Bottom panel: sech(x0/b) vs. y0/x0. The blue curve corresponds to a global minimum of A[y(x)], and the red curve to a local minimum or saddle point. where a and b = H 2π are constants of integration. Note there are two such constants, as the original equation was second order. This shape is called a catenary. As we shall later ﬁnd, it is also the shape of a uniformly dense rope hanging between two supports, under the inﬂuence of gravity. To ﬁx the constants a and b, we invoke the boundary conditions y(x1) = y1 and y(x2) = y2. Consider the case where −x1 = x2 ≡x0 and y1 = y2 ≡y0. Then clearly a = 0, and we have y0 = b cosh x0 b ⇒ γ = κ−1 cosh κ , (3.28) with γ ≡y0/x0 and κ ≡x0/b. One ﬁnds that for any γ > 1.5089 there are two solutions, one of which is a global minimum and one of which is a local minimum or saddle of A[y(x)]. The solution with the smaller value of κ (i.e. the larger value of sech κ) yields the smaller value of A, as shown in ﬁg. 3.4. Note that y y0 = cosh(x/b) cosh(x0/b) , (3.29) so y(x = 0) = y0 sech(x0/b). When extremizing functions that are deﬁned over a ﬁnite or semi-inﬁnite interval, one must take care to evaluate the function at the boundary, for it may be that the boundary yields a global extremum even though the derivative may not vanish there. Similarly, when extremizing functionals, one must investigate the functions at the boundary of function space. In this case, such a function would be the 8 CHAPTER 3. LAGRANGIAN MECHANICS discontinuous solution, with y(x) =                y1 if x = x1 0 if x1 < x < x2 y2 if x = x2 . (3.30) This solution corresponds to a surface consisting of two discs of radii y1 and y2, joined by an inﬁnitesi-mally thin thread. The area functional evaluated for this particular y(x) is clearly A = π(y2 1 + y2 2). In ﬁg. 3.4, we plot A/2πy2 0 versus the parameter γ = y0/x0. For γ > γc ≈1.564, one of the catenary solutions is the global minimum. For γ < γc, the minimum area is achieved by the discontinuous solution. Note that the functional derivative, K1(x) = δA δy(x) = ( ∂L ∂y −d dx ∂L ∂y′ ) = 2π 1 + y′2 −yy′′ (1 + y′2)3/2 , (3.31) indeed vanishes for the catenary solutions, but does not vanish for the discontinuous solution, where K1(x) = 2π throughout the interval (−x0, x0). Since y = 0 on this interval, y cannot be decreased. The fact that K1(x) > 0 means that increasing y will result in an increase in A, so the boundary value for A, which is 2πy2 0, is indeed a local minimum. We furthermore see in ﬁg. 3.4 that for γ < γ∗≈1.5089 the local minimum and saddle are no longer present. This is the familiar saddle-node bifurcation, here in function space. Thus, for γ ∈[0, γ∗) there are no extrema of A[y(x)], and the minimum area occurs for the discontinuous y(x) lying at the boundary of function space. For γ ∈(γ∗, γc), two extrema exist, one of which is a local minimum and the other a saddle point. Still, the area is minimized for the discontinuous solution. For γ ∈(γc, ∞), the local minimum is the global minimum, and has smaller area than for the discontinuous solution. 3.3.2 Example 2 : geodesic on a surface of revolution We use cylindrical coordinates (ρ, φ, z) on the surface z = z(ρ). Thus, ds2 = dρ2 + ρ2 dφ2 + dx2 = n 1 + z′(ρ) 2o dρ + ρ2 dφ2 , (3.32) and the distance functional D φ(ρ) is D φ(ρ) = ρ2 Z ρ1 dρ L(φ, φ′, ρ) , (3.33) where L(φ, φ′, ρ) = q 1 + z′2(ρ) + ρ2 φ′2(ρ) . (3.34) 3.3. EXAMPLES FROM THE CALCULUS OF VARIATIONS 9 The Euler-Lagrange equation is ∂L ∂φ −d dρ ∂L ∂φ′ = 0 ⇒ ∂L ∂φ′ = const. (3.35) Thus, ∂L ∂φ′ = ρ2 φ′ p 1 + z′2 + ρ2 φ′2 = a , (3.36) where a is a constant. Solving for φ′, we obtain dφ = a q 1 + z′(ρ) 2 ρ p ρ2 −a2 dρ , (3.37) which we must integrate to ﬁnd φ(ρ), subject to boundary conditions φ(ρi) = φi, with i = 1, 2. On a cone, z(ρ) = λρ, and we have dφ = a p 1 + λ2 dρ ρ p ρ2 −a2 = p 1 + λ2 d tan−1 r ρ2 a2 −1 , (3.38) which yields φ(ρ) = β + p 1 + λ2 tan−1 r ρ2 a2 −1 , (3.39) which is equivalent to ρ cos φ −β √ 1 + λ2 = a . (3.40) The constants β and a are determined from φ(ρi) = φi. 3.3.3 Example 3 : brachistochrone Problem: ﬁnd the path between (x1, y1) and (x2, y2) which a particle sliding frictionlessly and under constant gravitational acceleration will traverse in the shortest time. To solve this we ﬁrst must invoke some elementary mechanics. Assuming the particle is released from (x1, y1) at rest, energy conservation says 1 2mv2 + mgy = mgy1 . (3.41) Then the time, which is a functional of the curve y(x), is T y(x) = x2 Z x1 ds v = 1 √2g x2 Z x1 dx s 1 + y′2 y1 −y ≡ x2 Z x1 dx L(y, y′, x) , (3.42) 10 CHAPTER 3. LAGRANGIAN MECHANICS with L(y, y′, x) = s 1 + y′2 2g(y1 −y) . (3.43) Since L is independent of x, eqn. 3.20, we have that H = y′ ∂L ∂y′ −L = − h 2g (y1 −y) 1 + y′2i−1/2 (3.44) is conserved. This yields dx = − r y1 −y 2a −y1 + y dy , (3.45) with a = (4gH2)−1. This may be integrated parametrically, writing y1 −y = 2a sin2(1 2θ) ⇒ dx = 2a sin2(1 2θ) dθ , (3.46) which results in the parametric equations x −x1 = a θ −sin θ y −y1 = −a (1 −cos θ) . (3.47) This curve is known as a cycloid. 3.3.4 Ocean waves Surface waves in ﬂuids propagate with a deﬁnite relation between their angular frequency ω and their wavevector k = 2π/λ, where λ is the wavelength. The dispersion relation is a function ω = ω(k). The group velocity of the waves is then v(k) = dω/dk. In a ﬂuid with a ﬂat bottom at depth h, the dispersion relation turns out to be ω(k) = p gk tanh kh ≈      √gh k shallow (kh ≪1) √gk deep (kh ≫1) . (3.48) Suppose we are in the shallow case, where the wavelength λ is signiﬁcantly greater than the depth h of the ﬂuid. This is the case for ocean waves which break at the shore. The phase velocity and group velocity are then identical, and equal to v(h) = √gh. The waves propagate more slowly as they approach the shore. Let us choose the following coordinate system: x represents the distance parallel to the shoreline, y the distance perpendicular to the shore (which lies at y = 0), and h(y) is the depth proﬁle of the bottom. We assume h(y) to be a slowly varying function of y which satisﬁes h(0) = 0. Suppose a disturbance in the ocean at position (x2, y2) propagates until it reaches the shore at (x1, y1 = 0). The time of propagation is T y(x) = Z ds v = x2 Z x1 dx s 1 + y′2 g h(y) . (3.49) 3.3. EXAMPLES FROM THE CALCULUS OF VARIATIONS 11 Figure 3.5: For shallow water waves, v = √gh. To minimize the propagation time from a source to the shore, the waves break parallel to the shoreline. We thus identify the integrand L(y, y′, x) = s 1 + y′2 g h(y) . (3.50) As with the brachistochrone problem, to which this bears an obvious resemblance, L is cyclic in the independent variable x, hence H = y′ ∂L ∂y′ −L = − h g h(y) 1 + y′2i−1/2 (3.51) is constant. Solving for y′(x), we have tan θ = dy dx = r a h(y) −1 , (3.52) where a = (gH)−1 is a constant, and where θ is the local slope of the function y(x). Thus, we conclude that near y = 0, where h(y) →0, the waves come in parallel to the shoreline. If h(y) = αy has a linear proﬁle, the solution is again a cycloid, with x(θ) = b (θ −sin θ) y(θ) = b (1 −cos θ) , (3.53) where b = 2a/α and where the shore lies at θ = 0. Expanding in a Taylor series in θ for small θ, we may eliminate θ and obtain y(x) as y(x) = 9 2 1/3 b1/3 x2/3 + . . . . (3.54) A tsunami is a shallow water wave that propagates in deep water. This requires λ > h, as we’ve seen, which means the disturbance must have a very long spatial extent out in the open ocean, where h ∼ 12 CHAPTER 3. LAGRANGIAN MECHANICS 10 km. An undersea earthquake is the only possible source; the characteristic length of earthquake fault lines can be hundreds of kilometers. If we take h = 10 km, we obtain v = √gh ≈310 m/s or 1100 km/hr. At these speeds, a tsunami can cross the Paciﬁc Ocean in less than a day. As the wave approaches the shore, it must slow down, since v = √gh is diminishing. But energy is conserved, which means that the amplitude must concomitantly rise. In extreme cases, the water level rise at shore may be 20 meters or more. 3.4 More on Functionals We remarked in section 3.2 that a function f is an animal which gets fed a real number x and excretes a real number f(x). We say f maps the reals to the reals, or f : R →R (3.55) Of course we also have functions g: C →C which eat and excrete complex numbers, multivariable functions h: RN →R which eat N-tuples of numbers and excrete a single number, etc. A functional F[f(x)] eats entire functions (!) and excretes numbers. That is, F : n f(x) x ∈R o →R (3.56) This says that F operates on the set of real-valued functions of a single real variable, yielding a real number. Some examples: F[f(x)] = 1 2 ∞ Z −∞ dx f(x) 2 F[f(x)] = 1 2 ∞ Z −∞ dx ∞ Z −∞ dx′ K(x, x′) f(x) f(x′) F[f(x)] = ∞ Z −∞ dx 1 2A f 2(x) + 1 2B d f dx 2 . (3.57) In classical mechanics, the action S is a functional of the path q(t): S[q(t)] = tb Z ta dt n 1 2m ˙ q2 −U(q) o . (3.58) We can also have functionals which feed on functions of more than one independent variable, such as S[y(x, t)] = tb Z ta dt xb Z xa dx ( 1 2µ ∂y ∂t 2 −1 2τ ∂y ∂x 2) , (3.59) 3.4. MORE ON FUNCTIONALS 13 Figure 3.6: A functional S[q(t)] is the continuum limit of a function of a large number of variables, S(q1, . . . , qM). which happens to be the functional for a string of mass density µ under uniform tension τ. Another example comes from electrodynamics: S[Aµ(x, t)] = − Z d3x Z dt 1 16π Fµν F µν + 1 c jµ Aµ , (3.60) which is a functional of the four ﬁelds {A0, A1, A2, A3}, where A0 = cφ. These are the components of the 4-potential, each of which is itself a function of four independent variables (x0, x1, x2, x3), with x0 = ct. The ﬁeld strength tensor is written in terms of derivatives of the Aµ: Fµν = ∂µAν −∂νAµ, where we use a metric gµν = diag(+, −, −, −) to raise and lower indices. The 4-potential couples linearly to the source term Jµ, which is the electric 4-current (cρ, J). We extremize functions by sending the independent variable x to x + dx and demanding that the varia-tion d f = 0 to ﬁrst order in dx. That is, f(x + dx) = f(x) + f ′(x) dx + 1 2f ′′(x)(dx)2 + . . . , (3.61) whence d f = f ′(x) dx + O (dx)2 and thus f ′(x∗) = 0 ⇐ ⇒ x∗an extremum. (3.62) We extremize functionals by sending f(x) →f(x) + δf(x) (3.63) and demanding that the variation δF in the functional F[f(x)] vanish to ﬁrst order in δf(x). The vari-ation δf(x) must sometimes satisfy certain boundary conditions. For example, if F[f(x)] only operates on functions which vanish at a pair of endpoints, i.e. f(xa) = f(xb) = 0, then when we extremize the 14 CHAPTER 3. LAGRANGIAN MECHANICS functional F we must do so within the space of allowed functions. Thus, we would in this case require δf(xa) = δf(xb) = 0. We may expand the functional F[f + δf] in a functional Taylor series, F[f + δf] = F[f] + Z dx1 K1(x1) δf(x1) + 1 2 ! Z dx1 Z dx2 K2(x1, x2) δf(x1) δf(x2) + 1 3 ! Z dx1 Z dx2 Z dx3 K3(x1, x2, x3) δf(x1) δf(x2) δf(x3) + . . . (3.64) and we write Kn(x1, . . . , xn) ≡ δnF δf(x1) · · · δf(xn) . (3.65) In a more general case, F = F {fi(x)} is a functional of several functions, each of which is a function of several independent variables.1 We then write F[{fi + δfi}] = F[{fi}] + Z dx1 Ki 1(x1) δfi(x1) + 1 2 ! Z dx1 Z dx2 Kij 2 (x1, x2) δfi(x1) δfj(x2) + 1 3 ! Z dx1 Z dx2 Z dx3 Kijk 3 (x1, x2, x3) δfi(x1) δfj(x2) δfk(x3) + . . . , (3.66) with Ki1i2···in n (x1, x2, . . . , xn) = δnF δfi1(x1) δfi2(x2) δfin(xn) . (3.67) Another way to compute functional derivatives is to send f(x) →f(x) + ǫ1 δ(x −x1) + . . . + ǫn δ(x −xn) (3.68) and then differentiate n times with respect to ǫ1 through ǫn. That is, δnF δf(x1) · · · δf(xn) = ∂n ∂ǫ1 · · · ∂ǫn ǫ1=ǫ2=···ǫn=0 F f(x) + ǫ1 δ(x −x1) + . . . + ǫn δ(x −xn) . (3.69) Let’s see how this works. As an example, we’ll take the action functional from classical mechanics, S[q(t)] = tb Z ta dt n 1 2m ˙ q2 −U(q) o . (3.70) To compute the ﬁrst functional derivative, we replace the function q(t) with q(t)+ǫ δ(t−t1), and expand in powers of ǫ: S q(t) + ǫδ(t −t1) = S[q(t)] + ǫ tb Z ta dt n m ˙ q δ′(t −t1) −U ′(q) δ(t −t1) o = −ǫ n m ¨ q(t1) + U ′q(t1) o , (3.71) 1It may be also be that different functions depend on a different number of independent variables. E.g. F = F[f(x), g(x, y), h(x, y, z)]. 3.4. MORE ON FUNCTIONALS 15 hence δS δq(t) = − n m ¨ q(t) + U ′q(t) o (3.72) and setting the ﬁrst functional derivative to zero yields Newton’s Second Law, m¨ q = −U ′(q), for all t ∈[ta, tb]. Note that we have used the result ∞ Z −∞ dt δ′(t −t1) h(t) = −h′(t1) , (3.73) which is easily established upon integration by parts. To compute the second functional derivative, we replace q(t) →q(t) + ǫ1 δ(t −t1) + ǫ2 δ(t −t2) (3.74) and extract the term of order ǫ1 ǫ2 in the double Taylor expansion. One ﬁnds this term to be ǫ1 ǫ2 tb Z ta dt n m δ′(t −t1) δ′(t −t2) −U ′′(q) δ(t −t1) δ(t −t2) o . (3.75) Note that we needn’t bother with terms proportional to ǫ2 1 or ǫ2 2 since the recipe is to differentiate once with respect to each of ǫ1 and ǫ2 and then to set ǫ1 = ǫ2 = 0. This procedure uniquely selects the term proportional to ǫ1 ǫ2, and yields δ2S δq(t1) δq(t2) = − n m δ′′(t1 −t2) + U ′′q(t1) δ(t1 −t2) o . (3.76) In multivariable calculus, the stability of an extremum is assessed by computing the matrix of second derivatives at the extremal point, known as the Hessian matrix. One has ∂f ∂xi x∗= 0 ∀i ; Hij = ∂2f ∂xi ∂xj x∗ . (3.77) The eigenvalues of the Hessian Hij determine the stability of the extremum. Since Hij is a symmetric matrix, its eigenvectors ηα may be chosen to be orthogonal. The associated eigenvalues λα, deﬁned by the equation Hij ηα j = λα ηα i , (3.78) are the respective curvatures in the directions ηα, where α ∈{1, . . . , n} where n is the number of vari-ables. The extremum is a local minimum if all the eigenvalues λα are positive, a maximum if all are negative, and otherwise is a saddle point. Near a saddle point, there are some directions in which the function increases and some in which it decreases. In the case of functionals, the second functional derivative K2(x1, x2) deﬁnes an eigenvalue problem for δf(x): xb Z xa dx2 K2(x1, x2) δf(x2) = λ δf(x1) . (3.79) 16 CHAPTER 3. LAGRANGIAN MECHANICS In general there are an inﬁnite number of solutions to this equation which form a basis in function space, subject to appropriate boundary conditions at xa and xb. For example, in the case of the action functional from classical mechanics, the above eigenvalue equation becomes a differential equation, − m d2 dt2 + U ′′q∗(t) δq(t) = λ δq(t) , (3.80) where q∗(t) is the solution to the Euler-Lagrange equations. As with the case of ordinary multivariable functions, the functional extremum is a local minimum (in function space) if every eigenvalue λα is positive, a local maximum if every eigenvalue is negative, and a saddle point otherwise. Consider the simple harmonic oscillator, for which U(q) = 1 2 mω2 0 q2. Then U ′′q∗(t) = m ω2 0; note that we don’t even need to know the solution q∗(t) to obtain the second functional derivative in this special case. The eigenvectors obey m(δ¨ q + ω2 0 δq) = −λ δq, hence δq(t) = A cos q ω2 0 + (λ/m) t + ϕ , (3.81) where A and ϕ are constants. Demanding δq(ta) = δq(tb) = 0 requires q ω2 0 + (λ/m) tb −ta) = nπ , (3.82) where n is an integer. Thus, the eigenfunctions are δqn(t) = A sin nπ · t −ta tb −ta , (3.83) and the eigenvalues are λn = m nπ T 2 −mω2 0 , (3.84) where T = tb −ta. Thus, so long as T > π/ω0, there is at least one negative eigenvalue. Indeed, for nπ ω0 < T < (n+1)π ω0 there will be n negative eigenvalues. This means the action is generally not a minimum, but rather lies at a saddle point in the (inﬁnite-dimensional) function space. To test this explicitly, consider a harmonic oscillator with the boundary conditions q(0) = 0 and q(T) = Q. The equations of motion, ¨ q + ω2 0 q = 0, along with the boundary conditions, determine the motion, q∗(t) = Q sin(ω0t) sin(ω0T) . (3.85) The action for this path is then S[q∗(t)] = T Z 0 dt n 1 2m ˙ q∗2 −1 2mω2 0 q∗2o = m ω2 0 Q2 2 sin2ω0T T Z 0 dt n cos2ω0t −sin2ω0t o = 1 2mω0 Q2 ctn (ω0T) . (3.86) 3.4. MORE ON FUNCTIONALS 17 Next consider the path q(t) = Q t/T which satisﬁes the boundary conditions but does not satisfy the equations of motion (it proceeds with constant velocity). One ﬁnds the action for this path is S[q(t)] = 1 2mω0 Q2 1 ω0T −1 3ω0T ! . (3.87) Thus, provided ω0T ̸= nπ, in the limit T →∞we ﬁnd that the constant velocity path has lower action. Finally, consider the general mechanical action, S q(t) = tb Z ta dt L(q, ˙ q, t) . (3.88) We now evaluate the ﬁrst few terms in the functional Taylor series: S q∗(t) + δq(t) = tb Z ta dt ( L(q∗, ˙ q∗, t) + ∂L ∂qi q∗ δqi + ∂L ∂˙ qi q∗ δ ˙ qi + 1 2 ∂2L ∂qi ∂qj q∗ δqi δqj + ∂2L ∂qi ∂˙ qj q∗ δqi δ ˙ qj + 1 2 ∂2L ∂˙ qi ∂˙ qj q∗ δ ˙ qi δ ˙ qj + . . . ) . (3.89) To identify the functional derivatives, we integrate by parts. Let Φ...(t) be an arbitrary function of time. Then tb Z ta dt Φi(t) δ ˙ qi(t) = − tb Z ta dt ˙ Φi(t) δqi(t) (3.90) and tb Z ta dt Φij(t) δqi(t) δ ˙ qj(t) = tb Z ta dt tb Z ta dt′ Φij(t) δ(t −t′) d dt′ δqi(t) δqj(t′) = tb Z ta dt tb Z ta dt′ Φij(t)) δ′(t −t′) δqi(t) δqj(t′) , (3.91) and tb Z ta dt Φij(t) d ˙ qi(t) δ ˙ qj(t) = tb Z ta dt tb Z ta dt′ Φij(t) δ(t −t′) d dt d dt′ δqi(t) δqj(t′) = − tb Z ta dt tb Z ta dt′ ˙ Φij(t) δ′(t −t′) + Φij(t) δ′′(t −t′) δqi(t) δqj(t′) . (3.92) 18 CHAPTER 3. LAGRANGIAN MECHANICS Thus, the ﬁrst two functional derivatives are given by δS δqi(t) = " ∂L ∂qi −d dt ∂L ∂˙ qi # q∗(t) (3.93) and δ2S δqi(t) δqj(t′) = ( ∂2L ∂qi ∂qj q∗(t) δ(t −t′) − ∂2L ∂˙ qi ∂˙ qj q∗(t) δ′′(t −t′) + " 2 ∂2L ∂qi ∂˙ qj −d dt ∂2L ∂˙ qi ∂˙ qj # q∗(t) δ′(t −t′) ) . (3.94) 3.5 Generalized Coordinates A set of generalized coordinates q1, . . . , qn completely describes the positions of all particles in a mechani-cal system. In a system with df degrees of freedom and k constraints, n = df −k independent generalized coordinates are needed to completely specify all the positions. A constraint is a relation among coordi-nates, such as x2 + y2 + z2 = a2 for a particle moving on a sphere of radius a. In this case, df = 3 and k = 1. In this case, we could eliminate z in favor of x and y, i.e. by writing z = ± p a2 −x2 −y2, or we could choose as coordinates the polar and azimuthal angles θ and φ. For the moment we will assume that n = df −k, and that the generalized coordinates are independent, satisfying no additional constraints among them. Later on we will learn how to deal with any remaining constraints among the {q1, . . . , qn}. The generalized coordinates may have units of length, or angle, or perhaps something totally different. In the theory of small oscillations, the normal coordinates are conventionally chosen to have units of (mass)1/2 × (length). However, once a choice of generalized coordinate is made, with a concomitant set of units, the units of the conjugate momentum and force are determined: pσ = ML2 T · 1 qσ , Fσ = ML2 T 2 · 1 qσ , (3.95) where A means ‘the units of A’, and where M, L, and T stand for mass, length, and time, respectively. Thus, if qσ has dimensions of length, then pσ has dimensions of momentum and Fσ has dimensions of force. If qσ is dimensionless, as is the case for an angle, pσ has dimensions of angular momentum (ML2/T) and Fσ has dimensions of torque (ML2/T 2). 3.6. HAMILTON’S PRINCIPLE 19 3.6 Hamilton’s Principle The equations of motion of classical mechanics are embodied in a variational principle, called Hamilton’s principle. Hamilton’s principle states that the motion of a system is such that the action functional S q(t) = t2 Z t1 dt L(q, ˙ q, t) (3.96) is an extremum, i.e. δS = 0. Here, q = {q1, . . . , qn} is a complete set of generalized coordinates for our mechanical system, and L = T −U (3.97) is the Lagrangian, where T is the kinetic energy and U is the potential energy. Setting the ﬁrst variation of the action to zero gives the Euler-Lagrange equations, d dt momentum pσ z }\| { ∂L ∂˙ qσ = force Fσ z}\|{ ∂L ∂qσ . (3.98) Thus, we have the familiar ˙ pσ = Fσ, also known as Newton’s second law. Note, however, that the {qσ} are generalized coordinates, so pσ may not have dimensions of momentum, nor Fσ of force. For example, if the generalized coordinate in question is an angle φ, then the corresponding generalized momentum is the angular momentum about the axis of φ’s rotation, and the generalized force is the torque. 3.6.1 Invariance of the equations of motion Suppose ˜ L(q, ˙ q, t) = L(q, ˙ q, t) + d dt G(q, t) . (3.99) Then ˜ S[q(t)] = S[q(t)] + G(qb, tb) −G(qa, ta) . (3.100) Since the difference ˜ S −S is a function only of the endpoint values {qa, qb}, their variations are identical: δ ˜ S = δS. This means that L and ˜ L result in the same equations of motion. Thus, the equations of motion are invariant under a shift of L by a total time derivative of a function of coordinates and time. 3.6.2 Remarks on the order of the equations of motion The equations of motion are second order in time. This follows from the fact that L = L(q, ˙ q, t). Using the chain rule, d dt ∂L ∂˙ qσ = ∂2L ∂˙ qσ ∂˙ qσ′ ¨ qσ′ + ∂2L ∂˙ qσ ∂qσ′ ˙ qσ′ + ∂2L ∂˙ qσ ∂t . (3.101) 20 CHAPTER 3. LAGRANGIAN MECHANICS That the equations are second order in time can be regarded as an empirical fact. It follows, as we have just seen, from the fact that L depends on q and on ˙ q, but on no higher time derivative terms. Suppose the Lagrangian did depend on the generalized accelerations ¨ q as well. What would the equations of motion look like? Taking the variation of S, δ tb Z ta dt L(q, ˙ q, ¨ q, t) = ∂L ∂˙ qσ δqσ + ∂L ∂¨ qσ δ ˙ qσ −d dt ∂L ∂¨ qσ δqσ tb ta + tb Z ta dt ( ∂L ∂qσ −d dt ∂L ∂˙ qσ + d2 dt2 ∂L ∂¨ qσ ) δqσ . (3.102) The boundary term vanishes if we require δqσ(ta) = δqσ(tb) = δ ˙ qσ(ta) = δ ˙ qσ(tb) = 0 ∀σ. The equations of motion would then be fourth order in time. 3.6.3 Lagrangian for a free particle For a free particle, we can use Cartesian coordinates for each particle as our system of generalized coordinates. For a single particle, the Lagrangian L(x, v, t) must be a function solely of v2. This is because homogeneity with respect to space and time preclude any dependence of L on x or on t, and isotropy of space means L must depend on v2. We next invoke Galilean relativity, which says that the equations of motion are invariant under transformation to a reference frame moving with constant velocity. Let V be the velocity of the new reference frame K′ relative to our initial reference frame K. Then x′ = x −V t, and v′ = v −V . In order that the equations of motion be invariant under the change in reference frame, we demand L′(v′) = L(v) + d dt G(x, t) . (3.103) The only possibility is L = 1 2mv2, where the constant m is the mass of the particle. Note: L′ = 1 2m(v −V )2 = 1 2mv2 + d dt 1 2mV 2 t −mV · x = L + dG dt . (3.104) For N interacting particles, L = 1 2 N X a=1 ma dxa dt 2 −U {xa}, { ˙ xa} . (3.105) Here, U is the potential energy. Generally, U is of the form U = X a U1(xa) + X a ω2 0. The stability of these equilibria is assessed by examining the sign of U ′′ eﬀ(θ∗). We have U ′′ eﬀ(θ) = mga n cos θ −ω2 ω2 0 2 cos2 θ −1 o . (3.247) Thus, U ′′ eﬀ(θ∗) =                    mga 1 −ω2 ω2 0 at θ∗= 0 −mga 1 + ω2 ω2 0 at θ∗= π mga ω2 ω2 0 −ω2 0 ω2 at θ∗= ± cos−1 ω2 0 ω2 . (3.248) 42 CHAPTER 3. LAGRANGIAN MECHANICS Figure 3.12: The effective potential Ueﬀ(θ) = mga 1 −cos θ −ω2 2ω2 0 sin2 θ . (The dimensionless potential ˜ Ueﬀ(x) = Ueﬀ/mga is shown, where x = θ/π.) Left panels: ω = 1 2 √ 3 ω0. Right panels: ω = √ 3 ω0. Thus, θ∗= 0 is stable for ω2 < ω2 0 but becomes unstable when the rotation frequency ω is sufﬁciently large, i.e. when ω2 > ω2 0. In this regime, there are two new equilibria, at θ∗= ± cos−1(ω2 0/ω2), which are both stable. The equilibrium at θ∗= π is always unstable, independent of the value of ω. The situation is depicted in ﬁg. 3.12. 3.13.4 Charged particle in a magnetic ﬁeld Consider next the case of a charged particle moving in the presence of an electromagnetic ﬁeld. The particle’s potential energy is U(r, ˙ r) = q φ(r, t) −q c A(r, t) · ˙ r , (3.249) which is velocity-dependent. The kinetic energy is T = 1 2m ˙ r2, as usual. Here φ(r) is the scalar potential and A(r) the vector potential. The electric and magnetic ﬁelds are given by E = −∇φ −1 c ∂A ∂t , B = ∇× A . (3.250) The canonical momentum is p = ∂L ∂˙ r = m ˙ r + q c A , (3.251) 3.13. THE HAMILTONIAN 43 and hence the Hamiltonian is H(r, p, t) = p · ˙ r −L = m ˙ r2 + q c A · ˙ r −1 2m ˙ r2 −q c A · ˙ r + q φ = 1 2m ˙ r2 + q φ = 1 2m p −q c A(r, t) 2 + q φ(r, t) . (3.252) If A and φ are time-independent, then H(r, p) is conserved. Let’s work out the equations of motion. We have d dt ∂L ∂˙ r = ∂L ∂r (3.253) which gives m ¨ r + q c dA dt = −q ∇φ + q c ∇(A · ˙ r) , (3.254) or, in component notation, m ¨ xi + q c ∂Ai ∂xj ˙ xj + q c ∂Ai ∂t = −q ∂φ ∂xi + q c ∂Aj ∂xi ˙ xj , (3.255) which is to say m ¨ xi = −q ∂φ ∂xi −q c ∂Ai ∂t + q c ∂Aj ∂xi −∂Ai ∂xj ˙ xj . (3.256) It is convenient to express the cross product in terms of the completely antisymmetric tensor of rank three, ǫijk: Bi = ǫijk ∂Ak ∂xj , (3.257) and using the result ǫijk ǫimn = δjm δkn −δjn δkm , (3.258) we have ǫijk Bi = ∂j Ak −∂k Aj, and m ¨ xi = −q ∂φ ∂xi −q c ∂Ai ∂t + q c ǫijk ˙ xj Bk , (3.259) or, in vector notation, m ¨ r = −q ∇φ −q c ∂A ∂t + q c ˙ r × (∇× A) = q E + q c ˙ r × B , (3.260) which is, of course, the Lorentz force law. 44 CHAPTER 3. LAGRANGIAN MECHANICS 3.14 Motion in Rapidly Oscillating Fields 3.14.1 Slow and fast dynamics Consider a free particle moving under the inﬂuence of an oscillating force F(t) = F0 cos(ωt). Newton’s second law is then m¨ q = F cos ωt, the solution to which is q(t) = a + b t −F0 cos ωt mω2 . (3.261) where qh(t) ≡a + b t is the solution to the homogeneous (unforced) equation of motion. Note that the amplitude of the response q −qh goes as ω−2 and is therefore small when ω is large. Now consider a general n = 1 system, with H(q, p, t) = H0(q, p) + e V (q) cos(ωt) , (3.262) where we will assume e V (q) is small. We also assume that ω is much greater than any natural oscillation frequency associated with H0. We separate the motion q(t) and p(t) into slow and fast components: q(t) = Q(t) + ζ(t) p(t) = P(t) + π(t) , (3.263) where ζ(t) and π(t) oscillate with the driving frequency ω. Since ζ and π will be small, we expand Hamilton’s equations in these quantities: ˙ Q + ˙ ζ = ∂H0 ∂P + ∂2H0 ∂P 2 π + ∂2H0 ∂Q ∂P ζ + 1 2 ∂3H0 ∂Q2 ∂P ζ2 + ∂3H0 ∂Q ∂P 2 ζπ + 1 2 ∂3H0 ∂P 3 π2 + . . . ˙ P + ˙ π = −∂H0 ∂Q −∂2H0 ∂Q2 ζ −∂2H0 ∂Q ∂P π −1 2 ∂3H0 ∂Q3 ζ2 − ∂3H0 ∂Q2 ∂P ζπ −1 2 ∂3H0 ∂Q ∂P 2 π2 −∂e V ∂Q cos(ωt) −∂2 e V ∂Q2 ζ cos(ωt) −. . . . (3.264) We now average over the fast degrees of freedom to obtain an equation of motion for the slow variables Q and P, which we here carry to lowest nontrivial order in averages of ﬂuctuating quantities: ˙ Q = H0 P + 1 2H0 QQP ⟨ζ2⟩+ H0 QP P ⟨ζπ⟩+ 1 2H0 P P P ⟨π2⟩ ˙ P = −H0 Q −1 2H0 QQQ ⟨ζ2⟩−H0 QQP ⟨ζπ⟩−1 2H0 QP P ⟨π2⟩−e VQQ ⟨ζ cos ωt⟩ , (3.265) where we now adopt the shorthand notation H0 QQP = ∂3H0 ∂2Q ∂P , etc. The fast degrees of freedom obey ˙ ζ = H0 QP ζ + H0 P P π ˙ π = −H0 QQ ζ −H0 QP π −e VQ cos(ωt) . (3.266) 3.14. MOTION IN RAPIDLY OSCILLATING FIELDS 45 We can solve these by replacing e VQ cos ωt with e VQ e−iωt, and writing ζ(t) = ζ0 e−iωt and π(t) = π0 e−iωt, resulting in H0 QP + iω H0 P P −H0 QQ −H0 QP + iω ζ0 π0 = 0 e VQ . (3.267) We now invert the matrix to obtain ζ0 and π0, then take the real part, which yields ζ(t) = ω−2 H0 P P e VQ cos ωt + O ω−4 π(t) = −ω−2 H0 QP e VQ cos ωt −ω−1 e VQ sin ωt + O ω−3 . (3.268) Invoking cos2(ωt) = sin2(ωt) = 1 2 and cos(ωt) sin(ωt) = 0, we substitute into eqns. 3.265 to obtain ˙ Q = H0 P + 1 4 ω−2 H0 P P P e V 2 Q + O ω−4 ˙ P = −H0 Q −1 4 ω−2 H0 QP P e V 2 Q −1 2 ω−2H0 P P e VQ e VQQ + O ω−4 . (3.269) These equations may be written compactly as ˙ Q = ∂K ∂P , ˙ P = −∂K ∂Q , (3.270) where K(Q, P) = H0(Q, P) + 1 4ω2 ∂2H0 ∂P 2 ∂e V ∂Q 2 + . . . . (3.271) 3.14.2 Example : pendulum with oscillating support Consider a pendulum with a vertically oscillating point of support. The coordinates of the pendulum bob are x = ℓsin θ , y = a(t) −ℓcos θ . (3.272) The Lagrangian is easily obtained: L = 1 2mℓ2 ˙ θ2 + mℓ˙ a ˙ θ sin θ + mgℓcos θ + 1 2m˙ a2 −mga = 1 2mℓ2 ˙ θ2 + m(g + ¨ a)ℓcos θ+ these may be dropped z }\| { 1 2m˙ a2 −mga −d dt mℓ˙ a cos θ . (3.273) Thus we may take the Lagrangian to be L = 1 2mℓ2 ˙ θ2 + m(g + ¨ a) ℓcos θ , (3.274) from which we derive the Hamiltonian H(θ, pθ) = p2 θ 2mℓ2 −mgℓcos θ −mℓ¨ a cos θ = H0(θ, pθ, t) + e V (θ) sin ωt . (3.275) 46 CHAPTER 3. LAGRANGIAN MECHANICS Figure 3.13: Dimensionless potential v(Θ) for r = 0 (black curve), r = 0.5 (red), and r = 2 (blue). We have assumed a(t) = a0 sin ωt, so e V (θ) = mℓa0 ω2 cos θ . (3.276) Writing θ ≡Θ + ζ and pθ ≡L + π, the effective Hamiltonian, per eqn. 3.271, is K(Θ, L) = L2 2mℓ2 −mgℓcos Θ + 1 4m a2 0 ω2 sin2Θ . (3.277) Let’s deﬁne the dimensionless parameter r ≡ω2a2 0 2gℓ . (3.278) The slow variable Θ executes motion in the effective potential Veﬀ(Θ) = mgℓv(Θ), with v(Θ) = −cos Θ + r 2 sin2Θ . (3.279) Differentiating, we ﬁnd that Veﬀ(Θ) is stationary when v′(Θ) = 0 ⇒ r sin Θ cos Θ = −sin Θ . (3.280) Thus, Θ = 0 and Θ = π, where sin Θ = 0, are equilibria. When r > 1 (note r > 0 always), there are two new solutions, given by the roots of cos Θ = −r−1. To assess stability of these equilibria, we compute the second derivative: v′′(Θ) = cos Θ + r cos 2Θ . (3.281) 3.15. FIELD THEORY: SYSTEMS WITH SEVERAL INDEPENDENT VARIABLES 47 From this, we see that Θ = 0 is stable, i.e. v′′(Θ = 0) > 0, always, but Θ = π is stable for r > 1 and unstable for r < 1. When r > 1, two new solutions appear, at cos Θ = −r−1, for which v′′(cos−1(−1/r)) = r−1 −r , (3.282) which is always negative since r > 1 in order for these equilibria to exist. The situation is sketched in ﬁg. 3.13, showing v(Θ) for three representative values of the parameter r. For r < 1, the equilibrium at Θ = π is unstable, but as r increases, a subcritical pitchfork bifurcation is encountered at r = 1, and Θ = π becomes stable, while the outlying Θ = cos−1(−1/r) solutions are unstable. 3.15 Field Theory: Systems with Several Independent Variables 3.15.1 Equations of motion and Noether’s theorem Suppose φa(x) depends on several independent variables: x = {x1, x2, . . . , xn}. Furthermore, suppose S {φa(x)} = Z Ω dnx L(φa ∂µφa, x) , (3.283) i.e. the Lagrangian density L is a function of the ﬁelds φa˙ , their partial derivatives ∂φa/∂xµ, and possibly the independent variables xµ as well. Here Ωis a region in Rn. In dynamical ﬁeld theories, we write x = (x0, x1, . . . , xd) where d is the dimension of space and x0 = ct, where t is time and c is a constant with dimensions of speed. In such cases n = d + 1 and we can identify x0 ≡xn. Then the ﬁrst variation of S is δS = Z Ω dnx ( ∂L ∂φa δφa + ∂L ∂(∂µφa) ∂δφa ∂xµ ) = I ∂Ω dΣ nµ ∂L ∂(∂µφa) δφa + Z Ω dnx ( ∂L ∂φa − ∂ ∂xµ ∂L ∂(∂µφa) ) δφa , (3.284) where ∂Ωis the (n−1)-dimensional boundary of Ω, dΣ is the differential surface area, and nµ is the unit vector normal to ∂Ω. If we demand ∂L/∂(∂µφa) ∂Ω= 0 or δφa ∂Ω= 0, the surface term vanishes, and we conclude δS δφa(x) = ∂L ∂φa − ∂ ∂xµ ∂L ∂(∂µφa) . (3.285) Next, consider the one-parameter family of ﬁeld transformations φa(x) →e φa φ(x), ζ (3.286) 48 CHAPTER 3. LAGRANGIAN MECHANICS such that e φa φ(x), ζ = 0 = φa(x). If the Lagrangian density L is independent of this transformation, then dL dζ ζ=0 = ∂L ∂φa ∂e φa ∂ζ ζ=0 + n X µ=1 ∂L ∂(∂µφa) ∂(∂µ e φa) ∂ζ ζ=0 = n X µ=1 ( ∂ ∂xµ ∂L ∂(∂µφa) ∂e φa ∂ζ ζ=0 + ∂L ∂(∂µφa) ∂ ∂xµ ∂e φa ∂ζ ζ=0 ) = n X µ=1 ∂ ∂xµ ∂L ∂(∂µφa) ∂e φa ∂ζ ζ=0 (3.287) We can write this as ∂µ Jµ = 0, where Jµ ≡ ∂L ∂(∂µφa) ∂e φa ∂ζ ζ=0 . (3.288) We call Λ = J0/c the total charge. If we assume J = 0 at the spatial boundaries of our system, then integrating the conservation law ∂µ Jµ (summation convention) over the spatial region Ωgives dΛ dt = Z Ω d3x ∂0 J0 = − Z Ω d3x ∇·J = − I ∂Ω dΣ ˆ n · J = 0 , (3.289) assuming J = 0 at the boundary ∂Ω. As an example, consider the case of a stretched string of linear mass density ρ and tension τ. The action is a functional of the height y(x, t), where the coordinate along the string, x, and time, t, are the two independent variables. The Lagrangian density is L = 1 2 ρ ∂y ∂t 2 −1 2 τ ∂y ∂x 2 . (3.290) The Euler-Lagrange equations are 0 = δS δy(x, t) = −∂ ∂x ∂L ∂y′ −∂ ∂t ∂L ∂˙ y = ∂ ∂x τ ∂y ∂x −ρ ∂2y ∂t2 , (3.291) where y′ = ∂y/∂x and ˙ y = ∂y/∂t. We’ve assumed boundary conditions where δy(xa, t) = δy(xb, t) = δy(x, ta) = δy(x, tb) = 0. At this point, ρ(x) and τ(x) may be position-dependent. For constant ρ and τ, we obtain the Helmholtz equation ρ¨ y = τy′′, where c = (τ/ρ)1/2 is the speed of wave propagation. For practice with the Minkowski notation, we deﬁne x0 ≡ct and x1 ≡x and the two-dimensional space-time coordinate vector is then xµ = (x0, x1) = (ct, x). The Lagrangian can then be written L = 1 2 τ(∂µy)(∂µy), where xµ = gµν xν = (ct, −x), in which case ∂µ = ∂/∂xµ and ∂µ = ∂/∂xµ. Clearly L 3.15. FIELD THEORY: SYSTEMS WITH SEVERAL INDEPENDENT VARIABLES 49 remains invariant under the one-parameter family of transformations y →y + ζ, and the conserved Noether current is Jµ = τ ∂y ∂xµ , (3.292) and we have ∂µJµ = 0, which is equivalent to ∂µJµ = 0. (Upper indices are called covariant while lower ones are contravariant.) Current conservation in this system is simply a restatement of the Helmholtz equation. Maxwell’s equations The Lagrangian density for an electromagnetic ﬁeld with sources is L = −1 16π Fµν F µν −1 c jµ Aµ . (3.293) The equations of motion are then ∂L ∂Aµ − ∂ ∂xν ∂L ∂(∂µAν) = 0 ⇒ ∂µ F µν = 4π c jν , (3.294) which are Maxwell’s equations. Relativistic complex scalar ﬁeld As an example, consider the case of a complex scalar ﬁeld, with Lagrangian density L(ψ, ψ∗, ∂µψ, ∂µψ∗) = 1 2K (∂µψ∗)(∂µψ) −U ψ∗ψ . (3.295) This is invariant under the transformation ψ →eiζ ψ, ψ∗→e−iζ ψ∗. Thus, ∂˜ ψ ∂ζ = i eiζ ψ , ∂˜ ψ∗ ∂ζ = −i e−iζ ψ∗ , (3.296) and, summing over both ψ and ψ∗ﬁelds, we have Jµ = ∂L ∂(∂µψ) · (iψ) + ∂L ∂(∂µψ∗) · (−iψ∗) = K 2i ψ∗∂µψ −ψ ∂µψ∗ . (3.297) The potential, which depends on \|ψ\|2, is independent of ζ. Hence, this form of conserved 4-current is valid for an entire class of potentials. 50 CHAPTER 3. LAGRANGIAN MECHANICS 3.15.2 Gross-Pitaevskii model As one ﬁnal example of a ﬁeld theory, consider the Gross-Pitaevskii model, with L = iℏψ∗∂ψ ∂t −ℏ2 2m ∇ψ∗·∇ψ −g \|ψ\|2 −n0 2 . (3.298) This describes a nonrelativistic Bose ﬂuid with repulsive short-ranged interactions. Here ψ(x, t) is again a complex scalar ﬁeld, and ψ∗is its complex conjugate. Using the Leibniz rule, we have δS[ψ∗, ψ] = S[ψ∗+ δψ∗, ψ + δψ] = Z dt Z ddx iℏψ∗∂δψ ∂t + iℏδψ∗∂ψ ∂t −ℏ2 2m ∇ψ∗· ∇δψ −ℏ2 2m ∇δψ∗· ∇ψ −2g \|ψ\|2 −n0 (ψ∗δψ + ψδψ∗) = Z dt Z ddx ( −iℏ∂ψ∗ ∂t + ℏ2 2m ∇2ψ∗−2g \|ψ\|2 −n0 ψ∗ δψ + iℏ∂ψ ∂t + ℏ2 2m ∇2ψ −2g \|ψ\|2 −n0 ψ δψ∗ ) , (3.299) where we have integrated by parts where necessary and discarded the boundary terms. Extremizing S[ψ∗, ψ] therefore results in the nonlinear Schr¨ odinger equation (NLSE), iℏ∂ψ ∂t = −ℏ2 2m ∇2ψ + 2g \|ψ\|2 −n0 ψ (3.300) as well as its complex conjugate, −iℏ∂ψ∗ ∂t = −ℏ2 2m ∇2ψ∗+ 2g \|ψ\|2 −n0 ψ∗ . (3.301) Note that these equations are indeed the Euler-Lagrange equations: δS δψ = ∂L ∂ψ − ∂ ∂xµ ∂L ∂∂µψ δS δψ∗= ∂L ∂ψ∗− ∂ ∂xµ ∂L ∂∂µψ∗ , (3.302) with xµ = (t, x)6. Plugging in ∂L ∂ψ = −2g \|ψ\|2 −n0 ψ∗ , ∂L ∂∂tψ = iℏψ∗ , ∂L ∂∇ψ = −ℏ2 2m ∇ψ∗ (3.303) and ∂L ∂ψ∗= iℏψ −2g \|ψ\|2 −n0 ψ , ∂L ∂∂tψ∗= 0 , ∂L ∂∇ψ∗= −ℏ2 2m ∇ψ , (3.304) 6In the nonrelativistic case, there is no utility in deﬁning x0 = ct, so we simply deﬁne x0 = t. 3.16. CONSTRAINTS 51 we recover the NLSE and its conjugate. The Gross-Pitaevskii model also possesses a U(1) or O(2) invariance, viz. ψ(x, t) →e ψ(x, t) = eiζ ψ(x, t) , ψ∗(x, t) →e ψ∗(x, t) = e−iζ ψ∗(x, t) . (3.305) Thus, the conserved Noether current is then Jµ = ∂L ∂∂µψ ∂e ψ ∂ζ ζ=0 + ∂L ∂∂µψ∗ ∂e ψ∗ ∂ζ ζ=0 J0 = −ℏ\|ψ\|2 J = −ℏ2 2im ψ∗∇ψ −ψ∇ψ∗ . (3.306) Dividing out by ℏ, taking J0 ≡−ℏρ and J ≡−ℏj, we obtain the continuity equation, ∂ρ ∂t + ∇·j = 0 , (3.307) where ρ = \|ψ\|2 , j = ℏ 2im ψ∗∇ψ −ψ∇ψ∗ . (3.308) are the particle density and the particle current, respectively. 3.16 Constraints 3.16.1 Introduction A mechanical system of N point particles in d dimensions possesses n = dN degrees of freedom7. To specify these degrees of freedom, we can choose any independent set of generalized coordinates {q1, . . . , qn}. Oftentimes, however, not all n coordinates are independent. Consider, for example, the situation in ﬁg. 3.14, where a cylinder of radius a rolls over a half-cylinder of radius R. If there is no slippage, then the angles θ1 and θ2 are not independent, and they obey the equation of constraint, R θ1 = a (θ2 −θ1) . (3.309) In this case, we can easily solve the constraint equation and substitute θ2 = 1 + R a θ1. In other cases, though, the equation of constraint might not be so easily solved (e.g. it may be nonlinear). How then do we proceed? 7For N rigid bodies, the number of degrees of freedom is n′ = 1 2d(d + 1)N, corresponding to d center-of-mass coordinates and 1 2d(d −1) angles of orientation for each particle. The dimension of the group of rotations in d dimensions is 1 2d(d −1), corresponding to the number of parameters in a general rank-d orthogonal matrix (i.e. an element of the group O(d)). 52 CHAPTER 3. LAGRANGIAN MECHANICS Figure 3.14: A cylinder of radius a rolls along a half-cylinder of radius R. When there is no slippage, the angles θ1 and θ2 obey the constraint equation Rθ1 = a(θ2 −θ1). 3.16.2 Constrained extremization of functions : Lagrange multipliers Given F(x1, . . . , xn) to be extremized subject to k constraints of the form Gj(x1, . . . , xn) = 0 where j = 1, . . . , k, construct F ∗x1, . . . , xn; λ1, . . . , λk ≡F(x1, . . . , xn) + k X j=1 λj Gj(x1, . . . , xn) (3.310) which is a function of the (n+k) variables x1, . . . , xn; λ1, . . . , λk , where the quantities {λ1, . . . , λk} are Lagrange undetermined multipliers. We now freely extremize the extended function F ∗: dF ∗= n X σ=1 ∂F ∗ ∂xσ dxσ + k X j=1 ∂F ∗ ∂λj dλj = n X σ=1  ∂F ∂xσ + k X j=1 λj ∂Gj ∂xσ  dxσ + k X j=1 Gj dλj = 0 (3.311) This results in the (n + k) equations ∂F ∂xσ + k X j=1 λj ∂Gj ∂xσ = 0 (σ = 1, . . . , n) Gj = 0 (j = 1, . . . , k) . (3.312) The interpretation of all this is as follows. The ﬁrst n equations in 3.312 can be written in vector form as ∇F + k X j=1 λj ∇Gj = 0 . (3.313) 3.16. CONSTRAINTS 53 This says that the (n-component) vector ∇F is linearly dependent upon the k vectors ∇Gj. Thus, any movement in the direction of ∇F must necessarily entail movement along one or more of the directions ∇Gj. This would require violating the constraints, since movement along ∇Gj takes us off the level set Gj = 0. Were ∇F linearly independent of the set {∇Gj}, this would mean that we could ﬁnd a differential displacement dx which has ﬁnite overlap with ∇F but zero overlap with each ∇Gj. Thus x + dx would still satisfy Gj(x + dx) = 0, but F would change by the ﬁnite amount dF = ∇F(x) · dx. Put another way, when we extremize F(x) without constraints, we identify points x ∈Rn where the gradient ∇F vanishes. However, when we have k constraints of the form Gj(x) = 0, the subset Υ = x ∈Rn \| Gj(x) = 0 ∀j ∈{1, . . . , k} (3.314) is a hypersurface of dimension n −k. Generically we should not expect any of the solutions to ∇F = 0 to lie within the subspace Υ. Extremizing F(x) subject to the k constraints Gj(x) = 0 means that we must ﬁnd the extrema of F(x) for x ∈Υ ⊂Rn. All such extrema satisfy that ∇F(x) is perpendicular to the hypersurface Υ, i.e. ∇F(x) must lie in the k-dimensional subspace spanned by the vectors ∇Gj(x). Example : volume of a cylinder To see how this formalism works in practice, let’s extremize the volume V = πa2h of a cylinder of radius a and height h, subject to the constraint G(a, h) = 2πa + h2 b −ℓ= 0 . (3.315) We therefore deﬁne V ∗(a, h, λ) ≡V (a, h) + λ G(a, h) , (3.316) and set ∂V ∗ ∂a = 2πah + 2πλ = 0 (3.317) ∂V ∗ ∂h = πa2 + 2λ h b = 0 (3.318) ∂V ∗ ∂λ = 2πa + h2 b −ℓ= 0 . (3.319) Solving these three equations simultaneously gives a = 2ℓ 5π , h = r bℓ 5 , λ = − 2 53/2π b1/2 ℓ3/2 , V ∗= 4 55/2 π ℓ5/2 b1/2 . (3.320) 54 CHAPTER 3. LAGRANGIAN MECHANICS 3.16.3 Constraints and variational calculus Before addressing the subject of constrained dynamical systems, let’s consider the issue of constraints in the broader context of variational calculus. Suppose we have a functional F[y(x)] = xb Z xa dx L(y, y′, x) , (3.321) which we want to extremize subject to some constraints. Here y stands for an n-component vector of functions {yσ(x)}. We assume that the endpoint values yσ(xa) and yσ(xb) are ﬁxed for each σ. There are two classes of constraints we will consider: 1. Integral constraints: These are of the form xb Z xa dx Nj(y, y′, x) = Cj , (3.322) where j labels the constraint. 2. Holonomic constraints: These are of the form Gj(y, x) = 0 . (3.323) The cylinders system in ﬁg. 3.14 provides an example of a holonomic constraint. There, G(θ, t) = R θ1 −a (θ2 −θ1) = 0. As an example of a problem with an integral constraint, suppose we want to know the shape of a hanging rope of ﬁxed length C. This means we minimize the rope’s potential energy, U[y(x)] = ρg xb Z xa ds y(x) = ρg xb Z xa dx y q 1 + y′2 , (3.324) where ρ is the linear mass density of the rope, subject to the ﬁxed-length constraint C = xb Z xa ds = xb Z xa dx q 1 + y′2 . (3.325) Note ds = p dx2 + dy2 is the differential element of arc length along the rope. To solve problems like these, we again use the method of Lagrange multipliers. 3.16.4 Extremization of functionals : integral constraints Given a functional F {yσ(x)} = xb Z xa dx L {yσ}, {y′ σ}, x (σ = 1, . . . , n) (3.326) 3.16. CONSTRAINTS 55 subject to boundary conditions δyσ(xa) = δyσ(xb) = 0 and k constraints of the form xb Z xa dx Nl {yσ}, {y′ σ}, x = Cl (l = 1, . . . , k) , (3.327) construct the extended functional F ∗ {yσ(x)}; {λj} ≡ xb Z xa dx L {yσ}, {y′ σ}, x + k X l=1 λl Nl {yσ}, {y′ σ}, x − k X l=1 λl Cl (3.328) and freely extremize over {y1, . . . , yn; λ1, . . . , λk}. This results in (n + k) equations ∂L ∂yσ −d dx ∂L ∂y′ σ + k X l=1 λl ( ∂Nl ∂yσ −d dx ∂Nl ∂y′ σ ) = 0 (σ = 1, . . . , n) xb Z xa dx Nl {yσ}, {y′ σ}, x = Cl (l = 1, . . . , k) . (3.329) 3.16.5 Extremization of functionals : holonomic constraints Given a functional F {yσ(x)} = xb Z xa dx L {yσ}, {y′ σ}, x (σ = 1, . . . , n) (3.330) subject to boundary conditions δyσ(xa) = δyσ(xb) = 0 and k constraints of the form Gj {yσ(x)}, x = 0 (j = 1, . . . , k) , (3.331) construct the extended functional F ∗ {yσ(x)}; {λj(x)} ≡ xb Z xa dx L {yσ}, {y′ σ}, x + k X j=1 λj Gj {yσ} (3.332) and freely extremize over the (n + k) functions y1(x), . . . , yn(x); λ1(x), . . . , λk(x)}: δF ∗= xb Z xa dx ( n X σ=1 ∂L ∂yσ −d dx ∂L ∂y′ σ + k X j=1 λj ∂Gj ∂yσ δyσ + k X j=1 Gj δλj ) = 0 , (3.333) resulting in the (n + k) equations d dx ∂L ∂y′ σ ! −∂L ∂yσ = k X j=1 λj ∂Gj ∂yσ (σ = 1, . . . , n) Gj = 0 (j = 1, . . . , k) . (3.334) 56 CHAPTER 3. LAGRANGIAN MECHANICS 3.16.6 Examples of functional extremization with constraints Hanging rope We minimize the potential energy functional U y(x) = ρg x2 Z x1 dx y q 1 + y′2 , (3.335) where ρ is the linear mass density, subject to the constraint of ﬁxed total length, C y(x) = x2 Z x1 dx q 1 + y′2 . (3.336) Thus, U ∗ y(x), λ = U y(x) + λC y(x) = x2 Z x1 dx L∗(y, y′, x) , (3.337) with L∗(y, y′, x) = (ρgy + λ) q 1 + y′2 . (3.338) Since ∂L∗/∂x = 0 we have that H = y′ ∂L∗ ∂y′ −L∗= −ρgy + λ p 1 + y′2 (3.339) is constant. Thus, dy dx = ±H−1 p (ρgy + λ)2 −H2 , (3.340) with solution y(x) = −λ ρg + H ρg cosh ρg H (x −a) . (3.341) Here, H, a, and λ are constants to be determined by demanding y(xi) = yi (i = 1, 2), and that the total length of the rope is C. Geodesic on a curved surface Consider next the problem of a geodesic on a curved surface. Let the equation for the surface be G(x, y, z) = 0 . (3.342) We wish to extremize the distance, D = b Z a ds = b Z a p dx2 + dy2 + dz2 . (3.343) 3.16. CONSTRAINTS 57 We introduce a parameter t deﬁned on the unit interval: t ∈[0, 1], such that x(0) = xa, x(1) = xb, etc. Then D may be regarded as a functional, viz. D x(t), y(t), z(t) = 1 Z 0 dt p ˙ x2 + ˙ y2 + ˙ z2 . (3.344) We impose the constraint by forming the extended functional, D∗: D∗ x(t), y(t), z(t), λ(t) ≡ 1 Z 0 dt p ˙ x2 + ˙ y2 + ˙ z2 + λ G(x, y, z) , (3.345) and we demand that the ﬁrst functional derivatives of D∗vanish: δD∗ δx(t) = −d dt ˙ x p ˙ x2 + ˙ y2 + ˙ z2 + λ ∂G ∂x = 0 δD∗ δy(t) = −d dt ˙ y p ˙ x2 + ˙ y2 + ˙ z2 + λ ∂G ∂y = 0 δD∗ δz(t) = −d dt ˙ z p ˙ x2 + ˙ y2 + ˙ z2 + λ ∂G ∂z = 0 δD∗ δλ(t) = G(x, y, z) = 0 . (3.346) Thus, λ(t) = v¨ x −˙ x ˙ v v2 ∂xG = v¨ y −˙ y ˙ v v2 ∂yG = v¨ z −˙ z ˙ v v2 ∂zG , (3.347) with v = p ˙ x2 + ˙ y2 + ˙ z2 and ∂x ≡ ∂ ∂x, etc. These three equations are supplemented by G(x, y, z) = 0, which is the fourth. 3.16.7 Constraints in Lagrangian mechanics Let us write our system of constraints in the differential form n X σ=1 gjσ(q, t) dqσ + hj(q, t) dt = 0 (j = 1, . . . , k) . (3.348) If the partial derivatives satisfy ∂gjσ ∂qσ′ = ∂gjσ′ ∂qσ , ∂gjσ ∂t = ∂hj ∂qσ , (3.349) then the k differentials can be integrated to give dGj(q, t) = 0 for each j ∈{1, . . . , k}, where gjσ = ∂Gj ∂qσ , hj = ∂Gj ∂t . (3.350) 58 CHAPTER 3. LAGRANGIAN MECHANICS The action functional is S[{qσ(t)}] = tb Z ta dt L {qσ}, { ˙ qσ}, t (σ = 1, . . . , n) , (3.351) subject to boundary conditions δqσ(ta) = δqσ(tb) = 0. The ﬁrst variation of S is given by δS = tb Z ta dt n X σ=1 ( ∂L ∂qσ −d dt ∂L ∂˙ qσ ) δqσ . (3.352) Since the {qσ(t)} are no longer independent, we cannot infer that the term in brackets vanishes for each index σ. What are the constraints on the variations δqσ(t)? The constraints are expressed in terms of virtual displacements which take no time: δt = 0. Thus, n X σ=1 gjσ(q, t) δqσ(t) = 0 , (3.353) where j = 1, . . . , k is the constraint index. We may now relax the constraint by introducing k un-determined functions λj(t), by adding integrals of the above equations with undetermined coefﬁcient functions to δS: n X σ=1 ( ∂L ∂qσ −d dt ∂L ∂˙ qσ + k X j=1 λj(t) gjσ(q, t) ) δqσ(t) = 0 . (3.354) Now we can demand that the term in brackets vanish for all σ. Thus, we obtain a set of (n+k) equations, d dt ∂L ∂˙ qσ −∂L ∂qσ = k X j=1 λj(t) gjσ(q, t) ≡Qσ n X σ=1 gjσ(q, t) ˙ qσ + hj(q, t) = 0 , (3.355) in (n + k) unknowns q1, . . . , qn, λ1, . . . , λk . Here, Qσ is the force of constraint conjugate to the generalized coordinate qσ. Thus, with pσ = ∂L ∂˙ qσ , Fσ = ∂L ∂qσ , Qσ = k X j=1 λj gjσ , (3.356) we write Newton’s second law as ˙ pσ = Fσ + Qσ . (3.357) Note that we can write δS δq(t) = ∂L ∂q −d dt ∂L ∂˙ q (3.358) 3.16. CONSTRAINTS 59 and that the instantaneous constraints may be written gj · δq = 0 (j = 1, . . . , k) . (3.359) Thus, by demanding δS δq(t) + k X j=1 λj gj = 0 (3.360) we require that the functional derivative be linearly dependent on the k vectors gj. 3.16.8 Constraints and conservation laws We have seen how invariance of the Lagrangian with respect to a one-parameter family of coordinate transformations results in an associated conserved quantity Λ, and how a lack of explicit time depen-dence in L results in the conservation of the Hamiltonian H. In deriving both these results, however, we used the equations of motion ˙ pσ = Fσ. What happens when we have constraints, in which case ˙ pσ = Fσ + Qσ? Let’s begin with the Hamiltonian. We have H = ˙ qσ pσ −L, hence dH dt = pσ −∂L ∂˙ qσ ¨ qσ + ˙ pσ −∂L ∂qσ ˙ qσ −∂L ∂t = Qσ ˙ qσ −∂L ∂t . (3.361) We now use Qσ ˙ qσ = λj gjσ ˙ qσ = −λj hj (3.362) to obtain dH dt = −λj hj −∂L ∂t . (3.363) We therefore conclude that in a system with constraints of the form gjσ ˙ qσ + hj = 0, the Hamiltonian is conserved if each hj = 0 and if L is not explicitly dependent on time. In the case of holonomic constraints, hj = ∂Gj ∂t , so H is conserved if neither L nor any of the constraints Gj is explicitly time-dependent. Next, let us rederive Noether’s theorem when constraints are present. We assume a one-parameter family of transformations qσ →˜ qσ(ζ) leaves L invariant. Then 0 = dL dζ = ∂L ∂˜ qσ ∂˜ qσ ∂ζ + ∂L ∂˙ ˜ qσ ∂˙ ˜ qσ ∂ζ = ˙ ˜ pσ −˜ Qσ ∂˜ qσ ∂ζ + ˜ pσ d dt ∂˜ qσ ∂ζ = d dt ˜ pσ ∂˜ qσ ∂ζ −λj ˜ gjσ ∂˜ qσ ∂ζ . (3.364) 60 CHAPTER 3. LAGRANGIAN MECHANICS Now let us write the constraints in differential form as ˜ gjσ d˜ qσ + ˜ hj dt + ˜ kj dζ = 0 . (3.365) We now have dΛ dt = λj ˜ kj , (3.366) which says that if the constraints are independent of ζ then Λ is conserved. For holonomic constraints, this means that Gj ˜ q(ζ), t = 0 ⇒ ˜ kj = ∂Gj ∂ζ = 0 , (3.367) i.e. Gj(˜ q, t) has no explicit ζ dependence. 3.17 Worked Examples Here we consider several example problems of constrained dynamics, and work each out in full detail. 3.17.1 One cylinder rolling off another As an example of the constraint formalism, consider the system in ﬁg. 3.14, where a cylinder of radius a rolls atop a cylinder of radius R. We have two constraints: G1(r, θ1, θ2) = r −R −a = 0 (cylinders in contact) (3.368) G2(r, θ1, θ2) = R θ1 −a (θ2 −θ1) = 0 (no slipping) , (3.369) from which we obtain the gjσ: gjσ = 1 0 0 0 R + a −a , (3.370) which is to say ∂G1 ∂r = 1 , ∂G1 ∂θ1 = 0 , ∂G1 ∂θ2 = 0 ∂G2 ∂r = 0 , ∂G2 ∂θ1 = R + a , ∂G2 ∂θ2 = −a . (3.371) The Lagrangian is L = T −U = 1 2M ˙ r2 + r2 ˙ θ2 1 + 1 2I ˙ θ2 2 −Mgr cos θ1 , (3.372) where M and I are the mass and rotational inertia of the rolling cylinder, respectively. Note that the kinetic energy is a sum of center-of-mass translation Ttr = 1 2M ˙ r2 + r2 ˙ θ2 1 and rotation about the center-3.17. WORKED EXAMPLES 61 of-mass, Trot = 1 2I ˙ θ2 2. The equations of motion are d dt ∂L ∂˙ r −∂L ∂r = M ¨ r −Mr ˙ θ2 1 + Mg cos θ1 = λ1 ≡Qr d dt ∂L ∂˙ θ1 −∂L ∂θ1 = Mr2¨ θ1 + 2Mr ˙ r ˙ θ1 −Mgr sin θ1 = (R + a) λ2 ≡Qθ1 d dt ∂L ∂˙ θ2 −∂L ∂θ2 = I ¨ θ2 = −a λ2 ≡Qθ2 . (3.373) To these three equations we add the two constraints, resulting in ﬁve equations in the ﬁve unknowns r, θ1, θ2, λ1, λ2 . We solve by ﬁrst implementing the constraints, which give r = (R + a) a constant (i.e. ˙ r = 0), and ˙ θ2 = 1 + R a ˙ θ1. Substituting these into the above equations gives −M(R + a) ˙ θ2 1 + Mg cos θ1 = λ1 (3.374) M(R + a)2¨ θ1 −Mg(R + a) sin θ1 = (R + a) λ2 (3.375) I R + a a ¨ θ1 = −aλ2 . (3.376) From eqn. 3.376 we obtain λ2 = −I a ¨ θ2 = −R + a a2 I ¨ θ1 , (3.377) which we substitute into eqn. 3.375 to obtain M + I a2 (R + a)2¨ θ1 −Mg(R + a) sin θ1 = 0 . (3.378) Multiplying by ˙ θ1, we obtain an exact differential, which may be integrated to yield 1 2M 1 + I Ma2 ˙ θ2 1 + Mg R + a cos θ1 = Mg R + a cos θ◦ 1 . (3.379) Here, we have assumed that ˙ θ1 = 0 when θ1 = θ◦ 1, i.e. the rolling cylinder is released from rest at θ1 = θ◦ 1. Finally, inserting this result into eqn. 3.374, we obtain the radial force of constraint, Qr = Mg 1 + α n (3 + α) cos θ1 −2 cos θ◦ 1 o , (3.380) where α = I/Ma2 is a dimensionless parameter (0 ≤α ≤1). This is the radial component of the normal force between the two cylinders. When Qr vanishes, the cylinders lose contact – the rolling cylinder ﬂies off. Clearly this occurs at an angle θ1 = θ∗ 1, where θ∗ 1 = cos−1 2 cos θ◦ 1 3 + α . (3.381) 62 CHAPTER 3. LAGRANGIAN MECHANICS The detachment angle θ∗ 1 is an increasing function of α, which means that larger I delays detachment. This makes good sense, since when I is larger the gain in kinetic energy is split between translational and rotational motion of the rolling cylinder. Note also that Qr(θ◦ 1) = Mg cos θ◦ 1 balances the initial radial component of the force of gravity. Finally, note that the differential equation dt = R + a 2g 1/2 dθ p cos θ◦ 1 −cos θ1 (3.382) may be integrated to yield θ1(t) for t ∈[0, t∗], where θ1(t∗) = θ∗ 1, i.e. t∗is the time to detachment. 3.17.2 Frictionless motion along a curve Consider the situation in ﬁg. 3.15 where a skier moves frictionlessly under the inﬂuence of gravity along a general curve y = h(x). The Lagrangian for this problem is L = 1 2m( ˙ x2 + ˙ y2) −mgy (3.383) and the (holonomic) constraint is G(x, y) = y −h(x) = 0 . (3.384) Accordingly, the Euler-Lagrange equations are d dt ∂L ∂˙ qσ −∂L ∂qσ = λ ∂G ∂qσ , (3.385) where q1 = x and q2 = y. Thus, we obtain m¨ x = −λ h′(x) = Qx m¨ y + mg = λ = Qy . (3.386) We eliminate y in favor of x by invoking the constraint. Since we need ¨ y, we must differentiate the constraint, which gives ˙ y = h′(x) ˙ x , ¨ y = h′(x) ¨ x + h′′(x) ˙ x2 . (3.387) Using the second Euler-Lagrange equation, we then obtain λ m = g + h′(x) ¨ x + h′′(x) ˙ x2 . (3.388) Finally, we substitute this into the ﬁrst E-L equation to obtain an equation for x alone: 1 + h′(x) 2 ¨ x + h′(x) h′′(x) ˙ x2 + g h′(x) = 0 . (3.389) Had we started by eliminating y = h(x) at the outset, writing L(x, ˙ x) = 1 2m 1 + h′(x) 2 ˙ x2 −mg h(x) , (3.390) 3.17. WORKED EXAMPLES 63 Figure 3.15: Frictionless motion under gravity along a curved surface. The skier ﬂies off the surface when the normal force vanishes. we would also have obtained this equation of motion. The skier ﬂies off the curve when the vertical force of constraint Qy = λ starts to become negative, because the curve can only supply a positive normal force. Suppose the skier starts from rest at a height y0. We may then determine the point x at which the skier detaches from the curve by setting λ(x) = 0. To do so, we must eliminate ˙ x and ¨ x in terms of x. For ¨ x, we may use the equation of motion to write ¨ x = − gh′ + h′ h′′ ˙ x2 1 + h′2 , (3.391) which allows us to write λ = m g + h′′ ˙ x2 1 + h′2 . (3.392) To eliminate ˙ x, we use conservation of energy, E = mgy0 = 1 2m 1 + h′2 ˙ x2 + mgh , (3.393) which ﬁxes ˙ x2 = 2g y0 −h 1 + h′2 . (3.394) Putting it all together, we have λ(x) = mg 1 + h′22 n 1 + h′2 + 2(y0 −h) h′′o . (3.395) The skier detaches from the curve when λ(x) = 0, i.e. when 1 + h′2 + 2(y0 −h) h′′ = 0 . (3.396) 64 CHAPTER 3. LAGRANGIAN MECHANICS Figure 3.16: Finding the local radius of curvature: z = η2/2R. There is a somewhat easier way of arriving at the same answer. This is to note that the skier must ﬂy off when the local centripetal force equals the gravitational force normal to the curve, i.e. m v2(x) R(x) = mg cos θ(x) , (3.397) where R(x) is the local radius of curvature. Now tan θ = h′, so cos θ = 1 + h′2−1/2. The square of the velocity is v2 = ˙ x2 + ˙ y2 = 1 + h′2 ˙ x2. What is the local radius of curvature R(x)? This can be determined from the following argument, and from the sketch in ﬁg. 3.16. Writing x = x∗+ ǫ, we have y = h(x∗) + h′(x∗) ǫ + 1 2h′′(x∗) ǫ2 + . . . . (3.398) We now drop a perpendicular segment of length z from the point (x, y) to the line which is tangent to the curve at x∗, h(x∗) . According to ﬁg. 3.16, this means ǫ y = η · 1 √ 1+h′2 1 h′ −z · 1 √ 1+h′2 −h′ 1 . (3.399) Thus, we have y = h′ ǫ + 1 2h′′ ǫ2 = h′ η + z h′ p 1 + h′2 + 1 2h′′ η + z h′ p 1 + h′2 2 = η h′ + z h′2 p 1 + h′2 + h′′ η2 2 1 + h′2 + O(ηz) = η h′ −z p 1 + h′2 , (3.400) 3.17. WORKED EXAMPLES 65 from which we obtain z = − h′′ η2 2 1 + h′23/2 + O(η3) (3.401) and therefore R(x) = − 1 h′′(x) · 1 + h′(x) 2 3/2 . (3.402) Thus, the detachment condition, mv2 R = −m h′′ ˙ x2 p 1 + h′2 = mg p 1 + h′2 = mg cos θ (3.403) reproduces the result from eqn. 3.392. 3.17.3 Disk rolling down an inclined plane A hoop of mass m and radius R rolls without slipping down an inclined plane. The inclined plane has opening angle α and mass M, and itself slides frictionlessly along a horizontal surface. Find the motion of the system. Figure 3.17: A hoop rolling down an inclined plane lying on a frictionless surface. Solution : Referring to the sketch in ﬁg. 3.17, the center of the hoop is located at x = X + s cos α −a sin α y = s sin α + a cos α , (3.404) where X is the location of the lower left corner of the wedge, and s is the distance along the wedge to the bottom of the hoop. If the hoop rotates through an angle θ, the no-slip condition is a ˙ θ+ ˙ s = 0. Thus, L = 1 2M ˙ X2 + 1 2m ˙ x2 + ˙ y2 + 1 2I ˙ θ2 −mgy = 1 2 m + I a2 ˙ s2 + 1 2(M + m) ˙ X2 + m cos α ˙ X ˙ s −mgs sin α −mga cos α . (3.405) 66 CHAPTER 3. LAGRANGIAN MECHANICS Since X is cyclic in L, the momentum PX = (M + m) ˙ X + m cos α ˙ s , (3.406) is preserved: ˙ PX = 0. The second equation of motion, corresponding to the generalized coordinate s, is 1 + I ma2 ¨ s + cos α ¨ X = −g sin α . (3.407) Using conservation of PX, we eliminate ¨ s in favor of ¨ X, and immediately obtain ¨ X = g sin α cos α 1 + M m 1 + I ma2 −cos2 α ≡aX . (3.408) The result ¨ s = − g 1 + M m sin α 1 + M m 1 + I ma2 −cos2 α ≡as (3.409) follows immediately. Thus, X(t) = X(0) + ˙ X(0) t + 1 2aXt2 s(t) = s(0) + ˙ s(0) t + 1 2ast2 . (3.410) Note that as < 0 while aX > 0, i.e. the hoop rolls down and to the left as the wedge slides to the right. Note that I = ma2 for a hoop; we’ve computed the answer here for general I. 3.17.4 Pendulum with nonrigid support A particle of mass m is suspended from a ﬂexible string of length ℓin a uniform gravitational ﬁeld. While hanging motionless in equilibrium, it is struck a horizontal blow resulting in an initial angular velocity ω0. Treating the system as one with two degrees of freedom and a constraint, answer the following: (a) Compute the Lagrangian, the equation of constraint, and the equations of motion. Solution : The Lagrangian is L = 1 2m ˙ r2 + r2 ˙ θ2 + mgr cos θ . (3.411) The constraint is r = ℓ. The equations of motion are m¨ r −mr ˙ θ2 −mg cos θ = λ mr2 ¨ θ + 2mr ˙ r ˙ θ −mg sin θ = 0 . (3.412) (b) Compute the tension in the string as a function of angle θ. 3.17. WORKED EXAMPLES 67 Solution : Energy is conserved, hence 1 2mℓ2 ˙ θ2 −mgℓcos θ = 1 2mℓ2 ˙ θ2 0 −mgℓcos θ0 . (3.413) We take θ0 = 0 and ˙ θ0 = ω0. Thus, ˙ θ2 = ω2 0 −2 Ω2 1 −cos θ , (3.414) with Ω= p g/ℓ. Substituting this into the equation for λ, we obtain λ = mg 2 −3 cos θ −ω2 0 Ω2 . (3.415) (c) Show that if ω2 0 < 2g/ℓthen the particle’s motion is conﬁned below the horizontal and that the tension in the string is always positive (deﬁned such that positive means exerting a pulling force and negative means exerting a pushing force). Note that the difference between a string and a rigid rod is that the string can only pull but the rod can pull or push. Thus, the string tension must always be positive or else the string goes “slack”. Solution : Since ˙ θ2 ≥0, we must have ω2 0 2Ω2 ≥1 −cos θ . (3.416) The condition for slackness is λ = 0, or ω2 0 2Ω2 = 1 −3 2 cos θ . (3.417) Thus, if ω2 0 < 2Ω2, we have 1 > ω2 0 2Ω2 > 1 −cos θ > 1 −3 2 cos θ , (3.418) and the string never goes slack. Note the last equality follows from cos θ > 0. The string rises to a maximum angle θmax = cos−1 1 −ω2 0 2Ω2 . (3.419) (d) Show that if 2g/ℓ< ω2 0 < 5g/ℓthe particle rises above the horizontal and the string becomes slack (the tension vanishes) at an angle θ∗. Compute θ∗. Solution : When ω2 > 2Ω2, the string rises above the horizontal and goes slack at an angle θ∗= cos−1 2 3 −ω2 0 3Ω2 . (3.420) This solution craps out when the string is still taut at θ = π, which means ω2 0 = 5Ω2. (e) Show that if ω2 0 > 5g/ℓthe tension is always positive and the particle executes circular motion. Solution : For ω2 0 > 5Ω2, the string never goes slack. Furthermore, ˙ θ never vanishes. Therefore, the pendulum undergoes circular motion, albeit not with constant angular velocity. 68 CHAPTER 3. LAGRANGIAN MECHANICS 3.17.5 Falling ladder A uniform ladder of length ℓand mass m has one end on a smooth horizontal ﬂoor and the other end against a smooth vertical wall. The ladder is initially at rest and makes an angle θ0 with respect to the horizontal. Figure 3.18: A ladder sliding down a wall and across a ﬂoor. (a) Make a convenient choice of generalized coordinates and ﬁnd the Lagrangian. Solution : I choose as generalized coordinates the Cartesian coordinates (x, y) of the ladder’s center of mass, and the angle θ it makes with respect to the ﬂoor. The Lagrangian is then L = 1 2 m ( ˙ x2 + ˙ y2) + 1 2 I ˙ θ2 + mgy . (3.421) There are two constraints: one enforcing contact along the wall, and the other enforcing contact along the ﬂoor. These are written G1(x, y, θ) = x −1 2 ℓcos θ = 0 G2(x, y, θ) = y −1 2 ℓsin θ = 0 . (3.422) (b) Prove that the ladder leaves the wall when its upper end has fallen to a height 2 3L sin θ0. The equations of motion are d dt ∂L ∂˙ qσ −∂L ∂qσ = X j λj ∂Gj ∂qσ . (3.423) Thus, we have m ¨ x = λ1 = Qx m ¨ y + mg = λ2 = Qy I ¨ θ = 1 2ℓ λ1 sin θ −λ2 cos θ = Qθ . (3.424) 3.17. WORKED EXAMPLES 69 We now implement the constraints to eliminate x and y in terms of θ. We have ˙ x = −1 2 ℓsin θ ˙ θ , ¨ x = −1 2 ℓcos θ ˙ θ2 −1 2 ℓsin θ ¨ θ ˙ y = 1 2 ℓcos θ ˙ θ , ¨ y = −1 2 ℓsin θ ˙ θ2 + 1 2 ℓcos θ ¨ θ . (3.425) We can now obtain the forces of constraint in terms of the function θ(t): λ1 = −1 2mℓ sin θ ¨ θ + cos θ ˙ θ2 λ2 = + 1 2mℓ cos θ ¨ θ −sin θ ˙ θ2 + mg . (3.426) We substitute these into the last equation of motion to obtain the result I ¨ θ = −I0 ¨ θ −1 2mgℓcos θ , (3.427) or (1 + α) ¨ θ = −2ω2 0 cos θ , (3.428) with I0 = 1 4mℓ2, α ≡I/I0 and ω0 = p g/ℓ. This may be integrated once (multiply by ˙ θ to convert to a total derivative) to yield 1 2(1 + α) ˙ θ2 + 2 ω2 0 sin θ = 2 ω2 0 sin θ0 , (3.429) which is of course a statement of energy conservation. This, ˙ θ2 = 4 ω2 0 (sin θ0 −sin θ) 1 + α ¨ θ = −2 ω2 0 cos θ 1 + α . (3.430) We may now obtain λ1(θ) and λ2(θ): λ1(θ) = −mg 1 + α 3 sin θ −2 sin θ0 cos θ λ2(θ) = mg 1 + α n (3 sin θ −2 sin θ0 sin θ + α o . (3.431) Demanding λ1(θ) = 0 gives the detachment angle θ = θd, where sin θd = 2 3 sin θ0 . (3.432) Note that λ2(θd) = mgα/(1 + α) > 0, so the normal force from the ﬂoor is always positive for θ > θd. The time to detachment is T1(θ0) = Z dθ ˙ θ = √1 + α 2 ω0 θ0 Z θd dθ √sin θ0 −sin θ . (3.433) 70 CHAPTER 3. LAGRANGIAN MECHANICS (c) Show that the subsequent motion can be reduced to quadratures (i.e. explicit integrals). Solution : After the detachment, there is no longer a constraint G1. The equations of motion are m ¨ x = 0 (conservation of x-momentum) m ¨ y + m g = λ I ¨ θ = −1 2 ℓλ cos θ , (3.434) along with the constraint y = 1 2 ℓsin θ. Eliminating y in favor of θ using the constraint, the second equation yields λ = mg −1 2mℓsin θ ˙ θ2 + 1 2mℓcos θ ¨ θ . (3.435) Plugging this into the third equation of motion, we ﬁnd I ¨ θ = −2 I0 ω2 0 cos θ + I0 sin θ cos θ ˙ θ2 −I0 cos2 θ ¨ θ . (3.436) Multiplying by ˙ θ one again obtains a total time derivative, which is equivalent to rediscovering energy conservation: E = 1 2 I + I0 cos2 θ ˙ θ2 + 2 I0 ω2 0 sin θ . (3.437) Figure 3.19: Plot of time to fall for the slipping ladder. Here x = sin θ0. By continuity with the ﬁrst phase of the motion, we obtain the initial conditions for this second 3.17. WORKED EXAMPLES 71 phase: θ = sin−1 2 3 sin θ0 ˙ θ = −2 ω0 s sin θ0 3 (1 + α) . Thus, E = 1 2 I + I0 −4 9 I0 sin2 θ0 · 4 ω2 0 sin θ0 3 (1 + α) + 1 3 mgℓsin θ0 = 2 I0 ω2 0 · 1 + 4 27 sin2 θ0 1 + α sin θ0 . (3.438) (d) Find an expression for the time T(θ0) it takes the ladder to smack against the ﬂoor. Note that, ex-pressed in units of the time scale p L/g, T is a dimensionless function of θ0. Numerically integrate this expression and plot T versus θ0. Solution : The time from detachment to smack is T2(θ0) = Z dθ ˙ θ = 1 2 ω0 θd Z 0 dθ s 1 + α cos2 θ 1 −4 27 sin2 θ0 1+α sin θ0 −sin θ . (3.439) The total time is then T(θ0) = T1(θ0) + T2(θ0). For a uniformly dense ladder, I = 1 12 mℓ2 = 1 3 I0, so α = 1 3. (e) What is the horizontal velocity of the ladder at long times? Solution : From the moment of detachment, and thereafter, ˙ x = −1 2 ℓsin θ ˙ θ = s 4 g ℓ 27 (1 + α) sin3/2θ0 . (3.440) (f) Describe in words the motion of the ladder subsequent to it slapping against the ﬂoor. Solution : Only a fraction of the ladder’s initial potential energy is converted into kinetic energy of horizontal motion. The rest is converted into kinetic energy of vertical motion and of rotation. The slapping of the ladder against the ﬂoor is an elastic collision. After the collision, the ladder must rise again, and continue to rise and fall ad inﬁnitum, as it slides along with constant horizontal velocity. 3.17.6 Point mass inside rolling hoop Consider the point mass m inside the hoop of radius R, depicted in ﬁg. 3.20. We choose as generalized coordinates the Cartesian coordinates (X, Y ) of the center of the hoop, the Cartesian coordinates (x, y) 72 CHAPTER 3. LAGRANGIAN MECHANICS for the point mass, the angle φ through which the hoop turns, and the angle θ which the point mass makes with respect to the vertical. These six coordinates are not all independent. Indeed, there are only two independent coordinates for this system, which can be taken to be θ and φ. Thus, there are four constraints: X −Rφ ≡G1 = 0 Y −R ≡G2 = 0 x −X −R sin θ ≡G3 = 0 y −Y + R cos θ ≡G4 = 0 . (3.441) Figure 3.20: A point mass m inside a hoop of mass M, radius R, and moment of inertia I. The kinetic and potential energies are easily expressed in terms of the Cartesian coordinates, aside from the energy of rotation of the hoop about its CM, which is expressed in terms of ˙ φ: T = 1 2M( ˙ X2 + ˙ Y 2) + 1 2m( ˙ x2 + ˙ y2) + 1 2I ˙ φ2 U = MgY + mgy . (3.442) The moment of inertia of the hoop about its CM is I = MR2, but we could imagine a situation in which I were different. For example, we could instead place the point mass inside a very short cylinder with two solid end caps, in which case I = 1 2MR2. The Lagrangian is then L = 1 2M( ˙ X2 + ˙ Y 2) + 1 2m( ˙ x2 + ˙ y2) + 1 2I ˙ φ2 −MgY −mgy . (3.443) Note that L as written is completely independent of θ and ˙ θ! Continuous symmetry Note that there is an continuous symmetry to L which is satisﬁed by all the constraints, under ˜ X(ζ) = X + ζ ˜ Y (ζ) = Y ˜ x(ζ) = x + ζ ˜ y(ζ) = y ˜ φ(ζ) = φ + ζ R ˜ θ(ζ) = θ . (3.444) 3.17. WORKED EXAMPLES 73 Thus, according to Noether’s theorem, there is a conserved quantity Λ = ∂L ∂˙ X + ∂L ∂˙ x + 1 R ∂L ∂˙ φ = M ˙ X + m ˙ x + I R ˙ φ . (3.445) This means ˙ Λ = 0. This reﬂects the overall conservation of momentum in the x-direction. Energy conservation Since neither L nor any of the constraints are explicitly time-dependent, the Hamiltonian is conserved. And since T is homogeneous of degree two in the generalized velocities, we have H = E = T + U: E = 1 2M( ˙ X2 + ˙ Y 2) + 1 2m( ˙ x2 + ˙ y2) + 1 2I ˙ φ2 + MgY + mgy . (3.446) Equations of motion We have n = 6 generalized coordinates and k = 4 constraints. Thus, there are four undetermined multipliers {λ1, λ2, λ3, λ4} used to impose the constraints. This makes for ten unknowns: X , Y , x , y , φ , θ , λ1 , λ2 , λ3 , λ4 . (3.447) Accordingly, we have ten equations: six equations of motion plus the four equations of constraint. The equations of motion are obtained from d dt ∂L ∂˙ qσ = ∂L ∂qσ + k X j=1 λj ∂Gj ∂qσ . (3.448) Taking each generalized coordinate in turn, the equations of motion are thus M ¨ X = λ1 −λ3 M ¨ Y = −Mg + λ2 −λ4 m¨ x = λ3 m¨ y = −mg + λ4 I ¨ φ = −R λ1 0 = −R cos θ λ3 −R sin θ λ4 . (3.449) 74 CHAPTER 3. LAGRANGIAN MECHANICS Along with the four constraint equations, these determine the motion of the system. Note that the last of the equations of motion, for the generalized coordinate qσ = θ, says that Qθ = 0, which means that the force of constraint on the point mass is radial. Were the point mass replaced by a rolling object, there would be an angular component to this constraint in order that there be no slippage. Implementation of constraints We now use the constraint equations to eliminate X, Y , x, and y in terms of θ and φ: X = Rφ , Y = R , x = Rφ + R sin θ , y = R(1 −cos θ) . (3.450) We also need the derivatives: ˙ x = R ˙ φ + R cos θ ˙ θ , ¨ x = R ¨ φ + R cos θ ¨ θ −R sin θ ˙ θ2 , (3.451) and ˙ y = R sin θ ˙ θ , ¨ y = R sin θ ¨ θ + R cos θ ˙ θ2 , (3.452) as well as ˙ X = R ˙ φ , ¨ X = R ¨ φ , ˙ Y = 0 , ¨ Y = 0 . (3.453) We now may write the conserved charge as Λ = 1 R(I + MR2 + mR2) ˙ φ + mR cos θ ˙ θ . (3.454) This, in turn, allows us to eliminate ˙ φ in terms of ˙ θ and the constant Λ: ˙ φ = γ 1 + γ Λ mR −˙ θ cos θ , (3.455) where γ = mR2 I + MR2 . (3.456) The energy is then E = 1 2(I + MR2) ˙ φ2 + 1 2m R2 ˙ φ2 + R2 ˙ θ2 + 2R2 cos θ ˙ φ ˙ θ + MgR + mgR(1 −cos θ) = 1 2mR2 (1 + γ sin2θ 1 + γ ˙ θ2 + 2g R (1 −cos θ) + γ 1 + γ Λ mR 2 + 2Mg mR ) . (3.457) The last two terms inside the big bracket are constant, so we can write this as 1 + γ sin2θ 1 + γ ˙ θ2 + 2g R (1 −cos θ) = 4gk R . (3.458) Here, k is a dimensionless measure of the energy of the system, after subtracting the aforementioned constants. If k > 1, then ˙ θ2 > 0 for all θ, which would result in ‘loop-the-loop’ motion of the point mass inside the hoop – provided, that is, the normal force of the hoop doesn’t vanish and the point mass doesn’t detach from the hoop’s surface. 3.17. WORKED EXAMPLES 75 Equation motion for θ(t) The equation of motion for θ obtained by eliminating all other variables from the original set of ten equations is the same as ˙ E = 0, and may be written 1 + γ sin2θ 1 + γ ¨ θ + γ sin θ cos θ 1 + γ ˙ θ2 = −g R . (3.459) We can use this to write ¨ θ in terms of ˙ θ2, and, after invoking eqn. 3.458, in terms of θ itself. We ﬁnd ˙ θ2 = 4g R · 1 + γ 1 + γ sin2θ k −sin2 1 2θ ¨ θ = −g R · (1 + γ) sin θ 1 + γ sin2θ 2 h 4γ k −sin2 1 2θ cos θ + 1 + γ sin2θ i . (3.460) Forces of constraint We can solve for the λj, and thus obtain the forces of constraint λ3 = m¨ x = mR ¨ φ + mR cos θ ¨ θ −mR sin θ ˙ θ2 = mR 1 + γ h ¨ θ cos θ −˙ θ2 sin θ i (3.461) and λ4 = m¨ y + mg = mg + mR sin θ ¨ θ + mR cos θ ˙ θ2 = mR h ¨ θ sin θ + ˙ θ2 sin θ + g R i (3.462) and λ1 = −I R ¨ φ = (1 + γ)I mR2 λ3 λ2 = (M + m)g + m¨ y = λ4 + Mg . (3.463) One can check that λ3 cos θ + λ4 sin θ = 0. The condition that the normal force of the hoop on the point mass vanish is λ3 = 0, which entails λ4 = 0. This gives −(1 + γ sin2θ) cos θ = 4(1 + γ) k −sin2 1 2θ . (3.464) Note that this requires cos θ < 0, i.e. the point of detachment lies above the horizontal diameter of the hoop. Clearly if k is sufﬁciently large, the equality cannot be satisﬁed, and the point mass executes a periodic ‘loop-the-loop’ motion. In particular, setting θ = π, we ﬁnd that kc = 1 + 1 4(1 + γ) . (3.465) 76 CHAPTER 3. LAGRANGIAN MECHANICS If k > kc, then there is periodic ‘loop-the-loop’ motion. If k < kc, then the point mass may detach at a critical angle θ∗, but only if the motion allows for cos θ < 0. From the energy conservation equation, we have that the maximum value of θ achieved occurs when ˙ θ = 0, which means cos θmax = 1 −2k . (3.466) If 1 2 < k < kc, then, we have the possibility of detachment. This means the energy must be large enough but not too large. 3.18 Appendix : Legendre Transformations A convex function of a single variable f(x) is one for which f ′′(x) > 0 everywhere. The Legendre transform of a convex function f(x) is a function g(p) deﬁned as follows. Let p be a real number, and consider the line y = px, as shown in ﬁg. 3.21. We deﬁne the point x(p) as the value of x for which the difference F(x, p) = px −f(x) is greatest. Then deﬁne g(p) = F x(p), p .8 The value x(p) is unique if f(x) is convex, since x(p) is determined by the equation f ′x(p) = p . (3.467) Note that from p = f ′x(p) we have, according to the chain rule, d dp f ′x(p) = f ′′x(p) x′(p) = ⇒ x′(p) = h f ′′x(p) i−1 . (3.468) From this, we can prove that g(p) is itself convex: g′(p) = d dp h p x(p) −f x(p) i = p x′(p) + x(p) −f ′x(p) x′(p) = x(p) , (3.469) hence g′′(p) = x′(p) = h f ′′x(p) i−1 > 0 . (3.470) In higher dimensions, the generalization of the deﬁnition f ′′(x) > 0 is that a function F(x1, . . . , xn) is convex if the matrix of second derivatives, called the Hessian, Hij(x) = ∂2F ∂xi ∂xj (3.471) is positive deﬁnite. That is, all the eigenvalues of Hij(x) must be positive for every x. We then deﬁne the Legendre transform G(p) as G(p) = p · x −F(x) (3.472) where p = ∇F . (3.473) 8Note that g(p) may be a negative number, if the line y = px lies everywhere below f(x). 3.18. APPENDIX : LEGENDRE TRANSFORMATIONS 77 Figure 3.21: Construction for the Legendre transformation of a function f(x). Note that dG = x · dp + p · dx −∇F · dx = x · dp , (3.474) which establishes that G is a function of p and that ∂G ∂pj = xj . (3.475) Note also that the Legendre transformation is self dual, which is to say that the Legendre transform of G(p) is F(x): F →G →F under successive Legendre transformations. We can also deﬁne a partial Legendre transformation as follows. Consider a function of q variables F(x, y), where x = {x1, . . . , xm} and y = {y1, . . . , yn}, with q = m + n. Deﬁne p = {p1, . . . , pm}, and G(p, y) = p · x −F(x, y) , (3.476) where pa = ∂F ∂xa (a = 1, . . . , m) . (3.477) These equations are then to be inverted to yield xa = xa(p, y) = ∂G ∂pa . (3.478) Note that pa = ∂F ∂xa x(p, y), y . (3.479) Thus, from the chain rule, δab = ∂pa ∂pb = ∂2F ∂xa ∂xc ∂xc ∂pb = ∂2F ∂xa ∂xc ∂2G ∂pc ∂pb , (3.480) 78 CHAPTER 3. LAGRANGIAN MECHANICS which says ∂2G ∂pa ∂pb = ∂xa ∂pb = K−1 ab , (3.481) where the m × m partial Hessian is ∂2F ∂xa ∂xb = ∂pa ∂xb = Kab . (3.482) Note that Kab = Kba is symmetric. And with respect to the y coordinates, ∂2G ∂yµ ∂yν = − ∂2F ∂yµ ∂yν = −Lµν , (3.483) where Lµν = ∂2F ∂yµ ∂yν (3.484) is the partial Hessian in the y coordinates. Now it is easy to see that if the full q × q Hessian matrix Hij is positive deﬁnite, then any submatrix such as Kab or Lµν must also be positive deﬁnite. In this case, the partial Legendre transform is convex in {p1, . . . , pm} and concave in {y1, . . . , yn}.
14740	https://blog.wolframalpha.com/2011/05/17/plotting-functions-and-graphs-in-wolframalpha/	Plotting Functions and Graphs in Wolfram\|Alpha—Wolfram\|Alpha Blog HOMEABOUTPRODUCTSBUSINESSRESOURCES The Wolfram\|Alpha Blog is now part of the Wolfram Blog. Join us there for the latest on Wolfram\|Alpha and other Wolfram offerings » Plotting Functions and Graphs in Wolfram\|Alpha May 17, 2011 — Sam Blake Comments Off Plotting functions in the Cartesian plane is such a simple task with Wolfram\|Alpha: just enter the function you are looking to graph, and within seconds you will have a beautiful result. If you are feeling daring, enter a multivariate function, and the result will be a 3D Cartesian graph. Wolfram\|Alpha is certainly not limited to Cartesian plotting; we have the functionality to make number lines, 2D and 3D polar plots, 2D and 3D parametric plots, 2D and 3D contour plots, implicit plots, log plots, log-linear plots, matrix plots, surface of revolution plots, region plots, list plots, pie charts, histograms, and more. Furthermore, in Wolfram\|Alpha we can generate specialized plots for illustrating asymptotes, cusps, maxima, minima, inflection points, saddle points, solutions of ordinary differential equations, poles, eigenvalues, series expansions, definite integrals, 2D inequalities, interpolating polynomials, least-squares best fits, and more. Let’s take a look at the plotting functionality in Wolfram\|Alpha, some of which is newly improved! We will start simple with 2D Cartesian plots. Here we plot sin(√7 x)+19cos(x) for x between -20 and 20. If we change √7 to √-7, then we get a plot of the real and imaginary parts. In both these examples we have given Wolfram\|Alpha a horizontal plot range. What happens if we don’t give Wolfram\|Alpha a range? We still get back a plot with all of its defining features. One of the unique features of Wolfram\|Alpha is the functionality to automatically guess an appropriate plot range for univariate and bivariate functions. Here is another example: So far we have told Wolfram\|Alpha that we’re specifically requesting a plot. If we simply enter a univariate expression without the prefix “plot”, then we’ll always get a Cartesian plot in addition to a number of other pieces of information. Try: Versus: One important difference is that the image sizes are larger if you specifically ask for a plot. You can also plot more than one function at a time. The underlying Mathematica function used in all the examples is Plot. By clicking the bottom left of the images and then “Copyable plaintext”, you can see the Mathematica code used to generate the plots. For example: The code can then be evaluated in Mathematica. Now let’s give Wolfram\|Alpha a challenge and plot bivariate functions. Start by plotting y^2 cos(x) for x between -6 and 6 and y between -2 and 2. As in the univariate case, Wolfram\|Alpha has the capability to find an appropriate plot range for a bivariate function, the code for which is under continuous development. If Wolfram\|Alpha fails to generate a plot, then it’s most likely because the code that determines a plot range has been unable to find a region where the function has interesting behavior. In such cases you can always manually enter a plot range like we did in the example above. Here are a couple of examples: plot sin (x cos (y)) plot (x^5 – 4 x^4 y^2 + x y – 1)/(y^11 – x^11 + 34 x^3 y + 1) What happens if you want to plot more than one bivariate function? Wolfram\|Alpha will return an individual plot of each function in the list. Here are a couple more examples to test out: plot (1 – x)/(2 x + 7 y), 5 x^2 – 3 y^2 + 7 x y, (x + 2 y)^4 plot sqrt (1 + x y), sqrt (x^2 – y^2 + 2 x y) A new feature in Wolfram\|Alpha is the functionality to plot the real and imaginary parts of complex-valued bivariate functions. Here are a couple of examples: plot sin (x + I y) plot sqrt (y^2 + 4 y) – sqrt (-I x^3 + 3 x) In all of these examples Wolfram\|Alpha returned a contour plot in addition to the 3D plot. A nice way to see the connection between the 3D and contour plots is to click the “Show contour lines” button. Note that the 3D and contour plots will always use the same plot range. All of the 3D plots were made using Mathematica‘s function Plot3D; the contour plots were made using ContourPlot. In both cases the Mathematica code for generating the images can be found by clicking the plots. You have now had the opportunity to view the plotting abilities of Wolfram\|Alpha in 2D and 3D planes. Still not convinced? Try to plot your favorite function in Wolfram\|Alpha, and be sure to share your results with us! 29 Comments amazing… Posted by thay May 17, 2011 at 3:06 pm I think you should put all Mathematica on WolframAlpha 😀 It is amazing to have all those features in such a free website. thanks wolfram. Posted by Holmez May 17, 2011 at 3:47 pm A boon to dress designers! Posted by Brian Gilbert May 18, 2011 at 3:46 am Sam, congrats … excellent job!! … The whole thing would be more than completed with some comments on plotting functions given implicitly. Cheers. Posted by German Titiv May 25, 2011 at 12:42 am Thanks! 2D implicit plotting is pretty straight forward. Here’s an example Sam Blake Wolfram Research Posted by Sam Blake May 26, 2011 at 5:51 am What about logarithmic scaling? Posted by mrego May 25, 2011 at 10:18 am Just give Alpha a hint that you would like a log scale. Here’s an example, Sam Blake Wolfram Research Posted by Sam Blake May 26, 2011 at 5:53 am See what an amazing plot I have got Posted by Harsh Gupta May 27, 2011 at 2:36 pm Can you explain your inequality plots? For example I do not understand Posted by Colleen Young May 29, 2011 at 6:09 am The shaded area evaluates to true for that function – looks pretty standard to me. Did you mean “how does it choose what range to display when you haven’t given it context” ? I guess the same magic that it uses for the other graphs. Posted by Ethan June 7, 2011 at 6:36 pm Hi, I was wondering if it is possible to plot complex graphs in Wolfram Alpha…ie arg(z-3i)-arg(z+4)=?/2 ? If so, how? If not, it’d be nice if you added the feature! 😀 Posted by RUGRLN June 17, 2011 at 12:01 pm Thanks for the reply. Am I missing something here? looks like x+y< 7.5 to me. I would expect x+y < 5 to look like this: ie to go through (5,0) and (0,5) Posted by Colleen Young July 5, 2011 at 11:44 am I wonder if it is possible to plot parametric surface here? I tried but did not succeed. I wanted a sphere with this input: parameric plot3D (cos(theta)sin(phi), sin(theta)sin(phi),cos(phi)) But no sphere is produced. Thanks for your help. Posted by haowire September 5, 2011 at 9:49 pm Can you find specific points on a graph? like at X=0.25 Y=? Posted by Volkan Ceylan September 8, 2011 at 3:09 am @ Volkan – Take a look at this example: Is this what you are looking for? Posted by The Wolfram\|Alpha Team September 8, 2011 at 10:57 am How’s this? Posted by Alex November 18, 2011 at 9:39 am Hi! How can I ask alpha to keep the range of y e.g y=0..1, however divide this interval to the finer grids. I mean instead of dividing y to 0, 0.02, 0.04, 0.06, 0.08, 1; divide it to 0, 0.002, 0.004,…,1 Posted by Maryam November 19, 2011 at 7:57 am Hi! Is there a way to force alpha plot the full vertical range? Similar to PlotRange->Full in real Mathematica? Thanks!! Posted by \yaroslav January 15, 2012 at 8:42 am plot ? sin(x cos(y))^2 Posted by Danuke March 1, 2012 at 9:03 pm In the 3D plot is it possible to plot the real function only i.e. for the domain, which guarantees that f(x, y) is real. The “real part” of the plot (2x-y)^(1/4) is for the whole R^2, I’d like it to be for 2x > y. Posted by Eve April 12, 2012 at 11:37 am How to change ratio of height and width of a plot in wolframalpha? AspectRatio function doesnt work. Posted by Scyler September 11, 2012 at 2:18 pm I was wondering how you got the axes labels to show up? Whenever I attempt to plot on Wolfram\|Alpha they do not show up.. Posted by Stephanie February 20, 2014 at 7:01 pm Is there a way to plot a two variable function and a 3 variable function on a single graph (ex: a line and a plane)? Posted by josh March 13, 2014 at 10:04 pm how can i plot some points on the XOY plane using the computer? secondly, how can i draw a pie chart using the computer? Posted by stansfield July 4, 2014 at 2:03 pm Why isn’t it possible for Wolfram Alpha to plot points or planes in 3D space? Posted by Spencer September 14, 2014 at 8:39 pm Hello, thank you for comment. Actually, it is possible for Wolfram\|Alpha to perform both of those functions. Please see these two examples: Plane through points Plot points Posted by The W\|A Team September 18, 2014 at 11:31 am actually i got the output for 3 variables( v[i,j] , x[i] , y[i,j] ) ,how can i draw surface plot using this output. Posted by varunkumar m September 15, 2014 at 11:32 pm actually i got the output for 3 variables( v[i,j] , x[i] , y[i,j] ) ,how can i draw surface plot using this output. and here i dont have expression for v[i,j] in terms of x[i] and y[i,j]. Posted by varunkumar m September 15, 2014 at 11:34 pm how to plot a time varying function in 3D???? ex f(x,y,t) Posted by Ankit November 7, 2014 at 3:52 am CATEGORIES Best of Blog Astronomy Chemistry Colors Computational Sciences Culture & Media Dates & Times Earth Sciences Education Food & Nutrition Future Perspectives Health & Medicine Life Sciences Materials Mathematics Money & Finance Music News News Organizations People & History Physics Places & Geography Socioeconomic Data Sports & Games Statistics & Data Analysis Stephen Wolfram Technological World Transportation Weather Web & Computer Systems Wolfram Cloud Wolfram Language Wolfram\|Alpha Apps Wolfram\|Alpha Widgets BY YEAR 2015 \| 2014 \| 2013 \| 2012 \| 2011 \| 2010 \| 2009 For May 2015 or later, see the Wolfram Blog » About \| Pro \| Products \| Mobile Apps \| Business Solutions \| For Developers \| Resources & Tools Blog \| Community \| Participate \| Contact \| Follow: © 2025 Wolfram Alpha LLC—A Wolfram Research Company \| Terms \| Privacy
14741	https://mathematica.stackexchange.com/questions/86805/how-to-solve-for-the-intersection-points-of-two-ellipses	graphics - How to solve for the intersection points of two ellipses? - Mathematica Stack Exchange Join Mathematica By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematica helpchat Mathematica Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to solve for the intersection points of two ellipses? Ask Question Asked 10 years, 3 months ago Modified9 years, 2 months ago Viewed 921 times This question shows research effort; it is useful and clear 11 Save this question. Show activity on this post. I have a parametric equation of two ellipses as follow {x=a 1 sin θ+b 1 cos θ+c 1 y=d 1 sin θ+e 1 cos θ+f 1 {x=a 2 sin θ+b 2 cos θ+c 2 y=d 2 sin θ+e 2 cos θ+f 2 Now I want to solve the intersection point coordinates and visualize them. My trial Here, mat1 owns the style {{a1,b1,c1},{d1,e1,f1}}. ```mathematica solvePoints0[mat1_, mat2_] := Module[{sol, θValue, θ, ptRule}, sol = Solve[ Thread[ mat1.{Sin[θ1], Cos[θ1], 1} == mat2.{Sin[θ2], Cos[θ2], 1}], {θ1, θ2}]; θValue = Mod[(# & /@ sol) /. C -> 1., 2 Pi]; ptRule = List /@ Thread[θ -> θValue]; Cases[ mat1.{Sin[θ], Cos[θ], 1} /. ptRule,{_Real, _Real}] ] showEllipse0[mat1_, mat2_, opts : OptionsPattern[]] := With[{pts = solvePoints0[mat1, mat2]}, ParametricPlot[ {mat1.{Sin[θ], Cos[θ], 1}, mat2.{Sin[θ], Cos[θ], 1}}, {θ, 0, 2 Pi}, Epilog -> {PointSize[Large], Red, Point[pts]}, Evaluate[ Sequence @@ FilterRules[{opts}, Options[ParametricPlot]]] ] ] /; MatrixQ[mat1, NumericQ] && MatrixQ[mat2, NumericQ] ``` Test mathematica showEllipse0[##, ImageSize -> 200] & @@@ {{{{1, 2, 1}, {2, 3, 4}}, {{2, 2, 2}, {3, 2, 4}}}, {{{3, 1, 1}, {2, 3, 4}}, {{4, 2, 2}, {3, 4, 4}}}, {{{3, 2, 1}, {2, 3, 4}}, {{2, 2, 2}, {3, 2, 4}}}, {{{2, 1, 1}, {2, 3, 4}}, {{4, 1, 2}, {3, 2, 4}}}} However, solvePoints0 has a bug mathematica solvePoints0[{{4, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}] solvePoints0[{{5, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}] ```mathematica {} {} ``` In fact, they have intersection point coordinates. mathematica showEllipse0 @@@ {{{{4, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}}, {{{5, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}}} To fixed this bug, I rewrite the function solvePoints0 using NSolve mathematica solvePoints[mat1_, mat2_] := Module[ {sol, θValue, θ, ptRule}, sol = Quiet@ NSolve[ Thread[ mat1.{Sin[θ1], Cos[θ1], 1.} == mat2.{Sin[θ2], Cos[θ2], 1.}], {θ1, θ2}] /. C -> 1.; θValue = # & /@ sol; ptRule = List /@ Thread[θ -> θValue]; Cases[ mat1.{Sin[θ], Cos[θ], 1} /. ptRule, {_Real, _Real}] ] /; MatrixQ[mat1, NumericQ] && MatrixQ[mat2, NumericQ] Now, mathematica solvePoints[{{4, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}] solvePoints[{{5, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}] Question I think my methods is fussy and bad, so I would like to know how to solve this question in a elegant way. Update Thanks for J.M's suggestions, with the help of GraphicsMeshFindIntersections mathematica newSolution[mat1_, mat2_] := Module[{graph, pts}, graph = ParametricPlot[ {mat1.{Sin[θ], Cos[θ], 1}, mat2.{Sin[θ], Cos[θ], 1}}, {θ, 0, 2 Pi},Epilog -> {Point[{.1, .2}]}]; pts = Graphics`Mesh`FindIntersections[First[graph]]; graph /. (Epilog -> _) -> Epilog -> {Red, PointSize[Medium], Point[pts]} ] Test mathematica newSolution[##] & @@@ {{{{1, 2, 1}, {2, 3, 4}}, {{2, 2, 2}, {3, 2, 4}}}, {{{3, 1, 1}, {2, 3, 4}}, {{4, 2, 2}, {3, 4, 4}}}, {{{3, 2, 1}, {2, 3, 4}}, {{2, 2, 2}, {3, 2, 4}}}, {{{2, 1, 1}, {2, 3, 4}}, {{4, 1, 2}, {3, 2, 4}}}} mathematica newSolution @@@ {{{{4, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}}, {{{5, 2, 1}, {2, 5, 6}}, {{6, 2, 2}, {3, 2, 4}}}} Obviously, this is not a good method owning to that it cannot find the intersection points correctly. graphics equation-solving geometry intersection Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Jul 19, 2016 at 14:04 J. M.'s missing motivation 126k 11 11 gold badges 411 411 silver badges 590 590 bronze badges asked Jun 25, 2015 at 7:45 xyzxyz 685 4 4 gold badges 40 40 silver badges 118 118 bronze badges 6 6 A sketch: use GroebnerBasis[] to produce the implicit Cartesian equations of the two ellipses, and feed those equations to Solve[]. Retain only the real solutions, and you're done.J. M.'s missing motivation –J. M.'s missing motivation 2015-06-25 08:03:59 +00:00 Commented Jun 25, 2015 at 8:03 Alternatively, if you do not need high accuracy, see this answer.J. M.'s missing motivation –J. M.'s missing motivation 2015-06-25 08:06:49 +00:00 Commented Jun 25, 2015 at 8:06 1 Your code cannot figure out what is real due to phantom imaginary parts arising in numerical approximations involving arctans and the like.Daniel Lichtblau –Daniel Lichtblau 2015-06-25 15:27:57 +00:00 Commented Jun 25, 2015 at 15:27 @J. M. Thanks for your link. However, the FindIntersections cannot find all the intersection points correctly(Namely,it has wrong points or lacks of some right points). Please see my update.xyz –xyz 2015-06-26 01:56:29 +00:00 Commented Jun 26, 2015 at 1:56 3 Shutao, note that the wrong points are precisely the points where the ellipse closes up, so you know where they are (θ=0), and you can remove them from the results (with some tolerance since the results are not exactly the same).J. M.'s missing motivation –J. M.'s missing motivation 2015-06-26 02:16:34 +00:00 Commented Jun 26, 2015 at 2:16 \|Show 1 more comment 1 Answer 1 Sorted by: Reset to default This answer is useful 15 Save this answer. Show activity on this post. Alright, I managed to borrow a computer. Here's an implementation of my suggestion: mathematica ellipseIntersections[mat1_?MatrixQ, mat2_?MatrixQ] /; Dimensions[mat1] == Dimensions[mat2] == {2, 3} := {\[FormalX], \[FormalY]} /. RootReduce[Solve[Flatten[Map[ GroebnerBasis[Append[Thread[{\[FormalX], \[FormalY]} == #], \[FormalC]^2 + \[FormalS]^2 == 1], {\[FormalX], \[FormalY]}, {\[FormalC], \[FormalS]}] &, {mat1, mat2}.{\[FormalC], \[FormalS], 1}]] == 0, {\[FormalX], \[FormalY]}, Reals]] where, apart from the use of GroebnerBasis[], I use formal symbols for safety, and add validity checks for the arguments. I use the combination RootReduce[Solve[( stuff )]] to generate exact coordinates; replace this with NSolve[] if you only want numerical approximations. Before demonstrating this function, allow me to show a different way to render your ellipses. Here, I have chosen to use the NURBS (that is, BSplineCurve[]) representation of a circle, along with a suitable affine transformation. Thus: mathematica makeEllipse[mat_?MatrixQ] /; Dimensions[mat] == {2, 3} := Module[{aff}, aff = AffineTransform[{Drop[mat, None, -1], mat}]; BSplineCurve[aff /@ {{1, 0}, {1, 1}, {-1, 1}, {-1, 0}, {-1, -1}, {1, -1}, {1, 0}}, SplineClosed -> True, SplineDegree -> 2, SplineKnots -> {0, 0, 0, 1/4, 1/2, 1/2, 3/4, 1, 1, 1}, SplineWeights -> {1, 1/2, 1/2, 1, 1/2, 1/2, 1}]] Now, pictures! mathematica e1 = {{5, 2, 1}, {2, 5, 6}}; e2 = {{6, 2, 2}, {3, 2, 4}}; Graphics[{{RGBColor[7/19, 37/73, 22/31], makeEllipse[e1]}, {RGBColor[59/67, 11/18, 1/7], makeEllipse[e2]}, {Red, AbsolutePointSize, Point[ellipseIntersections[e1, e2]]}}, Frame -> True] Here's the case e1 = {{1, 2, 1}, {2, 3, 4}}; e2 = {{2, 2, 2}, {3, 2, 4}};: which shows that ellipseIntersections can deal with tangencies. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Jun 26, 2015 at 9:39 J. M.'s missing motivationJ. M.'s missing motivation 126k 11 11 gold badges 411 411 silver badges 590 590 bronze badges 0 Add a comment\| 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. Start asking to get answers Find the answer to your question by asking. 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14742	https://www.teacherspayteachers.com/browse/free?search=carry%20over%20addition%20worksheet	Carry Over Addition Worksheet "Crazy Math" - Double Digit Addition (With Carrying Over) Worksheet (2 pages) FREE - TWO DIGIT ADDITION WORKSHEETS 3-Addend Vertical Math Practice Problems Get more with resources under $5 Two Digit Addition with Regrouping (Carrying Over) Worksheet Adding Two-Digit Numbers with Carrying Over Two-digit Addition Carrying Over Worksheet \| Accessible Dyslexic Elementary Math St Patricks Day Addition With Regrouping Worksheet, Adding Carry Over Activities FREE - TWO DIGIT ADDITION WORKSHEETS 4-Addend Vertical Math Practice Problems FREE 4-Digit Addition With Regrouping \| 5 Worksheets - 250 Problems + Answer Safari 2 Digit Addition with & without Regrouping Worksheets and Activities FREE 3-Digit Addition With Regrouping \| 5 Worksheets × 50 Problems + Answer Keys 2-Digit Addition No Regrouping Worksheets : Free 5 pages Addition With and Without Regrouping Worksheets \| FREE 2-Digit Addition with Regrouping Worksheets for 1st and 2nd Grade - FREE Double Digit Addition Worksheets: With and Without Regrouping Addition With Carrying Over - Add Your Own Student's Names - FREEBIE - 7 pages Math Worksheet 2 Digit Addition Addition Of Decimals -Interactive Worksheet FREE - 4 DIGIT ADDITION WORKSHEETS 3-Addend Number Problems Mix Carrying Over Carrying Over/ Borrowing STEPS (Addition/ Subtraction) FREE ADDITION 3 Digit 4-Addend Mix Carrying Over Math Numbers Practice Problems Spider Web 3 Digit Addition without Carry Over "Simple Math" - Double Digit Addition worksheet (2 pages) FREE - ADDITION 3 Digit 3-Addend Mix Carrying Over Math Add Numbers Problems FREE - 4 DIGIT ADDITION WORKSHEETS 4-Addend Numbers Problems Mix Carrying Over Digit Addition 1-99 Worksheets - Free Spider Web 3 Digit Addition with Carry Over Addition Practice - Carry over Cut & Paste Double Digit Addition - No Carryover
14743	https://pubmed.ncbi.nlm.nih.gov/6263950/	Selectivity of dobutamine for adrenergic receptor subtypes: in vitro analysis by radioligand binding - 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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Selectivity of dobutamine for adrenergic receptor subtypes: in vitro analysis by radioligand binding R S Williams,T Bishop PMID: 6263950 PMCID: PMC370747 DOI: 10.1172/jci110208 Item in Clipboard Selectivity of dobutamine for adrenergic receptor subtypes: in vitro analysis by radioligand binding R S Williams et al. J Clin Invest.1981 Jun. Show details Display options Display options Format J Clin Invest Actions Search in PubMed Search in NLM Catalog Add to Search . 1981 Jun;67(6):1703-11. doi: 10.1172/jci110208. Authors R S Williams,T Bishop PMID: 6263950 PMCID: PMC370747 DOI: 10.1172/jci110208 Item in Clipboard Full text links Cite Display options Display options Format Abstract The cardiovascular responses elicited by dobutamine are distinctly different from those produced by other adrenergic or dopaminergic agonists. To test the hypothesis that dobutamine could have differential affinities for adrenergic receptor subtypes, and that such subtype selectivity could be related to its relatively unique pharmacologic properties, we assessed the ability of dobutamine to displace adrenergic radioligands from membrane receptors in a number of tissues of previously characterized adrenergic receptor subtype. For beta adrenergic receptors identified by (-) [(3)H]dihydroalprenolol (DHA), dobutamine had significantly greater affinity for the beta(1) subtype (K(D) = 2.5 muM in rat heart and 2.6 muM in turkey erythrocyte) than for the beta(2) subtype (K(D) = 14.8 muM in frog heart and 25.4 muM in rat lung) (P < 0.001). For alpha adrenergic receptors, dobutamine had markedly greater affinity for the alpha(1)-subtype identified by [(3)H]prazosin (K(D) = 0.09 muM in rat heart and 0.14 muM in rabbit uterus) than for the alpha(2)-subtype identified by [(3)H]dihydroergocryptine (DHE) (K(D) = 9.3 muM in human platelet) or by [(3)H]yohimbine (K(D) = 5.7 muM in rabbit uterus) (P < 0.001). Like other beta(1)-agonists, in the absence of guanine nucleotide, dobutamine competition curves for DHA binding in rat heart demonstrated two classes of binding sites, with one site of significantly higher affinity (K(D) = 0.5 muM, P = 0.008) than the single class of binding sites (K(D) = 5.2 muM) identified in the presence of guanine nucleotide. However, unlike beta(2)- or alpha(2)-agonists, dobutamine displacement of DHA binding in rat lung or of DHE binding in human platelets demonstrated only a single class of binding sites, and guanine nucleotide had only minimal effects. We conclude that dobutamine is selective for beta(1) as opposed to beta(2), and for alpha(1) as opposed to alpha(2) adrenergic receptors. Furthermore, guanine nucleotide effects on dobutamine binding, and biochemical response data in vitro suggest that dobutamine is a beta(1)-agonist, but has little intrinsic activity at beta(2) and alpha(2)-receptors. This selectivity for adrenergic receptor subtypes may be part of the basis for dobutamine's distinctive pharmacologic properties in vivo. PubMed Disclaimer Similar articles [3H]Dihydroergocryptine binding to alpha-adrenergic receptors of human platelets. A reassessment using the selective radioligands [3H]prazosin, [3H]yohimbine, and [3H]rauwolscine.Motulsky HJ, Insel PA.Motulsky HJ, et al.Biochem Pharmacol. 1982 Aug 15;31(16):2591-7. doi: 10.1016/0006-2952(82)90705-5.Biochem Pharmacol. 1982.PMID: 6291538 Effects of celiprolol (REV 5320), a new cardioselective beta-adrenoceptor antagonist, on in vitro adenylate cyclase, alpha- and beta-adrenergic receptor binding and lipolysis.Van Inwegen RG, Khandwala A, Weinryb I, Pruss TP, Neiss E, Sutherland CA.Van Inwegen RG, et al.Arch Int Pharmacodyn Ther. 1984 Nov;272(1):40-55.Arch Int Pharmacodyn Ther. 1984.PMID: 6151380 Adrenergic receptors in human liver plasma membranes: predominance of beta 2- and alpha 1-receptor subtypes.Kawai Y, Powell A, Arinze IJ.Kawai Y, et al.J Clin Endocrinol Metab. 1986 May;62(5):827-32. doi: 10.1210/jcem-62-5-827.J Clin Endocrinol Metab. 1986.PMID: 3007555 Catecholamine binding to CNS adrenergic receptors.U'Prichard DC, Snyder SH.U'Prichard DC, et al.J Supramol Struct. 1978;9(2):189-206. doi: 10.1002/jss.400090205.J Supramol Struct. 1978.PMID: 219293 Review. Heterogeneity of alpha-2 adrenergic receptors.Bylund DB.Bylund DB.Pharmacol Biochem Behav. 1985 May;22(5):835-43. doi: 10.1016/0091-3057(85)90536-2.Pharmacol Biochem Behav. 1985.PMID: 2989948 Review. See all similar articles Cited by Treatments targeting inotropy.Maack C, Eschenhagen T, Hamdani N, Heinzel FR, Lyon AR, Manstein DJ, Metzger J, Papp Z, Tocchetti CG, Yilmaz MB, Anker SD, Balligand JL, Bauersachs J, Brutsaert D, Carrier L, Chlopicki S, Cleland JG, de Boer RA, Dietl A, Fischmeister R, Harjola VP, Heymans S, Hilfiker-Kleiner D, Holzmeister J, de Keulenaer G, Limongelli G, Linke WA, Lund LH, Masip J, Metra M, Mueller C, Pieske B, Ponikowski P, Ristić A, Ruschitzka F, Seferović PM, Skouri H, Zimmermann WH, Mebazaa A.Maack C, et al.Eur Heart J. 2019 Nov 21;40(44):3626-3644. doi: 10.1093/eurheartj/ehy600.Eur Heart J. 2019.PMID: 30295807 Free PMC article. Evidence for the existence of postsynaptic beta-1 and beta-2 adrenoceptors in isolated simian facial veins.Chiba S, Tsukada M.Chiba S, et al.Heart Vessels. 1991;6(3):168-74. doi: 10.1007/BF02058282.Heart Vessels. 1991.PMID: 1680847 Sympathetic β1-adrenergic signaling contributes to regulation of human bone metabolism.Khosla S, Drake MT, Volkman TL, Thicke BS, Achenbach SJ, Atkinson EJ, Joyner MJ, Rosen CJ, Monroe DG, Farr JN.Khosla S, et al.J Clin Invest. 2018 Nov 1;128(11):4832-4842. doi: 10.1172/JCI122151. Epub 2018 Oct 2.J Clin Invest. 2018.PMID: 30153111 Free PMC article.Clinical Trial. Disruption of aminergic signalling reveals novel compounds with distinct inhibitory effects on mosquito reproduction, locomotor function and survival.Fuchs S, Rende E, Crisanti A, Nolan T.Fuchs S, et al.Sci Rep. 2014 Jul 2;4:5526. doi: 10.1038/srep05526.Sci Rep. 2014.PMID: 24984706 Free PMC article. The impact of intravenous dobutamine on spirometry with bronchodilator test.Romantowski J, Janowiak P, Wabich E, Kuziemski K.Romantowski J, et al.Respir Med Case Rep. 2020 Oct 15;31:101264. doi: 10.1016/j.rmcr.2020.101264. eCollection 2020.Respir Med Case Rep. 2020.PMID: 33101902 Free PMC article.No abstract available. See all "Cited by" articles References Life Sci. 1977 Sep 1;21(5):595-606 - PubMed Cardiovasc Res. 1977 Jan;11(1):17-25 - PubMed Naunyn Schmiedebergs Arch Pharmacol. 1977 Nov;300(2):101-5 - PubMed J Clin Invest. 1978 Feb;61(2):395-402 - PubMed J Biol Chem. 1978 Jul 25;253(14):5090-102 - PubMed Show all 40 references MeSH terms Animals Actions Search in PubMed Search in MeSH Add to Search Anura Actions Search in PubMed Search in MeSH Add to Search Binding, Competitive / drug effects Actions Search in PubMed Search in MeSH Add to Search Blood Platelets / metabolism Actions Search in PubMed Search in MeSH Add to Search Catecholamines / metabolism Actions Search in PubMed Search in MeSH Add to Search Dobutamine / metabolism Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Guanine Nucleotides / pharmacology Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search In Vitro Techniques Actions Search in PubMed Search in MeSH Add to Search Myocardium / metabolism Actions Search in PubMed Search in MeSH Add to Search Rabbits Actions Search in PubMed Search in MeSH Add to Search Radioligand Assay Actions Search in PubMed Search in MeSH Add to Search Rats Actions Search in PubMed Search in MeSH Add to Search Receptors, Adrenergic / metabolism Actions Search in PubMed Search in MeSH Add to Search Receptors, Adrenergic, beta / metabolism Actions Search in PubMed Search in MeSH Add to Search Species Specificity Actions Search in PubMed Search in MeSH Add to Search Turkeys Actions Search in PubMed Search in MeSH Add to Search Uterus / metabolism Actions Search in PubMed Search in MeSH Add to Search Substances Catecholamines Actions Search in PubMed Search in MeSH Add to Search Guanine Nucleotides Actions Search in PubMed Search in MeSH Add to Search Receptors, Adrenergic Actions Search in PubMed Search in MeSH Add to Search Receptors, Adrenergic, beta Actions Search in PubMed Search in MeSH Add to Search Dobutamine Actions Search in PubMed Search in MeSH Add to Search Related information PubChem Compound (MeSH Keyword) LinkOut - more resources Full Text Sources PubMed Central Full text links[x] Free PMC article [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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14744	https://www.socs.uoguelph.ca/~sawada/papers/pnf.pdf	On Preﬁx Normal Words and Preﬁx Normal Forms Péter Burcsia, Gabriele Ficib, Zsuzsanna Liptákc,∗, Frank Ruskeyd, Joe Sawadae aDept. of Computer Algebra, Eötvös Loránd Univ., Budapest, Hungary bDip. di Matematica e Informatica, University of Palermo, Italy cDip. di Informatica, University of Verona, Italy dDept. of Computer Science, University of Victoria, Canada eSchool of Computer Science, University of Guelph, Canada Abstract A 1-preﬁx normal word is a binary word with the property that no factor has more 1s than the preﬁx of the same length; a 0-preﬁx normal word is deﬁned analogously. These words arise in the context of indexed binary jumbled pattern matching, where the aim is to decide whether a word has a factor with a given number of 1s and 0s (a given Parikh vector). Each binary word has an associated set of Parikh vectors of the factors of the word. Using preﬁx normal words, we provide a characterization of the equivalence class of binary words having the same set of Parikh vectors of their factors. We prove that the language of preﬁx normal words is not context-free and is strictly contained in the language of pre-necklaces, which are preﬁxes of powers of Lyndon words. We give enumeration results on pnw(n), the number of preﬁx normal words of length n, showing that, for suﬃciently large n, 2n−4√n lg n ≤pnw(n) ≤2n−lg n+1. For ﬁxed density (number of 1s), we show that the ordinary generating function of the number of preﬁx normal words of length n and density d is a rational function. Finally, we give experimental results on pnw(n), discuss further properties, and state open problems. Keywords: preﬁx normal words, preﬁx normal forms, binary languages, ∗Corresponding author Email addresses: bupe@compalg.inf.elte.hu (Péter Burcsi), gabriele.fici@unipa.it (Gabriele Fici), zsuzsanna.liptak@univr.it (Zsuzsanna Lipták), ruskey@cs.uvic.ca (Frank Ruskey), jsawada@uoguelph.ca (Joe Sawada) Preliminary versions of parts of this article have appeared in , and . Preprint submitted to Elsevier Saturday 15th October, 2016 1 INTRODUCTION 2 binary jumbled pattern matching, pre-necklaces, Lyndon words, enumeration. 1. Introduction A binary word is called 1-preﬁx normal if no factor (substring) has more 1s than the preﬁx of the same length. For example, 11010 is 1-preﬁx normal, but 10110 is not. Similarly, a binary word is called 0-preﬁx normal if no factor has more 0s than the preﬁx of the same length. When not further speciﬁed, by preﬁx normal we mean 1-preﬁx normal. In , we gave an algorithm for generating all preﬁx normal words of ﬁxed length n. As we will see later, to each binary word, a 1-preﬁx normal word and a 0-preﬁx normal word can be associated in a unique way, which we will call its preﬁx normal forms. The Parikh vector of a binary word u is the pair (x, y), where x is the number of 1s in u, and y is the number of 0s in u. The set of Parikh vectors of factors of a word w is called the Parikh set of w. For binary words, the problem of deciding whether a particular pair (x, y) lies in the Parikh set of a word w is known as Binary Jumbled Pattern Matching (BJPM). There has been much interest recently in the indexed version of this problem (IBJPM), where an index for the Parikh set is created in a preprocessing step, which can then be used to answer queries fast. The Parikh set can be represented in linear space due to the following interval property of binary strings: If w has k-length substrings with x1 resp. x2 occurrences of 1, where x1 < x2, then it also has a k-length substring with y occurrences of 1, for every x1 ≤y ≤x2. Thus the Parikh set can be represented by storing, for every 1 ≤k ≤\|w\|, the minimum and maximum number of 1s in a substring of length k. Much recent research has focused on how to compute these numbers eﬃciently [14, 29, 30, 16, 2, 23, 22]. The problem has also been extended to graphs and trees [22, 15], to the streaming model , and to approximate indexes . There is also interest in the non-binary variant [20, 17, 11, 14, 7, 8, 26], as well as in reconstruction from the Parikh multi-set of a string . Applications in computational biology include SNP discovery, alignment, gene clusters, pattern discovery, and mass spectrometry data interpretation [4, 3, 5, 19, 33]. The current best construction algorithm for the linear size index for IB-JPM runs in O(n1.864) time , for a word of length n. As we will see later, computing the preﬁx normal forms of a word w is equivalent to creating an index for the Parikh set of w. Currently, we know no faster computation 2 BASICS 3 algorithms for the preﬁx normal forms than already exist for the linear-size index. However, should better algorithms be discovered, these would imme-diately carry over to the problem of IBJPM. It is worthwhile noting that some relevant sequences have made it into the On-Line Encyclopedia of Integer Sequences (OEIS ): A194850 is the number of preﬁx normal words of length n, A238109 is a list of preﬁx normal words (over the alphabet {1, 2}), and A238110 is the maximum size of a class of binary words of length n having the same preﬁx normal form. The paper is organized as follows: Section 2 contains basic deﬁnitions and results about preﬁx normal words; in particular that there are unique 0-preﬁx normal and 1-preﬁx normal words associated with every word, and thus the set of words can be partitioned according to this association. In Section 3 we consider the set of preﬁx normal words, giving several properties and characterizations and showing that their language is not context free. One of these properties is then used in Section 4, which is concerned with counting the number of preﬁx normal words of a given length. Finally, the paper concludes with some open problems in Section 5. 2. Basics A binary word (or string) w = w1 · · · wn over Σ = {0, 1} is a ﬁnite sequence of elements wi ∈Σ, for i = 1, . . . , n. Its length n is denoted by \|w\|. We denote by Σn the set of words over Σ of length n, by Σ∗= ∪n≥0Σn the set of ﬁnite words over Σ, and the empty word by ε. Let w ∈Σ∗. If w = uv for some u, v ∈Σ∗, we say that u is a preﬁx of w and v is a suﬃx of w. A factor or substring of w is a preﬁx of a suﬃx of w. We denote the set of factors of w by Fact(w). Let w = w1 · · · wn ∈Σ∗, then the word ˜ w = wnwn−1 · · · w1 is called the reversal of w. A word w s.t. w = ˜ w is called a palindrome. A binary language is any subset L of Σ∗. We denote by \|w\|1 the number of 1s in the word w; similarly, \|w\|0 is the number of 0s in w. The Parikh vector of a word w over Σ is deﬁned as p(w) = (\|w\|0, \|w\|1). The Parikh set of w is Π(w) = {p(v) \| v ∈Fact(w)}, the set of Parikh vectors of the factors of w. For example p(011) = p(101) = (1, 2) and Π(011) = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} = Π(101) ∪{(0, 2)}. Given a binary word w, we denote by P1(w, i) the number of 1s in the preﬁx of length i and by pos1(w, i) the position of the ith 1 in the word w, i.e. P1(w, i) = \|w1 · · · wi\|1 and pos1(w, i) = min{k : \|w1 · · · wk\|1 = i}. The functions P0 and pos0 are deﬁned similarly. Note that in the context of 2 BASICS 4 k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 F1 0 1 2 3 3 4 4 4 5 6 6 7 7 7 8 8 9 10 10 10 11 11 12 F0 0 1 2 3 3 3 4 4 5 5 6 6 7 7 7 8 8 9 9 10 10 10 10 Table 1: The sequences F1 and F0 for the word w = 1010011011000111001011. succint indexing, these functions are frequently called rank and select, cf. : We have, for x = 0, 1, Px(w, i) = rankx(w, i) and posx(w, i) = selectx(w, i). 2.1. Preﬁx normal words Deﬁnition 1 (Maximum-ones and maximum-zeros functions). Let w ∈ Σ∗. We deﬁne, for each 0 ≤k ≤\|w\|: F1(w, k) = max{\|v\|1 \| v ∈Fact(w) ∩Σk}, the maximum number of 1s in a factor of w of length k. When no confusion can arise, we also write F1(k) for F1(w, k). The function F0(w, k) is deﬁned analogously by taking 0 in place of 1. For a word w, we denote by F1(w) the function k 7→F1(w, k) (and similarly with other functions taking arguments w and k). Example 1. Take w = 1010011011000111001011. In Table 1, we give the values of F1 and F0 for w. Deﬁnition 2 (Preﬁx normal words). A word w ∈{0, 1}∗is called 1-preﬁx normal if P1(w) = F1(w). It is called 0-preﬁx normal if P0(w) = F0(w). In other words, w is 1-preﬁx normal (0-preﬁx normal) if and only if it does not have any factors with more 1s (more 0s) than the preﬁx of the same length. When not speciﬁed, then by preﬁx normal we mean 1-preﬁx normal. Example 2. The word w = 1100110 is 1-preﬁx normal, but the word w1 = 11001101 is not 1-preﬁx normal because the factor 1101 has three 1s, while the preﬁx of length 4 has only two. Also, w is not 0-preﬁx normal since every 0-preﬁx normal word, except those of the form 1∗, must start with a 0. We will soon see that it is possible to ﬁnd, for every word w, a 1-preﬁx normal word which has the same maximum-ones function F1 as w; and analogously for 0. These will be called the preﬁx normal forms of w. To this end, we deﬁne the following equivalence; we will then see that equivalent words have the same preﬁx normal form. 2 BASICS 5 Deﬁnition 3 (Preﬁx equivalence). Two words v, w ∈Σ∗are called 1-preﬁx equivalent if F1(v) = F1(w). They are called 0-preﬁx equivalent if F0(v) = F0(w). Example 3. The words 11010, 10110, 01101, 01011 are all 1-preﬁx equiva-lent, but not 0-preﬁx equivalent. When considering 0, we have that {01011, 11010, 10101} constitute one equivalence class, and {01101, 10110} another one (note that in the ﬁrst class, there is an additional word not present in the 1-preﬁx equivalence class). Next we will show that every equivalence class contains exactly one preﬁx normal word (Theorem 2), which can thus be used as its representative. This will allow us to associate two preﬁx normal words to every word w (Deﬁnition 4). First we need the following lemma. Lemma 1. Let w ∈Σ∗. Then, for all 0 ≤i ≤j ≤\|w\|: F1(j) −F1(i) ≤ F1(j −i). Proof. Observe that if v = yz, then \|v\|1 ≤F1(\|y\|) + F1(\|z\|). Thus if v is a length j word such that \|v\|1 = F1(j) and \|y\| = i, then F1(j) ≤ F1(i) + F1(j −i). 2 Theorem 2. For every w ∈Σ∗there is a unique 1-preﬁx normal word w′ such that F1(w′) = F1(w); similarly, there is a unique 0-preﬁx normal word w′′ such that F0(w′′) = F0(w). Proof. We only give the proof for w′. The construction of w′′ is analogous. First note that if the 1-preﬁx normal words u and v are 1-preﬁx equiv-alent, then necessarily u = v. This holds because the preﬁx function P1 determines the word, i.e. P1(u) = P1(v) implies u = v for any u, v. But since u and v are 1-preﬁx normal words, their preﬁx and maximum-ones functions coincide, and since they are 1-preﬁx equivalent, we have P1(u) = F1(u) = F1(v) = P1(v). This proves uniqueness. Next, we will construct w′, given w. It is easy to see that for 1 ≤k ≤\|w\|, one has either F1(w, k) = F1(w, k −1) or F1(w, k) = 1 + F1(w, k −1). Now deﬁne the word w′ by w′ k = ( 1 if F1(w, k) = 1 + F1(w, k −1) 0 if F1(w, k) = F1(w, k −1) for every 1 ≤k ≤\|w\|. 2 BASICS 6 By construction, we have P1(w′, k) = F1(w, k) for every 1 ≤k ≤\|w\|. We still need to show that P1(w′, k) = F1(w′, k) for all k. This will prove that w′ is 1-preﬁx normal, as well as that it is 1-preﬁx equivalent to w. By deﬁnition, P1(w′, k) ≤F1(w′, k) for all k. Now let v ∈Fact(w′), \|v\| = k, and v = wi+1 · · · wj. Then \|v\|1 = P1(w′, j) −P1(w′, i) = F1(w, j) − F1(w, i) ≤F1(w, j−i) = P1(w′, j−i) = P1(w′, k), where the inequality holds by Lemma 1. We have thus proved that F1(w′, k) ≤P1(w′, k), and hence w′ is 1-preﬁx normal. 2 2.2. Normal forms and Parikh sets Deﬁnition 4 ((Preﬁx) normal forms). Let w ∈Σ∗. Then we denote by PNF1(w) the unique 1-preﬁx normal word which is 1-preﬁx equivalent to w, and by PNF0(w) the unique 0-preﬁx normal word which is 0-preﬁx equivalent to w. We refer to PNF1(w) and PNF0(w) as the preﬁx normal form w.r.t. 1 (resp. w.r.t. 0) or just normal form w.r.t. 1 (resp. w.r.t. 0) of w. Example 4. Let w = 1010011011000111001011. The normal forms of w are the words PNF1(w) = 1110100110100101100101, PNF0(w) = 0001101010101101010111. Refer to Example 1 for the values of the two functions F1(w) and F0(w). The operators PNF1 and PNF0 are idempotent operators; i.e., if u = PNFx(w) then PNFx(u) = u, for x = 0, 1. This gives us an equivalent deﬁnition of preﬁx normality: a word w is x-preﬁx normal if PNFx(w) = w. Also, for any w ∈Σ∗and x ∈Σ, it holds that PNFx(w) = PNFx( ˜ w). Note further that if the equivalence class of w contains only one element, then w is necessarily preﬁx normal and a palindrome. In Table 2 we list all eight 1-preﬁx equivalence classes for words of length 4. The normal forms of a word allow us to determine the Parikh vectors of the factors of the word, as we will show in Theorem 4. We ﬁrst recall the following lemma from (which also appears to be folklore). We say that a Parikh vector q occurs in a word w if w has a factor v with p(v) = q. Lemma 3 (Interval Lemma ). Let w ∈Σ∗. Fix 1 ≤k ≤\|w\|. If the Parikh vectors (x1, k −x1) and (x2, k −x2) both occur in w, then so does (y, k −y) for any x1 ≤y ≤x2. 2 BASICS 7 PNF1 class cardinality 1111 {1111} 1 1110 {1110, 0111} 2 1101 {1101, 1011} 2 1100 {1100, 0110, 0011} 3 1010 {1010, 0101} 2 1001 {1001} 1 1000 {1000, 0100, 0010, 0001} 4 0000 {0000} 1 Table 2: The sets of 1-preﬁx equivalent words of length 4. The lemma can be proved with a simple sliding window argument, ex-ploiting the fact that when a ﬁxed size window is shifted by one, then the number of 1s in the window changes by at most one. Theorem 4. Let w, w′ be words over Σ. Then Π(w) = Π(w′) if and only if PNF1(w) = PNF1(w′) and PNF0(w) = PNF0(w′). Proof. Let f1(w, k) denote the minimum number of 1s in a factor of w of length k. As a direct consequence of Lemma 3, we have that for a Parikh vec-tor q = (x, y), q ∈Π(w) if and only if f1(w, x + y) ≤x ≤F1(w, x + y). Thus for two words w, w′, we have Π(w) = Π(w′) if and only if F1(w) = F1(w′) and f1(w) = f1(w′). It is easy to see that for all k, f1(w, k) = k −F0(w, k), thus the last statement is equivalent to F1(w) = F1(w′) and F0(w) = F0(w′). This holds if and only if PNF1(w) = PNF1(w′) and PNF0(w) = PNF0(w′), and the claim is proved. 2 Deﬁne I(w) = {(P0(w, k), P1(w, k)) \| 0 ≤k ≤\|w\|}, the set of Parikh vectors of all preﬁxes of w. The following lemma is immediate. Lemma 5. For all w ∈Σ∗, Π(w) = n [ i=1 I(wi · · · wn). There is an interesting geometrical way to view Lemma 5 which we de-scribe now. Imagine each Parikh pair as the coordinates of a point in the 2 BASICS 8 Euclidean plane that has been rotated clockwise π/4 radians. Each word w can be interpreted as a polygonal path in this plane going up and to the right for each 1 (↗) or down and to the right for each 0 (↘), for each successive bit of w. To obtain Π(w) imagine grabbing the polygonal path for w and pulling it one step at a time through the origin, keeping track of the integer lattice points that are hit after each pull (and ignoring the stuﬀto the left of the origin). The normal forms PNF1(w) and PNF0(w) are obtained by forming polygonal paths starting at the origin, and connecting the uppermost and the lowermost points of the region, respectively. w PNF1(w) PNF0(w) 1s 0s 1 1 2 2 3 3 Figure 1: The word w = 1010011011000111001011 (dark line), its normal forms PNF1(w) = 1110100110100101100101 and PNF0(w) = 0001101010101101010111 (lighter lines); the region between the two is the Parikh set of w; e.g. w has a substring containing 5 ones and 6 zeros (black dot). Note that the axes giving the number of 0s and 1s are rotated by 45 degrees clockwise. 2.3. Indexing for binary jumbled pattern matching Theorem 4 is relevant for the problem known as Indexed Binary Jumbled Pattern Matching, which has attracted much interest recently. Recall that a Parikh vector over {0, 1} is a multiplicity vector of a string, i.e. it has non-negative integer entries. Indexed Binary Jumbled Pattern Matching (IBJPM) Given a string w of length n over {0, 1}, create an index which allows fast answers to queries of the following form: Input: a Parikh vector q, Output: return yes if q occurs in Π(w), and no otherwise. For 1 ≤k ≤n, let f1(w, k) be the minimum number of 1s in a factor of length k, and F1(w, k), as before, the maximum number of 1s in a factor 2 BASICS 9 k 0 1 2 3 4 5 6 7 F1(w, k) 0 1 2 2 3 3 3 4 f1(w, k) 0 0 0 1 2 2 3 4 Table 3: The maximum and minimum number of 1s for the the word w = 1001101. k 0 1 2 3 4 5 6 7 F1(w, k) 0 1 2 2 3 3 3 4 F0(w, k) 0 1 2 2 2 3 3 3 Table 4: The maximum number of 1s and 0s for the the word w = 1001101. The normal forms of w are PNF1(w) = 1101001 and PNF0(w) = 0011011. of length k. It follows from Lemma 3 that the answer for query q = (x, y) is yes if and only if F1(w, x + y) ≥x ≥f1(w, x + y). Therefore, it suﬃces to store, for every 1 ≤k ≤n, the two numbers F1(w, k) and f1(w, k), and queries can be answered in constant time. The size of this data structure is O(n). All current solutions for IBJPM are based on this observation. The crux is how to construct this linear size data structure. The construction time of the index has steadily decreased since its ﬁrst introduction: from O(n2) to O(n2/ log n) [6, 29], to O(n2/ log2 n) in the word RAM-model , to n2/2Ω(log n/ log log n)1/2 . The fastest solution at present is due to Chan and Lewenstein and has running time O(n1.859) . Normal forms are in eﬀect an encoding of this linear size index. We have already seen that the F-function can be viewed as a binary string, namely PNF1(w). We have observed in the proof of Theorem 4 how the function f1(w) is determined by F0(w) and thus also by PNF0(w), thus we have shown the following lemma. Lemma 6. The answer for an IBJPM query q = (x, y) is yes if and only if P1(PNF1(w), x + y) ≥x ≥P1(PNF0(w), x + y). Note that P1 can be computed in constant time with constant time rank-queries on bit vectors, using only o(n) bits of extra space [31, 18]. Example 5. Let w = 1001101. Then the linear size data structure is given in the Table 3, and the F1 and F0 functions in Table 4. At present, no faster computation of the normal forms is known than the algorithms cited above for the IBJPM problem. But the connection shown 3 THE LANGUAGE OF PREFIX NORMAL WORDS 10 here implies that, should a fast normal form computation be found, it would immediately translate into a new solution for IBJPM. 3. The language of preﬁx normal words In this section, we take a closer look at preﬁx normal words. We give several equivalent characterizations of preﬁx normality, explore some prop-erties of preﬁx normal words, and then look at the language of preﬁx normal words. We denote by LPN1 ⊂Σ∗the language of 1-preﬁx normal words, and by LPN0 ⊂Σ∗the language of 0-preﬁx normal words. Note that these are exactly complemented, i.e. replacing every 1 by a 0 and vice versa, in each word of LPN1, yields LPN0. Therefore, every result about LPN1 has an equivalent formulation for LPN0, as well. Recall that whenever not fur-ther speciﬁed, we refer to 1-preﬁx normality. In Section 3.2 only, we will talk about 0-preﬁx normal words, and we will show that LPN0 is strictly contained in the language of pre-necklaces, when adopting the usual order 0 < 1 on the alphabet. 3.1. General observations about preﬁx normal words We start with several characterizations of preﬁx normal words. Proposition 7. Let w ∈Σ∗. The following properties are equivalent: 1. w is a preﬁx normal word; 2. ∀i, j where 0 ≤i ≤j ≤\|w\|, we have P1(j) −P1(i) ≤P1(j −i); 3. ∀v ∈Fact(w) such that \|v\|1 = i, we have \|v\| ≥pos1(i); 4. ∀i, j such that i + j −1 ≤\|w\|1, we have pos1(i) + pos1(j) −1 ≤ pos1(i + j −1). Proof. (1) ⇒(2). Follows from Lemma 1, since P1(w) = F1(w). (2) ⇒(3). Assume otherwise. Then there exists v ∈Fact(w) s.t. \|v\| < pos1(k), where k = \|v\|1. Let v = wi+1 · · · wj, thus j −i = k. Then P1(j) − P1(i) = k. But P1(j −i) = P1(\|v\|) ≤k −1 < k = P1(j) −P1(i), a contradiction. (3) ⇒(4). Again assume that the claim does not hold. Then there are i, j s.t. pos1(i + j −1) < pos1(i) + pos1(j) −1. Let k = pos1(j) and l = pos1(i + j −1) and deﬁne v = wk · · · wl. Then v has i many 1s. But \|v\| = pos1(i+j−1)−pos1(j)+1 < pos1(i)+pos1(j)−1−pos1(j)+1 = pos1(i), in contradiction to (3). 3 THE LANGUAGE OF PREFIX NORMAL WORDS 11 (4) ⇒(1). Let v ∈Fact(w), \|v\|1 = i. We have to show that P1(\|v\|) ≥i. This is equivalent to showing that pos1(i) ≤\|v\|. Let v = wl+1 · · · wr, thus P1(r) −P1(l) = i. Let j = P1(l) + 1, thus the ﬁrst 1 in v is the j’th 1 of w. Note that we have l < pos1(j) and r ≥pos1(i + j −1). By the assumption, we have pos1(i) ≤pos1(i + j −1) −pos1(j) + 1 ≤r −l = \|v\|. 2 Next we formulate a characterization of the preﬁx normal property that will be useful in the enumeration of ﬁxed-length preﬁx normal words (Sec-tion 4). Lemma 8. Let w ∈1Σ∗. For some sequence of positive integers r1, r2, . . ., rd−1 we can write w = 10r1−110r2−1 · · · 10rd−1. The word w is preﬁx normal if and only if the following inequalities hold. r1 ≤ rj j = 2, 3, . . . , d −1 r1 + r2 ≤ rj + rj+1 j = 2, 3, . . . , d −2 . . . . . . r1 + r2 + · · · + rd−2 ≤ rj + rj+1 + · · · + rd−1 j = 2 Proof. Note that for k = 1, 2, . . . d −1, we have pos1(k) = 1 + Pk−1 j=1 rj. The statement of the lemma then follows by property (4) of Proposition 7. 2 We now give some simple facts about the language LPN1. Proposition 9. Let LPN1 be the language of preﬁx normal words. 1. LPN1 is preﬁx-closed, that is, any preﬁx of a word in LPN1 is a word in LPN1. 2. If w ∈LPN1, then any word of the form 1kw or w0k, k ≥0, also belongs to LPN1. 3. Let \|w\|1 < 3. Then w ∈LPN1 iﬀeither w = 0n for some n ≥0 or the ﬁrst letter of w is 1. 4. Let w ∈Σ∗. Then there exist inﬁnitely many v ∈Σ∗such that vw ∈ LPN1. Proof. The claims 1., 2., 3. follow easily from the deﬁnition. For 4., note that for any n ≥\|w\|, the word 1nw belongs to LPN1. 2 We now deal with the question of how a preﬁx normal word can be extended to the right into another preﬁx normal word. 3 THE LANGUAGE OF PREFIX NORMAL WORDS 12 Lemma 10. Let w ∈LPN1. Then w1 ∈LPN1 if and only if for every 0 ≤k < \|w\| the suﬃx of w of length k has less 1s than the preﬁx of w of length k + 1. Proof. Note that for all 1 ≤k ≤\|w\|, P1(w1, k) = P1(w, k). Now if w1 ∈ LPN1, then for the k-length suﬃx u of w: \|u\|1 < \|u1\|1 ≤P1(w1, k + 1) = P1(w, k + 1). Conversely, let u be a factor of w1. If u is a factor of w, then \|u\|1 ≤P1(w, \|u\|) = P1(w1, \|u\|). Else u = u′1, with u′ a suﬃx of w, and \|u\|1 = \|u′\|1 + 1 < P1(w, \|u′\| + 1) + 1 = P1(w1, \|u\|) + 1 = P1(w1, \|u\|) + 1, and thus \|u\|1 ≤P1(w1, \|u\|). Therefore, w1 ∈LPN1. 2 We close this section by proving that LPN1 is not context-free. Theorem 11. LPN1 is not context-free. Proof. Recall that the intersection of a CFL with a regular language is a CFL. We will show that L′ = LPN1 ∩1∗01∗01∗is not a CFL by using the pumping lemma. Let n be the constant of the pumping lemma and let z = 1n01n01n ∈L′. Let z = uvwxy be the usual factorization of the pumping lemma, where we may assume that \|vx\| ≥1, \|vwx\| ≤n, and for all i ≥0 we have uviwxiy ∈L′. Clearly vx can not contain 0s. If vx contains some 1s from the ﬁrst block of 1s in z, then taking i = 0 give a contradiction since the third block of 1s is too long. If vx contains no 1s from the ﬁrst block of 1s then taking i = 2 makes the second or third block of 1s too long. 2 3.2. Connection with Lyndon words and pre-necklaces In this section we explore the relationship between the language LPN0 of preﬁx normal words w.r.t. 0 and some known classes of words deﬁned by means of lexicographic properties. Note that in this section, when referring to preﬁx normality, we mean with respect to 0. We assume the usual order 0 < 1 on the alphabet. A Lyndon word is a word which is lexicographically strictly smaller than any of its proper non-empty suﬃxes. Equivalently, w is a Lyndon word if it is the strictly smallest, in the lexicographic order, among its conjugates, i.e., for any factorization w = uv, with u, v non-empty words, one has that the word vu is lexicographically greater than w . A word w is a power if it can be obtained by concatenating two or more copies of another word, i.e. if there exists a non-empty v and a k > 1 such that w = vk. A word that is not a power is called primitive. Note that, by deﬁnition, a Lyndon word 3 THE LANGUAGE OF PREFIX NORMAL WORDS 13 is primitive. Let us denote by Lyn the set of Lyndon words over Σ. One has that Lyn ̸⊆LPN0 and LPN0 ̸⊆Lyn. For example, the word w = 0101 belongs to LPN0 but is not a Lyndon word since it is not primitive. An example of a Lyndon word which is not in normal form is w = 00110100111. A necklace is a Lyndon word or a power of a Lyndon word. A pre-necklace is a preﬁx of a necklace (also called preprime word , or sesquipower or fractional power of a Lyndon word ). Let us denote by PL the language of pre-necklaces. The next proposition shows that every preﬁx normal word diﬀerent from a power of the letter 1 is a preﬁx of a Lyndon word. Proposition 12. Let w ∈LPN0 with \|w\|0 > 0. Then the word w1\|w\| is a Lyndon word. Proof. We have to prove that every rotation of w′ = w1\|w\| is strictly greater than w′. If the rotation starts at a position within the second half of w′, then this is clearly true, since then its ﬁrst character is 1, while w′ starts with a 0, w being a preﬁx normal word containing at least one 0. So let v be a suﬃx of w′ of length at least \|w\| + 1, and let u be the longest common preﬁx of v and w′. If u = v, then v is a border (both a preﬁx and suﬃx) of w′, of length more than half its length, and thus w′ has a period of length i = \|w′\| −\|v\| < \|w\|, i.e., every character is the same as the one which follows i positions later. Since the second half of w′ consists of 1s only, this implies that so does the ﬁrst half, contrary to our assumption. So v is not a preﬁx of w′, and therefore u is followed by two diﬀerent characters in v and in w′. Let us write v = v′1\|w\|. If \|u\| ≥\|v′\|, then u1 is a preﬁx of v, implying that u0 is a preﬁx of w′, and thus w′ is smaller than v. If \|u\| < \|v′\|, assume that u0 is a preﬁx of v and u1 of w′. Then w has a substring (u0) which has more 0s than the preﬁx of the same length (u1), a contradiction to w being preﬁx normal. Therefore, again we have that w′ is smaller than v. 2 We can now state the following result: Theorem 13. Every preﬁx normal word is a pre-necklace. Proof. If w is of the form 1n, n ≥1, then w is a power of the Lyndon word 1, hence it is a pre-necklace. Otherwise, w contains at least one 0, thus by by Proposition 12, it is the preﬁx of a Lyndon word. 2 The languages LPN0 and PL, however, do not coincide. A shortest word in PL that does not belong to LPN0 is w = 00110100. Below we give the table of the number of words in LPN0 of each length n ≤16, compared with that 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 14 of pre-necklaces. Both sequences are listed in the On-Line Encyclopedia of Integer Sequences (sequences A062692 and A194850), where the reader can ﬁnd further terms. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 LPN0 ∩Σn 2 3 5 8 14 23 41 70 125 218 395 697 1273 2279 4185 7568 PL ∩Σn 2 3 5 8 14 23 41 71 127 226 412 747 1377 2538 4720 8800 Table 5: The number of words in LPN0 and in PL for each length up to 16. 4. Enumeration results about preﬁx normal words Let pnw(n) denote the number of preﬁx normal words of length n. It is an easy consequence both of Lemma 8 and of Proposition 9 that pnw(n) grows exponentially. To see this, note that the conditions of Lemma 8 are always satisﬁed if r1 ≤r2 ≤. . . ≤rk, and thus the number of partitions of n is a lower bound for pnw(n). On the other hand, Proposition 9 states that for all w, 1\|w\|w is preﬁx normal, so pnw(2n) ≥2n. In Table 5, we give pnw(n) for n up to 16, the sequence for n up to 50 can be found in the On-Line Encyclopedia of Integer Sequences , sequence A194850. In Fig. 2 we show the growth ratio for small values of n. Two interesting phenomena can be observed: the values seem to approach 2 slowly, i.e., the number of preﬁx normal words almost doubles as we increase the length by 1. Second, the values show on oscillation pattern between even and odd values. Figure 2: The value of pnw(n)/pnw(n −1) for preﬁx normal words w of length n, for n ≤50 (loglinear scale). 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 15 4.1. Asymptotic bounds on the number of preﬁx normal words We give lower and upper bounds on the number pnw(n) of preﬁx normal words of length n. Theorem 14. For n suﬃciently large pnw(n) ≥2n−4√n log n. (1) Proof. Let k = k(n) be a positive integer to be ﬁxed later. First we only consider words whose length, n, is a multiple of 2k, whose ﬁrst 4k letters are 1s, and in each of the following blocks of length 2k, there are exactly k 1s and k 0s. The number of such words is 2k k (n−4k)/2k and by construction, they are all preﬁx normal. We use the inequality 2k k ≥22k/(2 √ k) and substitute k = √n log n in the third step. 2k k (n−4k)/2k ≥ 22k 2 √ k n/(2k)−2 = 2n (2 √ k)n/(2k) 4k 24k = 2n 24√n log n (2 √ k)1−n/(2k) ≥ 2n 24√n log n for suﬃciently large n. The last inequality follows from the fact that limn→∞(2 √ k)1−n/(2k) = 0 if k = √n log n. 2 Next we show how to obtain an upper bound on pnw(n), considering the length of the ﬁrst 1-run. Theorem 15. For n suﬃciently large, we have pnw(n) ≤2n−lg n+1. Proof. This will follow from enumeration results about pre-necklaces since every 0-preﬁx normal word is a pre-necklace. Let PL(n) be the number of pre-necklaces of length n. In it is shown (top of page 424) that PL(n) ≤ n X i=1 2i i + n X i=1 √ 2i. 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 16 They also show that (Lemma 5 of ) lim n→∞ n 2n n X i=1 2i i = 2. Thus, for large enough n, and ﬁxed ε > 0, PL(n) ≤(1 + ε) n X i=1 2i i ≤(1 + 2ε)2n/n ≤2n−lg n+1. 2 4.2. Exact formulas for words with ﬁxed density. For a binary word w, its density is deﬁned as the number of 1s in w, i.e. as \|w\|1. If we count the number of preﬁx normal words of length n with a given ﬁxed number of 1s, we get exact results in a few cases. Let us denote by pnw(n, d) the cardinality of the set {w ∈LPN1 ∩Σn \| \|w\|1 = d}. Proposition 16. For d = 0, 1, . . . , 6, we have the generating functions fd(x) = P∞ n=0 pnw(n, d)xn: f0(x) = 1 1 −x f1(x) = x 1 −x f2(x) = x2 (1 −x)2 f3(x) = x3 (1 −x2)(1 −x)2 f4(x) = x4 (1 −x3)(1 −x)3 f5(x) = x5(1 + x + x2) (1 −x4)(1 −x2)2(1 −x)2 f6(x) = x6(1 + x + x2 + x3) (1 −x5)(1 −x3)(1 −x2)(1 −x)3 Proof. For d ≤3, one easily checks pnw(n, 0) = pnw(n, 1) = 1, pnw(n, 2) = n −1 and pnw(n, 3) = ⌊(n + 1)2/4⌋, giving the desired functions. 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 17 For d = 4, we calculate the number of positive solutions r1, r2, r3, r4 to the inequalities in Lemma 8. Let q1 = r1 −1, q4 = r4 −1, d2 = r2 −r1 and d3 = r3 −r1. We are counting the nonnegative solutions of 3q1 + d2 + d3 + q4 + 4 = n, which give generating function f4(x) by equating the coeﬃcients of xn in the expansion of the following product: (1 + x3 + x6 + · · · )(1 + x + x2 + · · · )3 · x4 (2) = x4 (1 −x3)(1 −x)3 . (3) More complicated but manageable case analysis leads to the results for d = 5 and 6. 2 Similar formulas can be derived for pnw(n, n −d) for small values of d. Unfortunately, no clear pattern is visible for fd(x) that we could use for calculating pnw(n). The inequalities in Lemma 8 deﬁne linear diophantine equations. The general theory for enumerating solutions of such equations guarantees that there is a closed rational function form for the generating functions with the observed denominators, in there are algorithms for calculating these functions (which, however are not eﬃcient enough to get results for much larger values of d). Above, we only discussed the ﬁrst few simple cases. We did not succeed in extending our list of concrete formulas for the rational functions fd for d > 6 using automated computation. 4.3. Exact formulas for words with a ﬁxed preﬁx. We now ﬁx a preﬁx w and give enumeration results on preﬁx normal words with preﬁx w. Our ﬁrst result indicates that we have to consider each w separately. Deﬁnition 5. If w is a binary word, let Lext(w) = {w′ : ww′ is preﬁx normal}, and Lext(w, m) = Lext(w) ∩Σm. Let ext(w, m, d) = \|{w′ : ww′ is preﬁx nor-mal of length \|w\| + m and density d}\|, and ext(w, m) = \|Lext(w, m)\|. Lemma 17. Let v, w ∈1{0, 1}∗be both preﬁx normal. If v ̸= w then Lext(v) ̸= Lext(w). 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 18 Proof. We may assume \|v\| ≤\|w\|. First case. v is not a preﬁx of w. Let i denote the ﬁrst position where they diﬀer. If vi = 1 and wi = 0, then for u = 0\|w\|v we have that vu is preﬁx normal while wu is not. If vi = 0 and wi = 1, then let u = 0\|w\|w. We have that vu is not preﬁx normal but wu is. Second case. v is a preﬁx of w. If w has a 1 in any position after \|v\|, then we can proceed as in the ﬁrst case. The remaining case is when w = v0m for some m > 0. If vv is preﬁx normal, then so must be vvv, but v0mvv cannot be. Otherwise, let k ≥1 be the smallest integer (which is sure to exist) such that v0kv is preﬁx normal. Then v0k−1v is not preﬁx normal while w0k−1v is. This completes the proof. 2 We were unable to prove that the growth of these two extension languages also diﬀer. Conjecture 18. Let v, w ∈1{0, 1}∗be both preﬁx normal. If v ̸= w then the inﬁnite sequences (ext(v, m))m≥1 and (ext(w, m))m≥1 are diﬀerent. The values ext(w, m, d) seem hard to analyze. We give exact formulas for a few special cases of interest. Using Lemma 8, it is possible to give formulas similar to those in Proposition 16 for ext(w, m, d) for ﬁxed w and d. We only mention one such result. Lemma 19. For 1 ≤d ≤n we have ext(10, n + d −3, d) = pnw(n, d). Proof. Consider the following map: let w be an arbitrary word of length n and density d > 1, starting with 1. Except for the starting 1, insert a 0 right before each subsequent occurrence of 1. This gives a word w′ of length n + d −1, starting with 10 that does not contain the factor 11. Clearly, the map is injective and all words of length n + d −1 starting with 10 and containing no factor 11 are obtained this way. In order to prove the lemma, we only need to show that preﬁx normality is preserved by the map and its inverse. For this, observe that there exists a preﬁx (resp. factor) of w of length k containing r 1s if and only if there exists a preﬁx (resp. factor) of w′ of length k + r −1 containing r 1s. 2 The following lemma lists exact values for ext(w, \|w\|) for some inﬁnite families of words w. Here F(n) denotes the nth Fibonacci number, i.e. F(1) = F(2) = 1 and F(n + 2) = F(n + 1) + F(n). Lemma 20. For all values of n where the exponents are nonnegative, we have the following formulas: 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 19 ext(0n, n) = 1 ext(1n, n) = 2n ext(1n−10, n) = 2n −1 ext(1n−201, n) = 2n −5 ext(1n−200, n) = 2n −(n + 1) ext((10) n 2 , n) = F(n + 2) if n is even ext((10) n−1 2 1, n) = F(n + 1) if n is odd ext(10n−21, n) = 3 ext(10n−1, n) = n + 1 Proof. For w = 1n, w = 1n−10, w = 1n−201 and w = 1n−200, it is easy to count those extensions that fail to give preﬁx normal words: None for w = 1n; only one for w = 1n−10, namely 1n−101n; for w = 1n−201, those extensions which contain a 1-run of length n −1, namely 1n−2 followed by any two characters, or 01n−1; and for w = 1n−200, those that contain at least n −1 many 1s in the second half, i.e. with second half 1n, 1n−10, 1n−201, . . . , 01n−1. Similarly, for w = 10n−21, w = 10n−1 and w = 0n, counting the ex-tensions that yield preﬁx normal words gives the result in a straightforward way. Let n be even. For w = (10) n 2 , note that ww′ is preﬁx normal if and only if w′ avoids 11. The number of such words is known to equal F(n + 2). For n odd, the argument is similar, with the preﬁx of interest, w1, being of length n + 1, hence the previous Fibonacci number. 2 4.4. Some experimental results about enumeration of preﬁx normal words We consider extensions of preﬁx normal words by a single symbol to the right. It turns out that this question has implications for the enumeration of preﬁx normal words. Deﬁnition 6 (Extension-critical words). We call a preﬁx normal word w extension-critical if w1 is not preﬁx normal. Let ecrit(n) denote the num-ber of extension-critical words in LPN1 ∩Σn. The lemma below applies to any family of words B for which ε ∈B and such that x ∈B implies x0 ∈B. 4 ENUMERATION RESULTS ABOUT PREFIX NORMAL WORDS 20 Lemma 21. For n ≥1 we have pnw(n) = 2pnw(n−1)−ecrit(n−1) = pnw(n−1) 2 −ecrit(n −1) pnw(n −1) . (4) From this it follows that pnw(n) = 2 n−1 Y i=1 2 −ecrit(i) pnw(i) . (5) Proof. The number of preﬁx normal words of length n ending in 0 is pnw(n −1), that of preﬁx normal words of length n ending in 1 is pnw(n − 1) −ecrit(n −1), hence we have (4). The product form follows if we use pnw(n) = pnw(1) Qn−1 i=1 pnw(i+1) pnw(i) . 2 Lemma 22. For n going to inﬁnity, lim inf ecrit(n)/pnw(n) = 0. Proof. Assume that there exist an integer N0 and a real number ε > 0 such that for n ≥N0 we have ecrit(n)/pnw(n) > ε. Then by (5) we would have pnw(n) = O((2 −ε)n), contradicting Theorem 14. 2 We conjecture that in fact the ratio of extension-critical words converges to 0. We study the behavior of ecrit(n)/pnw(n) for n ≤49. The left plot in Fig. 3 shows the ratio of extension-critical words for n ≤49. These data support the conjecture that the ratio tends to 0. Interestingly, the values decrease monotonically for both odd and even values, but we have ecrit(n + 1)/pnw(n + 1) > ecrit(n)/pnw(n) for even n. We were unable to ﬁnd an explanation for this. The right plot in Fig. 3 shows the ratio of extension-critical words mul-tiplied by n/ log n. Apart from a few initial data points, the values for even n increase monotonically and the values for odd n decrease monotonically, and the values for odd n stay above those for even n. Conjecture 23. Based on empirical evidence, we conjecture the following: ecrit(n) = pnw(n)Θ(log n/n), (6) pnw(n) = 2n−Θ((log n)2). (7) Note that the second estimate follows from the ﬁrst one by (5). 5 CONCLUSION AND OPEN PROBLEMS 21 Length of word 0 10 20 30 40 50 0 0,1 0,2 0,3 0,4 0,5 Length of word 0 10 20 30 40 50 0 0,2 0,4 0,6 0,8 1,0 1,2 Figure 3: The ratio ecrit(n) pnw(n) (left), and the value ecrit(n) pnw(n) · n ln n (right). 5. Conclusion and open problems We introduced two new normal forms of binary words, the preﬁx normal forms with respect to 1 and 0, and showed how they arise naturally in the investigation of Parikh sets of binary words and jumbled pattern matching. We introduced preﬁx normal words (w.r.t. 1 or 0), words which equal their own normal form, and discussed several properties of these words. We showed results about the language of preﬁx normal words, among these that 0-preﬁx normal are strictly contained in the language of pre-necklaces. We also discussed extensively the growth behaviour of the number of ﬁxed-length preﬁx normal words. Many open problems remain. It would be nice to have exact, or at least more precise asymptotic formulas for the enumeration of preﬁx normal words. Related to the enumeration, the strange oscillating behavior in Figures 2 and 3 between odd and even values calls for an explanation. Another question is testing binary words for preﬁx normality. Currently, no faster method is known (in worst-case running time), then calculating the normal form. It would be an interesting direction to explore the connection between the normal forms w.r.t. 1 and 0, for example how many diﬀerent values can PNF0(w) take (and what can we say about them) if we ﬁx PNF1(w). Finally, preﬁx normality could also be deﬁned over non-binary alphabets. In this case however, we do not obtain an index directly applicable to jumbled pattern matching. Combinatorial or formal language theoretic investigation 5 CONCLUSION AND OPEN PROBLEMS 22 and enumeration of preﬁx normal words for general alphabets is subject of future work. Acknowledgements Gabriele Fici was partially supported by the PRIN 2010/2011 project “Automi e Linguaggi Formali: Aspetti Matematici e Applicativi” of the Ital-ian Ministry of Education (MIUR). The research of Joe Sawada and Frank Ruskey was partially funded by grants from the National Engineering Re-search Council of Canada. We thank an anonymous referee for a very careful reading and helpful suggestions. References J. Acharya, H. Das, O. Milenkovic, A. Orlitsky, and S. Pan. Reconstruct-ing a string from its substring compositions. In Proc. of IEEE International Symposium on Information Theory (ISIT 2010), pages 1238–1242, 2010. G. Badkobeh, G. Fici, S. Kroon, and Zs. Lipták. Binary jumbled string match-ing for highly run-length compressible texts. Inf. Process. Lett., 113(17):604– 608, 2013. G. Benson. Composition alignment. In Proc. of the 3rd International Workshop on Algorithms in Bioinformatics (WABI 2003), pages 447–461, 2003. S. Böcker. Simulating multiplexed SNP discovery rates using base-speciﬁc cleavage and mass spectrometry. Bioinformatics, 23(2):5–12, 2007. S. Böcker, K. Jahn, J. Mixtacki, and J. Stoye. Computation of median gene clusters. In Proc. of the Twelfth Annual International Conference on Com-putational Molecular Biology (RECOMB 2008), pages 331–345, 2008. LNBI 4955. P. Burcsi, F. Cicalese, G. Fici, and Zs. Lipták. On Table Arrangements, Scrab-ble Freaks, and Jumbled Pattern Matching. In Proc. of the 5th International Conference on Fun with Algorithms (FUN 2010), volume 6099 of LNCS, pages 89–101, 2010. P. Burcsi, F. Cicalese, G. Fici, and Zs. Lipták. Algorithms for jumbled pattern matching in strings. Int. J. Found. Comput. Sci., 23(2):357–374, 2012. P. Burcsi, F. Cicalese, G. Fici, and Zs. Lipták. On approximate jumbled pattern matching in strings. Theory Comput. Syst., 50(1):35–51, 2012. 5 CONCLUSION AND OPEN PROBLEMS 23 P. Burcsi, G. Fici, Zs. Lipták, F. Ruskey, and J. Sawada. Normal, abby nor-mal, preﬁx normal. In Proc. of the 7th International Conference on Fun with Algorithms (FUN 2014), volume 8496 of LNCS, pages 74–88, 2014. P. Burcsi, G. Fici, Zs. Lipták, F. Ruskey, and J. Sawada. On combinatorial generation of preﬁx normal words. In Proc. 25th Ann. Symp. on Comb. Pattern Matching (CPM 2014), volume 8486 of LNCS, pages 60–69, 2014. A. Butman, R. Eres, and G. M. Landau. Scaled and permuted string matching. Inf. Process. Lett., 92(6):293–297, 2004. J. Champarnaud, G. Hansel, and D. Perrin. Unavoidable sets of constant length. Internat. J. Algebra Comput., 14:241–251, 2004. T. M. Chan and M. Lewenstein. Clustered integer 3SUM via additive com-binatorics. In Proc. of the 47th Annual ACM on Symposium on Theory of Computing (STOC 2015), pages 31–40, 2015. F. Cicalese, G. Fici, and Zs. Lipták. Searching for jumbled patterns in strings. In Proc. of the Prague Stringology Conference (PSC 2009), pages 105–117. Czech Technical University in Prague, 2009. F. Cicalese, T. Gagie, E. Giaquinta, E. S. Laber, Zs. Lipták, R. Rizzi, and A. I. Tomescu. Indexes for jumbled pattern matching in strings, trees and graphs. In Proc. of the 20th String Processing and Information Retrieval Symposium (SPIRE 2013), volume 8214 of LNCS, pages 56–63, 2013. F. Cicalese, E. S. Laber, O. Weimann, and R. Yuster. Near linear time con-struction of an approximate index for all maximum consecutive sub-sums of a sequence. In Proc. 23rd Annual Symposium on Combinatorial Pattern Match-ing (CPM 2012), volume 7354 of LNCS, pages 149–158, 2012. M. Cieliebak, T. Erlebach, Zs. Lipták, J. Stoye, and E. Welzl. Algorithmic complexity of protein identiﬁcation: combinatorics of weighted strings. Dis-crete Appl. Math., 137(1):27–46, 2004. D. Clark. Compact PAT trees. PhD thesis, University of Waterloo, Canada, 1996. K. Dührkop, M. Ludwig, M. Meusel, and S. Böcker. Faster mass decomposi-tion. In Proc. of the 13th International Workshop on Algorithms in Bioinfor-matics, (WABI 2013), pages 45–58, 2013. R. Eres, G. M. Landau, and L. Parida. Permutation pattern discovery in biosequences. Journal of Computational Biology, 11(6):1050–1060, 2004. G. Fici and Zs. Lipták. On preﬁx normal words. In Proc. of the 15th In-tern. Conf. on Developments in Language Theory (DLT 2011), volume 6795 of LNCS, pages 228–238. Springer, 2011. 5 CONCLUSION AND OPEN PROBLEMS 24 T. Gagie, D. Hermelin, G. M. Landau, and O. Weimann. Binary jumbled pattern matching on trees and tree-like structures. In Proc. of the 21st Annual European Symposium on Algorithm (ESA 2013), pages 517–528, 2013. E. Giaquinta and Sz. Grabowski. New algorithms for binary jumbled pattern matching. Inf. Process. Lett., 113(14-16):538–542, 2013. D. Hermelin, G. M. Landau, Y. Rabinovich, and O. Weimann. Binary jumbled pattern matching via all-pairs shortest paths. CoRR, abs/1401.2065, 2014. D. E. Knuth. Generating All Tuples and Permutations. The Art of Computer Programming, Vol. 4, Fascicle 2. Addison-Wesley, 2005. T. Kociumaka, J. Radoszewski, and W. Rytter. Eﬃcient indexes for jumbled pattern matching with constant-sized alphabet. In Proc. of the 21st Annual European Symposium on Algorithm (ESA 2013), pages 625–636, 2013. L.-K. Lee, M. Lewenstein, and Q. Zhang. Parikh matching in the stream-ing model. In Proc. of 19th International Symposium on String Processing and Information Retrieval, (SPIRE 2012), volume 7608 of Lecture Notes in Computer Science, pages 336–341. Springer, 2012. M. Lothaire. Algebraic Combinatorics on Words. Encyclopedia of Mathematics and its Applications. Cambridge Univ. Press, 2002. T. M. Moosa and M. S. Rahman. Indexing permutations for binary strings. Inf. Process. Lett., 110:795–798, 2010. T. M. Moosa and M. S. Rahman. Sub-quadratic time and linear space data structures for permutation matching in binary strings. J. Discrete Algorithms, 10:5–9, 2012. J. I. Munro. Tables. In Proc. of Foundations of Software Technology and Theoretical Computer Science (FSTTCS 1996), pages 37–42, 1996. G. Navarro and V. Mäkinen. Compressed full-text indexes. ACM Comput. Surv., 39(1), 2007. L. Parida. Gapped permutation patterns for comparative genomics. In Proc. of the 6th International Workshop on Algorithms in Bioinformatics, (WABI 2006), pages 376–387, 2006. F. Ruskey, C. Savage, and T. M. Y. Wang. Generating necklaces. J. Algo-rithms, 13(3):414 – 430, 1992. N. J. A. Sloane. The On-Line Encyclopedia of Integer Sequences. Available electronically at 5 CONCLUSION AND OPEN PROBLEMS 25 R. P. Stanley. Enumerative Combinatorics. Wadsworth Publ. Co., Belmont, CA, USA, 1986. D. Zeilberger. Lindiophantus: A maple package that ﬁnds generating functions representating solutions of systems of linear diophantine equations. http:// Ac-cessed: 2015-05-30.
14745	https://www.reddit.com/r/MathHelp/comments/ywm39u/how_did_he_get_this_answer/	How did he get this answer? : r/MathHelp Skip to main contentHow did he get this answer? : r/MathHelp Open menu Open navigationGo to Reddit Home r/MathHelp A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to MathHelp r/MathHelp r/MathHelp A sub for helping you with your mathematics problems! If you're willing to learn, we're willing to teach. 52K Members Online •3 yr. ago SaucyWench813 How did he get this answer? So basically in this show (My Little Monster), the characters are studying, and one of the characters solves one of the math problems. The problem is: ab(a-b)+bc(b-c)+ca(c-a) And the answer he got is: -(a-b)(b-c)(c-a) I tested his answer by making a=1, b=2, and c=3, and it's the same as what I got doing the same thing to the original problem, so I know it's correct. But I don't know why. Read more Archived post. New comments cannot be posted and votes cannot be cast. Share Related Answers Section Related Answers Math problem identities and solutions Integer solutions for a/(b+c)+b/(a+c)+c/(a+b) Formulas for a x b x c Properties of a(b+c)=ab+ac Tips for mastering calculus concepts New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of November 16, 2022 Reddit reReddit: Top posts of November 2022 Reddit reReddit: Top posts of 2022 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
14746	https://artofproblemsolving.com/wiki/index.php/Induction?srsltid=AfmBOooeqX_pE0KiaPCOfsF455LDa-9NlbixYtXOQjD_a6voKyCI6yWV	Art of Problem Solving Induction - 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 Induction Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Induction Induction is a method of proof in which the desired result is first shown to hold for a certain value (the Base Case); it is then shown that if the desired result holds for a certain value, it then holds for another, closely related value. Typically, this means proving first that the result holds for (in the Base Case), and then proving that having the result hold for implies that the result holds for . In this way, we can show that the result holds for all positive integers; we will have shown that it works for , and that implies that it works for , which in turn means it works for , and so on. Other, odder inductions are possible. If a problem asks you to prove something for all integers greater than 3, you can use as your base case instead. You might have to induct over the even positive integers numbers instead of all of them; in this case, you would take as your base case, and show that if gives the desired result, so does . If you wish, you can similarly induct over the powers of 2. Contents [hide] 1 Example 2 Uses 3 Video Lecture 4 Problems 4.1 Introductory 4.2 Intermediate 4.3 Olympiad 5 See also Example Here is a simple example of how induction works. Below is a proof (by induction, of course) that the th triangular number is indeed equal to (the th triangular number is defined as ; imagine an equilateraltriangle composed of evenly spaced dots). Base Case: If then and So, for Inductive Step: Suppose the conclusion is valid for . That is, suppose we have . Adding to both sides, we get so the conclusion holding for implies that it holds for , and our induction is complete. Uses Induction can be useful in almost any branch of mathematics. Often, problems in number theory and combinatorics are especially susceptible to induction solutions, but that's not to say that there aren't any problems in other areas, such as Inequalities, that can be solved with induction. Induction is also useful in any level of mathematics that has an emphasis on proof. Induction problems can be found anywhere from the Power Round of the ARML up through the USAMTS all the way up to the USAMO and IMO. A good example of an upper-level problem that can be solved with induction is USAMO 2006/5. Video Lecture Problems Introductory The Fibonacci numbers are a sequence of numbers that satisfy , , and the recursion when . Prove that . Prove that for all positive integers (From The Art and Craft of Problem Solving) Prove that . Prove Bernoulli's inequality. Intermediate For any set whose elements are positive integers, define to be the square of the product of the elements of . For example, if , then . For any positive integer , consider all nonempty subsets of that do not contain two consecutive integers. Prove that the sum of all the f(S)’s of these subsets is . A function defined on the positive integers satisfies and . Calculate . (United Kingdom 1996/2) Olympiad Prove Fermat's Little Theorem. Prove AM-GM Inequality by the first principle of mathematical induction. Prove that the number of odd binomial coefficients in any row of the Pascal triangle is a power of 2. (1956 Putnam Competition) See also Proof writing Cauchy Induction Retrieved from " Categories: Definition Proofs 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.
14747	https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/resources/lecture-1-derivatives/	Browse Course Material Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2006 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types assignment Problem Sets grading Exams with Solutions notes Lecture Notes theaters Lecture Videos Download Course search GIVE NOW about ocw help & faqs contact us 18.01 \| Fall 2006 \| Undergraduate Single Variable Calculus Video Lectures Lecture 1: Rate of Change Topics covered:: Derivatives, slope, velocity, rate of change Instructor: Prof. David Jerison Download video Download transcript Lecture Notes (PDF - 1.1MB) Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2006 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types assignment Problem Sets grading Exams with Solutions notes Lecture Notes theaters Lecture Videos Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
14748	https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_002B/UCD_Chem_2B/Text/Unit_IV%3A_Acids_and_Bases/16%3A_Additional_Aspects_of_Acid-Base_Equilibria/16.4%3A_Neutralization_Reactions_and_Titration_Curves	Skip to main content 16.4: Neutralization Reactions and Titration Curves Last updated : Feb 7, 2016 Save as PDF 16.3: Acid–Base Indicators 16.5: Solutions of Salts of Polyprotic Acids Page ID : 42199 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To calculate the pH at any point in an acid–base titration. In an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration. Titrations of Strong Acids and Bases Figure shows a plot of the pH as 0.20 M is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of as shown in Figure . As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid. Suppose that we now add 0.20 M to 50.0 mL of a 0.10 M solution of . Because is a strong acid that is completely ionized in water, the initial is 0.10 M, and the initial pH is 1.00. Adding decreases the concentration of H+ because of the neutralization reaction (Figure ): Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the ions originally present have been consumed. For the titration of a monoprotic strong acid () with a monobasic strong base (), we can calculate the volume of base needed to reach the equivalence point from the following relationship: If 0.20 M is added to 50.0 mL of a 0.10 M solution of , we solve for : At the equivalence point (when 25.0 mL of solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M . As shown in Figure , the titration of 50.0 mL of a 0.10 M solution of with 0.20 M produces a titration curve that is nearly the mirror image of the titration curve in Figure . The pH is initially 13.00, and it slowly decreases as is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M . The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities. The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities. Example : Hydrochloric Acid Calculate the pH of the solution after 24.90 mL of 0.200 M has been added to 50.00 mL of 0.100 M . Given: volumes and concentrations of strong base and acid Asked for: pH Strategy: Calculate the number of millimoles of and to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction. Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH. Solution A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of in 50.00 mL of 0.100 M can be calculated as follows: The number of millimoles of added is as follows: Thus is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of to the solution. Because only 4.98 mmol of has been added, the amount of excess is 5.00 mmol − 4.98 mmol = 0.02 mmol of . B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of is as follows: Hence, This is significantly less than the pH of 7.00 for a neutral solution. Exercise Calculate the pH of a solution prepared by adding of to of a solution of . Answer : 11.6 pH after the addition of 10 ml of Strong Base to a Strong Acid: (opens in new window) pH at the Equivalence Point in a Strong Acid/Strong Base Titration: Titrations of Weak Acids and Bases In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding or . As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its and its concentration. Because only a fraction of a weak acid dissociates, is less than . Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Figure shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M superimposed on the curve for the titration of 0.100 M shown in part (a) in Figure . Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the titration; the magnitude of the pH change at the equivalence point depends on the of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess present, regardless of whether the acid is weak or strong. The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the or . The titration curve in Figure was created by calculating the starting pH of the acetic acid solution before any is added and then calculating the pH of the solution after adding increasing volumes of . The procedure is illustrated in the following subsection and Example for three points on the titration curve, using the of acetic acid (4.76 at 25°C; . Calculating the pH of a Solution of a Weak Acid or a Weak Base As explained discussed, if we know or and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a ICE table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define as due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows: table of concentrations for the ionization of 0.100 M acetic acid \| ICE \| \| \| \| \| initial \| 0.100 \| \| 0 \| \| change \| −x \| +x \| +x \| \| final \| 0.100 − x \| x \| x \| In this and all subsequent examples, we will ignore and due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified. Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations), Solving this equation gives . Thus the pH of a 0.100 M solution of acetic acid is as follows: pH at the Start of a Weak Acid/Strong Base Titration: Calculating the pH during the Titration of a Weak Acid or a Weak Base Now consider what happens when we add 5.00 mL of 0.200 M to 50.00 mL of 0.100 M (part (a) in Figure ). Because the neutralization reaction proceeds to completion, all of the ions added will react with the acetic acid to generate acetate ion and water: All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine of the resulting solution. Step 1 To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of in the original solution and the amount of in the solution that was added. The acetic acid solution contained The solution contained 5.00 mL=1.00 mmol Comparing the amounts shows that is in excess. Because reacts with in a 1:1 stoichiometry, the amount of excess is as follows: 5.00 mmol − 1.00 mmol = 4.00 mmol Each 1 mmol of reacts to produce 1 mmol of acetate ion, so the final amount of is 1.00 mmol. The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations. ICE table \| ICE \| \| \| \| \| initial \| 5.00 mmol \| 1.00 mmol \| 0 mmol \| \| change \| −1.00 mmol \| −1.00 mmol \| +1.00 mmol \| \| final \| 4.00 mmol \| 0 mmol \| 1.00 mmol \| This ICE table gives the initial amount of acetate and the final amount of ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of in equilibrium is insignificant compared to the amount of added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of , but the amount of due to the autoionization of water is insignificant compared to the amount of added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem. Step 2 To calculate at equilibrium following the addition of , we must first calculate [] and using the number of millimoles of each and the total volume of the solution at this point in the titration: Knowing the concentrations of acetic acid and acetate ion at equilibrium and for acetic acid ( ), we can calculate at equilibrium: Calculating gives Comparing the titration curves for and acetic acid in Figure , we see that adding the same amount (5.00 mL) of 0.200 M to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example , we calculate another point for constructing the titration curve of acetic acid. pH Before the Equivalence Point of a Weak Acid/Strong Base Titration: Example What is the pH of the solution after 25.00 mL of 0.200 M is added to 50.00 mL of 0.100 M acetic acid? Given: volume and molarity of base and acid Asked for: pH Strategy: Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of and . Determine which species, if either, is present in excess. Tabulate the results showing initial numbers, changes, and final numbers of millimoles. If excess acetate is present after the reaction with , write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present. Calculate using the relationship . Calculate [OH−] and use this to calculate the pH of the solution. Solution A Ignoring the spectator ion (), the equation for this reaction is as follows: The initial numbers of millimoles of and are as follows: 25.00 mL(0.200 mmol OH−mL=5.00 mmol The number of millimoles of equals the number of millimoles of , so neither species is present in excess. B Because the number of millimoles of added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form. results of the neutralization reaction \| ICE \| \| \| \| \| initial \| 5.00 mmol \| 5.00 mmol \| 0 mmol \| \| change \| −5.00 mmol \| −5.00 mmol \| +5.00 mmol \| \| final \| 0 mmol \| 0 mmol \| 5.00 mmol \| C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction: The equilibrium reaction of acetate with water is as follows: The equilibrium constant for this reaction is where is the acid ionization constant of acetic acid. We therefore define x as produced by the reaction of acetate with water. Here is the completed table of concentrations: completed table of concentrations \| \| \| \| \| \| initial \| 0.0667 \| 0 \| 1.00 × 10−7 \| \| change \| −x \| +x \| +x \| \| final \| (0.0667 − x) \| x \| x \| D We can obtain by substituting the known values into Equation : Substituting the expressions for the final values from the ICE table into Equation and solving for : Thus and the pH of the final solution is 8.794 (Figure ). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce . Exercise Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M solution to 125.0 mL of a 0.150 M solution of ammonia. The of ammonia is 4.75 at 25°C. Answer : 9.23 As shown in part (b) in Figure , the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid. The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure illustrates the shape of titration curves as a function of the or the . As the acid or the base being titrated becomes weaker (its or becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point. One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures and for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows: If , this reduces to . Taking the negative logarithm of both sides, From the definitions of and pH, we see that this is identical to Thus the pH at the midpoint of the titration of a weak acid is equal to the of the weak acid, as indicated in part (a) in Figure for the weakest acid where we see that the midpoint for = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the (or the ) of a weak acid (or a weak base). The pH at the midpoint of the titration of a weak acid is equal to the of the weak acid. Titrations of Polyprotic Acids or Bases When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the values are separated by at least three units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid with is illustrated in Figure and shows two well-defined steps: the first midpoint corresponds to 1, and the second midpoint corresponds to 2. Because HPO42− is such a weak acid, 3 has such a high value that the third step cannot be resolved using 0.100 M as the titrant. The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure . The initial pH is high, but as acid is added, the pH decreases in steps if the successive values are well separated. Table E1 lists the ionization constants and values for some common polyprotic acids and bases. Example Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M solution to 100.0 mL of a 0.0510 M solution of oxalic acid (), a diprotic acid (abbreviated as ). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion (, abbreviated ).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids. Given: volume and concentration of acid and base Asked for: pH Strategy: Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution. Calculate the concentrations of all the species in the final solution. Determine and convert this value to pH. Solution: A Table E5 gives the values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present: The strongest acid () reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of to react with Hox−, forming ox2− and H2O. The reactions can be written as follows: In tabular form, Solutions to Example 17.3.3 \| \| \| \| \| \| \| initial \| 5.10 mmol \| 6.60 mmol \| 0 mmol \| 0 mmol \| \| change (step 1) \| −5.10 mmol \| −5.10 mmol \| +5.10 mmol \| 0 mmol \| \| final (step 1) \| 0 mmol \| 1.50 mmol \| 5.10 mmol \| 0 mmol \| \| change (step 2) \| — \| −1.50 mmol \| −1.50 mmol \| +1.50 mmol \| \| final \| 0 mmol \| 0 mmol \| 3.60 mmol \| 1.50 mmol \| B The equilibrium between the weak acid () and its conjugate base () in the final solution is determined by the magnitude of the second ionization constant, . To calculate the pH of the solution, we need to know , which is determined using exactly the same method as in the acetic acid titration in Example : Thus the concentrations of and are as follows: We can now calculate [H+] at equilibrium using the following equation: Rearranging this equation and substituting the values for the concentrations of and , So This answer makes chemical sense because the pH is between the first and second values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than ), but we added only enough to titrate less than half of the second, less acidic proton, with . If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to . Exercise : Piperazine Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine ( = 4.27, = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine. Answer Indicators In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful. We can describe the chemistry of indicators by the following general equation: where the protonated form is designated by and the conjugate base by . The ionization constant for the deprotonation of indicator is as follows: The (its ) determines the pH at which the indicator changes color. Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, Hydrangea macrophylla flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure ). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers. Irrespective of the origins, a good indicator must have the following properties: The color change must be easily detected. The color change must be rapid. The indicator molecule must not react with the substance being titrated. To minimize errors, the indicator should have a that is within one pH unit of the expected pH at the equivalence point of the titration. Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values. Figure : Some Common Acid–Base Indicators. Approximate colors are shown, along with values and the pH range over which the color changes. (CC BY-SA-NC; Anonymous by request) It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units. We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure . This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M . The pH ranges over which two common indicators (methyl red, , and phenolphthalein, ) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the titration, the phenolphthalein indicator will turn pink when about 50 mL of has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of will therefore cause the methyl red indicator to change color, resulting in a huge error. Figure : Choosing the Correct Indicator for an Acid–Base Titration. (CC BY-SA-NC; Anonymous by request) If the pH of the solution is between 4.95 and 6 then methyl red should be used. If the pH is between 8,2 and 10 then phenolphthalein should be used. The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M . Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point. In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used. The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure ). Figure : pH Paper. pH paper contains a set of indicators that change color at different pH values. The approximate pH of a solution can be determined by simply dipping a paper strip into the solution and comparing the color to the standards provided. (CC BY-SA-NC; Anonymous by request) pH Indicators: pH Indicators(opens in new window) [youtu.be] Summary and Takeaway Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the , and the of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the of the weak acid or the of the weak base. Thus titration methods can be used to determine both the concentration and the (or the ) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. 16.3: Acid–Base Indicators 16.5: Solutions of Salts of Polyprotic Acids
14749	https://scikit-learn.org/stable/modules/lda_qda.html	GitHub 1.2. Linear and Quadratic Discriminant Analysis# Linear Discriminant Analysis (LinearDiscriminantAnalysis) and Quadratic Discriminant Analysis (QuadraticDiscriminantAnalysis) are two classic classifiers, with, as their names suggest, a linear and a quadratic decision surface, respectively. These classifiers are attractive because they have closed-form solutions that can be easily computed, are inherently multiclass, have proven to work well in practice, and have no hyperparameters to tune. The plot shows decision boundaries for Linear Discriminant Analysis and Quadratic Discriminant Analysis. The bottom row demonstrates that Linear Discriminant Analysis can only learn linear boundaries, while Quadratic Discriminant Analysis can learn quadratic boundaries and is therefore more flexible. Examples Linear and Quadratic Discriminant Analysis with covariance ellipsoid: Comparison of LDA and QDA on synthetic data. 1.2.1. Dimensionality reduction using Linear Discriminant Analysis# LinearDiscriminantAnalysis can be used to perform supervised dimensionality reduction, by projecting the input data to a linear subspace consisting of the directions which maximize the separation between classes (in a precise sense discussed in the mathematics section below). The dimension of the output is necessarily less than the number of classes, so this is in general a rather strong dimensionality reduction, and only makes sense in a multiclass setting. This is implemented in the transform method. The desired dimensionality can be set using the n_components parameter. This parameter has no influence on the fit and predict methods. Examples Comparison of LDA and PCA 2D projection of Iris dataset: Comparison of LDA and PCA for dimensionality reduction of the Iris dataset 1.2.2. Mathematical formulation of the LDA and QDA classifiers# Both LDA and QDA can be derived from simple probabilistic models which model the class conditional distribution of the data (P(X\|y=k)) for each class (k). Predictions can then be obtained by using Bayes’ rule, for each training sample (x \in \mathcal{R}^d): [P(y=k \| x) = \frac{P(x \| y=k) P(y=k)}{P(x)} = \frac{P(x \| y=k) P(y = k)}{ \sum_{l} P(x \| y=l) \cdot P(y=l)}] and we select the class (k) which maximizes this posterior probability. More specifically, for linear and quadratic discriminant analysis, (P(x\|y)) is modeled as a multivariate Gaussian distribution with density: [P(x \| y=k) = \frac{1}{(2\pi)^{d/2} \|\Sigma_k\|^{1/2}}\exp\left(-\frac{1}{2} (x-\mu_k)^t \Sigma_k^{-1} (x-\mu_k)\right)] where (d) is the number of features. 1.2.2.1. QDA# According to the model above, the log of the posterior is: [\begin{split}\log P(y=k \| x) &= \log P(x \| y=k) + \log P(y = k) + Cst \ &= -\frac{1}{2} \log \|\Sigma_k\| -\frac{1}{2} (x-\mu_k)^t \Sigma_k^{-1} (x-\mu_k) + \log P(y = k) + Cst,\end{split}] where the constant term (Cst) corresponds to the denominator (P(x)), in addition to other constant terms from the Gaussian. The predicted class is the one that maximises this log-posterior. Note Relation with Gaussian Naive Bayes If in the QDA model one assumes that the covariance matrices are diagonal, then the inputs are assumed to be conditionally independent in each class, and the resulting classifier is equivalent to the Gaussian Naive Bayes classifier naive_bayes.GaussianNB. 1.2.2.2. LDA# LDA is a special case of QDA, where the Gaussians for each class are assumed to share the same covariance matrix: (\Sigma_k = \Sigma) for all (k). This reduces the log posterior to: [\log P(y=k \| x) = -\frac{1}{2} (x-\mu_k)^t \Sigma^{-1} (x-\mu_k) + \log P(y = k) + Cst.] The term ((x-\mu_k)^t \Sigma^{-1} (x-\mu_k)) corresponds to the Mahalanobis Distance between the sample (x) and the mean (\mu_k). The Mahalanobis distance tells how close (x) is from (\mu_k), while also accounting for the variance of each feature. We can thus interpret LDA as assigning (x) to the class whose mean is the closest in terms of Mahalanobis distance, while also accounting for the class prior probabilities. The log-posterior of LDA can also be written as: [\log P(y=k \| x) = \omega_k^t x + \omega_{k0} + Cst.] where (\omega_k = \Sigma^{-1} \mu_k) and (\omega_{k0} = -\frac{1}{2} \mu_k^t\Sigma^{-1}\mu_k + \log P (y = k)). These quantities correspond to the coef_ and intercept_ attributes, respectively. From the above formula, it is clear that LDA has a linear decision surface. In the case of QDA, there are no assumptions on the covariance matrices (\Sigma_k) of the Gaussians, leading to quadratic decision surfaces. See for more details. 1.2.3. Mathematical formulation of LDA dimensionality reduction# First note that the K means (\mu_k) are vectors in (\mathcal{R}^d), and they lie in an affine subspace (H) of dimension at most (K - 1) (2 points lie on a line, 3 points lie on a plane, etc.). As mentioned above, we can interpret LDA as assigning (x) to the class whose mean (\mu_k) is the closest in terms of Mahalanobis distance, while also accounting for the class prior probabilities. Alternatively, LDA is equivalent to first sphering the data so that the covariance matrix is the identity, and then assigning (x) to the closest mean in terms of Euclidean distance (still accounting for the class priors). Computing Euclidean distances in this d-dimensional space is equivalent to first projecting the data points into (H), and computing the distances there (since the other dimensions will contribute equally to each class in terms of distance). In other words, if (x) is closest to (\mu_k) in the original space, it will also be the case in (H). This shows that, implicit in the LDA classifier, there is a dimensionality reduction by linear projection onto a (K-1) dimensional space. We can reduce the dimension even more, to a chosen (L), by projecting onto the linear subspace (H_L) which maximizes the variance of the (\mu^_k) after projection (in effect, we are doing a form of PCA for the transformed class means (\mu^_k)). This (L) corresponds to the n_components parameter used in the transform method. See for more details. 1.2.4. Shrinkage and Covariance Estimator# Shrinkage is a form of regularization used to improve the estimation of covariance matrices in situations where the number of training samples is small compared to the number of features. In this scenario, the empirical sample covariance is a poor estimator, and shrinkage helps improving the generalization performance of the classifier. Shrinkage LDA can be used by setting the shrinkage parameter of the LinearDiscriminantAnalysis class to 'auto'. This automatically determines the optimal shrinkage parameter in an analytic way following the lemma introduced by Ledoit and Wolf . Note that currently shrinkage only works when setting the solver parameter to 'lsqr' or 'eigen'. The shrinkage parameter can also be manually set between 0 and 1. In particular, a value of 0 corresponds to no shrinkage (which means the empirical covariance matrix will be used) and a value of 1 corresponds to complete shrinkage (which means that the diagonal matrix of variances will be used as an estimate for the covariance matrix). Setting this parameter to a value between these two extrema will estimate a shrunk version of the covariance matrix. The shrunk Ledoit and Wolf estimator of covariance may not always be the best choice. For example if the distribution of the data is normally distributed, the Oracle Approximating Shrinkage estimator sklearn.covariance.OAS yields a smaller Mean Squared Error than the one given by Ledoit and Wolf’s formula used with shrinkage="auto". In LDA, the data are assumed to be gaussian conditionally to the class. If these assumptions hold, using LDA with the OAS estimator of covariance will yield a better classification accuracy than if Ledoit and Wolf or the empirical covariance estimator is used. The covariance estimator can be chosen using the covariance_estimator parameter of the discriminant_analysis.LinearDiscriminantAnalysis class. A covariance estimator should have a fit method and a covariance_ attribute like all covariance estimators in the sklearn.covariance module. Examples Normal, Ledoit-Wolf and OAS Linear Discriminant Analysis for classification: Comparison of LDA classifiers with Empirical, Ledoit Wolf and OAS covariance estimator. 1.2.5. Estimation algorithms# Using LDA and QDA requires computing the log-posterior which depends on the class priors (P(y=k)), the class means (\mu_k), and the covariance matrices. The ‘svd’ solver is the default solver used for LinearDiscriminantAnalysis, and it is the only available solver for QuadraticDiscriminantAnalysis. It can perform both classification and transform (for LDA). As it does not rely on the calculation of the covariance matrix, the ‘svd’ solver may be preferable in situations where the number of features is large. The ‘svd’ solver cannot be used with shrinkage. For QDA, the use of the SVD solver relies on the fact that the covariance matrix (\Sigma_k) is, by definition, equal to (\frac{1}{n - 1} X_k^tX_k = \frac{1}{n - 1} V S^2 V^t) where (V) comes from the SVD of the (centered) matrix: (X_k = U S V^t). It turns out that we can compute the log-posterior above without having to explicitly compute (\Sigma): computing (S) and (V) via the SVD of (X) is enough. For LDA, two SVDs are computed: the SVD of the centered input matrix (X) and the SVD of the class-wise mean vectors. The 'lsqr' solver is an efficient algorithm that only works for classification. It needs to explicitly compute the covariance matrix (\Sigma), and supports shrinkage and custom covariance estimators. This solver computes the coefficients (\omega_k = \Sigma^{-1}\mu_k) by solving for (\Sigma \omega = \mu_k), thus avoiding the explicit computation of the inverse (\Sigma^{-1}). The 'eigen' solver is based on the optimization of the between class scatter to within class scatter ratio. It can be used for both classification and transform, and it supports shrinkage. However, the 'eigen' solver needs to compute the covariance matrix, so it might not be suitable for situations with a high number of features. References On this page This Page Show Source
14750	https://www.sciencedirect.com/topics/engineering/common-tangent	Skip to Main content My account Sign in Common Tangent In subject area:Engineering The common tangent refers to the condition where the slopes of the free energy changes associated with two surfaces at distinct orientations must be equal, indicating a stable coexistence of the two phases under variations in their geometries. AI generated definition based on: Materials Fundamentals of Molecular Beam Epitaxy, 1993 How useful is this definition? Add to Mendeley Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Surface Morphology 1993, Materials Fundamentals of Molecular Beam EpitaxyJeffrey Y. Tsao The Common Tangent Criterion Now that we have derived a criterion for stability of a surface against facetting, let us ask the opposite question. Suppose that the original surface is unstable with respect to breakup? What will be the two straddling orientations, which can be considered two “phases,” that will coexist stably in its stead? To answer this question, we imagine making two distinct concerted variations in the geometries of the two surfaces, and require the total free energy change to vanish. First, we imagine varying the projected vertical height of surface 1 by dh1, while at the same time varying the projected vertical height of surface 2 by dh2 = −dh1, so that the two surfaces continue to join perfectly. If each surface is thought of loosely as made up of steps and “missing steps,” then this variation can be thought of as moving steps from surface 2 to surface 1, and at the same time moving missing steps from surface 1 to surface 2. Since x1 and x2 are unchanged during this variation, the free energy changes associated with surfaces 1 and 2 are (6.39) (6.40) where f1 and f2 are the surface free energies per unit area projected onto the reference surface. If the sum of these changes is to vanish, then we must have (6.41) In other words, the slopes of the f(tan θ) plot at the two orientations θ1 and θ2 must be equal. Second, we imagine varying the projected area of surface 1 by dx1, while at the same time varying the projected area of surface 2 by dx2 = −dx1, again so that the two surfaces continue to join perfectly. If each surface is thought of loosely as composed of steps and “missing steps,” then this variation can be thought of as moving missing steps from surface 1 to surface 2. Since h1 and h2 are unchanged during this variation, the free energy changes associated with surfaces 1 and 2 are (6.42) If the sum of these changes is to vanish, then we must have (6.43) In other words, the tan θ = 0 intercepts of the tangents to the f(tan θ) plot at the two orientations θ1 and θ2 must be equal. Altogether, Equations 6.41 and 6.43 combined tell us that both the slopes and intercepts of the two tangents must be equal, and so the tangents themselves must coincide. Therefore, the condition for coexistence of two surfaces of different orientation is that their f(tan θ) plots share a common tangent.22 Another way of viewing the origin of this construction is to think of the steps as particles. Then, equilibrium between surfaces of different orientation is analogous to equilibrium with respect to interchange of particles, hence equality of chemical potentials.23 In a sense, s ≡ tan θ is an extensive, rather than an intensive, variable, and can vary inhomogeneously within an equilibrium system.24 View chapterExplore book Read full chapter URL: Book1993, Materials Fundamentals of Molecular Beam EpitaxyJeffrey Y. Tsao Chapter Transient EHL analysis of helical gears 2014, International Gear Conference 2014: 26th–28th August 2014, LyonH.U. Jamali, ... R.W. Snidle 3EHL MODELLING OF HELICAL GEAR CONTACT The EHL model for the contacting gear teeth is developed in the plane containing the contact line that is perpendicular to the common normal of the contacting teeth. This is referred to as the common tangent plane as illustrated in Figure 1. It is perpendicular to the plane of contact which is tangential to the base cylinders of both gears and contains the contact line at all contact positions of the gear pair. The motion of the gear tooth surfaces relative to the contact line takes place in the tangent plane and the lubrication mechanism must be considered with regard to axes xyz in Figure 1, where z is the common normal direction, y is the contact line direction, and xy is the common tangent plane. The two-dimensional non-Newtonian Reynolds equation relating lubricant pressure, p, and film thickness, h, is (1) The lubricant entrainment velocity is the mean velocity of the surfaces normal to the (y,z) plane, i.e. in the x direction. The elastic deflection equation is written in a differential form as (2) where fk,l are weighting factors for the effect of pressure on the film thickness Laplacian. The time-varying EHL problem described by equations (1) and (2) is analysed using the technique described in suitably modified to include the variation of load, kinematic and geometrical conditions during the meshing cycle of the gears. In equation (6) the terms σx and σx are (3) Lubricant density and viscosity, ρ and η, are functions of pressure, and the non-Newtonian parameters Sx and Sy depend on h, η, ∂ p/ ∂x, ∂ p/ ∂ y, and the sliding speed, Us. To apply these equations to the gears the undeformed gap between the contacting surfaces is required to give hu(x,y) in equation 2. This is obtained by considering the distance, s, from the pitch line to each point of the contact line measured in the direction of z’, and establishing the local radii of curvature of the involute profiles which are used to obtain the undeformed gap hu. Micro geometry corrections such as axial crowning (to prevent contact extending to the face boundaries and consequent edge effects and stress concentrations) and involute profile tip relief (to prevent premature engagement of the teeth under loaded conditions) are added to hu as discussed in . With helical gears the motion is transmitted gradually and smoothly between the mating gears in comparison to spur gears where contact occurs along a straight line parallel to the gear axis. Contact starts as a point at the tooth face end and, as the gears rotate, this extends to become a line increasing steadily in length (e.g. line EE′ in Figure 1) until it starts to contract, finally ending as a point at the other tooth face. This gradual engagement and disengagement leads to the gradual, even action of the tooth and the distribution of the load. The lines of contact act diagonally between the face ends of the teeth and there are at least two pairs of teeth in contact during the meshing cycle. These factors allow helical gears to have increased load capacity compared with the corresponding spur gear drive. View chapterExplore book Read full chapter URL: Book2014, International Gear Conference 2014: 26th–28th August 2014, LyonH.U. Jamali, ... R.W. Snidle Chapter Cams and gears 2020, Manual of Engineering Drawing (Fifth Edition)Colin H. Simmons, ... Neil Phelps Stage 1 (Fig. 31.17) (a) : Draw the pitch circle and the common tangent. (b) : Mark out the pressure angle and the normal to the line of action. (c) : Draw the base circle. Note that the length of the normal is the base-circle radius. View chapterExplore book Read full chapter URL: Book2020, Manual of Engineering Drawing (Fifth Edition)Colin H. Simmons, ... Neil Phelps Chapter Electromagnetic waves 1993, Telecommunications Engineer's Reference BookJ H Causebrook BSc PhD CEng MIEE, R V Goodman CEng MIEE 6.4.8Approximating a land path to a cylindrical surface On a land path with many diffraction edges, a useful prediction is made with a cylindrical surface stylisation of 4 radii joining each other with common tangents, as shown in Figure 6.21. Point ‘a’ is the transmitter horizon, and ‘b’ is the receiver horizon. Arc ac just clears the terrain between point ‘a’ and the transmitter site, and has a tangent to the horizon line Ta at point ‘a’. This condition defines the radius r1. Radius rr is defined similarly at the receiver end. The radii r1 and r2 are defined such that arcs ae and be form tangents with horizon lines Ta and Rb at points a and b, respectively, and have a common tangent at e. The curved surface attenuation can then be approximated by Equations 6.73 to 6.77. (6.73) (6.74) (6.75) (6.76) (6.77) The calculation does not allow for the roughness of the terrain and, as a result, the loss of signal on the path is usually over estimated. A better result is obtained from an interpolation between this and the knife edge calculation. The form of interpolation can be determined by obtaining a best fit with a set of measured data. In the UK it was found that an averaging of losses, expressed in decibels, gave good results. View chapterExplore book Read full chapter URL: Book1993, Telecommunications Engineer's Reference BookJ H Causebrook BSc PhD CEng MIEE, R V Goodman CEng MIEE Review article Hilbert transform in vibration analysis 2011, Mechanical Systems and Signal ProcessingMichael Feldman 8.1.2Local level estimations Traditionally, analysis of a deterministic vibration uses a concept of the peak value as an absolute value of the maximum or minimum of the oscillating parameter during the time segment. For a random vibration the peak value characterizes only a quasi-maximum level, above which the signal is possible with a certain probability. For a random normal vibration the most common maximum amplitude holds the value between plus and minus three times the RMS value of the signal xmax/min=±3RMS called by its statistical name three sigma peaks. Thus, 99.73% values of the vibration signal fall within these defined limits, and only 0.27% out of them. A very large data sample size can yield signal random peaks out of the ±3RMS limits. For a deterministic vibration and, in particular, for the mono-harmonic signal, the peak value is equal to its amplitude, and the peak to peak magnitude is a double amplitude. Note, that the peak amplitude found for digital data is approximate, since the true peak of the output sinusoid generally occurs between samples. The initial signal and its envelope have common tangents at points of contact, but the signal never crosses the envelope. The common points of the contact between the signal and its envelope do not always correspond to the local extrema of a multicomponent signal (see an example in Fig. 15). The local extrema always have a zero tangent slope, but the common points of the contact can have a nonzero value of the tangent slope. Every maximum (top extremum) and minimum (bottom extremum) point of the function, known collectively as a set of the local extrema points xextr, is uniquely defined by the first derivative when the slope of the tangent is equal to zero. The distance between the common envelope points and the signal extrema points plays a dominant role in explaining the Empirical Mode Decomposition (EMD) mechanism . In effect, without such a difference between the envelope and the local extrema, the sum of maxima and minima curves required by the EMD would be always equal to zero, just like the zero sum of the upper and the lower envelopes. View article Read full article URL: Journal2011, Mechanical Systems and Signal ProcessingMichael Feldman Chapter Network comparisons and their applications in connectomics 2023, Connectome AnalysisNiharika S. D’Souza, Archana Venkataraman 8.5.2Geometric approaches Connectome representations are known to possess geometric properties, for example, FC matrices Γn belong to the Riemannian manifold of symmetric positive semidefinite matrices. This structure in the data is lost when vectorizing the features and when using classical dimensionality reduction. Thus methods have been developed to directly leverage these properties that are specific to FC geometry. One class of methods relies on Reimannian manifolds to analyze FC. From a theoretical standpoint, defining common operations on manifolds (e.g., addition and subtraction) requires additional machinery in the form of tangent spaces, Riemannian distance metrics, exponential and logarithmic maps, and Lie group theory [132,133]. From an application standpoint, Riemannian operations provide an efficient geometrically-informed way to study the connectome. One approach is to define population-level FC shrinkage operators (e.g., covariance shrinkage operators in [134–136]) for improved biomarker development, which has been shown to improve rs-fMRI test–retest reliability [134,136], and further aid discriminative group-wise comparisons among subjects . Likewise, parallel transport along the geodesic curves on symmetric positive definite manifolds may be used for classifying FC patterns before and after intervention in longitudinal studies . For example, Ref. maps longitudinal data to sample points on such a manifold trajectory using geodesic regression . Furthermore, they identify ROI level differences between males and females by mapping the resulting trajectories to a common tangent space. Ref. presents a geometric regression to predict cognitive measures, while Ref. takes an unsupervised geometric k-means clustering to identify dominant subnetworks. An alternative to manifold-based techniques is graph signal processing. Recall that in Section 8.3, we described graph theoretical measures for analyzing local and global connectivity properties. In lieu of these specific feature definitions, one may wish to compare two sets of brain networks simultaneously at the local and global levels while still accounting for the underlying graph structure. Here, kernel methods offer a natural framework to quantify network similarity. Mathematically, a graph kernel ψ(.,.) implicitly defines an inner product between a pair of graphs {Ai,Aj}. Once defined, one may use a kernel-based learning algorithm, such as kernel SVM, to facilitate classification or regression. The choice of kernel can be based on Riemannian geometry (e.g., the log-Euclidean Gaussian kernel or the Stein-Gaussian kernel ), while others may be defined by subgraphs or covariance-based distance metrics . Graph kernels have found applications for the classification of disorders, such as ASC [141,144], schizophrenia , Alzheimer’s disease , and ADHD . Topological data analysis is another emerging research field for connectome comparisons. Here, one of the most common analysis frameworks for FC is persistent homology, which is built on the theory of simplicial complexes, filtration, and homology. A simplicial complex is a collection of points (0-simplex), lines (1-simplex), triangles (2-simplex), and other higher dimensional counterparts. Homology refers to groups built on top of such simplicial complexes, while a filtration refers to a nested family of simplicial complexes. At a high level, instead of examining the FC Γn at a fixed threshold, persistent homology tracks changes in the underlying connected components over multiple resolutions. As the threshold increases, homologies in the homology group appear (birth) and disappear (death) and are charted using persistence diagrams. This procedure reveals the “persistent” topological features most robust to noise. The works of [147,148] build sets of hierarchical and nested subgraphs, as an alternative to threshold-based multiresolution network models . Various topological statistics derived from this process, such as persistence diagrams, landscapes, barcodes, and distances, have been useful for statistical inference . Combined with predictive models, these statistics have been used for classifying ASC versus controls [144,151], group-level comparison of ADHD versus controls , and for predicting behavioral measures in ASC . View chapterExplore book Read full chapter URL: Book2023, Connectome AnalysisNiharika S. D’Souza, Archana Venkataraman Chapter Force and Stress 2019, Continuum Mechanics Modeling of Material BehaviorMartin H. Sadd 4.2Cauchy Stress Principle: Stress Vector In order to quantify the nature of the internal distribution of forces within a continuum, consider a general body subject to arbitrary (concentrated and distributed) external loadings as shown in Fig. 4.2. For now, we will assume that the body is in its current deformed configuration. To investigate the internal forces, a section is made through the body as shown. On this section, consider a small area Δa with a unit normal vector n. In general, the resultant of the surface forces acting on Δa would be a force ΔF and a moment ΔM. We now wish to determine the pointwise value of the force per unit area and thus the stress or traction vector is defined by (4.2.1) For the resultant moment case, we assume that in the limit (4.2.2) which is consistent with our earlier discussion that no resultant surface couple will be included in classical continuum mechanics. Notice that the traction vector depends on both the spatial location and the unit normal vector to the surface under study. Thus, even though we may be investigating the same point, the traction vector will still vary as a function of the orientation of the surface normal. This concept is often called the Cauchy Stress Principle which can be stated as the stress or traction vector at any given place and time has a same value on all parts of material having a common tangent plane and lying on the same side of it. Note also the expected action–reaction principle (Newton’s third law) which is often expressed as (4.2.3) View chapterExplore book Read full chapter URL: Book2019, Continuum Mechanics Modeling of Material BehaviorMartin H. Sadd Chapter Rolling bearing types and applications 2023, Rolling Bearing TribologyGary L. Doll 2.2Radial bearings The discussion of rolling bearings begins with those bearings designed to primarily support radial loads; however, the geometries of the rolling element/raceway contacts in these bearings also allow the bearings to support axial or thrust loads. 2.2.1Radial ball bearings 2.2.1.1Deep-groove ball bearings A single-row, deep-groove ball bearing such as that shown schematically in Fig. 2.3 is the most widely used rolling element bearing. These bearings are commonly used in electric motors, compressors, fans, and conveyers, and are utilized in mechanical applications across most industries. This bearing type has low friction and can operate over a wide range of speeds. Deep-groove ball bearings can accommodate radial and axial loads in both directions, are easy to mount, and require less maintenance than other bearing types. The ratio of the inner (ri) and outer (ro) raceway radii to the diameter of the ball (D) ranges between 0.51 ≤ f ≤ 0.54, with a typical value of f = 0.52. In geometric terms, osculation describes the location where two curves or surfaces come into contact or where their common tangent exists. Deep-groove ball bearings have a high degree of osculation, so they have a relatively high load-bearing capacity. Bearings of this type are used in most of the mechanical industries. Deep-groove ball bearings can be distinguished by the method in which they are assembled. Conrad-assembly, deep-groove ball bearings were named after its inventor, Robert Conrad, who was awarded British patent 12,206 in 1903 and US patent 822,723 in 1906 . As illustrated in Fig. 2.4, Conrad-style bearings are assembled by placing the inner ring into an eccentric position relative to the outer ring, with the two rings in contact at one point, resulting in a large gap opposite the point of contact. The balls are inserted through the gap and then evenly distributed around the bearing assembly, causing the rings to become concentric. A stamped and riveted two-piece cage, piloted on the ball set, or a machined two-piece cage, ball piloted or race piloted, is usually used in a Conrad bearing. Illustrations of single-row, deep-groove filling-slot assembly ball bearings are shown in Fig. 2.5. A slot machined in the side walls of the inner and outer rings permits the insertion of more balls than the Conrad-style does, so a filling-slot assembly can have a greater radial-load-carrying capacity. Since the slot is present in the inner and outer rings, this style of deep-groove ball bearing cannot be used in applications requiring axial load support; otherwise, the bearing has characteristics similar to those of the Conrad-style bearing. A double-row, deep-groove ball bearing is illustrated in Fig. 2.6. It is used in applications where radial load-carrying capacity exceeds that of single-row types. Instrument ball bearings are a special class of deep-groove ball bearings. These bearings range in size from rings with 1.5 mm bores and 4 mm outer diameters, to rings with 9 mm bores and 26 mm outer diameters. Because the slightest amount of contamination can adversely affect the performance of these ball bearings, they are usually fabricated from stainless steel alloys and assembled in an ultra-clean environment. 2.2.1.2Angular-contact ball bearings Angular-contact ball bearings are designed to support combined radial and axial loads or heavy axial loads, depending on the magnitude of the contact angle. An illustration of an angular-contact ball bearing and its contact angle is shown in Fig. 2.7. Angular contact bearings are commonly used in gearboxes, pumps, electric motors, and clutches and other high-speed applications. Bearings having larger contact angles can support larger axial loads, and the contact angle does not usually exceed about 40 degrees. The ratio of the inner and outer raceway radii to ball diameter resides between 0.52 ≤ f ≤ 0.54. In most applications, the bearings are mounted in pairs, such as the configurations shown in Fig. 2.8. Tandem mounting is employed when greater axial load-carrying capacity is required. Double-row angular-contact ball bearings such as those shown in Fig. 2.9 can carry axial loads in either direction, or a combination of radial and axial loads. Rigid type versions are very effective in accommodating moment loading from misalignment, for example. 2.2.1.3Self-aligning double row ball bearing The outer raceway of a self-aligning double-row ball bearing is a portion of a sphere (Fig. 2.10), which makes the bearings internally self-aligning. This attribute also means that these bearings cannot support moment loads. Since the balls do not conform well to the outer raceway, the outer raceway has a reduced load-carrying capacity, which is somewhat compensated by employing a large ball complement that minimizes the load carried by each ball. Self-aligning double-row ball bearings are particularly useful in applications in which it is difficult to obtain exact parallelism between the shaft and housing bores, such as those found in equipment used in the pulp and paper, material handling, and agricultural industries. 2.2.1.4Magneto ball bearings Magneto bearings have an internal design similar to the single-row deep-groove ball bearings except that the outer raceway has a shoulder on only one side, which allows it to be easily removed from the rest of the bearing, and the groove on the inner raceway is a little shallower than that of deep-groove bearings so that, with the outer race already removed, the balls in their cage can be removed from the inner race. The bearing can be separated into three component parts which makes easy assembly of the magneto bearing. 2.2.2Thrust ball bearings Ball bearings with contact angles greater than 45 degrees are referred to as thrust ball bearings due to their ability to support larger axial loads. The amount of radial load that a thrust bearing can accommodate decreases with an increasing contact angle. For example, the thrust ball bearing illustrated in Fig. 2.11 has a contact angle of 90 degrees and cannot support any amount of radial load. Typical applications of thrust ball bearings include lathe tailstocks, powered jacks, and turn tables. View chapterExplore book Read full chapter URL: Book2023, Rolling Bearing TribologyGary L. Doll Chapter Elements of contact mechanics 1990, Tribology in Machine DesignT.A. STOLARSKI MSc, PhD, DSc, DIC, CEng, MIMechE 3.9.Representation of machine element contacts Many contacts between machine components can be represented by cylinders which provide good geometrical agreement with the profile of the undeformed solids in the immediate vicinity of the contact. The geometrical errors at some distance from the contact are of little importance. For roller-bearings the solids are already cylindrical as shown in Fig. 3.13. On the inner race or track the contact is formed by two convex cylinders of radii r and R1, and on the outer race the contact is between the roller of radius r and the concave surface of radius (R1 + 2r). For involute gears it can readily be shown that the contact at a distance s from the pitch point can be represented by two cylinders of radii, R1,2 sin ψ ± s, rotating with the angular velocity of the wheels. In this expression R represents the pitch radius of the wheels and ψ is the pressure angle. The geometry of an involute gear contact is shown in Fig. 3.14. This form of representation explains the use of disc machines to simulate gear tooth contacts and facilitate measurements of the force components and the film thickness. From the point of view of a mathematical analysis the contact between two cylinders can be adequately described by an equivalent cylinder near a plane as shown in Fig. 3.15. The geometrical requirement is that the separation of the cylinders in the initial and equivalent contact should be the same at equal values of x. This simple equivalence can be adequately satisfied in the important region of small x, but it fails as x approaches the radii of the cylinders. The radius of the equivalent cylinder is determined as follows: (3.59) Using approximations and For the equivalent cylinder Hence, the separation of the solids at any given value of x will be equal if The radius of the equivalent cylinder is then (3.60) If the centres of the cylinders lie on the same side of the common tangent at the contact point and Ra > Rb, the radius of the equivalent cylinder takes the form (3.61) From the lubrication point of view the representation of a contact by anequivalent cylinder near a plane is adequate when pressure generation is considered, but care must be exercised in relating the force components on the original cylinders to the force components on the equivalent cylinder. The normal force components along the centre-lines as shown in Fig. 3.15 are directly equivalent since, by definition The normal force components in the direction of sliding are defined as Hence and For the friction force components it can also be seen that where τo,h represents the tangential surface stresses acting on the solids. View chapterExplore book Read full chapter URL: Book1990, Tribology in Machine DesignT.A. STOLARSKI MSc, PhD, DSc, DIC, CEng, MIMechE Chapter Data collection in wireless sensor networks by ground robots with full freedom 2022, Wireless Communication Networks Supported by Autonomous UAVs and Mobile Ground RobotsHailong Huang, ... Chao Huang 4.3Shortest viable path planning Formulated as the DTSPN like [13,14], problem (4.11) is also NP-hard. But unlike the DTSPN, having an infinite number of possible configurations in each goal region, the candidate configuration number on , , in problem (4.11) is finite due to the tangent graph. So more appropriately, problem (4.11) is a sampled DTSPN. On the other hand, in the context of sensor networks, sensor nodes are usually deployed apart from each other. Then our considered problem is a standard version of the sampled DTSPN. We propose an algorithm called SVPP. The basic idea is similar to , where the first stage is to determine the permutation Σ based on . In the second stage, we simplify into and convert to a tree-like graph T, and then search the shortest path in T. The main steps of SVPP are outlined as Algorithm 1. The details are given below. In Step 1, to compute Σ, we construct a directed graph. The reason for choosing a directed graph instead of an undirected graph lies in condition (5) of Definition 4.2.1, i.e., different sensor nodes may require different contacting times. We take , , as the vertex set of the directed graph. We construct the edge set as follows. The length of the edge between two vertices takes into account two aspects: the length of the valid path between their visiting circles and the length of the adjusted arc on the latter vertex. Note that here, since there may be multiple valid paths between two visiting circles (see and in Fig. 4.2 for an example), we use the average length of them. Thus, the length of the edge from to equals the summation of the average length and the length of the adjusted arc on the visiting circle of . In contrast, the length of the edge from to equals the summation of the average length and the length of the adjusted arc on the visiting circle of . With such a directed graph, we use an ATSP solver to calculate the permutation Σ. Having Σ, can be simplified by retaining the tangent edges connecting two successful visiting circles in Σ and the corresponding arc edges. When any obstacle blocks any pair of visiting circles, the edges passing the obstacle safety boundaries are also retained. By doing this, we can get the blocking number which counts the occurrences of blocking two successful visiting circles in Σ. Since one obstacle can block more than one pair of successful visiting circles, the number of blocking obstacles in may be smaller than K. We insert these boundaries into the proper positions in Σ and get an extended permutation , whose length is . Obviously, . We call the new graph the simplified tangent graph , where and . Given , the problem (4.11) can be reformulated as (4.12) where . It is worth mentioning that if we name () as the arrival configuration on (), there is a departure configuration, say , on such that the heading constraint (4.5) is satisfied. Then always consists of two edges: one arc edge and one tangent edge . With such characteristic of , the objective of problem (4.12) can be stated to find the minimum-length path by selecting two configurations from each element in , subject to the heading constraint (4.5). Before we describe how to select such configurations, we analyze the number of possible paths in . We generate tangents between any pair of elements to construct . As the robots only need to move along the boundary of the convex hull, we have four common tangents and eight tangent points for each pair of successful elements in . Then one element has eight configurations, four of which are arrival configurations and the other four are departure configurations. For example, in Fig. 4.2, , , , and are the four arrival configurations of and , , , and are the four departure configurations. Also, we note that from one arrival configuration of one element, we have two options to reach the arrival configurations of the next element considering the heading constraint (4.5). Then, from a given we have paths to reach . Finally, the robot needs to return to from . Again due to the heading constraint (4.5), half of the arrival configurations on cannot reach , such as and in Fig. 4.2. Therefore, the total number of paths starting and ending at is . To better demonstrate the viable paths in , we convert it to a tree-like graph. Given , we cut into two parts: one part contains the departure configurations and , and the other contains the arrival configurations and . Then the circle-like graph turns out to be a tree-like graph, which is called tree-like graph T. Since elements are on and is divided into two parts, T consists of layers, where () and . An example of T is shown in Fig. 4.3. Evolved from , T inherits the feature that given a departure configuration, moving from one layer to the next one has only one path. Define () as an arrival configuration variable of layer . Define as the path length from an arrival configuration on to an arrival configuration on . Then we have (4.13) Suppose is the shortest path length from an arrival configuration on to an arrival configuration on , . Since the initial value of is prescribed, i.e., , it follows that is only a function of the viable . Then (4.14) where . Let , , denote the optimal configurations to problem (4.14). We use a dynamic programming-based method, given by Algorithm 2, to solve problem (4.14). To end this section, we analyze the time complexity of SVPP. In Step 1, the complexity of the ATSP algorithm is . In Step 2, we check n pairs of visiting circles to see whether they are blocked by any boundary of the convex hull. In each checking, the worst case is to check all the m obstacles and then the time complexity is . In Step 3, we perform a constant number of operations to each element in . Then the time complexity of the simplifying procedure is . Converting to T costs in Step 4. Finally, because the maximum number of arrival configurations is four on each layer of T, searching the shortest path in T costs . Therefore, the overall time complexity of SVPP is . View chapterExplore book Read full chapter URL: Book2022, Wireless Communication Networks Supported by Autonomous UAVs and Mobile Ground RobotsHailong Huang, ... Chao Huang Related terms: Energy Engineering Chemical Potential Eutectics Alpha Phase Beta Phase Straight Line Tangent Line Tangent Plane Gibbs Free Energy Phase Boundary View all Topics
14751	https://www.cuemath.com/algebra/mathematical-induction/	Mathematical Induction Mathematical induction is a concept that helps to prove mathematical results and theorems for all natural numbers. The principle of mathematical induction is a specific technique that is used to prove certain statements in algebra which are formulated in terms of n, where n is a natural number. Any mathematical statement, expression is proved based on the premise that it is true for n = 1, n = k, and then it is proved for n = k + 1. Let us understand the concept of the principle of mathematical induction, its statement, and its application for proving various mathematical theorems and statements for natural numbers. \| \| \| --- \| \| 1. \| What is Mathematical Induction? \| \| 2. \| Principle of Mathematical Induction Statement \| \| 3. \| Mathematical Induction - Steps To Solve \| \| 4. \| Application of Mathematical Induction \| \| 5. \| FAQs on Mathematical Induction \| What is Mathematical Induction? Mathematical Induction is a technique used to prove that a mathematical statements P(n) holds for all natural numbers n = 1, 2, 3, 4, ... It is often referred as the principle of mathematical induction. To prove a result P(n) using the principle of mathematical induction, we prove that P(1) holds. If P(1) is true, then we assume that P(k) holds for some natural number k, and using this hypothesis, we prove that P(k+ 1) is true. If P(k+1) holds true, then the statement P(n) becomes true for all natural numbers. Principle of Mathematical Induction Statement Now, let us state the principle of mathematical induction and understand how it is used to prove statements step-wise: Suppose there is a given statement P(n) involving the natural number n such that Then, P(n) is true for all natural numbers n. Now before we move on to solve a few examples using the principle of mathematical induction, let us go through some points that are important to understand: Mathematical Induction - Steps To Solve Now, each step that is used to prove the theorem or statement using mathematical induction has a defined name. Each step is named as follows: After proving these 3 steps, we can say that "By the principle of mathematical induction, P(n) is true for all n in N". The assumption that we make in the second step that P(n) holds for some natural number n = k is called induction hypothesis. Application of Mathematical Induction Now that we have understood the concept of mathematical induction, let us solve an example to understand its application better. Example 1: Prove that the formula for the sum of n natural numbers holds true for all natural numbers, that is, 1 + 2 + 3 + 4 + 5 + .... + n = n(n+1)/2 using the principle of mathematical induction. Solution: Suppose P(n): 1 + 2 + 3 + 4 + 5 + .... + n = n(n+1)/2 Here we use the concept of mathematical induction and prove this across the following three steps. Base Step: To prove P(1) is true. For n = 1, LHS = 1 RHS = 1(1+1)/2 = 2/2 = 1 Hence LHS = RHS ⇒ P(1) is true. Assumption Step: Assume that P(n) holds for n = k, i.e., P(k) is true ⇒ 1 + 2 + 3 + 4 + 5 + .... + k = k(k+1)/2 --- (1) Induction Step: Now we will prove that P(k+1) is true. To prove: 1 + 2 + 3 + 4 + ... + (k+1) = (k+1)(k+2)/2 Consider LHS = 1 + 2 + 3 + 4 + ... + (k+1) = 1 + 2 + 3 + 4 + ... k + (k+1) = (1 + 2 + 3 + 4 + ... + k) + k+1 = k(k+1)/2 + k+1 [Using (1)] = [k(k+1) + 2(k+1)]/2 = (k+1)(k+2)/2 = RHS ⇒ P(n) is true for n = k+1 Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n. Important Notes on Principle of Mathematical Induction Topics Related to Mathematical Induction Examples on Mathematical Induction Example 1: Prove the following formula using the Principle of Mathematical Induction. 12 + 32 + 52 + ... + (2n - 1)2 = n(2n-1)(2n+1)/3 Solution: Assume P(n): 12 + 32 + 52 + ... + (2n - 1)2 = n(2n-1)(2n+1)/3 Here we use the concept of mathematical induction across the following three steps. Base Step: To prove P(1) is true. For n = 1, LHS = 12 = 1 RHS = 1(2×1-1)(2×1+1)/3 = [1(2-1)(2+1)]/3 = 3/3 = 1 Hence LHS = RHS ⇒ P(1) is true. Assumption Step: Assume that P(n) holds for n = k, i.e., P(k) is true ⇒ 12 + 32 + 52 + ... + (2k - 1)2 = k(2k-1)(2k+1)/3 --- (1) Induction Step: Now we will prove that P(k+1) is true. To prove: 12 + 32 + 52 + ... + (2k - 1)2 + [2(k+1) - 1]2= (k+1)[2(k+1)-1][2(k+1)+1]/3 Consider LHS = 12 + 32 + 52 + ... + (2k - 1)2 + [2(k+1) - 1]2 = [12 + 32 + 52 + ... + (2k - 1)2] + [2(k+1) - 1]2 = k(2k-1)(2k+1)/3 + [2(k+1) - 1]2 [Using (1)] = k(2k-1)(2k+1)/3 + (2k+1)2 = [k(2k-1)(2k+1) + 3(2k+1)2]/3 = (2k+1)[k(2k-1)+3(2k+1)]/3 = (2k+1)[2k2 - k + 6k + 3]/3 = (2k+1)[2k2 + 5k + 3]/3 = (2k+1)[2k2 + 2k + 3k + 3]/3 = (2k+1)[2k(k+1) + 3(k+1)]/3 = (2k+1)(2k+3)(k+1)/3 = (k+1)[2(k+1)-1][2(k+1)+1]/3 = RHS ⇒ P(n) is true for n = k+1 Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n. Answer: 12 + 32 + 52 + ... + (2n - 1)2 = n(2n-1)(2n+1)/3 is true for all positive integers n. Example 2: Prove that 2n > n for all positive integers n. Solution: We will prove the result using the principle of mathematical induction. Assume P(n): 2n > n Base Step: To prove P(1) is true. For n = 1, we have 21 = 2 > 1 ⇒ P(1) is true. Assumption Step: Assume that P(n) holds for n = k, i.e., P(k) is true ⇒ 2k > k --- (1) Induction Step: Now we will prove that P(k+1) is true. To prove: 2k+1 > k + 1 Consider 2k+1 = 2.2k 2k [Using (1)] = k + k k + 1 [Because any natural number other than 1 is greater than 1.] ⇒ P(n) is true for n = k+1 Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n. Answer: 2n > n is true for all positive integers n. Example 3: Show that 102n-1 + 1 is divisible by 11 for all natural numbers. Solution: Assume P(n): 102n-1 + 1 is divisible by 11 Base Step: To prove P(1) is true. For n = 1, 102×1-1 + 1 = 101 + 1 = 11, which is divisible by 11. ⇒ P(1) is true. Assumption Step: Assume that P(n) holds for n = k, i.e., P(k) is true ⇒ 102k-1 + 1 is divisible by 11 ⇒ 102k-1 + 1 = 11a, for some integer 'a' --- (1) Induction Step: Now we will prove that P(k+1) is true. To prove: 102(k+1)-1 + 1 is divisible by 11 Consider 102(k+1)-1 + 1 = 102k+2-1 + 1 = 102.102k-1 + 1 = 102.(102k-1 + 1 - 1) + 1 = 102.(11a - 1) + 1 [Using (1)] = 102.11a - 102 + 1 = 102.11a - 100 + 1 = 102.11a - 99 = 11(102a - 9) = 11(100a - 9), which is divisible by 11. ⇒ P(n) is true for n = k+1 Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n. Answer: 102n-1 + 1 is divisible by 11 for all natural numbers. go to slidego to slidego to slide Book a Free Trial Class Practice Questions on Mathematical Induction go to slidego to slide FAQs on Mathematical Induction What Is Mathematical Induction In Maths? Mathematical induction is the process of proving any mathematical theorem, statement, or expression, with the help of a sequence of steps. It is based on a premise that if a mathematical statement is true for n = 1, n = k, n = k + 1 then it is true for all natural numbrs. What is the Principle of Mathematical Induction? The Principle of Mathematical Induction is a technique used to prove that a mathematical statements P(n) holds for all natural numbers n = 1, 2, 3, 4, ... It helps to solve or find proof for any mathematical expression over a sequence of steps. It is proved for n = 1, n = k, and n = k + 1, and then it is said to be true for all n natural numbers. What is the Use of Mathematical Induction? The Principle of Mathematical Induction is important because we can use it to prove a mathematical equation statement, (or) theorem based on the assumption that it is true for n = 1, n = k, and then finally prove that it is true for n = k + 1. What is the Principle of Mathematical Induction in Matrices? The Principle of Mathematical Induction in matrices is a specific technique of proving statements or theorems based on matrices using mathematical induction. How to Apply The Principle of Mathematical Induction? To prove a result P(n) using the principle of mathematical induction, we prove that P(1) holds. If P(1) is true, then we assume that P(k) holds for some natural number k, and using this hypothesis, we prove that P(k+ 1) is true. If P(k+1) holds true, then the statement P(n) becomes true for all natural numbers. What Are The Steps To Solve A Problem Using Mathematical Induction? There are mainly two steps to prove a statement using the Principle of Mathematical Induction. The first step is to prove that P(1) is true and the second step is to prove P(k+1) is true using the truth of P(k). Then we can say that P(n) is true for all natural numbers n. What are the Names of Each Of The Steps Used In Mathematical Induction? Each step that is used to prove the theorem or statement using the principle of mathematical induction has a defined name. Each step is named as follows:
14752	https://www.ajkd.org/article/S0272-6386(21)00796-4/fulltext	The Role of the Nephrologist in Management of Poisoning and Intoxication: Core Curriculum 2022 - American Journal of Kidney Diseases Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Articles provided by NKF in this session. Core Curriculum in NephrologyVolume 79, Issue 6p877-889 June 2022 Download Full Issue Download started Ok The Role of the Nephrologist in Management of Poisoning and Intoxication: Core Curriculum 2022 Michael E.Mullins Michael E.Mullins Affiliations Division of Emergency Medicine, Washington University School of Medicine, St Louis, Missouri Search for articles by this author 1 ∙ Jeffrey A.Kraut Jeffrey A.Kraut Correspondence Address for Correspondence: Jeffrey A. Kraut, MD, Division of Nephrology, VHAGLA Healthcare System, 11301 Wilshire Blvd, Los Angeles, CA 90073. jkraut@ucla.edu Affiliations Medical and Research Services, VA Greater Los Angeles Healthcare System, Los Angeles, California UCLA Membrane Biology Laboratory, Division of Nephrology, David Geffen School of Medicine, University of California–Los Angeles, Los Angeles, California Search for articles by this author 2,3jkraut@ucla.edu Affiliations & Notes Article Info 1 Division of Emergency Medicine, Washington University School of Medicine, St Louis, Missouri 2 Medical and Research Services, VA Greater Los Angeles Healthcare System, Los Angeles, California 3 UCLA Membrane Biology Laboratory, Division of Nephrology, David Geffen School of Medicine, University of California–Los Angeles, Los Angeles, California Publication History: Published online December 9, 2021 Footnotes: Complete author and article information provided at end of article. DOI: 10.1053/j.ajkd.2021.06.030 External LinkAlso available on ScienceDirect External Link Copyright: Published by Elsevier Inc. on behalf of the National Kidney Foundation, Inc. This is a US Government Work. There are no restrictions on its use. Download PDF Download PDF Outline Outline Abstract Index Words Introduction Toxic Alcohols Salicylate Intoxication Acetaminophen (Paracetamol) Metformin Lithium Article Information Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Abstract Index Words Introduction Toxic Alcohols Salicylate Intoxication Acetaminophen (Paracetamol) Metformin Lithium Article Information Article metrics Related Articles Abstract Poisoning is a common problem in the United States. Acid-base disturbances, electrolyte derangements, or acute kidney injury result from severe poisoning from toxic alcohols, salicylates, metformin, and acetaminophen. Lithium is highly sensitive to small changes in kidney function. These poisonings and drug overdoses often require the nephrologist’s expertise in diagnosis and treatment, which may require correction of acidosis, administration of selective enzyme inhibitors, or timely hemodialysis. The clinical and laboratory abnormalities associated with the poisonings and drug overdoses can develop rapidly and lead to severe cellular dysfunction and death. Understanding the pathophysiology of the disturbances and their clinical and laboratory findings is essential for the nephrologist to rapidly recognize the poisonings and establish an effective treatment plan. This installment of AJKD’s Core Curriculum in Nephrology presents illustrative cases of individual poisonings and drug overdoses and summarizes up to date information on their prevalence, clinical and laboratory findings, pathophysiology, diagnosis, and treatment. Index Words Acetaminophen acute kidney injury (AKI) extracorporeal toxin removal (ECTR) hemodialysis lithium metformin-associated lactic acidosis osmolal gap paracetamol poisoning review salicylate intoxication toxic alcohol FEATURE EDITOR Asghar Rastegar ADVISORY BOARD Ursula C. Brewster Michael Choi Ann O’Hare Biff F. Palmer The Core Curriculum aims to give trainees in nephrology a strong knowledge base in core topics in the specialty by providing an overview of the topic and citing key references, including the foundational literature that led to current clinical approaches. Introduction In 2019, the American Association of Poison Control Centers (AAPCC) recorded over 2.1 million cases of human exposure to poisons. Although emergency physicians, hospitalists, and toxicologists manage many of these poisonings, the nephrologist is often necessary to manage severe acid-base disorders, electrolyte abnormalities, or kidney dysfunction. Further, hemodialysis can be an important component of treatment, facilitating effective removal of the parent toxin and injurious metabolites. In this installment of AJKD’s Core Curriculum in Nephrology, we discuss the pathophysiology, diagnosis, and treatment of acute and chronic poisonings encountered by the nephrologist. These include toxic alcohol poisoning (methanol, ethylene glycol, diethylene glycol, propylene glycol, and isopropanol), salicylate intoxication, metformin-associated lactic acidosis, severe acetaminophen (paracetamol) poisoning, and lithium poisoning. Additional Readings ➢ Ghannoum M, Nolin TD, Lavergne V, Hoffman RS; EXTRIP Workgroup. Blood purification in toxicology: nephrology’s ugly duckling. Adv Chronic Kidney Dis. 2011;18(3):160-166. ★ESSENTIAL READING ➢ Gummin DD, Mowry JB, Beuhler MC, et al. 2019 Annual report of the American Association of Poison Control Centers’ National Poison Data System (NPDS): 37th annual report. Clin Toxicol (Phila.) 2020;57(12):1220-1413. ➢ Nelson LS, Howland MA, Lewin NA, et al. eds. Goldfrank’s Toxicologic Emergencies. 11th ed. McGraw-Hill;2018. Toxic Alcohols Case 1: Family members discover a 35-year-old man in his garage confused and poorly responsive. Paramedics find a suicide note and an open jug nearby. In the emergency department, he appears inebriated, with tachypnea, tachycardia, and blood pressure of 120/60 mm Hg. His blood chemistry values reveal the following concentrations: serum sodium ([Na+]), 136 mEq/L; potassium ([K+]), 3.2 mEq/L; [total CO 2], 10 mEq/L; and chloride ([Cl−]), 100 mEq/L . The calculated serum osmolarity is 290 mOsm/L, and the measured serum osmolality by freezing point depression is 350 mOsm/kg H 2 O. Question 1:What would be your next step(s) in managing this patient? a) Normal saline infusion at 150 mL/h. b) Infusion of sodium bicarbonate 150 mEq added to 1 L of 5% dextrose in water at 150 mL/h. c) Fomepizole at 15 mg/kg intravenously. d) Obtain ethylene glycol and methanol results before selecting treatment. e) Both (b) and (c). For the answer to the question, see the following text. Epidemiological Features The prevalence of toxic alcohol poisoning varies greatly from compound to compound. In the 2019 AAPCC report, isopropanol was the most frequent cause of toxic alcohol poisoning (16,000 cases reported) followed by ethylene glycol and methanol. Diethylene glycol was a rare cause, but outbreaks may occur in developing countries without strong pharmaceutical regulatory oversight. The prevalence of propylene glycol intoxication is unknown, but it is likely to be infrequent. The intoxications occur through different means, as summarized in Table 1. \| Alcohol \| MW (Da) \| Change in Serum Osmolality \| Common Sources \| Common Clinical Features \| Notable Laboratory Features \| Half-life, h \| --- --- --- \| −EtOH \| +EtOH \| \| Methanol \| 32.04 \| 3.09 mOsm/L per 10 mg/dL of alcohol \| Windshield washer fluid, carburetor cleaner, octane boosters, racing fuels, adulterated ethanol (“moonshine”) \| Abdominal pain; decreased vision, blindness; rarely, Parkinson-like features \| Osmolal gap; HAGMA \| 14-30 \| 43-96 \| \| Ethylene glycol \| 62.07 \| 1.60 mOsm/L per 10 mg/dL of alcohol \| Antifreeze, engine coolants, deicing fluids \| Inebriation, AKI \| Osmolal gap, HAGMA, calcium oxalate crystalluria, elevated lactate with point-of-care analyzer \| 2-6 \| 17-18 \| \| Isopropyl alcohol \| 60.10 \| 1.66 mOsm/L per 10 mg/dL of alcohol \| Rubbing alcohol, hand sanitizers \| Inebriation, depressed sensorium, abdominal pain \| Osmolal gap, acetonemia, spurious increase in creatinine (Jaffe reaction) \| NA \| NA \| \| Propylene glycol \| 76.09 \| 1.31 mOsm/L per 10 mg/dL of alcohol \| Diluent in parenteral medications, “nontoxic” automotive antifreeze \| Liver or kidney disease may increase likelihood of more severe toxicity \| Increased osmolal gap; lactic acidosis; rarely, AKI \| 1.4-3.3 \| 17 \| \| Diethylene glycol \| 106.12 \| 0.9 mOsm/L per 10 mg/dL of alcohol \| Automotive brake fluids, hydraulic fluids, adulterated liquid medications \| Abdominal pain, nausea/vomiting, acute pancreatitis, AKI \| Osmolal gap, HAGMA, AKI \| 4-6 \| Unknown \| Table 1 Clinical and Laboratory Features of the Toxic Alcohols Abbreviations: AKI, acute kidney injury;+EtOH, with coingested alcohol;−EtOH, without coingested alcohol; HAGMA, high anion gap metabolic acidosis; MW, molecular weight; NA, not applicable. Open table in a new tab In adults, methanol intoxication and ethylene glycol intoxication may develop after ingestion of adulterated liquids with a toxic alcohol substituted for ethanol. Methanol intoxication also occurs with ingestion or inhalation of automotive windshield-washer fluid or fuel-line antifreeze. Adults typically ingest ethylene glycol (antifreeze) in a suicide attempt. In children, most methanol and ethylene glycol ingestions unintentionally result from exploratory behavior. Isopropanol intoxication is generally the consequence of ingesting rubbing alcohol, hand sanitizer, or various industrial products. Intoxication also occurs by inhalation or absorption through dermal or rectal routes. Diethylene glycol intoxication can occur sporadically after ingestion of automotive brake fluids or industrial products. However, it most frequently occurs in outbreaks, particularly in children who have ingested consumer products or oral medications containing diethylene glycol improperly used as a diluent in lieu of propylene glycol. The earliest epidemic of diethylene poisoning due to contamination of drugs occurred in the United States in the 1930s, and it spurred the establishment of the US Food and Drug Administration (FDA). Outbreaks have subsequently occurred in several countries outside the United States including South Africa, India, Bangladesh, Haiti, and Panama. Children may be at higher risk than adults because of their smaller body surface area. Propylene glycol intoxication may occur in the hospital setting with high-dose infusions of relaxants such as lorazepam or diazepam, both of which contain 40% propylene glycol. Propylene glycol is also the principal ingredient of automotive antifreeze products marketed as being “nontoxic” or “environmentally friendly.” The actual prevalence of propylene glycol intoxication is unknown, but it is likely uncommon. Pathogenesis Except for isopropanol, the injurious effects of the toxic alcohols primarily result from accumulation of their toxic acid metabolites. Figure 1 illustrates the metabolism of the toxic alcohols. Alcohol dehydrogenase (ADH) is critical to the process, catalyzing oxidation of the toxic alcohols. This produces aldehydes (aside from acetone produced by metabolism of isopropanol) that then undergo further oxidation by aldehyde dehydrogenase to form carboxylic acid metabolites: methanol produces formic acid, ethylene glycol forms oxalic and glycolic acid, diethylene glycol forms 2-hydroxyethoxyacetic acid and diglycolic acid, and propylene glycol forms d-lactic and l-lactic acid. Figure viewer Figure 1(A) Metabolic pathways of toxic alcohols. Alcohol dehydrogenase catalyzes the first oxidation of the toxic alcohols and is an important target for antidotal therapy. The enclosed boxes highlight the putative toxic metabolites. Methanol is metabolized to formic acid, ethylene glycol to oxalic and glycolic acid, diethylene glycol to 2-hydroxyethoxyacetic acid and glycolic acid, and propylene glycol to d-lactic and l-lactic acid. (B) Time course of changes in the osmolal and anion gaps with and without coingested ethanol. An increased osmolal gap is prominent early owing to the accumulation of the un-ionized alcohols. As metabolism proceeds, the osmolal gap declines with the formation of ionized metabolites. Conversely, the serum anion gap is lowest before the alcohol is metabolized and increases with the formation of ionized metabolites. The time course of these changes in both parameters varies among the alcohols. They typically evolve over several hours to over a day. Coingested ethanol impedes metabolism (dashed lines) and delays the onset of the high anion-gap acidosis. Original graphic ©2018 Massachusetts Medical Society; reproduced with permission of the copyright holder from Kraut & Mullins, 2018 (N Engl J Med. The onset of toxicity after exposure to the toxic alcohols depends upon the rate of metabolism and the presence of coingested ethanol (Table 1). Coingestion of ethanol, the natural substrate of ADH, can markedly delay production of the toxic metabolites, making their recognition difficult. Spurious increments in blood lactate can occur in ethylene glycol poisoning due to glycolate interfering with the lactate measurement when point-of-care instruments are used. Clinical Features To varying degrees, alcohols—particularly ethylene glycol and isopropanol—produce some inebriation. Accumulation of their toxic metabolites produces organ dysfunction. Methanol intoxication frequently impairs vision and can produce permanent blindness in some cases. Pulmonary dysfunction, abdominal pain, coma, and, rarely, Parkinson-like symptoms can occur. The clinical abnormalities usually evolve over 6 to 24 hours, but coingested ethanol can delay the toxic effects. Rarely, neurologic sequelae may occur days or weeks after exposure. Ethylene glycol metabolism forms glycolic acid and then oxalate crystals. Glycolic acid is the principal cause of acidosis. Oxalate crystals produce organ dysfunction including acute kidney injury (AKI). Cranial nerve damage, sometimes delayed for days, can also occur. Typically, the signs and symptoms develop in a characteristic fashion: neurologic dysfunction develops within the first 12 hours, followed by cardiac and pulmonary dysfunction in the next 12 hours, and AKI at 48 to 72 hours after exposure. However, dysfunction of all 4 organ systems can occur concomitantly. Coingested ethanol delays the accumulation of toxic metabolites and the appearance of clinical abnormalities. Isopropanol intoxication depresses the sensorium and can cause respiratory dysfunction, cardiovascular collapse, acute pancreatitis, and hypotension-induced lactic acidosis. Serum isopropanol concentrations above 500 mg/dL (83 mmol/L) are clinically significant; those in excess of 1,500 mg/dL (250 mmol/L) result in deep coma. The major metabolite acetone can produce a spurious increase in serum creatinine concentration due to its interference with laboratory measurements using the Jaffe reaction. Diethylene glycol ingestion can cause abdominal pain, nausea, vomiting, diarrhea, acute pancreatitis, altered mental status, hepatic disease, central and peripheral neuropathy (occasionally causing quadriplegia), AKI, and death. The AKI often appears many hours after exposure (8 to 24 hours), may require hemodialysis, and is a major cause of death. Indeed, diethylene glycol poisoning should be excluded in instances in which a cohort of children present concurrently with severe AKI requiring hemodialysis. Coingestion of ethanol can delay toxicity by as much as 48 to 72 hours. Cranial nerve palsies and other neurologic complications can appear several days after exposure. Propylene glycol intoxication often leads only to an increase in the osmolal gap, but it can cause lactic acidosis and AKI. The predisposing factors are preexisting kidney disease, hepatic disease, or both. Patients receiving a continuous infusion for more than 48 hours of high-dose lorazepam (>10 mg/h), which contains 40% propylene glycol, are at higher risk. Diagnosis Because a history of the alcohol ingestion is often lacking, the presumptive diagnosis of toxic alcohol poisoning usually rests on a report of possible exposure in association with the aforementioned symptoms and physical findings and characteristic blood chemistry abnormalities. Chief among them are high serum osmolality and high anion gap metabolic acidosis. High concentrations of the parent alcohol elevate the osmolal gap (measured osmolality in mOsm/kg H 2 O minus the estimated osmolarity in mOsm/L) early in the clinical course. Accumulation of organic acid metabolites increases the serum anion gap as the poisoning progresses. The clinician must understand the utility and limitations of the serum osmolal and anion gaps for the diagnosis of toxic alcohol poisoning. The normal serum osmolality is the sum of the activity of all the osmotically active particles in the blood. When concentrations are low, the serum osmolarity and osmolality are basically equal. There are a number of formulas to estimate the osmolarity of the blood. The most common is: Serum osmolarity= (2× [Na+])+ ([SUN]/2.8)+ ([Glucose]/18) where serum osmolarity and [Na+] are in mmol/L and concentrations of serum urea nitrogen (SUN) and glucose are in mg/dL (note that the equation includes factors to convert the non-International System of Units [SI] units usually used in the United States to SI units). The 2 methods of measuring serum osmolality are freezing point depression and vapor pressure osmometry. Freezing point depression is more reliable. With vapor pressure osmometry, the volatile alcohols can evaporate, leading to a measured serum osmolality that is lower than it should be. If there are no exogenous osmotically active substances, the measured osmolality and estimated serum osmolarity should be essentially equivalent. By contrast, when there are osmotically active substances other than sodium, potassium, urea, and glucose, the measured serum osmolality should be greater than the estimated serum osmolarity (ie, an osmolal gap). But even if no other osmotically active substances are apparent, an osmolal gap averaging 8-9 mOsm/kg H 2 O is common. An osmolal gap of less than 10-15 mOsm/kg H 2 O is nondiagnostic and does not reliably exclude toxic alcohol poisoning. An osmolal gap of greater than 10-15 mOsm/kg H 2 O suggests foreign osmotically active substances in the blood. The normal osmolal gap reported by various clinical laboratories can vary substantially from 2 to 11 mOsm/kg H 2 O. These differences have important clinical implications because the baseline osmolal gap is an important influence on the actual osmolal gap found when alcohols accumulate in blood. Consider the effect of an osmolal gap of 2, 0, 5, and 9 mOsm/kg H 2 O baseline osmolal gap on the final osmolal gap noted with accumulation of different alcohols. A methanol or ethylene glycol concentration of 20 mg/dL is a widely accepted threshold for antidotal treatment. If the baseline osmolal gap were low (2 mOsm/kg H 2 O), the osmolal gap produced by a concentration of 20 mg/dL will remain below 10 mOsm/kg H 2 O. Even if the baseline osmolal gap were 5 mOsm/kg H 2 O, only methanol at a concentration of 20 mg/dL will cause the osmolal gap to exceed 10 mOsm/kg H 2 O. For diethylene glycol (which has a molecular weight of 106 Da), a concentration of 20 mg/dL will increase the osmolal gap by just 2 mOsm/kg H 2 O. A diethylene glycol concentration above 50 mg/dL would be required to raise the serum osmolality by more than 5 mOsm/kg H 2 O. In addition, as listed in Table 1, each alcohol has a different metabolic rate (the half-life is as short as 8 hours for methanol and 3 hours for ethylene glycol). As a result, if there has been substantial time since ingestion, much of the parent alcohol could have been metabolized to its organic acid metabolite, lowering the serum osmolality and thereby the osmolal gap. Thus, if the baseline osmolal gap is low or even negative, the accumulation alcohol has a high molecular weight, or significant metabolism of the alcohol has occurred, the osmolal gap may be normal. For these reasons, serum osmolal gap alone may not be a sensitive test for detecting exposure to toxic alcohol. Despite these caveats, serial changes in the osmolal gap reflect changes in the parent alcohol’s concentration, particularly during hemodialysis. An osmolal gap≤10 mOsm/kg H 2 O may be one indicator to discontinue therapy. Disorders other than toxic alcohols such as lactic acidosis, diabetic ketoacidosis, alcoholic ketoacidosis, chronic kidney disease, and sickle cell syndrome may increase in the serum osmolal gap, but it rarely exceeds 20 mOsm/kg. To this point, in one study, 77% of 341 patients from a single hospital with an elevated serum osmolal gap of 14 mOsm/kg or greater had disorders other than toxic alcohol intoxication. Similar limitations affect in the use of the serum anion gap. Although an elevated serum anion gap is frequently important in indicating that there has been a toxic alcohol exposure, an increase in the serum anion gap can be absent for several reasons. First, the span of lowest to highest serum anion gap values in a population of normal individuals is about 10 mEq/L. For patients with a baseline serum anion gap at the low end of range, the serum anion gap might remain in the normal range even in the presence of substantial accumulation of the organic acid anions. Also, the magnitude of the serum anion gap will depend on when it is sampled. If blood is sampled early after the toxic alcohol exposure before extensive metabolism of the toxic alcohol, the serum anion gap might be normal. Figure 1 illustrates the typical changes in the falling serum osmolal gap and rising serum anion gap during the course of toxic alcohol exposure. Many hospital laboratories can perform a “volatile screen” using gas chromatography to detect methanol, ethanol, isopropanol, and acetone but not ethylene glycol. However, this is not a universally available methodology. Lactate measurements by point-of-care analyzers (using lactate oxidase reaction) will give falsely elevated lactate concentrations in the presence of glycolate (acid metabolite of ethylene glycol). Lactate oxidase does not distinguish between lactate and glycolate, which differ by only 1 methyl group. This is a useful clue in cases of suspected ethylene glycol poisoning. Lactate measurements using lactate dehydrogenase (LDH) are specific for lactate. In hospitals with both methods available, a large difference between the rapid, point-of-care lactate measurement and a lactate measurement by LDH strongly suggests ethylene glycol poisoning. Automotive antifreeze usually contains fluorescein to detect radiator leaks. Case reports suggest that examination of the urine under ultraviolet light using a Wood’s lamp might detect antifreeze exposure. However, other substances in normal urine will fluoresce under ultraviolet light, so this method is highly unreliable. Oxalate crystals in the urine also suggest ethylene glycol poisoning but may be present in other conditions without ethylene glycol poisoning. The presence or absence of oxalate crystals in the urine does not reliably detect or exclude ethylene glycol poisoning. Definitive diagnosis of methanol and ethylene glycol uses high performance gas or liquid chromatography. This process is labor intensive, expensive, and not available in most clinical laboratories. Therefore, results can take several hours or even days. Thus, there is an unmet need for simple, quick, and inexpensive tests to detect and measure the toxic alcohols. One such test to detect ethylene glycol in serum is a modification of a veterinary assay using glycerol dehydrogenase to oxidize ethylene glycol to glycoaldehyde while reducing the cofactor nicotinamide adenine dinucleotide (NAD) to NADH. This has no interference by propylene glycol, ethanol, methanol, diethylene glycol, or fomepizole. It correlates closely with results using gas chromatography, can detect levels as low as 5 mg/dL, and is linear up to 300 mg/dL. This test, available from Catachem, currently lacks FDA approval. A similar enzymatic test measures methanol in serum samples with alcohol oxidase to convert methanol to formaldehyde and then formaldehyde dehydrogenase to reduce NAD to NADH. The assay is adaptable to most spectrophotometric clinical analyzers, is linear to a methanol concentration of 100 mg/dL (31 mmol/L), and shows no interference from ethanol up to a concentration of 690 mg/dL (a human is usually comatose at 400 mg/dL ethanol). This test, also from Catachem, currently lacks FDA approval. Treatment Delayed treatment of toxic alcohol poisoning inevitably results in worse outcomes. Therefore, some experts recommend early treatment when toxic alcohol poisoning is strongly suspected or there is unexplained metabolic acidosis. Treatment of toxic alcohol poisonings primarily includes antidotal use of fomepizole or ethanol (inhibitors of ADH) to delay or prevent metabolism to their toxic metabolites, and hemodialysis to remove the parent alcohol and its toxic byproducts. Figure 2 shows an algorithm consistent with our approach and recommendations from the literature. Box 1 shows indications for emergency hemodialysis. Figure viewer Figure 2 Algorithm for the diagnosis and treatment of methanol, ethylene glycol, and isopropanol intoxications. This algorithm provides an approach to the diagnosis and treatment of the 3 most common poisonings. A similar approach might be useful for diethylene glycol poisoning, although this poisoning is rare. One criterion for dialysis (serum ethylene glycol concentration > 300 mg/dL after antidote administration) reflects the practice of the second author. Conversion factors for methanol, ethylene glycol, and isopropanol in mmol/L,×0.3121,×0.1611, and×0.1664, respectively. Abbreviations: AKI, acute kidney injury; IV, intravenous; SUN, serum urea nitrogen. Original graphic ©2018 Massachusetts Medical Society; reproduced in modified form with permission of the copyright holder from Kraut & Mullins, 2018 (N Engl J Med. Box 1 Indications for Hemodialysis Based on information from www.extrip-workgroup.org and additional readings. Abbreviations: ADH, alcohol dehydrogenase; AKI, acute kidney injury; CKD, chronic kidney disease; IV, intravenous. Toxic Alcohols • Ethylene glycol or methanol concentration > 50 mg/dL without ADH inhibitor (fomepizole or ethanol) • Ethylene glycol concentration > 200-300 mg/dL with ADH inhibitor and normal kidney function • Methanol concentration > 70 mg/dL with ADH inhibitor and normal kidney function • Isopropanol concentration > 400-500 mg/dL • Any toxic alcohol: severe acidemia (pH< 7.2) or AKI Salicylate • Concentration > 7.2 mmol/L (100 mg/dL) • Concentration > 6.5 mmol/L (90 mg/dL) with AKI or CKD • Concentration > 6.5 mmol/L (90 mg/dL) after IV fluids, sodium bicarbonate, and potassium • Concentration > 5.8 mmol/L (80 mg/dL) after IV fluids, sodium bicarbonate, and potassium and with AKI or CKD • Altered mental status • Respiratory distress or new hypoxemia requiring supplemental oxygen • pH≤ 7.2 Metformin • Lactate > 10 mmol/L • pH< 7.2 • Shock • Failure of standard supportive measures (IV fluids, sodium bicarbonate) • Decreased level of consciousness Lithium • Concentration > 5.0 mEq/L • Concentration > 4.0 mEq/L with AKI or CKD • Decreased level of consciousness, seizures, or life-threatening dysrhythmias at any lithium concentration • Estimated time to reach lithium concentration< 1 mEq/L exceeds 36 hours Acetaminophen • Concentration > 1,000 mg/L (6,620 μmol/L) • Concentration > 700 mg/L (4,630 μmol/L) with altered mental status, metabolic acidosis, or elevated lactate Methanol and Ethylene Glycol Gastric decontamination is usually not helpful because the absorption of methanol and ethylene glycol in the gastrointestinal tract is so rapid. Intravenous administration of base (sodium bicarbonate) corrects metabolic acidosis and increases methanol’s ionization to formic acid. This promotes its urinary excretion and reduces its penetration into the optic nerve. Treatment guidelines recommend an ADH inhibitor when the serum methanol or ethylene glycol concentration exceeds 20 mg/dL (6 mmol/L of methanol or 3 mmol/L of ethylene glycol). Other indications include a high suspicion of toxic alcohol ingestion with either an osmolal gap greater than 10 mOsm/kg H 2 O or metabolic acidosis of unknown cause. Intravenous ethanol was the major therapy before FDA approval of fomepizole (4-methylpyrazole) 2 decades ago. It remains an alternative, particularly when fomepizole is not available. It is an effective competitive inhibitor of ADH. The target ethanol concentration is 100 mg/dL (22 mmol/L). Ethanol is generally available and inexpensive but requires compounding by a pharmacist for intravenous (IV) use. The serum ethanol concentration requires careful monitoring, so patients usually require hospitalization in the intensive care unit. It may be difficult to discern the degree of inebriation attributable to the toxic alcohol and to the ethanol used as an antidote. Fomepizole is a strong inhibitor of ADH with very high enzyme affinity (8,000 times that of ethanol). It is effective at micromolar concentrations, does not have serious side effects, and patients do not need to be monitored in an intensive care unit. The loading dose is 15 mg per kilogram of body weight and the subsequent maintenance dose is 10 mg/kg every 12 hours. Given that the drug may induce its own metabolism by cytochrome P450 enzymes, the maintenance dose is raised to 15 mg/kg every 12 hours after 48 hours. Because hemodialysis removes fomepizole, patients undergoing hemodialysis should receive it immediately after a hemodialysis session. Both fomepizole and ethanol are effective inhibitors of ADH, but there is no controlled clinical trial directly comparing their outcomes. In the United States, fomepizole is the most common antidote to treat methanol and ethylene glycol poisonings. By contrast, outside the United States, fomepizole is less readily available, and ethanol (either IV or oral) is more frequent. Although fomepizole is more expensive than ethanol, fomepizole results in lower mortality and fewer adverse effects. Oral ethanol is an effective first aid treatment until a poisoned patient can reach a hospital with access to fomepizole or hemodialysis. Methanol and ethylene glycol both have low molecular weights, high water solubility, low protein binding, and small volumes of distribution. These characteristics favor their rapid removal by extracorporeal toxin removal (ECTR). The Extracorporeal Treatments in Poisoning Workgroup (EXTRIP, www.extrip-workgroup.org) has guidelines for the use of hemodialysis in the treatment of methanol poisoning. They recommend hemodialysis with severe metabolic acidosis, serum methanol concentrations higher than 50 mg/dL (16 mmol/L), deteriorating vital signs despite supportive care, and AKI or problems with vision. Intermittent hemodialysis with a large (>2.1) surface area dialyzer and high-flux membrane is highly effective in the rapid removal of the toxic alcohols. Continuous kidney replacement therapy (CKRT) removes toxic alcohols more slowly but may be safer for hemodynamically unstable patients. Antidotal treatment of methanol or ethylene glycol intoxication without hemodialysis has been reported without adverse consequences. However, fomepizole (without hemodialysis) extends the half-lives for elimination of methanol and ethylene glycol to as long as 71 hours and 16 hours, respectively (these values are only 2.5 and 2.7 hours, respectively, with hemodialysis). Hemodialysis may shorten the total hospital stay but requires more critical care resources. The comparative costs of the 2 treatment strategies (ADH inhibition with or without hemodialysis) varies according to several factors, including absorbed dose of the toxic alcohol, drug costs, cost of hemodialysis, and room costs. Although controlled studies are not available, treatment in children is similar to that in adults. EXTRIP has reviewed ECTR for ethylene glycol. We expect that a published guideline for ECTR in ethylene glycol poisoning is forthcoming. Before the advent of fomepizole, any concentration of a toxic alcohols above 50 mg/dL warranted urgent hemodialysis in addition to antidotal treatment. Clinical experience in the 2 decades since the FDA approval of fomepizole has demonstrated that higher hemodialysis thresholds are possible when using fomepizole. This is more clearly true with ethylene glycol, which undergoes renal clearance, but less so for methanol, which has much lower renal clearance and little clearance in exhaled breath. Experts suggest that fomepizole allows a hemodialysis threshold as high as 300 mg/dL for ethylene glycol when acidosis is mild or not present. EXTRIP guidelines indicate a hemodialysis threshold of 70 mg/dL for methanol poisoning treated with fomepizole. Adjunctive treatments may promote conversion of toxic metabolites to less toxic metabolites. For ethylene glycol poisoning, thiamine and magnesium promote conversion of glyoxylate to α-ketoadipate. Pyridoxine (vitamin B 6) promotes conversion of glycolate to glycine. For methanol poisoning, high-dose folic acid or folinic acid (1 mg/kg of either) promotes conversion of formic acid to carbon dioxide and water. No prospective trials show the magnitude of these effects. Propylene Glycol Toxicity of propylene glycol is low. If hyperosmolality and lactic acidosis occur, discontinuation of the medication containing propylene glycol and administration of IV fluids are generally sufficient treatment. Fomepizole is generally unnecessary. Hemodialysis may be helpful in significant lactic acidosis with AKI. Diethylene Glycol Most experts recommend fomepizole. However, because AKI in patients with this poisoning is often severe enough to produce marked kidney failure, intermittent hemodialysis along with administration of fomepizole is often necessary. Isopropanol Generally, supportive measures are sufficient. Intermittent hemodialysis has been recommended when the serum isopropanol concentration is≥500 mg/dL (83 mmol/L), or if hypotension is present or severe lactic acidosis develops. In contrast to the other toxic alcohols, alcohol dehydrogenase inhibitors are unnecessary. Review of Case 1 Returning to question 1, (e) is the best answer. Correction of acidemia and inhibition of ADH are both early critical actions. This is true for both ethylene glycol and methanol. It has been suggested that provision of sdoium bicarbonate will increase the urinary excretion of formate and glycolate. Timely ethylene glycol concentrations are seldom available, so critical treatment must precede diagnostic certainty. The garage in the example may have antifreeze (ethylene glycol), windshield washer fluid (methanol), or brake fluid (diethylene glycol, glycol ethers). Additional Readings ➢ Hassanian-Moghaddam H, Zamani N, Roberts DM, et al. Consensus statements on the approach to patients in a methanol poisoning outbreak. Clin Toxicol (Phila.) 2019;57(12):1129-1136. ➢ Kraut JA. Diagnosis of toxic alcohols: limitations of present methods. Clin Toxicol (Phila.) 2015;53(7):589-595. ➢ Kraut JA, Kurtz I. Toxic alcohol ingestions: clinical features, diagnosis, and treatment. Clin J Am Soc Nephrol. 2008;3(1):208-225. ➢ Kraut JA, Mullins ME. Toxic alcohols. N Engl J Med. 2018;378(3):270-280. ★ESSENTIAL READING ➢ Kraut JA, Xing SX. Approach to the evaluation of a patient with an increased serum osmolal gap and high-anion-gap metabolic acidosis. Am J Kidney Dis. 2011;58(3):480-484. ➢ Roberts DM, Yates C, Megarbane B, et al. Recommendations for the role of extracorporeal treatments in the management of acute methanol poisoning: a systematic review and consensus statement. Crit Care Med. 2015;43(2): 461-472. Salicylate Intoxication Case 2: A 48-year-old man arrives by ambulance after a large dose of unknown medication. In the emergency department, he has altered mental status; blood pressure, 110/70 mm Hg; respiratory rate of 30; heart rate of 130, and temperature of 39°C. He complains of hearing a loud “buzzing” sound. His blood chemistry values reveal the following: [Na+], 135 mEq/L; [K+], 3.1 mEq/L; [total CO 2], 12 mEq/L; [Cl−], 102 mEq/L; pH 7.40; and P co 2, 18 mm Hg. Question 2:Which of the items presented above is most suggestive of salicylate intoxication rather than some other intoxication? a) Altered mental status b) Buzzing sound in his ears c) Hypokalemia d) Tachycardia e) Tachypnea For the answer to the question, see the following text. Epidemiology Salicylate intoxication can be either acute or chronic. Acute salicylate intoxication occurs after ingestion of≥100 to 150 mg/kg salicylate or ingestion of small amounts of methyl salicylate (as low as 5 mL of oil of wintergreen). Rarely, repeated topical use of topical analgesic cream (up to 30% methyl salicylate) may cause serious poisoning. Headache powders (most commonly used in the southeastern United States) provide a rapidly absorbable form of acetylsalicylic acid with caffeine. The most common source of salicylate poisoning is acetylsalicylic acid or aspirin (the latter is a trade name in some countries but a generic name in others, including the United States). Acetylsalicylic acid undergoes rapid hydrolysis to salicylate in the gastrointestinal tract, liver, and bloodstream. Acute salicylate intoxication most commonly occurs in adults taking salicylates in a suicide attempt or in children after unintentional exposure. US poison centers record approximately 25,000 salicylate exposures annually. Chronic poisoning is more common in elderly individuals. Preexisting kidney disease or compromise of kidney function produced by the salicylate itself can lead to an increase in blood salicylate concentrations and worsen toxicity. Pathogenesis Salicylates directly stimulate the respiratory center of the medulla, producing an increase in both the rate and depth of respiration resulting in respiratory alkalosis. They also uncouple oxidative phosphorylation and inhibit citric acid cycle dehydrogenases, causing a shift in metabolism due to increased glycolysis. This leads to generation of lactic acid and stimulation of hormone-sensitive lipase, leading to increased lipolysis and increased ketone production. Organic acid transporters in the proximal tubule of the kidney excrete these products, which compete for uric acid excretion. The kidney freely excretes only about 10% of unchanged salicylic acid. Clinical Features Patients with acute salicylate intoxication can present with confusion, agitation, disorientation that can progress to coma, shortness of breath, and tinnitus (or other hearing disturbances). Physical findings can include hyperventilation, evidence of volume depletion, noncardiogenic pulmonary edema, hematemesis, and petechiae. The latter may result from platelet dysfunction. A chest radiograph can reveal pulmonary opacities. Symptoms and signs can be subtle and insidious and may initially be misattributed to other causes such as sepsis. When toxicity occurs in individuals taking salicylates for treatment of chronic diseases, there may be no overt evidence of excess ingestion. Agitation, confusion, hallucinations, slurred speech, seizures, and coma appear to be more frequent in those with chronic salicylate poisoning than with acute intoxication. These patients often have a delay in diagnosis of salicylate intoxication. A delay in diagnosis and initiation of therapy may explain, in part, the higher morbidity reported for chronic intoxication than with acute intoxication. Diagnosis Prominent laboratory abnormalities include acid-base disturbances. In children, there is often an early, transient respiratory alkalosis followed by metabolic acidosis. In adults, approximately 20% will have respiratory alkalosis alone, and 56% will have combined respiratory alkalosis and high anion gap metabolic acidosis. In both groups, the respiratory alkalosis results from stimulation of the respiratory center by salicylate. Less commonly, a normal anion gap metabolic acidosis can develop due to excretion of sodium and potassium salts in the urine with subsequent retention of chloride. The normal anion gap metabolic acidosis can also result in a false hyperchloremia due to salicylate interference with the measurement of chloride. The latter effect can rarely cause a negative anion gap. Hypokalemia results from increased losses of potassium in the urine due to increased excretion of the organic acid anions, augmented aldosterone concentrations, and increased distal sodium delivery. Measurement of salicylate concentration best confirms the diagnosis. Tests for salicylate are simple and generally available in all clinical laboratories. Treatment Question 3:Which of the following is true? a) Hemodialysis is usually necessary only if the salicylate concentration is greater than 100 mg/dL. b) Tachypnea is a sign of respiratory distress, and the patient requires intubation before considering hemodialysis. c) Hemodialysis should commence as quickly as possible. d) Hemodialysis is only necessary if a trial of sodium bicarbonate fails to lower the salicylate concentration. e) Hemodialysis is only necessary if the serum potassium concentration is high in a salicylate-poisoned patient. For answer to the question, see the following text. Aggressive volume resuscitation with normal saline or lactated Ringer solution is important as fluid losses are common. However, once euvolemia is achieved, large quantities of fluid to induce forced diuresis is not recommended. Oral activated charcoal reduces further salicylate absorption when given within 1 to 2 hours of ingestion and may be useful beyond 2 hours if persistently high salicylate concentrations suggest the possibility of a gastric bezoar of acetylsalicylic acid. Salicylic acid is a weak acid (pKa 2.97), and therefore alkalinization of the blood and urine with IV sodium bicarbonate is important. Alkalinizing the serum decreases salicylate concentrations in the central nervous system. Alkalinizing the urine (target urine pH of 7.5 or greater) will increase excretion of salicylate. Increasing urine pH by 1 unit from 6.5 to 7.5 can triple urinary salicylate clearance. Oral bicarbonate should be avoided because it might enhance gastrointestinal absorption. Because bicarbonate administration can exacerbate any systemic alkalemia present, blood gases should be monitored carefully during therapy. Also, alkalemia can worsen any hypocalcemia, making monitoring of ionized calcium important. Salicylate-poisoned patients generally present with mild hypokalemia. Potassium replacement (both IV and orally) should accompany sodium bicarbonate administration. This prevents worsening hypokalemia and facilitates urinary alkalinization. Hemodialysis is the fastest and most effective method of eliminating salicylate from the body. Salicylate has low molecular weight, low volume of distribution, high water solubility, and limited protein binding. Hemodialysis should occur early when indications are present. Delaying hemodialysis increases mortality. Box 1 shows indications for emergency hemodialysis. Absolute indications for hemodialysis include a salicylate blood concentration of 90 mg/dL (6.5 mmol/L), regardless of the presence of signs or symptoms. Other indications for hemodialysis regardless of concentration include altered mental status, decreased kidney function, or acute respiratory distress. Promptly removing the drug at this stage can limit accumulation in tissue and prevent severe toxic effects. Intermittent hemodialysis is the preferred method, but hemoperfusion and CKRT are acceptable should intermittent hemodialysis not be available or if the patient is hemodynamically unstable. Observation of the patient with close monitoring of serum salicylate concentrations and blood pH are important until the patient has clinically improved and serum salicylate concentration has fallen considerably. Review of Case 2 Returning to question 2, the best answer is (b) the buzzing sound in his ears (tinnitus). The other findings are typical of salicylate poisoning but overlap with other conditions such as diabetic ketoacidosis, sepsis, methylxanthine poisoning, and alcoholic ketoacidosis. The best answer to question 3 is also (b): hemodialysis should commence as soon as possible because delays in hemodialysis can result in a fatal outcome. Tachypnea, altered mental status, and hyperthermia indicate severe toxicity in this patient. Additional Readings ➢ American College of Medical Toxicology. Guidance document: management priorities in salicylate toxicity. J Med Toxicol. 2015;11:140-152. ➢ Anderson RJ, Potts DE, Gabow PA, Rumack BH, Schrier RW. Unrecognized adult salicylate intoxication. Ann Intern Med. 1976;85(6):745-748. ➢ Gabow PA, Anderson RJ, Potts DE, Schrier RW. Acid-base disturbances in the salicylate intoxicated adult. Arch Intern Med. 1978;138(10):1481-1484. ➢ Juurlink DN, Gosselin S, Kielstein JT, et al. Extracorporeal treatment for salicylate poisoning: systematic review and recommendations from the EXTRIP Workgroup. Ann Emerg Med. 2015;266(2):165-181. ➢ McCabe DJ, Lu JJ. The association of hemodialysis and survival in intubated-salicylate-poisoned patients. Am J Emerg Med. 2017;35(6):899-903. ➢ Palmer BF, Clegg DJ. Salicylate toxicity. N Engl J Med. 2020;382(26):2544-2555. ★ESSENTIAL READING ➢ Pearlman BL, Gambhir R. Salicylate intoxication: a clinical review. Postgrad Med. 2009;121(4):162-168. Acetaminophen (Paracetamol) Case 3: A 29-year-old woman comes to the emergency department after texting a friend that she had ingested “handfuls” of acetaminophen from a container bought earlier in the day at a “big box” retailer. She arrives approximately 6 hours after ingestion. Her blood chemistry values reveal the following: [Na+], 140 mEq/L; [K+] 3.5 mEq/L; [Cl−], 100 mEq/L; [total CO 2], 10 mEq/L; [SUN], 25 mg/dL; serum creatinine concentration ([Scr]), 1.5 mg/dL (134 μmol/L); pH 7.21; and P co 2, 26 mm Hg. Her serum acetaminophen concentration is 980 mg/L (6,500 μmol/L). Her serum aspartate aminotransferase (AST) and alanine aminotransferase (ALT) are 122 and 115 IU/L, respectively. Question 4:Which one of the following regarding hemodialysis in acetaminophen overdose is true? a) Acetaminophen has a high protein binding in the plasma and is not easily dialyzable. b) Hemodialysis is only indicated if antidotal treatment with acetylcysteine fails to reduce the acetaminophen concentration. c) Indications for hemodialysis include serum acetaminophen concentration > 900 mg/L (6,000 μmol/L) or the presence of severe metabolic acidosis. d) AST activity greater than 1,000 IU/L is an indication for hemodialysis. For answer to the question, see the following text. Epidemiology Acetaminophen (paracetamol) is the most frequent pharmaceutical agent involved in human poisonings. It accounts for approximately 5% of the over 100,000 cases reported annually to US poison centers and remains a leading cause of poisoning death (overshadowed only by street drugs such as heroin or illicit fentanyl). Pathogenesis The principal toxic effects occur in the liver. Overdoses of acetaminophen saturate the glucuronide and sulfate conjugation pathways, and a larger fraction of the drug undergoes oxidation (mainly by cytochrome P450 2E1 isozyme [CYP2E1]) to N-acetyl parabenzoquinonimine (NAPQI). NAPQI is a potent oxidant that binds sulfhydryl (-SH) groups on intracellular proteins and denatures them, causing hepatocellular injury. Glutathione, an antioxidant sulfhydryl donor, binds NAPQI to prevent cellular injury. This process consumes glutathione, and hepatotoxicity results when NAPQI depletes available glutathione. Three aspects of acetaminophen poisoning may come to the attention of a nephrologist. The first is severe metabolic acidosis. Extremely large overdoses (eg, >30 g) cause early metabolic acidosis due to mitochondrial dysfunction and impaired oxidative phosphorylation developing well before the hepatic transaminase activities reach their peaks. These cases involve acetaminophen concentrations above 700 mg/L (4,600 μmol/L). The second is AKI after acute acetaminophen overdose. This occurs in approximately 2% of all acetaminophen overdoses but may occur more frequently in severe overdoses. In most cases, Scr slowly rises to a peak within 1 week of the overdose. Scr gradually returns to normal; patients seldom require hemodialysis. The third and rarest is high anion gap metabolic acidosis due to 5-oxoproline (pyroglutamic) in a small number of patients with chronic acetaminophen exposure (including therapeutic doses). The actual prevalence of this complication is unknown. Formation of 5-oxoproline reflects disordered glutathione production and metabolism. Most of these cases likely reflect pyroglutamic acidosis due to inherited enzyme deficiencies or malnutrition. In case of unexplained high anion gap metabolic acidosis (HAGMA) with concern for 5-oxoproline, a reference laboratory that screens newborns for inherited metabolic defects can detect and measure 5-oxoproline. Although some authors have suggested using N-acetylcysteine (NAC) for this problem, supportive care and discontinuing acetaminophen exposure are likely the mainstays of treatment. Clinical Findings On the first day after an acetaminophen ingestion, the patient may have nausea and abdominal pain or may be asymptomatic. Rising AST and ALT activities will become apparent on day 2, with peak values around day 3. The international normalized ratio (INR) may rise about 1 day after the rise in AST and ALT. Peak toxicity occurs around day 3 or 4. Severe cases may have hepatic encephalopathy and cerebral edema. AKI with acute tubular necrosis may appear. Following this, patients will either recover or die (unless they receive a liver transplant). Kings College Criteria for liver transplantation include a blood pH< 7.30 at any time or a composite of Scr > 3.3 mg/dL, prothrombin time >100 seconds, and severe (grade III or IV) hepatic encephalopathy. Lactic acidosis and hypoglycemia are sensitive indicators of severe hepatotoxicity. Blood lactate concentrations above 3.0 mmol/L on initial assessment or above 2.5 mmol/L after fluid resuscitation are highly sensitive for acute liver failure. Diagnosis Because there are no early signs or symptoms that reliably indicate toxicity, the diagnosis depends upon the serum acetaminophen concentration. The Rumack-Matthew nomogram determines the risk of hepatotoxicity by plotting serum acetaminophen concentration versus time after ingestion for any single serum acetaminophen concentration drawn at least 4 hours after ingestion. If the acetaminophen concentration exceeds the treatment line starting at 150 mg/L (993 μmol/L) at 4 hours, the patient should receive antidotal NAC either IV as Acetadote or orally as Mucomyst. Treatment The main therapeutic measures are supportive care and the administration of NAC, which as a sulfhydryl donor directly reduces NAPQI and repletes glutathione. Severe cases with profound metabolic acidosis and very high acetaminophen concentrations (exceeding 700 mg/L or 4,630 μmol/L) warrant hemodialysis. Acetaminophen is water soluble and has a low molecular weight (151.2 Da), low protein binding, and a low volume of distribution (0.9 L/kg). Hemodialysis removes acetaminophen and corrects acidosis. Intermittent hemodialysis is the preferred modality, but CKRT is acceptable if the patient is hemodynamically unstable or if hemodialysis is unavailable. Fomepizole, the ADH antagonist used in toxic alcohol poisoning, also blocks CYP2E1, the main enzyme that oxidizes acetaminophen to NAPQI. Case reports and case series suggest that fomepizole may be useful in patients with extremely high acetaminophen concentrations (>700 mg/L or >4,630 μmol/L) with metabolic acidosis. Review of Case 3 The best answer for question 4 is (c), the indications for hemodialysis include serum acetaminophen concentration > 900 mg/L or the presence of severe metabolic acidosis. One of the major indications for initiation of hemodialysis is a markedly elevated serum concentration of acetaminophen or the presence of a severe metabolic acidosis. Additional Readings ➢ Bernal W, Donaldson N, Wyncoll D, Wendon U. Blood lactate as an early predictor of outcome in paracetamol-induced acute liver failure: a cohort study. Lancet. 2002;359:558-563. ➢ Gosselin S, Juurlink DN, Kielstein JT, et al. Extracorporeal treatment for acetaminophen poisoning: recommendations from the EXTRIP Workgroup. Clin Toxicol (Phila.). 2014;52(8):856-867. ★ESSENTIAL READING ➢ Megarbane B, Oberlin M, Alvarez JC, et al. Management of pharmaceutical and recreational drug poisoning. Ann Intensive Care. 2020;10(1):157. ➢ Mullins ME, Jones MS, Nerenz RD, Schwarz ES, Dietzen DJ. 5-Oxoproline concentrations in acute acetaminophen overdose. Clin Toxicol (Phila.). 2020;58(1):62-64. ➢ Mullins ME, Yeager LH, Freeman WE. Metabolic and mitochondrial treatments for severe paracetamol poisoning: a systematic review. Clin Toxicol (Phila.). 2020;58:1284-1296. Metformin Case 4: A 63-year-old man comes to the emergency department reporting a headache. The patient has a history of type 2 diabetes mellitus for which he takes metformin and small doses of insulin. He is awake and responsive. His blood pressure is 110/70 mm Hg without orthostatic changes. His blood chemistry values reveal the following: [Na+], 138 mEq/L; [K+], 3.0 mEq/L; [total CO 2], 8 mEq/L; [Cl−], 100 mEq/L; [SUN], 25 mg/dL; [Scr], 2.5 mg/dL; pH, 7.15; and P co 2, 24 mm Hg. His initial lactate concentration is 15 mmol/L. A point-of-care ketone measurement (β -hydroxybutyrate) is 0.5 mmol/L. Question 5:Which of the following statements is true? a) This patient has diabetic ketoacidosis and requires insulin in addition to IV fluid resuscitation. b) Lactic acidosis only occurs in patients with metformin overdose. c) Lactic acidosis can develop in patients taking therapeutic doses of metformin. d) Hemodialysis does not remove metformin. For the answer to the question, see the following text. Epidemiology Metformin is one of the most frequently prescribed diabetic medication in the world and is first-line therapy in the treatment of type 2 diabetes mellitus. Complications appearing in as many as 20% to 30% of patients are not life threatening and usually include nausea, vomiting, and decreased appetite. Metformin-associated lactic acidosis (MALA) occurs in 3 to 10 cases per 100,000 patient-years. Mortality can be as high as 61% in some cases. Predisposing conditions include hepatic and acute or chronic kidney disease. For this reason, metformin is contraindicated in patients with an estimated glomerular filtration rate (eGFR) of≤30 mL/min/1.73 m 2 (CKD 4 or 5). Also, clinicians should be cautious in administering metformin to individuals with an eGFR≤45 mL/min/1.73 m 2 (CKD 3b) or who have potentially unstable kidney function. Pathogenesis Metformin inhibits glycerol-3-phosphate dehydrogenase and the glycerophosphate shuttle. This decreases the mitochondrial redox state and increases the cytosolic redox state, reducing conversion of lactate to pyruvate in the cytosol. Also, inhibition of the mitochondrial respiratory chain complex in peripheral tissues augments lactic acid production. Clinical Findings Gastrointestinal symptoms are common in patients with metformin-related lactic acidosis, but nonspecific signs may dominate. In severe cases, patients may have serious hemodynamic instability and depressed consciousness. Abdominal tenderness may mimic an acute abdomen. Laboratory findings include elevated lactate concentrations (>5 mmol/L) and acidemia. Because hepatic and kidney disease are predisposing factors, it is common to find evidence of both kidney and liver disease. Treatment Treatment includes sodium bicarbonate to treat the acidemia and supportive therapy to stabilize the blood pressure. Mortality approaches 40% with high blood concentrations (above 50 mg/L or 388 μmol/L), but most clinical laboratories cannot measure metformin concentrations. Early hemodialysis is the most effective therapy to remove metformin and to correct the acidosis. Box 1 shows indications for emergency hemodialysis. Intermittent hemodialysis is effective in hemodynamically stable patients with a drug clearance of 200 mL/min. CKRT is acceptable in hemodynamically challenged patients. Review of Case 4 The best answer to question 5 is (c), lactic acidosis occurs most often among patients taking therapeutic doses of metformin. Lactic acidosis does not require acute overdose of metformin. Hemodialysis removes metformin and corrects acidosis. Additional Readings ➢ Calello D, Liu KD, Wiegand TJ, et al. Extracorporeal treatment for metformin poisoning: systematic review and recommendations from the Extracorporeal Treatments in Poisoning Workgroup. Crit Care Med. 2015;43(8):1716-1730. ➢ Nguyen HL, Concepcion L. Metformin intoxication requiring dialysis. Hemodial Int. 2011;15:S68-S71. ➢ DeFronzo R, Gleming GA, Chen K, et al. Metformin-associated lactic acidosis: current perspectives on causes and risks. Metabolism. 2015;65(2):20-29. ★ESSENTIAL READING ➢ Seidowsky A, Nseir S, Houdret N, Fourrier F. Metformin-associated lactic acidosis: a prognostic and therapeutic study. Crit Care Med. 2009;37(7):2191-2196. Lithium Case 5: A 55-year-old woman arrives to the emergency department via the emergency medical service after neighbors found her with confusion. She had been well until she developed an acute diarrheal illness for the past 2 days. Her basic metabolic panel reveals AKI with a [Scr] of 2.4 mg/dL (compared with 1.2 mg/dL 1 month earlier). Her anion gap is 6 mmol/L (compared with 10 mmol/L 1 month earlier). Further history reveals that she takes lithium carbonate for bipolar disorder and that her psychiatrist increased her dose 1 month earlier. Question 6:Which of the following statements are true? a) Forced diuresis with normal saline and IV furosemide is the appropriate treatment. b) Sodium polystyrene resin (Kayexalate) is effective in removing lithium. c) The indication for hemodialysis depends solely on the serum concentration of lithium. d) Central nervous system dysfunction is an indication for hemodialysis regardless of the lithium concentration. For the answer to the question, see the following text. Lithium salts were first shown to have benefit in mania and bipolar disorder in 1949 in Australia. Subsequent studies in the 1950s to the 1970s corroborated its use. The exact mechanism of action remains incompletely understood but may involve inositol metabolism or modulation of serotonin release. Lithium (usually as lithium carbonate) has a very narrow therapeutic range (serum lithium concentration [Li+] usually between 0.6 and 1.3 mmol/L) and is sensitive to modest changes in kidney function. Acute-on-chronic lithium toxicity most often results from AKI from other causes (such as dehydration from diarrhea) or from rapid escalation of the dose. Acute overdose can rapidly produce high lithium concentrations. Hemodialysis effectively clears lithium, which has a molecular weight of 7 Da. Its volume of distribution is near 1 L/kg. Due to high intracellular concentrations, [Li+] often rebounds after hemodialysis due to redistribution from the intracellular space. Indications for hemodialysis depend upon the [Li+], kidney function, and neurological symptoms. The EXTRIP Workgroup recommends hemodialysis when [Li+] > 4.0 mEq/L or if the patient has a decreased level of consciousness, seizures, or life-threatening dysrhythmias, regardless of the [Li+]. CKRT is an acceptable alternative if hemodialysis is unavailable or inadvisable. When measuring [Li+], using the wrong tube produces an unexpectedly high apparent concentration. Green-top tubes (both Kelly green and mint green) contain lithium heparin. The heparin prevents coagulation of blood in the analyzer. Heparin has many negative charges and requires cations to neutralize the charge without interfering with chemistry measurements. Because lithium is an unmeasured cation in the basic metabolic panel, it is the cation used (instead of sodium) to neutralize the charge on heparin in the tubes. If the laboratory uses a green-top tube to measure lithium, the lithium heparin will produce a factitious elevation in the apparent [Li+]. The magnitude of the apparent [Li+] varies inversely with the amount of blood in the tube, but the range of concentrations overlaps the toxic range. Blood specimens for lithium measurement should go in red-top tubes. Returning to question 6, (d) is the best answer. Central nervous system dysfunction is an indication for hemodialysis regardless of [Li+]. This is one EXTRIP criterion for hemodialysis. Other EXTRIP criteria include [Li+] > 5.0 mmol/L, [Li+] > 4.0 with AKI, estimated time to [Li+]< 1.0 mmol/L exceeds 36 hours, confusion, seizure, or cardiac arrhythmia. Additional Readings ➢ Baird-Gunning J, Lea-Henry T, Hoegberg LCG, et al. Lithium poisoning. J Intensive Care Med. 2017;32(4):249-263. ➢ Buckley, NA, Cheng, S, Isoardi, K, et al. Haemodialysis for lithium poisoning: translating EXTRIP recommendations into practical guidelines. Br J Clin Pharmacol. 2020;86:999–1006. ➢ Decker B, Goldfarb DS, Dargan PI, et al. Extracorporeal treatment for lithium poisoning: systematic review and recommendations from the EXTRIP Workgroup. Clin J Am Soc Nephrol. 2015;10(5):875-887. ★ESSENTIAL READING ➢ Timmer RT, Sands JM. Lithium intoxication. J Am Soc Nephrol. 1999;10: 665-674. Article Information Authors’ Full Names and Academic Degrees Michael E. Mullins, MD, and Jeffrey A. Kraut, MD. Support This review was supported in part by unrestricted funds from UCLA School of Medicine (to JAK). Financial Disclosure The authors declare that they have no relevant financial interests. Peer Review Received March 9, 2021, in response to an invitation from the journal. Evaluated by 3 external peer reviewers and a member of the Feature Advisory Board, with direct editorial input from the Feature Editor and a Deputy Editor. Accepted in revised form June 10,2021. 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14753	https://www.vedantu.com/question-answer/colligative-properties-are-applicable-to-aideal-class-11-chemistry-cbse-61161cac8965da43e2a8a754	Talk to our experts 1800-120-456-456 Colligative properties are applicable to- A.Ideal dilute solutions B.Concentrated solutions C.Non ideal solutions D.All of these Repeaters Course for NEET 2022 - 23 © 2025.Vedantu.com. All rights reserved
14754	https://pubmed.ncbi.nlm.nih.gov/28718516/	Control of the Fluid Viscosity in a Mock Circulation - 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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Control of the Fluid Viscosity in a Mock Circulation Stefan Boës1,Gregor Ochsner1,Raffael Amacher2,Anastasios Petrou1,Mirko Meboldt1,Marianne Schmid Daners1 Affiliations Expand Affiliations 1 Product Development Group Zurich, Department of Mechanical and Process Engineering, ETH Zurich, Zurich, Switzerland. 2 Wyss Zurich, ETH Zurich and University of Zurich, Zurich, Switzerland. PMID: 28718516 DOI: 10.1111/aor.12948 Item in Clipboard Control of the Fluid Viscosity in a Mock Circulation Stefan Boës et al. Artif Organs.2018 Jan. Show details Display options Display options Format Artif Organs Actions Search in PubMed Search in NLM Catalog Add to Search . 2018 Jan;42(1):68-77. doi: 10.1111/aor.12948. Epub 2017 Jul 17. Authors Stefan Boës1,Gregor Ochsner1,Raffael Amacher2,Anastasios Petrou1,Mirko Meboldt1,Marianne Schmid Daners1 Affiliations 1 Product Development Group Zurich, Department of Mechanical and Process Engineering, ETH Zurich, Zurich, Switzerland. 2 Wyss Zurich, ETH Zurich and University of Zurich, Zurich, Switzerland. PMID: 28718516 DOI: 10.1111/aor.12948 Item in Clipboard Full text links Cite Display options Display options Format Abstract A mock circulation allows the in vitro investigation, development, and testing of ventricular assist devices. An aqueous-glycerol solution is commonly used to mimic the viscosity of blood. Due to evaporation and temperature changes, the viscosity of the solution drifts from its initial value and therefore, deviates substantially from the targeted viscosity of blood. Additionally, the solution needs to be exchanged to account for changing viscosities when mimicking different hematocrits. This article presents a method to control the viscosity in a mock circulation. This method makes use of the relationship between temperature and viscosity of aqueous-glycerol solutions and employs the automatic control of the viscosity of the fluid. To that end, an existing mock circulation was extended with an industrial viscometer, temperature probes, and a heating nozzle band. The results obtained with different fluid viscosities show that a viscosity controller is vital for repeatable experimental conditions on mock circulations. With a mixture ratio of 49 mass percent of aqueous-glycerol solution, the controller can mimic a viscosity range corresponding to a hematocrit between 29 and 42% in a temperature range of 30-42°C. The control response has no overshoot and the settling time is 8.4 min for a viscosity step of 0.3 cP, equivalent to a hematocrit step of 3.6%. Two rotary blood pumps that are in clinical use are tested at different viscosities. At a flow rate of 5 L/min, both show a deviation of roughly 15 and 10% in motor current for high rotor speeds. The influence of different viscosities on the measured head pressure is negligible. Viscosity control for a mock circulation thus plays an important role for assessing the required motor current of ventricular assist devices. For the investigation of the power consumption of rotary blood pumps and the development of flow estimators where the motor current is a model input, an integrated viscosity controller is a valuable contribution to an accurate testing environment. Keywords: Ventricular assist device; -Blood pump; -Flow estimator; -Flow probe calibration; -Viscosity controller. © 2017 International Center for Artificial Organs and Transplantation and Wiley Periodicals, Inc. PubMed Disclaimer Similar articles The effect of fluid viscosity on the hemodynamic energy changes during operation of the pulsatile ventricular assist device.Ahn CB, Son KH, Lee JJ, Choi J, Song SJ, Jung JS, Lee SH, Son HS, Sun K.Ahn CB, et al.Artif Organs. 2011 Nov;35(11):1123-6. doi: 10.1111/j.1525-1594.2011.01350.x. Epub 2011 Sep 29.Artif Organs. 2011.PMID: 21954946 Noninvasive average flow estimation for an implantable rotary blood pump: a new algorithm incorporating the role of blood viscosity.Malagutti N, Karantonis DM, Cloherty SL, Ayre PJ, Mason DG, Salamonsen RF, Lovell NH.Malagutti N, et al.Artif Organs. 2007 Jan;31(1):45-52. doi: 10.1111/j.1525-1594.2007.00339.x.Artif Organs. 2007.PMID: 17209960 Sensorless flow and head estimation in the VentrAssist rotary blood pump.Ayre PJ, Vidakovic SS, Tansley GD, Watterson PA, Lovell NH.Ayre PJ, et al.Artif Organs. 2000 Aug;24(8):585-8. doi: 10.1046/j.1525-1594.2000.06586.x.Artif Organs. 2000.PMID: 10971241 Blood viscosity in tube flow: dependence on diameter and hematocrit.Pries AR, Neuhaus D, Gaehtgens P.Pries AR, et al.Am J Physiol. 1992 Dec;263(6 Pt 2):H1770-8. doi: 10.1152/ajpheart.1992.263.6.H1770.Am J Physiol. 1992.PMID: 1481902 Review. Mock circulatory test rigs for the in vitro testing of artificial cardiovascular organs.Shi Y, Yang H.Shi Y, et al.J Med Eng Technol. 2019 May;43(4):223-234. doi: 10.1080/03091902.2019.1653390. Epub 2019 Aug 29.J Med Eng Technol. 2019.PMID: 31464556 Review. See all similar articles Cited by Ultrasound vector flow imaging during veno-arterial extracorporeal membrane oxygenation in a thoracic aorta model.Yambe K, Ishii T, Yiu BYS, Yu ACH, Endo T, Saijo Y.Yambe K, et al.J Artif Organs. 2024 Sep;27(3):230-237. doi: 10.1007/s10047-023-01413-z. Epub 2023 Jul 20.J Artif Organs. 2024.PMID: 37474830 Free PMC article. Development of Inspired Therapeutics Pediatric VAD: Benchtop Evaluation of Impeller Performance and Torques for MagLev Motor Design.Tompkins LH, Prina SR, Gellman BN, Morello GF, Roussel T, Kopechek JA, Williams SJ, Petit PC, Slaughter MS, Koenig SC, Dasse KA.Tompkins LH, et al.Cardiovasc Eng Technol. 2022 Apr;13(2):307-317. doi: 10.1007/s13239-021-00578-z. Epub 2021 Sep 13.Cardiovasc Eng Technol. 2022.PMID: 34518953 Free PMC article. Evaluating compliance in three-dimensional-printed polymeric vascular grafts compared to human arteries and commercial grafts in a mock circulation loop compliance in three-dimensional-printed polymeric vascular grafts.Hong W, Tewari V, Yu H, Chen J, Sawchuk AP.Hong W, et al.JVS Vasc Sci. 2025 May 23;6:100291. doi: 10.1016/j.jvssci.2025.100291. eCollection 2025.JVS Vasc Sci. 2025.PMID: 40641618 Free PMC article. A Review of the Advancements in the in-vitro Modelling of Acute Ischemic Stroke and Its Treatment.Johnson S, Dwivedi A, Mirza M, McCarthy R, Gilvarry M.Johnson S, et al.Front Med Technol. 2022 Jun 8;4:879074. doi: 10.3389/fmedt.2022.879074. eCollection 2022.Front Med Technol. 2022.PMID: 35756535 Free PMC article.Review. Mechanical and functional validation of a perfused, robot-assisted partial nephrectomy simulation platform using a combination of 3D printing and hydrogel casting.Melnyk R, Ezzat B, Belfast E, Saba P, Farooq S, Campbell T, McAleavey S, Buckley M, Ghazi A.Melnyk R, et al.World J Urol. 2020 Jul;38(7):1631-1641. doi: 10.1007/s00345-019-02989-z. Epub 2019 Nov 2.World J Urol. 2020.PMID: 31679063 Free PMC article. 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14755	https://mindyourdecisions.com/blog/2016/07/12/the-pirate-game-game-theory-tuesdays/	The Pirate Game – Game Theory Tuesdays – Mind Your Decisions Skip to content Mind Your Decisions Math Videos, Math Puzzles, Game Theory. By Presh Talwalkar Menu and widgets About Me: Presh Talwalkar I run the MindYourDecisions channel on YouTube, which has over 2 million subscribers and 400 million views. I am also the author of The Joy of Game Theory: An Introduction to Strategic Thinking, and several other books which are available on Amazon. (As you might expect, the links for my books go to their listings on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.) By way of history, I started the Mind Your Decisions blog back in 2007 to share a bit of math, personal finance, personal thoughts, and game theory. It's been quite a journey! 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(I send it 1 or 2 times a year, and I only collect your email to send this news). You can sign up for the newsletter here: Support Most YouTube channels of my size have a staff of 5 people with a large budget and sponsors. As of 2019, I make most of the videos myself and have declined all sponsors. This has only been possible thanks to tremendous support from everyone that watches and shares my videos and blog posts. I am also grateful to donations through Patreon. Right now MindYourDecisions is going ad-free on new blog posts thanks to generous support from patrons. It costs thousands of dollars to run the blog and your support matters. If you like the posts and videos, please consider a monthly pledge on Patreon. You may also consider a one-time donation to support my work. I post to the following sites, mostly with updates for new content. My Blog YouTube Facebook Twitter Instagram Patreon Contact me by email: presh@mindyourdecisions.com Send me your favorite puzzles/suggestions by email. I get so many emails that I may not reply, but I save all suggestions. I am not pursuing sponsors/guest posts/partnerships at this time. But I may in the future, and feel free to email me if there's an offer I couldn't possibly pass up ;) The Pirate Game – Game Theory Tuesdays If you buy from a link in this post, I may earn a commission. This does not affect the price you pay. As an Amazon Associate I earn from qualifying purchases. Learn more. Posted July 12, 2016 By Presh Talwalkar. Read about me, or email me. image by Elliott Brown. CC BY 2.0 This is a fun puzzle I have posted about before with 3 pirates. This is a variation with 5 pirates accompanied by a video explanation. A group of 5 pirates is dividing up 500 gold coins. How will they split the treasure? The pirates are a disciplined and logical group, and they have a custom of how to split up treasure. The 5 pirates (A, B, C, D, and E) have a strict organization by strength: pirate A is the strongest, followed by B, then C, then D, and then E. The voting process is a series of proposals with a lethal twist. Here are the rules: The strongest pirate (A) offers a split of the gold. An example split is: “450 to A (me), 10 to B, 20 to C, 10 to D, and 10 to E.” All of the pirates, including the proposer, vote on whether to accept the split. In the case of a tie, the proposer holds the casting vote to break the tie. If the pirates agree to the split, it happens. Otherwise, the pirate who proposed the plan gets thrown overboard from the ship and perishes. The next strongest pirate takes over and then offers a split of the money. The process is repeated until a proposal is accepted. Pirates care first and foremost about living, then about getting gold. If a pirate is indifferent between saying “Yes” and “No” to a split in terms of money, then the pirate votes “No” because the pirate prefers to eliminate stronger pirates. How does the game play out? . . "All will be well if you use your mind for your decisions, and mind only your decisions." It costs thousands of dollars to run a website and your support matters. If you like the posts and videos, please consider a monthly pledge on Patreon. You may also consider a one-time donation to support my work. . . Video explanation The YouTube channel EckoChamb3r offers a careful and detailed explanation of how to find the solution. The Pirate Problem – Famous Game Theory Puzzle If you cannot watch the video now, keep reading for an explanation. Solution to game The game can be solved by backwards induction. You want to think ahead and reason backwards. Start at the end of the game. What would happen if the game continued so only pirate E remained? This is a trivial case as pirate E would take all 500 coins for himself. Now reason backwards one step. What would happen if the game got to pirate D proposing a split? Similarly, pirate D would take all 500 coins for himself. While pirate E would oppose the split, pirate D is in favor, so the vote would be tied at 1-1. Pirate D could then cast the tie-breaking vote and make the proposal go through. Now comes the interesting part when we reason one more step backwards. What would happen if pirate C is offering the split? Pirate C needs to buy 1 vote to make the plan go through. If pirate C dies, then pirate D would take all 500 coins and pirate E ends up with nothing. All the pirates know this. This presents an opportunity to buy the vote of pirate E. Pirate C does not take all 500 coins for himself. Instead pirate C offers 1 coin to pirate E. Now pirate E can either vote for this split and get 1 coin, or pirate E can vote against it which leads to getting nothing when pirate D is in charge. So pirate E prefers this plan and would vote for it. Therefore, pirate C would offer, “499 to C, 0 to D, and 1 to E.” Pirates C and E would vote for this plan and it would go through. Now let’s reason one more step. What would happen if pirate B was in charge? Pirate B similarly needs to buy 1 vote. The easiest vote to buy is pirate D, who ends up with nothing if the split fails and pirate C ends up in charge. Accordingly, pirate B would offer, “499 to B, 0 to C, 1 to D, and 0 to E.” Pirates B and D vote affirmatively against C and E, and pirate B holds the tie-breaking vote so the plan goes through. Now we return to the original situation. What does pirate A do? Pirate A needs to buy two votes in order to make a proposal pass. If pirate A dies, then pirate B ends up in charge and that would leave pirates C and E with nothing. Pirate A can therefore buy the votes of pirates C and E by offering each 1 coin. Pirate A offers, “498 to A, 0 to B, 1 to C, 0 to D, and 1 to E.” Pirates A, C, and E vote for the plan and it passes. The surprising part of the problem is the strongest pirate can end up with most of the money, even though the other pirates hold the power to toss him overboard. The reason is some of the weaker pirates can be bought off for very little, knowing they would end up with nothing if the original split failed. Game theory in The Dark Knight One of my first posts and videos is about how this relates to the opening scene of the Dark Knight. The Joker plans a bank heist and uses similar planning to buy off weaker criminals. The scene plays out quite like the game theory outcome. Blog post: Game Theory in The Dark Knight: The Bank Robbery And The Pirate Game (Spoilers) Video: Dark Knight Game Theory: The Robbery Scene And The Pirate Game On a concluding note, the pirate game can be extended to more pirates. What would happen if 202 pirates are dividing up 100 gold coins? What would happen if there are 500 pirates dividing up 100 gold coins? It is a crazy logical problem to figure out what happens here. Check out the puzzle by Stephen Omohundro in Ian Stewart’s article “A Puzzle For Pirates”: A Puzzle For Pirates (500 pirates with 100 coins) Published by PRESH TALWALKAR I run the MindYourDecisions channel on YouTube, which has over 1 million subscribers and 200 million views. I am also the author of The Joy of Game Theory: An Introduction to Strategic Thinking, and several other books which are available on Amazon. (As you might expect, the links for my books go to their listings on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.) By way of history, I started the Mind Your Decisions blog back in 2007 to share a bit of math, personal finance, personal thoughts, and game theory. It's been quite a journey! I thank everyone that has shared my work, and I am very grateful for coverage in the press, including the Shorty Awards, The Telegraph, Freakonomics, and many other popular outlets. I studied Economics and Mathematics at Stanford University. People often ask how I make the videos. Like many YouTubers I use popular software to prepare my videos. You can search for animation software tutorials on YouTube to learn how to make videos. Be prepared--animation is time consuming and software can be expensive! Feel free to send me an email presh@mindyourdecisions.com. I get so many emails that I may not reply, but I save all suggestions for puzzles/video topics. MY BOOKS If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay. Book ratings are from January 2025. (US and worldwide links) Mind Your Decisions is a compilation of 5 books: (1) The Joy of Game Theory: An Introduction to Strategic Thinking (2) 40 Paradoxes in Logic, Probability, and Game Theory (3) The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias (4) The Best Mental Math Tricks (5) Multiply Numbers By Drawing Lines The Joy of Game Theory shows how you can use math to out-think your competition. (rated 4.2/5 stars on 564 reviews) 40 Paradoxes in Logic, Probability, and Game Theory contains thought-provoking and counter-intuitive results. (rated 4.2/5 stars on 81 reviews) The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 4.2/5 stars on 55 reviews) The Best Mental Math Tricks teaches how you can look like a math genius by solving problems in your head (rated 4.3/5 stars on 148 reviews) Multiply Numbers By Drawing Lines This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.5/5 stars on 57 reviews) Mind Your Puzzles is a collection of the three "Math Puzzles" books, volumes 1, 2, and 3. The puzzles topics include the mathematical subjects including geometry, probability, logic, and game theory. Math Puzzles Volume 1 features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 138 reviews. Math Puzzles Volume 2 is a sequel book with more great problems. (rated 4.2/5 stars on 45 reviews) Math Puzzles Volume 3 is the third in the series. 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14756	https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_4?srsltid=AfmBOopoCySYMl8u2Z9iwM6PM3XaCDbltrejQPHSeTZ66J1XizeKgQAJ	Art of Problem Solving 1969 Canadian MO Problems/Problem 4 - 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 1969 Canadian MO Problems/Problem 4 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1969 Canadian MO Problems/Problem 4 Problem Let be an equilateral triangle, and be an arbitrary point within the triangle. Perpendiculars are drawn to the three sides of the triangle. Show that, no matter where is chosen, . Solution 1 Let a side of the triangle be and let denote the area of Note that because Dividing both sides by , the sum of the perpendiculars from equals (It is independant of point ) Because the sum of the sides is , the ratio is always Solution 2 Let a be the side of the triangle. By Viviani's theorem (which has a similar proof to solution 1) PD + PE + PF = h where h is the height of the equilateral triangle. Now it is not hard to see that the ratio h: 3 a = 1 / 2 root 3 from 30 - 60 - 90 triangles or trigonometry. ~AK2006 1969 Canadian MO (Problems) Preceded by Problem 31•2•3•4•5•6•7•8•Followed by Problem 5 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.
14757	https://literariness.org/2020/07/05/analysis-of-t-s-eliots-what-is-a-classic/	Literary Theory and Criticism Home › Analysis of T.S. Eliot’s What Is a Classic? Analysis of T.S. Eliot’s What Is a Classic? By NASRULLAH MAMBROL on July 5, 2020 • ( 1 ) This essay was presented in 1944 as the Presidential Address to the Virgil Society, then published by Faber & Faber in 1945, and finally collected in On Poetry and Poets in 1956. Eliot begins his remarks by moving straight to the point. SYNOPSIS By classic, Eliot means a work that reflects the maturity of a culture. Indeed, he argues that “[a] classic can occur only when a civilization is mature; when a language and a literature are mature; and it must be the work of a mature mind.” Eliot had at this same time been preparing the preliminary essays from which his Notes towards the Definition of Culture would eventually emerge in 1948, and he had also discussed in the 1943 essay “The Social Function of Poetry” the integral relationship that exists between a people and their national language. It is no doubt that it is these same considerations that are now making him identify what is called a classic with the maturation of a people and of their language as they are realized in a single work, itself the product of a mind capable of wholly embracing that same measure of cultural maturity. Given the context of his remarks—an address to the members of the Virgil Society—it makes further sense that, along with the obligatory passing reference to William Shakespeare, Eliot should use as his model the first-century B.C. Roman poet Virgil, whose epic of the founding of Rome, the Aeneid, is one of the great masterpieces of the Classical Age. That, however, is not what makes Virgil’s epic poem a classic. Using the Aeneid as his running example, Eliot requires that a classic foremost cannot be manufactured with that aim in mind: “it is only by hindsight, and in historical perspective, that a classic can be known as such.” By the same token, there has to be a history behind it; that is to say, the literature of a people and their culture must have arrived at a point that the writer of genius has already in place the tools and traditions by which a classic can be achieved. Furthermore, with Virgil’s Rome still as his model, Eliot observes that a common literary style must also have emerged because the “society has achieved a moment of order and stability, of equilibrium and harmony.” Out of that kind of a culture can come that final ingredient: maturity of mind. That sort of maturity, Eliot feels, requires both “history, and the consciousness of history.” In other words, a Virgil must find all these conditions available to him but must also be able to avail himself of them. Thereby, the characters and situations that Virgil manipulates are not in any manner provincial but ready, as it were, to step onto the world stage. That these conditions obtained for Virgil is shown in the expansive way that he operates his material, one that “never appears to be according to some purely local or tribal code of manners: it is in its time, both Roman and European.” The result is not just great poetry by a great poet but a classic. The great poet, such as a Shakespeare or a Milton, may exhaust a form as it has come to be developed in that culture and for that language. When, by contrast, the poet in question is, like Virgil, a great classic poet, “he exhausts, not a form only, but the language of his time; and the language of his time, as used by him, will be the language in its perfection.” It follows, then, that the great classic poet will ultimately “express the maximum possible of the whole range of feeling which represents the character of the people who speak that language.” If, however, the resulting work is truly to achieve the status of a classic, it must not only have seized the right historical moment and exhausted the possibilities of the language but transcended even purely literary values. “If Virgil is thus the consciousness of Rome and the supreme voice of her language, he must have a significance for us which cannot be expressed wholly in terms of literary appreciation and criticism.” What such a poet leaves behind, rather than a critical legacy, is itself a criterion by which other works in its category may be judged. That would indeed be the very definition of a classic—that it is so much a product of its time, place, and cultural, linguistic, and historical conditions that it does not so much exhaust its form as set a new standard for what that form might achieve, should all the other conditions be propitious. In Virgil’s case, however, Eliot does not stop there. Because Virgil’s language happened to have been Latin, which, thanks to the influence of the Roman Imperium that Virgil himself celebrated, “came to be the universal means of communication between peoples of all tongues and cultures,” Virgil writes in a language to which no modern language can ever hope to aspire in terms of that very universality. As a result, no modern language, Eliot asserts, can achieve a classic on the order of Virgil’s. Indeed, on the basis of his preceding argument, he can say with some confidence that “[o]ur classic, the classic of all Europe, is Virgil.” Eliot concludes by strongly implying that in celebrating Rome and by placing Roman culture and values and language at the center of human history, Virgil unconsciously paved the way for a new epoch in history. His vision of a common order, an ideal harmony, for humanity “led Europe towards the Christian culture which he [Virgil] could never know.” CRITICAL COMMENTARY Eliot’s rationale for such broad claims is explained in detail in a subsequent essay, “Virgil and the Christian World,” which Eliot presented as a radio address on the BBC in 1951. There he makes a convincing case that Virgil’s Roman virtues found a hospitable soil in the ethics that the teachings of Christ inspired. It may seem that Eliot, who had started out “What Is a Classic” by making rather modest claims for his intentions, ends with extravagant ones instead. In fact, however, he goes from arguing that a classic must, in effect, summarize a whole people to establishing that classics are, in and of themselves, a summarization of even greater cultural and historical developments. Eliot also was likely aware that any reference on his part to “classic” might call up memories of the romanticism versus classicism debate that engaged much English literary thinking during the 1910s and the 1920s and in which Eliot himself had been a passionate partisan on the side of the Classicist agenda. This debate, in which Eliot most famously took issue with J. Middleton Murry, whom he characterized as marching under the banner of “Muddle Through,” centered primarily around the larger issue of the place and importance of tradition in the face of the constant, rapid, and dramatic social change that, in turn, characterized the modern scene at that time. In prose works as early as the 1923 essay “The Function of Criticism” and as late as his book-length diatribe, After Strange Gods, which was subtitled A Primer of Modern Heresy and published in 1934, Eliot had pretty much excoriated those whom he perceived to be representing the enemy camp, either in their professed thinking or their creative endeavors. Even from this vantage point, the matter would not necessarily strike any informed person as one to be taken lightly. Whether a society founds itself on long-established values or on self-corrective evaluations of contemporary needs remains a source for intellectual conflict. In the case of Eliot, then, who had in 1928 declared himself a Catholic, a classicist, and a royalist—that is to say, a traditionalist or conservative on all counts—his defense of valued traditions and traditional values was not petulance but a moral imperative. By 1944, however, the time of these present remarks, England along with most of the rest of Europe and virtually the entire globe as well had been engaged for five years in that armed conflict known to history as World War II. It would be hard to imagine anyone still harboring old intellectual animosities in any life-and-death struggle such as global warfare portends. At the very least, the far more pressing requirements of that conflict seem to have had an ameliorative effect on Eliot’s intellectual and moral largesse when it came to principled conflicts. If his definition of a classic as a summary work holds true, then it would be equally true that, for European history and culture, no classic can ever equal Virgil’s Aeneid for the simple reason that Europe would never again realize such cultural and linguistic cohesion as it did during the time of Caesar Augustus, whose reign Virgil celebrates. Eliot, who had been fighting a holding action to maintain the coherence of Christian Europe in the face of 20th-century secularism and who was writing in the midst of a European conflict that would, by its conclusion in May 1945, leave 50 million dead, knew whereof he was speaking. Share this: Like this: Categories: Uncategorized Tags: American Literature, Analysis of Eliot's What Is a Classic?, Analysis of T.S. Eliot’s What Is a Classic?, Analysis of What Is a Classic?, Bibliography of T.S. Eliot’s What Is a Classic?, Bibliography of What Is a Classic?, Character Study of T.S. Eliot’s What Is a Classic?, Character Study of What Is a Classic?, Criticism of T.S. Eliot’s What Is a Classic?, Criticism of What Is a Classic?, eliot, Eliot’s What Is a Classic?, Essays of T.S. Eliot’s What Is a Classic?, Essays of What Is a Classic?, Literary Criticism, Literary Theory, Modernism, Notes of T.S. Eliot’s What Is a Classic?, Notes of What Is a Classic?, Plot of T.S. Eliot’s What Is a Classic?, Plot of What Is a Classic?, Poetry, Simple Analysis of T.S. Eliot’s What Is a Classic?, Simple Analysis of What Is a Classic?, Study Guides of T.S. Eliot’s What Is a Classic?, Study Guides of What Is a Classic?, Summary of Eliot's What Is a Classic?, Summary of T.S. Eliot’s What Is a Classic?, Summary of What Is a Classic?, Synopsis of T.S. Eliot’s What Is a Classic?, Synopsis of What Is a Classic?, T. S. Eliot, Themes of T.S. Eliot’s What Is a Classic?, Themes of What Is a Classic?, What Is a Classic? Related Articles 1 reply This was a great read. The undersigned of a Classic, like Literature itself, has evolved over time and that has happened through open and free flowing discourse such as exactly this piece. Thank you! Leave a Reply Cancel reply You must be logged in to post a comment. Blog Stats Articles/Ebooks/Lectures Scholarly Articles Categories Archives Top Articles You must be logged in to post a comment.
14758	https://www.youtube.com/watch?v=f5qr1lH6ViU	Definite integral of piecewise function \| AP Calculus AB \| Khan Academy Khan Academy 9090000 subscribers 359 likes Description 93450 views Posted: 28 Jul 2016 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Sal evaluates definite integral of a piecewise function over an interval that goes through the two cases of the function. Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? AP Calculus AB on Khan Academy: Bill Scott uses Khan Academy to teach AP Calculus at Phillips Academy in Andover, Massachusetts, and heÕs part of the teaching team that helped develop Khan AcademyÕs AP lessons. Phillips Academy was one of the first schools to teach AP nearly 60 years ago. About Khan Academy: Khan Academy is a nonprofit with a mission to provide a free, world-class education for anyone, anywhere. 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Forever. #YouCanLearnAnything Subscribe to Khan AcademyÕs AP Calculus AB channel: Subscribe to Khan Academy: 14 comments Transcript: so we have an F ofx right over here and it's defined piecewise for X less than zero f ofx is x + one for X is greater than or equal to zero f ofx is cosine of pix and we want to evaluate the definite integral from 1 to one of f ofx DX and you might immediately say well which of these which of these versions of f ofx am I going to take the anti-derivative from because from negative 1 to 0 I would think about x + 1 but then from 0 to 1 I would think about cine pix and if you were thinking that you're thinking in the right direction and the way that we can make this a little bit more straightforward is to actually split up this definite integral this is going to be equal to the definite integral from -1 to 0 of f ofx DX plus plus the integral from 0 to one of f ofx DX now why was it useful for me to split it up this way and in particular to split it up split the interval from Nega - 1 to one split into two intervals from 1 to 0 and 0 to 1 well I did that because 0 x equals 0 is where we switch where F ofx switches from being x + 1 to cine pix so if you look at the interval from -1 to 0 f ofx is x + 1 so F ofx here is x + 1 and then when you go from 0 to 1 f ofx is cosine pix so cosine of cosine of pix and so now we just have to evaluate each of these separately and add them together so let's take the definite integral from -1 to 0 of x + 1 DX well let's see the anti-derivative of X+ one is anti-derivative of X is x^2 over 2 I'm just incrementing the exponent and then dividing by that value and then plus X and you could view it as I'm doing the same thing if this is X to the 0o it'll be x to the 1st x to the 1st over one which is just X and I'm going to evaluate that at zero and subtract from that it evaluated at one sorry it evaluated at negative 1 and so this is going to be equal to if I evaluate it at zero let me do this in another color if I evaluate it at zero it's going to be 0^ 2ar over 2 which is well I'll just write it 0^ 2ar over 2 + 0 well all of that's just going to be equal to 0er minus it evaluated at it evaluated at -1 so minus -1 21^ 2 over 2 + -1 so 1^ 2ar is just 1 so it's 12 +1 12 +1 is or 1 12 minus 1 is ne2 so all of that is -2 but then we're subtracting - 1/2 0 - 12 is going to be equal to positive2 so this is going to be equal to positive2 so this first part right over here is positive one2 and now let's evaluate the integral from 0er to one of cosine Pi I don't need that first parenthesis of cosine of pix DX what is this equal to now now if we were just trying to find the anti-derivative of cosine of x it's pretty straightforward we know that the derivative with respect to X of sin of X is equal to cosine of x cosine of x but that's not what we have here we have cosine of pi over pix so there is a technique here you could call it U substitution you could say U is equal to pix if you don't know how to do that you could still try to think about think this through where we could say all right well maybe it involves sine of pix somehow so the derivative with respect to X of sine of pix would be what well we would use the chain rule it would be the derivative of the outside function with respect to the inside or S of pix with respect to pix which would be cosine of pix and then times the derivative of the inside function with respect to X so it would be times pi or you could say the derivative of sin pix is pi cine of pix now we almost have that here except we just need a pi so what if we were to throw a PI right over here but so we don't change the value we also multiply by 1 over Pi so if you divide and multiply by the same number you're just you're not changing its value 1 over Pi pi is just equal to one but this is useful this is useful because we now know that Pi cosine pix is the derivative of sin pix so this is all going to be equal to this is equal to 1 let me let me take that 1 over Pi so this is equal to 1 over pi times now we're going to evaluate so the anti-derivative here we just said is s s of pi x and we're going to evaluate that at one and at zero so this is going to be equal to 1 over Pi 1 over Pi not pi hand is not listening to my mouth 1 over pi times sine of Pi s of Pi minus s of minus s of Pi 0 which is just 0o well sine of Pi that's 0o s of 0 is 0er so you're going to have 1 over Pi 0 - 0 so this whole thing is just all going to be equal to zero so this first part was 1/2 this second second part right over here is equal to 0er so the whole definite integral is going to be 1/2 + 0 which is equal to 1/2 so all of that together is equal to2
14759	https://artofproblemsolving.com/community?srsltid=AfmBOopgOZKaCBWyb3-ABOsC1gA6VTjOs874C1KqHSadZDooOUx8khKw	AoPS Community Math Message Boards & Community Forums \| AoPS 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 1-12. 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Community Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 3 V New Topic k Locked Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 3 V New Topic k Locked t 51,943 topics w 3 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H 9 Pythagorean Triples ZMB038 146 Please put some of the ones you know, and try not to troll/start flame wars! Thank you :D 146 replies ZMB038 May 19, 2025 melloncandy an hour ago J H Percent & Saline Mix Kushagra2012 2 How many liters of a % saline solution must be added to liters of a % saline solution to produce a % saline solution? %? %. Any pattern? 2 replies Kushagra2012 Saturday at 9:32 PM evt917 2 hours ago J H Two Liar Problems matharama 11 1 Frederick never tells the truth on Tuesdays, Thursdays, and Saturdays. He always tells the truth on Sundays, Mondays, Wednesdays, and Fridays. One day Matthew had this conversation with Frederick: "What day is it today?" asked Matthew. "Saturday," Frederick replied. "And what day will it be tomorrow?" "Wednesday." On which day did this conversation take place? First to answer correctly goes on the leaderboard! Leaderboard 2 The Liar Paradox Important Note! There is no correct solution. Therefore, there is no leaderboard! A man say that he is lying. Is what he says true or false? 11 replies matharama Sep 26, 2025 Wax_Wax 2 hours ago J H Mathcounts prep as I qualed into my school MATHCOUNTS team HoneyHap 13 Hey guys, With somewhat great difficulty, I was able to make it into the school MATHCOUNTS team. Now, my next step is to strengthen my rookie problem-solving skills and level it up. I am thinking because I don’t I have a lot of time, to take Po Shen Loh’s self paced course “Competition Booster Pack: Module 2+4+5” - , so I can be ready to perform well on the Mathcounts chapter round. Do you guys recommend taking this course with not that much time left? I am saying as I am in 8th grade and have never qualed for MATHCOUNTS team until now. So, I am thinking I might be way behind than the rest of my MATHCOUNTS peers in PA. Also, since I live in Pennsylvania, I want to know how competitive is PA in Mathcounts so I can see where I am at and how much practice I need to put in. If you guys have any helpful info, do let me know. 13 replies HoneyHap Yesterday at 9:41 PM quasar1 3 hours ago J High School Math Grades 9-12, Ages 13-18 Grades 9-12, Ages 13-18 3 V New Topic k Locked High School Math Grades 9-12, Ages 13-18 Grades 9-12, Ages 13-18 3 V New Topic k Locked t 122,697 topics w 6 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H IOQM 2025 Region 04 and 05 P16 Lunatic_Lunar7986 1 Find the number of ordered pairs , where and are positive integers such that and is a perfect square. 1 reply Lunatic_Lunar7986 an hour ago Lunatic_Lunar7986 13 minutes ago J H IOQM 2025 Region 04 and 05 P12 Lunatic_Lunar7986 1 In a convex quadrilateral , the lengths of the diagonals are 12 and 16 and the line segments joining the midpoints of the opposite sides are of equal length. What is the maximum possible area of the quadrilateral . 1 reply Lunatic_Lunar7986 an hour ago Lunatic_Lunar7986 14 minutes ago J H IOQM 2025 Region 04 and 05 P14 Lunatic_Lunar7986 1 The side of a square is 1 and it is also a chord of circle . The side does not intersect . The length of the tangent , drawn from to at the point is 2. If is the diameter of , then calculate . 1 reply 1 viewing Lunatic_Lunar7986 an hour ago Lunatic_Lunar7986 15 minutes ago J H IOQM 2025 Region 04 and 05 P9 Lunatic_Lunar7986 2 Find the largest positive integer for which the inequality holds. 2 replies Lunatic_Lunar7986 an hour ago Lunatic_Lunar7986 16 minutes ago J Contests & Programs AMC and other contests, summer programs, etc. AMC and other contests, summer programs, etc. 3 V New Topic k Locked Contests & Programs AMC and other contests, summer programs, etc. AMC and other contests, summer programs, etc. 3 V New Topic k Locked t 35,028 topics w 6 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H Help!! Why did MAA change the weighting Jessepinkman10 4 I feel like I have a decent shot at making JMO before they released that new change. I was mocking 135-141 with usually 1 silly on amc10 and 9 on aime but now i honestly think im cooked bc i need like an 11 on aime and idk if i can do that in 5 months. Does anyone have any tips for me? 10th grade btw :( 4 replies Jessepinkman10 Sep 26, 2025 Jessepinkman10 33 minutes ago J H AIME Qualification reeva28 7 Im currently a sophmore and I want to get better at competitive math. I know its a little late but I want to qualify for the AIME. Would anyone have any tips or recommendations or resources for what I should do to qualify? 7 replies reeva28 5 hours ago melloncandy an hour ago J H Computational Difficulty Measurement imbadatmath1233 4 We all now the infamous mohs which detects the difficulty of olympiad style problems(invented by evan chen). I think we should have one for computational problems as well! I will call it cohs(computational olympiad hardness scale) and the difficulty rating is much like what we see on mohs(I was thinking about raising the scale from 0-100 but idk it was just a thought). For example 2024 amc10a p23 might be an example of a 0-5 cmohs problem whereas aime 15's might be good 20-30 cmohs. Let me know your thoughts and also should we expand it to 100? 4 replies imbadatmath1233 Yesterday at 3:34 AM NamelyOrange 2 hours ago J H Nice substitution Owjebra 65 Source: 2020 AMC 10A #14 Real numbers and satisfy and . What is the value of 65 replies Owjebra Jan 31, 2020 clod 2 hours ago J High School Olympiads Regional, national, and international math olympiads Regional, national, and international math olympiads 3 V New Topic k Locked High School Olympiads Regional, national, and international math olympiads Regional, national, and international math olympiads 3 V New Topic k Locked t 317,092 topics w 12 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H Olympiad problem - I can't solve it pls help kjhgyuio 7 Source: smo 2016 It is given that x and y are positive integers such that x>y and √x + √y=√2000 How many different possible values can x take? 7 replies kjhgyuio Apr 1, 2025 kjhgyuio 9 minutes ago J H Mod3 complete system of residue MathMaxGreat 1 Source: 2020 Winter NSMO Prove that there exists infinite many positive integer pairs such that all contains a complete system of residue and . 1 reply MathMaxGreat 3 hours ago soryn 11 minutes ago J H ab+bc+ca=k^2 inequality jokehim 5 Source: my problem Let then When does equality hold? 5 replies jokehim Saturday at 7:35 AM lbh_qys 27 minutes ago J H Wilson's theorem on mod P^2 Not__Infinity 3 Source: Generalised question Elementary number theory by David M Burton Find primes for which following expression hold true. 3 replies Not__Infinity Yesterday at 3:07 PM soryn 44 minutes ago J College Math Topics in undergraduate and graduate studies Topics in undergraduate and graduate studies 3 V New Topic k Locked College Math Topics in undergraduate and graduate studies Topics in undergraduate and graduate studies 3 V New Topic k Locked t 114,928 topics w 2 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H Fourier Series EthanWYX2009 1 Source: 2025 Spring NSTE(2)-3 Let be real numbers. Define . Prove that: Proposed by Site Mu 1 reply EthanWYX2009 Aug 2, 2025 yofro 4 hours ago J H Question on Goldbach Conjecture Verification MDB001 5 Hello, I was looking at the AoPS page on the Goldbach Conjecture, and something came to mind. When people say the conjecture has been verified up to 10^{18}, what exactly is the verification method? Is it done by generating all primes up to that bound and checking their sums to cover all even numbers in the interval? Or is it done the other way around, starting with an even number and directly decomposing it into two primes? If the method is the first, then the second seems more intriguing. Because if one could start from an even number, especially a large one, say 100 digits or more, and directly retrieve its two primes, what would that imply for the problem itself? Thank you for clarifying. I would be very curious about your perspectives. 5 replies MDB001 Yesterday at 9:23 AM Tintarn 6 hours ago J H Group theory frr12 1 We consider the set of square matrices with real coefficients of the form Show that the group is isomorphic to the multiplicative group . First of all I proved that is a group but I have no idea how can I suggest a isomophic for this goup. 1 reply frr12 Yesterday at 8:14 PM Etkan Yesterday at 8:30 PM J H Fourier series We2592 4 Q) Expand the function in a Fourier series of period . please help for integration hint(upper lower limit and the ) 4 replies We2592 Sep 26, 2025 Mathzeus1024 Yesterday at 3:23 PM J Site Support Tech support and questions about AoPS classes and materials Tech support and questions about AoPS classes and materials 3 V New Topic k Locked Site Support Tech support and questions about AoPS classes and materials Tech support and questions about AoPS classes and materials 3 V New Topic k Locked t 18,906 topics w 5 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H Emoji rule O33ochan 18 Why does every post have a rule where there can only be 5 emojis? Can't it have more? I don't know why this rule exists, but if Idid, I would understand why this rule is not meant to be annoying. Thank you! 18 replies O33ochan Saturday at 3:13 PM doglover37 12 minutes ago J H Topics turning white RollingPanda4616 12 Summary of the problem: When rapidly clicking down topics, some of the topics turn white. Page URL: (found it here but might work elsewhere) Steps to reproduce: Go onto Site Support. View the top ~10 threads, turning them gray. Click the first topic "Suggestion Form". Go down the list of topics as fast as you can, using the things on the left. Click "Read me first/ How to...", then immediately click "Community Safety", and so on. Some topics will turn white randomly. Expected behavior: The topics will stay gray, unless someone else posts. Frequency: ~25% Operating system(s): macOS Sequoia 15.4 Browser(s), including version: Google Chrome 140 Additional information: It only happens on some threads. Refreshing fixes it, but it is still reproduceable. 12 replies RollingPanda4616 Sep 20, 2025 Staragon 20 minutes ago J H Is this a Glitch? ZuzabKoit 5 While I was looking at a topic it suddenly started scrolling. I was doing nothing, but I had to close the tab or hit the return to top button for it to stop. Even when I tried scrolling, it overrided my mouse and kept scrolling. Is my account getting hacked or is this a glitch? 5 replies ZuzabKoit Sep 26, 2025 KangarooPrecise 3 hours ago J H [RESOLVED] 8char limit wrongfully imposed jkim0656 12 Hi SS, I noticed that if you write a message and surround it in brackets [], it says that the message is shorter than eight characters even when it is definitely longer. Chrome Browser HP laptop I did notice that if you write at least 8 characters after the text in [], it lets the message through. 12 replies jkim0656 Yesterday at 1:05 AM SpeedCuber7 Yesterday at 1:47 PM J Other Forums and Collections ? AoPS Forums Forums for AoPS books, courses, and other resources. Forums for AoPS books, courses, and other resources. 3 V New Topic k Locked AoPS Forums Forums for AoPS books, courses, and other resources. Forums for AoPS books, courses, and other resources. 3 V New Topic k Locked Forums for AoPS books, courses, and other resources. AoPS Online Classes Questions and information about AoPS classesAoPS Books Questions and information about AoPS booksAlcumus The free AoPS online learning systemFor the Win! 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14760	https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-7-for-ccss/section/1.6/related/lesson/direct-variation-bsc-alg/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up Back To Using Equations to Represent Proportional RelationshipsBack 1.6 Direct Variation Written by:Andrew Gloag \| Melissa Kramer \| Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Suppose a person's annual salary is directly proportional to the number of years he or she has spent in school. What do you think this means? Could you determine the constant of proportionality? What information would you need to do so? If you wrote a linear equation to represent this relationship, what would be the slope and @$\begin{align}y-\end{align}@$intercept? Direct Variation At the local farmer’s market, you saw someone purchase 5 pounds of strawberries and pay $12.50. You want to buy strawberries too, but you want only 2 pounds. How much would you expect to pay? This situation is an example of a direct variation. You would expect that the strawberries are priced on a “per pound” basis, and that if you buy two-fifths of the amount of strawberries, you would pay two-fifths of $12.50 for your strawberries, or $5.00. Similarly, if you bought 10 pounds of strawberries (twice the amount), you would pay $25.00 (twice $12.50), and if you did not buy any strawberries you would pay nothing. Direct variation can be expressed as the equation @$\begin{align}y=(k)x\end{align}@$, where @$\begin{align}k\end{align}@$ is called the constant of proportionality. Direct variation occurs when: The fraction @$\begin{align}\frac{rise}{run}\end{align}@$ or @$\begin{align}\frac{change \ in \ y}{change \ in \ x}\end{align}@$ is always the same, and The ordered pair (0, 0) is a solution to the equation. Let's use direct variation to solve the following problem: If @$\begin{align}y\end{align}@$ varies directly with @$\begin{align}x\end{align}@$ according to the relationship @$\begin{align}y=k \cdot x\end{align}@$, and @$\begin{align}y=7.5\end{align}@$ when @$\begin{align}x=2.5\end{align}@$, determine the constant of proportionality, @$\begin{align}k\end{align}@$. We can solve for the constant of proportionality using substitution. Substitute @$\begin{align}x=2.5\end{align}@$ and @$\begin{align}y=7.5\end{align}@$ into the equation @$\begin{align}y=k \cdot x\end{align}@$. @$$\begin{align}7.5 & = k(2.5) && \text{Divide both sides by} \ 2.5.\ \frac{7.5}{2.5}& =k=3\end{align}@$$ The constant of variation (or the constant of proportionality) is 3. You can use this information to graph this direct variation situation. Remember that all direct variation situations cross the origin. You can plot the ordered pair (0, 0) and use the constant of variation as your slope. Now, explain why each of the following equations are not examples of direct variation: @$$\begin{align}y& =\frac{2}{x}\ y& =5x-1\ 2x+y& =6\end{align}@$$ In equation 1, the variable is in the denominator of the fraction, violating the definition. In equation 2, there is a @$\begin{align}y-\end{align}@$intercept of –1, violating the definition. In equation 3, there is also a @$\begin{align}y-\end{align}@$intercept, violating the definition. Translating a Sentence into a Direct Variation Equation Direct variation equations use the same phrase to give the reader a clue. The phrase is either “directly proportional” or “varies directly.” Let's translate the following sentence into an equation: The area of a square varies directly as the square of its side. The first variable you encounter is “area.” Think of this as your @$\begin{align}y\end{align}@$. The phrase “varies directly” means @$\begin{align}= (k)\times\end{align}@$. The second variable is “square of its side.” Call this letter @$\begin{align}s\end{align}@$. Now translate into an equation: @$\begin{align}y=(k)\times s^2\end{align}@$. You’ve written your first direct variation equation. Example Example 1 Earlier, you were asked to suppose that a person's annual salary is directly proportional to the number of years he or she has spent in school. What does this mean? Could you determine the constant of proportionality? What would the slope and @$\begin{align}y-\end{align}@$intercept be? That a person's salary is directly proportional to the number of years that they have spent in school means that as the number of years a person is in school increases, so does the person's annual salary. The increase is constant and described by the constant of proportionality. This would be found by calculating the slope between two data points relating to salary and years in school. The slope in an equation describing this situation would be the constant of proportionality and the @$\begin{align}y-\end{align}@$intercept would be the salary of a person with 0 years of school. Example 2 The distance you travel is directly proportional to the time you have been traveling. Write this situation as a direct variation equation. The first variable is distance; call it @$\begin{align}d\end{align}@$. The second variable is the time you have been traveling; call it @$\begin{align}t\end{align}@$. Apply the direct variation definition: @$$\begin{align}d=(k) \times t\end{align}@$$ Review Describe direct variation. What is the formula for direct variation? What does @$\begin{align}k\end{align}@$ represent? Translate the following direct variation situations into equations. Choose appropriate letters to represent the varying quantities. The amount of money you earn is directly proportional to the number of hours you work. The weight of an object on the Moon varies directly with its weight on Earth. The volume of a gas is directly proportional to its temperature in Kelvin. The number of people served varies directly with the amount of ground meat used to make burgers. The amount of a purchase varies directly with the number of pounds of peaches. Explain why each equation is not an example of direct variation. @$\begin{align}\frac{4}{x}=y\end{align}@$ @$\begin{align}y=9\end{align}@$ @$\begin{align}x=-3.5\end{align}@$ @$\begin{align}y=\frac{1}{8} x+7\end{align}@$ @$\begin{align}4x+3y=1\end{align}@$ Graph the following direct variation equations. @$\begin{align}y=\frac{4}{3}x\end{align}@$ @$\begin{align}y=-\frac{2}{3}x\end{align}@$ @$\begin{align}y=-\frac{1}{6}x\end{align}@$ @$\begin{align}y=1.75x\end{align}@$ Is @$\begin{align}y=6x-2\end{align}@$ an example of direct variation? Explain your answer. Mixed Review Graph @$\begin{align}3x+4y=48\end{align}@$ using its intercepts. Graph @$\begin{align}y=\frac{2}{3} x-4\end{align}@$. Solve for @$\begin{align}u: 4(u+3)=3(3u-7)\end{align}@$. Are these lines parallel? @$\begin{align}y=\frac{1}{2} x-7\end{align}@$ and @$\begin{align}2y=x+2\end{align}@$ In which quadrant is (–99, 100)? Find the slope between (2, 0) and (3, 7). Evaluate if @$\begin{align}a=-3\end{align}@$and@$\begin{align}b=4\end{align}@$: @$\begin{align}\frac{1+4b}{2a-5b}\end{align}@$. Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. 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14761	https://publish.obsidian.md/doctorstudio/2024+~.+Economics/4.2+Indifference+Curves+and+Budget+Constraints	4.2 Indifference Curves and Budget Constraints - Dr. Ju's Blog - Obsidian Publish Dr. Ju's Blog 2013 ~. Medicine 2023 ~. Medical Tourism 2023 ~. Storytelling 2023 ~. Study of Marketing 2024 ~. Case Studies 2024 ~. Economics 0. Economics Chapter Outline 0. Economics Study Guide 1.1 Definition and Scope 1.2 Scarcity, Choice, and Opportunity Cost 1.3 The Economic Way of Thinking 1.4 The History of Economic Thought 2.1 Types of Economic Systems - Capitalism, Socialism, Mixed Economies 2.2 Role of Government in Economic Decision-Making 2.3 Market vs. Command Economies - Pros and Cons 2.4 Emerging Economic Models - Gig Economy, Digital Currencies, Decentralization 3.1 Market Demand, Supply, and Equilibrium 3.2 Shifts in Demand and Supply Curves 3.3 Price Elasticity of Demand and Supply 4.1 Utility and Utility Maximization 4.2 Indifference Curves and Budget Constraints 4.3 Consumer Choice and Demand 4.4 Behavioral Economics Insights into Consumer Behavior 5.1 The Production Process - Short Run and Long Run 5.2 Cost Curves and Their Shapes 5.3 Economies and Diseconomies of Scale 6.1 Characteristics of Perfect Competition, Monopoly, Monopolistic Competition 6.2 Oligopoly and Game Theory 6.3 Digital Markets and Platform Economics 7.1 Labor Markets - Demand and Supply of Labor 7.2 Capital Markets and Investment Decisions 7.3 Income Distribution and Factor Pricing 8.1 Measuring GDP and National Income 8.2 Economic Growth and Development Indicators 8.3 Real vs. Nominal GDP 9.1 Functions of Money 9.2 How Banks Create Money 9.3 Central Banking and Monetary Policy 9.4 Financial Technology and Innovations - Blockchain, Cryptocurrencies 10.1 Tools of Monetary Policy 10.2 Fiscal Policy - Government Spending and Taxes 10.3 The Impacts of Policies on Inflation and Unemployment 11.1 Principles of Trade and Comparative Advantage 11.2 Exchange Rates and Their Determination 11.3 Balance of Payments and Trade Policies 11.4 Globalization's Effects on Macroeconomics 12.1 Characteristics of Developing Economies 12.2 Strategies for Economic Development 12.3 Challenges to Economic Growth 13.1 Externalities and Market Failure 13.2 Environmental Policies and Regulations 13.3 Sustainable Development and Green Economics 14.1 Labor Supply and Demand 14.2 Wage Determination and Labor Market Equilibriums 14.3 The Economics of Work and Leisure Anti-trust laws Blizzard의 2nd Hand Market 도입 - 행동 경제학과 Gamification 개념의 활용 Bounded Rationality Budget Constraints of Consumers Carbon Pricing Classic example of a positive externality where the social benefits exceed the private benefits Comparative Advantage Models Consumer and Producer Theory Consumer Price Index (CPI) Default Effects and Nudging Deflation Dependency Theory Economies of Scale Efficient market hypothesis (EMH) Fiscal policy, monetary policy General Equilibrium Theory Impulse Buying and Self-Control Issues Japan's Abenomics policies Key Concepts of Inflation from Michael Saylor Libertarian Paternalism Marginal revenue product (MRP) of labor Modern portfolio theory (MPT) Neoclassical Economics New Trade Theory Open Market Optimization Rule Pareto Efficiency Paris Agreement on climate change Paul Krugman’s Model Private Sector Demand Producer Price Index (PPI) Prospect Theory Quotas Ricardo’s Law of Rent Social Preferences Sustainable Development Goals (SDGs) The Brundtland Report The Coase Theorem The Double Coincidence of Wants The rule of 70 The Tax Cuts and Jobs Act of 2017 Time Inconsistency and Hyperbolic Discounting What’s the ROI of covering employee childcare? World-Systems Theory 2024 ~. Ethics and Philosophy 2024 ~. Psychology Aesthetic Medicine Hair transplantation SMP Aetherhealth 0000. Welcome! Dr. Ju's Blog 4.2 Indifference Curves and Budget Constraints 날짜 : 2024-03-24 15:13 주제 : Indifference Curves and Budget Constraints #economics 4.2 Indifference Curves and Budget Constraints Indifference Curves and Budget Constraints The concepts of indifference curves and budget constraints are essential to understanding how consumers make choices between different combinations of goods and services. These concepts help to graphically represent consumer preferences and the limitations imposed by income and prices. Indifference Curves An indifference curve represents a set of bundles of two goods that provide the consumer with the same level of utility, which means the consumer is indifferent between them. Here are key characteristics of indifference curves: Downward Sloping: Since you have to give up some of one good to get more of another while staying at the same utility level, indifference curves slope downwards. Higher Curves Represent Higher Utility: A curve that lies further from the origin reflects a higher utility level because it consists of bundles with more of at least one of the goods. Never Cross: Indifference curves cannot cross because that would imply inconsistent preferences. Bowed Inward: This shape reflects the principle of diminishing marginal rate of substitution (MRS), meaning as you consume more of one good, you are willing to give up less of the other good to keep the same level of utility. Budget Constraints A budget constraint represents all combinations of goods that a consumer can afford given their income and the prices of goods. Graphically, it's a straight line where: The Slope: The slope is determined by the relative prices of the two goods. Intercepts: The intercepts on the axes represent the maximum quantity of each good that can be purchased with the available budget. Income Changes: A change in income shifts the budget line inward (decrease in income) or outward (increase in income) without changing the slope. Price Changes: A change in the price of a good rotates the budget line around the axis intercept of the other good. Bringing Indifference Curves and Budget Constraints Together Optimal consumption occurs at the point where the highest attainable indifference curve is tangent to the budget constraint. This point of tangency indicates the best possible combination of two goods that maximizes utility given the consumer's income and the prices of the goods. Examples of Indifference Curves and Budget Constraints Example 1: Choosing Between Movies and Concerts Suppose you enjoy both movies and concerts, and you have a fixed entertainment budget. Let's say a movie ticket costs $10 and a concert ticket costs $50. Your budget constraint will show how many of each ticket you can afford. Now, imagine your indifference curves show that you're equally satisfied with 4 movies and no concerts as you are with 1 concert and no movies. If you had 5 such indifference curve options, each would represent a different level of utility, but within each curve, you're indifferent to the combination of concerts and movies. If there's one combination that sits just where the highest indifference curve you can reach touches your budget line, that's the optimal combination of movies and concerts for you given your budget. Example 2: Coffee and Croissant Consider that you have a budget of $10 for your morning snack. A coffee costs $2 and a croissant costs $4. Your budget line will show you can afford either 5 coffees, 2.5 croissants, or a combination of the two. Your indifference curves might show different combinations of coffee and croissants that give you the same satisfaction. For instance, you might be equally happy with 3 coffees and 1 croissant, or with 1 coffee and 2 croissants. The optimal choice will be where the budget line is tangent to the highest indifference curve, which might be at 2 coffees and 1 croissant. Practice Task: Drawing the Concepts A helpful study exercise would be to draw these concepts. Take a sheet of graph paper and sketch: A budget line based on an example budget and prices. Several indifference curves representing different utility levels. Find the optimal consumption bundle by illustrating the tangency between the highest reachable indifference curve and the budget constraint. Remember, the combination at which the budget constraint and the highest attainable indifference curve touch tangentially will give you the quantities of both goods that the consumer should choose to maximize utility. Understanding how changes in income and price alter the budget constraint and, subsequently, affect the consumer's choice is crucial to predicting how changes in economic conditions can impact consumer behavior. 4.2 Indifference Curves and Budget Constraints Not found This page does not exist Interactive graph On this page 날짜 : 2024-03-24 15:13 주제 : Indifference Curves and Budget Constraints #economics 4.2 Indifference Curves and Budget Constraints Indifference Curves and Budget Constraints Indifference Curves Budget Constraints Bringing Indifference Curves and Budget Constraints Together Examples of Indifference Curves and Budget Constraints Example 1: Choosing Between Movies and Concerts Example 2: Coffee and Croissant Practice Task: Drawing the Concepts Powered by Obsidian Publish
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Back to Forum Create Topic Reply 3 Deadly mistakes you must avoid in Permutation and Combination Important Topic 1 2 EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL Updated on: 08 Aug 2018, 07:09 Originally posted by EgmatQuantExpert on 18 Apr 2018, 05:23. Last edited by EgmatQuantExpert on 08 Aug 2018, 07:09, edited 10 times in total. Added the PDF of the article at the end of the post! 3 Deadly mistakes you must avoid in Permutation and Combination This is the 3rd and the final article in the series of Permutation and Combination. If you did not read our previous article, we recommend that you read them. The first 2 article will help you to get your basic in place.Article-1: Learn when to “Add” and “Multiply” in Permutation & Combination questionsArticle-2: Fool-proof method to Differentiate between Permutation & Combination QuestionsA brief summary of the previous articles • The logical way to apply addition or multiplication with the help of keywords• Frequently used keywords to identify a combination-selection question o Such as AND, OR, SELECT, CHOOSE, PICK, ARRANGE, ORDER etc • How to approach a PnC question when keywords are not given. In this article, we will explain 3 most common mistakes that a student makes in PnC questions and along with that we will also explain how to avoid these mistakes.Multiple Counting: Common Error Type 1Let us look at a very simple example to understand a common mistake.There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback? (OG question) A) 75B) 120C) 210D) 246E) 252Common Approach Used By Students:We need to select 5 books such that there is at least one paperback book and at least one hardback book.Hmm…so let us select 1 paperback and 1 hardback first. That will help us ensure that I have one book of each type. Then I will worry about selecting the extra 1 book from the remaining 8 books. • Hence, the answer should be (^4C_1)(^6C_1)(^8C_2). This is how usually a lot of student approach this question and then wonder if this is the correct way to solve the question or not. So, think, is this the right way to solve this question?NO!!!!!!!! It is not. The answer is incorrect, and it is incorrect because we double counted some cases.What is Double Counting or Multiple Counting?Let us take a small example to understand what double counting is. • We have 4 books of which 2 are paperbacks, A and B, and 2 are hardbacks, C and D. We have to find in how many ways we can select 3 books such that we have at least 1 paperback book and at least 1 hardback book. Let us list down the various combination when we solve the above example by (^2C_1)(^2C_1)(^2C_1)=8Notice that same color-coded cells in the 4th column. • The selection books ACB and BCA are same.• The selection books ACD and ADC are same.• The selection books ADB and BDA are same.• The selection books BCD and BDC are same. Thus, we repeated 4 cases and that is why we were getting the wrong answer to the actual question.We call this counting of some extra cases as double or multiple counting.How to avoid Double Counting in such cases?In questions like these, we have two categories: Hardback and Paperback.And we need a “mix” of both the types. In scenarios like these where there are multiple categories and we need a mix of all the types, in such situations, we will first jot down all the possible cases in which all the types can be mixed or collected together. • Thus, in this case we will write down all the possible ways of having paperback and hardback books (keeping in mind the constraint: we need at least one book of each type) o 3 books can be • 2 Paperback and 1 Hardback• 1 Paperback and 2 Hardback o Thus, Total possible ways of collecting 3 books =  (2 paperback AND 1 hardback) OR (1 paperback AND 2 hardback) • Possible way= (^2c_2)(^2c_1)+ (^2c_1) (^2c_2)= 12 + 21= 4 Wow!!! We now know the correct approach to solve this type of question.Now here is a small exercise for you: Let’s say you need to select 3 vehicles from 5 cars and 4 bicycles, in which we need at least 1 car and 1 bicycle. In how many ways can you do it? Will you write it as: a. (^5C_1) (^4C_1) (^7C_1) OR Will you write is as: b. Cases Possible: 2 Cars & 1 Bicycle OR 1 Car & 2 Bicycles = (^5C_2) (^4C_1) + (^5C_1) (^4C_2) If you have chosen Option B. Then Congrats you have successfully learned how to avoid double counting in such cases!Let us now solve our actual question again.There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback? (OG question)A) 75B) 120C) 210D) 246E) 252SolutionWe need to select 5 books such that there is at least one paperback book and at least one hardback book.We can select 5 books in various ways: • 1 Paperback book and 4 hardback books Or,• 2 Paperback books and 3 hardback books Or,• 3 Paperback books and 2 hardback books Or,• 4 Paperback books and 1 hardback book. Key Takeaways This simple example elaborates the most common mistake by a majority of students i.e. Double counting.2. In scenarios like these where there are multiple categories and we need a mix of all the types, in such situations, we will first jot down all the possible cases in which all the types can be mixed or collected together. Let us see another case where most of the students apply double counting: The case when certain objects are identical.The Curious Case of Identical Objects – Common Error Type 2While solving PnC questions, we may come across a few questions where we might be asked to select and arrange a few objects which are identical in nature. For example, you might be asked to arrange 5 identical books, or you might be asked to arrange the word “MISSISSIPPI”, in which there are few letters which are identical (the letters I, S and P are present in this word multiple times).Students have the tendency to make mistake whenever they come across such cases. And thus, we should learn the best and full-proof way to solve such questions!Let us understand how to tackle these situations with the help of an example.Assume a scenario where we have 3 different letters- A, B, C and we need to form 3 letter words. • Total 3 different letter words = 3! = 6 words• And these are the following cases: o ABCo ACBo BACo BCAo CABo CBA Now, let’s play with these words!Tell, me how many unique words will we get if we replace C by A??If we replace C with A in the above example, we will get: • Total words=ABA+AAB+BAA+BAA+AAB+ABA= 2(AAB+ABA+BAA) = 2! (AAB+ABA+BAA) = 2! 3 Now, notice that there are only 3 unique words possible: AAB, ABA, and BAA. However, we have 2! multiplied with it in above equation. Now, what should we do to get the actual answer which is 3?? • We are getting multiple case, because we wrote AAB twice, ABA twice and BBA twice.• And we wrote the twice, because while jotting the cases, we arrange AA in AAB, ABA and BBA twice or 2! times. • Thus, to get the actual answer, we need to divide 2! 3 by 2!.• AAB+ABA+BAA= (6/2!) = Total ways when all the letters were different/ repetitive counting of the identical letters. Now, let us extend this idea further to find the number of different words by using three A’s, two B’s.Since we know that we will again get repetitive cases because of three identical A’s and two identical B’s. • Total words= (Total ways when all the 5 letters are different) / (multiple counting of the identical letters)• Total ways= 5! / (Repetitive counting of three A’s and repetitive counting of Two B’s) o Three A give only 1 word, but we counted extra cases by assuming all three A’s to be different. Hence, we will divide by 3!o Similarly, we counted extra cases by assuming the two B’s to be different. Hence, we will divide by 2! • Total ways= (5! /{3! 2!}) Extending the idea further, we can generalize that: The number of arrangements of n objects with p1 identical objects of one kind, p2 identical objects of second kind, p3 identical objects of third kind and so on is equal to:(\frac{n!}{{(p1! p2! p3!…..)}}). e-GMAT Example 1Q--In how many different ways can the letters of the word SINISTER be arranged such that two S are never together?SolutionSINISTER is an 8-letter word with S and I repeated two times.We want the cases in which two S are never together and how can we do that??We can solve the question in two ways: • By adding all the cases when two S are separated has at least 1 letter between them. o Ways when (Two S separated by 1 letter+ Two S separated by 2 letter+…..+Two S separated by 6 letters) • By removing all the cases in which Two S are together from the total cases. o Total cases- Number of cases in which both the S are together. It is very clear that 2nd method is easy. Thus, we only need to find: • Total cases in which SINISTER can be arranged and,• Number of cases in which both the S are together. Total cases:SINISTER can be arranged in (8! / {2! 2!}) = 10080 waysNumber of cases in which both the S are together:Two S always has to be together, so we can group together as SS.We can assume SS to be one letter as they will only be arranged like a letter with the other 6 letters. • Thus, total words = 6 letters other than S+ 1 group of two S However, SS is actually a 2-letter word which can arrange in itself. Thus, Total ways when both the S are together= arrangement of 7 letters arrangements in the group SS • Total ways= 7!/2! 1(SS can be arranged in only 1 way)• Total ways when both the S are together= 7!/2! = 2520 ways [we are dividing by 2! because there are 2 Is which will give us repetitive cases] Thus, total cases in which two S are never together= 10080- 2520= 7560 ways.e-GMAT Example 2Q-- If we have 9 different points in a plane out of which 4 are collinear. How many different straight lines we can draw.SolutionTo make a straight line we need to select 2 points among 9 points. However, only 1 straight line can pass through from the 4 collinear points. • Can you visualize that this is similar to the case of identical objects??• In case of similar object, we only have 1 arrangement and in case of collinear points we only have 1 line. In this case, all the lines are not repeating, only the lines whose both the points lies on 4 collinear points are repeating.Thus, we can get the line in 3 different ways: • Select any two points from 5 non-collinear points Or,• Select one point from 5 non-collinear points and another from 4 collinear points Or,• 1 line from the 4 collinear points Total ways= (^5c_2)+(^5c_1)(^4c_1)+1 = 10+20+1= 31 Thus, we will get 31 different straight lines. Key Takeaways • Identical objects can be arranged in 1 way.• To find the arrangement of n different things of which some are identical then we divide by the arrangement of identical things assuming the identical thigs to be different.• The number of arrangements of n objects with p1 identical objects of one kind, p2 identical objects of second kind, p3 identical objects of third kind and so on is equal to:(\frac{n!}{{(p1! p2! p3!…..)}}).• From n collinear points, only 1 line can be drawn. Let us now move to the 3rd and one of most common type of mistake.Confusion of Powers: Common Error Type 3To understand the third type of mistake, let us try to find the answer to one simple question.e-GMAT Example 2Q--In how many ways, 3 different prizes can be distributed to 2 students where each is eligible for all the 3 prizes?Can you find the answer to this interesting question??? • Is the answer (2^3) or (3^2)??? SolutionIf your answer is (3^2) then You just did a very common mistake.So, we should figure out why (3^2) is wrong.Let us delve into the details of the question find this.Let us suppose the two students are: X and Y, and three prizes are A, B, and C.Now, when we say that total ways to distribute the prizes=33, we mean that: • X can get either 1 prize or two prizes or all the three prizes.• Similarly, Y can get either 1 prize or two prizes or all the three prizes. This can be shown in the tabular form as:Observe Case 1: • If, say, X gets prize A and Y gets prize B, then none of them got prize C.• • There is also a possibility that both X and Y might be given the same price A o Did we take any precaution that both get different prizes, while writing the first case? No, we did not. ☹ Observe Case 3, 5, 6, 7, 8:We are distributing more than 3 prizes to both of them, is it possible?No, right!!!!This implies that there is a mistake when we are distributing 3 prizes in 33 ways.This is common mistake that students make. This is known as counting invalid cases!Now, let us come to the correct approach to distribute 3 prizes in 2 students.Correct approachWe need to distribute all the three prizes, right?Thus, let us take Prize A and decide in how many ways can this be given to X and Y. • Prize A can be given to either X OR Y = 2 cases• Similarly, Prize B can be given to either X OR Y = 2 cases and• Prize C can be given to either X OR Y = 2 cases. Also, think, we need to distribute all the prizes, right? • Thus, total ways to distribute prize = Give Prize A AND Give Prize B AND Give Prize C = 2 x 2 x 2 = 8 ways. All the possible 8 cases can be shown as:Thus, there are (2^3)=8 possible ways.Key Takeaways from the exampleSolving the question in this way ensure 2 things: All the prizes are distributed: A, B and C was considered one by one and distributed.2. Eliminates the chances of double or invalid counting, since A will go to either X or Y, similarly B and C goes to either X or Y and not both!3. If we replace 3 by ‘n’ and 2 by ‘r’ in the above example, then we can infer that:• The total number of ways to distribute ‘n- different object’ in to ‘r- different things’ = r^n Let us solve another example to get 100% clarity over the concept.e-GMAT Example 2Q--There are 10 different envelopes and 3 post boxes. In how many ways can we post 4 envelopes in 3 post boxes such that each post box can send any number of envelopes.SolutionWe have 10 envelopes and we need to send any 4 envelopes from 3 post-boxes. • Thus, we first need to find select the 4 envelopes from the 10 envelopes then we can put the letter into post boxes. o Thus, total ways= Ways to select 4 envelopes from 10 envelopes ways to post 4 envelopes in 4 post boxes. Ways to select 4 envelopes • 4 envelopes can be selected from 10 envelopes in (^{10}C_4)= 210 ways. Now, we have 4 selected envelopes and we need to put the selected envelopes in to 3 post boxes.Way to post 4 envelopes in 3 post boxes • Method-1) o Since each envelope can go in to any of the post boxes thus each envelope can be posted into 3 ways.o For example, envelope one can go in PB1 or PB2 or PB3 = 3 ways.o And the same will be done for all the other envelopes.o Hence, total ways to post 4 envelopes in to 3 boxes= 3333= 3^4=81 • Method-2) – Use of Direct Formula: o Here Objects are envelopes and we have 4 envelopes. Thus n = 4o Since the three post boxes are the different things, in which they will be posted, r = 3.o Hence, by the application of r^n, total ways to post 4 envelopes in to 3 boxes= 3^4= 81 ways. Thus, total ways= 21081=17010Key Takeaways from the article In scenarios where there are multiple categories and we need a mix of all the types, in such situations, we will first jot down all the possible cases in which all the types can be mixed or collected together.2. The number of arrangements of n objects with p1 identical objects of one kind, p2 identical objects of second kind, p3 identical objects of third kind and so on is equal to:(\frac{n!}{{(p1! p2! p3!…..)}}).3. If r objects out of n objects are identical then we divide the total number of different cases by the number of repetitive case.4. The number of ways to distribute n different objects to r different things is (r^n). Attachments 3mistakes_P&C.pdf [651.54 KiB] Downloaded 381 times \| \| \| To download please login or register as a user \| _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More 8 Kudos Add Kudos 46 Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL Updated on: 13 Aug 2018, 06:23 Originally posted by EgmatQuantExpert on 18 Apr 2018, 22:55. Last edited by EgmatQuantExpert on 13 Aug 2018, 06:23, edited 1 time in total. Hey Everyone, Here is a list of questions specifically designed to help you apply the learnings from this article.Exercise QuestionsQuestion 1Question 2Question 3Question 4Detailed solutions will be posted soon.Happy Learning! _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos 1 Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL Updated on: 23 Apr 2018, 03:35 Originally posted by EgmatQuantExpert on 18 Apr 2018, 23:09. Last edited by EgmatQuantExpert on 23 Apr 2018, 03:35, edited 1 time in total. Hey everyone,4 practice questions have been added. Please check the post above, for the links to the questions.Thanks,AshutoshSubject Matter Experte-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post calculusbro calculusbro Joined: 29 Aug 2016 Last visit: 09 Sep 2019 Posts: 24 Posts: 24 Post URL20 Apr 2018, 00:46 Kindly provide PDF if there is a copy version.Sincerely, Kudos Add Kudos Bookmarks Bookmark this Post GMAT215 GMAT215 Joined: 01 Feb 2018 Last visit: 20 Jul 2022 Posts: 53 Location: India Concentration: Entrepreneurship, Marketing GPA: 4 WE:Consulting (Consulting) Posts: 53 Post URL20 Apr 2018, 06:32 in Question related to SINISTER : I think there is an error...Total cases:SINISTER can be arranged in 8!/2!∗2!8!/2!∗2! = 10080 ways OKNumber of cases in which both the S are together:Two S always has to be together, so we can group together as SS.We can assume SS to be one letter as they will only be arranged like a letter with the other 6 letters.• Thus, total words = 6 letters other than S+ 1 group of two SHowever, SS is actually a 2-letter word which can arrange in itself. Thus, Total ways when both the S are together= arrangement of 7 letters arrangements in the group SS• Total ways= 7! 1(SS can be arranged in only 1 way)• Total ways when both the S are together= 7! = 5040 ways Repetition of N is not considered .Thus, total cases in which two S are never together= 10080- 5040/2!= 10080 - 2520 = 7560 ways. Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL20 Apr 2018, 22:44 GMAT215 in Question related to SINISTER : I think there is an error...Total cases:SINISTER can be arranged in 8!/2!∗2!8!/2!∗2! = 10080 ways OKNumber of cases in which both the S are together:Two S always has to be together, so we can group together as SS.We can assume SS to be one letter as they will only be arranged like a letter with the other 6 letters.• Thus, total words = 6 letters other than S+ 1 group of two SHowever, SS is actually a 2-letter word which can arrange in itself. Thus, Total ways when both the S are together= arrangement of 7 letters arrangements in the group SS• Total ways= 7! 1(SS can be arranged in only 1 way)• Total ways when both the S are together= 7! = 5040 ways Repetition of N is not considered .Thus, total cases in which two S are never together= 10080- 5040/2!= 10080 - 2520 = 7560 ways. Show more Hey GMAT215I think you meant, repetition of I is not considered! We have updated the article. Thanks for pointing this out. I guess this clearly shows that anyone can fall for these deadly traps! _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL23 Apr 2018, 04:59 Hey Everyone,Official solution to all the practice questions has been posted.Regards,Ashutosh _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL24 Apr 2018, 06:04 Hey everyone,We have added the pdf of the article. Happy learning,Regards,AshutoshQuant Experte-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL27 Apr 2018, 06:39 Dear Students,We have added a new article to help you get clarity on how to avoid the 3 common mistakes in probability.You can go through the article from this 3 deadly mistakes you must avoid in ProbabilityStay tuned for more articles.Happy learning. Regards,e-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL29 May 2018, 02:07 Hey Everyone, We have added 4 practice questions on the concepts of Permutation and Combination.P&C Practice Question 1P&C Practice Question 2P&C Practice Question 3P&C Practice Question 4Detailed solutions will be posted soonHappy Learning! _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL31 May 2018, 02:07 Hey everyone,The solutions to all the P&C Practice Questions have been posted. Regards,e-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post laknaouiwarda1 laknaouiwarda1 Joined: 20 Dec 2017 Last visit: 11 Dec 2018 Posts: 3 Posts: 3 Post URL19 Jul 2018, 15:12 Thank you for the post, so helpful Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL20 Jul 2018, 03:47 Hey laknaouiwarda1,It feels great to hear that the article helped you.To read similar articles: Must Read Articles and Practice Questions to score Q51 !!!!You can solve the practice questions to test your understanding.Regards,e-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL20 Jul 2018, 05:35 Hey Everyone, We have added three more question for your practice.Do solve the questions and post your solutions.Practice QuestionsPractice Question 1Practice Question 2Practice Question 3Detailed solutions will be posted soon.Happy Learning! _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL23 Jul 2018, 06:00 Hello everyone,We have posted the official answer to the practice questions.Regards,Ashutoshe-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL12 Nov 2018, 21:45 Hey Everyone, We've added 3 practice questions in which you can apply your learning of Probability concepts.Exercise QuestionsQuestion 1Question 2Question 3Detailed solutions will be posted soon.Happy Learning! _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL14 Nov 2018, 05:17 Hey everyone,The official answers to all the practice questions have been posted.Regards,Sandeepe-GMAT _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL31 Jan 2019, 02:59 Hey Everyone, We've added 3 practice questions in which you can apply your learning of Probability concepts.Exercise QuestionsQuestion 1Question 2Question 3Detailed solutions will be posted soon.Happy Learning! _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL01 Feb 2019, 02:08 Hey everyone,The official answers to all the practice questions have been posted.Regards,Sandeep. _________________ Free Trial: Get 25 AI powered videos \| 400 practice questions \| 8 free webinarsCase studies: How a personalized study plan can save you 60+ hoursMBA admits: M7 admits \| INSEAD admits \| ISB admits [40+ case studies]Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post EgmatQuantExpert EgmatQuantExpert e-GMAT Representative Joined: 04 Jan 2015 Last visit: 02 Apr 2024 Posts: 3,672 Expert Expert reply Posts: 3,672 Post URL05 Mar 2019, 22:37 Hey Everyone, We've added 3 practice questions in which you can apply your learning of P&C and Probability concepts.Practice Exercise QuestionsQuestion 1Question 2Question 3Detailed solutions will be posted soon.Happy Learning! 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14763	https://www.geeksforgeeks.org/maths/what-is-hyperbola/	Hyperbola - Equation, Definition & Properties Hyperbola is one of the fundamental shapes in geometry formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. It is often encountered in both mathematics and real-world applications. Defined as the set of all points where the difference of the distances to two fixed points (called foci) is constant. It is a smooth curve in a plane with two branches that mirror each other, resembling two infinite bows. Hyperbolas are closely related to ellipses and parabolas, yet they possess distinct properties and applications. From the design of satellite dishes to the paths of celestial bodies, hyperbolas play a critical role in various scientific and engineering fields. Table of Content Hyperbola Hyperbola is the focus of points whose difference in the distances from two foci is constant. This difference is obtained by subtracting the distance of the nearer focus from the distance of the farther focus. If P (x, y) is a point on the hyperbola and F, F' are two foci, then the locus of the hyperbola is PF - PF' = 2a In analytic geometry, a hyperbola is a type of conic section created when a plane cuts through both halves of a double right circular cone at an angle. This intersection results in two separate, unbounded curves that are mirror images of each other. Standard Equation of Hyperbola The standard equations of a hyperbola are: \bold{\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1} OR \bold{\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}= 1} A hyperbola has two standard equations. These equations are based on its transverse axis and conjugate axis. \bold{\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}= 1} \bold{\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}= 1 } Real-Life Application of Hyperbola Parts of Hyperbola Hyperbola is a conic section that is developed when a plane cuts a double right circular cone at an angle such that both halves of the cone are joined. It can be described using concepts like foci, directrix, latus rectum, and eccentricity. The following table represents the parts of the hyperbola: \| Parts \| Description \| --- \| \| Foci \| Two foci with coordinates F(c, 0) and F'(-c, 0) \| \| Centre \| The midpoint of the line joining the two foci, denoted as O \| \| Major Axis \| Length of the major axis is 2a units \| \| Minor Axis \| Length of the minor axis is 2b units \| \| Vertices \| Intersection points with the axis, (a, 0) and (-a, 0) \| \| Transverse Axis \| Line that passes through the two foci and center of the hyperbola \| \| Conjugate Axis \| The line that passes throughcenterentre and is perpendicular to the transverse axis \| \| Asymptotes \| Equations of asymptotes are y = (b/a)x and y = -(b/a)x, lines that approach the hyperbola but never touch it \| \| Directrix \| Fixed straight line perpendicular to the axis of a hyperbola \| Eccentricity & Latus Rectum of Hyperbola Eccentricity The eccentricity of a hyperbola is the ratio of the distance of a point from the focus to its perpendicular distance from the directrix. It is denoted by the letter 'e'. e = √[1 + (b2/a2)] where, Read More: Eccentricity Latus Rectum Latus rectum of a hyperbola is a line passing through any of the foci of a hyperbola and perpendicular to the transverse axis of the hyperbola. The endpoints of a latus rectum lie on the hyperbola, and its length is 2b2/a. Read more: Latus Rectum Derivation of Equation of Hyperbola Let us consider a point P on the hyperbola whose coordinates are (x, y). From the definition of the hyperbola, we know that the difference between the distance of point P from the two foci F and F' is 2a, i.e., PF'-PF = 2a. Let the coordinates of the foci be F (c, o) and F '(-c, 0). Now, by using the coordinate distance formula, we can find the distance of point P (x, y) to the foci F (c, 0) and F '(-c, 0). √[(x + c)2 + (y - 0)2] - √[(x - c)2 + (y - 0)2] = 2a ⇒ √[(x + c)2 + y2] = 2a + √[(x - c)2 + y2] Now, by squaring both sides, we get (x + c)2 + y2 = 4a2 + (x - c)2 + y2 + 4a√[(x - c)2 + y2] ⇒ 4cx - 4a2 = 4a√[(x - c)2 + y2] ⇒ cx - a2 = a√[(x - c)2 + y2] Now, by squaring on both sides and simplifying, we get [(x2/a2) - (y2/(c2 - a2))] = 1 We have, c2 = a2 + b2, so by substituting this in the above equation, we get x2/a2 - y2/b2 = 1 Hence, the standard equation of the hyperbola is derived. Hyperbola Formula Following formulas are widely used in finding the various parameters which include, the equation of hyperbola, the major and minor axis, eccentricity, asymptotes, vertex, foci, and semi-latus rectum. \| Property \| Formula \| --- \| \| Equation of Hyperbola \| (x-xo)2 / a2 - (y-yo)2 / b2 = 1 \| \| Major Axis \| y = y0; Length = 2a \| \| Minor Axis \| x = x0; Length = 2b \| \| Eccentricity \| e = √(1 + b2/a2) \| \| Asymptotes \| y= y0 ±(b/a)(x − x0) \| \| Vertex \| (a, y0) and (−a, y0) \| \| Focus (Foci) \| (a, √(a2 + b2)y0) and (−a, √(a2 + b2)y0) \| \| Semi-Latus Rectum (p) \| p = b2/a \| \| Equation of Tangent \| (xx1)/a2 - (yy1)/b2 = 1, \| \| Equation of Normal \| y−y1=(−y1a2)(x−x1) / (x1b2), at point (x1,y1) where, x1 ≠ 0 \| Where, Graph of Hyperbola A hyperbola is a curve that has two unbounded curves that are mirror images of each other. The graph shows that curve in the 2-D plane. We can observe the different parts of a hyperbola in the hyperbola graphs for standard equations given below: \| Equation of the Hyperbola \| Graph of Hyperbola \| Parameters of Hyperbola \| --- \| \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1 \| Graph of Hyperbola 1 \| Coordinates of the center: (0, 0) Coordinates of the vertex: (a, 0) and (-a, 0) Coordinates of foci: (c, 0) and (-c, 0) The length of the transverse axis = 2a The length of the conjugate axis = 2b The length of the latus rectum = 2b2/a Equations of asymptotes: y = (b/a) x and y = -(b/a) x Eccentricity (e) = √[1 + (b2/a2)] \| \| \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}= 1 \| Graph of Hyperbola 2 \| Coordinates of the center: (0, 0) Coordinates of the vertex: (0, a) and (0, -a) Coordinates of foci: (0, c) and (0, -c) The length of the transverse axis = 2b The length of the conjugate axis = 2a The length of the latus rectum = 2b2/a Equations of asymptotes: y = (a/b) x and y = -(a/b) x Eccentricity (e) = √[1 + (b2/a2)] \| Equation of the Hyperbola Graph of Hyperbola Parameters of Hyperbola \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1 Coordinates of the center: (0, 0) Coordinates of the vertex: (a, 0) and (-a, 0) Coordinates of foci: (c, 0) and (-c, 0) The length of the transverse axis = 2a The length of the conjugate axis = 2b The length of the latus rectum = 2b2/a Equations of asymptotes: y = (b/a) x and y = -(b/a) x Eccentricity (e) = √[1 + (b2/a2)] \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}= 1 Coordinates of the center: (0, 0) Coordinates of the vertex: (0, a) and (0, -a) Coordinates of foci: (0, c) and (0, -c) The length of the transverse axis = 2b The length of the conjugate axis = 2a The length of the latus rectum = 2b2/a Equations of asymptotes: y = (a/b) x and y = -(a/b) x Eccentricity (e) = √[1 + (b2/a2)] Conjugate Hyperbola Conjugate Hyperbola are 2 hyperbolas such that the transverse and conjugate axes of one hyperbola are the conjugate and transverse axis of the other hyperbola respectively. Conjugate hyperbola of (x2 / a2) – (y2 /b2) = 1 is, (x2 / a2) - (y2 / b2) = 1 Where, Properties of Hyperbola (1 / e12) + (1 / e22) = 1 Auxiliary Circles of Hyperbola Auxiliary Circle is a circle that is drawn with center C and diameter as a transverse axis of the hyperbola. The auxiliary circle of the hyperbola equation is, x2 + y2 = a2 Rectangular Hyperbola A hyperbola with a transverse axis of 2a units and a conjugate axis of 2b units of equal length is called the Rectangular Hyperbola. i.e. in rectangular hyperbola, 2a = 2b ⇒ a = b The equation of a Rectangular Hyperbola is given as follows: x2 – y2 = a2 Note: Eccentricity of Rectangular hyperbola is √2. Parametric Representation of Hyperbola The parametric Representation of auxiliary circles of the hyperbola is: x = a sec θ, y = b tan θ People Also Read Hyperbola Class 11 In Class 11 mathematics, the study of hyperbolas forms a part of the conic sections in analytic geometry. Understanding hyperbolas at this level involves exploring their definition, standard equations, properties, and various elements associated with them. Conclusion Solved Examples on Hyperbola Question 1: Determine the eccentricity of the hyperbola x2/64 - y2/36 = 1. Solution: Equation of hyperbola is x2/64 - y2/36 = 0 By comparing given equation with standard equation of the hyperbola x2/a2 - y2/b2 = 1, we get a2 = 64, b2 = 36 ⇒ a = 8, b = 6 We have, Eccentricity of a hyperbola (e) = √(1 + b2/a2) ⇒ e = √(1 + 62/82) ⇒ e = √(1 + 36/64) ⇒ e = √(64 + 36)/64) = √(100/64) ⇒ e = 10/8 = 1.25‬ Hence, Eccentricity of given hyperbola is 1.25‬. Question 2: If the equation of the hyperbola is [(x-4)2/25] - [(y-3)2/9] = 1, find the lengths of the major axis, minor axis, and latus rectum. Solution: Equation of hyperbola is [(x-4)2/25] - [(y-3)2/9] = 1 By comparing given equation with the standard equation of the hyperbola, (x - h)2/a2 - (y - k)2/b2 = 1 Here, x = 4 is the major axis and y = 3 is the minor axis. a2 = 25 a = 5 b2 = 9 b = 3 Length of major axis = 2a = 2 × (5) = 10 units Length of minor axis = 2b = 2 × (3) = 6 units Length of latus rectum = 2b2/a = 2(3)2/5 = 18/5 = 3.6 units Question 3: Find the vertex, asymptote, major axis, minor axis, and directrix if the hyperbola equation is [(x-6)2/72]-[(y-2)2/42] = 1. Solution: Equation of hyperbola is [(x-6)2/72] - [(y-2)2/42] = 1 By comparing given equation with standard equation of hyperbola, (x - h)2/a2 - (y - k)2/b2 = 1 h = 6, k = 2, a = 7, b = 4 Vertex of a Hyperbola: (h + a, k) and (h - a, k) = (13, 2) and (-1, 2) Major axis of Hyperbola is x = h x = 6 Minor axis of Hyperbola is y = k y = 2 Equations of asymptotes of hyperbola are y = k − (b / a)x + (b / a)h and y = k+ (b / a)x - (b / a)h ⇒ y = 2 - (4/7)x + (4/7)6 and y = 2 + (4/7)x - (4/7)6 ⇒ y = 2 - 0.57x + 3.43 and y = 2 + 0.57x - 3.43 ⇒ y = 5.43 - 0.57x and y = -1.43 + 0.57x Equation of the directrix of a hyperbola is x = ± a2/√(a2 + b2) ⇒ x = ± 72/√(72 + 42) ⇒ x= ± 49/√65 ⇒ x = ± 6.077 Question 4: Find the eccentricity of the hyperbola whose latus rectum is half of its conjugate axis. Solution: Length of latus rectum is half of its conjugate axis Let, Equation of hyperbola be [(x2 / a2) – (y2 / b2)] = 1 Conjugate axis = 2b Length of Latus rectum = (2b2 / a) From given data, (2b2 / a) = (1/2) × 2b 2b = a We have, Eccentricity of Hyperbola (e) = √[1 + (b2/a2)] Now, substitute a = 2b in the formula of eccentricity ⇒ e = √[1 + (b2/(2b)2] ⇒ e = √[1 + (b2/4b2)] = √(5/4) ⇒ e = √5/2 Hence, required eccentricity is √5/2. Practice Problems on Hyperbola Question 1: Find the standard form equation of the hyperbola with vertices at (-3, 2) and (1, 2), and a focal length of 5. Question 2: Determine the center, vertices, and foci of the hyperbola with the equation 9x2 - 4y2 = 36. Question 3: Given the hyperbola with the equation (x - 2)2/16 - (y + 1)2/9 = 1, find the coordinates of its center, vertices, and foci. Question 4: Write the equation of the hyperbola with a horizontal major axis, center at (0, 0), a vertex at (5, 0), and a focus at (3, 0). 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14764	https://www.ams.org/bookstore/pspdf/stml-65-prev.pdf	Chapter 7 The Stirling Numbers of the Second Kind 7.1. Introduction This chapter considers the Stirling numbers of the second kind with special emphasis on their arithmetic properties. The other kind of Stirling numbers, those of the ﬁrst kind, are not considered here. Deﬁnition 7.1.1. Let n, k ≥1. The Stirling numbers of the second kind S(n, k) count the ways to divide a set of n objects into k nonempty subsets. Note 7.1.2. From the deﬁnition it is clear that S(n, k) = 0 if n < k. The extension S(0, k) = 1 if k = 0, 0 if k > 0 deﬁnes S(0, k). Example 7.1.3. The number S(n, 1) = 1 since the only option is to place all the objects into the single subset. Similarly, S(n, n) = 1 since then you must place each object in a diﬀerent subset. 191 192 7. The Stirling Numbers of the Second Kind Example 7.1.4. The list {{1, 2, 3}, {4}}, {{1, 2, 4}, {3}}, {{1, 3, 4}, {2}}, {{2, 3, 4}, {1}}, {{1, 2}, {3, 4}}, {{1, 3}, {2, 4}}, {{1, 4}, {2, 3}} shows that S(4, 2) = 7. Exercise 7.1.5. Give a combinatorial proof of the value (7.1.1) S(n, n −1) = n 2 . Exercise 7.1.6. Prove that (7.1.2) S(n, 2) = 2n−1 −1. 7.2. A recurrence In the partition of {1, 2, . . . , n} into k nonempty subsets, there are cases in which n appears as a singleton and others in which n is part of one of the k nonempty subsets with more than one element. In the ﬁrst case, the partition is {n} together with a partition of {1, 2, . . . , n−1} into k−1 nonempty parts. There are S(n−1, k−1) of them. In the second case, taking n out of the partition yields a collection of k nonempty subsets partitioning {1, 2, . . . , n−1}. There are S(n −1, k) of them. The number n can be placed back into any of the k parts. This proves the next result. Theorem 7.2.1. The Stirling numbers of the second kind satisfy the recurrence (7.2.1) S(n, k) = S(n −1, k −1) + kS(n −1, k) for n ≥2. Exercise 7.2.2. The theorem and S(n, 1) = n state that (7.2.2) S(n, 2) −2S(n −1, 2) = S(n −1, 1) = 1. Solve this recurrence to conﬁrm the value S(n, 2) = 2n−1−1 stated in Exercise 7.1.6. Discuss the consistency of the value S(0, k) and this recurrence. 7.2. A recurrence 193 Exercise 7.2.3. Let f(x) = 1/(1 + ex). Prove that (7.2.3) f (n)(x) = n+1 k=1 a(n, k) (1 + ex)k where (7.2.4) a(n, k) = (−1)n+k+1(k −1)!S(n + 1, k). Hint: Diﬀerentiate (7.2.3) to obtain a recurrence for a(n, k). Exercise 7.2.4. Use the recurrence (7.2.1) to establish the generat-ing function (7.2.5) ∞ n=0 S(n, k)xn = xk (1 −x)(1 −2x)(1 −3x) · · · (1 −kx) for the Stirling numbers S(n, k). Exercise 7.2.5. Iterate the recurrence (7.2.1) to obtain the identity S(n + j, k) = j i=0 pj,i(k)S(n, k −j + i) where the pj,i(k) are polynomials in k of degree i that satisfy the recurrence pj+1,i(k) = pj,i(k) + (k −j + i −1)pj,i−1(k) and have the initial conditions pj,0(k) = 1 and pj,j(k) = kj. Write the polynomials pj,i in terms of the falling factorials (k)r = k(k −1) · · · (k −r + 1), in the form pj,i(k) = i r=0 cj,i(r)(k)r, and check the recurrence cj+1,i(r) = cj,i(r) + (r −j + i −1)cj,i−1(r) + cj,i−1(r). The next theorem presents a combinatorial proof of a diﬀerent type of recurrence satisﬁed by the Stirling numbers. 194 7. The Stirling Numbers of the Second Kind Theorem 7.2.6. The Stirling numbers S(n, k) satisfy the recurrence (7.2.6) k!S(n, k) = kn − k−1 i=1 k(k −1) · · · (k −i + 1)S(n, i). Proof. Suppose there are n balls (numbered 1 to n) and k boxes (numbered 1 to k). Let t(n, k) be the number of ways to place the n balls into the k boxes in such a way that no box is empty. Then t(n, k) = k!S(n, k) and t(n, 1) = 1. Think about this. Assume that t(n, 1), . . . , t(n, k −1) is known. There are kn ways to place the n balls into the k boxes if the requirement that no box should be empty is dropped. Now, for each i in the range 1 ≤i ≤k −1, there are k i conﬁgurations for which exactly i boxes are empty, and for each such conﬁguration, there are t(n, k −i) ways to place the n balls into the other k −i boxes. Then t(n, k) = kn − k−1 i=1 k i t(n, k −1). Multiply by k! to get the stated recurrence. □ Exercise 7.2.7. Use (7.2.6) to conﬁrm the value S(n, 2) = 2n−1 −1 given as Exercise 7.1.6. 7.3. An explicit formula This section produces an explicit formula for S(n, k). The proofs employ the inclusion-exclusion principle. A short explanation is presented ﬁrst. Consider a collection of sets {Aj : 1 ≤j ≤n} and denote the cardinality of Aj by \|Aj\|. If the sets are disjoint, that is, Ai ∩Aj = ∅for i ̸= j, then (7.3.1) n j=1 Aj = n j=1 \|Aj\|. The proof is clear: every element in the union of the sets belongs to a unique set Aj. Thus, in (7.3.1), every element is counted exactly once. The inclusion-exclusion principle, stated below, describes how to count unions in case the sets have elements in common. 7.3. An explicit formula 195 Theorem 7.3.1. Let {Aj : 1 ≤j ≤n} be a collection of sets. Then n j=1 Aj = i1 \|Ai1\| − i1<i2 \|Ai1 ∩Ai2\| + i1<i2 n, since both sides vanish in this case. The inclusion-exclusion principle is now employed to produce a second count of the onto functions f : A →B. Let X be the set of all functions f : A →B. It is clear that \|X\| = kn. The value of the image for each element in A has exactly k choices. Now, for each i in the range 1 ≤i ≤k, deﬁne (7.3.6) Xi = {f : A →B : f omits i in its range.}. Then \|f : A →B that are onto \| = k 5 i=1 (X −Xi) = X − k i=1 Xi = kn − k i=1 Xi . Now, with the notation [n] = {1, 2, . . . , n}, observe that \|Xi\| = \|{f : [n] →[k −1]}\| = (k −1)n. Similarly \|Xi1 ∩Xi2 ∩· · · ∩Xij\| = \|{f : [n] →[k −j]}\| = k j (k −j)n. The inclusion-exclusion principle gives \|f : A →B that are onto \| = kn − ⎛ ⎝ k j=1 (−1)j−1 k j (k −j)n ⎞ ⎠ = k j=0 (−1)j k j (k −j)n. Comparing both computations gives the result. □ 7.4. The valuations of Stirling numbers 197 Exercise 7.3.3. Conﬁrm the values (7.3.7) S(n, 3) = 1 2(3n−1 −2n + 1) and (7.3.8) S(n, 4) = 1 6(3 · 2n−1 −3n + 22n−2 −1). Exercise 7.3.4. Use the result of Theorem 7.3.2 to check that the numbers a(n, k) in Exercise 7.2.3 can be written as (7.3.9) a(n, k) = (−1)n k−1 j=0 (−1)j k −1 j (j + 1)n. 7.4. The valuations of Stirling numbers This section discusses the sequence {ν2(S(n, k)) : n ≥k} for k ﬁxed. The cases k = 1, 2 are elementary since S(n, 1) = 1 and S(n, 2) = 2n −1. Therefore ν2(S(n, k)) = 0 for these values of k. The next theorem deals with k = 3. Theorem 7.4.1. The 2-adic valuations of the Stirling numbers of order 3 (7.4.1) S(n, 3) = 1 2(3n−1 −2n + 1), for n ≥3, are given by (7.4.2) ν2(S(n, 3)) = 0 if n is odd, 1 if n is even. Proof. The explicit formula comes from Exercise 7.3.3. Iterate the recurrence (7.4.3) S(n, 3) = S(n −1, 2) + 3S(n −1, 3) to produce (7.4.4) S(n, 3) = 2n−2 −1 + n−3 k=1 3k(2n−k−2 −1), for n ≥3. This shows that if n is odd, then S(n, 3) is odd. Thus ν2(S(n, 3)) = 0. 198 7. The Stirling Numbers of the Second Kind To treat the case of n even, iterate (7.4.3) to obtain (7.4.5) S(n, 3) = 2n−2 + 3 · 2n−3 −4 + 9S(n −2, 3). As an inductive step write S(n −2, 3) = 2Tn−2 with Tn−2 odd. Then (7.4.6) 1 2S(n, 3) = 2n−3 + 3 · 2n−4 −2 + 9Tn−2 is an odd integer. This completes the induction. □ The case of k = 4 can be decided in a similar manner. Theorem 7.4.2. The Stirling numbers of order 4 (7.4.7) S(n, 4) = 1 6(4n−1 −3n + 3 · 2n−1 −1) satisfy (7.4.8) ν2(S(n, 4)) = 1 −ν2(S(n, 3)) = 1 if n is odd, 0 if n is even. Proof. The expression for S(n, 4) comes from Theorem 7.3.2. The recurrence (7.2.1) gives (7.4.9) S(n, 4) = S(n −1, 3) + 4S(n −1, 4). For n even, the value S(n−1, 3) is odd, so that S(n, 4) is odd. There-fore ν2(S(n, 4)) = 0. For n odd, S(n, 4) is even since S(n −1, 3) is even. The relation (7.4.9) is now written as (7.4.10) 1 2S(n, 4) = 1 2S(n −1, 3) + 2S(n −1, 4). The value ν2(S(n −1, 3)) = 1 shows that the right-hand side is odd, yielding ν2(S(n, 4)) = 1. □ 7.4.1. The valuation of S(n, 5). The ﬁrst nontrivial case occurs when k = 5. The sequence of values for ν2(S(n, 5)) is computed by the formula (7.4.11) S(n, 5) = 1 24(5n−1 −4n + 2 · 3n −2n+1 + 1), n ≥5, coming from Theorem 7.3.2 or using the recurrence (7.4.12) S(n, 5) = S(n −1, 4) + 5S(n −1, 5). 7.4. The valuations of Stirling numbers 199 The ﬁrst few values of ν2(S(n, 5)) are given by ν2(S(n, 5)) = {0, 0, 2, 1, 0, 0, 1, 3, 0, 0, 3, 1, 0, 0, 1, 2}, and the pattern, shown in Figure 7.4.1, is now not easy to predict. 50 100 150 200 n 2 4 6 8 10 Figure 7.4.1. The 2-adic valuation of S(n, 5). The analysis of the valuation ν2(S(n, 5)) begins with some ele-mentary observations. Lemma 7.4.3. The Stirling numbers S(n, 5) satisfy (7.4.13) ν2(S(4n + 1, 5)) = ν2(S(4n + 2, 5)) = 0. Proof. The recurrence (7.4.12) yields S(4n + 1, 5) = S(4n, 4) + 5S(4n −1, 4) + 52S(4n −2, 4) +53S(4n −3, 4) + 54S(4n −3, 5) and the result for ν2(S(4n+1, 5)) follows by induction using the parity (7.4.14) S(n, 4) ≡ 1 mod 2 if n ≡0 mod 2, 0 mod 2 if n ≡1 mod 2. The case of S(4n + 2, 5) is similar. □ It remains to describe ν2(S(4n, 5)) and ν2(S(4n+3, 5)). The ﬁrst few values of ν2(S(4n, 5)), for n ≥2, are given by ν2(S(4n, 5)) = {1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1}. 200 7. The Stirling Numbers of the Second Kind The next step in the analysis of ν2(S(n, 5)) is to show that every other entry in the values of ν2(S(4n, 5)) is 1. Lemma 7.4.4. The Stirling numbers S(n, 5) satisfy ν2(S(8n, 5)) = 1 and ν2(S(8n + 4, 5)) ≥2 and also ν2(S(8n + 3, 5)) = 1 and ν2(S(8n + 7, 5)) ≥2. Proof. The identity 24S(8n, 5) = 58n−1 −48n + 2 · 38n −28n+1 + 1 is considered modulo 32. Using 58 ≡1 and 57 ≡13, it follows that 58n−1 ≡13. Also, 48n ≡0 and 38n ≡1. Therefore 58n−1 −48n + 2 · 38n −28n+1 + 1 ≡16 mod 32. This gives 24S(8n, 5) = 32t+16 for some t ∈N leading to 3S(8n, 5) = 2(2t + 1). Therefore ν2S(8n, 5) = 1. The valuation of S(8n + 4, 5) comes from the relation 24S(8n + 4, 5) = 58n+3 −48n+4 + 2 · 38n+4 −28n+5 + 1 modulo 32. Proceeding as before, it follows that 24S(8n + 4, 5) ≡ 0 mod 32. Therefore 24S(8n + 4, 5) = 32t for some t ∈N. This yields ν2(S(8n + 4, 5)) ≥2. The proof of the remaining cases is similar. □ Exercise 7.4.5. Give proofs of Theorems 7.4.1 and 7.4.2 in the style of the proof of Lemma 7.4.4. Note 7.4.6. The results described in Lemmas 7.4.3 and 7.4.4 can be described in terms of a tree similar to the example in Note 1.7.3. The procedure described here generates the valuation tree associated to the p-adic valuation of a sequence {xn : n ∈N}. Each of these trees has a branching number that depends on the prime p and the sequence {xn}. The branching number is denoted by b = b(p; xn). For clarity, in the construction described next, the prime p and the branching number b will be assumed to be equal to 2. The construction of the tree begins with a root vertex. This vertex represents the whole set N and it forms the 0th level of the tree. Now assume that the kth level has been formed. This level 7.4. The valuations of Stirling numbers 201 represents some of the modular classes modulo 2k. The transition to the next level is achieved by the following rules: let n0 be the label of a vertex at the kth level. The label indicates that the vertex corresponds to the class an0,k = {n ∈N : n ≡n0 mod 2k}. Then the following question is asked: Does the valuation {ν2(xn) : n ∈an0,k} reduce to a single value? If the answer is yes, the vertex n0 is declared to be a terminal vertex and the common value of the valuation is attached to n0. If the answer is no, then the vertex n0 is split into two classes modulo 2k+1, namely, {n ∈N : n ≡n0 mod 2k+1} and {n ∈N : n ≡n0 + 2k mod 2k+1}. The vertices produced by this splitting form the (k + 1)st level. This process is now applied to the sequence xn = S(n, 5) and the prime p = 2. The valuation tree starts with a root vertex representing all N. At this point, the question is whether {ν2(S(n, 5)) : n ∈N} is independent of n. The values S(5, 5) = 1 and S(7, 5) = 140 = 22 · 5 · 7 show that ν2(S(5, 5)) = 0 and ν2(S(7, 5)) = 2. Therefore the valuation ν2(S(n, 5)) depends on n and the answer is no. This leads to a splitting of the root vertex into two vertices: one labeled 0, which represents the class a0,1 = {n ∈N : n ≡0 mod 2}, and the second one, labeled 1, representing the class a1,1 = {n ∈N : n ≡1 mod 2}. These two classes form the ﬁrst level. The reader can check that each of these two classes do not have a constant 2-adic valuation and the process continues to the next level. 2n 2n Figure 7.4.2. The ﬁrst level of the tree for S(n, 5). 202 7. The Stirling Numbers of the Second Kind The class a0,1 now splits into a0,2 = {n ∈N : n ≡0 mod 22} and a2,2 = {n ∈N : n ≡2 mod 22} and a1,1 splits into a1,2 = {n ∈N : n ≡1 mod 22} and a3,2 = {n ∈N : n ≡3 mod 22}. The construction of the ﬁrst two levels is depicted in Figure 7.4.3. 2n 2n−1 4n 4n+2 4n+1 4n+3 0 0 Figure 7.4.3. The ﬁrst two levels of the tree for S(n, 5) 2n 2n 4n 4n 4n 4n 8n 8n 8n 8n 0 0 1 1 + + + + + + Figure 7.4.4. The ﬁrst three levels of the tree for S(n, 5). The vertices marked with 4n + 1 and 4n + 2 terminate at level 2. These correspond to the classes a1,2 and a2,2, respectively. They are both marked with 0, according to Lemma 7.4.3. The vertices marked 4n and 4n+3, corresponding to the classes a0,2 and a3,2, respectively, are split to produce the vertices 8n, 8n + 4 and 8n + 3, 8n + 7. These form the third level. Lemma 7.4.4 states that the vertices labeled 7.4. The valuations of Stirling numbers 203 8n and 8n + 3 terminate at this level and 8n + 4, 8n + 7 split into the four vertices 16n, 16n + 8, 16n + 7, 16n + 15. These vertices form the fourth level. 0 0 1 1 2 2 3 3 4 4 5 5 6 6 Figure 7.4.5. The continuation of the tree for S(n, 5). The main result of T. Amdeberhan, D. Manna, and V. Moll is stated next. Theorem 7.4.7. The valuation tree associated to the 2-adic valuation of S(n, 5) has, at every level starting with the third one, four vertices. At each level, two of these vertices terminate and the other two split to form the next level. Conjecture 7.4.8. The same type of behavior occurs for the tree associated to S(n, k) for k ≥6. The corresponding tree begins as before, with a double splitting. At some point this process changes and half of the vertices terminate and the other half split. Eventually the number of vertices per level remains constant. Note 7.4.9. For k ﬁxed, the sequence ν2(S(n, k)) oﬀers a wide variety of proﬁles. The next sequence of ﬁgures oﬀers a sample of them. It is unknown how to predict the form of the graph in terms of the ﬁxed index k. 204 7. The Stirling Numbers of the Second Kind 50 100 150 200 n 2 4 6 8 10 0 1 0 2 1 3 0 4 2 6 1 5 3 7 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 0 0 0 0 1 1 1 1 Figure 7.4.6. The 2-adic valuation of S(n, 20). 7.4. The valuations of Stirling numbers 205 100 200 300 400 500 n 1 2 3 4 5 6 100 200 300 400 500 n 2 4 6 8 10 12 Figure 7.4.7. The 2-adic valuation of S(n, 33) and S(n, 48). 206 7. The Stirling Numbers of the Second Kind 100 200 300 400 500 n 2 4 6 8 100 200 300 400 500 n 5 10 15 20 Figure 7.4.8. The 2-adic valuation of S(n, 99) and S(n, 128). 7.4. The valuations of Stirling numbers 207 100 200 300 400 500 n 2 4 6 8 100 200 300 400 500 n 2 4 6 8 10 12 14 Figure 7.4.9. The 2-adic valuation of S(n, 194) and S(n, 215). Note 7.4.10. There are many naturally occuring sequences whose p-adic valuations are capable of a complete analytic description. The 208 7. The Stirling Numbers of the Second Kind case of the ASM-numbers, deﬁned in (1.3.10) by (7.4.15) An = n−1 j=0 (3j + 1)! (n + j)! , has been described completely. Figure 7.4.10 shows the 2-adic valua-tion of An. 50 100 150 200 250 300 n 20 40 60 80 Figure 7.4.10. The 2-adic valuation of ASM-numbers. The sequence An counts a famous class of matrices. An alter-nating sign matrix is an array of 0, 1, and −1 such that the entries of each row and column add up to 1 and the nonzero entries of a given row/column alternate. After a fascinating sequence of events, D. Zeilberger proved that the numbers of such matrices is given by (7.4.15). In particular, the numbers An are integers—not an ob-vious fact. The story behind this formula and its many combinatorial inter-pretations are given in D. Bressoud’s book . The ﬁrst step in the analysis of ν2(An) was to characterize the indices n for which An is odd. These are the values where the graph in Figure 7.4.10 achieves its minimum. This sequence of indices starts as (7.4.16) 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 7.4. The valuations of Stirling numbers 209 and by looking into Sloane’s Encyclopedia of Integer Sequences, we see that they were recognized as the Jacobsthal numbers, with label A001045. These numbers satisfy the recurrence Jn = Jn−1 + 2Jn−2, J0 = 1, J1 = 1. The many interpretations of these numbers include the number of ways to tile a 3 × (n −1) rectangle with squares of size 1 or 2 and also as the numerators in the reduced fraction (7.4.17) 1 2 −1 4 + 1 8 −1 16 + 1 32 −· · · . The complete description of the p-adic valuation of ASM-numbers can be found in the paper by E. Beyerstedt, V. Moll, and X. Sun as well as in the paper by V. Moll and X. Sun . The main result is an expression for νp(An) similar to the content of Exercise 2.6.3, where the classical series for the valuation νp(n) is expressed as a series in which each summand is a periodic function of period pj. Theorem 7.4.11. Let n ∈N and let p ≥5 be a prime. Deﬁne Perj,p(n) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0 if 0 ≤n ≤ $ pj+1 3 % , n − $ pj+1 3 % if $ pj+1 3 % + 1 ≤n ≤pj−1 2 , $ 2pj+1 3 % −n if pj+1 2 ≤n ≤ $ 2pj+1 3 % , 0 if $ 2pj+1 3 % + 1 ≤n ≤pj −1. Then νp(An) = ∞ j=1 Perj,p n mod pj . Each summand in the series is of period pj. Note 7.4.12. The arithmetical statements about An can be extended to the sequence An(q) := n−1 j=0 (qj + 1)! (n + j)! , for q ∈N with q ≥3. The case of q = 3 corresponds to the ASM-numbers. An interesting question is to ﬁnd a combinatorial interpre-tation of An(q); that is, what do these numbers count? Chapter 15 Landen Transformations 15.1. Introduction The transformation of variables plays an important role in the theory of deﬁnite integrals. From the beginning, the reader has been exposed to some common changes of variables, motivated mainly by the fact that they work. For example, the basic knowledge of trigonometry presented in Chapter 12 shows that confronted with a problem of the type (15.1.1) I(a, b) = b a dt √ 1 −t2 , the change of variables (15.1.2) t = sin x leads to a simpler form of the integral. Naturally, this change of variable, presented in the form (15.1.3) x = sin−1 t manifests the new variable of integration as the inverse of a transcen-dental function. A diﬀerent type of map that leaves certain integrals invariant is deﬁned next. 411 412 15. Landen Transformations Deﬁnition 15.1.1. A Landen transformation for an integral (15.1.4) I = x1 x0 f(x; p) dx, which depends on a set of parameters p, is a map Φ, deﬁned on the parameters of I, such that (15.1.5) x1 x0 f(x; p) dx = Φ(x1) Φ(x0) f(x; Φ(p)) dx. Example 15.1.2. The classical example of a Landen transformation is given by (15.1.6) E(a, b) = a + b 2 , √ ab , which preserves the elliptic integral (15.1.7) G(a, b) = π/2 0 dx , a2 cos2 x + b2 sin2 x , that is, (15.1.8) G(a, b) = G a + b 2 , √ ab . It turns out that the sequence (an, bn) deﬁned inductively by (15.1.9) (an, bn) = E(an−1, bn−1) with (a0, b0) = (a, b) has the property that (15.1.10) lim n→∞an = lim n→∞bn. This limit is the arithmetic-geometric mean of a and b, denoted by AGM(a, b). The invariance of the elliptic integral shows that (15.1.11) G(a, b) = π 2AGM(a, b). This may be used to compute the elliptic integral G(a, b) by iteration. This chapter discusses Landen transformations in the case when the integrand is a rational function. The idea is to produce an appro-priate change of variables that leaves the rational integral invariant. The special example (15.1.12) x = R2(t) = t2 −1 2t 15.2. An elementary example 413 appearing in Example 8.1.6 leads to the simplest nontrivial class of Landen transformations. This chapter contains details of the eﬀect of this map on two special kinds of integrands. The ﬁrst one establishes the formula (15.1.13) ∞ 0 dx (x4 + 2ax2 + 1)m+1 = π 2 1 [2(1 + a)]m+1/2 Pm(a), where Pm(a) is a polynomial in a. Warning: The symbol Pm has been used to denote other polynomials in this text, for instance the Legendre polynomials. In this chapter, this refers only to the polynomial (15.4.16). Many properties of its coeﬃcients are presented (not proved) in this chapter. The second example provides a Landen transformation for the integral (15.1.14) U6(a, b; c, d, e) = ∞ 0 cx4 + dx2 + e x6 + ax4 + bx2 + 1 dx, which leads to an interesting nonlinear transformation. 15.2. An elementary example The goal of this section is to establish the following result. Theorem 15.2.1. Let f be a function, with ﬁnite integral over R. Deﬁne (15.2.1) f±(x) = f(x + , x2 + 1) ± f(x − , x2 + 1) and (15.2.2) L(f)(x) = f+(x) + xf−(x) √ x2 + 1 . Then (15.2.3) ∞ −∞ f(t) dt = ∞ −∞ L(f)(x) dx. Proof. The map x = R2(t) has two branches separated by the pole at t = 0. The inverses are given by (15.2.4) t = x ± , x2 + 1. 414 15. Landen Transformations Each branch maps a half-line onto R. Therefore it is natural to con-sider integrals over the whole line. The change of variables (15.2.4) leads to I = ∞ −∞ f(t) dt = 0 −∞ f(t) dt + ∞ 0 f(t) dt = ∞ −∞ f(x − , x2 + 1) 1 − x √ x2 + 1 dx + ∞ −∞ f(x + , x2 + 1) 1 + x √ x2 + 1 dx. Now collect terms to obtain the claim. □ Example 15.2.2. Take f(t) = 1/(t2 + 1). Then L(f)(x) = f(x) and the function f is ﬁxed by L. Now take f(t) = 1/(t2 + 2) to obtain (15.2.5) L(f)(x) = 6 8x2 + 9. The theorem gives the elementary identity (15.2.6) ∞ −∞ dx x2 + 2 = ∞ −∞ 6 dx 8x2 + 9. Both sides evaluate to π/ √ 2. Example 15.2.3. The function f(t) = sin t t and the value (15.2.7) ∞ −∞ sin t t dt = π obtained in (12.11.1) lead to the nontrivial integral (15.2.8) ∞ −∞ cos x sin √ x2 + 1 √ x2 + 1 dx = π 2 . The current version of Mathematica (the 8th) is unable to evaluate this integral. 15.3. The case of rational integrands 415 15.3. The case of rational integrands In the special case that the integrand f(x) is a rational function, Theorem 15.2.1 gives an identity among two rational integrals. This is the content of the next theorem. Theorem 15.3.1. Assume f(x) is a rational function. Then L(f(x)) is also rational. Proof. If f(x) is a rational function, then f(x + , x2 + 1) + f(x − , x2 + 1) and f(x + √ x2 + 1) −f(x − √ x2 + 1) √ x2 + 1 are also rational functions. Indeed, let y = √ x2 + 1 and assume f(x) = A(x)/B(x). Then f(x + y) + f(x −y) = A(x + y) B(x + y) + A(x −y) B(x −y) = A(x + y)B(x −y) + A(x −y)B(x + y) B(x + y)B(x −y) . The numerator is a polynomial in x and y, invariant under y →−y. Therefore it is a polynomial in y2 = x2 + 1, thus a polynomial in x. The same argument applies to the denominator. □ Example 15.3.2. Let (15.3.1) f(x) = 1 4x2 + 12x + 21. Then (15.3.2) L(f(x)) = 50 336x2 + 408x + 481. Both integrals may be evaluated in elementary terms to produce the common value (15.3.3) ∞ −∞ f(x) dx = ∞ −∞ L(f(x)) dx = π 4 √ 3. 416 15. Landen Transformations Note 15.3.3. A Landen transformation of a rational function has been deﬁned as a map of the function’s coeﬃcients. For instance, applying L to the function (15.3.4) f(x) = 1 ax2 + bx + c yields (15.3.5) L(f(x)) = 1 a1x2 + b1x + c1 with (15.3.6) a1 = 2 c a c + a, b1 = b(c −a) c + a , c1 = (c + a)2 −b2 2(c + a) . Exercise 15.3.4. Check that the discriminant of the denominator is preserved, that is, b2 −4ac = b2 1 −4a1c1. Note 15.3.5. Deﬁne (15.3.7) Φ2(a, b, c) = (a1, b1, c1) with (a1, b1, c1) given in (15.3.6). Iteration of Φ2 gives a sequence (an, bn, cn) that preserves the original integral (15.3.8) ∞ −∞ dx anx2 + bnx + cn = ∞ −∞ dx ax2 + bx + c. Exercise 15.3.6. Prove that (an, bn, cn) converges to (L, 0, L), for some L ∈R, under the assumption b2 −4ac < 0. Passing to the limit in (15.3.8) shows that (15.3.9) ∞ −∞ dx ax2 + bx + c = π L. In the case a > 0, the limiting value (15.3.10) L = 1 2 , 4ac −b2 is obtained from (15.3.9) by computing the integral. The case of a < 0 is similar. The map Φ2 is the rational analog of the classical arithmetic-geometric mean presented at the beginning of this chapter. A discussion of this case is presented in Section 15.6. 15.4. The evaluation of a quartic integral 417 Exercise 15.3.7. It is possible for a single integral to admit a variety of Landen transformations. For instance the quadratic integral (15.3.11) ∞ −∞ dx ax2 + bx + c is also invariant under the change of parameters an+1 = an (an+3cn)2−3b2 n (3an+cn)(an+3cn)−b2 n , (15.3.12) bn+1 = bn 3(an−cn)2−b2 n (3an+cn)(an+3cn)−b2 n , cn+1 = cn (3an+cn)2−3b2 n (3an+cn)(an+3cn)−b2 n , with a0 = a, b0 = b, and c0 = c. This is described in complete detail in the paper by D. Manna and V. Moll . Follow the steps given there to show that iteration of this converges to the stated limit (L, 0, L). 15.4. The evaluation of a quartic integral This section contains an application of the transformation L intro-duced in Theorem 15.2.1 to the evaluation of the deﬁnite integral (15.4.1) N0,4(a; m) = ∞ 0 dx (x4 + 2ax2 + 1)m+1 . The ﬁrst theorem describes the eﬀect of L on the integrand. Theorem 15.4.1. For m ∈N, let (15.4.2) Q(x) = 1 (x4 + 2ax2 + 1)m+1 . Then (15.4.3) Q1(y) := L(Q(x)) = Tm(2y) 2m(1 + a + 2y2)m+1 , where (15.4.4) Tm(y) = m k=0 m + k m −k y2k. 418 15. Landen Transformations Proof. Introduce the variable φ = y+ , y2 + 1. Then y− , y2 + 1 = −φ−1 and 2y = φ −φ−1. Moreover, Q1(y) = -Q(φ) + Q(φ−1) . + φ2 −1 φ2 + 1 Q(φ) −Q(φ−1) = 2 φ2 + 1 -φ2Q(φ) + Q(φ−1) . := Sm(φ). Then (15.4.3) is equivalent to (15.4.5) 2m 1 + a + 1 2(φ −φ−1)2m+1 Sm(φ) = Tm(φ −φ−1). A direct (but lengthy) simpliﬁcation of the left-hand side of (15.4.5) shows that this identity is equivalent to proving (15.4.6) φ2m+1 + φ−(2m+1) φ + φ−1 = Tm(φ −φ−1). Observe that the parameter a has disappeared. First proof. One simply checks that both sides of (15.4.6) satisfy the second-order recurrence (15.4.7) cm+2 −(φ2 + φ−2)cm+1 + cm = 0 and that the values for m = 0 and m = 1 match. This is straight-forward for the expression on the left-hand side, while the WZ-method settles the right-hand side. □ Second proof. In the textbook by R. Graham, D. Knuth, and O. Patashnik , one ﬁnds the generating function (15.4.8) Bt(z) = k≥0 (tk)k−1 zk k! , where (a)k = a(a + 1) · · · (a + k −1) is the Pochhammer symbol. The special values (15.4.9) B−1(z) = 1 + √1 + 4z 2 and B2(z) = 1 −√1 −4z 2z are combined to produce the identity 1 √1 + 4z B−1(z)n+1 −(−z)n+1B2(−z)n+1 = n k=0 n −k k zk. 15.4. The evaluation of a quartic integral 419 Replace n by 2m and z by (4y2)−1 to produce 1 2 , 1 + y2 (2y)2m (φ2m+1 + φ−(2m+1)) = m k=0 2m −k k zk. The sum on the right-hand side is simpliﬁed by the identity Tm(y) = m k=0 m + k m −k y2k = y2m m k=0 2m −k k y−2k. Thus, Tm(φ −φ−1) = Tm(2y) = (2y)2m m k=0 2m −k k zk, and it follows that Tm(φ −φ−1) = 1 2 , 1 + y2 (φ2m+1 + φ−(2m+1)), and the result is obtained from φ + φ−1 = 2 , y2 + 1. Evaluation of the integral N0,4(a; m). The identity in Theorem 15.2.1 shows that (15.4.10) ∞ 0 Q(x) dx = ∞ 0 Q1(y) dy, and this last integral can be evaluated in elementary form. Indeed, ∞ 0 Q1(y) dy = ∞ 0 Tm(2y) dy 2m(1 + 2y2)m+1 = 1 2m m k=0 m + k m −k ∞ 0 (2y)2k dy (1 + a + 2y2)m+1 . The change of variables y = t √1 + a/ √ 2 gives ∞ 0 Q1(y) dy = 1 [2(1 + a)]m+1/2 m k=0 m + k m −k 2k(1+a)k ∞ 0 t2k dt (1 + t2)m+1 . Exercise 15.4.2. Prove the Wallis-type identity (15.4.11) ∞ 0 t2k dt (1 + t2)m+1 = π 22m+1 2k k 2m −2k m −k m k −1 . 420 15. Landen Transformations The previous exercise now produces ∞ 0 Q1(y) dy = π 22m+1 1 [2(1 + a)]m+1/2 × m k=0 m + k m −k 2k 2k k 2m −2k m −k m k −1 (1 + a)k. This can be simpliﬁed further using (15.4.12) m + k m −k 2k k = m + k m m k , 0 ≤k ≤m, and (15.4.10) to produce ∞ 0 Q(y) dy = π 22m+1 1 [2(1 + a)]m+1/2 m k=0 2k m + k m 2m −2k m −k (1+a)k. Theorem 15.4.3. The integral N0,4(a; m), deﬁned in (15.4.1), is given by (15.4.13) N0,4(a; m) = π 2 1 [2(1 + a)]m+1/2 m j=0 dj,maj, where (15.4.14) dj,m = 2−2m m k=j 2k 2m −2k m −k m + k m k j . Note 15.4.4. The literature contains a variety of proofs of the for-mula given in Theorem 15.4.3. This is written here as (15.4.15) N0,4(a; m) = ∞ 0 dx (x4 + 2ax2 + 1)m+1 = π 2 Pm(a) [2(a + 1)]m+1/2 where (15.4.16) Pm(a) = m j=0 dj,maj. This note discusses some of these proofs and highlights the remarkable properties of the coeﬃcients dj,m. The ﬁrst proof. This proof is due to George Boros, a former stu-dent of the author. The idea is remarkably simple but has profound 15.4. The evaluation of a quartic integral 421 consequences. The change of variables x = tan θ yields N0,4(a; m) = π/2 0 cos4 θ sin4 θ + 2a sin2 θ cos2 θ + cos4 θ m+1 × dθ cos2 θ . Observe that the denominator of the integrand is a polynomial in cos 2θ. In terms of the double-angle u = 2θ, the original integral becomes N0,4(a; m) = 2−(m+1) π 0 (1 + cos u)2 (1 + a) + (1 −a) cos2 u m+1 × du 1 + cos u. Expanding the binomial (1 + cos u)2m+1, symmetry implies that π 0 (cos u)j du [(1 + a) + (1 −a) cos2 u]m+1 = 0, for j odd. The remanining integrals, those with j even, can be eval-uated by using the double-angle trick one more time. This leads to N0,4(a; m) = m j=0 2−j 2m + 1 2j π 0 (1 + cos v)j dv [(3 + a) + (1 −a) cos v]m+1 , where v = 2u and the symmetry of cosine about v = π has been used to reduce the integrals from [0, 2π] to [0, π]. The familiar change of variables z = tan(v/2) produces the form (15.4.15). The expression obtained for the coeﬃcients dj,m is not very pretty: dj,m = j r=0 m−j s=0 m k=j+s (−1)k−j−s 23k 2k k 2m + 1 2s + 2r m −s −r m −k × s + r r k −s −r j −r . A detour into the world of Ramanujan. The search for a simpler expression for the coeﬃcients dj,m began with the observation that they appear to be positive. Indeed, a symbolic calculation shows that for m = 5, these are {dj,5 : 0 ≤j ≤5} = 4389 256 , 8589 128 , 7161 64 , 777 8 , 693 16 , 63 8 . 422 15. Landen Transformations The second proof of (15.4.15) begins with the value of the elementary integral (15.4.17) ∞ 0 dx bx4 + 2ax2 + 1 = π 2 √ 2 1 , a + √ b and the functions h(c) = , a + √1 + c and g(c) = ∞ 0 dx x4 + 2ax2 + 1 + c. Then (15.4.17) gives g′(c) = π √ 2 h′(c). In particular, h′(0) = 1 π √ 2N0,4(a; 0). Further diﬀerentiation gives the higher-order derivatives of h in terms of the integrals N0,4. This is expressed as Theorem 15.4.5. The Taylor expansion of h(c) = , a + √1 + c is given by < a + √ 1 + c = √ a + 1 + 1 π √ 2 ∞ k=1 (−1)k−1 k N0,4(a; k −1)ck. The evaluation of the integrals N0,4(a; m) is now ﬁnished by using the Ramanujan master theorem stated below. Theorem 15.4.6. Suppose F has a Taylor expansion around c = 0 of the form F(c) = ∞ k=0 (−1)k k! ϕ(k) ck. Then, the moments of F, deﬁned by Mn = ∞ 0 cn−1F(c) dc, can be computed via Mn = Γ(n)ϕ(−n). B. Berndt , in the ﬁrst volume of Ramanujan’s Notebooks, provides a proof of the exact hypothesis for the validity of this the-orem. Applications to the evaluation of a large variety of deﬁnite integrals are given in the paper by T. Amdeberhan, O. Espinosa, I. Gonzalez, M. Harrison, V. Moll, and A. Straub . It turns out 15.4. The evaluation of a quartic integral 423 that, in the case considered here, the moments can be evaluated ex-plicitly, leading to the proof. Details can be found in the paper by G. Boros and V. Moll . A nice short proof. The following argument was communicated to the author by M. Hirschhorn . Start with I = ∞ 0 dx x4 + 2ax2 + 1. Make the substitution x →1/x and add the two forms of the integral I to obtain 2I = ∞ 0 (x2 + 1) dx x4 + 2ax2 + 1. The second substitution y = x −1/x gives 2I = ∞ −∞ dy y2 + 2a + 2 = π √2a + 2. Now, for an appropriate value of c (to guarantee convergence), ∞ 0 dx x4 + 2ax2 + c2 = π 2 √ 2 √a + c . Diﬀerentiation with respect to c leads to the identity ∞ 0 dx (x4 + 2ax2 + c2)m+1 = π 23m+3/2 c2m+1 (a + c)m+1/2 × m k=0 2m−k 2k k 2m −k m ck(a + c)m−k. The result now follows by taking c = 1. Proofs in other styles. There are several other proofs of Theorem 15.4.3 in the literature. The paper by G. Boros and V. Moll produced a proof based on elementary properties of the hypergeo-metric function plus an entry from the table by I. S. Gradshteyn and I. M. Ryzhik . A new proof based on a method for the evaluation of integrals coming from Feynman diagrams appears in the paper by T. Amdeberhan, V. Moll, and C. Vignat . An automatic proof has appeared in the work of C. Koutschan and V. Levandovskyy and one more based on the study of statistical densities can be found in the work by C. Berg and C. Vignat . Finally, a nice evaluation combining classical and automatic methods appears in the paper by 424 15. Landen Transformations M. Apagodu . The reader is encouraged to produce his/her own. The coeﬃcients dj,m. These numbers have remarkable properties. A related family of polynomials. Start with Pm(a) = 2 π [2(a + 1)]m+ 1 2 ∞ 0 dx (x4 + 2ax2 + 1)m+1 , and compute dj,m as coming from the Taylor expansion at a = 0 of the right-hand side. This yields (15.4.18) dj,m = 1 j!m!2m+j αj(m) m k=1 (4k −1) −βj(m) m k=1 (4k + 1) + , where αj and βj are polynomials in m of degrees j and j −1, respec-tively. The explicit expressions αj(m) = ⌊j/2⌋ t=0 j 2t m+t ν=m+1 (4ν −1) m ν=m−j+2t+1 (2ν + 1) t−1 ν=1 (4ν + 1) and βj(m) = ⌊(j+1)/2⌋ t=1 j 2t −1 m+t−1 ν=m+1 (4ν+1) m ν=m−j+2t (2ν+1) t−1 ν=1 (4ν−1) are given in the paper by G. Boros, V. Moll, and J. Shallit . Trying to obtain more information about the polynomials αj and βj directly proved diﬃcult. One uninspired day, the author decided to compute their roots numerically. It was a pleasant surprise to discover the following property. Theorem 15.4.7. For all j ≥1, all the roots of αj(m) = 0 lie on the line Re m = −1 2. Similarly, the roots of βj(m) = 0 for j ≥2 lie on the same vertical line. The proof of this theorem, due to J. Little , starts by writing (15.4.19) Aj(s) := αj((s −1)/2) and Bj(s) := βj((s −1)/2) 15.4. The evaluation of a quartic integral 425 and proving that Aj is equal to j! times the coeﬃcient of uj in f(s, u)g(s, u), where f(s, u) = (1 + 2u)s/2 and g(s, u) is the hyperge-ometric series (15.4.20) g(s, u) = 2F1 s 2 + 1 4, 1 4; 1 2; 4u2 . A similar expression is obtained for Bj(s). From here it follows that Aj and Bj each satisfy the three-term recurrence (15.4.21) xj+1(s) = 2sxj(s) −(s2 −(2j −1)2)xj−1(s). Little then establishes a version of Sturm’s theorem about interlacing zeros to prove the ﬁnal result. The location of the zeros of αj(m) now suggests studying the behavior of this family as j →∞. In the best of all worlds, one will obtain an analytic function of m with all the zeros on a vertical line. Perhaps some number theory will enter and . . . one never knows. Arithmetical properties. The expression (15.4.18) gives (15.4.22) m!2m+1 d1,m = (2m + 1) m k=1 (4k −1) − m k=1 (4k + 1), from which it follows that the right-hand side is an even number. This led naturally to the problem of determining the 2-adic valuation of Aj,m := j!m!2m+jdj,m = αj(m) m k=1 (4k −1) −βj(m) m k=1 (4k + 1) = j!m! 2m−j m k=j 2k 2m −2k m −k m + k k k j . The main result of is that ν2(Aj,m) = ν2(m(m+1))+1. This was extended in the work of T. Amdeberhan, D. Manna, and V. Moll to the next theorem. Theorem 15.4.8. The 2-adic valuation of Aj,m satisﬁes (15.4.23) ν2(Aj,m) = ν2((m + 1 −j)2j) + j, where (a)k = a(a + 1) · · · (a + k −1) is the Pochhammer symbol for k ≥1, with (a)0 = 1. 426 15. Landen Transformations The proof is an elementary application of the WZ-method. Deﬁne the numbers Bj,m := Aj,m 2j(m + 1 −j)2j , and use the WZ-method to obtain the recurrence Bj−1,m = (2m+1)Bj,m −(m−j)(m+j +1)Bj+1,m, 1 ≤j ≤m−1. Since the initial values Bm,m = 1 and Bm−1,m = 2m + 1 are odd, it follows inductively that Bj,m is an odd integer. The reader will also ﬁnd in a WZ-free proof of the theorem. The combinatorics of the valuations. The sequence of valuations {ν2(Aj,m) : m ≥j} increases in complexity with j. Some of the combinatorial nature of this sequence is described next. The ﬁrst feature of this sequence is that it has a block structure, reminiscent of the simple functions of real analysis. Deﬁnition 15.4.9. Let s ∈N, s ≥2. The sequence {aj : j ∈N} has block structure if there is an s ∈N such that for each t ∈ {0, 1, 2, . . .}, (15.4.24) ast+1 = ast+2 = · · · = as(t+1). The sequence is called s-simple if s is the largest value for which (15.4.24) occurs. Theorem 15.4.10. For each j ≥1, the set X(j) := {ν2(Aj,m) : m ≥j } is an s-simple sequence, with s = 21+ν2(j). Valuation patterns encoded in binary trees. The goal is to describe precisely the graph of the sequence {ν2(Aj,m) : m ≥j}. The reader is referred to the paper by X. Sun and V. Moll for complete details. In view of the block structure described earlier, it suﬃces to consider the sequences {ν2(Cj,m) : m ≥j}, which are deﬁned by Cj,m = Aj,j+(m−1)·21+ν2(j), so that the sequence {Cj,m : m ≥j} reduces each block of Aj,m to a single point. The emerging patterns are still very complicated. For instance, Figure 15.4.1 shows the case of j = 13 and j = 59. The 15.4. The evaluation of a quartic integral 427 remarkable fact is that in spite of the complexity of ν2(Cj,m) there is an exact formula for it. We now describe how to ﬁnd it. 20 40 60 80 100 m 37 38 39 40 41 42 ν2(C13, m) 10 20 30 40 50 60 70 m 173 174 175 176 177 178 ν2(C59, m) Figure 15.4.1. The valuations ν2(C13,m) and ν2(C59,m). The construction of the decision tree associated to the index j starts with a root v0 at level k = 0. To this vertex attach the sequence {ν2(Cj,m) : m ≥1} and ask whether ν2(Cj,m)−ν2(m) has a constant value independent of m. If the answer is yes, then it is said that v0 is a terminal vertex and we label it with this constant. The tree is complete. If the answer is negative, split the integers modulo 2 and produce two new vertices, v1, v2, connected to v0 and attach the classes {ν2(Cj,2m−1) : m ≥1} and {ν2(Cj,2m) : m ≥1} to these vertices. Now ask whether ν2(Cj,2m−1) −ν2(m) is independent of m and the same for ν2(Cj,2m)−ν2(m). Each vertex that yields a positive answer is considered terminal and the corresponding constant value 428 15. Landen Transformations is attached to it. Every vertex with a negative answer produces two new ones at the next level. Assume that the vertex v corresponding to the sequence {2k(m− 1)+a : m ≥1} produces a negative answer. Then it splits in the next generation into two vertices corresponding to the sequences {2k+1(m− 1) + a : m ≥1} and {2k+1(m −1) + 2k + a : m ≥1}. For instance, in Figure 15.4.2, the vertex corresponding to {4m : m ≥1}, which is not terminal, splits into {8m : m ≥1} and {8m −4 : m ≥1}. These two edges lead to terminal vertices. Theorem 15.4.11 shows that this process ends in a ﬁnite number of steps. root 2m 4m 8m 13 14 13 16 16 Figure 15.4.2. The decision tree for j = 5. Theorem 15.4.11. Let j ∈N and let T(j) be its decision tree. Deﬁne k∗(j) := ⌊log2 j⌋. Then (1) T(j) depends only on the odd part of j; that is, if r ∈N, then T(j) = T(2rj), up to the labels. (2) The generations of the tree are labelled starting at 0; that is, the root is generation 0. Then, for 0 ≤k ≤k∗(j), the kth generation of T(j) has 2k vertices. Up to that point, T(j) is a complete binary tree. (3) The k∗th generation contains 2k∗+1 −j terminal vertices. The constants associated with these vertices are given by the following algorithm. Deﬁne j1(j, k, a) := −j + 2(1 + 2k −a) and γ1(j, k, a) = j + k + 1 + ν2 ((j1 + j −1)!) + ν2 ((j −j1)!) . Then, for 1 ≤a ≤2k∗+1 −j, ν2 Cj,2k(m−1)+a = ν2(m) + γ1(j, k, a). 15.4. The evaluation of a quartic integral 429 Thus, the vertices at the k∗th generation have constants given by γ1(j, k, a). (4) The remaining terminal vertices of the tree T(j) ap-pear in the next generation. There are 2(j −2k∗(j)) of them. The constants attached to these vertices are deﬁned as follows: let j2(j, k, a) := −j +2(1+2k+1 −a) and j3(j, k, a) := j2(j, k, a+2k). Deﬁne γ2(j, k, a) := j + k + 2 + ν2 ((j2 + j −1)!) + ν2 ((j −j2)!) and γ3(j, k, a) := j + k + 2 + ν2 ((j3 + j −1)!) + ν2 ((j −j3)!) . Then, for 2k∗(j)+1 −j + 1 ≤a ≤2k∗(j), ν2 Cj,2k∗(j)+1(m−1)+a = ν2(m) + γ2(j, k∗(j), a) and ν2 Cj,2k∗(j)+1(m−1)+a+2k∗(j) = ν2(m) + γ3(j, k∗(j), a) give the constants attached to these remaining terminal vertices. The theorem is now employed to produce an analytic formula for ν2(C3,m). The value k∗(3) = 1 shows that the ﬁrst level contains 21+1−3 = 1 terminal vertex. This corresponds to the sequence 2m−1 and has constant value 7. Thus, (15.4.25) ν2 (C3,2m−1) = 7. The next level has 2(3 −21) = 2 terminal vertices. These correspond to the sequences 4m and 4m −2, with constant value 9 for both of them. This tree produces (15.4.26) ν2 (C3,m) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 7 + ν2 m+1 2 if m ≡1 mod 2, 9 + ν2 m 4 if m ≡0 mod 4, 9 + ν2 m+2 4 if m ≡2 mod 4. 430 15. Landen Transformations The complexity of the graph for j = 13 is reﬂected in the analytic formula for this valuation. The theorem yields (15.4.27) ν2 (C13,m) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 36 + ν2 m+7 8 if m ≡1 mod 8, 37 + ν2 m+6 8 if m ≡2 mod 8, 36 + ν2 m+5 8 if m ≡3 mod 8, 40 + ν2 m+12 16 if m ≡4 mod 16, 38 + ν2 m+11 16 if m ≡5 mod 16, 39 + ν2 m+10 16 if m ≡6 mod 16, 38 + ν2 m+9 16 if m ≡7 mod 16, 40 + ν2 m+8 16 if m ≡8 mod 16, 40 + ν2 m+4 16 if m ≡12 mod 16, 38 + ν2 m+3 16 if m ≡13 mod 16, 39 + ν2 m+2 16 if m ≡14 mod 16, 38 + ν2 m+1 16 if m ≡15 mod 16, 40 + ν2 m 16 if m ≡16 mod 16. Note. The p-adic valuations of Aj,m for p odd have diﬀerent behavior from the case p = 2. Figure 15.4.3 shows the plot of ν17(A1,m) where linear growth is observed. Experimental data suggest that, for any odd prime p, one has (15.4.28) νp(Aj,m) ∼ m p −1. The error term ν17(A1,m) −m/16 is also shown in the ﬁgure. The structure of the error remains to be explored. Unimodality and logconcavity. A ﬁnite sequence of real numbers {a0, a1, . . . , am} is said to be unimodal if there exists an index 0 ≤ j ≤m such that a0 ≤a1 ≤· · · ≤aj and aj ≥aj+1 ≥· · · ≥am. A polynomial is said to be unimodal if its sequence of coeﬃcients is unimodal. The sequence {a0, a1, . . . , am} with aj ≥0 is said to be logarithmically concave (or logconcave for short) if aj+1aj−1 ≤ a2 j for 1 ≤j ≤m −1. A polynomial is said to be logconcave if its sequence of coeﬃcients is logconcave. It is easy to see that if a sequence is logconcave, then it is unimodal. See the book by H. S. Wilf for an introduction to these ideas. 15.4. The evaluation of a quartic integral 431 50 100 150 200 250 m 5 10 15 ν17(A1, m) 200 400 600 800 1000 1200 1400 m 0.5 1.0 1.5 error Figure 15.4.3. The valuation ν17(A1,m) and the error term. Unimodal polynomials arise often in combinatorics, geometry, and algebra and have been the subject of considerable research in recent years. The reader is referred to the papers by F. Brenti and R. Stanley for surveys of the diverse techniques employed to prove that speciﬁc families of polynomials are unimodal. 432 15. Landen Transformations For m ∈N, the sequence {dj,m : 0 ≤j ≤m} is unimodal. This is a consequence of the following criterion established in the paper by G. Boros and V. Moll . Theorem 15.4.12. Let ak be a nondecreasing sequence of positive numbers and let A(x) = m k=0 akxk. Then A(x + 1) is unimodal. This theorem was applied to the polynomial (15.4.29) A(x) := 2−2m m k=0 2k 2m −2k m −k m + k m xk that satisﬁes Pm(x) = A(x + 1). The criterion was extended in a project at SIMU (Summer Institute in Mathematics for Undergrad-uates), an REU program in Puerto Rico. The result was the paper by J. Alvarez, M. Amadis, G. Boros, D. Karp, V. Moll, and L. Ros-ales to include the shifts A(x + j) and the paper by Yi Yang and Yeong-Nan Yeh for arbitrary shifts. The original proof of the unimodality of Pm(a) can be found in the paper by G. Boros and V. Moll . The author conjectured in the logconcavity of {dj,m : 0 ≤ j ≤m}. This turned out to be a more diﬃcult question. Some of our failed attempts are described next. (1) A result of F. Brenti states that if A(x) is logconcave, then so is A(x + 1). Unfortunately this does not apply in this case since (15.4.29) is not logconcave. Indeed, 24m−2k a2 k −ak−1ak+1 = 2m m −k 2m + k m 2 × 1 − k(m −k)(2m −2k + 1)(m + k + 1) (k + 1)(m + k)(2m −2k −1)(m −k + 1) and this last factor could be negative—for example, for m = 5 and j = 4. The number of negative terms in this sequence is small, so perhaps there is a way out of this. (2) The coeﬃcients dj,m satisfy many recurrences. For example, dj+1,m = 2m + 1 j + 1 dj,m −(m + j)(m + 1 −j) j(j + 1) dj−1,m. 15.4. The evaluation of a quartic integral 433 This can be found by a direct application of the WZ-method. There-fore, dj,m is logconcave provided j(2m + 1)dj−1,mdj,m ≤(m + j)(m + 1 −j)d2 j−1,m + j(j + 1)d2 j,m. The author conjectured that the smallest value of the expression (m + j)(m + 1 −j)d2 j−1,m + j(j + 1)d2 j,m −j(2m + 1)dj−1,mdj,m is 22mm(m + 1) 2m m 2 and it occurs at j = m. This has been es-tablished by W. Y. C. Chen and E. X. W. Yia . It implies the logconcavity of {dj,m : 0 ≤j ≤m}. Actually, the author has conjectured that the dj,m satisfy a stronger version of logconcavity. Given a sequence {aj} of positive numbers, deﬁne a map L ({aj}) := {bj} by bj := a2 j −aj−1aj+1. Thus {aj} is logconcave if {bj} has positive coeﬃcients. The nonnegative sequence {aj} is called inﬁnitely log-concave if any number of applications of L produces a nonnegative sequence. Conjecture 15.4.13. For each ﬁxed m ∈N, the sequence {dj,m : 0 ≤j ≤m} is inﬁnitely logconcave. The logconcavity of {dj,m : 0 ≤j ≤m} has recently been estab-lished by M. Kauers and P. Paule as an applications of their work on establishing inequalities by automatic means. The starting point is the triple sum expression for dj,m written as dj,m = j,s,k (−1)k+j−l 23(k+s) 2m + 1 2s m −s k 2(k + s) k + s s j k l −j . Using the RISC (Research Institute for Symbolic Computation) pack-age Multisim developed by K. Wegschaider , Kauers and Paule derived the recurrence 2(m + 1)dj,m+1 = 2(j + m)dj−1,m + (2j + 4m + 3)dj,m. 434 15. Landen Transformations The positivity of dj,m follows directly from this. To establish the logconcavity of dj,m, the new recurrence 4j(j + 1)dj+1,m = −2(2j −4m −3)(j + m + 1)dj,m + 4(j −m −1)(m + 1)dj,m+1 is derived automatically and the logconcavity of dj,m is reduced to establishing the inequality d2 j,m ≥4(m + 1) 4(j −m −1)(m + 1) −(2j2 −4m2 −7m −3)dj,m+1dj,m 16m3 + 16jm2 + 40m2 + 28jm + 33m + 9j + 9 . This is now accomplished in automatic fashion. The 2-logconcavity of {dj,m : 0 ≤j ≤m} is not achievable by these methods. At the end of , M. Kauers and P. Paule state that “...we have little hope that a proof of 2-logconcavity could be completed along these lines, not to mention that a human reader would have a hard time digesting it.” Actually, 2-logconcavity has been established by W. Y. C. Chen and E. X. W. Xia in . The general concept of inﬁnite logconcavity has generated some interest. D. Uminsky and K. Yeats have studied the action of L on sequences of the form (15.4.30) {. . . , 0, 0, 1, x0, x1, . . . , xn, . . . , x1, x0, 1, 0, 0, . . .} and (15.4.31) {. . . , 0, 0, 1, x0, x1, . . . , xn, xn, . . . , x1, x0, 1, 0, 0, . . .} and have established the existence of a large unbounded region in the positive orthant of Rn that consists only of inﬁnitely logconcave se-quences {x0, . . . , xn}. P. R. McNamara and B. Sagan have con-sidered sequences satisfying the condition a2 k ≥rak−1ak+1. Clearly this implies logconcavity if r ≥1. Their techniques apply to the rows of the Pascal triangle. Choosing appropriate r-factors and a computer veriﬁcation procedure, they obtain the following. Theorem 15.4.14. The sequence { n k : 0 ≤k ≤n} is inﬁnitely logconcave for ﬁxed n ≤1450. 15.5. An integrand of degree six 435 Newton began the study of logconcave sequences by establish-ing the following result (paraphrased in Section 2.2 of the book by G. H. Hardy, J. E. Littlewood, and G. Polya ). Theorem 15.4.15. Let {ak} be a ﬁnite sequence of positive real num-bers. Assume all the roots of the polynomial (15.4.32) P[ak; x] := a0 + a1x + · · · + anxn are real. Then the sequence {ak} is logconcave. P. R. McNamara and B. Sagan and, independently, R. Stan-ley (personal communication) and S. Fisk have proposed the next problem. This was settled by P. Br¨ and` en . See for the complete details on the conjecture. Theorem 15.4.16. Let {ak} be a ﬁnite sequence of positive real numbers. If P[ak; x] has only real roots, then the same is true for P[L(ak); x]. The polynomials Pm(a) are the generating function for the se-quence {dj,m} described here. It is an unfortunate fact that they do not have real roots, as established by G. Boros and V. Moll . Thus, the previous theorem does not apply to Conjecture 15.4.13. In spite of this, the asymptotic behavior of these zeros has remarkable properties. D. Dimitrov has shown that, in the right scale, the zeros converge to a lemniscate. The inﬁnite logconcavity of {dj,m} has resisted all our eﬀorts. It remains to be established. 15.5. An integrand of degree six The result of Theorem 15.2.1 is now applied to an integral where the integrand is an even rational function of degree six. The result is given in the next theorem. Theorem 15.5.1. Deﬁne I(a, b) = ∞ 0 a4x4 + a2x2 + a0 b6x6 + b4x4 + b2x2 + b0 dx. 436 15. Landen Transformations Then I(a, b) = I(a∗, b∗), where a∗ 4 = 32(a4b0 + a0b6), a∗ 2 = 8(a2b0 + 3a4b0 + a4b2 + a0b4 + 3a0b6 + a2b6), a∗ 0 = 2(a0 + a2 + a4)(b0 + b2 + b4 + b6) and b∗ 6 = 64b0b6, b∗ 4 = 16(b0b4 + 6b0b6 + b2b6), b∗ 2 = 4(b0b2 + 4b0b4 + b2b4 + 9b0b6 + 4b2b6 + b4b6), b∗ 0 = (b0 + b2 + b4 + b6)2. The next exercise suggests a scaling that reduces the number of parameters in the problem. Exercise 15.5.2. Prove that the integral (15.5.1) U6(a, b; c, d, e) := ∞ 0 cx4 + dx2 + e x6 + ax4 + bx2 + 1 dx is invariant under the transformation an+1 = anbn + 5an + 5bn + 9 (an + bn + 2)4/3 , (15.5.2) bn+1 = an + bn + 6 (an + bn + 2)2/3 , cn+1 = cn + dn + en (an + bn + 2)2/3 , dn+1 = (bn + 3)cn + 2dn + (an + 3)en an + bn + 2 , en+1 = cn + en (an + bn + 2)1/3 . The ﬁrst two equations in (15.5.2) are independent of the vari-ables c, d, and e so they deﬁne a map Φ6(a, b) = ab + 5a + 5b + 9 (a + b + 2)4/3 , a + b + 6 (a + b + 2)2/3 (15.5.3) 15.5. An integrand of degree six 437 that is well-deﬁned on R2 minus the line a + b + 2 = 0. The main result of M. Chamberland and V. Moll is stated below. Theorem 15.5.3. The set of points in R2 that converge to the ﬁxed point (3, 3) of the dynamical system an+1 = anbn + 5an + 5bn + 9 (an + bn + 2)4/3 , (15.5.4) bn+1 = an + bn + 6 (an + bn + 2)2/3 is the region Λ of the (a, b)-plane for which the integral (15.5.1) con-verges. Note 15.5.4. The region Λ is given in terms of the resolvent curve R deﬁned by (15.5.5) R(a, b) := 4a3 + 4b3 −18ab −a2b2 + 27 = 0. This function appeared in Theorem 4.7.12. The set R(a, b) = 0 is a real algebraic curve with two connected components R±. The com-ponent R+ is contained in the ﬁrst quadrant and contains the point (3, 3) as a cusp. The second component R−is disjoint from the ﬁrst quadrant. The region Λ is deﬁned as the points on the ab-plane that are above the curve R−. 10 10 5 5 0 0 -5 -10 -5 -10 b a Figure 15.5.1. The resolvent curve. 438 15. Landen Transformations The identity R(a1, b1) = (a −b)2R(a, b) (a + b + 2)4 (15.5.6) plays an important role in the dynamics of (15.5.4). In particular it follows from (15.5.6) that the resolvent curve R(a, b) = 0, and the regions {(a, b) : R(a, b) > 0}, located between the two branches in Figure 15.5.1, and {(a, b) : R(a, b) < 0} are preserved by Φ6. The identity (15.5.6) also shows that the diagonal Δ = {(a, b) : a = b} of R2 is mapped onto the resolvent curve R. This yields the parametriza-tion a(t) = t + 9 24/3(t + 1)1/3 and b(t) = 21/3(t + 3) (t + 1)2/3 of this curve. This parametrization may be employed to analyze the behavior on the resolvent curve. Note 15.5.5. The relation between the resolvent curve and the dis-criminant of a cubic polynomial is clariﬁed in Exercise 4.7.10. 15.6. The original elliptic case Among the many beautiful results in the theory of elliptic integrals, a calculation of Gauss stands among the best: take two positive real numbers a and b, with a > b, and form a new pair by replacing a with the arithmetic mean (a + b)/2 and b with the geometric mean √ ab. Then iterate: (15.6.1) an+1 = an + bn 2 , bn+1 = , anbn starting with a0 = a and b0 = b. K. F. Gauss was interested in the initial conditions a = 1 and b = √ 2. The iteration generates a sequence of algebraic numbers which rapidly become impossible to describe explicitly; for instance, (15.6.2) a3 = 1 23 (1 + 4 √ 2)2 + 2 √ 2 8 √ 2 < 1 + √ 2 is a root of the polynomial G(a) = 16777216a8 −16777216a7 + 5242880a6 −10747904a5 +942080a4 −1896448a3 + 4436a2 −59840a + 1. 15.6. The original elliptic case 439 The numerical behavior is surprising; a6 and b6 agree to 87 digits. It is simple to check that (15.6.3) lim n→∞an = lim n→∞bn. This common limit is called the arithmetic-geometric mean and is denoted by AGM(a, b). It is the explicit dependence on the initial condition that is hard to discover. Gauss computed some numerical values and observed that (15.6.4) a11 ∼b11 ∼1.198140235 and then he recognized the reciprocal of this number as a numerical approximation to the elliptic integral (15.6.5) I = 2 π 1 0 dt √ 1 −t4 . It is unclear to the authors how Gauss recognized this number—he simply knew it. (Stirling’s tables may have been a help; the book by J. M. Borwein and D. H. Bailey contains a reproduction of the original notes and comments.) He was particularly interested in the evaluation of this deﬁnite integral as it provides the length of a lemniscate. In his diary Gauss remarked, “This will surely open up a whole new ﬁeld of analysis.” More details can be found in the book by J. M. Borwein and P. B. Borwein and the paper by D. Cox . Gauss’ procedure to ﬁnd an analytic expression for AGM(a, b) began with the elementary observation (15.6.6) AGM(a, b) = AGM a + b 2 , √ ab and the homegeneity condition (15.6.7) AGM(λa, λb) = λAGM(a, b). He used (15.6.6) with a = (1+ √ k)2 and b = (1− √ k)2, with 0 < k < 1, to produce AGM(1 + k + 2 √ k, 1 + k −2 √ k) = AGM(1 + k, 1 −k). 440 15. Landen Transformations He then used the homogeneity of AGM to write AGM(1 + k + 2 √ k, 1 + k −2 √ k) = AGM((1 + k)(1 + k∗), (1 + k)(1 −k∗)) = (1 + k)AGM(1 + k∗, 1 −k∗), with (15.6.8) k∗= 2 √ k 1 + k . This resulted in the functional equation (15.6.9) AGM(1 + k, 1 −k) = (1 + k) AGM(1 + k∗, 1 −k∗). In his analysis of (15.6.9), Gauss substituted the power series (15.6.10) 1 AGM(1 + k, 1 −k) = ∞ n=0 ank2n into (15.6.9) and solved an inﬁnite system of nonlinear equations, to produce (15.6.11) an = 2−2n 2n n 2 . Then he recognized the series as that of an elliptic integral, to obtain (15.6.12) 1 AGM(1 + k, 1 −k) = 2 π π/2 0 dx , 1 −k2 sin2 x . This is a remarkable tour de force. The function (15.6.13) K(k) = π/2 0 dx , 1 −k2 sin2 x is the elliptic integral of the ﬁrst kind. It can also be written in the algebraic form (15.6.14) K(k) = 1 0 dt , (1 −t2)(1 −k2t2) . In this notation, (15.6.9) becomes (15.6.15) K(k∗) = (1 + k)K(k). 15.6. The original elliptic case 441 This is the Landen transformation for the complete elliptic in-tegral. J. Landen , the namesake of the transformation, studied related integrals: for example, (15.6.16) κ := 1 0 dx , x2(1 −x2) . He derived identites such as (15.6.17) κ = ε , ε2 −π, where ε := π/2 0 , 2 −sin2 θ dθ , proven mainly by suitable changes of variables in the integral for ε. In the paper by G. N. Watson , the reader will ﬁnd a historical account of Landen’s work, including the above identities. The reader will ﬁnd proofs in a variety of styles in the books by J. M. Borwein and P. B. Borwein and H. McKean and V. Moll . In trigonometric form, the Landen transformation states that (15.6.18) G(a, b) = π/2 0 dθ , a2 cos2 θ + b2 sin2 θ is invariant under the change of parameters (a, b) → a+b 2 , √ ab . D. J. Newman presents a very clever proof: the change of vari-ables x = b tan θ yields (15.6.19) G(a, b) = 1 2 ∞ −∞ dx , (a2 + x2)(b2 + x2) . Now let x →x + √ x2 + ab to complete the proof. Many of the above identities can now be searched for and proven on a computer; see the book by J. M. Borwein and D. H. Bailey . Note 15.6.1. The reader will ﬁnd a survey of the many aspects of Landen transformations in the paper by D. Manna and V. Moll . As an intriguing open problem, the question of producing a Landen transformation for the integral U + 2 (a, b, c) := ∞ 0 dx ax2 + bx + c remains a challenge. 442 15. Landen Transformations Note 15.6.2. The dynamics of the ﬁrst two equations in (15.5.2) with initial data below the resolvent curve is quite complicated. Fig-ure 15.6.1 shows the ﬁrst 50000 iterates starting at a = −5.0 and b = −20.4. Figure 15.6.2 shows 50000 iterates of the dynamics starting at (−25.0, −2.4) and (11.0, −13.7). These are identical to the naked eye. Figure 15.6.3 shows 500000 iterates starting at (11.0, −13.7). This ﬁgure illustrates the following conjecture by the author: Conjecture 15.6.3. The orbit of any point below the resolvent curve is dense in the open region below this curve. 10 10 a(n) 10 10 20 b(n) Figure 15.6.1. The dynamics below the resolvent curve. 15.6. The original elliptic case 443 10 10 a(n) 10 10 20 b(n) 10 a(n) 10 10 20 b(n) Figure 15.6.2. Two more examples of dynamics below the resolvent curve. 444 15. Landen Transformations 50 40 30 20 10 10 a(n) 10 10 20 b(n) Figure 15.6.3. Illustration of the density conjecture.
14765	https://www.geeksforgeeks.org/dsa/find-the-middle-element-of-an-array-or-list/	Find the Middle Element of an Array or List - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In DSA Tutorial Array Strings Linked List Stack Queue Tree Graph Searching Sorting Recursion Dynamic Programming Binary Tree Binary Search Tree Heap Hashing Sign In ▲ Open In App Find the Middle Element of an Array or List Last Updated : 18 Apr, 2024 Comments Improve Suggest changes Like Article Like Report Given an array or a list of elements. The task is to find the middle element of given array or list of elements. If the array size is odd, return the single middle element. If the array size is even, return the two middle elements. Example Input:arr = {1, 2, 3, 4, 5} Output: 3 Input:arr = {7, 8, 9, 10, 11, 12} Output:9 10 Approach: Check the length of array is even or odd. For odd array length the middle element would be arr[length/2]. Otherwise, two middle element that is: arr[length/2] and arr[length/2 -1] Below is the implementation of the above approach: Java ```java import java.util.; public class Main { // Function to find middle element(s) of a list static List findMiddleElements(List arr) { List result = new ArrayList<>(); int n = arr.size(); // Check if the size of the list is even or odd if (n % 2 == 0) { // If even, push two middle elements into the // result list result.add(arr.get(n / 2 - 1)); result.add(arr.get(n / 2)); } else { // If odd, push the single middle element into // the result list result.add(arr.get(n / 2)); } return result; } public static void main(String[] args) { // Sample list List<Integer> arr = new ArrayList<>(); arr.add(1); arr.add(2); arr.add(3); arr.add(4); arr.add(5); // Find middle elements using the function List<Integer> middleElements = findMiddleElements(arr); // Display the middle element(s) System.out.print("Middle Element(s): "); for (int num : middleElements) { System.out.print(num + " "); } System.out.println(); } } ``` import java.util.; public class Main { // Function to find middle element(s) of a list static List findMiddleElements(List arr) { List result = new ArrayList<>(); int n = arr.size(); // Check if the size of the list is even or odd if (n % 2 == 0) { // If even, push two middle elements into the // result list result.add(arr.get(n / 2 - 1)); result.add(arr.get(n / 2)); } else { // If odd, push the single middle element into // the result list result.add(arr.get(n / 2)); } return result; } public static void main(String[] args) { // Sample list List arr = new ArrayList<>(); arr.add(1); arr.add(2); arr.add(3); arr.add(4); arr.add(5); // Find middle elements using the function List middleElements = findMiddleElements(arr); // Display the middle element(s) System.out.print("Middle Element(s): "); for (int num : middleElements) { System.out.print(num + " "); } System.out.println(); }} Python3 ```python3 def find_middle_elements(arr): result = [] n = len(arr) # Check if the size of the list is even or odd if n % 2 == 0: # If even, append two middle elements to the result list result.append(arr[n // 2 - 1]) result.append(arr[n // 2]) else: # If odd, append the single middle element to the result list result.append(arr[n // 2]) return result if name == "main": # Sample list arr = [1, 2, 3, 4, 5] # Find middle elements using the function middle_elements = find_middle_elements(arr) # Display the middle element(s) print("Middle Element(s):", end=" ") for num in middle_elements: print(num, end=" ") print() C#csharp using System; using System.Collections.Generic; public class Program { // Function to find middle element(s) of a list static List FindMiddleElements(List arr) { List result = new List(); int n = arr.Count; // Check if the size of the list is even or odd if (n % 2 == 0) { // If even, push two middle elements into the // result list result.Add(arr[n / 2 - 1]); result.Add(arr[n / 2]); } else { // If odd, push the single middle element into // the result list result.Add(arr[n / 2]); } return result; } public static void Main(string[] args) { // Sample list List<int> arr = new List<int>{ 1, 2, 3, 4, 5 }; // Find middle elements using the function List<int> middleElements = FindMiddleElements(arr); // Display the middle element(s) Console.Write("Middle Element(s): "); foreach(int num in middleElements) { Console.Write(num + " "); } // Keep the console window open in debug mode. Console.WriteLine(); } } JavaScriptjavascript function findMiddleElements(arr) { let result = []; let n = arr.length; // Check if the size of the array is even or odd if (n % 2 === 0) { // If even, append two middle elements to the result array result.push(arr[n / 2 - 1]); result.push(arr[n / 2]); } else { // If odd, append the single middle element to the result array result.push(arr[Math.floor(n / 2)]); } return result; } // Sample array let arr = [1, 2, 3, 4, 5]; // Find middle elements using the function let middleElements = findMiddleElements(arr); // Display the middle element(s) console.log("Middle Element(s):", middleElements.join(" ")); C++14cpp14 include include using namespace std; // Function to find middle element(s) of a vector vector findMiddleElements(const vector& arr) { vector result; int n = arr.size(); // Check if the size of the vector is even or odd if (n % 2 == 0) { // If even, push two middle elements into the result vector result.push_back(arr[n / 2 - 1]); result.push_back(arr[n / 2]); } else { // If odd, push the single middle element into the result vector result.push_back(arr[n / 2]); } return result; } int main() { // Sample vector vector arr = { 1, 2, 3, 4, 5 }; // Find middle elements using the function vector<int> middleElements = findMiddleElements(arr); // Display the middle element(s) cout << "Middle Element(s): "; for (int num : middleElements) { cout << num << " "; } return 0; } ``` OutputMiddle Element(s): 3 Time Complexity: O(1) Auxiliary Space: O(1) Approach 2 (Slow and fast pointer approach) We use the concept of slow and fast pointer to find the middle element of array or list. We declare two pointers slow and fast for linked list and two variables slow and fast for arrays. Initially both slow and fast points to starting element of array or list. Increase slow pointer by 1 and increase fast pointer by 2. Traverse until fast reaches to end of array or list. then traverse again fast pointing to start and Return the slow pointer node as middle element of list or array. Example Below is the Implementation of the above approach: C++ ```cpp include include using namespace std; // Definition of a ListNode for linked list (if using linked // list) struct ListNode { int val; ListNode next; ListNode(int x) : val(x) , next(NULL) { } }; // Function to find the middle element of an array using // slow and fast pointers int findMiddle(vector& nums) { int slow = 0, fast = 0; while (fast < nums.size() && fast + 1 < nums.size()) { slow++; fast += 2; } return nums[slow]; } // Function to find the middle element of a linked list // using slow and fast pointers int findMiddle(ListNode head) { ListNode slow = head; ListNode fast = head; while (fast != NULL && fast->next != NULL) { slow = slow->next; fast = fast->next->next; } return slow->val; } int main() { // Example usage with array vector nums = { 1, 2, 3, 4, 5 }; cout << "Middle element of array: " << findMiddle(nums) << endl; // Example usage with linked list ListNode head = new ListNode(1); head->next = new ListNode(2); head->next->next = new ListNode(3); head->next->next->next = new ListNode(4); head->next->next->next->next = new ListNode(5); cout << "Middle element of linked list: " << findMiddle(head) << endl; return 0; } ``` include include using namespace std;// Definition of a ListNode for linked list (if using linked// list)struct ListNode { int val; ListNode next; ListNode(int x) : val(x) , next(NULL) { }};// Function to find the middle element of an array using// slow and fast pointers int findMiddle(vector& nums){ int slow = 0, fast = 0; while (fast < nums.size() && fast + 1 < nums.size()) { slow++; fast += 2; } return nums[slow];}// Function to find the middle element of a linked list// using slow and fast pointers int findMiddle(ListNode head){ ListNode slow = head; ListNode fast = head; while (fast != NULL && fast->next != NULL) { slow = slow->next; fast = fast->next->next; } return slow->val;}int main(){ // Example usage with array vector nums = { 1, 2, 3, 4, 5 }; cout << "Middle element of array: " << findMiddle(nums) << endl; // Example usage with linked list ListNode head = new ListNode(1); head->next = new ListNode(2); head->next->next = new ListNode(3); head->next->next->next = new ListNode(4); head->next->next->next->next = new ListNode(5); cout << "Middle element of linked list: " << findMiddle(head) << endl; return 0;} Java ```java import java.util.; // Definition of a ListNode for linked list (if using linked // list) class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } public class Main { // Function to find the middle element of an array using // slow and fast pointers public static int findMiddle(List nums) { int slow = 0, fast = 0; while (fast < nums.size() && fast + 1 < nums.size()) { slow++; fast += 2; } return nums.get(slow); } // Function to find the middle element of a linked list // using slow and fast pointers public static int findMiddle(ListNode head) { ListNode slow = head; ListNode fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } return slow.val; } public static void main(String[] args) { List<Integer> nums = Arrays.asList(1, 2, 3, 4, 5); System.out.println("Middle element of array: " + findMiddle(nums)); ListNode head = new ListNode(1); head.next = new ListNode(2); head.next.next = new ListNode(3); head.next.next.next = new ListNode(4); head.next.next.next.next = new ListNode(5); System.out.println("Middle element of linked list: " + findMiddle(head)); } } Python3python3 Python program for the above approach Definition of a ListNode for linked list class ListNode: def init(self, x): self.val = x self.next = None Function to find the middle element of an array using slow and fast pointers def findMiddle(nums): slow = 0 fast = 0 while fast < len(nums) and fast + 1 < len(nums): slow += 1 fast += 2 return nums[slow] Function to find the middle element of a linked list using slow and fast pointers def findMiddleLinkedList(head): slow = head fast = head while fast is not None and fast.next is not None: slow = slow.next fast = fast.next.next return slow.val Main function if name == "main": nums = [1, 2, 3, 4, 5] print("Middle element of array:", findMiddle(nums)) head = ListNode(1) head.next = ListNode(2) head.next.next = ListNode(3) head.next.next.next = ListNode(4) head.next.next.next.next = ListNode(5) print("Middle element of linked list:", findMiddleLinkedList(head)) This code is contributed by Susobhan Akhuli C#csharp using System; using System.Collections.Generic; public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; } } public class Program { // Function to find the middle element of an array using // slow and fast pointers static int FindMiddle(List nums) { int slow = 0, fast = 0; while (fast < nums.Count && fast + 1 < nums.Count) { slow++; fast += 2; } return nums[slow]; } // Function to find the middle element of a linked list // using slow and fast pointers static int FindMiddle(ListNode head) { ListNode slow = head; ListNode fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } return slow.val; } public static void Main(string[] args) { // Example usage with array List<int> nums = new List<int>{ 1, 2, 3, 4, 5 }; Console.WriteLine("Middle element of array: " + FindMiddle(nums)); // Example usage with linked list ListNode head = new ListNode(1); head.next = new ListNode(2); head.next.next = new ListNode(3); head.next.next.next = new ListNode(4); head.next.next.next.next = new ListNode(5); Console.WriteLine("Middle element of linked list: " + FindMiddle(head)); } } JavaScriptjavascript // JavaScript program for the above approach // Definition of a ListNode for linked list class ListNode { constructor(val) { this.val = val; this.next = null; } } // Function to find the middle element of an array function findMiddleArray(nums) { let slow = 0, fast = 0; while (fast < nums.length && fast + 1 < nums.length) { slow++; fast += 2; } return nums[slow]; } // Function to find the middle element of a linked list function findMiddleLinkedList(head) { let slow = head; let fast = head; while (fast !== null && fast.next !== null) { slow = slow.next; fast = fast.next.next; } return slow.val; } // Example usage with array let nums = [1, 2, 3, 4, 5]; console.log("Middle element of array:", findMiddleArray(nums)); // Example usage with linked list let head = new ListNode(1); head.next = new ListNode(2); head.next.next = new ListNode(3); head.next.next.next = new ListNode(4); head.next.next.next.next = new ListNode(5); console.log("Middle element of linked list:", findMiddleLinkedList(head)); // This code is contributed by Susobhan Akhuli ``` OutputMiddle element of array: 3 Middle element of linked list: 3 This approach of finding middle element is contributed by Ayush Mishra. Comment More info A abhiisaxena09 Follow Improve Article Tags : DSA Arrays Basic Coding Problems Explore DSA Fundamentals Logic Building Problems 2 min readAnalysis of Algorithms 1 min read Data Structures Array Data Structure 3 min readString in Data Structure 2 min readHashing in Data Structure 2 min readLinked List Data Structure 2 min readStack Data Structure 2 min readQueue Data Structure 2 min readTree Data Structure 2 min readGraph Data Structure 3 min readTrie Data Structure 15+ min read Algorithms Searching Algorithms 2 min readSorting Algorithms 3 min readIntroduction to Recursion 14 min readGreedy Algorithms 3 min readGraph Algorithms 3 min readDynamic Programming or DP 3 min readBitwise Algorithms 4 min read Advanced Segment Tree 2 min readBinary Indexed Tree or Fenwick Tree 15 min readSquare Root (Sqrt) Decomposition Algorithm 15+ min readBinary Lifting 15+ min readGeometry 2 min read Interview Preparation Interview Corner 3 min readGfG160 3 min read Practice Problem GeeksforGeeks Practice - Leading Online Coding Platform 6 min readProblem of The Day - Develop the Habit of Coding 5 min read Like Corporate & Communications Address: A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305) Registered Address: K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305 Company About Us Legal Privacy Policy Contact Us Advertise with us GFG Corporate Solution Campus Training Program Explore POTD Job-A-Thon Community Blogs Nation Skill Up Tutorials Programming Languages DSA Web Technology AI, ML & Data Science DevOps CS Core Subjects Interview Preparation GATE Software and Tools Courses IBM Certification DSA and Placements Web Development Programming Languages DevOps & Cloud GATE Trending Technologies Videos DSA Python Java C++ Web Development Data Science CS Subjects Preparation Corner Aptitude Puzzles GfG 160 DSA 360 System Design @GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved Improvement Suggest changes Suggest Changes Help us improve. 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14766	https://www.sciencedirect.com/science/article/pii/S1097276520307723	Skip to article My account Sign in View PDF ## Molecular Cell Volume 81, Issue 1, 7 January 2021, Pages 166-182.e6 Article Distinct Structures and Dynamics of Chromatosomes with Different Human Linker Histone Isoforms Author links open overlay panel, , , , , , , , rights and content Under an Elsevier user license Open archive Highlights ¢ 2.83.1 Ã resolution structures of chromatosomes containing human H1 isoforms ¢ Human somatic H1 isoforms bind to the nucleosome on the dyad ¢ C-terminal tails of the H1 isoforms control distinct flanking DNA orientations ¢ Chromatosome H3 N-tails relocate and inhibit ISWI remodeler binding and activity Summary The repeating structural unit of metazoan chromatin is the chromatosome, a nucleosome bound to a linker histone, H1. There are 11 human H1 isoforms with diverse cellular functions, but how they interact with the nucleosome remains elusive. Here, we determined the cryoelectron microscopy (cryo-EM) structures of chromatosomes containing 197 bp DNA and three different human H1 isoforms, respectively. The globular domains of all three H1 isoforms bound to the nucleosome dyad. However, the flanking/linker DNAs displayed substantial distinct dynamic conformations. Nuclear magnetic resonance (NMR) and H1 tail-swapping cryo-EM experiments revealed that the C-terminal tails of the H1 isoforms mainly controlled the flanking DNA orientations. We also observed partial ordering of the core histone H2A C-terminal and H3 N-terminal tails in the chromatosomes. Our results provide insights into the structures and dynamics of the chromatosomes and have implications for the structure and function of chromatin. Graphical Abstract Keywords chromatosome nucleosome linker histone isoform Cryo-EM NMR chromatin structure single-chain antibody chromatosome structure chromatosome dynamics linker histone tail Cited by (0) 6 : Lead Contact Published by Elsevier Inc.
14767	https://math.stackexchange.com/questions/2288464/is-this-derivation-of-the-path-length-formula-actually-correct	Is this "derivation" of the path length formula actually correct? - 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 Is this "derivation" of the path length formula actually correct? Ask Question Asked 8 years, 4 months ago Modified7 years, 5 months ago Viewed 3k times This question shows research effort; it is useful and clear 13 Save this question. Show activity on this post. Saw this in a physics lecture. This all assumes we have some function, y=f(x). First he defined d s=√d x 2+d y 2, where the professor drew a picture and seemed to be using dx and dy to mean a very small change in x (or y). I'm not sure what this really means. What does it mean to square a differential, or to have it under the square root sign? But alright, for now I'll say it's an abuse of notation, and just consider them tiny distances. He then says you can factor out a dx from under the square-root sign as follows: d s=d x√1+(d y d x)2. Ok, assuming we're interpreting dx and dy as tiny lengths, ok. Now he says interpret the dy/dx as taking the derivative of y with respect to x. What?!? How can what was once considered a length all of a sudden become an operator? If dx and dy are not just tiny distances, how can you do algebra with differentials? What sort of black magic is this? Can someone please explain what's going on? Thanks. Edit: Maybe a better, and more general question is this. If we replace all the d's with Δ's and the equal sign with an approximation sign, the algebra becomes correct. Why and when, if ever, is it okay to "take the limit" and replace a Δ y Δ x with a derivative, and the approximation sign with an equal sign? Because I feel like physicists do this a lot. arc-length Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited May 19, 2017 at 23:26 DargscisyhpDargscisyhp asked May 19, 2017 at 23:15 DargscisyhpDargscisyhp 829 7 7 silver badges 16 16 bronze badges 6 21 Welcome to the world of physics, I guess. This is par for the course in that camp.Arthur –Arthur 2017-05-19 23:17:28 +00:00 Commented May 19, 2017 at 23:17 1 You might want to assume d x≥0...user251257 –user251257 2017-05-19 23:38:45 +00:00 Commented May 19, 2017 at 23:38 3 A physicist is going on!Henrik supports the community –Henrik supports the community 2017-05-20 08:06:59 +00:00 Commented May 20, 2017 at 8:06 You may want to take a look at this question as well.polfosol –polfosol 2017-05-20 15:47:54 +00:00 Commented May 20, 2017 at 15:47 You might enjoy Sylvanus Thompson's Calculus made Easy (gratis public-domain edition from Project Gutenberg).Andrew D. Hwang –Andrew D. Hwang 2017-05-20 18:23:15 +00:00 Commented May 20, 2017 at 18:23 \|Show 1 more comment 4 Answers 4 Sorted by: Reset to default This answer is useful 18 Save this answer. Show activity on this post. What he did is correct, though the reasons for doing so seem to be glossed over. I'll give you a rigorus version of what he did. Let Δ x>0, represent the length of some horizontal line segment, and for now let y=f(x), where f is some function. (If y is not a function, simply break it up into several pieces, where each piece is a function.) Then define Δ y=f(x+Δ x)−f(x) Now, Δ x and Δ y are the lengths of legs of a right triangle, so we should probably talk about the hypotenuse, as well, whose length I will denote by Δ s. Now, by the Pythagorean Theorem, (Δ s)2=(Δ x)2+(Δ y)2=(Δ x)2[1+(Δ y Δ x)2] (where I pulled out (Δ x)2 from both terms on the RHS). Substituting our two expression from above, (Δ s)2=(Δ x)2[1+(f(x+Δ x)−f(x)Δ x)2]⟹(Δ s Δ x)2=1+(f(x+Δ x)−f(x)Δ x)2 Now, take limits of both sides as Δ x→0: lim Δ x→0(Δ s Δ x)2=lim Δ x→0[1+(f(x+Δ x)−f(x)Δ x)2] ⟹(lim Δ x→0 Δ s Δ x)2=1+(lim Δ x→0 f(x+Δ x)−f(x)Δ x)2⟹(d s d x)2=1+[f′(x)]2=1+(d y d x)2⟹\|d s d x\|=√1+(d y d x)2 If you assume s increases as x increases, i.e. the length s of your path increases as you move from left to right, we can drop the absolute values: d s d x=√1+(d y d x)2 or, in differential form, d s=d x√1+(d y d x)2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 20, 2017 at 0:10 answered May 19, 2017 at 23:45 d4rk_1nf1n1tyd4rk_1nf1n1ty 590 1 1 gold badge 4 4 silver badges 17 17 bronze badges 3 1 Thank you for your answer. In your second line of your proof (Δ s)2=(Δ x)2+(Δ y)2, unless f is a linear function, for noninfinitesimal Δ x and Δ y, shouldn't the equality be an approximation, and if so, where in the proof does the approximation become an equality again? Also, why is Delta s/Delta x ds/dx in the limit? Thanks again.Dargscisyhp –Dargscisyhp 2017-05-20 01:53:39 +00:00 Commented May 20, 2017 at 1:53 1 Why would it be an approximation? Perhaps the only thing missing from this proof is that x is fixed and delta x is a variable.chx –chx 2017-05-20 03:10:57 +00:00 Commented May 20, 2017 at 3:10 2 @chx I'm sorry, my mistake. I was interpreting Δ s as the arclength along the curve, not the straight-line distance between the points. I see my error now, and again, thank you for this answer.Dargscisyhp –Dargscisyhp 2017-05-20 16:14:00 +00:00 Commented May 20, 2017 at 16:14 Add a comment\| This answer is useful 15 Save this answer. Show activity on this post. This is what helped me stay afloat during general relativity classes; it might help you too. First note, that this is always in the context of some (given or arbitrary) path S. Say your path is parametrised by some parameter t, that the path goes from a to b as t goes from 0 to 1, just to have some concretes down. Then d s=√d x 2+d y 2 can be translated into d s d t=√(d x d t)2+(d y d t)2 or, correspondingly (using the appropriate form of the fundamental theorem of calculus) ∫b a d s=∫1 0√(d x d t)2+(d y d t)2 d t Now you can factor out d x d t from the square root, and apply the chain rule to get what I have learned to think when I see d s=d x√1+(d y d x)2, namely ∫b a d s=∫1 0\|d x d t\|√1+(d y d x)2 d t which is valid as long as the path isn't vertical (d y d x is evaluated along the path, where y can be a function of x locally, as long as it's not vertical). But if the path is vertical, then you can factor out d y d t from the square root instead, and everything works out nicely. The integral above can, of course, with simple substitution also be made into an integral over x if that makes the problem easier. Note that in general relativity (and perhaps elsewhere in physics) the form d s 2=d x 2+d y 2 is not unheard of. As far as I can tell that's just because they don't want to bother with the square root sign when writing down formulae. You still definitely have to put the square root back into the expression before any calculations can be done. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 20, 2017 at 8:21 answered May 19, 2017 at 23:35 ArthurArthur 205k 14 14 gold badges 186 186 silver badges 332 332 bronze badges 3 1 "As far as I can tell that's just because they don't want to bother with the square root sign when writing down formulae." That's not quite right -- it's compact notation for writing the metric which is 2-tensor and has meaning beyond computing arc lengths along curves through spacetime. It encodes everything there is to know about the local structure of the spacetime.joshphysics –joshphysics 2017-05-20 02:40:49 +00:00 Commented May 20, 2017 at 2:40 1 @joshPhysics I know that. But you get exactly the same information from d s=√d x 2+d y 2. So why bother with the second form at all? That's what I was referring to, and the only difference I can see is the lack of a square root.Arthur –Arthur 2017-05-20 06:05:35 +00:00 Commented May 20, 2017 at 6:05 1 @Arthur Because in general relativity, the metric is not positive-definite; it is only non-degenerate. Taking the square root is thus awkward because for certain curves, e.g. timelike curves if you are using (−,+,+,+) signature, this will lead to d s being "imaginary."joshphysics –joshphysics 2017-05-20 17:10:52 +00:00 Commented May 20, 2017 at 17:10 Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. Although mathematicians go nuts over this stuff, abusing notation and working with symbols in a way that relies on intuition is often critical to a physicist and useful to a mathematician who wants to build deeper understanding and intuition. We should all be grateful that not everyone is paralyzed by rigor-mortis because scientific discovery benefits from the audacity to break formal rules. A mathematician would say that for a parameterized curve γ(t)=(x(t),y(t)), the arc length along the curve between parameter values t a and t b is ∫t b t a\|˙γ(t)\|d t=∫t b t a√˙x(t)2+˙y(t)2 d t This definition is motivated by thinking about little pieces of time d t and the fact that \|˙γ(t)\| is the speed at time t. For little pieces of time, the speed times time will give a little straight line piece of distance traveled, and then one simply adds up all the distances of these little pieces. In other words, there is a very nice physical motivation for this definition of arc length along a parameterized curve. Now as a special case, suppose that there is a parameterization of the curve whose arc length we're trying to compute taking the following form: α(s)=(s,y(s)). Since the x-coordinate function is simply the identity function, we might as well call this parameter x (it's a dummy variable anyway), in which case we get α(x)=(x,y(x)). Now plug this into the original arc length definition to obtain ∫x b x a\|α′(x)\|d x=∫x b x a√1+y′(x)2 d x. This is precisely the formula written down by the physicist. Comment on the Edit. Treating derivatives as difference quotients of small quantities often works because, well, that really is what a derivative is doing. Look at the definition of the derivative as a limit of a difference quotient. If Δ x is small, then replacing the derivative by the difference quotient won't generally incur a large error, so it's not always such a bad way to look at things. This should, of course, be taken with a grain of salt when you want to rigorously clean everything up in the end, but it's often unproductive to tie your hands and not think about these things intuitively, especially when you're first learning them. I enjoy and appreciate rigor as much as any other respectable citizen off the street, but I've also learned to appreciate that working loosely with mathematical quantities can often lead to great intuition and insight. Take, for example, path integrals in physics. No one really knows how to define these beasts in a way that would satisfy a modern mathematician (especially path integrals in quantum field theory), but nonetheless physicists' formal manipulations have led to some of the most accurately predicted measurements in human history. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 9, 2018 at 5:20 answered May 19, 2017 at 23:30 joshphysicsjoshphysics 1,927 1 1 gold badge 14 14 silver badges 37 37 bronze badges 0 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. your professor is probably not meaning dx and dy to be tiny numbers. He was probably saying that as minor justification of how he set up the original equation. I do not have the full and complete ability to prove what your professor did was valid. After all, I am willing to assume that his first equation was valid. As many answers point out, that is his 'definition' of arc length. Of course, one can go deeper and somehow prove that is arc length, but let's be frank. Arc length is a human defined term. We have to accept that as the starting point. Moving on from that; however, I believe I can clarify what your professor was trying to do in a mathematical manner with my own words. Proposition: For every pair of real valued function f and g, there exist differential functors P f and P g such that P f P g=D g(f), f⋅P g=∫(f)d g, and P f P f=1 What this all does is establish that the handwavy math that the professor did was valid. Things like "dx" (in my eyes) are neither functions nor numbers. They are something else. I don't know what to call them, but the algebra of them appears to work just fine. Think of it this way. The professor is not doing algebra of numbers or an algebra of functions. He is doing an algebraic manipulation of something else, and I wouldn't write it off as garbage. All the professor truly needs to do is set forth what kind of objects these things are. Are they elements of a function space? Are they infinitesimally small numbers? Are they some kind of number-operator hybrid (never heard of such a thing but physics people are weird)? Honestly, the only one who can answer this question is your professor. Just tell the professor that you recognize that dx and dy are not numbers nor functions nor operators and that you wish to know what they are. If he just says "they are numbers", then you have your answer. He is most certainly doing some non-rigorous stuff and messing up in math (but it is physics so that shouldn't be an issue). Chances are, he'll point you in the direction of some other mathematical construct and you'll learn something new. Or maybe he'll just say that's how he learned it and he was simply attempting to 'justify' the definition. Answering your question in your edit, I think it really like I said. You can write any equation you want and claim they are the objects I set forth. The issue is whether or not it is well defined. I see no reason that wouldn't occur. Worst is that you get something that is neither an integral nor a derivative and in that situation I'd argue that the equation is simply unknown in meaning and not just undefined. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 20, 2017 at 4:08 user64742user64742 2,225 7 7 gold badges 31 31 silver badges 58 58 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. 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14768	https://dcc.dickinson.edu/grammar/latin/tenses-subjunctive-indirect-discourse	Toggle menu Allen and Greenough / Latin Grammar edited by Meagan Ayer Tenses of the Subjunctive in Indirect Discourse 585. The tenses of the Subjunctive in Indirect Discourse follow the rule for the Sequence of Tenses (§ 482). They depend for their sequence on the verb of saying, etc., by which the Indirect Discourse is introduced. Thus in the sentence, dīxit sē Rōmam itūrum ut cōnsulem vidēret (He said he should go to Rome in order that he might see the consul), vidēret follows the sequence of dīxit without regard to the Future Infinitive, itūrum [esse], on which it directly depends. Note— This rule applies to the Subjunctive in subordinate clauses, to that which stands for the imperative, etc., (see examples, § 588), and to that in questions (§ 586). a. A Subjunctive depending on a Perfect Infinitive is often in the Imperfect or Pluperfect, even if the verb of saying, etc., is in a primary tense (cf. § 485.j); so regularly when these tenses would have been used in Direct Discourse. Tarquinium dīxisse ferunt tum exsulantem sē intellēxisse quōs fīdōs amīcōs habuisset (Lael. 53) They tell us that Tarquin said that then in his exile he had found out what faithful friends he had had. [Here the main verb of saying, ferunt, is primary, but the time is carried back by dīxisse and intellēxisse, and the sequence then becomes secondary.] tantum prōfēcisse vidēmur ut ā Graecīs nē verbōrum quidem cōpiā vincerēmur. (N. D. 1.8) We seem to have advanced so far that even in abundance of words we ARE not surpassed by the Greeks. Note 1— The proper sequence may be seen, in each case, by turning the Perfect Infinitive into that tense of the Indicative which it represents. Thus, if it stands for an Imperfect or an Historical Perfect, the sequence will be secondary; if it stands for a Perfect Definite, the sequence may be either primary or secondary (§ 485.a). Note 2— The so-called Imperfect Infinitive after meminī (§ 584.a, Note) takes the secondary sequence. Ad mē adīre quōsdam meminī, quī dīcerent, etc. (Fam. 3.10.6) I remember that some persons visited me, to tell me, etc. b. The Present and Perfect Subjunctive are often used in dependent clauses of the Indirect Discourse even when the verb of saying etc. is in a secondary tense. dīcēbant . . . totidem Nerviōs (pollicērī) quī longissimē absint. (B. G. 2.4) They said that the Nervii, who live farthest off, promised as many. Note— This construction comes from the tendency of language to refer all time in narration to the time of the speaker (repraesentātiō). In the course of a long passage in the Indirect Discourse the tenses of the subjunctive often vary, sometimes following the sequence, and sometimes affected by repraesentātiō. Examples may be seen in B. G. 1.13, 7.20, etc. Certain constructions are never affected by repraesentātiō. Such are the Imperfect and Pluperfect Subjunctive with cum temporal, antequam, and priusquam. XML Files Chapter-585.xml Suggested Citation Meagan Ayer, Allen and Greenough’s New Latin Grammar for Schools and Colleges. Carlisle, Pennsylvania: Dickinson College Commentaries, 2014. ISBN: 978-1-947822-04-7.
14769	https://www.ncbi.nlm.nih.gov/books/NBK470386/	An official website of the United States government 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 in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation 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-. Hyponatremia Helbert Rondon; Madhu Badireddy. Author Information and Affiliations Authors Helbert Rondon1; Madhu Badireddy2. Affiliations 1 University of Pittsburgh 2 Christus Santa Rosa Hospitals Last Update: June 14, 2023. Continuing Education Activity Hyponatremia is defined as a serum sodium concentration of less than 135 mEq/L but can vary to a small extent in different laboratories. Hyponatremia is a common electrolyte abnormality caused by an excess of total body water when compared to total body sodium content. Edelman discovered that serum sodium concentration does not depend on total body sodium but the ratio of total body solutes (e.g., total body sodium and total body potassium) to total body water. Hyponatremia represents an imbalance in this ratio where total body water is more than total body solutes. This activity explains when this condition should be considered on differential diagnosis, articulates how to properly evaluate for this condition, and highlights the role of the interprofessional team in caring for patients with this condition. Objectives: Review the causes of hyponatremia. Describe the pathophysiology of the syndrome of inappropriate antidiuretic hormone secretion (SIADH) and its association with hyponatremia. Summarize the treatment of hyponatremia. Outline the importance of enhancing care coordination among the interprofessional team to ensure proper evaluation and management of hyponatremia. Access free multiple choice questions on this topic. Introduction Hyponatremia is defined as a serum sodium concentration of less than 135 mEq/L but can vary to some extent depending upon the set values of varied laboratories. Hyponatremia is a common electrolyte abnormality caused by an excess of total body water in comparison to that of the total body sodium content. Edelman approved of the fact that serum sodium concentration does not depend on total body sodium but is determined by the ratio of total body solutes (e.g., total body sodium and total body potassium) to total body water. Hyponatremia represents an imbalance in this ratio where total body water is more than total body solutes. Total body water (TBW) has two main compartments, extracellular fluid (ECF), accounting for one-third, and intracellular fluid (ICF), accounting for the remaining two-thirds. Sodium is the major solute of ECF, and potassium for ICF. Etiology The etiology of hyponatremia can be classified based on the volume status of the extracellular fluid. As mentioned earlier, sodium is the major solute of extracellular fluid (ECF). Based on the volume of ECF, a patient can be classified into hypovolemic, euvolemic, or hypervolemic. Physiological stimuli that cause vasopressin release in adjunct with increased fluid intake can cause hyponatremia. Hypothyroidism and adrenal insufficiency may contribute to an increased release of vasopressin. Physiological stimuli for vasopressin release include loss of intravascular volume (hypovolemic hyponatremia) and the loss of effective intravascular volume (hypervolemic hyponatremia). Causes of Hypovolemic Hyponatremia (TBW decreases less than a decrease in total body sodium) Gastrointestinal fluid loss (diarrhea or vomiting) The third spacing of fluids (pancreatitis, hypoalbuminemia, small bowel obstruction) Diuretics Osmotic diuresis (glucose, mannitol) Salt-wasting nephropathies Cerebral salt-wasting syndrome (urinary salt wasting, possibly caused by increased brain natriuretic peptide) Mineralocorticoid deficiency Causes of Hypervolemic Hyponatremia (TBW increases greater than an increase in total body sodium) Renal causes (acute renal failure, chronic renal failure, nephrotic syndrome) Extrarenal causes (congestive heart failure, cirrhosis) Iatrogenic Causes of Euvolemic Hyponatremia (TBW increase with stable total body sodium) Nonosmotic, pathologic vasopressin release may occur in the setting of normal volume status, as with euvolemic hyponatremia. Causes of euvolemic hyponatremia include: Drugs, as mentioned below. Syndrome of inappropriate antidiuretic hormone (SIADH) Addison's disease Hypothyroidism High fluid intake in conditions like primary polydipsia; or potomania, caused by a low intake of solutes with relatively high fluid intake Medical testing related to excessive fluids, such as a colonoscopy or cardiac catheterization Iatrogenic Many drugs cause hyponatremia, and the most common include: Vasopressin analogs such as desmopressin and oxytocin Medications that stimulate vasopressin release or potentiate the effects of vasopressin, such as selective serotonin-reuptake inhibitors and other antidepressants, morphine, and other opioids Medications that impair urinary dilution, such as thiazide diuretics Medications that cause hyponatremia, such as carbamazepine or its analogs, vincristine, nicotine, antipsychotics, chlorpropamide, cyclophosphamide, nonsteroidal anti-inflammatory drugs Illicit drugs such as methylenedioxymethamphetamine (MDMA or ecstasy). Epidemiology Hyponatremia is the most common electrolyte disorder, with a prevalence of 20% to 35% among hospitalized patients. The incidence of hyponatremia is high among critical patients in the intensive care unit (ICU) and also in postoperative patients. This is more common in elderly patients due to multiple comorbidities, multiple medications, and a lack of access to food and drinks. Pathophysiology Thirst stimulation, antidiuretic hormone (ADH) secretion, and handling of filtered sodium by kidneys maintain serum sodium and osmolality. Normalplasma osmolality is around 275 mOsm/kg to 290 mOsm/kg. To maintain normal osmolality, water intake should be equal to water excretion. The imbalance of water intake and excretion causes hyponatremia or hypernatremia. Water intake is regulated by the thirst mechanism, where osmoreceptors in the hypothalamus trigger thirst when body osmolality reaches 295 mOsm/kg. Water excretion is tightly regulated by antidiuretic hormone (ADH), synthesized in the hypothalamus, and stored in the posterior pituitary gland. Changes in tonicity lead to either enhancement or suppression of ADH secretion. Increased ADH secretion causes reabsorption of water in the kidney, and suppression causes the opposite effect. Baroreceptors in the carotid sinus can also stimulate ADH secretion, but it is less sensitive than the osmoreceptors. Baroreceptors trigger ADH secretion due to decreased effective circulating volume, nausea, pain, stress, and drugs. Hypertonic Hyponatremia (Serum osmolality of greater than 290 mOsm/kg) Hyperglycemia Mannitol Isotonic Hyponatremia (Serum osmolality between 275 mOsm/kg and 290 mOsm/kg) Pseudo-hyponatremia is a laboratory artifact. It is usually caused by hypertriglyceridemia, cholestasis (lipoprotein X), and hyperproteinemia (monoclonal gammopathy, intravenous immunoglobulin [IVIG]). Two-thirds of clinical labs in use still use indirect ion-selective electrode technology, and therefore this problem is still present. Nonconductive irrigant solutions: these solutions contain mannitol, glycine, or sorbitol and are used in urological and gynecological procedures such as transurethral resection of the prostate (TURP). Hypotonic Hyponatremia (Serum osmolality of less than 275 mOsm/kg) Hypotonic hyponatremia represents an excess of free water. This excess free water can be caused by two mechanisms: Increased free water intake: The patient drinks a large volume of free water (greater than 18 L/day or greater than 750 mL/h) that overwhelms the kidney's capacity to excrete free water. Examples of this are psychogenic polydipsia, marathon runners, water drinking competitions, and ecstasy. Decreased free water excretion: Patients drink a normal volume of free water, but the kidneys cannot excrete the water for some reason. There are three mechanisms involved in the inability of kidneys to excrete water: High ADH activity: Three different mechanisms can cause high ADH: Decreased effective arterial blood volume (EABV): antidiuretic hormone (ADH) is released when there is a reduction of 15% or more of the EABV. This occurs with hypovolemia (e.g., vomiting, diarrhea), decreased cardiac output (e.g., heart failure), or vasodilation (e.g., cirrhosis). SIADH: ADH is secreted autonomously. Four general causes of this are brain disorders, lung disorders, drugs (e.g., SSRI), and miscellanea (e.g., nausea and pain). Cortisol deficiency: Cortisol exerts an inhibitory effect on ADH release. When cortisol is decreased, ADH is released in large amounts. Adrenal insufficiency is the cause of this mechanism. Low glomerular filtration rate (GFR): a low glomerular filtration rate would impair the kidney's ability to get rid of water. Typical examples are acute kidney injury (AKI), chronic kidney disease (CKD), and end-stage renal disease (ESRD). Low solute intake: Patients on a regular diet consume 600 mOsm to 900 mOsm of solute per day. Solutes are defined as substances that are freely filtered by the glomeruli but have relative or absolute difficulty in being reabsorbed by the tubules in relationship to water. The main solutes are urea (which comes from the metabolism of proteins) and electrolytes (e.g., salt). Carbohydrates do not contribute to solute load. In steady-state conditions, solute intake is equal to urine solute load. Therefore, it is expected that these patients also excrete 600 mOsm to 900 mOsm of solute in the urine. Urine volume, and hence water excretion, is dependent on the urine solute load. The more solute one needs to excrete, the larger the urine volume one needs to produce. The less solute one needs to excrete, the smaller the urine volume one needs to produce. Patients who eat a low amount of solute per day (eg., 200 mOsm/day) on steady-state conditions will also excrete a low amount of solute in the urine, and therefore they will do it in a smaller volume of urine. This decreased urine volume will limit the capacity of the kidneys to excrete water. Typical examples of this are beer potomania and the tea-and-toast diet. SIADH (Syndrome of inappropriate antidiuretic hormone secretion) This is a condition where inappropriate secretion of ADH despite normal or increased plasma volume causes impaired water excretion by the kidney leading to hyponatremia. SIADH is a diagnosis of exclusion, as there is no single test to confirm the diagnosis. Patients are hyponatremic and euvolemic. Causes of SIADH include Any central nervous system (CNS) disorder, Ectopic production of ADH (most commonly small cell carcinoma of the lung), Drugs (carbamazepine, oxcarbazepine, chlorpropamide, and multiple other drugs), HIV, Pulmonary diseases (pneumonia, tuberculosis), Postoperative patients (pain medicated) Treatment includes fluid restriction and the use of vasopressin 2 receptor inhibitors. History and Physical Symptoms depend upon the degree and chronicity of hyponatremia. Patients with mild-to-moderate hyponatremia (greater than 120 mEq/L) or a gradual decrease in sodium (greater than 48 hours) have minimal symptoms. Patients with severe hyponatremia (less than 120 mEq/L) or rapid decrease in sodium levels have multiple varied symptoms. Symptoms can range from anorexia, nausea and vomiting, fatigue, headache, and muscle cramps to altered mental status, agitation, seizures, and even coma. Apart from symptoms, a detailed history taking to include a history of pulmonary and CNS disorders, all home medications, and social history (increased beer intake or use of MDM or ecstasy) is very important. Physical examination includes assessing volume status and neurological status. Patients with neurological symptoms and signs need to be treated promptly to prevent permanent neurological damage. Evaluation The following steps may be performed while evaluating a patient with suspected hyponatremia: Step 1: Plasma Osmolality (275 mOsm to 290 mOsm/kg) It can help differentiate between hypertonic, isotonic, and hypotonic hyponatremia. True hyponatremic patients are hypotonic. If the patient is hypotonic, then go to step 2. Step 2: Urine Osmolality Urine osmolality less than 100 mOsm/kg indicates primary polydipsia or reset osmostat. Urine osmolality greater than 100 mOsm/kg usually indicates a high ADH state; go to step 3. Step 3: Volume Status (ECF status) Hypovolemic vs euvolemic vs hypervolemic. If the patient is hypovolemic, then proceed to step 4. Step 4: Urine Sodium Concentration Urine sodium less than 10 mmol/L indicates extrarenal loss of fluid (remote diuretic use and remote vomiting). Urine sodium greater than 20 mmol/L suggests renal loss of urine (diuretics, vomiting, cortisol deficiency, and salt-wasting nephropathies). Other tests that might help in differentiating the causes include Serum thyroid-stimulating hormone (TSH) Serum adrenocorticotropic hormone (ACTH) Serum urea Liver function tests Chest X-ray or computed tomography (CT) scan of the chest CT scan of the head Treatment / Management Treatment of hyponatremia depends upon the degree of hyponatremia, duration of hyponatremia, severity of symptoms, and volume status. Acute Symptomatic Hyponatremia Severely symptomatic hyponatremia: Administer 3% sodium chloride; 100 mL intravenous (IV) bolus (repeat up to twice if symptoms persist). Mild to moderately symptomatic hyponatremia: 3% Sodium chloride, slow infusion (use sodium deficit formula to calculate the rate of infusion but recalculate rate with frequent sodium monitoring). Chronic Asymptomatic Hyponatremia Hypovolemic hyponatremia: Isotonic fluids administration and holding of any diuretics. Hypervolemic hyponatremia: Treat underlying condition, restrict salt and fluids, and administer loop diuretics. Euvolemic hyponatremia: Fluid restriction to less than 1 liter per day. Drugs: Selective vasopressin 2 receptor antagonists are being used recently. They increase the excretion of water in the kidneys without affecting sodium, thereby increasing serum sodium levels. These medications are used in patients with euvolemic and hypervolemic conditions (except liver failure) if the above measure does not help. The goal of correction: Correct sodium by no more than 10 mEq/L to 12 mEq/L in any 24-hour period. Risk factors for osmotic demyelination syndrome (ODS): Hypokalemia, liver disease, malnutrition, and alcohol use. Limits of Correction High-risk for ODS: less than 8 mEq/L in any 24-hour period Average risk for ODS: less than 10 mEq/L in any 24-hour period In the absence of false laboratory hyponatremia, pseudo hyponatremia, and a lack of hypovolemic state, including postural hypotension, the next step is to measure urine sodium and osmolarity. In the low urinary sodium of less than 100 mOsm/kg and absence of rapid water consumption, the potential for a high fluid, low protein diet, including beer potomania, should be examined. In patients with severe hyponatremia of less than 120 mEq/L, the chronicity of the hyponatremia should be considered. Accordingly, in severe, chronic hyponatremia, intravenous 3 percent saline at a rate of 15 to 30 mL/hour should be initiated. In some patients, desmopressin (dDAVP) should also be administered to prevent overly rapid correction. Three percent saline can be safely infused via a peripheral vein, and so far, vascular thrombosis and extravasation injuries have not been reported. However, some centers have policies against the peripheral infusion of hypertonic saline. In these circumstances, central vein infusions or infusions of a lower concentration with higher infusion rates are required. Tolvaptan is indicated in hyponatremia associated with high anti-diuretic hormone (ADH) activity. Fluid restriction is adequate for patients who have normovolemic hypotonic hyponatremia. Some patients with the syndrome of inappropriate antidiuretic hormone secretion (SIADH) who are malnourished may need a high protein intake, which increases the solute load for renal excretion, resulting in more free water removal. Laboratory findings in patients with SIADH reveal hyponatremia (plasma sodium level of less than 135 mEq/L) and low serum osmolality (less than 280 mOsm/kg). Moreover, patients with SIADH have increased urinary sodium levels (greater than 20 mMol/L) and urine osmolality (generally above 100 mOsm/L). Differential Diagnosis True hyponatremia is associated with hypo osmolality. Conditions causing hyperosmolar hyponatremia and iso-osmolar hyponatremia (pseudo-hyponatremia) should be differentiated first. Hyperglycemia Mannitol overdose Hyperlipidemia Hyperproteinemia Differential Diagnosis for Hypo-Osmolar Hyponatremia Gastroenteritis Diuretic use Congestive heart failure Liver failure Psychogenic polydipsia Renal causes SIADH Adrenal crisis Hypothyroidism Prognosis The prognosis in patients with hyponatremia depends on the severity of hyponatremia and the underlying condition causing it. The prognosis is poor in patients with severe hyponatremia, acute hyponatremia, and older patients. Complications If left untreated or inadequately treated, patients with hyponatremia can develop rhabdomyolysis, altered mental status, seizures, and even coma. Rapid correction of chronic hyponatremia (greater than 10 mEq/L to 12 mEq/L of sodium in 24 hrs) can lead to osmotic demyelination syndrome. Osmotic demyelination syndrome, formerly known as central pontine myelinolysis, is a complication of rapid correction of sodium in patients with chronic hyponatremia. In patients with hyponatremia, the brain adapts to a fall in serum sodium level, without developing cerebral edema, in about 48 hours. As a result, patients with chronic hyponatremia are mostly asymptomatic. Once the brain adapts to low serum sodium, the rapid correction of sodium leads to osmotic demyelination syndrome. Clinical manifestations are typically delayed by a few days and comprise several irreversible neurological symptoms, including seizures, disorientation, and even coma. "Locked-in" syndrome occurs in severely affected patients. These patients are awake but unable to move or can communicate with the help of their eyes only. Consultations It is imperative to consult a nephrologist in a patient with severe hyponatremia or a rapid decrease in sodium, or persistent hyponatremia. Cardiology and gastroenterology consultation might be necessary for patients with congestive heart failure and hepatic failure, respectively. Deterrence and Patient Education Patients with hyponatremia should be followed closely at discharge by both the primary care provider and nephrology. Follow-up labs are ordered as needed, and patients needing fluid restriction should be educated appropriately. Pearls and Other Issues Hyponatremia is a common electrolyte abnormality. Hyponatremia can range from an asymptomatic condition to a life-threatening condition. Hyponatremia can occur with hypovolemic or hypervolemic or euvolemic states. Common causes include diuretics, vomiting, diarrhea, congestive heart failure, renal and liver disease. Degree, and duration of hyponatremia, along with the severity of symptoms, determine the management algorithm and the rapidity to correct sodium. Do not correct the hyponatremia by more than 10 mEq/L to 12 mEq/L in 24 hours, except in patients with severe symptoms and rapidly decreased sodium levels. Too rapid correction of the sodium levels can lead to osmotic demyelinating syndrome. Enhancing Healthcare Team Outcomes Hyponatremia is a common electrolyte abnormality. Sodium levels need to be closely monitored, as this could lead to life-threatening complications if left untreated. This is even more important in patients with renal disease and those who are on diuretics. Good interprofessional communication between the primary care provider and a nephrologist is imperative to keep a close eye on sodium levels and their proper correction as and when needed. Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. References 1. : Overgaard-Steensen C. Initial approach to the hyponatremic patient. Acta Anaesthesiol Scand. 2011 Feb;55(2):139-48. [PubMed: 21029052] 2. : Overgaard-Steensen C, Larsson A, Bluhme H, Tønnesen E, Frøkiaer J, Ring T. Edelman's equation is valid in acute hyponatremia in a porcine model: plasma sodium concentration is determined by external balances of water and cations. 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Lixivaptan, a New Generation Diuretic, Counteracts Vasopressin-Induced Aquaporin-2 Trafficking and Function in Renal Collecting Duct Cells. Int J Mol Sci. 2019 Dec 26;21(1) [PMC free article: PMC6981680] [PubMed: 31888044] 22. : Krisanapan P, Vongsanim S, Pin-On P, Ruengorn C, Noppakun K. Efficacy of Furosemide, Oral Sodium Chloride, and Fluid Restriction for Treatment of Syndrome of Inappropriate Antidiuresis (SIAD): An Open-label Randomized Controlled Study (The EFFUSE-FLUID Trial). Am J Kidney Dis. 2020 Aug;76(2):203-212. [PubMed: 32199708] 23. : Girot H, Déhais M, Fraissinet F, Wils J, Brunel V. Atypical Pseudohyponatremia. Clin Chem. 2018 Feb;64(2):414-415. [PubMed: 29378746] 24. : C A, Badiger R. Epidemiology of Hyponatraemia Among Elderly Patients with Lower Respiratory Tract Infection. J Assoc Physicians India. 2020 Jan;68(1):80. [PubMed: 31979779] 25. : Sterns RH. Adverse Consequences of Overly-Rapid Correction of Hyponatremia. Front Horm Res. 2019;52:130-142. [PubMed: 32097948] 26. : Reijnders TDY, Janssen WMT, Niamut SML, Kramer AB. Role of Risk Factors in Developing Osmotic Demyelination Syndrome During Correction of Hyponatremia: A Case Study. Cureus. 2020 Jan 02;12(1):e6547. [PMC free article: PMC6996461] [PubMed: 32042522] 27. : Golestaneh L, Neugarten J, Southern W, Kargoli F, Raff A. Improving the diagnostic workup of hyponatremia in the setting of kidney disease: a continuing medical education (CME) initiative. Int Urol Nephrol. 2017 Mar;49(3):491-497. [PubMed: 28091865] : Disclosure: Helbert Rondon declares no relevant financial relationships with ineligible companies. : Disclosure: Madhu Badireddy declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. 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14770	http://www.gutenberg.org/ebooks/18603	What Social Classes Owe to Each Other by William Graham Sumner \| Project Gutenberg [x] Skip to main content X Go! Donate About▼ About Project GutenbergContact UsHistory & PhilosophyKindle & eReadersHelp PagesOffline CatalogsDonate Frequently DownloadedMain CategoriesReading ListsSearch Options Frequently DownloadedMain Categories Project Gutenberg 76,768 free eBooks 4 by William Graham Sumner What Social Classes Owe to Each Other by William Graham Sumner "What Social Classes Owe to Each Other" by William Graham Sumner is a sociopolitical treatise written in the late 19th century. The work critically examines the interactions and responsibilities between different social classes, specifically addressing the notion of duty among the rich and poor in society. Sumner aims to challenge the prevailing sentiments of his time regarding wealth redistribution and the perceived obligations of the affluent to support the less fortunate. The - [x] ... Read More opening of the text sets the stage for an exploration of the social problems that plague society, particularly the confusion surrounding class definitions and rights. Sumner begins by interrogating who has the right to demand solutions for societal issues, establishing a dichotomy between the prosperous and those less fortunate. He critiques the notion that the wealthy owe their comforts to the labor of the poor, suggesting instead that individuals must take responsibility for their own lives. The introduction foreshadows a discourse on economic principles, individual responsibility, and the potential pitfalls of class-based sentiments and policies, which will be further dissected in the subsequent chapters. (This is an automatically generated summary.) Show Less Read or download for free \| \| How to read \| Url \| Size \| \| \| \| --- --- --- \| \| Read now! \| \| 229 kB \| \| \| \| \| \| EPUB3 (E-readers incl. Send-to-Kindle) \| \| 175 kB \| \| \| \| \| \| EPUB (older E-readers) \| \| 177 kB \| \| \| \| \| \| EPUB (no images, older E-readers) \| \| 151 kB \| \| \| \| \| \| Kindle \| \| 334 kB \| \| \| \| \| \| older Kindles \| \| 313 kB \| \| \| \| \| \| Plain Text UTF-8 \| \| 205 kB \| \| \| \| \| \| Download HTML (zip) \| \| 167 kB \| \| \| \| \| \| There may be more files related to this item. \| Similar Books Readers also downloaded… In Category: Sociology In Category: Politics In Category: Economics About this eBook \| Author \| Sumner, William Graham, 1840-1910 \| \| Title \| What Social Classes Owe to Each Other \| \| Credits \| Produced by Jeff G., Jeannie Howse and the Online Distributed Proofreading Team at www.pgdp.net \| \| Reading Level \| Reading ease score: 56.1 (10th to 12th grade). Somewhat difficult to read. \| \| Language \| English \| \| LoC Class \| HN: Social sciences: Social history and conditions, Social problems \| \| Subject \| Economics \| \| Subject \| Social ethics \| \| Category \| Text \| \| EBook-No. \| 18603 \| \| Release Date \| Jun 16, 2006 \| \| Copyright Status \| Public domain in the USA. \| \| Downloads \| 350 downloads in the last 30 days. \| \| Project Gutenberg eBooks are always free! \| About Project Gutenberg Privacy policy Permissions Terms of Use Contact Us Help
14771	https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Difference_Between_K_And_Q	Difference Between K And Q - Chemistry 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 Chemical Equilibria Equilibria { } { Balanced_Equations_and_Equilibrium_Constants : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Calculating_an_Equilibrium_Concentration : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Calculating_An_Equilibrium_Concentrations : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Calculating_an_Equilibrium_Constant_Kp_with_Partial_Pressures : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Determining_the_Equilibrium_Constant : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Difference_Between_K_And_Q : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Dissociation_Constant : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Effect_of_Pressure_on_Gas-Phase_Equilibria" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Equilibrium_Calculations : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Kc : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Kp : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Law_of_Mass_Action : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Mass_Action_Law : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Principles_of_Chemical_Equilibria : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", The_Equilibrium_Constant : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", The_Reaction_Quotient : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "Acid-Base_Equilibria" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Chemical_Equilibria : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Dynamic_Equilibria : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Heterogeneous_Equilibria : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Le_Chateliers_Principle : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Physical_Equilibria : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Solubilty : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 30 Jan 2023 07:04:33 GMT Difference Between K And Q 1378 1378 admin { } Anonymous Anonymous User 2 false false [ "article:topic", "showtoc:no", "license:ccbyncsa", "licenseversion:40" ] [ "article:topic", "showtoc:no", "license:ccbyncsa", "licenseversion:40" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Bookshelves 3. Physical & Theoretical Chemistry 4. Supplemental Modules (Physical and Theoretical Chemistry) 5. Equilibria 6. Chemical Equilibria 7. Difference Between K And Q Expand/collapse global location Difference Between K And Q Last updated Jan 30, 2023 Save as PDF Determining the Equilibrium Constant 2 Dissociation Constant Page ID 1378 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Introduction 2. Q vs. K: What Does It Mean? 1. Situation 1: Q = K 2. Situation 2: Q < K 1. Q Equals Zero 3. Situation 3: Q > K/Equilibria/Chemical_Equilibria/Difference_Between_K_And_Q#Situation_3:_Q_.3E_K) 1. Q Equals Infinity/Equilibria/Chemical_Equilibria/Difference_Between_K_And_Q#Q_Equals_Infinity) Remembering the Relationship Between K and Q Handy Chart Outlining the Relationships of Q and K Predicting the Shift of a Reaction Without Calculations Example: Putting It All Together Problems Solutions References Contributors and Attributions Sometimes it is necessary to determine in which direction a reaction will progress based on initial activities or concentrations. In these situations, the relationship between the reaction quotient, Q c, and the equilibrium constant, K c, is essential in solving for the net change. With this relationship, the direction in which a reaction will shift to achieve chemical equilibrium, whether to the left or the right, can be easily calculated. Introduction K c can be used calculate the final concentrations at equilibriumfor a reaction using an ICE table and the natural progression of the reaction, from left to right or from right to left. However, what if you do not know which way the reaction will progress? A simple relationship between K c and the reaction quotient, known as Q c, can help. The reaction quotient, Q, expresses the relative ratio of products to reactants at a given instant. Using either the initial concentrations or initial activities of all the components of the reaction, the progression of an reaction can easily be determined. Given the general chemical reaction (1)a⁢A+b⁢B⇌g⁡G+h⁡H Q may be expressed as the following equations: (2)Q=a i⁢n⁢i⁢t g⁢a i⁢n⁢i⁢t h a i⁢n⁢i⁢t a⁢a i⁢n⁢i⁢t b or (3)Q c=[G]i⁢n⁢i⁢t g⁢[H]i⁢n⁢i⁢t h[A]i⁢n⁢i⁢t a⁢[B]i⁢n⁢i⁢t b Remember that the concentrations of liquids and solids do not change, so they are excluded from the expression. As shown above, the value of Q can be found by raising the products to the power of their coefficients, or stoichiometric factors, divided by the reactants raised to their coefficients. If the concentration of products in the numerator is much larger than that of reactants in the denominator, Q will be a large value. On the other hand, a small amount of products (small numerator) divided by a large value for the concentration of reactants (large denominator) would result in a small value for Q. The expressions for Q are very similar to those for K: (4)K=a G g⁢a H h a A a⁢a B b or (5)K c=[G]g⁢[H]h[A]a⁢[B]b To determine which direction a reaction will go towards, simply compare Q c, the initial concentration ratio, to K c, the equilibrium constant, and evaluate the results. Q vs. K: What Does It Mean? When you set Q against K, there are five possible relationships: Q=K Q=0 Q<K Q=∞ and Q>K. To properly predict which way a reaction will progress, you must know these relationships. Situation 1: Q = K When Q=K, the system is at equilibrium and there is no shift to either the left or the right. Take, for example, the reversible reaction shown below: (6)C⁢O(g)+2⁢H 2⁢(g)⇌C⁢H 3⁡O⁢H(g) The value of K c at 483 K is 14.5. If Q=14.5, the reaction is in equilibrium and will be no evolution of the reaction either forward or backwards. Situation 2: Q < K When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right. Consider again: (7)C⁢O(g)+2⁢H 2⁢(g)⇌C⁢H 3⁡O⁢H(g) For Q<K: (8)C⁢O(g)+2⁢H 2⁢(g)⟶C⁢H 3⁡O⁢H(g) so that equilibrium may be established. Q Equals Zero If Q=0, then Q is less than K. Therefore, when Q=0, the reaction shifts to the right (forward). An easy way to remember this relationship is by thinking, “once you have nothing, the only thing left to do is to move forward.” If Q equals to zero, the reaction will shift forward (to the right): (9)C⁢O(g)+2⁢H 2⁢(g)⟶C⁢H 3⁡O⁢H(g) Situation 3: Q > K When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K: (10)C⁢O(g)+2⁢H 2⁢(g)⟵C⁢H 3⁡O⁢H(g) Q Equals Infinity When Q=∞, the reaction shifts to the left (backwards). This is a variation of when Q>>>K. (11)C⁢O(g)+2⁢H 2⁢(g)⟵C⁢H 3⁡O⁢H(g) Remembering the Relationship Between K and Q An easy way to remember these relationships is by thinking of the > or < as the mouth of an alligator. The alligator will "eat" in the direction that the reaction shifts as long as Q is written before K. Handy Chart Outlining the Relationships of Q and K Remembering these simple relationships will aid you to solving for the progression of a reaction. A chart outlining them can be found below. Predicting the Shift of a Reaction Without Calculations Depending on what a problem asks of you, sometimes it is unnecessary to make any calculations at all. Take, for example, the now familiar reversible reaction listed below: (12)C⁢O(g)+2⁢H 2⁢(g)⇌C⁢H 3⁡O⁢H(g) What do you think will happen if more of the product, methanol (CH3OH), is added? Equilibrium will be disrupted, and the increase in products mean that Q>K. In order to re-establish equilibrium, the reaction will progress to the left, towards the reactants. This means some of the added methanol will break down into carbon monoxide and hydrogen gas. (13)C⁢O(g)+2⁢H 2⁢(g)⟵C⁢H 3⁡O⁢H(g) Now, what if more of the reactants, carbon monoxide and hydrogen gas, are a? You should realize that this would upset the equilibrium. Q<K, because the value for the amount of reactants, or the denominator of the Q expression, has increased. To establish equilibrium again, the reaction will favor the product, so the reaction will progress to the right. (14)C⁢O(g)+2⁢H 2⁢(g)⟶C⁢H 3⁡O⁢H(g) The ideas illustrated above show Le Chatelier's Principle whereby when an equilibrated system is subjected to a change in temperature, pressure, or concentration of a species in the reaction, the system responds by achieving a new equilibrium that partially offsets the impact of the change. Predicting which way a reaction will go can be the easiest thing that you will ever do in chemistry! Example: Putting It All Together To properly use the relationship between Q and K, you must know how to set it up. Take, for example, the reaction below: If you start with 4.00M CH 4, 2.00M C 2 H 2, and 3.00M H 2, which direction will the reaction progress to reach equilibrium? Problems 1) Consider this reaction: 2⁢N⁢O⁢B⁢r(g)⇋2⁢N⁢O(g)+B⁢r 2 If K c= 0.0142 and the initial concentrations are 1.0 M NOBr, 0.2M NO, and 0.8M Br 2, which way will the reaction progress to reach equilibrium? 2) What is Q and its purpose? 3) Consider the reaction following in equilibrium: N 2⁢O 4⁢(g)⇋2⁢N⁢O 2⁢(g) If more N 2 O 4 is added, which way will the reaction proceed? 4) Consider the following reaction: C⁢O(g)+C⁢l 2⁢(g)⇋C⁢O⁢C⁢l 2⁢(g) With a K c of 1.2 x 10 3 at 668 K, is the reaction in equilibrium when there are 5.00 mol CO(g), 2.00 mol Cl 2(g), and 6.00 mol of COCl 2(g) in a 3.00L flask? If not, which direction will the reaction progress to reach equilibrium? 5) Consider the following reaction: H 2⁢(g)+I 2⁢(g)⇋2⁢H⁡I(g) If K c=50.2 at 718 K and the initial concentrations are 0.5 M H 2, 0.15M I 2, and 0.05M HI, which way will the reaction progress? 6) Consider the following reaction: 2⁢C⁢O⁢F 2⁢(g)⇋C⁢O 2⁢(g)+C⁢F 4⁢(g) If K c= 2.00 at 473 K and the initial concentrations are 2.0 M CO 2, 4.0 M CF 4, and 0.5 M COF 2, which way will the reaction progress? 7) Consider the following reaction: 2⁢S⁢O 2⁢(g)+O 2⁢(g)⇋2⁢S⁢O 3⁢(g) K c=100. With the initial masses of 20 g SO 2, 13 g O 2, and 25 g SO 3 in a 5.0 L container, which way will the reaction progress. Solutions 1) The reactions shifts to the left, towards the reactants. NOBr= 1M, NO= 0.2M, Br 2= 0.8M Q c=[0.2]2⁢[0.8]2 Q c=0.032 Therefore, Q c> K c and the reactions shifts towards the reactants. 2) Q is a reaction quotient, which helps determine if a reaction will shift forward or backwards. As a system approaches towards equilibrium, Q approaches towards K. 3) The reaction will proceed to the right. 4) No, it is not at equilibrium. Since Q<K, the reaction will shift to the right to reach equilibrium. 5) Q = 0.033, so Q<K. The reaction will shift to the right. 6) Q = 32.0, so Q>K. The reaction will shift to the left. 7) Q = 12, so Q<K. The reaction will shift to the right. References Alberty, R., A. Cornish-Bowden, et al. (1994). "Recommendations for nomenclature and tables in biochemical thermodynamics." Pure Appl. Chem66: 1641–1666. Gold, J. and V. Gold (1985). "Le Chatelier's Principle and the Laws of van 't Hoff." Education in Chemistry22: 82-85. Petrucci, Harwood, Herring, Madura. General Chemistry: Principles & MOdern Application. Ninth Edition. Pages 636-638. Contributors and Attributions Rubi Medrano (UCD), Irene Ly (UCD) Difference Between K And Q is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. 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14773	https://math.stackexchange.com/questions/4141449/intuition-behind-the-max-min-inequality	Skip to main content Intuition behind the max-min inequality Ask Question Asked Modified 3 years, 9 months ago Viewed 2k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Here is the statement of the max-min inequality: I understand the proof, but I'm struggling to visualize this inequality. Here is my attempt at intuition. Let f:R×R→R. The first slot of f is the z-coordinate and the second is the w-coordinate. Then to find the value of supz∈Zinfw∈Wf(z,w), we first walk along the z-axis. At each point along the z-axis, we stop and look out over the line that is parallel to the w-axis and intersects the point where we are currently standing. We record the elevation of the lowest valley created by f along this line. We continue this process until we've walked along the entirety of the z-axis. At the end, we scour our notebook and return the largest number we recorded. In other words, supz∈Zinfw∈Wf(z,w) is the highest, lowest valley found through this process. infw∈Wsupz∈Zf(z,w) can be found through an analogous process. This time we walk along the w-axis and look out over the lines parallel to the z-axis. We note the highest peak along each line parallel to the z-axis. At the end, we return the lowest of the numbers we recorded, i.e., the lowest, highest peak. I am struggling to intuitively see why the highest, lowest valley will always be lower than the lowest, highest peak. Is there a better interpretation of this inequality? Thanks! inequality game-theory Share CC BY-SA 4.0 Follow this question to receive notifications asked May 17, 2021 at 2:34 nkyraf33nkyraf33nkyraf33 16566 bronze badges 2 I am also looking for an 'intuitive' explanation, in addition to the formal proof. Nathan G – Nathan G 2023-04-25 07:10:58 +00:00 Commented Apr 25, 2023 at 7:10 1 See here an intuitive example: nadavb.com/max_min_inequality_intuitive_explanation Nathan G – Nathan G 2023-11-12 09:37:11 +00:00 Commented Nov 12, 2023 at 9:37 Add a comment \| 3 Answers 3 Reset to default This answer is useful 2 Save this answer. Show activity on this post. You can translate each step of the proof into your analogy. For intuition's sake, I've assumed in some places that the infima/suprema are actually attained, but the argument holds even if the they are not attained. If you stand at a fixed z=z0 and look at the line parallel to the w-axis, by definition, the lowest valley infwf(z0,w) will be less than or equal to any other point on this line f(z0,w0) (for any w0). So far, this leads to infwf(z0,w)≤f(z0,w0). This is true for any z0, so we can take a supremum on both sides to obtain supzinfwf(z,w)≤supzf(z,w0). The intuition is that the left-hand side is the "highest lowest valley" in your words. If we fix z at the maximizer z∗, the lowest valley is still lower than any other point on that line where z=z∗, i.e. supzinfwf(z,w)=infwf(z∗,w)≤f(z∗,w0). But this last value is smaller than the highest peak when fixing w=w0, i.e. f(z∗,w0)≤supzf(z,w0). The above is true for any w0, so we can take an infimum over the right-hand side to obtain supzinfwf(z,w)≤infwsupzf(z,w). If you like, you can let w∗ be the minimizer on the right-hand side, and then step through the intuition above with w∗ instead of w0. Share CC BY-SA 4.0 Follow this answer to receive notifications answered May 17, 2021 at 3:10 angryavianangryavianangryavian 93.8k77 gold badges7474 silver badges145145 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I think it makes sense when you step through the proof heuristically. So using your analogy, imagine first walking across the z axis and for every z finding the value of w that minimizes f(z,w). Now clearly this is going to be smaller or equal to any f(z,w) for all z and w because we essentially just did that operation and picked the smallest value. Now suppose we take that list of w that gives us the minimum value for any z and we now only consider the value of z that gives us the largest value of f(z,w) for that list of minimizing w values. This is the max-min value. And note that because we already knew that no matter the choice of z or w this would be less than or equal to the corresponding f(z,w) we have the inequality, supzinfwf(z,w)≤supzf(z,w) Now we are almost done. All that is left is to realize is that this inequality holds for any choice of w on the right-hand side and so it must also hold for the minimizing w of supzf(z,w). And so, supzinfwf(z,w)≤infwsupzf(z,w) So what this is saying is that when we found the lowest valley for every z, even when we then found the z so that this valley is the "highest lowest valley". It has to be as low or lower than the "lowest highest peak" because we know the highest lowest valley is lower than every peak from the first expression. But then it must also be lower or at least as low as the lowest peak. Share CC BY-SA 4.0 Follow this answer to receive notifications answered May 17, 2021 at 3:22 ArielArielAriel 40033 silver badges1010 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Another path towards understanding the intuition of this statement is the game theory interpretation. Let's say we have players Z and W and a function f(z,w) which is the result/score of the game. Player Z wants a high score while W wants a low score. Say one player chooses a value first, then the other player chooses their value, then you get the resulting score f(z,w). The intuition is that going second is always better for the player. if W can choose their move based off Z's move, then this always gives a better (smaller) outcome compared with Z choosing their move based off W's move. Case in point, winning in rock-paper-scissors is much easier if you know your opponent's move ahead of time rather than vice versa. supzinfwf(z,w) translates into "first z is chosen, then w is chosen." (Z goes first) infwsupzf(z,w) translates into "first w is chosen, then z is chosen." (W goes first) To go along with the proof, we could say: g(z) chooses the best strategy for W given z will be played. For any choice of z or w, g(z) will be better: ∀z,w:g(z)≤f(z,w) So if player Z chooses something, knowing that player W will choose after seeing this (i.e. supzg(z)) then this is going to be worse for Z than if Z could choose given any alternative w (i.e supzf(z,w)): ∀w:supzg(z)≤supzf(z,w) This is also still worse no matter what w is played, since Z chooses based on the w chosen. So we can turn the "for all" into an inf over w: supzg(z)≤infwsupzf(z,w) It seems like the crux of the intuition here is being comfortable converting the universal quantifiers into inf's and sup's. In the visual analogy with the grid, this represents transferring between thinking about all possible points (z,w) in the grid to a given row/column. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Nov 30, 2021 at 22:38 Simon AlfordSimon AlfordSimon Alford 15311 silver badge44 bronze badges Add a comment \| You must log in to answer this question. 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14774	https://www.ebsco.com/research-starters/science/chi-squared-distribution	Chi-squared Distribution \| Research Starters \| EBSCO Research Skip to navigationSkip to main contentSkip to article heroSkip to research detailsSkip to ot pc lstSkip to ot fltr modalSkip to ot lst cntSkip to footerSkip to chi-squared distributionSkip to overviewSkip to bibliographySkip to privacy preference center Company Markets We Serve Start Your Research Contact Us Log in to MyEBSCO Menu Company Who We AreLeadershipProducts and ServicesNews CenterPublishers Directory Markets We Serve Academic LibrariesPublic LibrariesSchoolsHealthcareCorporationsPublishers Start Your Research Contact Us Contact UsCareersLocationsSocialNewsletters Log in to MyEBSCO Research Starters Home EBSCO Knowledge Advantage TM Chi-squared Distribution The chi-squared distribution is a family of continuous probability distribution functions widely used in statistical hypothesis testing across various fields, including mathematics, science, and social science. It plays a crucial role in tests like the chi-square test of independence, which assesses the association between two categorical variables, and the chi-square goodness-of-fit test, used to determine if a chosen population model properly fits observed data. The distribution is generated by squaring and summing independent standard normal distributions, with its shape influenced by a parameter known as degrees of freedom (df). As the degrees of freedom increase, the distribution approaches a bell-shaped curve. Originating from the need to evaluate the fit of statistical models to data, the chi-squared distribution was shaped by contributions from several researchers, notably Karl Pearson, who formalized its application in 1900. This distribution is also related to the gamma distribution and has variations like the generalized and noncentral chi-squared distributions, which find uses in fields such as medical imaging and wireless communication. The mean and variance of the chi-squared distribution are directly linked to its degrees of freedom, aiding researchers in validating models and testing hypotheses effectively. Published in:2022 By:Thomley, Jill E., PhD<br />Greenwald, Sarah J., PhD Go to EBSCOhost and sign in to access more content about this topic. Chi-squared Distribution Researchers in mathematics, science, social science, and a variety of other fields use the chi-squared distribution for statistical hypothesis tests, such as the chi-square test of independence, which determines whether or not two categorical variables are associated, and the chi-square goodness-of-fit test, which helps determine whether a chosen population model is a good fit for an observed set of data. It is also used in applications such as signal processing. The chi-squared distribution is a family of continuous probability distribution functions (pdfs), created by squaring and adding together one or more independent standard normal distributions—a particular variety of bell curve. The function’s shape (see Figure 1) is determined by a parameter called its degrees of freedom (df). For large df values (approximately fifty and above), the chi-squared distribution itself is almost bell-shaped. The number of degrees of freedom is determined by the total number of standard normal distributions that were added together. Standard statistical notation uses the Greek letter chi: (2(df). For example, (2(7) is a chi-squared distribution with seven degrees of freedom. The mean of any chi-squared distribution is equal to its number of degrees of freedom and its variance is two times its number of degrees of freedom. Overview The development of the chi-squared distribution resulted from mathematicians and scientists trying to solve the problem of whether a particular model, such as the normal distribution, was a good fit for their observed data. For example, were measurement errors made by astronomers really bell-shaped? This was important to assess the accuracy of measurements taken by different scientists, especially when data were combined to locate objects or derive astronomical constants. Mathematician Pierre-Simon Laplace and physicist Auguste Bravais were among those whose research on errors in the nineteenth century contributed to the chi-squared distribution. Geodesist Friedrich Helmert derived the distribution in 1876, as part of his research on the distribution of gravity on the surface of the earth. However, as was often the case with people working independently, Helmert’s contribution was not well-known until later. Instead, the distribution was named by English statistician Karl Pearson, who published an article on his chi-squared test of goodness of fit in 1900. He used data from experiments on flipping coins and spinning roulette wheels to determine whether they produced the expected distributions of heads and roulette numbers. Although his paper contained errors, some of which were later corrected by statistician Ronald Fisher, it was a significant achievement in statistics and motivated research on goodness-of-fit testing and chi-squared properties. Chi-square goodness-of-fit testing and other chi-square tests are used to validate models and test hypotheses. The F distribution, commonly used by researchers in many disciplines for hypothesis testing, is the ratio of two chi squares. Chi squared is special case of the gamma distribution, which is used in applications like modeling insurance claims, rainfall, and genomics. The generalized and noncentral chi-squared distributions are variations of the chi-squared distribution that are useful in applications such as medical imaging, multi-antenna wireless communication, and power analysis of statistical tests. Bibliography "Chi-Square Distribution." NIST/SEMATECH e-Handbook of Statistical Methods. National Institute of Standards and Technology, 2013. Web. 10 December 2014. Forbes, Catherine, Merran Evans, Nicholas Hastings, and Brian Peacock. Statistical Distributions. Hoboken, NJ: Wiley, 2011. Print. Freedman, David, Robert Pisani, and Roger Purves. Statistics. 4th ed. London: Norton, 2011. Print. Glantz, Stanton A. Primer of Biostatistics. 7th ed. New York,: McGraw, 2011. Print. Moore, David, and William I. Notz. Statistics: Concepts and Controversies. 8th ed. New York: Freeman, 2012. Print. Salsburg, David. The Lady Tasting Tea: How Statistics Revolutionized Science in the Twentieth Century. New York: Freeman, 2001. Print. On This Page OverviewFull Article Bibliography Related Topics Signal ProcessingDistribution (mathematics)Medical imaging and mathematicsWireless communication and mathematicsData analysis and probability in society Related Topics Signal ProcessingDistribution (mathematics)Medical imaging and mathematicsWireless communication and mathematicsData analysis and probability in society Company About EBSCO Leadership Offices Careers Contact Us Commitments Open for Research Open Access Accessibility Artificial Intelligence (AI) Values Carbon Neutrality Trust & Security Corporate Responsibility Our People & Community Log In and Support EBSCOhost EBSCONET EBSCOadmin EBSCOhost Collection Manager EBSCO Experience Manager Support Center © 2025 EBSCO Information Services, Inc.Privacy PolicyCookies SettingsCookie PolicyTerms We use cookies to give you a better experience on our sites. To learn more about how we use cookies, please read ourCookie Policy. 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14775	https://www.chowhound.com/1759832/replace-salmon-with-steelhead-trout/	Why Steelhead Trout Is The Perfect Fish Replacement For Salmon Recipes- [x] Course Dish Type Main Ingredients Drink Recipes Dietary Considerations Preparation Cuisine Cooking- [x] Baking Grilling & Smoking Drinks- [x] Coffee & Tea Cocktails & Spirits Beer & Wine Mixology & Techniques Kitchen- [x] Tips How-To Kitchen Tools Storage & Preservation Cleaning Design & Decor Shopping- [x] Shopping Tips Stores & Chains Facts- [x] Food Science Food History Restaurants- [x] Fast Food Pizzerias Coffee Shops Gardening Features About Editorial Policies Our Experts Privacy Policy Terms of Use © 2025 Static Media. All Rights Reserved Why Steelhead Trout Is The Perfect Fish Replacement For Salmon RecipesCookingDrinksKitchenShoppingFactsRestaurantsGardeningFeatures Cooking Why Steelhead Trout Is The Perfect Fish Replacement For Salmon ByJonathan KeshJan. 19, 2025 6:05 am EST Kirabahamut/Shutterstock The taste of salmon, even with thedifferences between wild and farmed salmon, is a fairly distinctive flavor. It's rich in an oddly subtle way, and it's meaty and oily and a little bit buttery. It's unique enough that salmon dishes aren't similar to other fish dishes, like the even more umami tuna or the mildly sweet tilapia. That said, if you're looking to adjust the flavor of a salmon fillet or if you're cooking on a tighter budget, plenty of chefs use steelhead trout as a salmon substitute, and it works well. Steelhead and salmon are vaguely close relatives, both being fish in the Salmonidae family that alternate between freshwater and the ocean during their lives. Beyond the fact that they look similar, both in the wild and on your plate, steelhead trout and salmon have a similarly delicate flakiness to them, and their differences in taste are minor enough that your average dinner guest likely won't notice the difference unless you point it out. Steelhead trout is also typically cheaper on top of that. It's not entirely the same, though; steelhead is softer and less fatty than salmon, and it's more mild and has a bit less of that fishy taste that salmon has in abundance. Steelhead is the cheaper salmon relative Fudio/Getty Images Steelhead's cheaper cost is a plus, but perhaps more importantly regarding the bigger picture, it's easier to find ethically sourced steelhead than salmon at the grocery store — which is certainly doable but requires some research. U.S. steelhead trout farming has a good rating from the Monterey Bay Aquarium's Seafood Watch, which is one of the better authorities on environmental impacts of seafood globally. Farm-raised trout are usually kept in highly regulated raceway systems which mimic freshwater rivers, which are considered low pollution risks that keep the more endangered wild species off diners' plates. Salmon can also only be safely farmed at certain parts of year to protect their breeding cycle, while fresh steelhead can be farmed all year. Any variety of either fish is fairly nutritious regarding omega-3 acids and other vitamins, although steelhead usually contains less fat. Steelhead is easy to substitute in restaurant-worthy grilled salmon fillets with potatoes and green beans and all the works on the side, but you'd also be surprised how well steelhead works in lox. It depends slightly on the type of salmon, too: steelhead may be noticeably less meaty than Chinook salmon, but could be a sneaky, cheap swap for coho salmon. Recommended
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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in University Physics Volume 2 10.4 Electrical Measuring Instruments University Physics Volume 210.4 Electrical Measuring Instruments Contents Contents Highlights Table of contents Preface Thermodynamics Electricity and Magnetism 5 Electric Charges and Fields 6 Gauss's Law 7 Electric Potential 8 Capacitance 9 Current and Resistance 10 Direct-Current Circuits Introduction 10.1 Electromotive Force 10.2 Resistors in Series and Parallel 10.3 Kirchhoff's Rules 10.4 Electrical Measuring Instruments 10.5 RC Circuits 10.6 Household Wiring and Electrical Safety Chapter Review 11 Magnetic Forces and Fields 12 Sources of Magnetic Fields 13 Electromagnetic Induction 14 Inductance 15 Alternating-Current Circuits 16 Electromagnetic Waves A \| Units B \| Conversion Factors C \| Fundamental Constants D \| Astronomical Data E \| Mathematical Formulas F \| Chemistry G \| The Greek Alphabet Answer Key Index Search for key terms or text. Close Learning Objectives By the end of this section, you will be able to: Describe how to connect a voltmeter in a circuit to measure voltage Describe how to connect an ammeter in a circuit to measure current Describe the use of an ohmmeter Ohm’s law and Kirchhoff’s method are useful to analyze and design electrical circuits, providing you with the voltages across, the current through, and the resistance of the components that compose the circuit. To measure these parameters require instruments, and these instruments are described in this section. DC Voltmeters and Ammeters Whereas voltmeters measure voltage, ammeters measure current. Some of the meters in automobile dashboards, digital cameras, cell phones, and tuner-amplifiers are actually voltmeters or ammeters (Figure 10.34). The internal construction of the simplest of these meters and how they are connected to the system they monitor give further insight into applications of series and parallel connections. Figure 10.34 The fuel and temperature gauges (far right and far left, respectively) in this 1996 Volkswagen are voltmeters that register the voltage output of “sender” units. These units are proportional to the amount of gasoline in the tank and to the engine temperature. (credit: Christian Giersing) Measuring Current with an Ammeter To measure the current through a device or component, the ammeter is placed in series with the device or component. A series connection is used because objects in series have the same current passing through them. (See Figure 10.35, where the ammeter is represented by the symbol A.) Figure 10.35(a) When an ammeter is used to measure the current through two resistors connected in series to a battery, a single ammeter is placed in series with the two resistors because the current is the same through the two resistors in series. (b) When two resistors are connected in parallel with a battery, three meters, or three separate ammeter readings, are necessary to measure the current from the battery and through each resistor. The ammeter is connected in series with the component in question. Ammeters need to have a very low resistance, a fraction of a milliohm. If the resistance is not negligible, placing the ammeter in the circuit would change the equivalent resistance of the circuit and modify the current that is being measured. Since the current in the circuit travels through the meter, ammeters normally contain a fuse to protect the meter from damage from currents which are too high. Measuring Voltage with a Voltmeter A voltmeter is connected in parallel with whatever device it is measuring. A parallel connection is used because objects in parallel experience the same potential difference. (See Figure 10.36, where the voltmeter is represented by the symbol V.) Figure 10.36 To measure potential differences in this series circuit, the voltmeter (V) is placed in parallel with the voltage source or either of the resistors. Note that terminal voltage is measured between the positive terminal and the negative terminal of the battery or voltage source. It is not possible to connect a voltmeter directly across the emf without including the internal resistance r of the battery. Since voltmeters are connected in parallel, the voltmeter must have a very large resistance. Digital voltmeters convert the analog voltage into a digital value to display on a digital readout (Figure 10.37). Inexpensive voltmeters have resistances on the order of R M=10 M Ω,R M=10 M Ω,R M=10 M Ω, whereas high-precision voltmeters have resistances on the order of R M=10 G Ω R M=10 G Ω R M=10 G Ω. The value of the resistance may vary, depending on which scale is used on the meter. Figure 10.37(a) An analog voltmeter uses a galvanometer to measure the voltage. (b) Digital meters use an analog-to-digital converter to measure the voltage. (credit: modification of works by Joseph J. Trout) Analog and Digital Meters You may encounter two types of meters in the physics lab: analog and digital. The term ‘analog’ refers to signals or information represented by a continuously variable physical quantity, such as voltage or current. An analog meter uses a galvanometer, which is essentially a coil of wire with a small resistance, in a magnetic field, with a pointer attached that points to a scale. Current flows through the coil, causing the coil to rotate. To use the galvanometer as an ammeter, a small resistance is placed in parallel with the coil. For a voltmeter, a large resistance is placed in series with the coil. A digital meter uses a component called an analog-to-digital (A to D) converter and expresses the current or voltage as a series of the digits 0 and 1, which are used to run a digital display. Most analog meters have been replaced by digital meters. Check Your Understanding 10.8 Digital meters are able to detect smaller currents than analog meters employing galvanometers. How does this explain their ability to measure voltage and current more accurately than analog meters? Interactive In the virtual lab simulation below, you may construct circuits with different voltage sources, resistors, and devices, then explore and compare currents and voltage with ammeters and voltmeters. Access multimedia content Ohmmeters An ohmmeter is an instrument used to measure the resistance of a component or device. The operation of the ohmmeter is based on Ohm’s law. Traditional ohmmeters contained an internal voltage source (such as a battery) that would be connected across the component to be tested, producing a current through the component. A galvanometer was then used to measure the current and the resistance was deduced using Ohm’s law. Modern digital meters use a constant current source to pass current through the component, and the voltage difference across the component is measured. In either case, the resistance is measured using Ohm’s law (R=V/I),(R=V/I),(R=V/I), where the voltage is known and the current is measured, or the current is known and the voltage is measured. The component of interest should be isolated from the circuit; otherwise, you will be measuring the equivalent resistance of the circuit. An ohmmeter should never be connected to a “live” circuit, one with a voltage source connected to it and current running through it. Doing so can damage the meter. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Samuel J. Ling, William Moebs, Jeff Sanny Publisher/website: OpenStax Book title: University Physics Volume 2 Publication date: Oct 6, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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14777	https://brilliant.org/wiki/telescoping-series-product/	Telescoping Series - Product Pi Han Goh, Abhinav Raichur, Anish Harsha, and Mahindra Jain Akeel Howell Scott Cambo Calvin Lin Jimin Khim contributed Contents Basic Form Advanced Form Determining How a Product Telescopes Common Pitfalls See Also Advanced Form In the previous section, we learned two similar unidirectional cancellations in a telescoping product: backward and forward cancellations. The next thing we should ask ourselves is, "What if there's a combination of these unidirectional cancellations? Is it still possible for the terms to cancel out in pairs?" Yes! We just need to identify and partition the expression into products of strictly backward and strictly forward cancellations. We call such expressions as bidirectional forms. Combination of 2 unidirectional forms We shall begin with a combination of 2 unidirectional forms by diving straight into an example: r=3∏100r2−4r2−1. The fraction in the product looks horrendous, doesn't it? Fear not! This fraction can be factorized. Note that both the expressions in the numerator (r2−1) and denominator (r2−4) are in the form of a difference of two squares identity: a2−b2=(a+b)(a−b). Thus, we have r=3∏100r2−4r2−1=r=3∏100(r−2)(r+2)(r−1)(r+1). Because ∏f and ∏g each have a finite non-zero value, we can use the property ∏(f⋅g)=∏f⋅∏g to get r=3∏100r2−4r2−1=r=3∏100r−2r−1×r=3∏100r+2r+1. What remains is to evaluate these two products and multiply them! For the first product, r=3∏100r−2r−1=12×23×34×45×⋯×9798×9899. Now this looks familiar, doesn't it? Yes it does! It's a forward cancellation, because r−2r−1 is in the form of S(r−k)S(r) for a positive integer k. We can now cancel out in pairs: r=3∏100r−2r−1=12×23×34×45×⋯×9798×9899=12×23×34×45×⋯×9798×9899=199=99. Analogously, we can see that the second product is a backward cancellation: r=3∏100r+2r+1=54×65×76×⋯×101100×102101=1024=512. Putting them all together yields r=3∏100r2−4r2−1=r=3∏100r−2r−1×r=3∏100r+2r+1=99×512=1766, and you're done! Food for thought: Can we still evaluate the same product above if we split the product ∏r=3100r2−4r2−1 as the multiplication of ∏r=3100r+2r−1 and ∏r=3100r−2r+1 instead? As you can see in the illustration above, the product represents a product of more than one unidirectional forms, so we call such products as bidirectional forms. Using the concepts that we've just learned, can you solve the following problems? (1−221)(1−321)(1−421)(1−521)⋯=? The correct answer is: 0.5 Given that f(x)=x2−1x2, compute 51×[f(50)×f(49)×⋯×f(3)×f(2)]. The correct answer is: 100 Combination of more than 2 unidirectional forms Now that you're capable of handling a combination of 2 unidirectional forms, is it possible to push further? That is, is it possible to find the combination of more than 2 unidirectional forms? The answer is still yes! Let's use a modified version of the very first product we used earlier in this section, ∏n=320n4(n2−1)(n2−4). Applying the difference of two squares identity, we have n2−1=(n−1)(n+1) and n2−4=(n−2)(n+2), so n4(n2−1)(n2−4)=n4(n−1)(n+1)(n−2)(n+2)=nn−1×nn+1×nn+2×nn+2. Now, we're back to the combination of backward and foward cancellations again: n=3∏20n4(n2−1)(n2−4)=n=3∏20n4(n−1)(n+1)(n−2)(n+2)=n=3∏20nn−1×n=3∏20nn+1×n=3∏20nn−2×n=3∏20nn+2. Then we just need to evaluate the 2 backward cancellations and 2 forward cancellations. Using the concepts we learned in the unidirectional forms, we have n=3∏20nn−1=202,n=3∏20nn+1=321,n=3∏20nn−2=19×201×2,n=3∏20nn+2=3×421×22. Calculating the product of these four values gives the desired answer of 3800539, and we're done again. As you can see in the above illustration, technically there's no difference in techniques whether it be 2, or more than 2 combinations of unidirectional forms. Now let's try out the following example that uses this very same idea: m=4∏42m2(m2−32)(m2−12)(m2−22) If the value of the product above is equal to BA, where A and B are coprime positive integers, find A−B. The correct answer is: 767 Determining How a Product Telescopes We have learned about the different forms which allow for a telescoping product. This is easy to determine if everything is presented clearly, as shown above. However, since that is rarely going to be the case, this section deals with determining if a given product allows for a telescoping form. To get started, test small cases. If the final result looks simple or seems to have a general form to it, then it provides some evidence that there is cancelling via a telescoping product approach. It remains to find suitable sequences of T(n)T(n+k) or S(n)S(n−k). We present 2 approaches of figuring out these terms. Approach 1: factoring of terms Factorize the general term as far as possible: suppose that ak=dk×ekbk×ck. List out terms of the form bk−2,bk−1,bk,bk+1,bk+2, and compare them with terms dk−2,dk−1,dk,dk+1,dk+2 and likewise for ck and ek. If two of these terms are equal, then these sequences will pair up and result in cancellation. Evaluate Mk=k=2∏nk3+1k3−1. Start by testing small cases: n=2:M2=97 n=3:M3=97×2826=1813 n=4:M4=1813×6563=107. Now, despite having denominators of 9, 28, 65, we see that the final term is extremely small, suggesting that we can cancel these terms. Since we have an algebraic expression, let's factorize the general term: ak=(k+1)(k2−k+1)(k−1)(k2+k+1). Let's label bk=k−1,ck=k2+k+1,dk=k+1,ek=k2−k+1. Then it should be clear how bk,dk cancel out, but let's follow the procedure listed above: nk−2k−1kk+1k+2bnk−3k−2k−1kk+1dnk−1kk+1k+2k+3 Now, it is apparent that bk=dk−2, which allows the canceling. Let's proceed similarly with ck,ek. Right now, it is not clear if, or how, they will cancel out, since they look so different. Once again, we list out the terms: nk−2k−1kk+1k+2cnk2−3k+1k2−k+1k2+k+1k2+3k+3k2+5k+7enk2−5k+3k2−3k+1k2−k+1k2+k+1k2+3k+3 Now, it is apparent that ck=ek+1, which allows the canceling. We now write ak=bk+2bk×ck−1ck, which allows us to telescope and obtain k=2∏nak=k=2∏nbk+2bk×k=2∏nck−1ck=bk+1bk+2b2b3×c1cn=3n(n+1)2(n2+n+1). □ Approach 2: non-obvious factoring The general term is a single term, because T(n+1)T(n) simplifies. It is most often explained as "multiply and divide by the same term." We have to make a guess at what T(n) is, guided by various identities that we are familiar with. Evaluate (1+x)(1+x2)(1+x4)(1+x8)(1+x16). Of course, one could expand out the 25 terms to see what we get. In the spirit of this section, let's consider instead how we could write this as a telescoping product. Using the identity that a−ba2−b2=a+b, we get [\begin{align} 1+ x & = \frac{ 1- x^2 } { 1-x } \\ 1+ x^2& = \frac{ 1- x^4 } { 1-x^2 } \\ 1+ x^4 & = \frac{ 1- x^8 } { 1-x^4 } \\ 1+ x^8 & = \frac{ 1- x^{16} } { 1-x^8 } \\ 1+ x^{16} & = \frac{ 1- x^{32} } { 1-x^{16} }. \end{align} ] We thus get a forward cancellation, and the product is equal to 1−x1−x32. □ Evaluate cosθ×cos2θ×cos4θ×cos8θ×cos16θ. In this case, it's harder to guess how to proceed. If we tried expressing cosnθ in terms of cosθ, we will end up with a mess, and it's not clear why we would have a nice answer. We use the identity sinθsin2θ=2cosθ. Then we get [ \begin{align} 2 \cos \theta & = \frac{ \sin 2 \theta } { \sin \theta } \\ 2 \cos 2 \theta & = \frac{ \sin 4 \theta } { \sin 2 \theta } \\ 2 \cos 4 \theta & = \frac{ \sin 8 \theta } { \sin 4 \theta } \\ 2 \cos 8 \theta & = \frac{ \sin 16 \theta } { \sin 8 \theta } \\ 2 \cos 16 \theta & = \frac{ \sin 32 \theta } { \sin 16 \theta }. \end{align} ] Thus, the product can be written as 32sinθsin32θ. □ Now, let's work on the following problems to practice identifying these forms: The value of the infinite product below can be expressed as ba for coprime positive integers a and b. Find a+b. k=2∏∞(k2−0.81)(k2−1.21) The correct answer is: 134 For x>1, define f(x)=n=2∏∞(1−nx1). Find the value of f2(3)f(6). The correct answer is: 1.5 Evaluate [ \frac {(8^4 + 64) ( 16^4 + 64)(24^4+64)(32^4+64)(40^4+64)(48^4 + 64)(56^4+64)(64^4+64) } { (4^4+64)(12^4+64)(20^4+64)(28^4+64)(36^4+64)(44^4+64)(52^4+64)(60^4+64) }. ] Details and Assumptions: Make sure you scroll right (if necessary) to see the full fraction. The dot on the extreme right is a full stop (punctuation mark). The correct answer is: 545 x4+64=(x2+8)2−16x2=(x2−4x+8)(x2+4x+8)=((x−2)2+4)((x+2)2+4). So the product is equal to (22+4)(62+4)(102+4)(142+4)…(584+4)(622+4)(62+4)(102+4)(142+4)(182+4)…(622+4)(662+4). Thus the product telescopes (terms cancel out) and we obtain 22+4662+4=545. Two infinite products A and B are defined as follows: A=n=2∏∞(1−n31),B=n=1∏∞(1+n(n+1)1). If BA=km, where m and k are relatively prime positive integers, determine 100m+k. The correct answer is: 103 n=0∏∞(22n+11−22n1+1) The above infinite product can be written as BA, where A and B are coprime positive integers. What is the value of A+B? The correct answer is: 11 Common Pitfalls In this section, we will be discussing common mistakes students make when solving telescoping products, as it is very easy for students to commit such errors, especially when dealing with infinities. (1) Not remembering what the last term is, especially for an infinite product: Often times, when we cancel off terms, we forget what terms actually remain. For example, the following calculation for the infinite product is wrong: r=1∏∞r+1r===21×32×43×54×⋯21×32×43×54×⋯1. Weird, isn't it? The product of fractions that are all smaller than 1 produces a number that is equal to 1? How is that possible? Can you spot the error? The mistake here is that we didn't consider its partial product and that's why we think that every term except the first number, 1, is being canceled out in pairs, whereas in actuality there should be more than one term remaining. By calculating the infinite product, we are in fact taking the limit of its partial product: r=1∏∞r+1r====R→∞limr=1∏Rr+1rR→∞lim(21×32×43×54×⋯×RR−1×R+1R)R→∞lim(21×32×43×54×⋯×RR−1×R+1R)R→∞lim(1×R+11)=1×0=0. Let us try a slight variation on the product above. Find the error in the following working: r=1∏∞r+3r===41×52×63×74×85×96×107×⋯41×52×63×74×85×96×107×⋯1. As before, it appears that we have found that the product of fractions that are smaller than 1 produces a number that is equal to 1. The mistake made here (that most students make) is that they cancel off the terms that they can see. That is, they knew that, for each new fraction they write up, all the numerators and denominators would be pairwise cancelled off. But what they didn't account for is that that they didn't treat it as a partial product; that is, it is not always true that the terms remaining are the first few terms that were left uncancelled. Here's the correct working (by partial product): r=1∏∞r+3r====R→∞limr=1∏Rr+3rR→∞lim(41×52×63×74×85×96×107×⋯×R+1R−2×R+2R−1×R+3R)R→∞lim(41×52×63×74×85×96×107×⋯×R+1R−2×R+2R−1×R+3R)R→∞lim(R+1)(R+2)(R+3)1×2×3=0. □ Now, should we find the partial product for the product in the following example as well? Why or why not? Yes No Does the infinite product ∏k=1∞kk+1 converge to a finite value? The correct answer is: No (2) Attempting to cancel far-away terms: Notice that for all the backward cancellation and forward cancellation that occur, the distance between each pair of terms is always fixed. A common mistake that students make when evaluating telescoping products is that they cancel off terms in pairs whose distances are arbitrary. For example, they may cancel off the denominators of the fraction in the nth term with the numerator of the fraction in the (2n)th term. So as n becomes larger and larger, the distance between the cancellation of terms becomes unboundedly large. To illustrate this point, let us consider the Wallis product n=1∏∞(2n−1)(2n+1)2n⋅2n=12⋅32⋅34⋅54⋅56⋅76⋅78⋅98⋅910⋅1110⋯=2π. Suppose we start cancelling off the denominators with values of 3 versus the numerators with values of 3×2=6, then we get 12⋅32⋅34⋅54⋅56 2⋅76 2⋅78⋅98⋅910⋅1110⋯. Likewise, suppose we also canceled off the denominators with values of 5 versus the numerators with values of 5×2=10, then we get 12⋅32⋅34⋅54⋅56 2⋅76 2⋅78⋅98⋅910 2⋅1110 2⋯. By repeating this process indefinitely, we will be cancelling off the denominators with values of n versus the numerators with values of 2n, and the resultant number is 12⋅32⋅34⋅54⋅56 2⋅76 2⋅78⋅98⋅910 2⋅1110 2⋯=∞, because when the fraction is simplified, the numerator consists of the product of infinitely many powers of 2, whereas the denominator is equal to 1 only. And we will make the absurd conclusion that 2π=1. The reason why we have committed a fallacy here is because we make the wrongful assumption that we think it's alright to cancel off terms that have no fixed distance. Another way to phrase it is this: we make the assumption that the logarithm of this telescoping product (which is a telescoping sum) is absolutely convergent when in fact it's not the case here. (3) Breaking up an infinite telescoping product into multiple infinite telescoping products: Another common mistake students make when attempting to simplify a telescoping product is that they make the assumption that the product can be expressed as a product of multiple products in attempts to calculate each of these simplified expressions in partition in hopes of making their work easier. To illustrate this point, consider the telescoping product ∏n=3∞n2n2−1, whose value is equal to 32 (we will leave it as a exercise for the reader). However, suppose we break the product as such: n=3∏∞n2n2−1===(n=3∏∞nn−1)×(n=3∏∞nn+1)R→∞lim(32×43×54×⋯)(34×45×56×⋯)0×∞, which is an indeterminate form, and is a wrong value because the product holds a finite value. The proper way to evaluate this product is by using the combination of more than 2 unidirectional forms in the section above, or alternatively by finding each of its partial products first before combining the two products, as shown below: n=3∏∞n2n2−1===R→∞lim(n=3∏Rnn−1)×(n=3∏Rnn+1)R→∞lim(32×43×54×⋯×RR−1)(34×45×56×⋯×RR+1)R→∞limR2×3R+1=32. To summarize, calculating telescoping products that deals with infinities can be tricky, and we should always be cautious of their partial products first. Now that you know how easy it is to make a mistake when dealing with infinities, let's try out a simple problem to ensure we won't make the same mistake twice! (4) Forgetting that the terms have shifted: Another common pitfall students make is to make the assumption that all terms can be rearranged. For example, in a previous example, we have shown that ∏r=1∞r+3r=0. Similarly, we will know that 61∏r=1∞r+3r=0×61=0, right? Then what is wrong with the following: 61r=1∏∞r+3r======1×2×31(41×52×63×74×85×⋯)1×2×31(41×52×63×74×85×⋯)1×2×31(41×52×63×74×85×⋯)1×2×31(41×52×63×74×85×⋯)1×2×31(41×52×63×74×85×⋯)1? The terms that are being cancelled out in the bracket is a backward cancellation and the cancellations are all of a fixed distance. Then why does the product (whose value is equal to 0) is equal to 1? The flaw behind this is that we have forgotten about writing it in terms of a its partial product again! That's all! In fact, the right working should consist of the final lines ⋯=R→∞lim1×2×31(41×52×63×74×85×⋯×R+1R−2×R+2R−1×R+3R)=0. See Also Telescoping Series - Sum Absolutely Convergent Cite as: Telescoping Series - Product. Brilliant.org. 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14778	https://cklixx.people.wm.edu/teaching/math430/Note-26.pdf	Chapter 26 Generators and Relations Chapter 26 Generators and Relations Motivation Construct the largest group satisfying some prescribed properties. For example: D4 is the only group generated by a, b, satisfying a4 = b2 = (ab)2 = e. One can show that any other group generated by two elements that satisfy the above relations is isomorphic to D4. The subgroup {R0, R180, H, V } of D4 is generated by a = R180 and b = H that satisfy a4 = b2 = (ab)2 = e and a2 = e. Chapter 26 Generators and Relations Deﬁnitions and Notation Let S = {a, b, c, . . . }. Create S−1 = {a−1, b−1, c−1, . . . }. Deﬁne the set W(S) of words of ﬁnite length x1 · · · xk with xi ∈S ∪S−1. Combine two words x1 · · · xk and y1 · · · yt by juxtaposition yielding x1 · · · xky1 · · · yt, and let e represents the empty word. Deﬁne an equivalence relation on W(S) by: two words are equivalent if one can be obtained from the other by adding or deleting words of the form xx−1 or x−1x, where x ∈S. Theorem 26.1 The set of equivalence classes of W(S) under the above relation form a group under the operation [u][v] = [uv]. The group is called a free group on S. Theorem 26.2 Every group is a homomorphic image of a free group. Proof. Let G be a group, and let S be a generating set. Then deﬁne φ : W(S)/∼→G by φ([x1 · · · xk]) = (x1 · · · xk)G ... Corollary Every group is isomorphic to a factor group of a free group. Chapter 26 Generators and Relations Presentations: Generators and relations Let G be a group generated by A = {a1, . . . , an}, and let F be the free group on A. Let W = {w1, . . . , wt} be a subset of F and N be the smallest normal subgroup of F containing W. Then G is given by the generators a1, . . . , an and the relations w1, . . . , wt = e if there is an isomorphism φ : F/N →G such that φ(aiN) = ai. In such a case, we write G = ⟨a1, . . . , an\|w1 = · · · = wt = e⟩. Chapter 26 Generators and Relations Example Z = ⟨a⟩. Example D4 = ⟨a, b\|a4 = b2 = (ab)2 = e⟩. Proof. Let F be the free group on {a, b}, and let N be the smallest subgroup containing {a4, b2, (ab)2}. Deﬁne φ : F →D4 such that φ(a) = R90, φ(b) = H. Then N ⊆ker(φ). Then F/ker(φ) is isomorphic to D4. Claim: F/N = K = {N, aN, a2N, a3N, bN, abN, a2bN, a3bN}. We need only show that (aN)K = K and (bN)K = K. The ﬁrst case is clear. One need to focus on the second case. For example, (bN)(aN) = baNb2 = babNb = a−1ababNb = a−1Nb = a−1a4Nb = a3Nb = s3bN. Other cases are similar. So, F/N has at most 8 elements. Now, F/ker(φ) is isomorphic to (F/N)/(kerφ/N). Thus, ker(φ)/N is trivial, i.e., ker(φ) = N, and hence F/N the same as F/ker(φ), which is isomorphic to D4. Chapter 26 Generators and Relations More results Theorem 16.3 Let G = ⟨a1, . . . , an\|w1 = · · · = wt = e⟩, and ˜ G = ⟨a1, . . . , an\|w1 = · · · = wt = wt+1 · · · = wt+k = e⟩. Then ˜ G = φ[G] for some group homomorphism φ. Proof. Exercise 5. Corollary If K is a group satisfying the deﬁning relations of a ﬁnite group G (with the same set of generators) and \|K\| ≥\|G\|, then K is isomorphic to G. Chapter 26 Generators and Relations More examples Example Quaternions. G = ⟨a, b\|a2 = b2 = (ab)2⟩. Let F be the free group on {a, b} and N is the smallest normal subgroup containing b−2a2, (ab)−2a2. Show that K = N, bN, b2N, b3N, aN, abN, ab2N, ab3N} is closed under multiplication. Then show that, say, by inspecting the group table, K ∼{±1, ±i, ±j, ±k} with i2 = j2 = k2 = −1, ij = k = −ji, jk = i = −kj, ki = j = −ik satisﬁes the relations and has 8 elements. Chapter 26 Generators and Relations Groups with eight elements Example G = ⟨a, b\|a3, b9, a−1ba−1b−1⟩implies that G = Z3. Note that b−1 = a−1ba imples that b = a−1b−1a. Then b = a−3ba3 = a−2b−1a2 = a−1ba1 = b−1. Thus, b2 = e. Because b9 = e. So, b = e, and G = Z3. Theorem 26.4 Up to isomorphism, there are ﬁve groups of order 8: Z8, Z4 ⊕Z,Z2 ⊕Z2 ⊕Z2, D4, the quaternions. Proof. Suppose G is non-Abelian. There is an element a of order 4. Else all elements has order 2 and is Abelian. Thus, G = H ∪Hb with H = ⟨a⟩. Now, b2 / ∈{b, ab, a2b, a3b}. Else, b ∈H. Also, b2 ̸= a because b2 commute with b, but a does not. Similarly, b2 ̸= a−1 = a3. So, b2 = e or a2. In the former case, we get D4; in the latter case, we get the quaternions. Chapter 26 Generators and Relations Characterization of Dihedral groups Theorem 26.5 Any group generated by a pair of order 2 elements is dihedral. Proof. Suppose G = ⟨a, b\|a2, b2⟩. If (ab) has inﬁnite order, then F = {e, a, b, ab, ba, aba, bab, abab, baba, . . . }. If G = F/H and H ̸= {e}. Then H contains (ab)i, (ab)ia, (ba)i, or (ba)ib. Then G cannot contain elements with word length larger than 2i + 2. Thus, G is ﬁnite and ab has ﬁnite order. If (ab) has order n, then G = {a, b, ab, ba, . . . , (ab)n = e = (ba)n}. Chapter 26 Generators and Relations
14779	https://math.stackexchange.com/questions/679622/intersection-between-conic-and-line-in-homogeneous-space	Intersection between conic and line in homogeneous space - 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 Intersection between conic and line in homogeneous space Ask Question Asked 11 years, 7 months ago Modified8 years, 1 month ago Viewed 3k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. In homogeneous space (so 3 coordinates for each point) I have: A conic C, defined by a symmetric 3x3 matrix of real values. The conic actually should have only imaginary points (don't know if this is important). A line l, defined by a vector of 3 real values How do I find the intersection of the two? I think I shoud be able to find the intersection (I expect two complex solutions), but I'm having troubles doing it. Solving it with the classic pen&paper leads me to a solution X=(x, y, 1) with x and y complex such that, when I try to verify that the point belongs to C (by cheching if XCX' = 0, where the ' stands for transposed), it seems that it does NOT belong to C. For those who know computer vision stuff: C is actually the image of the absolute conic, estimated from a picture, while l is a vanishing line of a plane. I'm trying to intersect the two in order to find the circular points, and then do a metric rectification of the plane in the image. conic-sections homogeneous-equation Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Feb 17, 2014 at 15:53 DennyDenny 11 1 1 silver badge 3 3 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. First find the pole (point) of the line L=(a b c)L=⎛⎝⎜a b c⎞⎠⎟ using the conic C=[A C D C B E D E F]. This is found using the inverse of the conic P=C−1 L=(u v w) Since we will use the inverse again, set C−1=[a c d c b e d e f] The pencil of lines through P is parametrically defined as T(ψ)=(−w sin ψ w cos ψ u sin ψ−v cos ψ) Now if you find a line T that is tangent to the conic, it will have T⊤C−1 T=0 and the tangent point Q=C−1 T lies on L. So Q is an intersection point. To solve T⊤C−1 T=0 for ψ involves solving an equation of the form K 0+K 1 sin(2 ψ)+K 2 cos(2 ψ)=0 with K 0=a w 2−2 d u w+f u 2 K 1=−2(c w 2−w(d v+e u)+f u v)K 2=w 2(b−a)+2 w(d u−e v)+f(v 2−u 2) There are two solutions to the above trig equation ψ={1 2(tan−1(K 1 K 2)−sin−1(2 K 0+K 2√K 2 1+K 2 2)−π 2)solution 1 1 2(tan−1(K 1 K 2)+sin−1(2 K 0+K 2√K 2 1+K 2 2)+π 2)solution 2 In the end the two intersection points are defined by Q=C−1 T(ψ) Q=((c w−d v)cos ψ+(d u−a w)sin ψ(b w−e v)cos ψ+(e u−c w)sin ψ(e w−f v)cos ψ+(f u−d w)sin ψ) Example Conic C=[1−7 6−3−7 6 4 1−3 1 13 6] and line L=(3−2−7). The polar point is P=C−1 L=(u v w)=(11208 5245 1134 5245−390 1049) The tangent lines are thus T(ψ)=(390 1049 sin ψ−390 1049 cos ψ 11208 5245 sin ψ−1134 5245 cos ψ) The tangency equation T⊤C−1 T=0 simplifis to the following: 1.26602894762909 cos 2 ψ+0.330351717237625 cos ψ sin ψ−1.24876309636214=0 with solution {ψ=0.299923260411840 solution 1 ψ=−0.0446792696983973 solution 2 The intersection points are Q=(−0.231100412938826−0.141550789771816−0.0585999513246922)Q=(0.136962490550640−0.0727785333232919 0.0794920768997864) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 25, 2017 at 14:18 John AlexiouJohn Alexiou 14.7k 1 1 gold badge 40 40 silver badges 78 78 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The matrix L=ˆ l=(0−l 3 l 2 l 3 0−l 1−l 2 l 1 0) can be used to describe a cross product with l: L g=l×g. Now consider D=L T⋅C⋅L. It is a degenerate conic which you best interpret dually as a pair of points, namely the points of intersection. A line g is tangent to that conic if its intersection with l, which can be computed as l×g=L g, lies on C, i.e. if (l×g)T C(l×g)=0. All you have to do is decompose that conic D into its components. You already have rank(D)=2. Now consider P=D+λ L. For some suitable λ, this matrix will have rank 1. Simply look at any 2×2 subdeterminant, and choose λ in such a way that it becomes zero. You will have two possible choices which only differ by sign. Either one will do. Once you have this, P=p q T is a matrix of rank 1. You can choose any non-zero column and call that p, and any non-zero row will be q (up to scalar multiples). So I'd look for the greatest absolute value in that matrix and choose its row and column as the two points of intersection. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 14, 2014 at 23:05 MvGMvG 44.2k 9 9 gold badges 94 94 silver badges 186 186 bronze badges 2 Great answer. I'm wondering about the why/how the decomposition of D works. I've asked that question here: scicomp.stackexchange.com/questions/26938/… Would love to get your input on it.CADJunkie –CADJunkie 2017-05-24 19:38:08 +00:00 Commented May 24, 2017 at 19:38 @CADJunkie For one thing, D is a scalar multiple of p q T+q T p.amd –amd 2017-07-17 20:09:18 +00:00 Commented Jul 17, 2017 at 20:09 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 conic-sections homogeneous-equation See similar questions with these tags. 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14780	https://www.quora.com/Whats-the-quickest-method-to-determine-if-a-number-is-rational-or-irrational	Something went wrong. Wait a moment and try again. Irrational Numbers Number Theory Rational Number Theory Irrational Number Theory 1 What's the quickest method to determine if a number is rational or irrational? Awnon Bhowmik Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views · 8y My limited knowledge tells me to first look at the number. If its simply a number with no decimals, I may be looking at natural or integer numbers. They are rational numbers . If I am given a small decimal number, then yes it is rational. Removing the decimals and writing it as a fraction gives us the ratio of two natural numbers, that is a rational number. If its a number inside a root, something like m √ x n then I can say its irrational provided that n ≠ m k where k ∈ Z Irrational numbers are non repeating, non terminating numbers. Example 1 9 = 0. ¯ 1 , \dfrac{1}{7}=0.\overline My limited knowledge tells me to first look at the number. If its simply a number with no decimals, I may be looking at natural or integer numbers. They are rational numbers . If I am given a small decimal number, then yes it is rational. Removing the decimals and writing it as a fraction gives us the ratio of two natural numbers, that is a rational number. If its a number inside a root, something like m √ x n then I can say its irrational provided that n ≠ m k where k ∈ Z Irrational numbers are non repeating, non terminating numbers. Example 1 9 = 0. ¯ 1 , 1 7 = 0. ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 142857 these are not irrational numbers. But 0.1 = 1 10 , 0.142857 = 142857 1000000 are rational numbers. √ 2 ≈ 1.414 , √ 3 ≈ 1.732 are irrational numbers. I can think of only this much for now. Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. Keith Ramsay Ph.D. in mathematics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 5.4K answers and 8.1M answer views · 5y Originally Answered: How do you determine whether a number is irrational? · In general it’s very hard. There are many numbers which we suspect are irrational but can’t prove are irrational. But there are some instances when we can do it. If the number is algebraic and you know a polynomial with an integer coefficients which it satisfies, then there is a standard procedure for determining whether the root is rational. Any polynomial with integer coefficients factors as a product of irreducible polynomials (ones that can’t be factored further). The root will be a root of one of the irreducible factors. A root of nx+m=0 where m and n are integers is rational of course, an In general it’s very hard. There are many numbers which we suspect are irrational but can’t prove are irrational. But there are some instances when we can do it. If the number is algebraic and you know a polynomial with an integer coefficients which it satisfies, then there is a standard procedure for determining whether the root is rational. Any polynomial with integer coefficients factors as a product of irreducible polynomials (ones that can’t be factored further). The root will be a root of one of the irreducible factors. A root of nx+m=0 where m and n are integers is rational of course, and if a root is rational, then a factor of that form can be found. The procedure for factoring polynomials requires more of an explanation than I’m ready to give right now. It has been observed that very often, when a number has been proved to be irrational, the way it worked is that some kind of approximation was found that is not exact, but for the number to be m/n where m and n are integers, the approximation would have to be exact. For example, e, the base of natural logarithms, can be expressed by an infinite series 1/0!+1/1!+1/2!+1/3!+… where n! is n factorial, the product of the integers 1,…,n. We define 0! to be 1. The sum of the first n+1 terms (ending with 1/n!) gives us an approximation with denominator n!, because all of the other denominators divide n!, so that they can be written with a common denominator of n!. If it were true that e=m/n, where m and n are integers, then multiplying e by n! would clear the denominator and give us an integer. Putting these two together, if e=m/n were true, then we would have n!e=n!(1/0!+1/1+…+1/n!)+n!(1/(n+1)!+1/(n+2)!+…) where the left-hand side is an integer, and so is the first part of the right-hand side. It would follow that n!(1/(n+1)!+1/(n+2)!+…) is also be an integer. We can estimate how big it is. It’s positive because each term in it is positive. The ratio n!/(n+1)!=1/(n+1), and each successive term is less than 1/(n+1) of the previous one. That makes the sum less than the sum 1/(n+1)+1/(n+1)2+1/(n+1)3+… which is 1/n. Unless n=1 this is <1. (We know that e can’t be written as m/1 already.) Any number between 0and 1 can’t be an integer, so it’s not possible for e=m/n to be true. It seems to be a kind of running theme in the field of Diophantine approximation (which is what mathematicians call the area where they study irrational numbers), that the proofs all use the fact that no integer is between 0 and 1. The cases which we don’t know how to do yet probably require some kind of unknown approximation to the value of the number. But in general it’s really hard. There is a constant γ known as Euler’s gamma, which is defined as the limit as n goes to infinity of 1+1/2+1/3+1/4+…−loge(n). The logarithm of n is the area under the graph of y=1/x for 1≤x≤n, while the sum is the area of rectangles inserted under the curve. The area between can be shown to have a limit, because all of the pieces can be slid over to the unit square, 0≤x,y≤1. We expect it to be an irrational number, but there is no proof yet. A proof probably would involve finding some way to approximate γ. There was a famous proof by Apéry that the sum of the series 1/1+1/8+1/27+…, where the denominators are the cubes, is irrational. I don’t know the details but it involved some carefully crafted approximations. We expect e+π to be irrational, but haven’t proven it, and I don’t think anybody expects to prove it any time soon. In a sense the problem is that although we have ways to prove that they are separately irrational, they don’t mesh together to give a proof that their sum is too. One could try for example to use the proof that e is irrational by assuming e+π=m/n and then considering n!(e+π). The proof above that e is irrational shows that the n!e term is close to an integer but just a bit above it. Unfortunately there is no known way to show that n!π is not close to an integer but slightly less than it. In general the field is pretty hard. Paul Caplin BSc and MA in mathematics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Raphael Laufer , PhD Mathematics, University of California, Berkeley (1997) · Author has 2.2K answers and 3.2M answer views · Updated 4y Originally Answered: How can a number be proven irrational? · Proving that a number is irrational means proving that it can't be equal to the ratio of two integers. This is easy to do for some irrational numbers, and much harder for others. For many numbers that are probably irrational it's so hard that no one has managed it yet. For example, no one has yet managed to prove that π+e is irrational – though almost every mathematician would bet almost any amount of money that it is. The usual approach is “proof by contradiction”, one of the most powerful and useful proof techniques in mathematics. You start by assuming that a number is rational, and then sho Proving that a number is irrational means proving that it can't be equal to the ratio of two integers. This is easy to do for some irrational numbers, and much harder for others. For many numbers that are probably irrational it's so hard that no one has managed it yet. For example, no one has yet managed to prove that π+e is irrational – though almost every mathematician would bet almost any amount of money that it is. The usual approach is “proof by contradiction”, one of the most powerful and useful proof techniques in mathematics. You start by assuming that a number is rational, and then show that this leads to a logical contradiction. This demonstrates that your initial assumption must be false, so the number must be irrational. The proof that √2 is irrational is very simple, and is a famous example of the proof-by-contradiction method. Here's how it goes. Start by assuming that √2 is rational. We're going to show that this leads to a contradiction, and so must be wrong. If it's rational, it must be possible to write it as a fraction in its lowest terms (just by cancelling out all the common factors in the numerator and denominator). Let's call this fraction p/q, where p and q are integers with no common factors. I'm going to use the standard notation a\|b to mean “a divides b” – in other words, b is an exact multiple of a. The logic goes like this: √2=p/q ⇒p2=2q2 ⇒2\|p (because the only way that p2 can be even is if p is even) ⇒4\|p2 ⇒2\|q2 (because p2=2q2) ⇒2\|q …so p and q are both divisible by 2, which contradicts our initial choice of p and q as numbers that aren't both divisible by anything. We have our contradiction, so our initial assumption must be wrong. Therefore √2 is irrational. Related questions What is the method to find whether a number is rational or irrational? What method is used to see if a number is rational or irrational? How do you determine whether a number is irrational? Is √ 3 a rational number or irrational number? Is there a method to determine if a number is rational or irrational without assuming it to be one or the other? If so, what is it? David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by Justin Rising , PhD in statistics and Daniel McLaury , Ph.D. Student in Mathematics at University of Illinois at Chicago · Author has 9.9K answers and 68.1M answer views · Updated 7y There is no known algorithm to determine whether a real number is rational or irrational. Recall that it took over 2000 years to determine whether π was rational or irrational. It is not known if the number ζ(5)=1+125+135+145+⋯ is rational or irrational. There are many other interesting numbers that are not known to be rational or irrational. It may be that someone has shown that such an algorithm cannot exist. One would first formally define a class of real numbers such as those given by convergent series of rational numbers (or perhaps those s There is no known algorithm to determine whether a real number is rational or irrational. Recall that it took over 2000 years to determine whether π was rational or irrational. It is not known if the number ζ(5)=1+125+135+145+⋯ is rational or irrational. There are many other interesting numbers that are not known to be rational or irrational. It may be that someone has shown that such an algorithm cannot exist. One would first formally define a class of real numbers such as those given by convergent series of rational numbers (or perhaps those series for which there exist proofs that they’re convergent). Then some argument—perhaps a diagonal argument like the one in Gödel’s incompleteness theorem or the halting problem—to show the existence of such an algorithm leads to a contradiction. Related questions What is the method to find whether a number is rational or irrational? What method is used to see if a number is rational or irrational? How do you determine whether a number is irrational? Is √ 3 a rational number or irrational number? Is there a method to determine if a number is rational or irrational without assuming it to be one or the other? If so, what is it? How do I find a rational number between two irrational numbers? Is there a way to determine if a given number is irrational or rational without doing long calculations in math? What is the method for determining if a number is rational or irrational without calculating its decimal value? Is 22/7 a rational number or irrational number? Is infinity a rational or irrational number? Is zero a rational or irrational number? What 2 properties determine if a number is rational or irrational? Is √10 rational or irrational? How can you determine if a number is rational or irrational without any operations being performed on it? What is the difference between rational and irrational numbers? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
14781	https://www.quora.com/How-do-you-prove-that-ln-e-x-x	How to prove that ln [e^x] = x - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Derivatives of Exponentia... Algebraic Proofs Logarithm Calculations Basic Algebra Exponential Equations Algebra Logarithmic and Exponenti... Proofs (mathematics) 5 How do you prove that ln [e^x] = x? All related (47) Sort Recommended Marc Murphy Studied at St. Cloud State University ·1y Originally Answered: How do you prove the identity \mathrm{e}^{\ln(x)} = x? Why does this work? · From Desmos for x, e x e x, and ln(x) John Napier is credited for logarithms. Using natural logarithms on both sides gets to l n(x)=l n(x)l n e l n(x)=l n(x)l n e Continue Reading From Desmos for x, e x e x, and ln(x) John Napier is credited for logarithms. Using natural logarithms on both sides gets to l n(x)=l n(x)l n e l n(x)=l n(x)l n e Upvote · 9 3 9 1 Sponsored by Reliex Looking to Improve Capacity Planning and Time Tracking in Jira? ActivityTimeline: your ultimate tool for resource planning and time tracking in Jira. Free 30-day trial! Learn More 99 24 Related questions More answers below Why does e^ln(x) = x? What is x if ln(x) = e^x? How do you prove that ln(e^x) = xln(e)? How do you prove that S(x) = e^ (ln x)? How do you prove that e^ [ln(x)] = x? Doug Dillon Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views ·6y Originally Answered: How do you prove e^{\ln x}=x? · Upvote · 9 5 Fabio García MSc in Mathematics, CIMAT (Graduated 2018) · Author has 247 answers and 623.3K answer views ·6y Originally Answered: How do you prove e^{\ln x}=x? · As is usual with questions asking “how do you prove…”, i reply with another question: prove from what? You’re probably aware that in mathematics we don’t prove stuff from thin air. We always have axioms, definitions, and prevously proved theorems to work from. This is our from what. The from what matters because there are various (though ultimately equivalent) vays to define each of e x e x and ln x ln⁡x. You might start off by defining e x e x as some function with some properties, and then define ln x ln⁡x as the inverse of e x e x, which makes the statement in your question true by definition. Or you could start Continue Reading As is usual with questions asking “how do you prove…”, i reply with another question: prove from what? You’re probably aware that in mathematics we don’t prove stuff from thin air. We always have axioms, definitions, and prevously proved theorems to work from. This is our from what. The from what matters because there are various (though ultimately equivalent) vays to define each of e x e x and ln x ln⁡x. You might start off by defining e x e x as some function with some properties, and then define ln x ln⁡x as the inverse of e x e x, which makes the statement in your question true by definition. Or you could start by defining ln x ln⁡x as some function with some properties, and then define e x e x as the inverse of ln x ln⁡x, which again makes the statement in your question true by definition. You could also take the roundabout route and start with the fundamental properties of e x e x and ln x ln⁡x. Let f f be some function such that f′=f,f(0)=1.f′=f,f(0)=1. (These two properties uniquely define e x e x). If can be shown that f f is injective and therefore has an inverse g g. Recall the trick for differentiating the inverse of a function: g′(x)=1 f′(g(x))=1 f(g(x))=1 x.g′(x)=1 f′(g(x))=1 f(g(x))=1 x. Now, g(1)=g(f(0))=0 g(1)=g(f(0))=0. But these two properties, g′(x)=1/x,g(1)=0,g′(x)=1/x,g(1)=0, uniquely define ln x ln⁡x. Hence g(x)=ln x g(x)=ln⁡x. And we’re done. Upvote · 99 15 9 1 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·6y Originally Answered: How do you prove e^{\ln x}=x? · This is always a troublesome identity. To show why it is true, you n... Upvote · 9 9 9 1 9 1 Related questions More answers below What is e−ln x e−ln⁡x? How do you prove e ln x=x e ln⁡x=x? Can you prove that ln(e x)=x ln⁡(e x)=x? How do you solve -ln (5^(e^x)) = -x? Why does e^x = exp(x)? How can we prove it algebraically or by solving some problems? Assistant Bot · 1y To prove that ln[e x]=x ln⁡[e x]=x, we can use the properties of logarithms and exponentials. Here’s a step-by-step explanation: Understand the Definitions: The natural logarithm function, ln(y)ln⁡(y), is defined as the inverse of the exponential function e x e x. This means that if y=e x y=e x, then ln(y)=x ln⁡(y)=x. Apply the Definition: Let y=e x y=e x. By the definition of the logarithm, we know that: ln(y)=x ln⁡(y)=x 3. Substitute Back: Since y=e x y=e x, we can substitute back into our equation: ln[e x]=x ln⁡[e x]=x 4. Conclusion: Thus, we have shown that: ln[e x]=x ln⁡[e x]=x This proof relies on the fundamental relati Continue Reading To prove that ln[e x]=x ln⁡[e x]=x, we can use the properties of logarithms and exponentials. Here’s a step-by-step explanation: Understand the Definitions: The natural logarithm function, ln(y)ln⁡(y), is defined as the inverse of the exponential function e x e x. This means that if y=e x y=e x, then ln(y)=x ln⁡(y)=x. Apply the Definition: Let y=e x y=e x. By the definition of the logarithm, we know that: ln(y)=x ln⁡(y)=x 3. Substitute Back: Since y=e x y=e x, we can substitute back into our equation: ln[e x]=x ln⁡[e x]=x 4. Conclusion: Thus, we have shown that: ln[e x]=x ln⁡[e x]=x This proof relies on the fundamental relationship between logarithms and exponents, confirming that the natural logarithm of e e raised to any power returns that power. Upvote · 9 1 Dipak Kambale 1y Originally Answered: How do you prove the identity \mathrm{e}^{\ln(x)} = x? Why does this work? · Let us say y=e l n(x)y=e l n(x) Taking log w.r.t. base e on both sides, we get l n(y)=l n(x)l n(y)=l n(x) ∴y=x∴y=x Hence, e l n(x)=x e l n(x)=x Hence proved. Upvote · 9 1 Sponsored by Sync.com Protect your files, photos and data with Sync.com. Treat your data carefully. Use Sync.com to protect it. Try 5 GB for free. Learn More 2.7K 2.7K Thomas Bell Ph.D. in Mathematics, University of Oregon (Graduated 2013) · Author has 3.9K answers and 687.3K answer views ·2y Uh, that’s the definition of the natural log. ln(b) = a means e^a = b. So ln(e^x) = y means e^x = e^y. Since f(x) = e^x is increasing, it is one-to-one, so e^x = e^y implies x = y. Upvote · Robert Sheraw Author has 1.3K answers and 215.1K answer views ·2y f(x)=e^x and g(x)=ln(x) are inverses of each other. Therefore their composition is the identity function so g(f(x)) = x. in this case g(f(x)) = ln [e^x]. Hence ln [e^x] = x. Upvote · Sponsored by SpeechText.AI AI-powered legal transcription service. Convert speech to text from any audio/video format, including TRM files. Fully compliant with GDPR. Start Now 9 3 Andy Baker Works at University of Glasgow · Author has 7.3K answers and 1.7M answer views ·2y Originally Answered: How do we prove that ln(e^x) =x for all x>0 using calculus? · This makes no sense unless you indicate what you are taking as the starting point. One definition of ln ln is as the inverse of the exponential function so I am assuming you don’t want to trivialise things by doing that. So maybe you mean to define ln ln by for example taking it to be the integral ln(x)=∫x 1 x−1 d x ln⁡(x)=∫1 x x−1 d x. Then the substitution x=e t x=e t gives ln(e x)=∫x 0 e−t e t d t=x ln⁡(e x)=∫0 x e−t e t d t=x. But that assumes you have defined e x e x with certain properties. I think you need to clarify what you really want. Upvote · Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·5y Related How do you prove that e log x=x e log⁡x=x ? This is always a troublesome identity. To show why it is true, you need to understand a few log ideas. Continue Reading This is always a troublesome identity. To show why it is true, you need to understand a few log ideas. Upvote · 9 6 Sponsored by Wordeep Proofreading Grammar correction solution using artificial intelligence. Have a look at our real-time AI based proofreading and grammar correction service. You will be amazed! Learn More 99 95 George Ivey Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views ·1y Originally Answered: How do you prove the identity \mathrm{e}^{\ln(x)} = x? Why does this work? · You prove it using the DEFINITIONS of “e^x” and “ln(x)”. Do you know what they are? Some texts first define ln(x) to be ∫x 1 d t t∫1 x d t t and then define e x e x to be its inverse. Others first define e x e x and then define ln(x) to be its inverse. In either case both e l n(x)=x e l n(x)=x and l n(e x)=x l n(e x)=x follow immediately from the definition of “inverse function”. Do you know what that is? Upvote · George Ivey Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views ·2y Originally Answered: Can you prove that \ln (e^x) = x? · How are you DEFINING “e[x e[x” and “[math]ln(x)[math]”? Typically one either defines “f(x)=e x f(x)=e x”, by showing that a x a x, for any positive a, has derivative C a a x C a a x, where C a C a is a constant, depending on a but not the variable, x. The “e” is defined as the value of a so that C e=1 C e=1 so the the derivative of e x e x is simply e x e x again. Now one can define “ln(x)” to be the function of “e x e x. OR one defines l n(x)=∫x 1 d x x[m a t h],f o r x p o s i t i v e,t h e n d e f i n e“[m a t h]e x l n(x)=∫1 x d x x[m a t h],f o r x p o s i t i v e,t h e n d e f i n e“[m a t h]e x” to be the inverse function to “ln(x)”. In either case, since they are inverse functions, l n(e x)=x l n(e x)=x and e l n(x)=x e l n(x)=x. Upvote · Anonymous 6y Originally Answered: How do you prove e^{\ln x}=x? · Let f:R→R>0 f:R→R>0 be defined as f(x)=e x f(x)=e x. Then by definition of an inverse function, and the definition of the logarithm, f−1:R>0→R f−1:R>0→R is f−1(x)=ln(x)f−1(x)=ln⁡(x). Hence, since for any invertible function g:A→B g:A→B, g(g−1(x))=I A(x)g(g−1(x))=I A(x), then e ln(x)=f(f−1(x))=x e ln⁡(x)=f(f−1(x))=x. Upvote · 9 2 Related questions Why does e^ln(x) = x? What is x if ln(x) = e^x? How do you prove that ln(e^x) = xln(e)? How do you prove that S(x) = e^ (ln x)? How do you prove that e^ [ln(x)] = x? What is e−ln x e−ln⁡x? How do you prove e ln x=x e ln⁡x=x? Can you prove that ln(e x)=x ln⁡(e x)=x? How do you solve -ln (5^(e^x)) = -x? Why does e^x = exp(x)? How can we prove it algebraically or by solving some problems? How can I solve ln(x) =-e^(x)? What does it mean if ln(x) = e^x? Can you explain why e^x = ex + x, but there's no such thing as ln(e^x)? What is the mathematical explanation for ln (e^x) = x? What's an algebraic proof for e ln(x)=x e ln⁡(x)=x? Related questions Why does e^ln(x) = x? What is x if ln(x) = e^x? How do you prove that ln(e^x) = xln(e)? How do you prove that S(x) = e^ (ln x)? How do you prove that e^ [ln(x)] = x? 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14782	https://books.google.com/books/about/Riemannian_Geometry.html?id=Wg-gQcvS25sC	Riemannian Geometry: A Modern Introduction - Isaac Chavel - Google Books Sign in Hidden fields Try the new Google Books Books View sample 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 Cambridge University Press 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 Riemannian Geometry: A Modern Introduction ========================================== Isaac Chavel Cambridge University Press, Jan 27, 1995 - Mathematics - 386 pages This book provides an introduction to Riemannian geometry, the geometry of curved spaces. Its main theme is the effect of the curvature of these spaces on the usual notions of geometry, angles, lengths, areas, and volumes, and those new notions and ideas motivated by curvature itself. Isoperimetric inequalities--the interplay of curvature with volume of sets and the areas of their boundaries--is reviewed along with other specialized classical topics. A number of completely new themes are created by curvature: they include local versus global geometric properties, that is, the interaction of microscopic behavior of the geometry with the macroscopic structure of the space. Also featured is an ambitious "Notes and Exercises" section for each chapter that will develop and enrich the reader's appetite and appreciation for the subject. More » Preview this book » Selected pages Title Page Table of Contents Index References Contents II 1 III 3 IV 7 V 8 VI 12 VII 14 VIII 17 IX 24 XXXI 178 XXXII 187 XXXIII 188 XXXIV 199 XXXV 203 XXXVI 213 XXXVII 222 XXXVIII 230 More X 27 XI 30 XII 49 XIII 22 XIV 37 XV 38 XVI 40 XVII 46 XVIII 85 XIX 94 XX 96 XXI 104 XXII 112 XXIII 123 XXIV 140 XXV 141 XXVI 146 XXVII 149 XXVIII 157 XXIX 167 XXX 170 XXXIX 232 XL 244 XLI 248 XLII 251 XLIII 256 XLIV 263 XLV 266 XLVI 273 XLVII 283 XLVIII 284 XLIX 285 L 288 LI 294 LII 297 LIII 300 LIV 304 LV 306 LVI 317 LVII 323 LVIII 339 Copyright Less Other editions - View all Riemannian Geometry: A Modern Introduction Isaac Chavel No preview available - 1993 Common terms and phrases 1-formsargumentassumeboundedcalculatechartcompact Riemanniancomparison theoremscomplete Riemannian manifoldconjugateconsiderconstant sectional curvatureconvexcoordinatesCorollarycovariant differentiationcut locusdefineDefinitiondenotediffeomorphismdiskdomaineigenvalueequalExerciseexistsexponential mapfinitefollowsfunction fgeodesic flowgeometrygivengrad fhomotopy classimplies the claimisometryisoperimetric inequalityJacobi fieldLemmalengthLevi-Civita connectionmetric spaceminimizing geodesicmoving framesn-dimensional Riemannian manifoldneighborhoodNoteorthogonalorthonormalparallel translationpathProofProveradiusrespectRicci curvatureRiemannian manifoldRiemannian measureRiemannian metricsatisfyingsectional curvatureShowsmooth boundaryspheresubmanifoldsubsetsymmetrict₁tangent spacetheorem Theoremtopologyvanishesvector fieldvolumeW₁δο References to this book Riemannian Geometry Peter Petersen Limited preview - 2006 The Concentration of Measure Phenomenon Michel Ledoux No preview available - 2001 All Book Search results » Bibliographic information Title Riemannian Geometry: A Modern Introduction Issue 108 of B.Bollohas P.Sarnak and C.T.C.Wall Volume 108 of Cambridge Tracts in Mathematics, ISSN 0950-6284 Volume 108 of Cambridge tracts in mathematics and mathematical physics AuthorIsaac Chavel Edition reprint Publisher Cambridge University Press, 1995 ISBN 0521485789, 9780521485784 Length 386 pages SubjectsMathematics › Geometry › General Mathematics / Geometry / Analytic Mathematics / Geometry / Differential Mathematics / Geometry / General Mathematics / Geometry / Non-Euclidean Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
14783	https://www.cnblogs.com/conpi/p/18951282	正态分布（Normal Distribution） - 华小电 - 博客园 Typesetting math: 100% 会员众包新闻博问闪存赞助商 HarmonyOS Chat2DB 所有博客当前博客我的博客我的园子账号设置会员中心简洁模式 ...退出登录注册登录华小电新随笔管理随笔 - 220 文章 - 0 评论 - 2 阅读 - 19万正态分布（Normal Distribution） 1. 正态分布的定义正态分布（Normal Distribution），又称为高斯分布（Gaussian Distribution），是一种在统计学和概率论中最重要的连续概率分布。它广泛应用于自然科学、社会科学、工程、金融等领域。正态分布的概率密度函数（PDF）如下： f(x\|μ,σ 2)=1√2 π σ 2 e−(x−μ)2 2 σ 2 f(x\|μ,σ 2)=1 2 π σ 2 e−(x−μ)2 2 σ 2 其中： x x：随机变量，表示数据点 μ μ：均值（mean），即数据的中心 σ 2 σ 2：方差（variance），表示数据的离散程度 σ σ（标准差，standard deviation）：σ=√σ 2 σ=σ 2 2. 正态分布的参数解释在正态分布中，有两个重要的参数：均值μ μ 和方差σ 2 σ 2。（1）均值 μ μ 决定正态分布的中心位置。直观来说，它表示数据的平均值，即数据的集中趋势。若 μ μ 变大，整个分布会向右平移；若 μ μ 变小，分布会向左平移。（2）方差 σ 2 σ 2 与标准差 σ σ 决定正态分布的宽度（离散程度）。方差越大（即标准差越大），数据的波动性越大，分布曲线越“扁平”；方差越小，数据越集中，分布曲线越“陡峭”。标准差的影响示意：当 σ σ 较小时，数据点更集中于均值附近，分布更窄。当 σ σ 较大时，数据点更分散，分布更宽。 3. 正态分布的性质正态分布有以下重要的数学性质：（1）对称性正态分布是关于均值 μ μ 对称的，即： P(X≤μ−c)=P(X≥μ+c)P(X≤μ−c)=P(X≥μ+c) 这意味着数据左右分布是均匀的。（2）68-95-99.7 经验法则对于任意正态分布：约 68% 的数据落在 μ±σ μ±σ 区间内。约 95% 的数据落在 μ±2 σ μ±2 σ 区间内。约 99.7% 的数据落在 μ±3 σ μ±3 σ 区间内。这说明大部分数据点会集中在均值附近，离均值越远的点出现的概率越小。（3）标准正态分布当正态分布的均值 μ=0，标准差 σ=1 时，我们称其为标准正态分布（Standard Normal Distribution），记作： Z∼N(0,1)Z∼N(0,1) 标准正态分布的概率密度函数为： ϕ(z)=1√2 π e−z 2 2 ϕ(z)=1 2 π e−z 2 2 其中，z=x−μ σ z=x−μ σ 为标准化变量。标准正态分布的分布曲线是对称的“钟形曲线”，其均值为 0，标准差为 1，广泛用于统计推断，如计算 z-score（标准分数）。 4. 正态分布的计算在实际应用中，我们经常需要计算某个数值 x 在正态分布中的概率。通常有以下两种方法：（1）直接计算概率密度使用公式： f(x\|μ,σ 2)=1√2 π σ 2 e−(x−μ)2 2 σ 2 f(x\|μ,σ 2)=1 2 π σ 2 e−(x−μ)2 2 σ 2 （2）标准化计算由于直接计算积分较难，我们可以使用标准正态分布表：先计算标准化变量（Z-score）： z=x−μ σ z=x−μ σ 然后查询标准正态分布表，获取累积分布函数（CDF）值，即： P(X≤x)=P(Z≤z)P(X≤x)=P(Z≤z) 对于非标准正态分布，可以通过变换 Z 来计算概率。 5. 应用案例假设我们测量一批产品的重量，重量的分布服从正态分布，均值为50克，标准差为5克。我们希望可视化这些产品的重量分布，并计算重量在45到55克范围内的概率。 ```python import numpy as np import matplotlib.pyplot as plt plt.rcParams['font.size'] = 14 plt.rcParams['font.sans-serif'] = ['Microsoft Yahei'] from scipy.stats import norm 参数设置 mu = 50 # 均值 sigma = 5 # 标准差生成正态分布数据 X = np.linspace(mu - 4 sigma, mu + 4 sigma, 1000) y = norm.pdf(X, mu, sigma) 可视化正态分布 plt.figure(figsize=(8, 6)) plt.plot(X, y) plt.title('正态分布曲线') plt.xlabel('重量(克)') plt.ylabel('概率密度') plt.show() ``` 分类: Python 好文要顶关注我收藏该文微信分享华小电粉丝 - 4关注 - 0 +加关注 0 0 升级成为会员 « 上一篇：主成分分析PCA » 下一篇： KMeans 算法 posted @ 2025-06-27 05:31华小电阅读(779) 评论(0)收藏举报刷新页面返回顶部登录后才能查看或发表评论，立即登录或者逛逛博客园首页【推荐】100%开源！大型工业跨平台软件C++源码提供，建模，组态！【推荐】HarmonyOS 专区 —— 闯入鸿蒙：浪漫、理想与「草台班子」【推荐】2025 HarmonyOS 鸿蒙创新赛正式启动，百万大奖等你挑战【推荐】天翼云爆款云主机2核2G限时秒杀，28.8元/年起！立即抢购相关博文： ·正态分布概率密度函数（PDF） ·sklearn之StandardScaler ·正态分布——“牛而B之” ·关于正态分布 ·统计基础篇之十二：怎么理解正态分布(一) HarmonyOS专区： · 闯入鸿蒙：浪漫、理想与「草台班子」 · 3天赚2万！开发者的梦想也可以掷地有声！ · Flutter 适配 HarmonyOS 5 开发知识地图 · HEIF：更高质量、更小体积，开启 HarmonyOS 图像新体验 · CodeGenie 的 AI 辅助调优让你问题定位效率大幅提升搜索随笔分类 Linux(38) Modbus(5) Python(147) SQL(15) 树莓派(2) 随笔档案 2025年6月(7) 2025年4月(4) 2025年3月(3) 2025年2月(3) 2025年1月(3) 2024年12月(2) 2024年11月(1) 2024年10月(1) 2024年9月(2) 2024年8月(3) 2024年7月(9) 2024年4月(3) 2024年3月(3) 2024年2月(9) 2023年12月(3) 2023年10月(2) 2023年9月(3) 2023年8月(2) 2023年7月(2) 2023年6月(1) 2023年4月(2) 2023年3月(2) 2023年2月(12) 2023年1月(8) 2022年12月(5) 2022年11月(4) 2022年10月(12) 2022年9月(5) 2022年8月(3) 2022年7月(5) 2022年6月(1) 2022年5月(5) 2022年4月(4) 2022年3月(6) 2022年1月(1) 2021年12月(3) 2021年11月(4) 2021年6月(2) 2021年5月(1) 2021年4月(2) 2021年3月(5) 2021年1月(8) 2020年11月(7) 2020年10月(1) 2020年9月(1) 2020年7月(3) 2020年6月(1) 2020年5月(5) 2020年3月(2) 2020年2月(1) 2020年1月(7) 2019年12月(3) 2019年11月(3) 2019年10月(7) 2019年6月(5) 2019年5月(4) 2019年4月(4) 更多阅读排行榜 1. Python 读取和输出到txt(14444) 2. 在PyQt5中显示matplotlib绘制的图形(8983) 3. Windows 10 LTSC 2019 密钥激活(7467) 4. 利用Python将多个PDF文件合并(6345) 5. Python报错：SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape(6044) 推荐排行榜 1. sublime_text 4143 for Linux 注册(2) 2. UserWarning: FixedFormatter should only be used together with FixedLocator(1) 3. PIP 更换国内安装源(1) 4. 在PyQt5中显示matplotlib绘制的图形(1) 5. Python报错：SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape(1) 最新评论 1. Re:Python报错：SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape 感谢！！ --youthYong 2. Re:sublime_text 4143 for Linux 注册牛的。直接运行那个脚本更方便。能找到linux的软件破解方法真是不容易啊 --仓葵与暮博客园© 2004-2025 浙公网安备 33010602011771号浙ICP备2021040463号-3 点击右上角即可分享
14784	http://www.exodusbooks.com/samples/aops/66779sample2.pdf	Chapter 17 Counting in the Twilight Zone In Volume 1 we examined a great many counting methods, but all were based on the rock of common sense. In this chapter we will look at counting methods which go far beyond common sense, and thus allow the counting of far more interesting things. 17.1 One to One We will escape the realm of common sense with the help of the obvious-seeming proposition that if a one to one correspondence can be drawn between the members of two groups, the two groups are equal in size. Recall that one to one (also written 1-1 or 1 : 1) means that each object in either group corresponds to one and only one object in the other. EXAMPLE 17-1 Prove that the number of integers greater than 0 and less than 100 equals the number of integers greater than 100 and less than 200. Proof: For any integer n in the ﬁrst group (so 0 < n < 100) the corresponding integer in the second group will be n + 100. Clearly 100 < n + 100 < 200. It is clear that every integer n in the ﬁrst group has a single counterpart n + 100 in the second, and that every integer m in the second group has a single counterpart m −100 in the ﬁrst group. Thus the correspondence is one to one, and the two groups of numbers are the same size. 17.2 Clever Correspondences Like any obvious statement, the one to one correspondence principle is useless taken by itself. It must be coupled with a clever correspondence if it is to have any power. A common example is of the form: A dog trainer wants to buy 8 dogs all of which are either cocker spaniels, Irish setters, or Russian wolfhounds. In how many ways can she make the choice? / 196 . Excerpt from "Art of Problem Solving Volume 2: and Beyond" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material the ART of PROBLEM SOLVING: Volume 2 / 197 The problem is an example of making selections with repetition, because we can choose as many as we wish from each category. At face value it doesn’t seem any harder than many problems we tackled in Volume 1. However, if you try it you’ll see it’s much harder. EXERCISE 17-1 Think about the problem until you see why it can’t easily be done with our previous methods. Our problem can be solved with a neat one to one correspondence. Namely, for each choice of 8 dogs we can write a sequence like ddd dd ddd. The number of d’s before the ﬁrst represents the number of spaniels, the number in the middle the number of setters, and the number after the second the number of wolfhounds. EXAMPLE 17-2 To what choice of dog varieties does the sequence above correspond? Solution: Since there are three d’s before the ﬁrst , we have three spaniels. Since there are two d’s in the middle, there are two setters. Since there are three d’s at the end, there are three wolfhounds. (Note that this is in fact a legal sequence since exactly eight dogs are accounted for.) EXERCISE 17-2 Prove that we can write exactly one sequence for each choice of dogs, and that each sequence corresponds to exactly one choice of dogs. What happens in the case that there are zero dogs of some variety? Is this OK? Since we have established in Exercise 17-2 a one to one correspondence between choices of dogs and sequences, all we have to count is the number of sequences. But counting the sequences is easy, since each sequence is just a matter of choosing two positions for the ’s out of 10 total positions. (There are 10 positions because we have 8 d’s and 2 ’s.) The number of ways to choose 2 positions out of 10 is 10 2 = 45. EXERCISE 17-3 Find (and prove) a simple formula for the number of ways to buy n dogs if there are r varieties to choose from. EXERCISE 17-4 Prove that the number you found in the previous exercise is equal to the number of solutions in nonnegative integers of x1 + x2 + · · · + xr = n. There are many other ways to set up one to one correspondences; in general, when a problem looks too diﬃcult by other methods, look for a correspondence to a simpler problem. It will in general require some creativity on your part to come up with the particular correspondences which do the job, but you’ll get a feel for what works with experience. EXAMPLE 17-3 In how many ways may ﬁve people be seated in a row of twenty chairs given that no two people may sit next to one another? Excerpt from "Art of Problem Solving Volume 2: and Beyond" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material 198 . CHAPTER 17. COUNTING IN THE TWILIGHT ZONE Solution: Consider some arrangement of the ﬁve people as speciﬁed, then take one chair out from between each pair of people. What you’re left with is a unique arrangement of 5 people in 16 chairs without restrictions. Similarly, starting with an unrestricted arrangement of 5 people in 16 chairs, adding a chair between each pair of people gives a unique arrangement of 5 non-adjacent people in 20 chairs. (Convince yourself of these two assertions.) Thus there is a one to one correspondence between the restricted 20-chair arrangements of the problem and unrestricted 16-chair arrangements. The number of unrestricted 16-chair arrangements is the number of ways to choose 5 chairs out of 16, or 16 5 . EXAMPLE 17-4 How many nonnegative integer solutions are there to x1 + x2 + x3 50? Solution: We have seen how to solve x1 + x2 + · · · + xr = n, but the inequality complicates the problem. We might be tempted to successively solve x1 + x2 + x3 = 50, then x1 + x2 + x3 = 49, and so on, then try to simplify the sum of the results. But creative thinking yields a di↵erent way: just put as much of the 50 as is desired into x1 + x2 + x3, and put the rest into a new variable y. With this idea, it becomes clear that a one to one correspondence exists between nonnegative solutions of our inequality and nonnegative solutions of the equality x1 + x2 + x3 + y = 50, which by Exercises 17-3 and 17-4 has 53 3 solutions. 17.3 Easy as . . . In the ﬁrst volume we used sets and Venn diagrams to attack problems like: There are 100 students taking language classes at Austin High School. If 60 are taking German and 75 are taking Spanish and these are the only languages taught, how many students take both Spanish and German? Let’s try a new approach. Say there are x students taking both languages. We could try counting the number of students taking languages by just adding the number of students in each language, or 60+75 = 135. Since we know there are 100 students taking languages, we have made a mistake. Our error is in counting students taking both languages twice, once for German and once for Spanish. Thus, we must subtract from 135 the number of students taking both languages so that we only count them once. Hence, the total number of students taking language classes is 135 −x. Setting this equal to 100 we ﬁnd x = 35. This is the heart of the Principle of Inclusion-Exclusion, or PIE: if we count something twice, subtract it once so we only count it once. It’s that simple! When we move from two classes to three, the counting becomes a bit trickier, but the concept is the same. For example, let’s call the classes A, B, and C and let the number of students in each be #(A), #(B), and #(C), respectively. To take care of overcounting students in both classes A and B, we must subtract the number of students in both classes, which we’ll call #(A \ B). Similarly, we subtract the number of students in both B and C and in both A and C, for a total of #(A) + #(B) + #(C) −#(A \ B) −#(A \ C) −#(B \ C) students. To convince yourself that this takes care of students enrolled in two classes, pretend you are in exactly two of the classes. How many times are you added to the total? Subtracted? How many times are you counted? Excerpt from "Art of Problem Solving Volume 2: and Beyond" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material the ART of PROBLEM SOLVING: Volume 2 / 199 Now what if you’re in all three classes? You’re added three times and subtracted three times, so you’re not counted at all! To take care of this, we must add back the number of students in all three classes. This makes our total number of students #(A) + #(B) + #(C) −#(A \ B) −#(A \ C) −#(B \ C) + #(A \ B \ C). EXERCISE 17-5 Again, pretend you are in 1, 2, or 3 classes and make sure that the aforementioned method only counts you once no matter how many times you are added or subtracted. EXERCISE 17-6 What if there were 4 classes? How about 5 classes? EXAMPLE 17-5 How many positive integers less than or equal to 1000 do not have 2, 3, or 5 among their prime factors? Solution: We count the number of positive integers less than or equal to 1000 which do have 2, 3, or 5 among their prime factors and subtract that number from 1000 to ﬁnd how many do not. There are 1000/2 = 500 multiples of 2, b1000/3c = 333 multiples of 3 and 1000/5 = 200 multiples of 5 in this range, for a total of 500 + 333 + 200 = 1033 positive integers less than or equal to 1000 with 2, 3, or 5 as a factor. We’ve obviously overlooked something. We’ve badly overcounted, because many numbers are multiples of more than just one of these three numbers. Applying the Principle of Inclusion-Exclusion, we subtract from 1033 the number of multiples of both 2 and 3, of both 2 and 5, and of both 3 and 5. Finally, we then add to the result the number of integers which are multiples of all three. Any number which is a multiple of 2 and 3 is a multiple of (2)(3) = 6, since 2 and 3 have no common nontrivial factors. Hence, we seek the multiples of 6, 10, 15, and 2(3)(5) = 30. There are b1000/6c = 166 multiples of 6, b1000/10c = 100 multiples of 10, b1000/15c = 66 multiples of 15, and b1000/30c = 33 multiples of 30. By PIE, there are 1033 −166 −100 −66 + 33 = 734 integers less than or equal to 1000 with 2, 3, or 5 among their factors. Hence, there are 1000−734 = 266 integers in that range which are not multiples of 2, 3, or 5. EXERCISE 17-7 Suppose there is some number of objects placed in n categories A1, A2,. . ., An, where each object may be in more than one category. Let #(Ai) be the number of objects in category Ai. Suppose #(Ai) is the same for all i, #(Ai \ Aj) is the same for all distinct pairs (i, j), #(Ai \ Aj \ Ak) is the same for all triples (i, j, k), etc. Show that the Principle of Inclusion-Exclusion gives #(A1 [ A2 [ A3 [ · · · [ An) = n 1 #(A1) − n 2 #(A1 \ A2) + n 3 #(A1 \ A2 \ A3) − n 4 #(A1 \ A2 \ A3 \ A4) + · · · + (−1)n n n #(A1 \ A2 \ · · · \ An). This application of the Principle of Inclusion-Exclusion is very useful in problems containing sym-metry. EXAMPLE 17-6 There are four baskets numbered from 1 to 4 and four balls numbered from 1 to 4. Each basket is allowed to have at most two balls. In how many ways can the balls be placed in the baskets such that no ball has the same number as the basket it is in? Excerpt from "Art of Problem Solving Volume 2: and Beyond" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material 200 . CHAPTER 17. COUNTING IN THE TWILIGHT ZONE Solution: This problem has symmetry, so we apply the concepts of the previous exercise. First we count the number of ways to put the balls in the baskets with no restrictions. There are 4! ways to put the balls in 4 di↵erent baskets. There are 4 1 4 2 (3)(2) ways to put two balls in one basket and the others in their own baskets (4 ways to pick the basket with two balls, 6 ways to pick the balls in that basket, (3)(2) ways to put the other balls in di↵erent baskets). Finally, we can put two balls each in two baskets in 4 2 4 2 ways (6 ways to pick the two non-empty baskets and 6 ways to distribute the balls among these baskets). Hence there are 24 + 144 + 36 = 204 ways to put the balls in the baskets. To solve the problem, we will count the ways the balls can be put in the baskets such that at least one ball has the same number as the basket that holds it. Let #(i) be the number of ways to ﬁll the baskets such that basket i holds ball i. From the previous exercise, we seek the quantity 4 1 #(1) − 4 2 #(1 \ 2) + 4 3 #(1 \ 2 \ 3) − 4 4 #(1 \ 2 \ 3 \ 4). For the ﬁrst, after we put ball 1 in basket 1, we can either put another ball in that basket or put all the balls in the other baskets. The former can be done in 3 1 (32) = 27 ways (3 ways to pick the other ball in basket 1, 32 ways to put the other balls in the other baskets); the latter can be done in 3 1 3 2 (2) + 3! = 24 ways, where we divide this into the case of two of the balls in one of the other three baskets and the case of the other three being in di↵erent baskets. For #(1 \ 2) we put balls 1 and 2 in baskets 1 and 2. The other two balls can be put in the baskets in 2 + (2)(2)(2) + (2)(2) = 14 ways, where we consider the cases of putting 2, 1, or 0 of the remaining balls in the ﬁrst two baskets. (Make sure you see this.) The last two are easy. After we put 3 balls in the right baskets, the other has four choices and we can only get them all right in 1 way. Putting these in our above expression, there are 4(27 + 24) −6(14) + 4(4) −1(1) = 135 ways to put the balls in the baskets so that at least one ball is in the right basket. This leaves 204 −135 = 69 ways of ﬁlling the baskets so that no ball has the same number as the basket that holds it. This is a very complicated counting problem, and you may argue that you would be better o↵ just listing the possibilities and counting. Using the Principle of Inclusion-Exclusion is better than listing and counting because it’s awfully hard to tell if we’ve listed all the possibilities, and as the number of possibilities gets large (what if there were 6 balls and 6 baskets), the listing and counting method becomes very unreliable. 17.4 Generating Functions When it comes to counting, generating functions are the cleverest thing there is. The idea of generating functions is that functions can be manipulated in various ways which combinatorial quantities cannot, so to examine the properties of some combinatorial function A(k), we instead look at the function A(0) + A(1)x + A(2)x2 + A(3)x3 + · · · , Excerpt from "Art of Problem Solving Volume 2: and Beyond" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material
14785	https://www.theplasticsfella.com/glomus-tumours/	Sign In Go Pro Sign up Sign in Log in Go Pro Glomus Tumours: Pathology, Diagnosis, & Management A glomus tumor is a benign lesion from glomus cells in the Sucquet-Hoyer canal, often causing severe pain in young adults’ nail beds or palms. It aids in thermoregulation. Diagnosis involves imaging and histology. Excision is the main treatment; for multiple lesions, consider laser or sclerotherapy. PlasticsFella 7 min read Summary Card DefinitionA glomus tumour is a benign lesion from glomus cells in the Sucquet-Hoyer canal, typically on the nail bed or palm. ClassificationThe glomus body, aiding thermoregulation and common in the finger dermis, can develop benign hamartomas called glomus tumours. DiagnosisA glomus tumour typically presents with fingertip pain, tenderness, cold sensitivity, and nail changes, diagnosed through clinical tests, imaging (US and MRI), and histology.ManagementComplete excision of glomus tumours is advised for pain relief and to reduce recurrence; laser or sclerotherapy is recommended for multiple, inoperable lesions. Primary Contributor: Dr Kurt Lee Chircop, Educational Fellow. Reviewer: Dr Waruguru Wanjau, Educational Fellow. Definition of Glomus Tumours Key PointGlomus tumour is a benign vascular neoplasm originating from the cells of the glomus body, a rich microvascular network of cells involved in thermoregulation. Glomus tumours are described as benign hamartomas of the glomus apparatus, originating from modified smooth muscle cells of the Sucquet-Hoyer canal (Morey, 2016). Demographic analysis shows: 1 to 5% of all hand tumours, usually in the subungual region. More common in females. Usually occurs between the 2nd-5th decade (Trehan, 2015). Extradigital glomus tumours show equal distribution between the genders (Chou, 2016). A glomus tumour is illustrated in the image below. 😎 Fun Fact:The glomus body, in the finger's reticular dermis, regulates skin temperature and includes an afferent arteriole, Sucquet-Hoyer canal, primary collecting vein, and intraglomerular reticulum. Classification of Glomus Tumours Key PointThe WHO classifies glomus tumors into three types: solid glomus tumors (glomus cells), glomangiomas (vascular cells), and glomangiomyomas (vascular and smooth muscle cells). WHO Classification The World Health Organization (WHO) classifies glomus tumours according to the predominance of glomus cells, vascular structures, and smooth muscle cells into 3 distinct entities (Sethu, 2016): Solid glomus tumours: Most common, with a predominance of glomus cells. Glomangioma: With a predominance of vascular cells. Glomangiomyoma: With a predominance of both vascular and smooth muscle cells. Benign vs. Malignant Glomus Tumors Glomus tumors are typically benign, but they can occasionally transform into malignant forms or arise as malignant glomus tumors, known as glomangiosarcomas. These malignant lesions have a high rate of local recurrence but a very low rate of metastasis. Risk factors for malignancy include: Deeper location Size greater than 2 cm Rapid growth Atypical mitosis High pathological grade 😎 Fun Fact:A familial variant of glomus tumour has been associated with chromosome 1p21–22 and involves truncating mutations in the glomulin gene (Brouillard, 2002). Diagnosis of Glomus Tumours Key PointGlomus Tumours usually exhibit a triad of symptoms: sharp pain in the fingertips, pinpoint tenderness, and heightened sensitivity to cold. It can be confirmed on ultrasound or MRI. Clinical Assessment Glomus tumors typically present as small (<2 cm), solitary lesions in the acral regions, often subungual. They appear as blue or red blanchable papules or nodules in the deep dermis or subcutis. Multiple tumors may be present in up to 10% of cases (Sethu, 2016) Typical Symptoms (Chou, 2016): Stabbing pain in the fingertips Pinpoint tenderness Cold hypersensitivity Diagnostic Signs (Tang, 2013): Love’s Maneuver: Pinpoint pressure applied to the nail elicits pain using a fine needle. Hildreth’s Sign: Pain and pinpoint tenderness reduce with proximal inflation of a tourniquet, causing transient ischemia. Subungual glomus tumors can cause nail changes such as ridging, erythronychia, splitting, and thickening. Any subungual nodule that alters nail color or structure should raise suspicion for a glomus tumor (Lu, 2018). 😎 Fun Fact:Glomus tumours are rare outside the fingers, with case reports noting occurrences in the gastrointestinal tract, mediastinum, trachea, mesentery, cervix, and vagina (Chou, 2010 & Filice, 2008 & Almaghrabi, 2017). Imaging Ultrasound: detection of lesions as small as 3mm, and helps with depth assessment (Matsunaga, 2007). MRI: Tumours show a high signal intensity on T2-weighted and STIR images (Drape, 1996). Histology Glomus tumor cells have round to oval nuclei, pale eosinophilic cytoplasm, distinct borders, and minimal mitotic activity & pleomorphism. They are immunoreactive for αSMA, MSA, and h-caldesmon, and contain type IV collagen. These markers distinguish glomus tumors from hemangiomas and S100-positive paragangliomas (Mravic, 2015). 😎 Fun Fact:Type I neurofibromatosis is associated with glomus tumors, showing similar sex distribution, tumor location, and burden as isolated glomus tumors (Harrison, 2014). Differential Diagnosis Benign and malignant lesions can resemble glomus tumors. The most common tumor presenting similarly is an epithelial inclusion cyst. Initially thought to be a subset of glomus tumors, it can be distinguished by the following characteristics: Prominent vascular components with dilated vessels, possible thrombosis, and phlebolith formation Positive for epithelial markers with focal ductal differentiation Naevus cells positive for S100 and Melan-A, unlike glomus tumors Other tumors to consider include: Blue Naevus: Rare in the subungual region but may cause diagnostic uncertainty. Angioleiomyoma Dermatologic Manifestation of Kaposi Sarcoma Glomangioma: More common in children and adolescents, typically multifocal, hereditary, and generally painless (Boon, 2004). Management of Glomus Tumours Key PointComplete excision is advised for solitary glomus tumours to alleviate pain and prevent recurrence, with varied techniques for different locations and low recurrence rates if completely excised. Solitary Glomus Tumours For solitary glomus tumours, complete removal of the tumour capsule is advised to alleviate pain and reduce the risk of recurrence. Most subungual lesions are treated by removing the entire nail and excising the tumour, though other techniques include: Excision via the nail bed margin The trap-door method A nail-conserving technique by Lee et al (Pahwa, 2010). In the transungual approach, the nail plate is removed, the tumour is excised, and the nail bed is repaired. Multiple Glomus Tumours For multiple glomus tumours, excision is challenging due to poor circumscription and numerous lesions, making laser therapy (argon, CO2, Nd) and sclerotherapy more effective options (Gould, 1991). These tumours have a low incidence of recurrence if completely excised. Conclusion Upon completing this article, you will have accomplished the following:1. Basic Characteristics of Glomus Tumors: You've gained a foundational understanding of glomus tumors, including their origin from glomus cells, typical locations, and the severe pain they can cause.2. Pathology and Classification: You are now familiar with the components of the glomus body, the classification of glomus tumors into different types based on cell predominance, and the potential for malignant transformation.3. Diagnostic Techniques: You've recognized the importance of clinical examination, imaging techniques like ultrasound and MRI, and histology in diagnosing glomus tumors.4. Management Strategies: You've learned that complete surgical excision is the preferred treatment for solitary tumors to alleviate pain and prevent recurrence, and alternative therapies like laser or sclerotherapy for multiple lesions.5. Complications and Outcomes: Understanding potential complications and outcomes of treatment for better preparation and management of patient expectations regarding glomus tumors. Further Reading Morey, Vivek Machhindra, Bhavuk Garg, and Prakash P. Kotwal. "Glomus tumours of the hand: review of literature."Journal of clinical orthopaedics and trauma 7.4 (2016): 286-291. Trehan, S. K., Soukup, D. S., Mintz, D. N., Perino, G., & Ellis, S. J. (2015). Glomus Tumors in the Foot: Case Series. Foot & ankle specialist, 8(6), 460–465. Chou, T., Pan, S. C., Shieh, S. J., Lee, J. W., Chiu, H. Y., & Ho, C. L. (2016). Glomus Tumor: Twenty-Year Experience and Literature Review. Annals of plastic surgery, 76 Suppl 1, S35–S40. Chou, H. P., Tiu, C. M., Chen, J. D., & Chou, Y. H. (2010). Glomus tumor in the stomach. Abdominal imaging, 35(4), 390–392. Filice, M. E., Lucchi, M., Loggini, B., Mussi, A., & Fontanini, G. (2008). Glomus tumour of the lung: case report and literature review. Pathologica, 100(1), 25–30. Almaghrabi, A., Almaghrabi, N., & Al-Maghrabi, H. (2017). Glomangioma of the Kidney: A Rare Case of Glomus Tumor and Review of the Literature. Case reports in pathology, 2017, 7423642. Sethu, C., & Sethu, A. U. (2016). Glomus tumour. Annals of the Royal College of Surgeons of England, 98(1), e1–e2. Mravic M, LaChaud G, Nguyen A, Scott MA, Dry SM, James AW. Clinical and histopathological diagnosis of glomus tumor: an institutional experience of 138 cases. Int J Surg Pathol. 2015 May;23(3):181-8. doi: 10.1177/1066896914567330. Epub 2015 Jan 22. PMID: 25614464; PMCID: PMC4498398. De Chiara A, Apice G, Mori S, Silvestro G, Losito SN, Botti G, Ninfo V. Malignant glomus tumour: a case report and review of the literature. Sarcoma. 2003;7(2):87-91. doi: 10.1080/1357714031000081207. PMID: 18521375; PMCID: PMC2395518. Harrison, B., & Sammer, D. (2014). Glomus tumors and neurofibromatosis: a newly recognized association. Plastic and reconstructive surgery. Global open, 2(9), e214. Brouillard P, Boon LM, Mulliken JB, Enjolras O, Ghassibé M, Warman ML, Tan OT, Olsen BR, Vikkula M. Mutations in a novel factor, glomulin, are responsible for glomuvenous malformations ("glomangiomas"). Am J Hum Genet. 2002 Apr;70(4):866-74. doi: 10.1086/339492. Epub 2002 Feb 13. PMID: 11845407; PMCID: PMC379115. Lu, H., Chen, L. F., & Chen, Q. (2018). Rupture of a subungual glomus tumor of the finger. BMC cancer, 18(1), 505. Tang CY, Tipoe T, Fung B. Where is the Lesion? Glomus Tumours of the Hand. Arch Plast Surg. 2013 Sep;40(5):492-5. doi: 10.5999/aps.2013.40.5.492. Epub 2013 Sep 13. PMID: 24086799; PMCID: PMC3785579. Matsunaga, A., Ochiai, T., Abe, I., Kawamura, A., Muto, R., Tomita, Y., & Ogawa, M. (2007). Subungual glomus tumour: evaluation of ultrasound imaging in preoperative assessment. European journal of dermatology : EJD, 17(1), 67–69. Drapé, J. L., Idy-Peretti, I., Goettmann, S., Guérin-Surville, H., & Bittoun, J. (1996). Standard and high resolution magnetic resonance imaging of glomus tumors of toes and fingertips. Journal of the American Academy of Dermatology, 35(4), 550–555. Boon LM, Mulliken JB, Enjolras O, Vikkula M. Glomuvenous malformation (glomangioma) and venous malformation: distinct clinicopathologic and genetic entities. Arch Dermatol. 2004 Aug;140(8):971-6. doi: 10.1001/archderm.140.8.971. PMID: 15313813. Lee, H. J., Kim, P. T., Kyung, H. S., Kim, H. S., & Jeon, I. H. (2014). Nail-preserving excision for subungual glomus tumour of the hand. Journal of plastic surgery and hand surgery, 48(3), 201–204. Pahwa, M., Pahwa, P., & Kathuria, S. (2010). Glomus tumour of the nail bed treated with the 'trap door' technique: a report of two patients. The Journal of dermatological treatment, 21(5), 298–300. Gould E. P. (1991). Sclerotherapy for multiple glomangiomata. The Journal of dermatologic surgery and oncology, 17(4), 351–352. 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14786	https://static.bigideasmath.com/protected/content/pe/hsim/int_math1_pe_01_04.pdf	Section 1.4 Solving Absolute Value Equations 27 Solving Absolute Value Equations 1.4 Essential Question Essential Question How can you solve an absolute value equation? Solving an Absolute Value Equation Algebraically Work with a partner. Consider the absolute value equation ∣ x + 2 ∣ = 3. a. Describe the values of x + 2 that make the equation true. Use your description to write two linear equations that represent the solutions of the absolute value equation. b. Use the linear equations you wrote in part (a) to fi nd the solutions of the absolute value equation. c. How can you use linear equations to solve an absolute value equation? Solving an Absolute Value Equation Graphically Work with a partner. Consider the absolute value equation ∣ x + 2 ∣ = 3. a. On a real number line, locate the point for which x + 2 = 0. −10−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 b. Locate the points that are 3 units from the point you found in part (a). What do you notice about these points? c. How can you use a number line to solve an absolute value equation? Solving an Absolute Value Equation Numerically Work with a partner. Consider the absolute value equation ∣ x + 2 ∣ = 3. a. Use a spreadsheet, as shown, to solve the absolute value equation. b. Compare the solutions you found using the spreadsheet with those you found in Explorations 1 and 2. What do you notice? c. How can you use a spreadsheet to solve an absolute value equation? Communicate Your Answer 4. How can you solve an absolute value equation? 5. What do you like or dislike about the algebraic, graphical, and numerical methods for solving an absolute value equation? Give reasons for your answers. MAKING SENSE OF PROBLEMS To be profi cient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution. A x -6 -5 -4 -3 -2 -1 0 1 2 B \|x + 2\| 4 2 1 3 4 5 6 7 8 9 10 11 abs(A2 + 2) 28 Chapter 1 Solving Linear Equations 1.4 Lesson What You Will Learn What You Will Learn Solve absolute value equations. Solve equations involving two absolute values. Identify special solutions of absolute value equations. Solving Absolute Value Equations An absolute value equation is an equation that contains an absolute value expression. You can solve these types of equations by solving two related linear equations. Property of Absolute Value Solving Absolute Value Equations Solve each equation. Graph the solutions, if possible. a. ∣ x − 4 ∣ = 6 b. ∣ 3x + 1 ∣ = −5 SOLUTION a. Write the two related linear equations for ∣ x − 4 ∣ = 6. Then solve. x − 4 = 6 or x − 4 = −6 Write related linear equations. x = 10 x = −2 Add 4 to each side. The solutions are x = 10 and x = −2. 0 −2 −4 2 4 6 8 10 12 Each solution is 6 units from 4. 6 6 b. The absolute value of an expression must be greater than or equal to 0. The expression ∣ 3x + 1 ∣ cannot equal −5. So, the equation has no solution. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation. Graph the solutions, if possible. 1. ∣ x ∣ = 10 2. ∣ x − 1 ∣ = 4 3. ∣ 3 + x ∣ = −3 absolute value equation, p. 28 extraneous solution, p. 31 Previous absolute value opposite Core Vocabulary Core Vocabulary Properties of Absolute Value Let a and b be real numbers. Then the following properties are true. 1. ∣ a ∣ ≥ 0 2. ∣ −a ∣ = ∣ a ∣ 3. ∣ ab ∣ = ∣ a ∣ ∣ b ∣ 4. ∣ a — b ∣ = ∣ a ∣ — ∣ b ∣ , b ≠ 0 Solving Absolute Value Equations To solve ∣ ax + b ∣ = c when c ≥ 0, solve the related linear equations ax + b = c or ax + b = − c. When c < 0, the absolute value equation ∣ ax + b ∣ = c has no solution because absolute value always indicates a number that is not negative. Core Core Concept Concept Section 1.4 Solving Absolute Value Equations 29 Solving an Absolute Value Equation Solve ∣ 3x + 9 ∣ − 10 = −4. SOLUTION First isolate the absolute value expression on one side of the equation. ∣ 3x + 9 ∣ − 10 = −4 Write the equation. ∣ 3x + 9 ∣ = 6 Add 10 to each side. Now write two related linear equations for ∣ 3x + 9 ∣ = 6. Then solve. 3x + 9 = 6 or 3x + 9 = −6 Write related linear equations. 3x = −3 3x = −15 Subtract 9 from each side. x = −1 x = −5 Divide each side by 3. The solutions are x = −1 and x = −5. Writing an Absolute Value Equation In a cheerleading competition, the minimum length of a routine is 4 minutes. The maximum length of a routine is 5 minutes. Write an absolute value equation that represents the minimum and maximum lengths. SOLUTION 1. Understand the Problem You know the minimum and maximum lengths. You are asked to write an absolute value equation that represents these lengths. 2. Make a Plan Consider the minimum and maximum lengths as solutions to an absolute value equation. Use a number line to fi nd the halfway point between the solutions. Then use the halfway point and the distance to each solution to write an absolute value equation. 3. Solve the Problem The equation is ∣ x − 4.5 ∣ = 0.5. 4. Look Back To check that your equation is reasonable, substitute the minimum and maximum lengths into the equation and simplify. Minimum Maximum ∣ 4 − 4.5 ∣ = 0.5 ✓ ∣ 5 − 4.5 ∣ = 0.5 ✓ Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation. Check your solutions. 4. ∣ x − 2 ∣ + 5 = 9 5. 4 ∣ 2x + 7 ∣ = 16 6. −2 ∣ 5x − 1 ∣ − 3 = −11 7. For a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines. Write an absolute value equation that represents the minimum and maximum lengths. ∣ x − 4.5 ∣ = 0.5 distance from halfway point halfway point ANOTHER WAY Using the product property of absolute value, \|ab\| = \|a\| \|b\|, you could rewrite the equation as 3\|x + 3\| − 10 = −4 and then solve. REASONING Mathematically profi cient students have the ability to decontextualize problem situations. 4.2 4.1 4.0 4.3 4.4 4.5 4.6 4.8 4.7 4.9 5.0 0.5 0.5 3 4 M S 30 Chapter 1 Solving Linear Equations Solving Equations with Two Absolute Values If the absolute values of two algebraic expressions are equal, then they must either be equal to each other or be opposites of each other. Solving Equations with Two Absolute Values Solve (a) ∣ 3x − 4 ∣ = ∣ x ∣ and (b) ∣ 4x − 10 ∣ = 2 ∣ 3x + 1 ∣ . SOLUTION a. Write the two related linear equations for ∣ 3x − 4 ∣ = ∣ x ∣ . Then solve. 3x − 4 = x or 3x − 4 = −x − x − x + x + x 2x − 4 = 0 4x − 4 = 0 + 4 + 4 + 4 + 4 2x = 4 4x = 4 2x — 2 = 4 — 2 4x — 4 = 4 — 4 x = 2 x = 1 The solutions are x = 2 and x = 1. b. Write the two related linear equations for ∣ 4x − 10 ∣ = 2 ∣ 3x + 1 ∣ . Then solve. 4x − 10 = 2(3x + 1) or 4x − 10 = 2[−(3x + 1)] 4x − 10 = 6x + 2 4x − 10 = 2(−3x − 1) − 6x − 6x 4x − 10 = −6x − 2 − 2x − 10 = 2 + 6x + 6x + 10 + 10 10x − 10 = −2 −2x = 12 + 10 + 10 −2x — −2 = 12 — −2 10x = 8 x = −6 10x — 10 = 8 — 10 x = 0.8 The solutions are x = −6 and x = 0.8. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation. Check your solutions. 8. ∣ x + 8 ∣ = ∣ 2x + 1 ∣ 9. 3 ∣ x − 4 ∣ = ∣ 2x + 5 ∣ Check ∣ 3x − 4 ∣ = ∣ x ∣ ∣ 3(2) − 4 ∣ = ? ∣ 2 ∣ ∣ 2 ∣ = ? ∣ 2 ∣ 2 = 2 ✓ ∣ 3x − 4 ∣ = ∣ x ∣ ∣ 3(1) − 4 ∣ = ? ∣ 1 ∣ ∣ −1 ∣ = ? ∣ 1 ∣ 1 = 1 ✓ Solving Equations with Two Absolute Values To solve ∣ ax + b ∣ = ∣ cx + d ∣ , solve the related linear equations ax + b = cx + d or ax + b = −(cx + d). Core Core Concept Concept Section 1.4 Solving Absolute Value Equations 31 Identifying Special Solutions When you solve an absolute value equation, it is possible for a solution to be extraneous. An extraneous solution is an apparent solution that must be rejected because it does not satisfy the original equation. Identifying Extraneous Solutions Solve ∣ 2x + 12 ∣ = 4x. Check your solutions. SOLUTION Write the two related linear equations for ∣ 2x + 12 ∣ = 4x. Then solve. 2x + 12 = 4x or 2x + 12 = −4x Write related linear equations. 12 = 2x 12 = −6x Subtract 2x from each side. 6 = x −2 = x Solve for x. Check the apparent solutions to see if either is extraneous. The solution is x = 6. Reject x = −2 because it is extraneous. When solving equations of the form ∣ ax + b ∣ = ∣ cx + d ∣ , it is possible that one of the related linear equations will not have a solution. Check ∣ 2x + 12 ∣ = 4x ∣ 2(6) + 12 ∣ = ? 4(6) ∣ 24 ∣ = ? 24 24 = 24 ✓ ∣ 2x + 12 ∣ = 4x ∣ 2(−2) + 12 ∣ = ? 4(−2) ∣ 8 ∣ = ? −8 8 ≠ −8 ✗ Solving an Equation with Two Absolute Values Solve ∣ x + 5 ∣ = ∣ x + 11 ∣ . SOLUTION By equating the expression x + 5 and the opposite of x + 11, you obtain x + 5 = −(x + 11) Write related linear equation. x + 5 = −x − 11 Distributive Property 2x + 5 = −11 Add x to each side. 2x = −16 Subtract 5 from each side. x = −8. Divide each side by 2. However, by equating the expressions x + 5 and x + 11, you obtain x + 5 = x + 11 Write related linear equation. x = x + 6 Subtract 5 from each side. 0 = 6 ✗ Subtract x from each side. which is a false statement. So, the original equation has only one solution. The solution is x = −8. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the equation. Check your solutions. 10. ∣ x + 6 ∣ = 2x 11. ∣ 3x − 2 ∣ = x 12. ∣ 2 + x ∣ = ∣ x − 8 ∣ 13. ∣ 5x − 2 ∣ = ∣ 5x + 4 ∣ REMEMBER Always check your solutions in the original equation to make sure they are not extraneous. 32 Chapter 1 Solving Linear Equations Exercises 1.4 Dynamic Solutions available at BigIdeasMath.com In Exercises 3−10, simplify the expression. 3. ∣ −9 ∣ 4. − ∣ 15 ∣ 5. ∣ 14 ∣ − ∣ −14 ∣ 6. ∣ −3 ∣ + ∣ 3 ∣ 7. − ∣ −5 ⋅ (−7) ∣ 8. ∣ −0.8 ⋅ 10 ∣ 9. ∣ 27 — −3 ∣ 10. ∣ − −12 — 4 ∣ In Exercises 11−24, solve the equation. Graph the solution(s), if possible. (See Examples 1 and 2.) 11. ∣ w ∣ = 6 12. ∣ r ∣ = −2 13. ∣ y ∣ = −18 14. ∣ x ∣ = 13 15. ∣ m + 3 ∣ = 7 16. ∣ q − 8 ∣ = 14 17. ∣ −3d ∣ = 15 18. ∣ t — 2 ∣ = 6 19. ∣ 4b − 5 ∣ = 19 20. ∣ x − 1 ∣ + 5 = 2 21. −4 ∣ 8 − 5n ∣ = 13 22. −3 ∣ 1 − 2 — 3 v ∣ = −9 23. 3 = −2 ∣ 1 — 4 s − 5 ∣ + 3 24. 9 ∣ 4p + 2 ∣ + 8 = 35 25. WRITING EQUATIONS The minimum distance from Earth to the Sun is 91.4 million miles. The maximum distance is 94.5 million miles. (See Example 3.) a. Represent these two distances on a number line. b. Write an absolute value equation that represents the minimum and maximum distances. 26. WRITING EQUATIONS The shoulder heights of the shortest and tallest miniature poodles are shown. 10 in. 15 in. a. Represent these two heights on a number line. b. Write an absolute value equation that represents these heights. USING STRUCTURE In Exercises 27−30, match the absolute value equation with its graph without solving the equation. 27. ∣ x + 2 ∣ = 4 28. ∣ x − 4 ∣ = 2 29. ∣ x − 2 ∣ = 4 30. ∣ x + 4 ∣ = 2 A. −10 −8 −6 −4 −2 0 2 2 2 B. −8 −6 −4 −2 0 2 4 4 4 C. −4 −2 0 2 4 6 8 4 4 D. −2 0 2 4 6 8 10 2 2 Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 1. VOCABULARY What is an extraneous solution? 2. WRITING Without calculating, how do you know that the equation ∣ 4x − 7 ∣ = −1 has no solution? Vocabulary and Core Concept Check Vocabulary and Core Concept Check Section 1.4 Solving Absolute Value Equations 33 In Exercises 31−34, write an absolute value equation that has the given solutions. 31. x = 8 and x = 18 32. x = −6 and x = 10 33. x = 2 and x = 9 34. x = −10 and x = −5 In Exercises 35−44, solve the equation. Check your solutions. (See Examples 4, 5, and 6.) 35. ∣ 4n − 15 ∣ = ∣ n ∣ 36. ∣ 2c + 8 ∣ = ∣ 10c ∣ 37. ∣ 2b − 9 ∣ = ∣ b − 6 ∣ 38. ∣ 3k − 2 ∣ = 2 ∣ k + 2 ∣ 39. 4 ∣ p − 3 ∣ = ∣ 2p + 8 ∣ 40. 2 ∣ 4w − 1 ∣ = 3 ∣ 4w + 2 ∣ 41. ∣ 3h + 1 ∣ = 7h 42. ∣ 6a − 5 ∣ = 4a 43. ∣ f − 6 ∣ = ∣ f + 8 ∣ 44. ∣ 3x − 4 ∣ = ∣ 3x − 5 ∣ 45. MODELING WITH MATHEMATICS Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance d (in feet) of the car from you after t seconds is given by the equation d = ∣ 300 − 48t ∣ . At what times is the car 60 feet from you? 46. MAKING AN ARGUMENT Your friend says that the absolute value equation ∣ 3x + 8 ∣ − 9 = −5 has no solution because the constant on the right side of the equation is negative. Is your friend correct? Explain. 47. MODELING WITH MATHEMATICS You randomly survey students about year-round school. The results are shown in the graph. Year-Round School Oppose Favor 0% 20% 40% 60% 80% 32% Error: ±5% 68% The error given in the graph means that the actual percent could be 5% more or 5% less than the percent reported by the survey. a. Write and solve an absolute value equation to fi nd the least and greatest percents of students who could be in favor of year-round school. b. A classmate claims that 1 — 3 of the student body is actually in favor of year-round school. Does this confl ict with the survey data? Explain. 48. MODELING WITH MATHEMATICS The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams. a. Write and solve an absolute value equation to fi nd the minimum and maximum acceptable soccer ball weights. b. A soccer ball weighs 423 grams. Due to wear and tear, the weight of the ball decreases by 16 grams. Is the weight acceptable? Explain. ERROR ANALYSIS In Exercises 49 and 50, describe and correct the error in solving the equation. 49. ∣ 2x − 1 ∣ = −9 2x − 1 = −9 or 2x − 1 = −(−9) 2x = −8 2x = 10 x = −4 x = 5 The solutions are x = −4 and x = 5. ✗ 50. ∣ 5x + 8 ∣ = x 5x + 8 = x or 5x + 8 = −x 4x + 8 = 0 6x + 8 = 0 4x = −8 6x = −8 x = −2 x = − 4 — 3 The solutions are x = −2 and x = − 4 — 3 . ✗ 51. ANALYZING EQUATIONS Without solving completely, place each equation into one of the three categories. No solution One solution Two solutions ∣ x − 2 ∣ + 6 = 0 ∣ x + 3 ∣ − 1 = 0 ∣ x + 8 ∣ + 2 = 7 ∣ x − 1 ∣ + 4 = 4 ∣ x − 6 ∣ − 5 = −9 ∣ x + 5 ∣ − 8 = −8 34 Chapter 1 Solving Linear Equations 52. USING STRUCTURE Fill in the equation ∣ x − ∣ = with a, b, c, or d so that the equation is graphed correctly. a b c d d ABSTRACT REASONING In Exercises 53−56, complete the statement with always, sometimes, or never. Explain your reasoning. 53. If x2 = a2, then ∣ x ∣ is _ equal to ∣ a ∣ . 54. If a and b are real numbers, then ∣ a − b ∣ is ___ equal to ∣ b − a ∣ . 55. For any real number p, the equation ∣ x − 4 ∣ = p will _ have two solutions. 56. For any real number p, the equation ∣ x − p ∣ = 4 will __ have two solutions. 57. WRITING Explain why absolute value equations can have no solution, one solution, or two solutions. Give an example of each case. 58. THOUGHT PROVOKING Describe a real-life situation that can be modeled by an absolute value equation with the solutions x = 62 and x = 72. 59. CRITICAL THINKING Solve the equation shown. Explain how you found your solution(s). 8 ∣ x + 2 ∣ − 6 = 5 ∣ x + 2 ∣ + 3 60. HOW DO YOU SEE IT? The circle graph shows the results of a survey of registered voters the day of an election. Democratic: 47% Republican: 42% Libertarian: 5% Error: ±2% Green: 2% Which Party’s Candidate Will Get Your Vote? Other: 4% The error given in the graph means that the actual percent could be 2% more or 2% less than the percent reported by the survey. a. What are the minimum and maximum percents of voters who could vote Republican? Green? b. How can you use absolute value equations to represent your answers in part (a)? c. One candidate receives 44% of the vote. Which party does the candidate belong to? Explain. 61. ABSTRACT REASONING How many solutions does the equation a ∣ x + b ∣ + c = d have when a > 0 and c = d? when a < 0 and c > d? Explain your reasoning. Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Identify the property of equality that makes Equation 1 and Equation 2 equivalent. (Section 1.1) 62. Equation 1 3x + 8 = x − 1 Equation 2 3x + 9 = x 63. Equation 1 4y = 28 Equation 2 y = 7 Use a geometric formula to solve the problem. (Skills Review Handbook) 64. A square has an area of 81 square meters. Find the side length. 65. A circle has an area of 36π square inches. Find the radius. 66. A triangle has a height of 8 feet and an area of 48 square feet. Find the base. 67. A rectangle has a width of 4 centimeters and a perimeter of 26 centimeters. Find the length. Reviewing what you learned in previous grades and lessons
14787	https://www.vedantu.com/question-answer/the-locus-of-the-midpoint-of-the-chords-of-the-class-9-maths-cbse-5f6ef0cddc480e2a5c673baf	Talk to our experts 1800-120-456-456 The locus of the midpoint of the chords of the hyperbola [\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1] passing through a fixed point [\left( {\alpha ,\beta } \right)] is a hyperbola with center is: A) [\left( {\dfrac{\alpha }{3},\dfrac{\beta }{3}} \right)] B) [\left( {\alpha ,\beta } \right)] C) [\left( {\dfrac{\alpha }{5},\dfrac{\beta }{5}} \right)] D) [\left( {\dfrac{\alpha }{2},\dfrac{\beta }{2}} \right)] © 2025.Vedantu.com. All rights reserved
14788	https://pubmed.ncbi.nlm.nih.gov/17032149/	Effect of enzyme replacement therapy with imiglucerase on BMD in type 1 Gaucher disease - 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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Effect of enzyme replacement therapy with imiglucerase on BMD in type 1 Gaucher disease Richard J Wenstrup1,Katherine A Kacena,Paige Kaplan,Gregory M Pastores,Ainu Prakash-Cheng,Ari Zimran,Thomas N Hangartner Affiliations Expand Affiliation 1 Cincinnati Children's Hospital Research Foundation, Ohio, USA. PMID: 17032149 DOI: 10.1359/jbmr.061004 Free article Item in Clipboard Clinical Trial Effect of enzyme replacement therapy with imiglucerase on BMD in type 1 Gaucher disease Richard J Wenstrup et al. J Bone Miner Res.2007 Jan. Free article Show details Display options Display options Format J Bone Miner Res Actions Search in PubMed Search in NLM Catalog Add to Search . 2007 Jan;22(1):119-26. doi: 10.1359/jbmr.061004. Authors Richard J Wenstrup1,Katherine A Kacena,Paige Kaplan,Gregory M Pastores,Ainu Prakash-Cheng,Ari Zimran,Thomas N Hangartner Affiliation 1 Cincinnati Children's Hospital Research Foundation, Ohio, USA. PMID: 17032149 DOI: 10.1359/jbmr.061004 Item in Clipboard Full text links Cite Display options Display options Format Abstract The effect of ERT with imiglucerase on BMD in type 1 GD was studied using BMD data from the International Collaborative Gaucher Group Gaucher Registry. Data were analyzed for 160 untreated patients and 342 ERT-treated patients. Imiglucerase significantly improves BMD in patients with GD, with 8 years of ERT leading to normal BMD. Introduction: The objective was to determine the effect of enzyme replacement therapy (ERT; Cerezyme, imiglucerase) on BMD in type 1 Gaucher disease (GD). Materials and methods: The study population included all adults (men, 18-70 years; women, 18-50 years) enrolled in the International Collaborative Gaucher Group (ICGG) Gaucher Registry for whom lumbar spine BMD measurements were available. BMD data with up to 8 years of follow-up were analyzed for 160 patients who received no ERT and 342 patients treated with ERT alone. BMD was assessed by DXA of the lumbar spine. Z scores for patients with GD were compared with a reference population. From the model's estimate, percent of patients by age and sex with osteoporosis (T score < or = -2.5) were calculated. Results: DXA Z scores for patients with GD in the no ERT (untreated) group were significantly below normal (y intercept = -0.80 Z score units, p < 0.001) and remained approximately 1 SD below the reference population over time (slope = -0.010 Z score units per year, p = 0.68). The DXA Z scores for patients with GD who received ERT at a dose of 60 U/kg/2 weeks were significantly lower than the reference population at baseline (y-intercept = -1.17 Z score units, p < 0.001), but improved significantly over time (slope = +0.132 Z score units per year, p < 0.001). A significant dose-response relationship was noted for the ERT group, with the slopes for the three main dosing groups of 15, 30, and 60 U/kg/2 weeks of +0.064, +0.086, and +0.132 Z score units per year, respectively. The BMD of patients with GD treated with ERT increased to -0.12 (60 U/kg/2 weeks), -0.48 (30 U/kg/2 weeks), and -0.66 (15 U/kg/2 weeks) SD of the mean of the reference population after 8 years of ERT, approaching the reference population. Estimated risk of osteoporosis of this GD population, if left untreated, ranged from approximately 10 to 30% in women and 10% to 25% in men. Conclusions: ERT with imiglucerase (Cerezyme) may increase BMD in patients with GD. Response to treatment with imiglucerase is slower for BMD than for hematologic and visceral aspects of GD. A normal (age- and sex-adjusted) BMD should be a therapeutic goal for patients with type 1 GD. PubMed Disclaimer Similar articles Velaglucerase alfa for the management of type 1 Gaucher disease.Morris JL.Morris JL.Clin Ther. 2012 Feb;34(2):259-71. doi: 10.1016/j.clinthera.2011.12.017. Epub 2012 Jan 20.Clin Ther. 2012.PMID: 22264444 Review. Osteopenia in Gaucher disease develops early in life: response to imiglucerase enzyme therapy in children, adolescents and adults.Mistry PK, Weinreb NJ, Kaplan P, Cole JA, Gwosdow AR, Hangartner T.Mistry PK, et al.Blood Cells Mol Dis. 2011 Jan 15;46(1):66-72. doi: 10.1016/j.bcmd.2010.10.011. Epub 2010 Nov 26.Blood Cells Mol Dis. 2011.PMID: 21112800 Free PMC article. Long-term hematological, visceral, and growth outcomes in children with Gaucher disease type 3 treated with imiglucerase in the International Collaborative Gaucher Group Gaucher Registry.El-Beshlawy A, Tylki-Szymanska A, Vellodi A, Belmatoug N, Grabowski GA, Kolodny EH, Batista JL, Cox GF, Mistry PK.El-Beshlawy A, et al.Mol Genet Metab. 2017 Jan-Feb;120(1-2):47-56. doi: 10.1016/j.ymgme.2016.12.001. Epub 2016 Dec 6.Mol Genet Metab. 2017.PMID: 28040394 Clinical Trial. Baseline characteristics of 32 patients with Gaucher disease who were treated with imiglucerase: South African data from the International Collaborative Gaucher Group (ICGG) Gaucher Registry.Sevittz H, Laher F, Varughese ST, Nel M, McMaster A, Jacobson BF.Sevittz H, et al.S Afr Med J. 2022 Feb 2;112(1):13518.S Afr Med J. 2022.PMID: 35140000 Bone complications in children with Gaucher disease.Bembi B, Ciana G, Mengel E, Terk MR, Martini C, Wenstrup RJ.Bembi B, et al.Br J Radiol. 2002;75 Suppl 1:A37-44. doi: 10.1259/bjr.75.suppl_1.750037.Br J Radiol. 2002.PMID: 12036831 Review. See all similar articles Cited by Timing of initiation of enzyme replacement therapy after diagnosis of type 1 Gaucher disease: effect on incidence of avascular necrosis.Mistry PK, Deegan P, Vellodi A, Cole JA, Yeh M, Weinreb NJ.Mistry PK, et al.Br J Haematol. 2009 Nov;147(4):561-70. doi: 10.1111/j.1365-2141.2009.07872.x. Epub 2009 Sep 3.Br J Haematol. 2009.PMID: 19732054 Free PMC article. Review of the safety and efficacy of imiglucerase treatment of Gaucher disease.Elstein D, Zimran A.Elstein D, et al.Biologics. 2009;3:407-17. doi: 10.2147/btt.2009.3497. Epub 2009 Sep 15.Biologics. 2009.PMID: 19774208 Free PMC article. Olipudase alfa for treatment of acid sphingomyelinase deficiency (ASMD): safety and efficacy in adults treated for 30 months.Wasserstein MP, Diaz GA, Lachmann RH, Jouvin MH, Nandy I, Ji AJ, Puga AC.Wasserstein MP, et al.J Inherit Metab Dis. 2018 Sep;41(5):829-838. doi: 10.1007/s10545-017-0123-6. Epub 2018 Jan 5.J Inherit Metab Dis. 2018.PMID: 29305734 Free PMC article.Clinical Trial. Long-term effectiveness of enzyme replacement therapy in adults with Gaucher disease: results from the NCS-LSD cohort study.Anderson LJ, Henley W, Wyatt KM, Nikolaou V, Hughes DA, Waldek S, Logan S.Anderson LJ, et al.J Inherit Metab Dis. 2014 Nov;37(6):953-60. doi: 10.1007/s10545-014-9680-0. Epub 2014 Feb 11.J Inherit Metab Dis. 2014.PMID: 24515873 Radiographic Cortical Thickness Index Predicts Fragility Fracture in Gaucher Disease.D'Amore S, Sano H, Chappell DDG, Chiarugi D, Baker O, Page K, Ramaswami U, Johannesdottir F, Cox TM, Deegan P, Poole KE; MRC Gaucherite Consortium; MRC Gaucherite Consortium Collaborators.D'Amore S, et al.Radiology. 2023 Apr;307(1):e212779. doi: 10.1148/radiol.212779. Epub 2022 Dec 20.Radiology. 2023.PMID: 36537898 Free PMC article.Clinical Trial. 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14789	https://energyeducation.ca/encyclopedia/Sensible_heat	Sensible heat - Energy Education EnglishFrançaisEspañol Energy Education Navigation menu ENERGY SOURCES Fuels Fossil Fuels Biofuels Nuclear Fuels Flows Hydro Solar Wind Geothermal ENERGY USE Carriers Electricity Gasoline Hydrogen Sectors Transportation Residential Industrial ENERGY IMPACTS Living standard Pollution Acid Rain Smog Pollutants Climate Change Climate Feedback Ocean Acidification Rising Sea Level INDEX Search Sensible heat Figure 1: Campfires emit radiant heat which is felt as sensible heat because it increases the temperature of your body. Sensible heat is literally the heat that can be felt. It is the energy moving from one system to another that changes the temperature rather than changing its phase. For example, it warms water rather than melting ice. In other words, it is the heat that can be felt standing near a fire, or standing outside on a [[sunny day. Sensible heat is used in contrast to latent heat (the heat needed to change from one form of matter to another, which doesn't change temperature), as the two are essentially opposite. For example, in a cooling system condensation forms due to removal of latent heat, and the refrigerant (cooling liquid) changes temperature due to sensible heat. The sensible heat capacity then describes the capacity required to lower the temperature whereas latent heat capacity is the capacity to remove the moisture from the air. In figure 2, very cold ice has heat added to it. The temperature goes up, so that's sensible heat, but once it starts melting, that heat is latent heat (and is represented by the flat parts of the line, during melting or evaporation). Figure 2. Adding heat to water can either raise the temperature or change the phase. The heat that changes the temperature is sensible heat, the heat that changes the phase is latent heat. References ↑Wikimedia Commons [Online], Available: ↑Space Air, What is the difference between sensible and latent heat? [Online], Available: ↑Wikimedia Commons [Online], Available: Retrieved from " Get Citation Contact usAbout usPrivacy policyTerms of use
14790	https://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.1008817	Supplementary Information for: Predicting microbial growth dynamics in response to nutrient availability Olga A. Nev 1, Richard J. Lindsay 1Y, Alys Jepson 1Y, Lisa Butt 1, Robert E. Beardmore 1,Ivana Gudelj 1 1 Biosciences and Living Systems Institute, University of Exeter, Exeter, UK YThese authors contributed equally to this work i.gudelj@exeter.ac.uk Appendix A. Supplementary Tables. The optimal estimates for the parameters Y , V , and K (A1 Table – A7 Table, A10 Table – A13 Table) are obtained by means of the non-linear least-squares fit of the model (2) from the main text to the experimental data using Mathematica ’s build in tools with a method Gradient . We observed that the reliability of the estimated parameters V , Y , K as well as the quality of the fit of the model (2) from the main text to the experimental data do not depend on the initial resource concentration (see SE for the fitted parameters in different resource concentrations in A1 Table – A7 Table, A10 Table – A13 Table and 95% confidence bands for the model fit shown in B1 Fig – B7 Fig). A8 Table and A9 Table contain RMSE values calculated for different approaches for predicting growth described in the main text under all considered environmental conditions by means of the following formula: RMSE = √√√√ n ∑ i=1 (xf,i − xo,i )2 n , (1) where xo,i is an experimentally observed value i, whereas xf,i is a value predicted by function f for the experimental observation i, thus the term ( xf,i − xo,i ) represents a residual corresponding to the experimental observation i, and n is the total number of observations. The RMSE is a measurement of how concentrated the experimental data is around a predictive curve, so quantitatively estimates which prediction better describes the data (the lower the RMSE value, the better the prediction). March 22, 2021 1/21 A1 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. albicans growth on glucose (see B1 Fig). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (a) 0.1 15.81 0.61 10 11 9.64 × 10 −11 10.6 0.09 (b) 0.3 7.05 0.16 10 11 1.09 × 10 −11 8.66 0.06 (c) 0.5 5.24 0.09 10 11 4.61 × 10 −12 6.56 0.04 (d) 1 3.89 0.03 10 11 1.21 × 10 −12 3.87 0.01 (e) 1.4 3.67 0.03 10 11 9.4 × 10 −13 2.85 0.01 A2 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. glabrata growth on glucose (see B2 Fig). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (a) 0.1 0.97 0.02 10 10 2.38 × 10 −12 17.36 0.13 (b) 0.3 0.3 0.004 10 10 1.06 × 10 −13 15.81 0.08 (c) 1.4 0.17 0.0005 10 10 7.49 × 10 −15 7.31 0.01 (d) 2 0.12 0.0005 10 10 5.33 × 10 −15 5.21 0.006 (e) 4 0.11 0.0009 10 10 9.02 × 10 −15 2.85 0.008 A3 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on S. cerevisiae growth on sucrose (see B3 Fig). Initial sucrose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of sucrose/cell/min) (pmol of sucrose/ml) (cells/pmol of sucrose) (a) 0.0625 14.57 0.11 10 11 1.63 × 10 −11 19.14 0.04 (b) 0.25 6.55 0.03 10 11 2.18 × 10 −12 12.05 0.02 (c) 1 5.28 0.03 10 11 1.35 × 10 −12 4.46 0.01 (d) 4 4.43 0.01 10 11 5.31 × 10 −13 1.18 0.001 March 22, 2021 2/21 A4 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on E. coli growth on glucose (see B4 Fig). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (a) 0.05 1.57 0.04 10 11 6.35 × 10 −13 140.26 1.67 (b) 0.1 1.4 0.02 10 11 3.03 × 10 −13 99.36 0.79 (c) 0.2 0.94 0.01 10 11 1.31 × 10 −13 67.76 0.55 (d) 0.4 0.85 0.01 10 11 7.44 × 10 −14 37.9 0.22 A5 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. glabrata growth on glucose. For each of the initial glucose concentrations N0,1 = 0 .025%, N0,2 = 1%, and N0,3 = 2%, we list optimal parameter values Vi, Ki, and Yi (i = 1 .. 3) obtained by fitting the model (2) from the main text to the experimental data on C. glabrata growth on glucose (see B5 Fig). Note that we refer to Vi, Ki, and Yi in the context of Example 1 from the main text, while the same parameters were labelled V i G , Ki G , and Y i G in Example 2 from the main text, with the subscript G distinguishing C. glabrata data used to parameterise the model of competition with C. albicans .Initial glucose Fitted value Vi SE Fitted value Ki SE Fitted value Yi SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (1) 0.025 2.35 0.14 10 10 3.21 × 10 −11 19.32 0.26 (2) 1 0.19 0.001 10 10 1.6 × 10 −14 8.96 0.03 (3) 2 0.12 0.002 10 10 1.14 × 10 −14 5.36 0.05 March 22, 2021 3/21 A6 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. glabrata growth on glucose. For each of the initial glucose concentrations N0,1 = 0 .2%, N0,2 = 1%, and N0,3 = 1 .6%, we list optimal parameter values Vi, Ki, and Yi (i = 1 .. 3) obtained by fitting the model (2) from the main text to the experimental data on C. glabrata growth on glucose (see B6 Fig). Initial glucose Fitted value Vi SE Fitted value Ki SE Fitted value Yi SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (1) 0.2 0.51 0.009 10 10 4.34 × 10 −13 17 0.12 (2) 1 0.19 0.001 10 10 1.6 × 10 −14 8.96 0.03 (3) 1.6 0.14 0.001 10 10 9.97 × 10 −15 6.3 0.03 A7 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. albicans growth on glucose. For each of the initial glucose concentrations N0,1 = 0 .025%, N0,2 = 0 .5%, and N0,3 = 1%, we list optimal parameter values V i A , Ki A , and Y i A (i = 1 .. 3) obtained by fitting the model (2) from the main text to the experimental data on C. albicans growth on glucose (see B7 Fig). Initial glucose Fitted value V i A SE Fitted value Ki A SE Fitted value Y i A SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (1) 0.025 28.46 2.69 10 11 7.67 × 10 −10 16.94 0.34 (2) 0.5 4.77 0.13 10 11 6.25 × 10 −12 6.54 0.07 (3) 1 3.62 0.03 10 11 1.18 × 10 −12 3.8 0.02 A8 Table. RMSE values calculated for two approaches used to predict microbial growth inside the interval of initial nutrient concentrations. Since the RMSE values for the approach based on variable growth parameters are less than the RMSE values for the approach based on fixed growth parameters, we can conclude that our approach performs better. Initial nutrient concentration (%) Approach based on fixed growth parameters Approach based on variable growth parameters 0.1 7.1 ×10 7 6.5 ×10 6 0.4 1.6 ×10 8 3.3 ×10 7 1.2 1.1 ×10 8 1.3 ×10 7 1.8 4.1 ×10 8 4.9 ×10 7 March 22, 2021 4/21 A9 Table. RMSE values calculated for two approaches used to predict microbial growth outside the interval of initial nutrient concentrations. Since the RMSE values for the approach based on variable growth parameters are less than the RMSE values for the approach based on fixed growth parameters, we can conclude that our approach performs better. Initial nutrient concentration (%) Approach based on fixed growth parameters Approach based on variable growth parameters 0.1 7.1 ×10 7 1.6 ×10 7 1.8 4.1 ×10 8 3.8 ×10 7 A10 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. glabrata growth on glucose in the control medium (see C1 Fig (a)). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (a) 0.1 1.04 0.01 10 10 1 × 10 −12 14.97 0.09 (b) 0.3 0.47 0.004 10 10 2 × 10 −13 10.94 0.05 (c) 1.4 0.16 0.001 10 10 8 × 10 −15 6.26 0.03 (d) 2 0.13 0.0004 10 10 4 × 10 −15 5.04 0.01 (e) 4 0.11 0.0006 10 10 6 × 10 −15 2.64 0.01 A11 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. glabrata growth on glucose in the nutrient enriched medium (see C1 Fig (b)). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (a) 0.1 0.72 0.004 10 10 3 × 10 −13 18.43 0.05 (b) 0.3 0.48 0.002 10 10 9 × 10 −14 11.5 0.03 (c) 1.4 0.18 0.001 10 10 1 × 10 −14 5.57 0.03 (d) 2 0.13 0.0005 10 10 5 × 10 −15 4.84 0.01 (e) 4 0.1 0.0006 10 10 5 × 10 −15 2.94 0.01 March 22, 2021 5/21 A12 Table. Optimal parameter values obtained by fitting model (2) from the main text to the experimental data on C. glabrata growth on glucose in the pH buffered medium (see C1 Fig (c)). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE concentration (%) (pmol of glucose/cell/min) (pmol of glucose/ml) (cells/pmol of glucose) (a) 0.1 0.81 0.005 10 10 4 × 10 −13 18.52 0.05 (b) 0.3 0.48 0.004 10 10 2 × 10 −13 11.08 0.05 (c) 1.4 0.21 0.002 10 10 5 × 10 −14 5.29 0.04 (d) 2 0.12 0.0003 10 10 3 × 10 −15 5.98 0.01 (e) 4 0.09 0.0003 10 10 2 × 10 −15 3.65 0.01 A13 Table. Optimal parameter values obtained by fitting model (2) to the experimental data on E. coli growth on glucose (see D1 Fig (b)). Initial glucose Fitted V value SE Fitted K value SE Fitted Y value SE Fitted l value SE concentration (pmol of glucose/ (pmol of glucose/ (cells/ (min) (%) cell/min) ml) pmol of glucose) (a) 0.05 3.41 0.2 10 11 1 × 10 −11 202 17.7 267 41.9 (b) 0.1 2.5 0.16 10 11 2 × 10 −11 163 19 381 58.7 (c) 0.2 1.92 0.04 10 11 3 × 10 −11 140 8 531 30.5 (d) 0.4 1.42 0.02 10 11 1 × 10 −11 53.5 1.1 344 12.2 March 22, 2021 6/21 Appendix B. Supplementary Figures. B1 Fig. 0 4 8 12 16 20 24 0.2 0.4 0.6 0.8 0 4 8 12 16 20 24 0.5 11.5 0 4 8 12 16 20 24 0.5 11.5 20 4 8 12 16 20 24 0.5 11.5 22.5 0 4 8 12 16 20 24 0.5 11.5 22.5 4 8 12 16 5 × 10 6 1×10 7 5×10 7 1×10 8 Time (h) Candida albicans density (cells/ml) ×10 8 (a) 0.1% ×10 8 (b) 0.3% ×10 8 (c) 0.5% ×10 8 (d) 1% ×10 8 (e) 1.4% (f ) Growth of C. albicans on glucose. Experimental data on C. albicans growth in SC media with glucose together with the optimal fit of the model (2) from the main text for the following initial glucose concentrations: (a) 0.1%, (b) 0.3%, (c) 0.5%, (d) 1%, (e) 1.4%. Dots with error bars (mean values ± SE, where SE might be obscured by mean values, n = 2) represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (a) V = 15 .81 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 10 .6 cells/pmol of glucose; (b) V = 7 .05 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 8 .66 cells/pmol of glucose; (c) V = 5 .24 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 6 .56 cells/pmol of glucose; (d) V = 3 .89 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 3 .87 cells/pmol of glucose; (e) V = 3 .67 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 2 .85 cells/pmol of glucose (see A1 Table). (f) Experimental data on C. albicans growth in SC media with glucose together with the optimal fit of the model (2) from the main text as in (a) – (e) plotted on a March 22, 2021 7/21 single semi-logarithmic plot using the following corresponding colors: (a) black; (b) purple; (c) brown; (d) orange; (e) gray. B2 Fig. 0 4 8 12 16 20 24 0.2 0.4 0.6 0.8 11.2 0 4 8 12 16 20 24 0.5 11.5 22.5 30 4 8 12 16 20 24 1234560 12 24 36 48 60 1234560 12 24 36 48 60 12345674 8 12 16 5 × 10 6 1×10 7 5×10 7 1×10 8 5×10 8 Time (h) Candida glabrata density (cells/ml) ×10 8 (a) 0.1% ×10 8 (b) 0.3% ×10 8 (c) 1.4% ×10 8 (d) 2% ×10 8 (e) 4% (f ) Growth of C. glabrata on glucose. Experimental data on C. glabrata growth in SC media with glucose together with the optimal fit of the model (2) from the main text for the following initial glucose concentrations: (a) 0.1%, (b) 0.3%, (c) 1.4%, (d) 2%, (e) 4%. Dots with error bars (mean values ± SE, where SE might be obscured by mean values, n = 2) represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (a) V = 0 .97 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 17 .36 cells/pmol of glucose; (b) V = 0 .3 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 15 .81 cells/pmol of glucose; (c) V = 0 .17 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 7 .31 cells/pmol of glucose; (d) V = 0 .12 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 5 .21 cells/pmol of glucose; (e) V = 0 .11 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 2 .85 cells/pmol of glucose (see A2 Table). (f) Experimental data on C. glabrata growth in SC media with glucose together March 22, 2021 8/21 with the optimal fit of the model (2) from the main text as in (a) – (e) plotted on a single semi-logarithmic plot using the following corresponding colors: (a) black; (b) purple; (c) brown; (d) orange; (e) gray. B3 Fig. 0 12 24 36 48 60 72 84 0.1 0.2 0.3 0.4 0 12 24 36 48 60 72 84 0.2 0.4 0.6 0.8 10 12 24 36 48 60 72 84 0.2 0.4 0.6 0.8 11.2 1.4 0 12 24 36 48 60 72 84 0.2 0.4 0.6 0.8 11.2 1.4 4 8 12 16 20 1 × 10 6 5×10 6 1×10 7 5×10 7 1×10 8 Time (h) Saccharomyces cerevisiae density (cells/ml) ×10 8 (a) 0.0625% ×10 8 (b) 0.25% ×10 8 (c) 1% ×10 8 (d) 4% (e) Growth of S. cerevisiae in sucrose. Experimental data on S. cerevisiae growth in SC media with sucrose (previously published by us in Fig. S2 of ) together with the optimal fit of the model (2) from the main text for the following initial sucrose concentrations: (a) 0.0625%, (b) 0.25%, (c) 1%, (d) 4%. Dots with error bars (mean values ± SE, where SE might be obscured by mean values, n = 3) represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (a) V = 14 .57 pmol of sucrose/cell/min, K = 10 11 pmol of sucrose/ml, Y = 19 .14 cells/pmol of sucrose; (b) V = 6 .55 pmol of sucrose/cell/min, K = 10 11 pmol of sucrose/ml, Y = 12 .05 cells/pmol of sucrose; (c) V = 5 .28 pmol of sucrose/cell/min, K = 10 11 pmol of sucrose/ml, Y = 4 .46 cells/pmol of sucrose; (d) V = 4 .43 pmol of sucrose/cell/min, K = 10 11 pmol of sucrose/ml, Y = 1 .18 cells/pmol of sucrose (see A3 Table). (e) Experimental data on March 22, 2021 9/21 S. cerevisiae growth in SC media with sucrose together with the optimal fit of the model (2) from the main text as in (a) – (d) plotted on a single semi-logarithmic plot using the following corresponding colors: (a) black; (b) brown; (c) orange; (d) gray. B4 Fig. 0 4 8 12 16 20 24 12340 4 8 12 16 20 24 2460 4 8 12 16 20 24 24680 4 8 12 16 20 24 24684 8 12 16 20 1 × 10 7 5×10 7 1×10 8 5×10 8 1×10 9 Time (h) Escherichia coli density (cells/ml) ×10 8 (a) 0.05% ×10 8 (b) 0.1% ×10 8 (c) 0.2% ×10 8 (d) 0.4% (e) Growth of E. coli on glucose. Experimental data on E. coli growth in DM media with glucose together with the optimal fit of the model (2) from the main text for the following initial glucose concentrations: (a) 0.05%, (b) 0.1%, (c) 0.2%, (d) 0.4%. Dots with error bars (mean values ± SE, where SE might be obscured by mean values, n = 3) represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (a) V = 1 .57 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 140 .26 cells/pmol of glucose; (b) V = 1 .4 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 99 .36 cells/pmol of glucose; (c) V = 0 .94 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 67 .76 cells/pmol of glucose; (d) V = 0 .85 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 37 .9 cells/pmol of glucose (see A4 Table). (e) Experimental data on E. coli growth in DM media with glucose together March 22, 2021 10/21 with the optimal fit of the model (2) from the main text as in (a) – (d) plotted on a single semi-logarithmic plot using the following corresponding colors: (a) black; (b) brown; (c) orange; (d) gray. B5 Fig. 0 6 12 18 24 0.1 0.2 0.3 0.4 0 6 12 18 24 123450 6 12 18 24 123456Time (h) Candida glabrata density (cells/ml ×10 8) (a) 0.025% (b) 1% (c) 2% Growth of C. glabrata on glucose. Experimental data on C. glabrata growth in SC media with glucose together with the optimal fit of the model (2) from the main text for the following initial glucose concentrations: (a) 0.025%; (b) 1%; (c) 2%. Dots represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (a) V = 2 .35 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 19 .32 cells/pmol of glucose; (b) V = 0 .19 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 8 .96 cells/pmol of glucose; (c) V = 0 .12 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 5 .36 cells/pmol of glucose (see A5 Table). March 22, 2021 11/21 B6 Fig. 0 6 12 18 24 0.5 11.5 20 6 12 18 24 123450 6 12 18 24 123456Time (h) Candida glabrata density (cells/ml ×10 8) (a) 0.2% (b) 1% (c) 1.6% Growth of C. glabrata on glucose. Experimental data on C. glabrata growth in SC media with glucose together with the optimal fit of the model (2) from the main text for the following initial glucose concentrations: (1) 0.2%; (2) 1%; (3) 1.6%. Dots represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (1) V = 0 .51 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 17 cells/pmol of glucose; (2) V = 0 .19 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 8 .96 cells/pmol of glucose; (3) V = 0 .14 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 6 .3 cells/pmol of glucose (see A6 Table). B7 Fig. 0 6 12 18 24 0.1 0.2 0.3 0.4 0 6 12 18 24 0.5 11.5 20 6 12 18 24 0.5 11.5 2Time (h) Candida albicans density (cells/ml ×10 8) (a) 0.025% (b) 0.5% (c) 1% Growth of C. albicans on glucose. Experimental data on C. albicans growth in SC March 22, 2021 12/21 media with glucose together with the optimal fit of the model (2) from the main text for the following initial glucose concentrations: (1) 0.025%; (2) 0.5%; (3) 1%. Dots represent experimental growth data, a solid line is the optimal non-linear least-squares fit of the model (2) from the main text to the data, and dashed lines show 95% confidence bands for the model fit. Estimated parameters values are as follows: (1) V = 28 .46 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 16 .94 cells/pmol of glucose; (2) V = 4 .77 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 6 .54 cells/pmol of glucose; (3) V = 3 .62 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 3 .8 cells/pmol of glucose (see A7 Table). B8 Fig. 0 1 2 3 4 5 C. albicans (cell/ml) 10 8 00.5 11.5 OD 0246 C. glabrata (cell/ml) 10 8 00.5 11.5 OD 0510 15 E. coli (cell/ml) 10 8 00.1 0.2 0.3 0.4 0.5 0.6 OD (a) (b) (c) Conversion from optical density to CFU of (a) C. albicans , (b) C. glabrata ,and (c) E. coli growing on glucose. (a) and (b): Serial dilutions of an initial culture were prepared in a 48-well plate and their OD measured. The number of cell/ml in the initial culture was calculated using a haemocytometer. The fitted curve uses the function cell/ml = −OD × c/ (OD − OD max ), where OD is the optical density measured, and c and OD max are fitted parameters. For C. albicans we found OD max = 2 .5 ± 0.2and c = (1 .5 ± 0.2) × 10 8 cell/ml. For C. glabrata we found OD max = 2 .4 ± 0.2and c = (4 .3 ± 0.8) × 10 8 cell/ml. (c): Serial dilutions were prepared in a 96-well plate and their OD measured. The CFU in each dilution was determined using platecounting. The fitted line gives cell/ml = OD × c, with c = (2 .3 ± 0.3) × 10 9 cell/ml. March 22, 2021 13/21 Appendix C. Media acidification and alternative limiting nutrients. The growth assays used to estimate the relationship between the initial nutrient concentration and parameters V and Y (see Fig 1 in the main text) were conducted in batch cultures using SC media for Candida spp. and S.cerevisiae or DM media for E.coli (for details see Materials and methods section in the main text). Glucose was assumed to be the limiting nutrient in Candida spp. and E.coli experiments, while in S.cerevisiae experiments it was sucrose. Since in batch cultures nutrients are not replenished nor waste products removed, two additional factors, other than the limiting sugar, could affect microbial growth. First, the media can become acidified due to the secretion of organic acids and second, individual nutrient components such as nitrogen, could become limiting. To test whether the observed relationships between the initial nutrient concentration and parameters V and Y could result from either of these factors we repeated the growth assays for C. glabrata in control (C1 Fig (a)) and modified growth media (C1 Fig (b) and (c)). While the control media was the same as in B2 Fig, in addition we used enriched media at 1 .8×the concentration of the control media as well as control media that was additionally buffered with 0.1M potassium phosphate (pH 6) (see Materials and methods section in the main text). We found that negative relationships between initial glucose concentration and parameters V and Y (see C1 Fig (d) and (e), respectively) were observed in control (black line), nutrient enriched (gray line), and pH buffered (brown line) media. But is glucose the only limiting nutrient in our experiments? While it has previously been shown that yeast completely deplete 2% glucose in SC media , this might not be the case for 4% glucose. Indeed, minor increases in the estimated values of Y at 4% glucose were observed in nutrient enriched and pH buffered media compared to the control media (see C1 Fig (e)). This could suggest that nutrients, other than the carbon source, might be limiting growth in control media and so yield might be restricted by factors other than metabolic inefficiencies . To rule this scenario out we carried out an experiment using an engineered strain of S. cerevisiae (TM6) that has fully respiratory metabolism resulting in a much higher metabolic efficiency compared to the wild-type’s (CEN.PK2-1C) respiro-fermentative metabolism . We found that the TM6 strain was able to reach a substantially higher yield than the wild-type in 4% March 22, 2021 14/21 sucrose media (see C1 Fig (f) and (g)). Thus we argue that the observed decline in yield from low to high glucose environments is the result of metabolic inefficiencies known to constrain microbial growth at high nutrient concentrations . Collectively, these results indicate that the negative relationships between initial nutrient concentration and parameters V and Y (Fig 1 from the main text) are not sensitive to media acidification or enrichment. C1 Fig. 4 8 12 16 20 24 10 6 10 7 10 8 10 94812 16 20 24 10 6 10 7 10 8 10 94812 16 20 24 10 6 10 7 10 8 10 9123400.4 0.8 12340510 15 012 24 36 48 60 72 Time (h) CEN.PK2-1C TM6 5 x 10 8 5 x 10 7 5 x 10 6 5 x 10 5 C.F.U. / ml 0 0.5 1 1.5 2 CEN.PK2-1C TM6 Estimated Y (cell/pmol of nutrient) Time (h) Candida glabrata density (cells/ml) (a) (b) (c) (d) (e) (f ) (g) Initial nutrient concentration (%) Estimated V values (pmol of nutrient/cell/min) Estimated Y values (cells/pmol of nutrient) C.F.U./ml Time (h) Estimated Y values (cells/pmol of nutrient) (a) Growth of C. glabrata on glucose in the control medium. Experimental data on C. glabrata growth in the SC medium with glucose together with the optimal fit of the model (2) from the main text plotted on a single semi-logarithmic plot for the following initial glucose concentrations: 0.1% (black), 0.3% (purple), 1.4% (brown), 2% March 22, 2021 15/21 (orange), 4% (gray). Estimated parameters values are as follows: (0.1%) V = 1 .04 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 14 .97 cells/pmol of glucose; (0.3%) V = 0 .47 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 10 .94 cells/pmol of glucose; (1.4%) V = 0 .16 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 6 .26 cells/pmol of glucose; (2%) V = 0 .13 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 5 .04 cells/pmol of glucose; (4%) V = 0 .11 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 2 .64 cells/pmol of glucose (see A10 Table). (b) Growth of C. glabrata on glucose in the nutrient enriched medium. Experimental data on C. glabrata growth at 1 .8×the concentration of the SC control medium with glucose together with the optimal fit of the model (2) from the main text plotted on a single semi-logarithmic plot for the following initial glucose concentrations: 0.1% (black), 0.3% (purple), 1.4% (brown), 2% (orange), 4% (gray). Estimated parameters values are as follows: (0.1%) V = 0 .72 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 18 .43 cells/pmol of glucose; (0.3%) V = 0 .48 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 11 .5 cells/pmol of glucose; (1.4%) V = 0 .18 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 5 .57 cells/pmol of glucose; (2%) V = 0 .13 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 4 .84 cells/pmol of glucose; (4%) V = 0 .1 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 2 .94 cells/pmol of glucose (see A11 Table). (c) Growth of C. glabrata on glucose in the pH buffered medium. Experimental data on C. glabrata growth in the SC control medium with glucose with the addition of 0.1M potassium phosphate together with the optimal fit of the model (2) from the main text plotted on a single semi-logarithmic plot for the following initial glucose concentrations: 0.1% (black), 0.3% (purple), 1.4% (brown), 2% (orange), 4% (gray). Estimated parameters values are as follows: (0.1%) V = 0 .81 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 18 .52 cells/pmol of glucose; (0.3%) V = 0 .48 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 11 .08 cells/pmol of glucose; (1.4%) V = 0 .21 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 5 .29 cells/pmol of glucose; (2%) V = 0 .12 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 5 .98 cells/pmol of glucose; (4%) V = 0 .09 pmol of glucose/cell/min, K = 10 10 pmol of glucose/ml, Y = 3 .65 cells/pmol of glucose (see A12 Table). March 22, 2021 16/21 (d) Relationships between the maximal uptake rate parameter V and the initial nutrient concentration N0 observed for C. glabrata growing on glucose. Dots with error bars (estimated values ± SE, where SE might be obscured by data points) represent optimal estimates for maximal uptake rate parameters Vi (i = 1 .. 5) obtained by fitting the model (2) from the main text in the control medium (black), the nutrient enriched medium (gray), and the pH buffered medium (brown) to the experimental data on growth at the following initial glucose concentrations N0,i (i = 1 .. 5): 0.1%, 0.3%, 1.4%, 2%, 4%, and a solid line is the optimal non-linear least-squares fit of the function (8) from the main text to the plotted data. (e) Relationships between the yield parameter Y and the initial nutrient concentration N0 observed for C. glabrata growing on glucose. Dots with error bars (estimated values ± SE, where SE might be obscured by data points) represent optimal estimates for yield parameters Yi (i = 1 .. 5) obtained by fitting the model (2) from the main text in the control medium (black), the nutrient enriched medium (gray), and the pH buffered medium (brown) to the experimental data on growth at the following initial glucose concentrations N0,i (i = 1 .. 5): 0.1%, 0.3%, 1.4%, 2%, 4%, and a solid line is the optimal non-linear least-squares fit of the function (3) from the main text to the plotted data. (f) Growth of S. cerevisiae in sucrose. Growth data for the engineered strain of S. cereisiae TM6 (red) and the wild-type strain CEN.PK2-1C (blue) in 4% sucrose SC media. The data is present as mean values (solid lines) ± SE (dashed lines, might be obscured by mean values, n = 3). (g) Estimated values of the yield parameter Y for S. cerevisiae growing in sucrose. Bars with error bars (estimated values ± SE, where SE might be obscured by values) represent optimal estimates for the yield parameter Y obtained by fitting the model (2) from the main text to the experimental data on growth for the engineered strain of S. cereisiae TM6 (left, Y = 1 .72 cells/pmol of sucrose) and the wild-type strain CEN.PK2-1C (right, Y = 1 .18 cells/pmol of sucrose) in 4% sucrose media (see C1 Fig (f)). March 22, 2021 17/21 Appendix D. An extended mathematical model with a lag term. Due to its simplicity, our model (2) from the main text cannot accurately capture the lag phase of microbial growth (e.g. as in the E. coli ’s example, see B4 Fig). Here we extend the model (2) from the main text to include a lag-phase term as suggested by Baranyi et al. to obtain:  ˙N (t) = −q(N (t)) × B(t)˙B(t) = Y × q(N (t)) × B(t) × t2/(l2 + t2) t ∈ [0 , T ] , (2) where q(N (t)) is as in Eq (1) from the main text, and l denotes the lag time. Numerical solutions of the extended model (2) were fitted to the experimental data on E. coli following the same procedure as before (see Appendix A), to estimate the parameters V , Y , K, and l. As in the case of the simple model (2) from the main text, we observed that the estimated values of V and Y decrease with increasing initial nutrient concentration, whereas the estimated value of K did not depend on the initial nutrient concentration. The results are presented in D1 Fig. March 22, 2021 18/21 D1 Fig. 4 8 12 16 20 24 1 × 10 7 5×10 7 1×10 8 5×10 8 1×10 94812 16 20 24 1×10 7 5×10 7 1×10 8 5×10 8 1×10 90.1 0.2 0.3 0.4 012340.1 0.2 0.3 0.4 0100 200 Time (h) Escherichia coli density (cells/ml) (a) (b) (c) (d) Initial nutrient concentration (%) Estimated V values (pmol of nutrient/cell/min) Estimated Y values (cells/pmol of nutrient) (a) Growth of E. coli on glucose. Experimental data on E. coli growth in DM media with glucose together with the optimal fit of the model (2) from the main text plotted on a single semi-logarithmic plot for the following initial glucose concentrations: 0.05% (black), 0.1% (brown), 0.2% (orange), 0.4% (gray). Estimated parameters values are as follows: (0.05%) V = 1 .57 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 140 .26 cells/pmol of glucose; (0.1%) V = 1 .4 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 99 .36 cells/pmol of glucose; (0.2%) V = 0 .94 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 67 .76 cells/pmol of glucose; (0.4%) V = 0 .85 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 37 .9 cells/pmol of glucose (see B4 Fig (e) and A4 Table). (b) Growth of E. coli on glucose. Experimental data on E. coli growth in DM media with glucose together with the optimal fit of the model (2) plotted on a single semi-logarithmic plot for the following initial glucose concentrations: 0.05% (black), 0.1% (brown), 0.2% (orange), 0.4% (gray). Estimated parameters values are as follows: March 22, 2021 19/21 (0.05%) V = 3 .41 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 202 cells/pmol of glucose, l = 267 min; (0.1%) V = 2 .5 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 163 cells/pmol of glucose, l = 381 min; (0.2%) V = 1 .92 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 140 cells/pmol of glucose, l = 531 min; (0.4%) V = 1 .42 pmol of glucose/cell/min, K = 10 11 pmol of glucose/ml, Y = 53 .5 cells/pmol of glucose, l = 344 min (see A13 Table). (c) Relationships between the maximal uptake rate parameter V and the initial nutrient concentration N0 observed for E. coli growing on glucose. Dots with error bars (estimated values ± SE, where SE might be obscured by values) represent optimal estimates for maximal uptake rate parameters Vi (i = 1 .. 4) obtained by fitting the model (2) from the main text (black) and the model (2) (gray) to the experimental data on growth at the following initial glucose concentrations N0,i (i = 1 .. 4): 0.05%, 0.1%, 0.2%, 0.4%, and a solid line is the optimal non-linear least-squares fit of the function (8) from the main text to the plotted data. (d) Relationships between the yield parameter Y and the initial nutrient concentration N0 observed for E. coli growing on glucose. Dots with error bars (estimated values ± SE, where SE might be obscured by values) represent optimal estimates for yield parameters Yi (i = 1 .. 4) obtained by fitting the model (2) from the main text (black) and the model (2) (gray) to the experimental data on growth at the following initial glucose concentrations N0,i (i = 1 .. 4): 0.05%, 0.1%, 0.2%, 0.4%, and a solid line is the optimal non-linear least-squares fit of the function (3) from the main text to the plotted data. March 22, 2021 20/21 References Lindsay RJ, Pawlowska BJ, Gudelj I. Privatization of public goods can cause population decline. Nat Ecol Evol. 2019;3:1206–1216. 2. Burtner CR, Murakami CJ, Kennedy BK, Kaeberlein M. A molecular mechanism of chronological aging in yeast. Cell cycle. 2009;8:1256–1270. 3. Pfeiffer T, Schuster S, Bonhoeffer S. Cooperation and Competition in the Evolution of ATP-Producing Pathways. Science. 2001;292(5516):504–507. 4. Otterstedt K, Larsson C, Bill RM, St˚ ahlberg A, Boles E, Hohmann S, Gustafsson L. Switching the mode of metabolism in the yeast Saccharomyces cerevisiae .EMBO Rep. 2004;5:532–537. 5. Postma E, Verduyn C, Scheffers W. A., Van Dijken JP. Enzymic analysis of the crabtree effect in glucose-limited chemostat cultures of Saccharomyces cerevisiae .Appl Environ Microbiol. 1989;55:468–477. 6. Baranyi J, Roberts TA, McClure P. A non-autonomous differential equation to model bacterial growth. Food Microbiol. 1993;10:43–59. March 22, 2021 21/21
14791	https://www.bcbsm.com/amslibs/content/dam/public/mpr/mprsearch/pdf/2185449.pdf	1 Medical Policy Joint Medical Policies are a source for BCBSM and BCN medical policy information only. These documents are not to be used to determine benefits or reimbursement. Please reference the appropriate certificate or contract for benefit information. This policy may be updated and is therefore subject to change. Current Policy Effective Date: 5/1/25 (See policy history boxes for previous effective dates) Title: GT-Maturity-Onset Diabetes of the Young (MODY) Description/Background Maturity-onset diabetes of the young (MODY) is a clinically and genetically heterogeneous form of diabetes that is characterized by impaired insulin secretion.1 MODY, which is inherited in an autosomal dominant manner, is estimated to account for between 1% and 5% of non-insulin-dependent diabetes cases.1,2 MODY is diagnosed clinically in patients with hyperglycemia or diabetes who have a family history of abnormal glucose metabolism in at least 2 consecutive generations, with the patient or 1 or more family members diagnosed before age 25.2-4 Currently, there are nine known subtypes of MODY that differ with regard to average age at disease presentation, pattern of hyperglycemia, response to the various treatment modalities, and the presence of extra pancreatic features.5 The penetrance of disease-causing variants may also vary depending on the subtype, but the overall penetrance of MODY gene variants is reported to be > 90%.4 The most common types of MODY are MODY2 and MODY3.2 MODY2 is caused by variants in the glucokinase gene (GCK), and is characterized by mild fasting hyperglycemia that is generally stable but persistent.3 It is suspected that GCK variants are relatively common among the general population, but that most carriers are asymptomatic and, thus, undiagnosed.2,6 Because of the mild manifestations, complications are rare in MODY2 patients.7,8 Furthermore, MODY2 does not typically require treatment with medication or insulin, although treatment may be warranted in affected women during pregnancy, depending on the growth of the fetus.1,2,8,9 MODY 3 is caused by mutations of the hepatocyte nuclear factor 1alpha (HNF1-alpha) gene, a homeobox gene on human chromosome 12.11 HNF1α is a transcription factor (also known as transcription factor 1, TCF1) that is thought to control a regulatory network (including, among other genes, HNF1α) important for differentiation of beta cells. One of the incentives for 2 diagnosing it is that insulin may be discontinued or deferred in favor of oral sulfonylureas. Some people treated with insulin for years due to a presumption of type 1 diabetes have been able to switch to oral medication and discontinue injections. Testing for the major types of MODY often involves analysis of the coding sequences, intron-exon boundaries, and minimal promoter of each MODY gene by direct sequencing.8,27 However, gene scanning techniques, such as denaturing high-performance liquid chromatography (dHPLC) and single-strand conformation polymorphism (SSCP) analysis, may be used prior to sequence analysis.28-32 Testing for deletions and duplications of the genes for MODY is performed by multiplex ligation-dependent probe amplification (MLPA).26,33 Regulatory Status Clinical laboratories may develop and validate tests in-house and market them as a laboratory service; laboratory-developed tests must meet the general regulatory standards of the Clinical Laboratory Improvement Amendments (CLIA). Laboratories that offer laboratory-developed tests must be licensed by the CLIA for high-complexity testing. To date, the U.S. FDA has chosen not to require any regulatory review of these tests. Therefore, no FDA approvals will be found or listed on their website. Medical Policy Statement The safety and effectiveness of genetic testing for maturity-onset diabetes of the young (MODY) have been established. It may be considered a useful diagnostic option when indicated for individuals meeting specified guidelines. Inclusionary and Exclusionary Guidelines Inclusions: For the diagnosis of MODY in individuals with: • Early-onset diabetes in children or young adulthood (typically age <45 years); AND • Have a family history of diabetes in successive generations (suggestive of an autosomal dominant pattern of inheritance); AND • Any one of the following atypical features for Type 1 diabetes: o Absence of pancreatic islet autoantibodies (e.g., GAD and IA2) o Evidence of endogenous insulin production beyond the honeymoon period (i.e., 3-5 years after the onset of diabetes) o Measurable C-peptide in the presence of hyperglycemia (C-peptide ≥0.60 ng/mL or 0.2 nmol/L) o Low insulin requirement for treatment (i.e., <0.5 U/kg/d) o Lack of ketoacidosis when insulin is omitted from treatment; OR • Any one of the following atypical features for Type 2 Diabetes: o Lack of significant obesity 3 o Lack of acanthosis nigricans o Normal triglyceride levels and/or normal or elevated high-density lipoprotein cholesterol (HDL-C) OR • Any one of the following: o Mild, stable fasting hyperglycemia that does not progress or respond appreciably to pharmacologic therapy o Extreme sensitivity to sulfonylureas o A personal history or family history of neonatal diabetes or neonatal hypoglycemia Exclusions: • MODY testing for all other indications not meeting the criteria above. CPT/HCPCS Level II Codes (Note: The inclusion of a code in this list is not a guarantee of coverage. Please refer to the medical policy statement to determine the status of a given procedure.) Established codes: 81405 81406 Other codes (investigational, not medically necessary, etc.): N/A Note: Individual policy criteria determine the coverage status of the CPT/HCPCS code(s) on this policy. Codes listed in this policy may have different coverage positions (such as established or experimental/investigational) in other medical policies. Rationale Evidence reviews assess the clinical evidence to determine whether the use of a technology improves the net health outcome. Broadly defined, health outcomes are length of life, quality of life, and ability to function including benefits and harms. Every clinical condition has specific outcomes that are important to patients and to managing the course of that condition. Validated outcome measures are necessary to ascertain whether a condition improves or worsens; and whether the magnitude of that change is clinically significant. The net health outcome is a balance of benefits and harms. To assess whether the evidence is sufficient to draw conclusions about the net health outcome of a technology, 2 domains are examined: the relevance and the quality and credibility. To be relevant, studies must represent 1 or more intended clinical use of the technology in the intended population and compare an effective and appropriate alternative at a comparable intensity. For some conditions, the alternative will be supportive care or surveillance. The quality and credibility of the evidence depend on study design and conduct, minimizing bias and confounding that can generate incorrect findings. The randomized controlled trial is preferred to assess efficacy; however, in some circumstances, nonrandomized studies may be adequate. Randomized controlled trials are rarely large enough or long enough to capture less 4 common adverse events and long-term effects. Other types of studies can be used for these purposes and to assess generalizability to broader clinical populations and settings of clinical practice. Maturity-Onset Diabetes of the Young (MODY) Clinical Context and Therapy Purpose The purpose of testing for MODY variants is to establish a specific diagnosis and whether that specific diagnosis has direct implications for the individual’s medical management? The following PICO was used to select literature to inform this review. Populations The relevant population of interest is individuals with hyperglycemia or non-insulin dependent diabetes. Interventions The testing being considered is MODY. Comparators The following are alternatives to MODY gene testing include evaluating fasting plasma glucose (FPG) levels, oral glucose tolerance testing, glycosylated hemoglobin (HbA1c) levels, glutamatedecarboxylase (GAD) and islet antigen-2 (IA-2) antibodies. Outcomes The general outcomes of interest are improved medical management. Study Selection Criteria Methodologically credible studies were selected using the following principles: • To assess efficacy outcomes, comparative controlled prospective trials were sought, with a preference for RCTs; • In the absence of such trials, comparative observational studies were sought, with a preference for prospective studies. • To assess long-term outcomes and adverse events, single-arm studies that capture longer periods of follow-up and/or larger populations were sought. • Consistent with a 'best available evidence approach,' within each category of study design, studies with larger sample sizes and longer durations were sought. • Studies with duplicative or overlapping populations were excluded. Review of Evidence Pearson et al (2003) conducted a randomized crossover trial to assess whether different causes for diabetes change the response to oral hypoglycaemic therapy.1 In a few cases, patients with diabetes caused by mutations in the hepatocyte nuclear factor 1alpha (HNF-1alpha) gene have been described as sensitive to the hypoglycaemic effects of sulphonylureas. In this trial, glicazide and metformin were used in 36 patients either with diabetes caused by HNF-1alpha mutations or type 2 diabetes, who were matched for body-mass index and fasting plasma glucose. The primary outcome was reduction in fasting plasma glucose. Analysis was by intention to treat. The authors assessed possible mechanisms for sulphonylurea sensitivity through insulin sensitivity, insulin secretory response to glucose and 5 tolbutamide, and tolbutamide clearance. Patients with HNF-1alpha diabetes had a 5.2-fold greater response to gliclazide than to metformin (fasting plasma glucose reduction 4.7 vs 0.9 mmol/L, p=0.0007) and 3.9-fold greater response to gliclazide than those with type 2 diabetes (p=0.002). Patients with HNF-1alpha diabetes had a strong insulin secretory response to intravenous tolbutamide despite a small response to intravenous glucose, and were more insulin sensitive than those with type 2 diabetes. Sulphonylurea metabolism was similar in both patient groups. Garin et al (2008) conducted a study to characterize glucokinase (GCK) alterations in maturity-on-set diabetes of the young 2 (MODY2)-suspected patients and to investigate their clinical characteristics in relation to the parental origin of the mutation.2 A group of 57 unrelated Spanish patients presenting with MODY2 phenotype were studied. Patients without mutation in the coding region of the GCK gene were screened for rearrangements by Multiplex Ligation-dependent Probe Amplification (MLPA). After classification according to the parental origin of the mutation, clinical characteristics were compared between the groups. A point mutation or small deletion or insertion of the GCK gene was detected in 47 patients (82.5%); 19 mutations were novel. In addition, a whole-gene deletion by MLPA was found. Patients carrying a GCK gene defect and those with MODY of unknown genetic origin shows similar phenotypes. Comparison of clinical parameters according to the origin of the mutation did not show any differences in the birth weight (BW) nor in age at diagnosis. Patients who inherited the mutation from the father had higher fasting glucose levels at diagnosis. Ellard et al (2008) along with 22 clinicians and scientists held a workshop to discuss clinical criteria for testing and the interpretation of molecular genetic test results in mutations in the GCK and HNF1A genes which are the most common cause of the MODY diabetic.3 GCK encodes the glucokinase enzyme, which acts as the pancreatic glucose sensor, and mutations result in stable, mild fasting hyperglycaemia. A progressive insulin secretory defect is seen in patients with mutations in theHNF1A and HNF4A genes encoding the transcription factors hepatocyte nuclear factor-1 alpha and -4alpha. A molecular genetic diagnosis often changes management, since patients with GCK mutations rarely require pharmacological treatment and HNF1A/4A mutation carriers are sensitive to sulfonylureas. These monogenic forms of diabetes are often misdiagnosed as type 1 or 2 diabetes. Best practice guidelines for genetic testing were developed to guide testing and reporting of results. Bellanne-Chantelot et al (2011) conducted a retrospective multicenter study including 487 unrelated patients referred because of suspicion of MODY3.4 Genetic analysis identified 196 MODY3 and 283 non-MODY3 cases. Criteria associated with MODY3 were assessed by multivariate analysis. The capacity of the model to predict MODY3 diagnosis was assessed by the area under the receiver-operating characteristic curve and was further validated in an independent sample of 851 patients (165 MODY3 and 686 non-MODY3). In the MODY3 patients, diabetes was revealed by clinical symptoms in 25% of the cases and was diagnosed by screening in the others. Age at diagnosis of diabetes was more than 25 years in 40% of the MODY3 patients. There was considerable variability and overlap of all assessed parameters in MODY3 and non-MODY3 patients. The best predictive model was based on criteria available at diagnosis of diabetes, including age, body mass index, number of affected generations, presence of diabetes symptoms, and geographical origin. The area under the curve of the receiver-operating characteristic analysis was 0.81. When sensitivity was set to 90%, specificity was 49%. 6 According to McDonald et al (2011), MODY is rare (∼1% diabetes) and may be misdiagnosed as Type 1 diabetes and inappropriately treated with insulin.5 Type 1 diabetes is characterized by the presence of islet autoantibodies, including glutamatedecarboxylase (GAD) and islet antigen-2 (IA-2) antibodies. The prevalence of islet autoantibodies is unknown in maturity-onset diabetes of the young and may have the potential to differentiate this form of diabetes from Type 1 diabetes. In this study, plasma GAD and IA-2 antibodies was measured in 508 patients with the most common forms of maturity-onset diabetes of the young (GCK: n = 227; HNF1A: n = 229; HNF4A: n = 52) and 98 patients with newly diagnosed Type 1 diabetes (diagnosed<6 months). Autoantibodies were considered positive if ≥99th centile of 500 adult control subjects. GAD and/or IA-2 antibodies were present in 80/98 (82%) patients with Type 1 diabetes and 5/508 (<1%) patients with maturity-onset diabetes of the young. In the cohort with Type 1 diabetes, both GAD and IA-2 antibodies were detected in 37.8% of patients, GAD only in 24.5% and IA-2 only in 19.4%. All five patients with maturity-onset diabetes of the young with detectable antibodies had GAD antibodies and none had detectable IA-2 antibodies. In 2019, GoodSmith et al. assessed the cost-effectiveness of genetic testing, preceded by biomarker screening and followed by cascade genetic testing of first-degree relatives, for subtypes of MODY in U.S. pediatric patients with diabetes.6 Simulation models of distinct forms of diabetes were used to forecast the clinical and economic consequences of a systematic genetic testing strategy compared with usual care over a 30-year time horizon. In the genetic testing arm, patients with MODY received treatment changes (sulfonylureas for HNF1A- and HNF4A-MODY associated with a 1.0% reduction in HbA1c; no treatment for GCK-MODY). Study outcomes included costs, life expectancy (LE), and quality-adjusted life years (QALY). The strategy of biomarker screening and genetic testing was cost-saving as it increased average quality of life (+0.0052 QALY) and decreased costs (-$191) per simulated patient relative to the control arm. Adding cascade genetic testing increased quality-of-life benefits (+0.0081 QALY) and lowered costs further (-$735). Urbanova et al (2020) presented a summary of the actual diagnostic possibilities and differentiation of MODY (Maturity-Onset Diabetes of the Young) from gestational diabetes (GDM) found during routine screening, and specific aspects of care and treatment of MODY during pregnancy and early postpartum period.7 Many patients with MODY, especially the glucokinase MODY, can be first diagnosed during pregnancy. It is estimated that MODY patients account for up to 5% of GDM cases found in routine screening of GDM. MODY should be considered in lean women around 25 years of age, with a positive family history of diabetes in one of the parents. The differentiation of MODY from GDM is of particular importance not only for the different management and goals of antidiabetic therapy and planning ultrasound controls of fetal growth during pregnancy, but also because of the risk of hyperinsulinemic hypoglycemia in newborns. The authors concluded that recognition of MODY during pregnancy and adherence to existing recommendations concerning specific care of these patients is essential for the optimal course of their pregnancy and proper care of the newborn in the early postpartum period. Summary of Evidence MODY is a dominantly inherited form of diabetes that is characterized by defective insulin secretion. The clinical criteria for a diagnosis of MODY are well-established and indicate that the diagnosis of MODY should be considered in certain patients. Studies of the clinical validity of MODY gene testing demonstrate that there is a clear association between variants in 7 the known MODY genes and the MODY phenotype. In addition, it is apparent that the phenotype of MODY patients, including the age at disease onset and the clinical manifestations, varies depending on the type of MODY each patient has. The evidence supporting the clinical utility of MODY gene testing suggests that the optimal treatment of MODY patients may depend on the subtype. For example, it is possible that MODY3 patients may be able to discontinue or avoid insulin treatment, since these individuals are often sensitive to sulphonylureas. Also, patients in whom a diagnosis of MODY2 is established may be able to discontinue treatment with either insulin or oral hypoglycemic agents, since they are often able to manage the disease with diet and lifestyle changes alone. In addition to the implications for treatment, families report that MODY 2 and 3 gene testing is useful for future or family planning and to reduce uncertainty and anxiety in those at risk of developing the disease. The evidence for MODY 2 & 3 testing is sufficient to determine that the technology results in an improvement in the net health outcome. SUPPLEMENTAL INFORMATION PRACTICE GUIDELINES AND POSITION STATEMENTS American Diabetes Association (ADA)9 In 2024, the ADA published Diagnosis and Classification of Diabetes guidelines. The following recommendations were made: • 2.24a Regardless of current age, all people diagnosed with diabetes in the first 6 months of life should have immediate genetic testing for neonatal diabetes. • A 2.24b Children and young adults who do not have typical characteristics of type 1 or type 2 diabetes and who often have a family history of diabetes in successive generations (suggestive of an autosomal dominant pattern of inheritance) should have genetic testing for maturity-onset diabetes of the young (MODY). • A 2.24c In both instances, consultation with a center specializing in diabetes genetics is recommended to understand the significance of genetic mutations and how best to approach further evaluation, treatment, and genetic counseling. European Molecular Genetics Quality Network (EMQN)3 The EMQN (2008) developed guidelines for diagnostic and predictive testing for MODY1, MODY2, and MODY3. MODY2 (due to GCK variants) should be considered in patients with an FPG level ≥ 5.5 mmol/L that is persistent and stable, an HbA1c that is just above the upper limit of normal but rarely exceeds 7.5%, and a small (< 4.6 mmol/L) 2-hour increment on oral glucose tolerance test. In addition, MODY2 patients may have a parent diagnosed with hyperglycemia or type 2 DM. MODY3 (due to HNF1A variants) should be considered in patients with young-onset non-insulin-dependent diabetes, a family history of DM in at least 2 consecutive generations, and an absence of autoimmune markers of DM. In addition, MODY3 patients often have glycosuria and may exhibit a marked sensitivity to sulphonylureas. Testing for MODY1 (caused by HNF4A variants) should be considered in patients who test negative for an HNF1A gene variant. Specific recommendations regarding the reporting of test results are also provided by the EMQN. ONGOING AND UNPUBLISHED CLINICAL TRIALS Some currently unpublished trials that might influence this policy are listed in Table 1. Table 1. Clinical Trials 8 NCT No. Trial Name Planned Enrollment Completion Date Ongoing NCT05586594 Identifying Maturity-onset Diabetes of the Young in Emirati Patients 150 Aug 2025 NCT06111833 Optimized Diagnosis and Precision Medicine of MODY 1500 May 2028 Unpublished NCT03246828 Glucagon in MODY (Maturity Onset Diabetes of the Young) 10 Aug 2022 NCT05918484 Usefulness of Continuous Glucose Monitoring in MODY Diagnosis (UCMODY) 4000 Dec 2023 Government Regulations National: There is no NCD for this testing. Local: There is no LCD for this testing. (The above Medicare information is current as of the review date for this policy. However, the coverage issues and policies maintained by the Centers for Medicare & Medicare Services [CMS, formerly HCFA] are updated and/or revised periodically. Therefore, the most current CMS information may not be contained in this document. For the most current information, the reader should contact an official Medicare source.) Related Policies N/A References 1. Pearson ER, Starkey BJ, Powell RJ, et al. Genetic cause of hyperglycemia and response to treatment in diabetes. Lancet. Oct 2003;362(9392):1275-1281. 2. Garin I, Rica I, Estalella I, et al. Haploinsufficiency at GCK gene is not a frequent event in MODY2 patients. Clin Endocrinol. June 2008;68(6):873-878. 3. Ellard S, Bellanne-Chantelot C, Hattersley AT. Best practice guidelines for the molecular genetic diagnosis of maturity-onset diabetes of the young. Diabetologia. April 2008;51(4):546-553. 4. Bellanne-Chantelot C, Levy DJ, Carette C, et al. Clinical characteristics and diagnostic criteria of maturity-onset diabetes of the young (MODY) due to molecular anomalies of the HNF1A gene. J Clin Endorinol Metab. Aug 2011;96(8):E1346-1351. 5. McDonald TJ, Colclough K, Brown R, et al. Islet autoantibodies can discriminate maturity-onset diabetes of the young (MODY) from type 1 diabetes. Diabet Med. Sep 2011;28(9):1028-1033. 6. GoodSmith MS, Skandari MR, Huang ES, Naylor RN. The impact of biomarker screening and cascade genetic testing on the cost-effectiveness of MODY genetic testing. Diabetes Care. Dec 2019; 42(12):2247-2255. 9 7. Urbanova J, Brunerova L, Nunes MA, Broz J. MODY diabetes and screening of gestational diabetes. Ceska Gynekol. 2020;85(2):124-130. 8. Balasubramanyam A, Nathan DM, et al. Classification of diabetes mellitus and genetic diabetic syndromes. UpToDate. 2024. 9. American Diabetes Association. Diagnosis and classification of diabetes: standards of care in diabetes-2024. Diabetes Care. 2024;47(1):S20-S42. 10. Frayling TM, Evans JC, Bulman MP, et al. Molecular and Clinical Characterization of Mutations in Transcription Factors. Diabetes. 2001;50(1):S94-S100. 11. Naylor R, Johnson AK, Gaudio D. Maturity-onset diabetes of the young overview. National Library of Medicine. May 2018. The articles reviewed in this research include those obtained in an Internet based literature search for relevant medical references through January 2025, the date the research was completed. 10 Joint BCBSM/BCN Medical Policy History Policy Effective Date BCBSM Signature Date BCN Signature Date Comments 5/1/24 3/4/24 Joint policy established 5/1/25 2/18/25 Routine policy maintenance, no change in policy status. Vendor managed: N/A (ds) Next Review Date: 1st Qtr., 2026 Pre-Consolidation Medical Policy History Original Policy Date Comments BCN: Revised: BCBSM: Revised: 11 BLUE CARE NETWORK BENEFIT COVERAGE POLICY: GT-MATURITY-ONSET DIABETES OF THE YOUNG (MODY) I. Coverage Determination: Commercial HMO (includes Self-Funded groups unless otherwise specified) Per policy BCNA (Medicare Advantage) See government section BCN65 (Medicare Complementary) Coinsurance covered if primary Medicare covers the service. II. Administrative Guidelines: • The member's contract must be active at the time the service is rendered. • Coverage is based on each member’s certificate and is not guaranteed. Please consult the individual member’s certificate for details. Additional information regarding coverage or benefits may also be obtained through customer or provider inquiry services at BCN. • The service must be authorized by the member's PCP except for Self-Referral Option (SRO) members seeking Tier 2 coverage. • Services must be performed by a BCN-contracted provider, if available, except for Self-Referral Option (SRO) members seeking Tier 2 coverage. • Payment is based on BCN payment rules, individual certificate and certificate riders. • Appropriate copayments will apply. Refer to certificate and applicable riders for detailed information. • CPT - HCPCS codes are used for descriptive purposes only and are not a guarantee of coverage.
14792	https://www.tiger-algebra.com/drill/m~2-2m-3=0/	Home Topics Add to Home Contact TigerMilk.Education GmbH Privacy policy Terms of service Copyright Ⓒ 2013-2025tiger-algebra.com ☰ Tiger Algebra Calculator Formatting help Add to Home Tiger Algebra Solver Are you looking for... Enter an equation or problem Camera input is not recognized! a x y / \|abs\| ( ) 7 8 9 4 5 6 - % 1 2 3 + < 0 , . = abc a b c d e f g h i j k l m n o p q r s t u v w x y z ␣ . Solution - Quadratic equations m=3 m=-1 Other Ways to Solve Quadratic equations Solving quadratic equations by factoring Solving quadratic equations using the quadratic formula Solving quadratic equations by completing the square Step by Step Solution Step by step solution : Step 1 : Trying to factor by splitting the middle term 1.1 Factoring m2-2m-3 The first term is, m2 its coefficient is 1 .The middle term is, -2m its coefficient is -2 .The last term, "the constant", is -3 Step-1 : Multiply the coefficient of the first term by the constant 1 • -3 = -3 Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is -2 . \| \| \| \| \| \| \| \| --- --- --- \| \| -3 \| + \| 1 \| = \| -2 \| That's it \| Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 1 m2 - 3m + 1m - 3Step-4 : Add up the first 2 terms, pulling out like factors : m • (m-3) Add up the last 2 terms, pulling out common factors : 1 • (m-3) Step-5 : Add up the four terms of step 4 : (m+1) • (m-3) Which is the desired factorization Equation at the end of step 1 : (m + 1) • (m - 3) = 0 Step 2 : Theory - Roots of a product : 2.1 A product of several terms equals zero. When a product of two or more terms equals zero, then at least one of the terms must be zero. We shall now solve each term = 0 separately In other words, we are going to solve as many equations as there are terms in the product Any solution of term = 0 solves product = 0 as well. Solving a Single Variable Equation : 2.2 Solve : m+1 = 0 Subtract 1 from both sides of the equation : m = -1 Solving a Single Variable Equation : 2.3 Solve : m-3 = 0 Add 3 to both sides of the equation : m = 3 Supplement : Solving Quadratic Equation Directly m2-2m-3 Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula Parabola, Finding the Vertex : 3.1 Find the Vertex of y = m2-2m-3Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero). Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. For any parabola,Am2+Bm+C,the m -coordinate of the vertex is given by -B/(2A) . In our case the m coordinate is 1.0000 Plugging into the parabola formula 1.0000 for m we can calculate the y -coordinate : y = 1.0 1.00 1.00 - 2.0 1.00 - 3.0 or y = -4.000 Parabola, Graphing Vertex and X-Intercepts : Root plot for : y = m2-2m-3 Axis of Symmetry (dashed) {m}={ 1.00} Vertex at {m,y} = { 1.00,-4.00} m -Intercepts (Roots) : Root 1 at {m,y} = {-1.00, 0.00} Root 2 at {m,y} = { 3.00, 0.00} Solve Quadratic Equation by Completing The Square 3.2 Solving m2-2m-3 = 0 by Completing The Square . Add 3 to both side of the equation : m2-2m = 3Now the clever bit: Take the coefficient of m , which is 2 , divide by two, giving 1 , and finally square it giving 1 Add 1 to both sides of the equation : On the right hand side we have : 3 + 1 or, (3/1)+(1/1) The common denominator of the two fractions is 1 Adding (3/1)+(1/1) gives 4/1 So adding to both sides we finally get : m2-2m+1 = 4Adding 1 has completed the left hand side into a perfect square : m2-2m+1 = (m-1) • (m-1) = (m-1)2 Things which are equal to the same thing are also equal to one another. Since m2-2m+1 = 4 and m2-2m+1 = (m-1)2 then, according to the law of transitivity, (m-1)2 = 4We'll refer to this Equation as Eq. #3.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of (m-1)2 is (m-1)2/2 = (m-1)1 = m-1Now, applying the Square Root Principle to Eq. #3.2.1 we get: m-1 = √ 4 Add 1 to both sides to obtain: m = 1 + √ 4 Since a square root has two values, one positive and the other negative m2 - 2m - 3 = 0 has two solutions: m = 1 + √ 4 or m = 1 - √ 4 Solve Quadratic Equation using the Quadratic Formula 3.3 Solving m2-2m-3 = 0 by the Quadratic Formula . According to the Quadratic Formula, m , the solution for Am2+Bm+C = 0 , where A, B and C are numbers, often called coefficients, is given by : - B ± √ B2-4AC m = ———————— 2A In our case, A = 1 B = -2 C = -3 Accordingly, B2 - 4AC = 4 - (-12) = 16Applying the quadratic formula : 2 ± √ 16 m = ————— 2Can √ 16 be simplified ?Yes! The prime factorization of 16 is 2•2•2•2 To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).√ 16 = √ 2•2•2•2 =2•2•√ 1 = ± 4 • √ 1 = ± 4 So now we are looking at: m = ( 2 ± 4) / 2Two real solutions:m =(2+√16)/2=1+2= 3.000 or:m =(2-√16)/2=1-2= -1.000 Two solutions were found : m = 3 m = -1 How did we do? Please leave us feedback. Why learn this Terms and topics Solving quadratic equations by completing the square Solving quadratic equations using the quadratic formula Parabola finding vertex and X intercepts Simplifying radicals Related links Simplifying radicals \| Khan Academy Radicals: Introduction & Simplification \| Purplemath Worked example: completing the square (leading coefficient ≠ 1) (video) \| Khan Academy Quadratic Formula Explained \| Purplemath Graphing Quadratic Functions: More Examples Latest Related Drills Solved
14793	https://wiki.lzhbrian.me/notes/differientiable-sampling-and-argmax	📝 Tzu-Heng's wiki Ctrl``K Differientiable Sampling and Argmax WIP, last updated: 2019.12.6 Introduction Softmax is a commonly used function for turning an unnormalized log probability into a normalized probability (or categorical distribution). π=softmax(o)=∑jeojeo,oj∈(−∞,+∞) Say o is the output of a neural network before softmax, we call o the unnormalized log probability. After softmax, we usually sample from this categorical distribution, or taking an argmax function to select the index. However, one can notice that neither the sampling nor the argmax is differientiable. Researchers have proposed several works to make this possible. I am going to discuss them here. Sampling I will introduce Gumbel Softmax [1611.01144], which have made the sampling procedure differentiable. Gumbel Max First, we need to introduce Gumbel Max. In short, Gumbel Max is a trick to use gumbel distribution to sample a categorical distribution. Say we want to sample from a categorical distribution π. The usual way of doing this is using π to separate [0,1] into intervals, sampling from a uniform distribution U∼[0,1], and see where it locates. The Gumbel Max trick provides an alternative way of doing this. It use Reparameterization Trick to avoid the stochastic node during backpropagation. y=argimax(oi+gi) where gi∼Gumbel(0,1), which can be sampled by −log(−log(Uniform[0,1])). We can prove that y is distributed according to π. Prove that y=argmaxi(oi+gi), where gi∼Gumbel(0,1) which can be sampled by −log(−log(Uniform[0,1])) is distributed with π=softmax(oi)=∑jeojeoi Prerequisites Gumbel Distribution (param by location μ, and scale β>0) (wikipedia) CDF: F(x;μ,β)=e−e(x−μ)/β PDF: f(x;μ,β)=β1e−(z+e−z),z=βx−μ Mean: E(X)=μ+γβ,γ≈0.5772is the Euler–Mascheroni constant. Quantile Function: Q(p)=μ−βlog(−log(p))(Quantile Function is used to sample random variables from a distribution given CDF, it is also called inverse CDF) Proof We actually want to prove that Gumbel(μ=oi,β=1) is distributed with πi=∑jeojeoi. We can find that Gumbel(μ=oi,β=1) has the following PDF and CDF f(x;μ,1)F(x;μ,1)=e−(x−μ)–e−(x−μ)=e−e−(x−μ) .Then, the probability that all other πj=i are less than πi is: Pr(πi is the largest∣πi,{oj})=j=i∏e−e−(πi−oj) We know the marginal distribution over πi and we are able to integrate it out to find the overall probability: (p(x)=∫yp(x,y)dy=∫yp(x∣y)p(y)dy) Pr(i is largest∣{oj})=∫e−(πi−oi)−e−(πi−oi)×j=i∏e−e−(πi−oj)dπi=∫e−πi+oi−e−πi∑jeojdπi=∑jeojeoi which is exactly a softmax probablity. QED. Reference: Gumbel Softmax Notice that there is still an argmax in Gumbel Max, which still makes it indifferentiable. Therefore, we use a softmax function to approximate this argmax procedure. y=∑je(oj+gj)/τe(oi+gi)/τ where τ∈(0,∞) is a temparature hyperparameter. We note that the output of Gumbel Softmax function here is a vector which sum to 1, which somewhat looks like a one-hot vector (but it's not). So by far, this does not actually replace the argmax function. To actually get a pure one-hot vector, we need to use a Straight-Through (ST) Gumbel Trick. Let's directly see an implementation of Gumbel Softmax in PyTorch (We use the hard mode, soft mode does not get a pure one-hot vector). def gumbel_softmax(logits, tau=1, hard=False, eps=1e-10, dim=-1): def gumbel_softmax(logits, tau=1, hard=False, eps=1e-10, dim=-1): # type: (Tensor, float, bool, float, int) -> Tensor # type: (Tensor, float, bool, float, int) -> Tensor r""" r""" Samples from the Gumbel-Softmax distribution (`Link 1`_ `Link 2`_) and optionally discretizes. Samples from the Gumbel-Softmax distribution (`Link 1`_ `Link 2`_) and optionally discretizes. Args: Args: logits: `[..., num_features]` unnormalized log probabilities logits: `[..., num_features]` unnormalized log probabilities tau: non-negative scalar temperature tau: non-negative scalar temperature hard: if ``True``, the returned samples will be discretized as one-hot vectors, hard: if ``True``, the returned samples will be discretized as one-hot vectors, but will be differentiated as if it is the soft sample in autograd but will be differentiated as if it is the soft sample in autograd dim (int): A dimension along which softmax will be computed. Default: -1. dim (int): A dimension along which softmax will be computed. Default: -1. Returns: Returns: Sampled tensor of same shape as `logits` from the Gumbel-Softmax distribution. Sampled tensor of same shape as `logits` from the Gumbel-Softmax distribution. If ``hard=True``, the returned samples will be one-hot, otherwise they will If ``hard=True``, the returned samples will be one-hot, otherwise they will be probability distributions that sum to 1 across `dim`. be probability distributions that sum to 1 across `dim`. .. note:: .. note:: This function is here for legacy reasons, may be removed from nn.Functional in the future. This function is here for legacy reasons, may be removed from nn.Functional in the future. .. note:: .. note:: The main trick for `hard` is to do `y_hard - y_soft.detach() + y_soft` The main trick for `hard` is to do `y_hard - y_soft.detach() + y_soft` It achieves two things: It achieves two things: - makes the output value exactly one-hot - makes the output value exactly one-hot (since we add then subtract y_soft value) (since we add then subtract y_soft value) - makes the gradient equal to y_soft gradient - makes the gradient equal to y_soft gradient (since we strip all other gradients) (since we strip all other gradients) Examples:: Examples:: >>> logits = torch.randn(20, 32) >>> logits = torch.randn(20, 32) >>> # Sample soft categorical using reparametrization trick: >>> # Sample soft categorical using reparametrization trick: >>> F.gumbel_softmax(logits, tau=1, hard=False) >>> F.gumbel_softmax(logits, tau=1, hard=False) >>> # Sample hard categorical using "Straight-through" trick: >>> # Sample hard categorical using "Straight-through" trick: >>> F.gumbel_softmax(logits, tau=1, hard=True) >>> F.gumbel_softmax(logits, tau=1, hard=True) .. _Link 1: .. _Link 1: .. _Link 2: .. _Link 2: """ """ if eps != 1e-10: if eps != 1e-10: warnings.warn("`eps` parameter is deprecated and has no effect.") warnings.warn("`eps` parameter is deprecated and has no effect.") gumbels = -torch.empty_like(logits).exponential_().log() # ~Gumbel(0,1) gumbels = -torch.empty_like(logits).exponential_().log() # ~Gumbel(0,1) gumbels = (logits + gumbels) / tau # ~Gumbel(logits,tau) gumbels = (logits + gumbels) / tau # ~Gumbel(logits,tau) y_soft = gumbels.softmax(dim) y_soft = gumbels.softmax(dim) if hard: if hard: # Straight through. # Straight through. index = y_soft.max(dim, keepdim=True) index = y_soft.max(dim, keepdim=True) y_hard = torch.zeros_like(logits).scatter_(dim, index, 1.0) y_hard = torch.zeros_like(logits).scatter_(dim, index, 1.0) ret = y_hard - y_soft.detach() + y_soft ret = y_hard - y_soft.detach() + y_soft else: else: # Reparametrization trick. # Reparametrization trick. ret = y_soft ret = y_soft return ret return ret When fowarding, the code use an argmax to get an actual one-hot vector. And it uses ret = y_hard - y_soft.detach() + y_soft, y_hard has no grad, and by minusing y_soft.detach() and adding y_soft, it achieves a grad from y_soft without modifying the forwarding value. So eventually, we are able to get a pure one-hot vector in forward pass, and a grad when back propagating, which makes the sampling procedure differientiable. Finally, let's look at how τaffects the sampling procedure. The below image shows the sampling distribution (which is also called the Concrete Distribution [1611.00712]) and one random sample instance when using different hyperparameter τ. when τ→0, the softmax becomes an argmax and the Gumbel-Softmax distribution becomes the categorical distribution. During training, we let τ>0 to allow gradients past the sample, then gradually anneal the temperature τ (but not completely to 0, as the gradients would blow up). from Eric Jang. Argmax How to make argmax differentiable? Intuitively, the Straight-Through Trick is also applicable for softmax+argmax (or softargmax + argmax). I am still not sure, needs more digging in the literature. Some have introduced the soft-argmax function. It doesn't actually makes it differentiable, but use a continuous function to approximate the softmax+argmax procedure. π=soft-argmax(o)=∑jeβojeβo where β can be a large value to make π very much "look like" a one-hot vector. Discussion Goal softmax + argmax is used for classification, we only want the index with the highest probability. gumbel softmax + argmax is used for sampling, we may want to sample an index not with the highest probability. Deterministic softmax + argmax is deterministic. Get the index with the highest probablity. gumbel softmax + argmax is stochastic. We need to sample from a gumbel distribution in the beginning. Output vector softmax and gumbel softmax aboth output a vector sum to 1. softmax outputs a normalized probability distribution. gumbel softmax outputs a sample somewhat more similar to a one-hot vector.(can be controlled by τ) Straight-Through Trick can actually be applied to both softmax + argmax and gumbel softmax + argmax, which can make both of them differentiable. (?) Gumbel Softmax [1611.01144] Concrete Distribution (Gumbel Softmax Distribution) [1611.00712] Eric Jang official blog: PyTorch Implementation of Gumbel Softmax: PreviousMeta LearningNextGAN theory Last updated Was this helpful?
14794	https://www.quora.com/Is-there-a-geometric-explanation-for-the-kinematics-equation-v_f-2-v_0-2-+-2ad	Something went wrong. Wait a moment and try again. Equation of Motion for a ... 3D Geometry Motion (physics) Motion Math Geometric Mathematics Kinematic Equations 5 Is there a geometric explanation for the kinematics equation v 2 f = v 2 0 + 2 a d ? Eric Neville Author has 56 answers and 180.9K answer views · 7y We first need two previous equations. From the definition of acceleration: a=v−ut⟹v=u+at And from average velocity: vavg=u+v2 vavgt=s=u+v2t And since we now know v=u+at: s=u+(u+at)2t s=2u+at2t⟹s=ut+12at2 Now knowing s=ut+12at2 and v=u+at: (v)2=(u+at)2 v2=u2+2uat+a2t2 v2=u2+2a(ut+12at2) From the given equation for s, we can replace this entire bracket: We first need two previous equations. From the definition of acceleration: a=v−ut⟹v=u+at And from average velocity: vavg=u+v2 vavgt=s=u+v2t And since we now know v=u+at: s=u+(u+at)2t s=2u+at2t⟹s=ut+12at2 Now knowing s=ut+12at2 and v=u+at: (v)2=(u+at)2 v2=u2+2uat+a2t2 v2=u2+2a(ut+12at2) From the given equation for s, we can replace this entire bracket: v2=u2+2as Which is the same equation you were asking about just with different labels for the variables. Brad Moffat Author has 4K answers and 11.6M answer views · Updated 7y This is a great question. I’ve been searching for a simple geometric interpretation for years that relates the graph of velocity vs time to that equation. Mr. Cosiglio’s answer is the closest thing I’ve seen. So I’ll try again. Here’s the graph: Let’s look at V1andV2squared . . . Now the big one is supposed to equal the little one plus 2ad. This sounds like it should represent the area of 2 rectangles. The problem is that “a” is a slope and “d” is the total area of the middle trapezoid, (V1,V2,t2,t1). I couldn’t make sense of that so I did what Mr. Consiglio did (kudos) and wrote 2∗a∗d as This is a great question. I’ve been searching for a simple geometric interpretation for years that relates the graph of velocity vs time to that equation. Mr. Cosiglio’s answer is the closest thing I’ve seen. So I’ll try again. Here’s the graph: Let’s look at V1andV2squared . . . Now the big one is supposed to equal the little one plus 2ad. This sounds like it should represent the area of 2 rectangles. The problem is that “a” is a slope and “d” is the total area of the middle trapezoid, (V1,V2,t2,t1). I couldn’t make sense of that so I did what Mr. Consiglio did (kudos) and wrote 2∗a∗d as 2∗ΔVΔt∗d. (since acceleration = difference in velocity / time.) Slide the denominator over and you get 2∗ΔV∗dΔt. Now recognize that dΔt is the average velocity and now we have something to work with. The big square is equal to the little square plus twice the rectangle made with the difference in velocity and the average velocity. Here’s what that looks like: Let’s do the math to prove it for this one. The big square is 8units x 8units = 64units2. The little one is 2units x 2units = 4units2. ΔV=6units and avg.V = 5units here, so the rectangle has an area of 30units2. It looks like 4 + 2 30 = 64 in this example. So it works - let’s take a look closer. Place the little square inside the big one like so: Clearly the rest of the area can be split in two equal parts by the diagonal shown. Last step, rotate one of the split areas about the centre point of the diagonal to form this nice rectangle (rotated shape plus non-rotated shape.) : The large square is equal to the little square plus the sum of this rectangle. The base of this rectangle is V2−V1. The height is V1+V2. I trust you can see why. The area of the rectangle is base times height. That’s, oh look, the difference in the velocity times the sum of the velocity. That star point around which we rotated was in fact the average velocity (half the sum.) So V22=V21+2∗avg.V∗ΔV which is the same as V22=V21+2∗a∗d So that tall rectangle represents 2ad and V2 square is made of that and the V1 square. Sadly, I can’t readily see why each shape that makes the tall rectangle is equal to ad , (beyond the belaboured explanation given above.) So I must stop here. I can’t see how that will help you understand the motion any better, but I enjoyed the exercise, and found a cute relationship that can be used with any trapezoid. Of course you first have to right-angle-ize the trapezoid like this: I marked the difference and the average with green hashes. Now you can test the square theorem: It seems that 64 must equal 16 plus 2 4 6. I can live with that. I’ve never seen this relationship in geometry class. (Maybe some Quorans out there have?) Do you remember how these squares were related to the area of the trapezoid? (Hint: the area of the trapezoid was what we called “d” above.) All this may not have answered your question (it certainly was not simple), nor helped your understanding of motion, but I had fun. Thanks for the question. Roy Narten Mechanical Engineer · Author has 2.2K answers and 4.9M answer views · 7y Originally Answered: Is there a geometric explanation for the kinematics equation v_f^2 = v_0^2 + 2ad? I can derive it, I’m looking for a more intuitive understanding. · Consider an object moving with initial velocity = Vi. It has kinetic energy = 12mV2i. Suppose we want to slow it down until its final velocity = Vf. So we apply a force F over a distance (d). The work done to slow the object is: Work = (force)(distance) or W=(F)(d) but F=ma If the motion remains horizontal, then or Related questions Is there any intuition behind the 4th kinematic equation v^2 = v(0) ^2 + 2ad ? I'm not looking for derivation from the previous 3 equations. What is the intuition behind the kinematic equation v^2-u^2=2as? Is there something special about the kinematic equation ? What's the intuition for why the kinematic equation, , is true? I've seen the proof based on , but I really want some intuition. What does mean? Peng Peng Zheng Ph.D in Physics, The University of Texas at Dallas (Graduated 2021) · Author has 427 answers and 10.8M answer views · 7y Originally Answered: Is there a geometric explanation for the kinematics equation v_f^2 = v_0^2 + 2ad? I can derive it, I’m looking for a more intuitive understanding. · Multiply the mass M throughout the equation and divide by two throughout. You will notice that the mass multiplied by acceleration multiplied by distance is equal to the difference in kinetic energy between initial state and final state. This becomes effectively a statement of conservation of energy and that work is a form of energy. 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The geometric interpretation can be understood through the context of a velocity-time graph. Velocity-Time Graph: In a velocity-time graph: The x-axis represents time. The y-axis represents velocity. For an object moving with constant acceleration, the graph will be a straight line. The slope of this line is equal to the acceleration ([math]a[/math]), and the area under Yes, there is a geometric explanation for the kinematic equation [math]v_f^2 = v_0^2 + 2ad.[/math] This equation relates the final velocity ([math]v_f[/math]), initial velocity ([math]v_0[/math]), acceleration ([math]a[/math]), and displacement ([math]d[/math]) of an object undergoing uniform acceleration. The geometric interpretation can be understood through the context of a velocity-time graph. Velocity-Time Graph: In a velocity-time graph: The x-axis represents time. The y-axis represents velocity. For an object moving with constant acceleration, the graph will be a straight line. The slope of this line is equal to the acceleration ([math]a[/math]), and the area under the line represents the displacement ([math]d[/math]). Finding Displacement: The area under the velocity-time graph from time [math]t_0[/math] to time [math]t_f[/math] can be calculated as follows: The area of the trapezoid formed by the initial and final velocities can be expressed as: [math]d = \text{Area} = \frac{1}{2} \times (v_0 + v_f) \times t,[/math] where [math]t[/math] is the time duration over which the acceleration occurs. Using the Average Velocity: The average velocity ([math]v_{avg}[/math]) during the acceleration can be expressed as: [math]v_{avg} = \frac{v_0 + v_f}{2}.[/math] Substituting this into the area equation gives: [math]d = v_{avg} \times t = \left(\frac{v_0 + v_f}{2}\right) t.[/math] Relating Time and Acceleration: From the definition of acceleration, we know: [math]a = \frac{v_f - v_0}{t} \implies t = \frac{v_f - v_0}{a}.[/math] Substituting for Time: Substituting this expression for [math]t[/math] back into the displacement equation: [math]d = \left(\frac{v_0 + v_f}{2}\right) \left(\frac{v_f - v_0}{a}\right).[/math] Multiplying through by [math]2a[/math] gives: [math]2ad = (v_0 + v_f)(v_f - v_0).[/math] Expanding and Rearranging: Expanding the right side: math(v_f - v_0) = v_f^2 - v_0^2.[/math] Now, equating both sides gives: [math]2ad = v_f^2 - v_0^2.[/math] Rearranging this leads to the desired kinematic equation: [math]v_f^2 = v_0^2 + 2ad.[/math] Conclusion: Thus, geometrically, this equation can be understood through the relationship between velocity, acceleration, and displacement depicted in a velocity-time graph. The area under the graph correlates with displacement, and the relationships between average velocity and acceleration provide the necessary connections to derive the equation. Anand Manikutty Independent stats & compsci consultant for companies in finance, s/w & hi tech · Author has 157 answers and 659.7K answer views · 7y Yes. Denote the final speed by vf and the initial speed by v0. The equation looks like this: (vf - v0)(vf + v0) = 2ad That is, (vf + v0)/2 times (vf - v0)/a = d That is, the average velocity times the time taken equals the distance. This is exactly what you would expect. Geometrically speaking, these two quantities are directly proportional to the distance traveled (by a body being accelerated by a constant force). Related questions In kinematics, when is it ok to use v=d/t and when do you use other equations like d=vt+1/2at^2 or vf=vi+at or vf^2=vi^2+2ad? When can I not use the fourth kinematic equation? How can I isolate time in the kinematic equation d= VIT + at^2/2? Could someone explain the physics equation [math]d=0.75r+0.03r^2[/math] to me? What are the two criteria needed to use kinematics equations? Alon Amit Math Ph.D., Physics and Computer Science B.Sc. · Upvoted by Ahmed Khalil , Master's degree Physics, University of Cologne (2024) and Erik Kofoed , M.Sc. Physics & Theoretical Physics, Lund University (2016) · Author has 8.8K answers and 173.8M answer views · Updated 6y Related What's the intuition for why the kinematic equation, [math]∆x=v_it+ \frac{1}{2}at^2[/math] , is true? I've seen the proof based on [math]v_f=v_i+a∆t[/math] , but I really want some intuition. The intuition behind most basic results in kinematics (and much else) can be gained by considering what happens when you partition the situation into small time slices. Let’s say you start at position [math]0[/math], with an initial speed of [math]0[/math], and constant acceleration [math]1\text{m/s$^2$}[/math]. What does this constant acceleration mean? It means your speed increases by [math]1\text{m/s}[/math] every second. Right? That’s what acceleration is: the change in speed. “one meter per second, per second” is [math]1\text{m/s$^2$}[/math]. So, what happens during the first second? Your speed is [math]0[/math] at the start of the second, and it’s [math]1\text{m/s}[/math] at the The intuition behind most basic results in kinematics (and much else) can be gained by considering what happens when you partition the situation into small time slices. Let’s say you start at position [math]0[/math], with an initial speed of [math]0[/math], and constant acceleration [math]1\text{m/s$^2$}[/math]. What does this constant acceleration mean? It means your speed increases by [math]1\text{m/s}[/math] every second. Right? That’s what acceleration is: the change in speed. “one meter per second, per second” is [math]1\text{m/s$^2$}[/math]. So, what happens during the first second? Your speed is [math]0[/math] at the start of the second, and it’s [math]1\text{m/s}[/math] at the end of the second. On average, your speed during this second was [math]\frac{1}{2}\text{m/s}[/math]. I hope this is non controversial. In the next second, your speed changed from [math]1\text{m/s}[/math] at the start to [math]2\text{m/s}[/math] at the end. So, on average, your speed in the [math]2^\text{nd}[/math] second was [math]\frac{3}{2}\text{m/s}[/math]. And this continues. In the [math]t^\text{th}[/math] second you’ll accelerate from a speed of math\text{m/s}[/math] to a speed of math\text{m/s}[/math], for an average speed of [math]\left(t-\frac{1}{2}\right)\text{m/s}[/math]. Great. How much distance will you gain? Simply, the sum of those speeds, each times [math]1[/math] second. Distance is speed times time. So we need [math]\displaystyle \frac{1}{2}+\frac{3}{2}+\ldots+\frac{2t-1}{2} = \frac{t^2}{2}[/math] …and that, essentially, is your intuition: over [math]t[/math] seconds, you’re moving about [math]\frac{1}{2}t^2[/math] meters, when your acceleration is [math]1\text{m/s$^2$}[/math]. If your acceleration is [math]a[/math] instead of [math]1[/math], that simply multiplies everything by [math]a[/math], that’s all. This is intuition, rather than proof, because we didn’t really determine how your distance grows within each second. We simply assumed that the movement within each second is one of constant speed, which it isn’t. But if the motion is long (many seconds), this isn’t a big deal. And if the whole thing took up just [math]2[/math] seconds, you can repeat the same argument by looking at hundredths of a second instead of seconds. Same thing. The rigorous version of this is to split time into intervals of length [math]\Delta t[/math] and then let [math]\Delta t \to 0[/math]. It’s useful for you to do so, and it will simply recover the reason why [math]t[/math] is the derivative of [math]\frac{1}{2}t^2[/math], or in other words why [math]\frac{1}{2}t^2[/math] is the integral from [math]0[/math] to [math]t[/math] of [math]t\,\mathrm{d}t[/math]. Sponsored by CDW Corporation How can AI help your teams make faster decisions? CDW’s AI solutions offer retrieval-augmented generation (RAG) to expedite info with stronger insights. Ron Brown Decades of teaching physics to undergrads · Author has 13.6K answers and 84.2M answer views · Updated 2y Related What's the intuition for why the kinematic equation, [math]∆x=v_it+ \frac{1}{2}at^2[/math] , is true? I've seen the proof based on [math]v_f=v_i+a∆t[/math] , but I really want some intuition. It can help to think about what each term is telling you. And in doing that, I like restating [math]∆x[/math] as the [math]x(t)-x_i[/math] and rearranging the equation to read [math]x(t)=x_i+v_it +(1/2)at^2[/math]. That is, the position at any time [math]t[/math] is its initial position ([math]x_i[/math]) plus how far it would have traveled in time [math]t[/math] had it not accelerated ([math]v_it[/math]} plus how much additional distance was accumulated due to its acceleration (mathat^2[/math]). If you graph the position as a function of time assuming it was at rest and never moved, its position would never change … so would be a horizontal line on the [math]x[/math] vs [math]t[/math] graph. If the object were jus It can help to think about what each term is telling you. And in doing that, I like restating [math]∆x[/math] as the [math]x(t)-x_i[/math] and rearranging the equation to read [math]x(t)=x_i+v_it +(1/2)at^2[/math]. That is, the position at any time [math]t[/math] is its initial position ([math]x_i[/math]) plus how far it would have traveled in time [math]t[/math] had it not accelerated ([math]v_it[/math]} plus how much additional distance was accumulated due to its acceleration (mathat^2[/math]). If you graph the position as a function of time assuming it was at rest and never moved, its position would never change … so would be a horizontal line on the [math]x[/math] vs [math]t[/math] graph. If the object were just moving at constant speed, that is [math]v[/math] didn’t change from its initial speed, that graph would just be a straight line with positive slope (if moving in the positive [math]x[/math]-direction). The third term simply says that in each increment of time, the speed has increased, so the slope of the [math]x[/math] vs [math]t[/math] graph must be increasing (if the acceleration is in the positive [math]x[/math]-direction), so the graph curves upward. First look at each term separately, then add the three terms. Norman Goldstein Studied at The University of British Columbia · Author has 83 answers and 23.6K answer views · 7y A physical explanation is: Multiply by 1/2 the mass: 1/2 m Vf^2 - 1/2 m V0^2 = ma d The RHS is the work done. The LHS is the change in kinetic energy. This is not a problem in geometry. Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Anonymous 6y Related What's the intuition for why the kinematic equation, [math]∆x=v_it+ \frac{1}{2}at^2[/math] , is true? I've seen the proof based on [math]v_f=v_i+a∆t[/math] , but I really want some intuition. Remember average speed? It’s just the distance traveled over time. [math]\displaystyle \Delta x= v_{average}t[/math] When acceleration is constant, the average speed is just going to be the average of the starting and ending speed. A simple arithmetic average: [math]\displaystyle v_{average} = \frac{v_0+v_t}{2}[/math] (That’s where the 1/2 comes from, right there.) The ending speed is simply the starting speed, plus whatever constant acceleration has occurred. [math]\displaystyle v_t=v_0+at[/math] Now put it all together: [math]\displaystyle \Delta x= \frac{v_0+(v_0+at)}{2}t[/math] In this form, you can see that the distance is the average speed, times Remember average speed? It’s just the distance traveled over time. [math]\displaystyle \Delta x= v_{average}t[/math] When acceleration is constant, the average speed is just going to be the average of the starting and ending speed. A simple arithmetic average: [math]\displaystyle v_{average} = \frac{v_0+v_t}{2}[/math] (That’s where the 1/2 comes from, right there.) The ending speed is simply the starting speed, plus whatever constant acceleration has occurred. [math]\displaystyle v_t=v_0+at[/math] Now put it all together: [math]\displaystyle \Delta x= \frac{v_0+(v_0+at)}{2}t[/math] In this form, you can see that the distance is the average speed, times the time. And you can see how the average is calculated. A little algebra will put this into the familiar form: [math]\displaystyle \Delta x= v_0t + \frac{1}{2}at^2[/math] When acceleration is not constant, the same principles apply, only the calculation of the average can get tricky. The average of a function is closely related to the definite integral, but for a constant acceleration, we can bury the calculus in the intuitive understanding of the average. Richard Muller Prof Physics, UCBerkeley, author of "Now—The Physics of Time" (2016) · Upvoted by Satya Parkash Sud , M.Sc. Physics & Nuclear Physics, University of Delhi (1962) and Allan Steinhardt , PhD, Author "Radar in the Quantum Limit",Formerly DARPA's Chief Scientist,Fellow · Author has 2.3K answers and 249M answer views · Updated 5y Related D = vt + 1/2at^2. What is the reason behind this equation? First, imagine that there is no acceleration, that is, a = 0. Then the equation reads D = vt OK. That makes sense. The distance you go is equal to the velocity times the time. Go 50 miles per hour (the velocity) for 2 hours (the time) and you will go 50 x 2 = 100 miles. Now let’s look at the second term. Assume, for the moment, that the initial velocity is zero. (That’s what v is in the equations, the initial velocity.) So you start with zero velocity. After a time t, your velocity is at. (So, for example, if your acceleration is 2 mph every second, then after 10 seconds your speed will be at = 2 First, imagine that there is no acceleration, that is, a = 0. Then the equation reads D = vt OK. That makes sense. The distance you go is equal to the velocity times the time. Go 50 miles per hour (the velocity) for 2 hours (the time) and you will go 50 x 2 = 100 miles. Now let’s look at the second term. Assume, for the moment, that the initial velocity is zero. (That’s what v is in the equations, the initial velocity.) So you start with zero velocity. After a time t, your velocity is at. (So, for example, if your acceleration is 2 mph every second, then after 10 seconds your speed will be at = 2 x 10 = 20 mph. That’s your velocity at the end. Your velocity at the beginning was zero. So for that time period t, your average velocity is the average between [math]0[/math] and [math]at[/math], that is, your average velocity V = 1/2 at. If you moved at this average velocity for the time t, then your distance would be D = V t = (1/2 at) t = 1/2 at^2 That’s the second term of the equation. If your initial velocity is not zero but v, then you gain distance from both terms, and the distance you go is D = v t + 1/2 at^2 Of course, this equation can be devised using calculus. But it doesn’t require calculus. Ron Brown Decades of teaching physics to undergrads · Author has 13.6K answers and 84.2M answer views · 5y Related Is there any intuition behind the 4th kinematic equation v^2 = v(0) ^2 + 2ad ? I'm not looking for derivation from the previous 3 equations. There is, but you won’t see it until you talk about Newton’s laws and then the work-energy theorem which leads to the conservation of energy. That is, the “fourth” kinematics equation you cite, as are the other three, are just special case equations associated with one-dimensional motion subject to constant acceleration. But as you will learn (if you haven’t already), an object’s acceleration cannot be constant if it is not subject to a constant net force. But if a constant force is applied to an object over a distance d, the force does work equal to the force times that distance. But the net fo There is, but you won’t see it until you talk about Newton’s laws and then the work-energy theorem which leads to the conservation of energy. That is, the “fourth” kinematics equation you cite, as are the other three, are just special case equations associated with one-dimensional motion subject to constant acceleration. But as you will learn (if you haven’t already), an object’s acceleration cannot be constant if it is not subject to a constant net force. But if a constant force is applied to an object over a distance d, the force does work equal to the force times that distance. But the net force also equals the objects mass times its acceleration. So without deriving anything, let’s just say that the above equation is equivalent (if you multiply every term by m/2) to saying, the final kinetic energy equals the initial kinetic energy plus the work done by the net force. Symbolically: Kf=Ki + W, where W represents the work done by the net force. The beauty of the work-kinetic energy theorem is that it does not have the special case considerations of one-dimensional motion with constant acceleration that the kinematics equation does - but leads to the same result. So it is more general. And many problems that you might struggle with using the “simpler” kinematics equations become easier using energy considerations. (For example, fire a projectile from a higher elevation to a lower elevation and the final kinetic energy equals the initial kinetic energy plus the work done by all the forces that act on it while it is in flight.) That is why learning physics is so much fun! David Townsend applied mathematics and theoretical physics · Author has 29.5K answers and 11M answer views · 8y Related Is there something special about the kinematic equation [math]2da= {V_ {f}} ^ {2}-{V_ {I}} ^{2}[/math] ? You can interpret this question in terms of change in kinetic energy and work done. Consider a object of mass being accelerated by a force. The work done by the force would be force x distance, since f=ma we can say mad. The body accelerates from velocity vf to vi, so the change in kinetic energy is 1/2mvf^2 - 1/2mvi^2 - under the principle of the conservation of energy, the two quantities are equal, hence the equation. Related questions Is there any intuition behind the 4th kinematic equation v^2 = v(0) ^2 + 2ad ? I'm not looking for derivation from the previous 3 equations. What is the intuition behind the kinematic equation v^2-u^2=2as? Is there something special about the kinematic equation ? What's the intuition for why the kinematic equation, , is true? I've seen the proof based on , but I really want some intuition. What does mean? In kinematics, when is it ok to use v=d/t and when do you use other equations like d=vt+1/2at^2 or vf=vi+at or vf^2=vi^2+2ad? When can I not use the fourth kinematic equation? How can I isolate time in the kinematic equation d= VIT + at^2/2? Could someone explain the physics equation to me? What are the two criteria needed to use kinematics equations? How could you graph the kinematic equation Vf^2 = Vi^2 + 2aΔx to show or solve for acceleration (a)? Why don't these two kinematic equations equal each other? What's the difference between these two kinematic equations, and in which cases should each be used “ and “ What are some useful kinematics equations besides the standard three (v=vo+at, x=xo+vot+1/at^2 and v^2=vo^2+2ax)? Apart from v = u + at, , and , is there any other kinematic equation? If yes, then what are they? 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14795	https://pmc.ncbi.nlm.nih.gov/articles/PMC12040606/	Clonorchis sinensis and Cholangiocarcinoma - 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. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Korean Med Sci . 2025 Apr 9;40(16):e145. doi: 10.3346/jkms.2025.40.e145 Search in PMC Search in PubMed View in NLM Catalog Add to search Clonorchis sinensis and Cholangiocarcinoma Eun-Min Kim Eun-Min Kim 1 Department of Microbiology, College of Medicine and Lee Gil Ya Cancer and Diabetes Institute, Gachon University, Incheon, Korea. Find articles by Eun-Min Kim 1, Sung-Tae Hong Sung-Tae Hong 2 Department of Tropical Medicine and Parasitology and Institute of Endemic Disease, Seoul National University College of Medicine, Seoul, Korea. Find articles by Sung-Tae Hong 2,✉ Author information Article notes Copyright and License information 1 Department of Microbiology, College of Medicine and Lee Gil Ya Cancer and Diabetes Institute, Gachon University, Incheon, Korea. 2 Department of Tropical Medicine and Parasitology and Institute of Endemic Disease, Seoul National University College of Medicine, Seoul, Korea. ✉ Address for Correspondence: Sung-Tae Hong, MD, PhD. Department of Tropical Medicine and Parasitology and Institute of Endemic Disease, Seoul National University College of Medicine, 103 Daehak-ro, Jongno-gu, Seoul 03080, Korea. hst@snu.ac.kr ✉ Corresponding author. Received 2025 Mar 11; Accepted 2025 Mar 24; Collection date 2025 Apr 28. © 2025 The Korean Academy of Medical Sciences. This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC12040606 PMID: 40296827 Abstract Clonorchiasis is a parasitic disease caused by Clonorchis sinensis, a trematode that inhabits the intrahepatic bile ducts of humans and mammals. C. sinensis is one of common food-borne trematodes, prevalent in East Asia including Korea. The International Agency for Research on Cancer reclassified C. sinensis as the Group 1 biological carcinogen of human cholangiocarcinoma (CCA). Evidence supporting the carcinogenicity of C. sinensis includes epidemiological studies showing increased prevalence and odds ratio (OR) of CCA in clonorchiasis patients, the development of CCA in experimental animals, and molecular studies. Approximately 10% of CCA in Korea are believed to be solely caused by clonorchiasis, with an OR of 4.7 for CCA risk among clonorchiasis patients. All hamsters exposed to both of C. sinensis and N-nitrosodimethylamine (NDMA) developed CCA while those exposed to either C. sinensis or NDMA alone did not. In vitro studies using cell models investigated carcinogenetic changes of the intracellular molecules and genes following stimulation with a soluble extract of C. sinensis. The in vitro stimulated cells showed a significant shift to G2/M phage, produced oncogenic molecules, changed expression of oncogenes, increased cell proliferation and suppressed apoptosis. Additionally, the gap-junction proteins between cells, such as connexin (Cx) 43, Cx 26, and Cx 32, were changed significantly, disrupting intercellular homeostasis. These findings suggest that C. sinensis and nitrogen compounds synergistically stimulate the cholangiocytes to become neoplastic. C. sinensis is a biological carcinogen of human CCA, and the World Health Organization guidelines enlist food-borne trematodes as one of target neglected tropical diseases to be eliminated by 2030. The present article reviews and updates perspectives on clonorchiasis, focusing on carcinogenesis. Keywords:Clonorchis sinensis; Clonorchiasis; Cholangiocarcinoma; Biocarcinogen, Group 1; N-nitrosodimethylamine; Epidemiology; Syrian Golden Hamster; H69 Cells; Food-borne Trematodes Graphical Abstract INTRODUCTION Clonorchis sinensis Looss, 1907 is one of common human infecting zoonotic trematodes in Family Opisthorchiidae, Class Trematoda. Since McConnell first discovered adult worms of C. sinensis from an infected Chinese man in 1874 and published his recovery in The Lancet in 1875,1 it has been targeted of research for biology, epidemiology, pathology, and clinical studies. Clonorchiasis, human infection by C. sinensis, is transmitted by eating raw freshwater fish, and it has been endemically transmitted in East Asia including China, Korea, East Russia, Japan, and northern Vietnam. Although clonorchiasis was eliminated in Japan, over 15 million infected population is still estimated in the endemic areas until present.2 The World Health Organization (WHO) is running many programs to control or eliminate neglected tropical diseases (NTDs) under the discipline of the United Nation’s Sustainable Development Goals (SDGs) by 2030.3 WHO enlists 20 target NTDs of priority by 2030 including food-borne trematodes, such as fascioliasis, clonorchiasis, opisthorchiasis, and paragonimiasis. According to the SDGs strategy, clonorchiasis is one of target NTDs for control or elimination by 2030. At present, clonorchiasis is the only one food-borne trematodiasis for which mass drug administration is recommended in the areas where its prevalence is > 10% by WHO Western Pacific Region.4 One evolutionary study reviewed DNA evidence and speculated that human liver flukes, such as Clonorchis, Fasciola, Opisthorchis, and Dicrocoelium, spread to human from animals in early prehistoric era.5 Several archaeology studies reported that clonorchiasis was widely and heavily endemic over thousand years in East Asia and Korea.6,7,8,9 In Korea, clonorchiasis had been widely prevalent in the past under favorite natural and social epidemiology environment. Total 8 national status surveys were made in Korea for intestinal parasites from 1971 to 2012 and reported egg positive rates of C. sinensis from 1.4% to 4.6% by fecal examination.10 The status of intestinal parasites in Korea has been characterized by prevailing endemicity of soil-transmitted helminths during 1970s, rapid declining during 1980s, and elimination of disease in 1990s.11 The soil-transmitted helminths, especially Ascaris, had been successfully eliminated over the country in Korea thanks to national efforts,12 but C. sinensis remained highly endemic by 2010s. In 1994, the International Agency for Research on Cancer (IARC) classified C. sinensis in Group 2A as a probable biological carcinogen to human with limited evidence while put Opisthorchis viverrini in the Group 1. Later in 2009, the IARC Monograph Working Group for Volume 100B discussed the issue and reclassified C. sinensis in the Group 1 biological carcinogen.13 The reclassification in 2009 meant that scientific evidence for carcinogenesis by C. sinensis was sufficient and facilitated research on mechanism of the carcinogenesis. The present review article updates basic knowledge on clonorchiasis and study findings of carcinogenesis by C. sinensis. In other words, O. viverrini, the taxonomically nearest fluke in the Family Opisthorchiidae, was grouped in the Group 1 already in 1994 with sufficient research evidence for its carcinogenesis of human cholangiocarcinoma (CCA). After the grouping in 1994, O. viverrini has been investigated actively for the pathogenesis and carcinogenic mechanism by several Thai and international researchers in Thailand.14 C. SINENSIS AND CLONORCHIASIS Biology C. sinensis is a slender, leaf-shaped digenetic trematode in the Family Opisthorchiidae, Class Trematoda. Mature worms measure 10–15 mm long, 4–5 mm wide, and 1 mm thick (Fig. 1). The anterior body, from ventral sucker to oral sucker, is sharp and slender, while the posterior end is round. The anterior part is moving actively while they are alive. The worms live in the intrahepatic or extrahepatic bile duct of mammals such as humans, dogs, cats, pigs, and rats. A few worms may be found in the intestine. Experimentally, rabbits and guinea pigs are highly susceptible.15 The worms may live in the bile duct of human about 30 years16 and one adult worm produces about 4,000 eggs a day.17 A rough estimate based on these data suggests that one single mature worm produces over 30 million eggs throughout its life. Fig. 1. Adult worm of C. sinensis from an experimentally infected rabbit, acetocarmine stained. Open in a new tab The C. sinensis worms recovered from humans after medication appeared in 3 different colors: red, black, and pale white (Fig. 2).17,18 Red worms are young and actively producing eggs, with the uterus filled with fresh eggs giving them their red color. Black worms are middle-aged, with lipofuscin pigmentation. The pigments make worms appear grossly black.19 Pale white worms are older with few eggs, atrophied reproductive organs, and depigmentation. The white worms are looking almost unable to reproduce. Some humans are infected with worms of same color but some heavily infected humans often have worms of mixed colors, which reflects different stages of infection (Fig. 2). Fig. 2. Adult worms of C. sinensis from 4 infected humans after praziquantel medication. (A) Twenty-two worms recovered from a man, mostly red worms and a few dark or white worms. (B) Twenty-eight worms from a man with mostly red worms. (C) Thirteen pale white worms from a man. (D) Forty-eight worms from a man with mostly dark and a few red worms. The red worms are young and actively producing eggs; dark worms are some aged and with lipofuscin pigments in their posterior part; pale white worms are in senile change, grossly pale with few eggs in the uterus and depigmented. Fig. 2C and Fig. 2D are reproduced from reference 18 and 17, respectively. Open in a new tab When eggs are discharged through feces and drained into nearby water bodies, they may have a chance of continuing life cycle. It is difficult to figure out what proportion of the eggs successfully complete life cycle and produce next generation, but the high egg productivity is likely a major factor contributing survival of the fluke. The eggs in water are ingested by freshwater snails, the first intermediate host. In the snail tissue, a miracidium hatches out of the egg shell’s operculum and metamorphoses into a sporocyst. The sporocyst produces 20–30 rediae, and each mature redia produces about 40–50 cercariae, which are shed from the snail into the water. Ultimately, the larval stages of C. sinensis in the snail host allow one egg to amplify to 1,000 to 1,500 cercariae.20 The snail hosts are mainly Parafossarulus and Bithynia species, which are commonly and widely distributed in rivers, small streams, and reservoirs in East Asia.20 The cercariae swim freely in the water and find the susceptible second intermediate host, mostly freshwater fish of the Family Cyprinidae.20 When a cercaria encounters a suitable fish, it penetrates the skin beneath the scales and encysts in the muscle, becoming a metacercaria. A total of 15 freshwater fish species in 10 Families have been listed as the second intermediate hosts of C. sinensis in Korea.21 The metacercaria infects the final hosts when the fish is ingested. In the final hosts, the metacercaria excysts rapidly in the duodenum through the action of trypsin and endogenous cysteine proteinases.22 The excysted juvenile worm migrates through the common bile duct and finally reaches the intrahepatic bile duct. An animal experiment demonstrated that it took 9–10 minutes for the worms to migrate from the duodenum to the intrahepatic bile duct in rabbits by bile chemotaxis.23 When the small size of the juvenile worms is considered, the initial migration into the intrahepatic bile duct is significantly rapid, taking only 10 minutes. In the bile duct of the final host, the juvenile worms grow and maturate rapidly. Two or more worms make close contacting and overlapping their bodies, when fresh adult worms were incubated in culture media (Fig. 3).24 This body contact may be a process of sperm exchange between mature worms. In mature worms, fertilized ova, which are commonly called eggs, are accumulated in the gravid uterus and the miracidium is maturated within the intrauterine eggs. The eggs with mature miracidium are discharged through the uterine pore into the bile flow. Finally, the eggs in the bile pass out the host body mixed in feces. The eggs first appear in feces after 26 days from the metacercarial infection, which is the incubation period of clonorchiasis. The infected bile duct undergoes pathological changes due to both mechanical and chemical stimulation of the worms, and eventually the worms reshape the bile duct into a nest that is suitable for their survival and reproduction.25 Fig. 3. Live worms of C. sinensis cultured in media.24 (A) Mature worms with a dark-looking uterus by eggs. (B) Worms are moving actively by extending their anterior end. (C, D) Some worms are overlapped actively. Open in a new tab Epidemiology Clonorchis and Opisthorchis are sister genera in the Family Opisthorchiidae. C. sinensis causes clonorchiasis, while O. viverrini and Opisthorchis felineus cause opisthorchiasis. These are human liver fluke diseases transmitted by fish. Clonorchiasis is primarily found in East Asia, from East Russia to Northern Vietnam, while O. viverrini is found in the Indochina Peninsula and O. felineus has a wide distribution, from Siberian Russia to Europe (Fig. 4).2 The three flukes do not overlap in their distribution, and this segregated distribution reflects different biological characteristics among them. In Korea, only clonorchiasis is transmitted in areas near rivers, streams, and lakes. Fig. 4. Global distribution of the 3 opisthorchid human liver flukes. Open in a new tab In the endemic areas, men in their 50s or 60s tend to have the highest and heaviest infection.25 Humans become infected with C. sinensis by eating raw freshwater fish which determines the epidemiological characteristics of clonorchiasis. Men are more likely to fish and eat raw fish while drinking alcohol, making them more likely infected than women in all endemic areas. In most endemic areas, older men or women are resistant to changing their behavior of eating raw fish. Many infected people in endemic areas have been infected for a lifetime. The life span of C. sinensis in human body may be quite long as about 30 years.16 Therefore, both the host and the parasite age together, and older individuals in endemic areas tend to have a higher prevalence and heavier worm burdens due to accumulated infections over their lifetimes. The accumulated infection means that new worms are superimposed on preexisting worms in the same host. In endemic communities, C. sinensis worms of different ages have been recovered from a single human host (Fig. 2).17,18 The co-infection of different aged worms in one host is indirect evidence of ineffective host immune responses in humans, meaning the body does not protect reinfection or superinfection. The host immune response to existing worms has no protective effect against newly infecting worms in humans. The host immune response has been shown to vary depending on the host species, but in rats and rabbits, it has been observed to provide protection against reinfection or superinfection of C. sinensis.26 In Korea, a total of 8 national surveys had been conducted to assess the status of intestinal parasitic infections.10 These surveys monitored the prevalence of soil-transmitted helminths, food-borne trematodes, and cestodes as national statistics. Based on the survey data, the WHO endorsed the elimination of soil-transmitted helminthiases in Korea in 2000.12 The first national status survey of parasitic infections, conducted in 1971, reported a 4.6% egg positive rate for C. sinensis.10 The population-based cross-sectional survey estimated at least 2.5 million individuals were infected with clonorchiasis throughout the country in 1970s, when no effective anthelmintics for clonorchiasis were available. After the successful elimination of soil-transmitted helminthiases in 1980s, clonorchiasis showed the highest egg positive rates among intestinal helminthiases in subsequent national surveys from the 1990s through 2012. The egg positive rates of C. sinensis in the national surveys were 4.6% in 1971, 1.8% in 1976, 2.6% in 1981, 2.7% in 1986, 2.2% in 1992, 1.4% in 1997, 2.4% in 2004, and 1.9% in 2012.10 Seo et al.27 investigated prevalence of food-borne trematodes among 13,373 residents along 7 river basins in Korea and demonstrated that all of the major river basin areas were endemic for clonorchiasis with an overall prevalence rate of 21.5%, estimating 830,000–890,000 infected individuals in these river basin areas. In 1981, the overall prevalence was 2.7% by the 3rd national survey,10 estimating 1.4 million cases of clonorchiasis in general population. The two datasets combined estimated 2.2 to 2.3 million individuals with clonorchiasis in Korea in 1981. Although praziquantel was introduced for treatment in 1980s, clonorchiasis remained endemic along the river basins in southern Korea, and an overall prevalence of 1.9% in 2012.10 The Ministry of Health and Welfare implemented a community program of screening and chemotherapy for clonorchiasis at river basin areas in southern Korea from 2011 to 2020.28 The program examined 23,000–42,000 residents annually and observed a slow decrease of the egg positive rates, from 11.1% in 2011 to 3.8% in 2020. Continuous screening, chemotherapy, and health education significantly reduced the prevalence of clonorchiasis < 5% in most endemic areas, serving as a good control model of clonorchiasis at the community level.28 These updated data suggest that Koreans with clonorchiasis must be far less than one million at present. China is a large country but clonorchiasis was mainly distributed in two regions; southeastern region along the Pearl River in Guangdong and Guangxi provinces, and northeastern region along the Songhuajiang River and Nen River in Heilongjiang and Jilin provinces.29,30 In 2016, a surveillance of clonorchiasis was implemented in 301 surveillance points in 30 provinces in China.30 The survey detected 6,226 cases (2.04%) infected with C. sinensis among the 305,081 screened people, from 70 counties in 15 provinces. However, 89.4% of the infected people were distributed in Jilin, Heilongjiang, Guangdong, and Guangxi provinces. Qian et al.2 made a meta-analysis and estimated that 10.82 million individuals have clonorchiasis in China. A study in a community in Guangxi province traced individuals with clonorchiasis and observed 40.6% of them were reinfected within one year. Of the reinfected individuals, 75.1% were reinfected once, 23.2% twice, and 1.7% 3 times.31 Many infected individuals are frequently reinfected in most endemic areas because they live in the liver fluke environment that is formed by both biological ecology and sociocultural tradition. A meta-analysis of food-borne trematodiases across the Great Mekong Subregion recorded prevalence of opisthorchiasis as 21.11% (95% confidence interval [CI], 17.74–24.47%) in Thailand, Laos, Cambodia, and Vietnam while clonorchiasis had a prevalence of 25.33% (95% CI, 18.32–32.34%) in China and Vietnam.32 In Northern Vietnam, the prevalence of clonorchiasis was estimated at 20.30% (95% CI, 9.13–31.47%). About 20 provinces in Northern Vietnam are known to be endemic of clonorchiasis, and Nam Dinh and Ninh Binh are heavily endemic, with prevalence rates over 20%.2 Additionally, several thousand residents with clonorchiasis are estimated along the Amur river in Russia, downflow basin of the Songhuajiang in China.2 In Japan, clonorchiasis was endemic in many regions of Honshu, northeast Shikoku, and north Kyushu but not in Hokkaido in the past. Industrialization and agricultural modernization after the World War II drastically changed freshwater ecology, making the first intermediate host vulnerable throughout Japan. Eventually, clonorchiasis rapidly disappeared from former endemic regions and no new cases have been reported recently.33 Clinical manifestations Most people with clonorchiasis have few symptoms or signs, especially when only a small number of flukes are involved, causing minimal tissue damage. A few flukes typically trigger little or no noticeable subjective symptoms because they live in the lumen of the hypertrophic bile duct. However, heavy infection leads to more severe manifestations. The symptoms, signs, and pathological findings in clonorchiasis depend on the duration of infection, number of infecting worms, the frequency of reinfection, and any associated microbial infections.25 Individuals with chronic and heavy infections may experience epigastric pain, feeling of indigestion, fever, and loose stools. Overtime, they may develop serious complications, such as gallstone formation, liver abscess, cholangitis, cholecystitis, pancreatitis, and CCA.25 A cross-sectional study in an endemic community in southeastern China investigated correlation between C. sinensis infection and hepatobiliary morbidity.34 The prevalence and infection intensity of C. sinensis were higher in male, elder people, and the individuals consuming alcohol, same as found in Korea. C. sinensis infection was associated with the increase of diarrhea (adjusted odds ratio [aOR], 2.2, 95% CI, 1.1–4.5) and with the increase of fatty liver (aOR, 2.7, 95% CI, 1.4–5.2). Moderate C. sinensis infection was associated with the increase of gallbladder stone (aOR, 3.0, 95% CI, 1.1–8.6), while moderate and heavy infections with the increase of intrahepatic bile duct dilatation (aOR, 2.2, 95% CI, 1.0–4.9 and aOR, 4.3, 95% CI, 1.9–9.9, respectively) and the development of periductal fibrosis (aOR, 3.2, 95% CI, 2.1–4.9).34 Pathogenesis When a human host swallows metacercariae, the juvenile C. sinensis worms excyst in the intestine and first crawl into the intrahepatic bile duct, where they grow rapidly. The worms stimulate the bile duct mucosal wall, leading to adenomatous hyperplasia of the mucosa.35,36,37,38,39 The hyperplastic bile duct becomes wider and tortuous, and eventually transforms into a comfortable nest, nurturing the fluke. Over time, the thin and narrow bile duct dilates up to 10 times in its original diameter, from 0.8 mm to 8 mm.40 The volume of the dilated intrahepatic bile duct may increase by about 1,000-times, providing more space for the worms in the liver. The volume expansion of the intrahepatic bile duct may create pressure on surrounding liver tissue. As the worm burden increases, the volume pressure can damage the periductal liver parenchyma. Histopathological findings demonstrate that the worms in the duct lumen are surrounded by and in close contact with hyperplastic bile duct epithelial cells, mixed with many metaplastic mucin-producing cells. The hyperplastic epithelial cells showed a 15-fold increase of cell division activity.40 The thick adenomatous hyperplastic cell layer, interspersed with mucin-secreting cells, helps minimize damage to the duct wall by the worms (Fig. 5). The worms feed on blood and epithelial cells in the bile duct and survive for a long time.20 Inflammatory changes are most evident at the duct wall and periductal tissue, with subsequent fibrosis gradually progressing around the infected duct.36,37,38,39,40 Dysplastic cells were noted among the hyperplastic cells.38 When the infected host is treated with chemotherapy, the pathological changes in the bile duct significantly resolve after the worms are eliminated. However, the ductal dilatation and periductal fibrosis persist in mild degree over one year.41 The remaining mild degree pathology of the intrahepatic bile duct can lead to false-positive results of the ultrasound scanning images of the liver after clonorchiasis has been treated. 42,43,44 Fig. 5. Adult worms of C. sinensis live in the intrahepatic bile duct inducing the duct into a nurturing nest in an infected cat. (A, B) The infected bile ducts show dilation, adenomatous hyperplasia, mucin-secreting metaplasia, inflammation, and fibrosis. Original magnification ×40, hematoxylin-eosin stained. Open in a new tab CARCINOGENESIS BY CLONORCHIASIS Notice of correlation between C. sinensis and human CCA The first patient of CCA associated with C. sinensis was reported in China in 1919,1 and after then several studies reported causal correlation between CCA and C. sinensis. In Korea, Mr. Ju Shik Lee, an officer of Ministry of Health and Social Affairs, claimed that there were many patients with liver cancer and related diseases of jaundice and ascites in Busan, Korea in 1948.45 He investigated those patients and noticed that many of them were heavily infected with C. sinensis. Some surveys and hospital data confirmed that C. sinensis was highly prevalent with egg positive rate 50–70% in Busan and neighboring localities along the Nakdong-gang river basin. However, his work was stopped by the Korean War in 1950 and could not continue thereafter. His finding suggested possible correlation between clonorchiasis and CCA in endemic areas of clonorchiasis. Epidemiological studies There are several reports from hospital case studies and epidemiological studies that report a correlation between clonorchiasis and CCA. In Hong Kong, studies based on autopsy series from the 1950s and 1960s described pathological observations of CCA without cirrhosis in association with C. sinensis.35,36,37 These articles reported a high prevalence of clonorchiasis in Hong Kong and described pathological characteristics of the intrahepatic bile duct in cases of CCA with clonorchiasis. The tumor masses spread along the bile duct in the liver causing bile duct obstruction. Histopathologically, the bile ducts were dilated with adenomatous hyperplasia of the epithelium and periductal fibrosis. Most cases showed multiple metastatic tumors. These case series studies described autopsy findings of CCA patients describing liver pathology of human clonorchiasis and suggesting causal relation with clonorchiasis in Hong Kong. In Korea, Kim et al.46 compared hospital data from patients with primary cancer and CCA in Seoul and Busan, where Seoul was non-endemic but Busan was endemic for clonorchiasis. The proportion of primary liver carcinoma was 26.7% in Seoul and 27.5% in Busan, showing no significant difference. However, CCA accounted for 8.6% of the primary liver carcinomas in Seoul but 19.3% in Busan, which was a significant difference. The prevalence rate of clonorchiasis was 38.9% in CCA patients, compared to 8.0% in hepatocellular carcinoma (HCA) patients. The study revealed a higher incidence of CCA in clonorchiasis endemic area compared to the non-endemic area, as well as a higher infection rate of clonorchiasis in CCA patients than that in HCA patients. Another study in Busan demonstrated that the relative risk of clonorchiasis for CCA was 5 times higher compared to HCA.47 The two clinical studies described a highly probable association between CCA and clonorchiasis in Korea. Later, Shin et al.48 investigated the epidemiological correlation between CCA and clonorchiasis in Busan, suggesting a relative risk of 2.7 (95% CI, 1.1–6.3) for CCA due to clonorchiasis. Shin et al.49 estimated the odds ratio (OR) of CCA attributable to clonorchiasis at 4.7 (95% CI, 2.2–9.8), and found that approximately 10% of CCA cases in Korea were associated with chronic clonorchiasis. A meta-analysis identified risk factors for CCA, revealing that liver flukes (both C. sinensis and O. viverrini), hepatitis B virus, and hepatitis C virus had relative risks (95% CI) 4.8 (2.8–8.4), 2.6 (1.5–4.6), and 1.8 (1.4–2.4), respectively.50 The study also demonstrated that past history of clonorchiasis is a significant risk factor of CCA, alongside the current infection. Lim et al. (2006)51 surveyed the prevalence of clonorchiasis in non-endemic (2.1%), lightly endemic (7.8%), and moderately endemic (31.3%) areas and compared the mortality and incidence rates of CCA in these regions using the national mortality data and cancer registry database in Korea. The age-standardized CCA mortality rates were 1.2, 1.1, and 2.6/100,000 persons and the incidence rate of CCA was 0.3, 1.8, and 5.5/100,000 persons in the matched non-endemic, lightly endemic, and moderately endemic areas, respectively. The OR of clonorchiasis was 14.7 (95% CI, 7.90–25.00) in the moderately endemic area compared to the non-endemic area. The proportion of CCA among primary liver cancers was 1.4%, 8.9%, and 13.2% in the non-endemic, lightly endemic, and moderately endemic areas, respectively.51 Choi et al.52 collected data from patients with CCA and clonorchiasis and analyzed radiologic images of the liver. They observed a strong etiological correlation of clonorchiasis and CCA and reviewed patients’ imaging data. Most clonorchiasis patients showed dilation of the intrahepatic bile ducts and dilation of the distal duct from the mass in CCA. These combined imaging features were considered pathognomonic for clonorchiasis-associated CCA.52 In a later study, retrospective analysis of clinical data from a tertiary hospital in Seoul confirmed that radiologic images of C. sinensis, positive history of eating raw freshwater fish, and positive serology were significant risk factors for CCA.53 The ORs for these risk factors were OR, 8.615 (95% CI, 5.045–16.062), OR, 2.385 (95% CI, 1.527–3.832), and OR, 2.272 (95% CI, 1.147–4.811), respectively. Another clinical study identified several significant risk factors for intrahepatic CCA, with clonorchiasis being one of the risk factors, showing an OR of 13.6 in a tertiary hospital in Korea.54 These findings of human CCA provided clear evidence of the significant causal correlation between clonorchiasis and CCA in human population, which contributed to the reclassification of C. sinensis from Group 2A to Group 1 biological carcinogen by the IARC in 2009.13 In China, Shi et al.55 analyzed 20 patients with primary liver cancer, including 8 with CCA and 11 with HCA. Seven of the 8 CCA patients were infected with C. sinensis and 2 of them were co-infected with hepatitis B virus. Among the 11 HCA patients, 10 were infected with C. sinensis and 4 were co-infected with hepatitis B virus. They suggested a high risk of both CCA and HCA due to co-infection of C. sinensis and hepatitis B virus. Shin et al.50 also proposed that hepatitis B and C viruses are oncogenic in relation to CCA. The synergistic oncogenic effect of C. sinensis and hepatitis viruses or other microbial pathogens, may warrant further research. One meta-analysis in China reported ORs of human liver flukes with CCA: 4.49 (95% CI, 3.43–5.87, P< 0.001) for C. sinensis and 3.69 for O. viverrini. The OR of clonorchiasis with cholelithiasis was 3.69.56 These studies are summarized in Table 1. Table 1. Studies on human clonorchiasis and CCA. \| Studies, Reference No. \| Study design, year, place \| No. of subjects with \| Key findings \| :---: :---: \| \| CCA \| Cs+ \| \| Hou (1956)35 \| Autopsy series, 7 yr, Hong Kong \| 30 \| 28 \| 15% of primary liver cancer with Cs \| \| Gibson (1971)37 \| Autopsy series, 1964–1966, Hong Kong \| 17 \| 11 \| Adenocarcinoma without cirrhosis but with Cs \| \| Belamaric (1972)36 \| Autopsy series, 1961–1966, Hong Kong \| 19 \| 18 \| Well-differentiated adenocarcinoma of intrahepatic bile duct with moderate to severe clonorchiasis \| \| Kim et al. (1974)46 \| Cross-sectional study, 1961–1972, hospitals in Busan and Seoul, Korea \| 54 \| 21 \| Higher rate of CCA (19.3%) and Cs associated (61.9%) in Busan \| \| Chung and Lee (1976)47 \| Cross-sectional study, 1963–1974, Busan, Korea \| 36 \| 19 \| RR of Cs for CCA: 5.68 ± 2.23 \| \| Shin et al. (1996)48 \| Case-control study in hospital, 1990–1993, Busan, Korea \| 36 \| 12 \| RR of Cs for CCA: 2.7 (95% CI, 1.1–6.3) \| \| Choi et al. (2006)52 \| Case-control study in hospital, 2003–2004, Seoul, Korea \| 122 \| 3 \| 3 by eggs, 13 by pathology, 25 by serology, 56 by radiology, 94 by history \| \| Choi et al. (2006)53 \| Case-control study in hospital, 2003–2004, Seoul, Korea \| 185 \| \| Radiologic images, history of eating raw fish, serology positive for Cs \| \| Lim et al. (2006)51 \| Field survey and national cancer registry \| \| \| OR 14.1 in heavy endemic area, positive correlation of Cs prevalence and CCA incidence \| \| Lee et al. (2008)54 \| Case-control study in hospital, 2000–2004, Seoul, Korea \| 622 \| 26 \| OR of Cs for CCA: 13.6 \| \| Shin et al. (2010)49 \| National cancer registry data analysis, 1999–2005, Korea \| 3,606 (in 2005) \| 9.5% (estimated) \| OR of CCA by clonorchiasis: 4.7 (95% CI, 2.2–9.8), 10% of CCA by chronic clonorchiasis \| \| Shin et al. (2010)50 \| Meta-analysis \| \| \| Risk factors for CCA: liver flukes (Cs and Ov) with RR 4.8 (95% CI, 2.8–8.4), HBV with RR 2.6 (95% CI, 1.5–4.6), HCV with RR 1.8 (95% CI, 1.4–2.4) \| \| KAHP (2014)44 \| Health history in Korea \| \| \| History of liver cancer patients before 1950 \| \| Shi et al. (2017)55 \| Case series in hospital, 2014–2015, Hengxian, China \| 8 \| 7 \| Cs and HBV: risk factors of both CCA & HCA \| \| Huang et al. (2024)56 \| Meta-analysis, China \| \| \| CCA OR 4.49 by Cs; 3.69 by Ov \| Open in a new tab CCA = cholangiocarcinoma, Cs = Clonorchis sinensis, RR = relative risk, OR = odds ratio, Ov = Opisthorchis viverrini, CI = confidence interval, KAHP = Korea Association of Health and Promotion, HBV = hepatitis B virus, HCV = hepatitis C virus, HCA = hepatocellular carcinoma. Animal models Several animal experiments have investigated immunopathology, host-parasite interactions, genetic alterations, and CCA pathogenesis by C. sinensis infection. C. sinensis-induced CCA is characterized continuous excessive remodeling of the bile duct cells through persistent stimulation and injury by the worms. The tumor-associated hepatic microenvironment consists of hepatocytes, hepatic sinusoidal endothelial cells, hepatic stellate cells, cholangiocytes, Kupffer cells and recruited immune cells, including bone-marrow-derived macrophages.57 Adult C. sinensis worms cause mechanical injury and inflammation on the bile duct wall, adenomatous hyperplasia, and metaplasia of mucin-producing cells in the biliary mucosa, as described above. In addition to these histopathological findings, changes in the bile duct epithelium in response to stimuli in acute and chronic clonorchiasis include infiltration of Kupffer cells, macrophages and cyst formation in mice. These pathological changes in mice tend to correlate with the longevity of C. sinensis infection, the worm burden, and the susceptibility of the host.58 For pathogen recognition, Toll-like receptor 2 (TLR2) and TLR4 were reportedly upregulated in a mouse model of clonorchiasis during C. sinensis infection, and TLR4 upregulation induced tumor necrosis factor (TNF)-α secretion in excretory-secretory product (ESP)-stimulated biliary epithelial cells in mice.59,60,61 Kim et al.62 reported that Kupffer cells most likely served as antigen-presenting cells and showed a 70-fold increase during the primary infection of C. sinensis. The recovered Kupffer cells produced significantly higher levels of cytokines, including TNF-α, interleukin (IL)-6, IL-10, and IL-13, in C. sinensis-infected FVB mice.63 However, the effects of persistent TNF production by Kupffer cells and cholangiocytes upon chronic liver injury and high levels of reactive oxygen species (ROS) remain unclear. Kupffer cells were concentrated in the periductal tissues and near the veins. Additionally, the livers of mice with cirrhosis showed infiltration of Kupffer cells around the portal areas.64 The number of recruited Kupffer cells increased in regions characterized by tissue remodeling during the cirrhotic stage of infection. Kupffer cells are activated by different components and can be differentiated into M1 (classical) or M2 (alternative) macrophages in the liver. M1 macrophages suppressed tumor growth, whereas M2 macrophages contributed to cancer progression.64 The fibrotic and cirrhotic stage following C. sinensis infection markedly increased the proportion of Arg-1-producing macrophages, the M2 phenotype, which modulate fibrosis and tissue repair, and also promote neoplastic transformation of bile duct cells.62 Different host animals exhibit varying susceptibility to C. sinensis infection, but hamsters, guinea pigs, and rabbits are considered good experimental laboratory animal hosts.15,65,66 Among them, the Syrian golden hamster is the suitable model for studies of CCA.67,68,69,70 Previous studies have reported that C. sinensis infection acts a promoter, and the initiator, such as N-nitrosodimethylamine (NDMA), is required to induce CCA.67,68 Carcinogens, including NMDA, initiate tumorigenesis through DNA methylation and induce a genotoxic effect, leading to structural changes in DNA. No definite neoplastic changes of CCA were observed in BALB/c, ICR, B6, and C3H/He mice infected with C. sinensis and exposure to NDMA.58 Instead of CCA, C3H/He mice developed extensive fibrosis when challenged with C. sinensis, NDMA, and dicyclanil together. Interestingly, adenomas developed only in FVB mice, accompanied cystic dilatation of the infected bile duct.58 No other animals developed CCA through experimental infection with C. sinensis. Features of hamster model Mass-forming lesions (MFLs) are a major pathologic finding of CCA induced by C. sinensis and NDMA in the Syrian golden hamster model, with 100% prevalence of MFLs in the C. sinensis + NDMA group of hamsters.71 Histological progressions of CCA occurred sequentially from well-differentiated CCA, followed by poorly differentiated CCA in a time-dependent manner, beginning at the 8th week of infection (Fig. 6). Fig. 6. Experimental carcinogenesis in Syrian golden hamsters induced by Cs and NDMA. Gross morphology of the livers: Ctrl (A), Cs only (B), NDMA only (C), Cs + NDMA (D). Histopathology of the livers: Ctrl (E), Cs only, boxes marking eggs in the bile duct (F), NDMA only (G), Cs + NDMA (H). Original magnification ×200 (E-H), hematoxylin-eosin stained. Reproduced from the reference 71. Open in a new tab Ctrl = control, Cs = C. sinensis, NDMA = N-nitrosodimethylamine. Another study using the Syrian golden hamster model reported that continuous mechanical and chemical irritation by C. sinensis and NDMA upregulated cancer-critical genes, including PSMD10, CDK4, and PCNA and downregulated tumor suppressor gene TP53, as well as the protein RB, caspase-9, and BAX.71 Downregulation of BAX and caspase-9 may further facilitate the survival of transformed cells with denatured DNA and those undergoing uncontrolled proliferation to form a CCA mass. PCNA overexpression can be detected in tumor tissue as a marker of CCA in tumor necrosis. The animal experiments on carcinogenesis are summarized in Table 2. Table 2. Animal studies on Cs and CCA. \| Studies, Reference No. \| Experimental animals \| Key findings \| :---: \| Animals mixed \| \| \| \| \| Sohn et al. (2006)65 \| Various animal hosts \| Various host susceptibility \| \| \| Loeuillard et al. (2019)70 \| \| Animal model of CCA \| \| Syrian golden hamster \| \| \| \| \| Lee et al. (1993)67 \| Syrian golden hamster \| Cs + 15 ppm NDMA induced CCA in Syrian golden hamsters \| \| \| Lee et al. (1994)68 \| Syrian golden hamster \| Cs as a promoter \| \| \| Lee et al. (1997)69 \| Syrian golden hamster \| Oval cell in CCA by Cs and NDMA \| \| \| Uddin et al. (2015)71 \| Syrian golden hamster \| Mass-forming lesions: 100% after 8–16 wk by Cs infection and NDMA ingestion \| \| Upregulation of cancer-critical genes: CDK4, p16INK4, PSMD10, PCNA) by Cs and NDMA \| \| Downregulation of tumor suppressor genes: RB, BAX, TP53, Caspase-9 \| \| Mouse \| \| \| \| \| Uddin et al. (2012)66 \| Mice, mixed strains \| Strain variation in mice \| \| \| Nguyen-Lefebvre & Horuzsko (2015)64 \| Mice \| Metabolism of Kupffer cells \| \| \| Yan et al. (2015)59 \| Mice, C3H/HeN \| Upregulation of TNF-α by activation of TLR2 and TLR4 \| \| \| Yan et al. (2015)60 \| Mice, BALB/c \| ESP of Cs stimulated expression of TLR4 via activation of MyD88 and NF-κB in primary biliary epithelial cells \| \| \| Uddin et al. (2016)58 \| Mice, BALB/c, ICR, and C3H/HeN \| No CCA in mice by Cs and NDMA \| \| Different sensitivities according to host species \| \| \| Yan et al. (2017)61 \| Mice, C3H/HeN \| TLR4 signal in pathogen-associated biliary fibrosis by Cs \| \| \| Kim et al. (2017)63 \| Mice, FVB \| Increased Kupffer cells in Cs infected mice \| \| \| Kim et al. (2017)62 \| Mice, BALB/c \| Kupffer cells to M2 tumor-associated macrophages by Cs \| \| \| Li et al. (2018)57 \| Mice \| Tumor-associated hepatic microenvironment \| Open in a new tab CCA = cholangiocarcinoma, Cs = Clonorchis sinensis, NDMA = N-nitrosodimethylamine, TNF = tumor necrosis factor, TLR = Toll-like receptor, ESP = excretory-secretory product of Clonorchis sinensis, NF-κB = nuclear factor kappa B. IN VITRO EXPERIMENTS WITH ESP OF C. SINENSIS The ESP is a mixture of various functional molecules secreted from the liver fluke. O. viverrini is known to secrete granulin-like growth factor, which stimulates angiogenesis in the host tissue, promoting rapid wound healing and accelerating neoplastic pathways.72 Additionally, thioredoxin from O. viverrini downregulates apoptosis of cholangiocytes, supporting its induction of CCA.73 Various proteinases, including granulin, have been observed from C. sinensis,74,75 and the ESPs of C. sinensis were suggested to be related with CCA,76,77 but the oncogenic molecules in the ESPs of C. sinensis still require further investigation. Several in vitro studies have cultured epithelial cells stimulated with ESPs of C. sinensis and investigated changes of tumor related intracellular molecules and genes. The stimulated cells exhibited various changes of the cell characteristics. The findings from these cell studies may help explain the mechanisms behind cellular changes that lead to CCA transformation due to C. sinensis infection. Cancer-associated microenvironment Several studies have reported that oxidative stress potentially mediates C. sinensis-associated carcinogenesis.77,78,79 Bile duct inflammation induces endogenous oxidative stress through ROS and reactive nitrogen species, leading to oxidative/nitrative DNA damage in cholangiocytes.77 In response to oxidative stress in hepatocytes, TNF-producing Kupffer cells are activated via JNK and ROS. TNF from Kupffer cells then induces JNK-mediated cholangiocyte proliferation and oncogenic transformation.80 Specifically, the levels of peroxiredoxin (Prdx) isomers (Prdx2, 3, and 6) and thioredoxin1, which are potential carcinogens, were found very high, similar to the levels of intracellular ROS in ESP-exposed HuCCT1 cells.80 Nam et al.77 and Pak et al.81 demonstrated that C. sinensis ESPs significantly induced matrix metalloproteinase-9 expression and activity by enhancing the activity of nuclear factor kappa B (NF-κB) and subsequent induction of IκBα. Through this signal pathway, C. sinensis ESPs regulate the migration and invasion of various CCA cell lines.82 Won et al.83 developed a three-dimensional (3D) in vitro model to understand the microenvironment by culturing HuCCT1 cells and normal H69 cholangiocytes together with C. sinensis ESP. This 3D system consisted of a type I collagen extracellular matrix, HuCCT1 cells treated with ESP, and quiescent biliary ductal plates formed by H69 cholangiocytes. Ultimately, the expression of local and cell-cell adhesion proteins such as vinculin, paxillin, and MMPs (1, 9 and 13) was increased in ESP-stimulated HuCCT1 cells within the 3D system. HuCCT1 cells migrated to the direction of the ESP by a gradient of ESP. Consequently, both IL-6 and tumor growth factor (TGF)-β1 levels were elevated in normal H69 cholangiocytes when they were stimulated with C. sinensis ESP, and a cadherin switch (decrease in E-cadherin/increase in N-cadherin expression) was observed in HuCCT1 cells. Won et al.84 also reported that HuCCT1 cells underwent epithelial-mesenchymal transition-like changes in response to ESP stimulation, promoting their migration through the interaction between cancer cells and normal cholangiocytes. These findings suggest that C. sinensis ESP promotes cholangiocytes to undergo neoplastic changes. Cell division cycle ESPs are highly immunogenic and may exert cytotoxic effects on the biliary epithelium, stimulating inflammation, promoting proliferation, and suppressing apoptosis.85,86 Kim et al.86 reported that C. sinensis ESPs increased cell proliferation and suppressed apoptosis in the human HEK293T cells. Furthermore, normal epithelial cells actively proliferated after treatment with ESP. Specifically, when normal epithelial cells (HEK293T, human kidney epithelium cell line; H69, human bile duct epithelium cell line) were treated together with ESP and NMDA, the G2/M phase of the cell cycle became dominant, while most cells remain at the G1/S phase in the control. In addition to the changes in cell population, cell cycle-associated molecules, including CDK2, cyclin B, Rb, pRb, and E2F1 were significantly upregulated.85,86,87 Co-stimulation with ESP and NDMA induced overexpression of E2F1 and downregulation of pRb, indicating the involvement of the Rb-E2F1 pathway.86 E2F1 expression is correlated with tumor proliferation, and overexpression of E2F1 potentially results in aggressive tumors with a high proliferation rate and suppressed apoptosis.87 The abnormal distribution of cells in the division cycle suggests that these cells may receive stress for division and acquire neoplastic characteristics. Cancer related molecules Studies using C. sinensis-infected mouse liver tissues reported that COX-2 and 5-LOX expression, along with increased 8-oxodG nuclear accumulation in inflammatory cells, were detected in the inflammatory nidus.78 Oxidative stress-mediated liver lesions were observed following C. sinensis infection of mice. COX-2 is known to be overexpressed in various inflammatory diseases and in CCA.88 In primary pancreatitis, biliary epithelial cells exhibited COX-2 upregulation, similar to carcinoma cells. At the intracellular level, combined treatment of normal H69 cholangiocytes with ESP of C. sinensis and NDMA significantly upregulated COX-2.89 Kim et al.89 reported that normal H69 cholangiocytes stimulated by C. sinensis ESP and NDMA overexpressed Cy19, Ki-67, and COX-2. These molecular changes support the idea that normal H69 bile duct cells can undergo neoplastic transformation when stimulated by ESP of C. sinensis and NDMA. C. sinensis was found to secrete C. sinensis granulin (CsGRN), a growth factor-like protein in the ESP. The CsGRN was localized in the testes and tegument of adult worms.74 The in vitro study with hepatocytes and biliary epithelial cells found enhancement of cell migration, upregulation of vimentin, N-cadherin, β-catenin, MMP2, and MMP9, and downregulation of ZO-1 in PLC-GRN/RBE-GRN cells. When the CsGRN was introduced into normal human hepatocytes, the cells underwent cell arrest and malignant transformation by regulating epidermal growth factor receptor-mediated RAS/MAPK/ERK and PI3K/Akt signaling pathways, which suggested that CsGRN could serve as a novel oncoprotein.75 ESPs of C. sinensis induce pro-inflammatory responses in primary biliary epithelial cells by upregulating TLR4 and its downstream signaling pathways, including MyD88-dependent IκB-α degradation, and NF-κB activation.59,60 A DNA modification may be potentially triggered by increased levels of proinflammatory cytokines and NF-κB, which regulate COX-2 and inducible nitric oxide synthase (iNOS) and disrupt homeostasis of oxidants/anti-oxidants and DNA repair enzymes.77 Additionally, free radicals generated enzymatically by C. sinensis ESPs can cause NF-κB-mediated inflammation in HuCCT1 cells.77 The HuCCT1 cells exposed to ESP of C. sinensis induce free radical production through the activation of NOX, XO, and iNOS, and xanthine oxidase leading to increased expression of the NF-κB-regulated proinflammatory genes such as IL-1β and IL-6. The ESP-induced stimulation in HuCCT cells was mediated through the TLRs, leading to production of free radicals that can damage DNA.90 Cancer related genes Neoplastic transformation of cells begins at the genetic level, and several studies have provided evidence of genetic changes. Using miRNA microarray chips containing 135 cancer-related miRNAs, 13 miRNAs were upregulated and 3 were downregulated in human HuCCT1 CCA stimulated by C. sinensis ESP.91 These dysregulated miRNAs were involved in cell proliferation, inhibition of tumor suppressor genes, neoplasia, cell migration, and invasion. Among these, 6% of miRNAs were associated with DNA methylation, an important factor in carcinogenesis. In particular, tumor-suppressor miRNA including let-7a, let7i, and miR-124a were downregulated in both cancerous and non-cancerous cells via ESP stimulation. Analysis of gene expression profiles across the three developmental stages of C. sinensis may provide insight into their pathogenesis and carcinogenesis. Yoo et al.92 reported that gene expression profiles in the C. sinensis transcriptome contained several contigs encoding proteins associated with inhibition and regulation of apoptosis and cell proliferation, such as granulin, epithermal growth factor, TGF interacting proteins, oncogenic laminins, JNK, cyclin-dependent kinase, and transcription factors in adult C. sinensis. These results suggest that apoptosis inhibitors and regulators identified in the C. sinensis transcriptome promote neoplastic transformation by preventing the death of DNA-damaged cells. Gap junction molecules and connexins (Cxs) In the homeostatic condition of the tissue, cells are networked tightly with gap-junction molecules, mainly Cxs.93 NF-κB may influence the production of Cx 43, a gap-junction protein, in liver cirrhosis.94 The formation and nitrosation of nitric oxide (NO) can contribute to C. sinensis-associated carcinogenesis. Induction of iNOS under inflammatory conditions suggests that NO is involved in Cx 43 upregulation.95 Among gap junction proteins, Cx 43 plays a key role in almost all steps of the inflammatory response of cells, inflammatory cell migration, and wound healing.93,94,95 Cx 43 upregulation has been shown to increase tumor cell invasion in melanoma, playing a key role in carcinogenesis.96 Expression profiles of Cx are altered in various cancers.97,98,99 The gap-junction proteins Cx 43 and Cx 26 were upregulated, while Cx 32 was downregulated in H69 cholangiocytes stimulated by C. sinensis ESP and NDMA.89 The study confirmed that silencing of Cx 43 reduced ESP- and NDMA-induced cell proliferation and COX-2 overexpression to control levels (Fig. 7). Consequently, exposure to ESPs of C. sinensis combined with low level NDMA promotes carcinogenesis in normal bile duct epithelial cells through dysregulated intercellular communication, mediated by altered gap-junction molecules. Fig. 7. Role of Cxs in carcinogenesis by C. sinensis and NDMA. Open in a new tab Cx = connexin, NDMA = N-nitrosodimethylamine, PBS = phosphate buffered saline, ESP = excretory-secretory product of C. sinensis, H69 = cell line of normal cholangiocytes. Summary of in vitro experiments The carcinogenetic mechanism of CCA associated with C. sinensis is multistage, beginning with initial stimulation of normal cholangiocytes to proliferate due to the worm and endogenous or exogenous radicals. The stimulation is followed by DNA damage, overexpression of oncogenes, downregulation of apoptosis inhibitory genes, disrupted communication between cells, and uncontrolled proliferation and migration of the damaged cells. Smout et al.100 suggested mechanistic insights that secretions of the liver flukes, such as Ov-GRN-1 granulin growth factor, combined with aberrant inflammation and repeated cycles of chronic wounding at the parasitic sites, promote biliary hyperplasia, fibrosis, and ultimately malignant transformation. The in vitro cell experiment studies are summarized in Table 3. Table 3. In vitro or in vivo studies for carcinogenesis with ESPs of C. sinensis . \| Studies, Reference No. \| Pathways \| Cells \| Major findings \| :---: :---: \| \| Cell cycle disruption \| \| \| \| \| \| Kim et al. (2008)85 \| Cell cycle/Proliferation \| HEK293T with ESP + NDMA \| p-RB↑, CDK2↑, Cyclin B↑, E2F1↑, RB↓, G2/M phase dominant cycle \| \| \| Kim et al. (2008)86 \| Cell cycle/Proliferation \| HEK293T with ESP \| Cyclin B↑, Cyclin E↑, E2F1↑, G2/M phase dominant cycle \| \| \| Kim et al. (2009)87 \| Cell cycle/Proliferation \| HuCCT1 with ESP \| CDK2↑, CdC2↑, COX-2↑, G2/M phase dominant cycle \| \| Molecule/Signal stimulation or inhibition \| \| \| \| \| \| Hayashi et al. (2001)88 \| COX-2 \| Human bile duct cells \| Different expression \| \| \| Fujimoto et al. (2005)99 \| PP1 \| Renal cancer cells \| Cx 32 suppressed tumor \| \| \| Pak et al. (2009)81 \| NF-κB signal pathway \| HuCCT1 with ESP \| ROS↑, peroxiredoxin (2, 3, and 6) ↑, thioredoxin1↑ \| \| \| Villares et al. (2009)96 \| Receptor-1 \| Melanoma \| Cx 43 ↑ \| \| \| Nam et al. (2012)77 \| Free radicals \| HuCCT1 CCA with ESP \| NOX↑, XO↑, iNOS↑, xanthine oxidase↑ \| \| \| Won et al. (2014)83 \| Intercellular adhesion \| HuCCT1 \| Vinculin↑, paxillin↑, MMPs (1, 9 and 13) ↑ \| \| \| Yan et al. (2015)60 \| Cell signal transduction \| Primary biliary epithelial cells with ESP \| TLR2↑, TLR4↑, NF-κB↑, MyD88↑, IkB degradation \| \| \| Matchimakul et al. (2015)72 \| Thioredoxin of Ov \| Cholangiocytes \| Apoptosis↓ \| \| \| Smout et al. (2015)100 \| Granulin of Ov \| Cholangiocytes, angiogenesis \| Wound healing↑, oncogenesis↑ \| \| \| Maeng et al. (2016)78 \| Oxidative stress \| HuCCT1 with ESP \| COX-2↑, 5-LOX↑ \| \| \| Bian et al. (2016)79 \| Oxidative stress \| Mouse liver \| Reactive nitrogen species↑ \| \| \| Bahk and Pak (2016)90 \| TLR signal \| HuCCT1 \| TLRs↑, intracellular free radicals↑ \| \| \| Wang et al. (2017)74 \| Granulin \| ESP \| Metastasis↑ \| \| \| Yuan et al. (2017)80 \| JNK-mediated \| Human, mouse \| Mitochondrial dysfunction, oxidative stress \| \| \| Pak et al. (2017)82 \| NF-κB \| Human CCA cells \| MMP-9 \| \| \| Won et al. (2019)84 \| Cytokines/EMT related proteins \| HuCCT1 with ESP \| IL-6↑, TGF-b↑, E-cadherin↓, N-cadherin↑ \| \| \| Kim et al. (2019)89 \| \| H69 with ESP + NDMA \| Cx 43↑, Cx 26↑, Cx 32↓ \| \| H69 with ESP + NDMA \| Ki67↑, COX-2↑, cytokeratin19↑ \| \| \| Wang et al. (2021)75 \| EGFR \| ESP \| Granulin↑, malignant transformation↑ \| \| \| Smout et al. (2024)73 \| Multi-stage \| Granulin \| Inflammation, chronic wound, fibrosis, hyperplasia, neoplasia \| \| Gene expression \| \| \| \| \| \| Yoo et al. (2011)92 \| Transcriptome \| Expressed sequence tags in cDNA library \| Stage specific expression \| \| \| Pak et al. (2014)91 \| Tumor suppressor genes \| HuCCT1, H69 with ESP \| Let-7a↓, Let-7i↓, miR-124a↓ \| \| Gap junction molecule/Cx alteration \| \| \| \| \| \| Bode et al. (2002)98 \| Gap junction \| Cholangiocytes \| Cx 43 synchronized Ca++ oscillations \| \| \| Li et al. (2011)95 \| Gap junctions \| Bladder smooth muscle cells \| Proinflammatory cytokine-induced iNOS↑, Cx 43↑ \| \| \| Balasubramaniyan et al. (2013)94 \| Gap junction \| Rat cirrhosis \| Cx 43↑, Cx 26/32↓ \| \| \| Faniku et al. (2018)93 \| Peptide gap27 \| Skin model \| Cx 43↑ in wound healing \| \| \| Kim et al. (2019)89 \| Gap junction \| H69 with ESP \| Cx 43↑, Cx 32↓ \| \| H69 with ESP + NDMA \| Cx 43↑, Cx 26↑, Cx 32↓ \| \| H69 with ESP + NDMA \| Ki67↑, COX-2↑, cytokeratin19↑ \| \| Review \| \| \| \| \| \| Kim et al. (2016)76 \| Apoptosis \| Review \| Apoptosis inhibition \| Open in a new tab ESP = excretory-secretory product of C. sinensis, NDMA = N-nitrosodimethylamine, Cx = connexin, NF-κB = nuclear factor kappa B, ROS = reactive oxygen species, MMP = matrix metalloproteinase, IL = interleukin, TGF = tumor growth factor, Ov = O. viverrini, TLR = Toll-like receptor, CCA = cholangiocarcinoma, EMT = epithelial-mesenchymal transition, EGFR = epidermal growth factor receptor, iNOS = inducible nitric oxide synthase. CONCLUSION There is substantial evidence supporting the carcinogenesis of C. sinensis from both epidemiological and experimental studies. Epidemiological research has shown an overall OR of 4.7 for CCA associated with clonorchiasis, and 10% of Korean CCA patients linked to the infection. The carcinogenic factors include ESPs with granulin of the parasite, a diet rich in nitrosamines, co-infections with microbes, and endogenous radicals produced by local chronic inflammation. These stimulants lead to expression of oncogenes and suppression of apoptosis. Eventually, oncogenic intracellular molecules accumulate, and the altered gap-junction proteins disrupt intercellular communication in the bile duct epithelial cells. These neoplastic cells then proliferate forming CCA masses. Fig. 8 draws the theoretical process of carcinogenesis by C. sinensis. Since C. sinensis was classified as a biological carcinogen for human CCA, clonorchiasis poses a significant public health threat in East Asia, where it still remains prevalent. 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14796	https://www.ck12.org/flexi/precalculus/systems-of-linear-equations-in-three-variables/	Systems of Linear Equations in Three Variables Concept Summary: \| \| \| A consistent system has at least one solution, while an inconsistent system has no solutions. Three equations are linearly independent if each equation cannot be produced by a linear combination of the other two. Common techniques for solving systems of equations involve swapping rows, dividing and multiplying a row by a constant, and adding or subtracting a multiple of one row to another. \| Add: @$\begin{align}3 a(2 b+5 c), 3 c(2 a+2 b)\end{align}@$ Add: @$\begin{align}3 a(2 b+5 c), 3 c(2 a+2 b)\end{align}@$ Add: @$\begin{align}3 a(a-b+c), 2 b(a-b+c)\end{align}@$ Add: @$\begin{align}3 a(a-b+c), 2 b(a-b+c)\end{align}@$ Add: @$\begin{align}x y^{2} z^{2}+3 x^{2} y^{2} z-4 x^{2} y z^{2},-9 x^{2} y^{2} z+3 x y^{2} z^{2}+x^{2} y z^{2}\end{align}@$ Add: @$\begin{align}x y^{2} z^{2}+3 x^{2} y^{2} z-4 x^{2} y z^{2},-9 x^{2} y^{2} z+3 x y^{2} z^{2}+x^{2} y z^{2}\end{align}@$ Add: @$\begin{align}9 a x,+3 b y-c z,-5 b y+a x+3 c z\end{align}@$ Add: @$\begin{align}9 a x,+3 b y-c z,-5 b y+a x+3 c z\end{align}@$ @$\begin{align}(-1) \times(-2) \times(-3)=1 \times 2 \times 3\end{align}@$ @$\begin{align}(-1) \times(-2) \times(-3)=1 \times 2 \times 3\end{align}@$ Solve the following pairs of equations: @$\begin{align}x+y=3.3\end{align}@$ @$\begin{align}\frac{0.6}{3 x-2 y}=-1, \quad 3 x-2 y \neq 0\end{align}@$ Solve the following pairs of equations: @$\begin{align}x+y=3.3\end{align}@$ @$\begin{align}\frac{0.6}{3 x-2 y}=-1, \quad 3 x-2 y \neq 0\end{align}@$ For which value(s) of @$\begin{align}k\end{align}@$ will the pair of equations @$$ \begin{aligned} & k x+3 y=k-3 \ & 12 x+k y=k \end{aligned} @$$ have no solution? For which value(s) of @$\begin{align}k\end{align}@$ will the pair of equations @$$ \begin{aligned} & k x+3 y=k-3 \ & 12 x+k y=k \end{aligned} @$$ have no solution? How to solve 3-variable equations? How to solve a 3-variable system of equations? How to solve a system of equations with 3 variables? How to solve a system of 3 equations? How do you solve a system of equations with 3 variables? Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is 36, find the numbers. Solve the system. If there is no solution or if there are infinitely many solutions and the system's equations are dependent, so state. @$\begin{align}\begin{array}{rr} 3x+y=&-2 \ x+y-z=&-6 \ 5x+2y+z=&0 \end{array}\end{align}@$ Select the correct choice below and fill in any answer boxes within your choice. A. There is one solution. The solution set is @$\begin{align}{(□,□,□)} (Simplify\end{align}@$ your answers.) B. There are infinitely many solutions. C. There is no solution. Determine the intersection, if any, of the planes with the equations x + y – z + 12 = 0 and 3x + 4y – 2z + 8 = 0? A test is worth 100 points. The test is made up of 40 items. Each item is worth either 2 points or 3 points. Which matrix equation and solution represent the situation? Find @$\begin{align}k\end{align}@$ such that the system of equations has no unique solution. @$\begin{align} \begin{array}{l} k x + 0 y + 0 z = k \ 6 x - y + z = 26 \ -6 x + k y - z = -26 \ x + 5 y - z = -29 \end{array} \end{align}@$ Find @$\begin{align}k\end{align}@$ such that the system of equations has no unique solution. Find three different integer solutions to the equation 3x - y = 1. 1st point: ( , ) 2nd point: ( , ) 3rd point: ( , ) @$\begin{align} \begin{array}{l} 0.2x + 0.4y + 0.3z = 30 \ 0.3x + 0.4y + 0.4z = 40 \ 0.5x + 0.2y + 0.3z = 35 \end{array} \end{align}@$ Find the values of x, y, and z. @$\begin{align} \begin{array}{l} 0.2x + 0.4y + 0.3z = 30 \ 0.3x + 0.4y + 0.4z = 40 \ 0.5x + 0.2y + 0.3z = 35 \end{array} \end{align}@$ Find the values of x, y, and z. Use Lagrange multipliers to solve the following exercise. Find three positive integers @$\begin{align} x \end{align}@$, @$\begin{align} y \end{align}@$, and @$\begin{align} z \end{align}@$ that satisfy the given conditions: The product is 64 and the sum is a minimum. What are @$\begin{align}(x, y, z) \textrm{, if } (x, y, z)=(24 \times)\end{align}@$? Find three different integer solutions to the equation 2x + 3y = 6. 1st point: ( , ) 2nd point: ( , ) 3rd point: ( , ) Three workers, X, Y, and Z, can complete a task in 3, 4, and 6 hours, respectively. How long will it take to complete the task if all three work together? Solve the system of equations: x+y=15, -y+3z=15, x+11z=14 Find an equation for the line that passes through the points (3, 3) and (−3, 6). Marie weighs 90 pounds, Alice weighs 70 pounds, and Josephine weighs 110 pounds. What is their average weight in pounds? Among A, B, C, D, E, and F, each one of them has a different weight. F is not the lightest. C is heavier than E and F. B is heavier than four people. E is heavier than D and F. A is heavier than B. Then, how many people are heavier than C? Polly is 47 years old. Alice is 59 years younger than Dana and 38 years younger than Polly. How old is Dana? A bow factory puts 4 bows in each box. It takes 4 1/3 inches of ribbon to make each bow. How many inches of ribbon does it take to make enough bows to fill a box? When is a vector considered to be linearly independent? When can vectors be classified as linearly independent? When are the vectors linearly independent? What defines a system as inconsistent? What does linearly independent mean? What is meant by linearly independent? How can you tell if a system is inconsistent? What is a consistent system of equations?
14797	http://hyperphysics.phy-astr.gsu.edu/hbase/Class/PhSciLab/frictioni.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| Work, Energy and Friction Frictional resistance between two objects which are at rest with respect to one another is called static friction, but if they have relative motion it is called kinetic friction. Generally speaking, if you push on a block at rest, it will resist with a force which exactly counters your applied force up until a point where you overcome the friction and the block starts to move. Then when the block starts to move, it usually takes less force to keep the block in motion at a constant speed. The maximum force at the point at which the block starts to move, and the force required to keep in constant velocity motion are characterized by static and kinetic coefficients of friction. These coefficients are just the applied forces under those two conditions divided by the force pressing the surfaces together. 1. Place the wooden track horizontally on the table, attach a string and weight hanger as indicated. Placing a wooden block at some point near the end of the board , find out how much mass must be added to start the block in motion from rest. Then find the mass of the block and the coefficient of static friction. Note: in all your measurements of the amount of hanging mass, be sure to include the mass of the hanger, which is typically 50 grams. Note that ms is a ratio of forces which are determined by the weights of the blocks, but since the masses are also proportional to the weights, the coefficient of friction is 2. Using the same setup, give the block a slight push on each trial and find out how much hanging mass is required to maintain the block in motion at a constant velocity. You will have to make a judgment by watching the motion to find the proper hanging mass so that, once bumped into motion, the block moves without speeding up or slowing down. \| \| \| mass added ________________________ \| \| \| \| \| --- \| coefficient \| \| = ____________________ \| 3. You now have some experience with how much mass it takes to put the block in motion. Choose a convenient mass significantly above that threshold mass, like the nearest 100 grams above the amount required to put it in motion. Place the block on the wooden track at about the same starting point and suspend the hanging mass at a height so that it can hit the floor and such that the wooden block will move and stop without hitting the pulley. Practice a bit until you have an experimental setup which works. \| \| \| --- \| \| Hanging mass ________________ \| Height of mass _______________ \| \| Time to floor ________________ \| Distance block moved _______________ \| Before you released the hanging mass, it had potential energy equal to its weight (mg) times its height from the floor. Evaluate that initial potential energy by converting the mass to kilograms, using 9.8 m/s2 for g, and multiplying times the initial height in meters. \| \| \| --- \| \| Initial potential energy PE = mgh = ____________ joules. Evaluate the final kinetic energy of the hanging mass, KE = 0.5 mv2 just before it hits the floor Distance = ________ meters, time = ___________ s, average velocity = __________m/s Final velocity __________ m/s, KE = 0.5 mv2 = _____________ joules. \| \| How does your measured kinetic energy compare with the original potential energy? Does this make sense in light of the conservation of energy principle? Why is the KE less than the PE? Now examine the amount of work done on the block. If the block moves and then stops on the horizontal surface, then its initial and final kinetic energies are zero. That means that the work done on it by the falling mass was counteracted by friction so that all the energy supplied to the block is dissipated by friction. The work done against friction is equal to the difference between the potential and kinetic energies you measured above. Work against friction = force of friction x distance = m mblock g dblock Setting the work done by friction against the motion equal to the work done on the block by the falling mass allows you to get another evaluation of the coefficient of friction m since you now know all the other parameters in the above equation. Work done on block by falling mass = ______________ joules. Coefficient of friction = ______________. Is this coefficient of friction closer to your static or kinetic friction coefficients from above? Which of the ways to determine the coefficient of friction would you consider to be most reliable and why? \| \| \| Example data and calculation. \| Friction puzzle: hold a meter stick horizontal, supported by one finger under each end. Move your fingers slowly toward each other. Describe the motion of the meter stick -- does it slide smoothly over both fingers? Describe its motion. Record your thoughts about why it moves that way. Equipment: Friction, Work and Energy \| \| \| \| --- \| Wooden track with pulley \| block \| meter stick \| \| mass hanger \| scale \| set of assorted masses \| \| Index Apparatus movie \| \| HyperPhysics NSCI 3001 NSCI 3001 Hands-on exercises \| Go Back \|
14798	https://nrhmmanipur.org/wp-content/uploads/2012/08/national_Population_Policy_2000.pdf	1 C O N T E N T S A. Introduction ................................................................................................ 1 B. Objectives.................................................................................................. 2 C. Strategic Themes ....................................................................................... 4 D. Legislation................................................................................................ 11 E. Public Support ......................................................................................... 12 F. New Structures ....................................................................................... 12 G. Funding .................................................................................................... 13 H. Promotional and Motivational Measures for ............................................. 13 Adoption of the Small Family Norm I. Conclusion ............................................................................................... 15 List of Appendices Appendix I .......................................................................................................... 16 Action Plan Appendix II ......................................................................................................... 30 Milestones in the Evolution of the Population Policy of India Appendix III ........................................................................................................ 32 Demographic Profile Appendix IV ....................................................................................................... 36 Unmet Needs & Deficiencies in Contraceptives Services, Health Infrastructure, Specialists and Trained Manpower with Implications for Funding 2 List of Tables 1. Population Projections for India (million) .................................................... 1 2. Anticipated Growth in Population (million) ................................................. 3 3 Projections of Crude Birth Rate, Infant Mortality Rate, ............................. 3 and TFR, if the NPP 2000 is fully implemented. 4 Age Composition as Percentage of .......................................................... 32 the Total Population 5. Population Profile of 9 States and Union Territories ............................... 33 of India with TFR less than or equal to 2.1 6. Population Profile of 11 States and Union Territories ............................. 33 of India with TFR > 2.1 but < 3 7. Population Profile of 11 States and Union Territories ............................. 34 of India with TFR greater than or equal to 3 8. Maternal Mortality Ratios in Asia............................................................ 34 9. Inter-State Differences within India in ...................................................... 35 Maternal Mortality Ratios 10. Infant Mortality Rates in Asia .................................................................. 35 List of Boxes 1. India's Demographic Achievement ............................................................. 1 2. National Socio-Demographic Goals for 2010.............................................. 2 3 A. INTRODUCTION 1. The overriding objective of economic and social development is to improve the quality of lives that people lead, to enhance their well-being, and to provide them with opportunities and choices to become productive assets in society. 2. In 1952, India was the first country in the world to launch a national programme, emphasizing family planning to the extent necessary for reducing birth rates "to stabilize the population at a level consistent with the requirement of national economy"1 . After 1952, sharp declines in death rates were, however, not accompanied by a similar drop in birth rates. The National Health Policy, 1983 stated that replacement levels of total fertility rate2 (TFR) should be achieved by the year 2000. 3. On 11 May, 2000 India is projected to have 1 billion3 (100 crore) people, i.e. 16 percent of the world's population on 2.4 percent of the globe's land area. If current trends continue, India may overtake China in 2045, to become the most populous country in the world. While global population has increased threefold during this century, from 2 billion to 6 billion, the population of India has increased nearly five times from 238 million (23 crores) to 1 billion in the same period. India's current annual increase in population of 15.5 million is large enough to neutralize efforts to conserve the resource endowment and environment. Box 1: India's Demographic Achievement Half a century after formulating the national family welfare programme, India has: v reduced crude birth rate (CBR) from 40.8 (1951) to 26.4 (1998, SRS); v halved the infant mortality rate (IMR) from 146 per 1000 live births (1951) to 72 per 1000 live births (1998, SRS); v quadrupled the couple protection rate (CPR) from 10.4 percent (1971) to 44 percent (1999); v reduced crude death rate (CDR) from 25 (1951) to 9.0 (1998, SRS); v added 25 years to life expectancy from 37 years to 62 years; v achieved nearly universal awareness of the need for and methods of family planning, and v reduced total fertility rate from 6.0 (1951) to 3.3 (1997, SRS). 4. India's population in 1991 and projections to 2016 are as follows: Table 1: Population Projections for India (million)3 March 1991 March 2001 March 2011 March 2016 846.3 1012.4 1178.9 1263.5 1 Milestones in the Evolution of the Population Policy are listed at Appendix II, page 30 2TFR: Average number of children born to a woman during her lifetime. 3Source: Technical Group on Population Projections,Planning Commission. 4 5. Stabilising population is an essential requirement for promoting sustainable development with more equitable distribution. However, it is as much a function of making reproductive health care accessible and affordable for all, as of increasing the provision and outreach of primary and secondary education, extending basic amenities including sanitation, safe drinking water and housing, besides empowering women and enhancing their employment opportunities, and providing transport and communications. 6. The National Population Policy, 2000 (NPP 2000) affirms the commitment of government towards voluntary and informed choice and consent of citizens while availing of reproductive health care services, and continuation of the target free approach in administering family planning services. The NPP 2000 provides a policy framework for advancing goals and prioritizing strategies during the next decade, to meet the reproductive and child health needs of the people of India, and to achieve net replacement levels (TFR) by 2010. It is based upon the need to simultaneously address issues of child survival, maternal health, and contraception, while increasing outreach and coverage of a comprehensive package of reproductive and child heath services by government, industry and the voluntary non-government sector, working in partnership. B. OBJECTIVES 7. The immediate objective of the NPP 2000 is to address the unmet needs for contraception, health care infrastructure, and health personnel, and to provide integrated service delivery for basic reproductive and child health care. The medium-term objective is to bring the TFR to replacement levels by 2010, through vigorous implementation of inter-sectoral operational strategies. The long-term objective is to achieve a stable population by 2045, at a level consistent with the requirements of sustainable economic growth, social development, and environmental protection. 8. In pursuance of these objectives, the following National Socio-Demographic Goals to be achieved in each case by 2010 are formulated: Box 2: National Socio-Demographic Goals for 2010 (1) Address the unmet needs for basic reproductive and child health services, supplies and infrastructure. (2) Make school education up to age 14 free and compulsory, and reduce drop outs at primary and secondary school levels to below 20 percent for both boys and girls. (3) Reduce infant mortality rate to below 30 per 1000 live births. (4) Reduce maternal mortality ratio to below 100 per 100,000 live births. (5) Achieve universal immunization of children against all vaccine preventable diseases. (6) Promote delayed marriage for girls, not earlier than age 18 and preferably after 20 years of age. 5 (7) Achieve 80 percent institutional deliveries and 100 percent deliveries by trained persons. (8) Achieve universal access to information/counseling, and services for fertility regulation and contraception with a wide basket of choices. (9) Achieve 100 per cent registration of births, deaths, marriage and pregnancy. (10) Contain the spread of Acquired Immunodeficiency Syndrome (AIDS), and promote greater integration between the management of reproductive tract infections (RTI) and sexually transmitted infections (STI) and the National AIDS Control Organisation. (11) Prevent and control communicable diseases. (12) Integrate Indian Systems of Medicine (ISM) in the provision of reproductive and child health services, and in reaching out to households. (13) Promote vigorously the small family norm to achieve replacement levels of TFR. (14) Bring about convergence in implementation of related social sector programs so that family welfare becomes a people centred programme. If the NPP 2000 is fully implemented, we anticipate a population of 1107 million (110 crores) in 2010, instead of 1162 million (116 crores) projected by the Technical Group on Population Projections: Table 2: Anticipated Growth in Population (million) Year If current trends continue If TFR 2.1 is achieved by 2010 Total Increase in Total Increase in Population population population population 1991 846.3 -846.3 -1996 934.2 17.6 934.2 17.6 1997 949.9 15.7 949.0 14.8 2000 996.9 15.7 991.0 14.0 2002 1027.6 15.4 1013.0 11.0 2010 1162.3 16.8 1107.0 11.75 Similarly, the anticipated reductions in the birth, infant mortality and total fertility rates are: Table 3: Projections of Crude Birth Rate, Infant Mortality Rate, and TFR, if the NPP 2000 is fully implemented. Year Crude Birth Rate Infant Mortality Rate Total Fertility Rate 1997 27.2 71 3.3 1998 26.4 72 3.3 2002 23.0 50 2.6 2010 21.0 30 2.1 Source for Tables 2 and 3: Ministry of Health and Family Welfare 6 9. Population growth in India continues to be high on account of : F The large size of the population in the reproductive age-group (estimated contribution 58 percent). An addition of 417.2 million between 1991 and 2016 is anticipated despite substantial reductions in family size in several states, including those which have already achieved replacement levels of TFR. This momentum of increase in population will continue for some more years because high TFRs in the past have resulted in a large proportion of the population being currently in their reproductive years. It is imperative that the the reproductive age group adopts without further delay or exception the "small family norm", for the reason that about 45 percent of population increase is contributed by births above two children per family. F Higher fertility due to unmet need for contraception (estimated contribution 20 percent). India has 168 million eligible couples, of which just 44 percent are currently effectively protected. Urgent steps are currently required to make contraception more widely available, accessible, and affordable. Around 74 percent of the population lives in rural areas, in about 5.5 lakh villages, many with poor communications and transport. Reproductive health and basic health infrastructure and services often do not reach the villages, and, accordingly, vast numbers of people cannot avail of these services. F High wanted fertility due to the high infant mortality rate (IMR) (estimated contribution about 20 percent). Repeated child births are seen as an insurance against multiple infant (and child) deaths and accordingly, high infant mortality stymies all efforts at reducing TFR. F Over 50 percent of girls marry below the age of 18, the minimum legal age of marriage, resulting in a typical reproductive pattern of "too early, too frequent, too many". Around 33 percent births occur at intervals of less than 24 months, which also results in high IMR. The country's demographic profile is given in Appendix III (pages 32-35). C. STRATEGIC THEMES 10. We identify 12 strategic themes which must be simultaneously pursued in "stand alone" or inter-sectoral programmes in order to achieve the national socio-demographic goals for 2010. These are presented below: (i) Decentralised Planning and Programme Implementation 11. The 73rd and 74th Constitutional Amendments Act, 1992, made health, family welfare, and education a responsibility of village panchayats. The panchayati raj institutions are an important means of furthering decentralised planning and programme implementation in the context of the NPP 2000. However, in order to realize their potential, they need strengthening by further delegation of administrative and financial powers, including powers of resource mobilization. 7 Further, since 33 percent of elected panchayat seats are reserved for women, representative committees of the panchayats (headed by an elected woman panchayat member) should be formed to promote a gender sensitive, multi-sectoral agenda for population stabilisation, that will "think, plan and act locally, and support nationally". These committees may identify area-specific unmet needs for reproductive health services, and prepare need-based, demand-driven, socio-demographic plans at the village level, aimed at identifying and providing responsive, people-centred and integrated, basic reproductive and child health care. Panchayats demonstrating exemplary performance in the compulsory registration of births, deaths, marriages, and pregnancies, universalizing the small family norm, increasing safe deliveries, bringing about reductions in infant and maternal mortality, and promoting compulsory education up to age 14, will be nationally recognized and honored. (ii) Convergence of Service Delivery at Village Levels 12. Efforts at population stabilisation will be effective only if we direct an integrated package of essential services at village and household levels. Below district levels, current health infrastructure includes 2,500 community health centres, 25,000 primary health centres (each covering a population of 30,000), and 1.36 lakh subcentres (each covering a population of 5,000 in the plains and 3,000 in hilly regions)4. Inadequacies in the existing health infrastructure have led to an unmet need of 28 percent for contraception services, and obvious gaps in coverage and outreach. Health care centres are over-burdened and struggle to provide services with limited personnel and equipment. Absence of supportive supervision, lack of training in inter-personal communication, and lack of motivation to work in rural areas, together impede citizens' access to reproductive and child health services, and contribute to poor quality of services and an apparent insensitivity to client's needs. The last 50 years have demonstrated the unsuitability of these yardsticks for provision of health care infrastructure, particularly for remote, inaccessible, or sparsely populated regions in the country like hilly and forested areas, desert regions and tribal areas. We need to promote a more flexible approach, by extending basic reproductive and child health care through mobile clinics and counseling services. Further, recognizing that government alone cannot make up for the inadequacies in health care infrastructure and services, in order to resolve unmet needs and extend coverage, the involvement of the voluntary sector and the non-government sector in partnership with the government is essential. 13. Since the management, funding, and implementation of health and education programmes has been decentralised to panchayats, in order to reach household levels, a one-stop, integrated and coordinated service delivery should be provided at village levels, for basic reproductive and child health services. A vast increase in the number of trained birth attendants, at least two per village, is necessary to universalise coverage and outreach of ante-natal, natal and post-natal health care. An equipped maternity hut in each village should be set up to serve as a delivery room, with functioning midwifery kits, basic medication for essential obstetric aid, and indigenous medicines and supplies for maternal and new born care. A key feature of the integrated service delivery will be the registration at village levels, of births, deaths, marriage, and pregnancies. Each village should maintain a list of community midwives and trained birth attendants, village health guides, panchayat sewa sahayaks, primary school teachers and aanganwadi workers who may be entrusted with various responsibilities in the implementation of integrated service delivery. 4 Source: MOHFW Statistics, 1998. 8 14. The panchayats should seek the help of community opinion makers to communicate the benefits of smaller, healthier families, the significance of educating girls, and promoting female participation in paid employment. They should also involve civil society in monitoring the availability, accessibility and affordability of services and supplies. Operational strategies are described in the Action Plan at Appendix I (pages16-17). (iii) Empowering Women for Improved Health and Nutrition 15. The complex socio-cultural determinants of women's health and nutrition have cumulative effects over a lifetime. Discriminatory childcare leads to malnutrition and impaired physical development of the girl child. Undernutrition and micronutrient deficiency in early adolescence goes beyond mere food entitlements to those nutrition related capabilities that become crucial to a woman's well-being, and through her, to the well-being of children. The positive effects of good health and nutrition on the labour productivity of the poor is well documented. To the extent that women are over-represented among the poor, interventions for improving women's health and nutrition are critical for poverty reduction. 16. Impaired health and nutrition is compounded by early childbearing, and consequent risk of serious pregnancy related complications. Women's risk of premature death and disability is highest during their reproductive years. Malnutrition, frequent pregnancies, unsafe abortions, RTI and STI, all combine to keep the maternal mortality ratio in India among the highest globally. 17. Maternal mortality is not merely a health disadvantage, it is a matter of social injustice. Low social and economic status of girls and women limits their access to education, good nutrition, as well as money to pay for health care and family planning services. The extent of maternal mortality is an indicator of disparity and inequity in access to appropriate health care and nutrition services throughout a lifetime, and particularly during pregnancy and child-birth, and is a crucial factor contributing to high maternal mortality. 18. Programmes for Safe Motherhood, Universal Immunisation, Child Survival and Oral Rehydration have been combined into an Integrated Reproductive and Child Health Programme, which also includes promoting management of STIs and RTIs. Women's health and nutrition problems can be largely prevented or mitigated through low cost interventions designed for low income settings. 19. The voluntary non-government sector and the private corporate sector should actively collaborate with the community and government through specific commitments in the areas of basic reproductive and child health care, basic education, and in securing higher levels of participation in the paid work force for women. Operational strategies are described in the Action Plan at Appendix I (pages 17-20). 9 (iv) Child Health and Survival 20. Infant mortality is a sensitive indicator of human development. High mortality and morbidity among infants and children below 5 years occurs on account of inadequate care, asphyxia during birth, premature birth, low birth weight, acute respiratory infections, diarrhoea, vaccine preventable diseases, malnutrition and deficiencies of nutrients, including Vitamin A. Infant mortality rates have not significantly declined in recent years. 21. Our priority is to intensify neo-natal care. A National Technical Committee should be set up, consisting principally of consultants in obstetrics, pediatrics (neonatologists), family health, medical research and statistics from among academia, public health professionals, clinical practitioners and government. Its terms of reference should include prescribing perinatal audit norms, developing quality improvement activities with monitoring schedules and suggestions for facilitating provision of continuing medical and nursing education to all perinatal health care providers. Implementation at the grass-roots must benefit from current developments in the fields of perinatology and neonatology. The baby friendly hospital initiative (BFHI) should be extended to all hospitals and clinics, up to subcentre levels. Additionally, besides promoting breast-feeding and complementary feeds, the BFHI should include updating of skills of trained birth attendants to improve new born care practices to reduce the risks of hypothermia and infection. Essential equipment for the new born must be provided at subcentre levels. 22. Child survival interventions i.e. universal immunisation, control of childhood diarrhoeas with oral rehydration therapies, management of acute respiratory infections, and massive doses of Vitamin A and food supplements have all helped to reduce infant and child mortality and morbidity. With intensified efforts, the eradication of polio is within reach. However, the decline in standards, outreach and quality of routine immunisation is a matter of concern. Significant improvements need to be made in the quality and coverage of the routine immunisation programme. Operational strategies are described in the Action Plan at Appendix I (pages 21-22). (v) Meeting the Unmet Needs for Family Welfare Services 23. In both rural and urban areas there continue to be unmet needs for contraceptives, supplies and equipment for integrated service delivery, mobility of health providers and patients, and comprehensive information. It is important to strengthen, energise and make accountable the cutting edge of health infrastructure at the village, subcentre and primary health centre levels, to improve facilities for referral transportation, to encourage and strengthen local initiatives for ambulance services at village and block levels, to increase innovative social marketing schemes for affordable products and services and to improve advocacy in locally relevant and acceptable dialects. Operational strategies are described in the Action Plan in Appendix I (page 22). 10 (vi) Under-Served Population Groups (a) Urban Slums 24. Nearly 100 million people live in urban slums, with little or no access to potable water, sanitation facilities, and health care services. This contributes to high infant and child mortality, which in turn perpetuate high TFR and maternal mortality. Basic and primary health care, including reproductive and child health care, needs to be provided. Coordination with municipal bodies for water, sanitation and waste disposal must be pursued, and targeted information, education and communication campaigns must spread awareness about the secondary and tertiary facilities available. Operational strategies are described in the Action Plan in Appendix I (pages 22-23). (b) Tribal Communities, Hill Area Populations and Displaced and Migrant Populations 25. In general, populations in remote and low density areas do not have adequate access to affordable health care services. Tribal populations often have high levels of morbidity arising from poor nutrition, particularly in situations where they are involuntarily displaced or resettled. Frequently, they have low levels of literacy, coupled with high infant, child, and maternal mortality. They remain under-served in the coverage of reproductive and child health services. These communities need special attention in terms of basic health, and reproductive and child health services. The special needs of tribal groups which need to be addressed include the provision of mobile clinics that will be responsive to seasonal variations in the availability of work and income. Information and counseling on infertility, and regular supply of standardised medication will be included. Operational strategies are described in the Action Plan at Appendix I (pages 23). (c) Adolescents 26. Adolescents represent about a fifth of India's population. The needs of adolescents, including protection from unwanted pregnancies and sexually transmitted diseases (STD), have not been specifically addressed in the past. Programmes should encourage delayed marriage and child-bearing, and education of adolescents about the risks of unprotected sex. Reproductive health services for adolescent girls and boys is especially significant in rural India, where adolescent marriage and pregnancy are widely prevalent. Their special requirements comprise information, counseling, population education, and making contraceptive services accessible and affordable, providing food supplements and nutritional services through the ICDS, and enforcing the Child Marriage Restraint Act, 1976. Operational strategies are described in the Action Plan in Appendix I (page 24). 11 (d) Increased Participation of Men in Planned Parenthood 27. In the past, population programmes have tended to exclude menfolk. Gender inequalities in patriarchal societies ensure that men play a critical role in determining the education and employment of family members, age at marriage, besides access to and utilisation of health, nutrition, and family welfare services for women and children. The active involvement of men is called for in planning families, supporting contraceptive use, helping pregnant women stay healthy, arranging skilled care during delivery, avoiding delays in seeking care, helping after the baby is born and, finally, in being a responsible father. In short, the active cooperation and participation of men is vital for ensuring programme acceptance. Further, currently, over 97 percent of sterilisations are tubectomies and this manifestation of gender imbalance needs to be corrected. The special needs of men include re-popularising vasectomies, in particular no-scalpel vasectomy as a safe and simple procedure, and focusing on men in the information and education campaigns to promote the small family norm. Operational strategies are described in the Action Plan in Appendix I (page 24). (vii) Diverse Health Care Providers 28. Given the large unmet need for reproductive and child health services, and inadequacies in health care infrastructure it is imperative to increase the numbers and diversify the categories of health care providers. Ways of doing this include accrediting private medical practitioners and assigning them to defined beneficiary groups to provide these services; revival of the system of licensed medical practitioner who, after appropriate certification from the Indian Medical Association (IMA), could provide specified clinical services. Operational strategies are described in the Action Plan at Appendix I (pages 24-25). (viii) Collaboration With and Commitments from Non-Government Organisations and the Private Sector 29. A national effort to reach out to households cannot be sustained by government alone. We need to put in place a partnership of non-government voluntary organizations, the private corporate sector, government and the community. Triggered by rising incomes and institutional finance, private health care has grown significantly, with an impressive pool of expertise and management skills, and currently accounts for nearly 75 percent of health care expenditures. However, despite their obvious potential, mobilising the private (profit and non-profit) sector to serve public health goals raises governance issues of contracting, accreditation, regulation, referral, besides the appropriate division of labour between the public and private health providers, all of which need to be addressed carefully. Where government interventions or capacities are insufficient, and the participation of the private sector unviable, focused service delivery by NGOs may effectively complement government efforts. Operational strategies are described in the Action Plan in Appendix I (pages 25-27). 12 (ix) Mainstreaming Indian Systems of Medicine and Homeopathy 30. India's community supported ancient but living traditions of indigenous systems of medicine has sustained the population for centuries, with effective cures and remedies for numerous conditions, including those relating to women and children, with minimal side effects. Utilisation of ISMH in basic reproductive and child health care will expand the pool of effective health care providers, optimise utilisation of locally based remedies and cures, and promote low-cost health care. Guidelines need to be evolved to regulate and ensure standardisation, efficacy and safety of ISMH drugs for wider entry into national markets. 31. Particular challenges include providing appropriate training, and raising awareness and skill development in reproductive and child health care to the institutionally qualified ISMH medical practitioners. The feasibility of utilising their services to fill in gaps in manpower at village levels, and at subcentres and primary health centres may be explored. ISMH institutions, hospitals and dispensaries may be utilised for reproductive and child health care programmes. At village levels, the services of the ISMH "barefoot doctors", after appropriate training, may be utilised for advocacy and counseling, for distributing supplies and equipment, and as depot holders. ISMH practices may be applied at village maternity huts, and at household levels, for ante-natal, natal and post natal care, and for nurture of the new born. Operational strategies are described in the Action Plan in Appendix I (page 27). (x) Contraceptive Technology and Research on Reproductive and Child Health 32. Government must constantly advance, encourage, and support medical, social science, demographic and behavioural science research on maternal, child and reproductive health care issues. This will improve medical techniques relevant to the country's needs, and strengthen programme and project design and implementation. Consultation and frequent dialogue by Government with the existing network of academic and research institutions in allopathy and ISMH, and with other relevant public and private research institutions engaged in social science, demography and behavioural research must continue. The International Institute of Population Sciences, and the population research centres which have been set up to pursue applied research in population related matters, need to be revitalised and strengthened. 33. Applied research relies upon constant monitoring of performance at the programme and project levels. The National Health and Family Welfare Survey provides data on key health and family welfare indicators every five years. Data from the first National Family Health Survey (NFHS-1), 1992-93, has been updated by NFHS-2, 1998-99, to be published shortly. Annual data is generated by the Sample Registration Survey, which, inter alia, maps at state levels the birth, death and infant mortality rates. Absence of regular feedback has been a weakness in the family welfare progamme. For this reason, the Department of Family Welfare is strengthening its management information systems (MIS) and has commenced during 1998, a system of ascertaining impacts and outcomes through district surveys and facility surveys. The district surveys cover 50% districts every year, so that every 2 years there is an update on every district in the country. The facility surveys ascertain the availability of infrastructure and services up to primary health centre level, covering one district per month. The feedback from both these surveys enable remedial action at district and sub-district levels. Operational strategies are described in the Action Plan in Appendix I (page 27). 13 (xi) Providing for the Older Population 34. Improved life expectancy is leading to an increase in the absolute number and proportion of persons aged 60 years and above, and is anticipated to nearly double during 1996-2016, from 62.3 million to 112.9 million5 . When viewed in the context of significant weakening of traditional support systems, the elderly are increasingly vulnerable, needing protection and care. Promoting old age health care and support will, over time, also serve to reduce the incentive to have large families. 35. The Ministry of Social Justice and Empowerment has adopted in January 1999 a National Policy on Older Persons. It has become important to build in geriatric health concerns in the population policy. Ways of doing this include sensitising, training and equipping rural and urban health centres and hospitals for providing geriatric health care; encouraging NGOs to design and implement formal and informal schemes that make the elderly economically self-reliant; providing for and routinising screening for cancer, osteoporosis, and cardiovascular conditions in primary health centres, community health centres, and urban health care centres at primary, secondary and tertiary levels; and exploring tax incentives to encourage grown-up children to look after their aged parents. Operational strategies are described in the Action Plan in Appendix I (page 28). (xii) Information, Education, and Communication 36. Information, education and communication (IEC) of family welfare messages must be clear, focused and disseminated everywhere, including the remote corners of the country, and in local dialects. This will ensure that the messages are effectively conveyed. These need to be strengthened and their outreach widened, with locally relevant, and locally comprehensible media and messages. On the model of the total literacy campaigns which have successfully mobilised local populations, there is need to undertake a massive national campaign on population related issues, via artists, popular film stars, doctors, vaidyas, hakims, nurses, local midwives, women's organizations, and youth organizations. Operational strategies are described in the Action Plan in Appendix I (pages 28-29). D. LEGISLATION 37. As a motivational measure, in order to enable state governments to fearlessly and effectively pursue the agenda for population stabilisation contained in the National Population Policy, 2000, one legislation is considered necessary. It is recommended that the 42nd Constitutional Amendment that freezes till 2001, the number of seats to the Lok Sabha and the Rajya Sabha based on the 1971 Census be extended up to 2026. 5 Source: Technical group on Population Projections, Planning Commission. 14 E. PUBLIC SUPPORT 38. Demonstration of strong support to the small family norm, as well as personal example, by political, community, business, professional and religious leaders, media and film stars, sports personalities, and opinion makers, will enhance its acceptance throughout society. The government will actively enlist their support in concrete ways. F. NEW STRUCTURES 39. The NPP 2000 is to be largely implemented and managed at panchayat and nagar palika levels, in coordination with the concerned state/Union Territory administrations. Accordingly, the specific situation in each state/UT must be kept in mind. This will require comprehensive and multisectoral coordination of planning and implementation between health and family welfare on the one hand, along with schemes for education, nutrition, women and child development, safe drinking water, sanitation, rural roads, communications, transportation, housing, forestry development, environmental protection, and urban development. Accordingly, the following structures are recommended: (i) National Commission on Population 40. A National Commission on Population, presided over by the Prime Minister, will have the Chief Ministers of all states and UTs, and the Central Minister in charge of the Department of Family Welfare and other concerned Central Ministries and Departments, for example Department of Woman and Child Development, Department of Education, Department of Social Justice and Empowerment in the Ministry of HRD, Ministry of Rural Development, Ministry of Environment and Forest, and others as necessary, and reputed demographers, public health professionals, and NGOs as members. This Commission will oversee and review implementation of policy. The Commission Secretariat will be provided by the Department of Family Welfare. (ii) State / UT Commissions on Population 41. Each state and UT may consider having a State / UT Commission on Population, presided over by the Chief Minister, on the analogy of the National Commission, to likewise oversee and review implementation of the NPP 2000 in the state / UT. (iii) Coordination Cell in the Planning Commission 42. The Planning Commission will have a Coordination Cell for inter-sectoral coordination between Ministries for enhancing performance, particularly in States/UTs needing special attention on account of adverse demographic and human development indicators. 15 (iv) Technology Mission in the Department of Family Welfare 43. To enhance performance, particularly in states with currently below average socio-demographic indices that need focused attention, a Technology Mission in the Department of Family Welfare will be established to provide technology support in respect of design and monitoring of projects and programmes for reproductive and child health, as well as for IEC campaigns. G. FUNDING 44. The programmes, projects and schemes premised on the goals and objectives of the NPP 2000, and indeed all efforts at population stabilisation, will be adequately funded in view of their critical importance to national development. Preventive and promotive services such as ante-natal and post-natal care for women, immunisation for children, and contraception will continue to be subsidised for all those who need the services. Priority in allocation of funds will be given to improving health care infrastructure at the community and primary health centres, subcentre and village levels. Critical gaps in manpower will be remedied through redeployment, particularly in under-served and inaccessible areas, and referral linkages will be improved. In order to implement immediately the Action Plan, it would be necessary to double the annual budget of the Department of Family Welfare to enable government to address the shortfall in unmet needs for health care infrastructure, services and supplies (in Appendix IV, page 36). 45. Even though the annual budget for population stabilisation activities assigned to the Department of Family Welfare has increased over the years, at least 50 percent of the budgetary outlay is deployed towards non-plan activities (recurring expenditures for maintenance of health care infrastructure in the states and UTs, and towards salaries). To illustrate, of the annual budget of Rs. 2920 crores for 1999-2000, nearly Rs 1500 crores is allocated towards non-plan activities. Only the remaining 50 percent becomes available for genuine plan activities, including procurement of supplies and equipment. For these reasons, since 1980 the Department of Family Welfare has been unable to revise norms of operational costs of health infrastructure, which in turn has impacted directly the quality of care and outreach of services provided. H. PROMOTIONAL AND MOTIVATIONAL MEASURES FOR ADOPTION OF THE SMALL FAMILY NORM 46. The following promotional and motivational measures will be undertaken: (i) Panchayats and Zila Parishads will be rewarded and honoured for exemplary performance in universalising the small family norm, achieving reductions in infant mortality and birth rates, and promoting literacy with completion of primary schooling. (ii) The Balika Samridhi Yojana run by the Department of Women and Child Development, to promote survival and care of the girl child, will continue. A cash incentive of Rs. 500 is awarded at the birth of the girl child of birth order 1 or 2. 16 (iii) Maternity Benefit Scheme run by the Department of Rural Development will continue. A cash incentive of Rs. 500 is awarded to mothers who have their first child after 19 years of age, for birth of the first or second child only. Disbursement of the cash award will in future be linked to compliance with ante-natal check up, institutional delivery by trained birth attendant, registration of birth and BCG immunisation. (iv) A Family Welfare-linked Health Insurance Plan will be established. Couples below the poverty line, who undergo sterilisation with not more than two living children, would become eligible (along with children) for health insurance (for hospitalisation) not exceeding Rs. 5000, and a personal accident insurance cover for the spouse undergoing sterilisation. (v) Couples below the poverty line, who marry after the legal age of marriage, register the marriage, have their first child after the mother reaches the age of 21, accept the small family norm, and adopt a terminal method after the birth of the second child, will be rewarded. (vi) A revolving fund will be set up for income-generating activities by village-level self help groups, who provide community-level health care services. (vii) Crèches and child care centres will be opened in rural areas and urban slums. This will facilitate and promote participation of women in paid employment. (viii) A wider, affordable choice of contraceptives will be made accessible at diverse delivery points, with counseling services to enable acceptors to exercise voluntary and informed consent. (ix) Facilities for safe abortion will be strengthened and expanded. (x) Products and services will be made affordable through innovative social marketing schemes. (xi) Local entrepreneurs at village levels will be provided soft loans and encouraged to run ambulance services to supplement the existing arrangements for referral transportation. (xii) Increased vocational training schemes for girls, leading to self-employment will be encouraged. (xiii) Strict enforcement of Child Marriage Restraint Act, 1976. (xiv) Strict enforcement of the Pre-Natal Diagnostic Techniques Act, 1994. (xv) Soft loans to ensure mobility of the ANMs will be increased. 17 (xvi) The 42nd Constitutional Amendment has frozen the number of representatives in the Lok Sabha (on the basis of population) at 1971 Census levels. The freeze is currently valid until 2001, and has served as an incentive for State Governments to fearlessly pursue the agenda for population stabilisation. This freeze needs to be extended until 2026. I. CONCLUSION 47. In the new millenium, nations are judged by the well-being of their peoples; by levels of health, nutrition and education; by the civil and political liberties enjoyed by their citizens; by the protection guaranteed to children and by provisions made for the vulnerable and the disadvantaged. 48. The vast numbers of the people of India can be its greatest asset if they are provided with the means to lead healthy and economically productive lives. Population stabilisation is a multi-sectoral endeavour requiring constant and effective dialogue among a diversity of stakeholders, and coordination at all levels of the government and society. Spread of literacy and education, increasing availability of affordable reproductive and child health services, convergence of service delivery at village levels, participation of women in the paid work force, together with a steady, equitable improvement in family incomes, will facilitate early achievement of the socio-demographic goals. Success will be achieved if the Action Plan contained in the NPP 2000 is pursued as a national movement. 18 Appendix I NATIONAL POPULATION POLICY, 2000 ACTION PLAN Operational Strategies (i) & (ii) Converge Service Delivery at Village Levels 1. Utilise village self help groups to organise and provide basic services for reproductive and child health care, combined with the ongoing Integrated Child Development Scheme (ICDS). Village self help groups are in existence through centrally sponsored schemes of: (a) Department of Women and Child Development, Ministry of HRD, (b) Ministry of Rural Development, and (c) Ministry of Environment and Forests. Organise neighbourhood acceptor groups, and provide them with a revolving fund that may be accessed for income generation activities. The groups may establish rules of eligibility, interest rates, and accountability for which capital may be advanced, usually to be repaid in installments within two years. The repayments may be used to fund another acceptor group in a nearby community, who would exert pressure to ensure timely repayments. Two trained birth attendants and the aanganwadi worker (AWW) should be members of this group. 2. Implement at village levels a one-stop integrated and coordinated service delivery package for basic health care, family planning and maternal and child health related services, provided by the community and for the community. Train and motivate the village self-help acceptor groups to become the primary contact at household levels. Once every fortnight, these acceptor groups will meet, and provide at one place 6 different services for (i) registration of births, deaths, marriage and pregnancy; (ii) weighing of children under 5 years, and recording the weight on a standard growth chart; (iii) counseling and advocacy for contraception, plus free supply of contraceptives; (iv) preventive care, with availability of basic medicines for common ailments: antipyretics for fevers, antibiotic ointments for infections, ORT /ORS1 for childhood diarrhoeas, together with standardised indigenous medication and homeopathic cures; (v) nutrition supplements; and (vi) advocacy and encouragement for the continued enrolment of children in school up to age 14. One health staff, appointed by the panchayat, will be suitably trained to provide guidance. Clustering services for women and children at one place and time at village levels will promote positive interactions in health benefits and reduce service delivery costs. 3. Wherever these village self-help groups have not developed for any reason, community midwives, practitioners of ISMH, retired school teachers and ex-defence personnel may be organised into neighbourhood groups to perform similar functions. 4. At village levels, the aanganwadi centre may become the pivot of basic health care activities, contraceptive counseling and supply, nutrition education and supplementation, as well as pre-school activities. The aanganwadi centres can also function as depots for ORS/basic medicines and contraceptives. 1 Oral Rehydration Therapy /Oral Rehydration Salts 19 5. A maternity hut should be established in each village to be used as the village delivery room, with storage space for supplies and medicines. It should be adequately equipped with kits for midwifery, ante-natal care, and delivery; basic medication for obstetric emergency aid; contraceptives, drugs and medicines for common ailments; and indigenous medicines/supplies for maternal and new-born care. The panchayat may appoint a competent and mature mid-wife, to look after this village maternity hut. She may be assisted by volunteers. 6. Trained birth attendants as well as the vast pool of traditional dais should be made familiar with emergency and referral procedures. This will greatly assist the Auxiliary Nurse Midwife (ANM) at the subcentres to monitor and respond to maternal morbidity/emergencies at village levels. 7. Each village may maintain a list of community mid-wives, village health guides, panchayat sewa sahayaks, trained birth attendants, practitioners of indigenous systems of medicine, primary school teachers and other relevant persons, as well as the nearest institutional health care facilities that may be accessed for integrated service delivery. These persons may also be helpful in involving civil society in monitoring availability, quality and accessibility of reproductive and child health services; in disseminating education and communication on the benefits of smaller and healthier families, with emphasis on education of the girl child; and female participation in the work force. 8. Provide a wider basket of choices in contraception, through innovative social marketing schemes to reach household levels. Comment : Meaningful decentralisation will result only if the convergence of the national family welfare programme with the ICDS programme is strengthened. The focus of the ICDS programme on nutrition improvement at village levels and on pre-school activities must be widened to include maternal and child health care services. Convergence of several related activities at service delivery levels with, in particular, the ICDS programme, is critical for extending outreach and increasing access to services. Intersectoral coordination with appropriate training and sensitisation among field functionaries will facilitate dissemination of integrated reproductive and child health services to village and household levels. People will willingly cooperate in the registration of births, deaths, marriages and pregnancies if they perceive some benefit. At the village level, this community meeting every fortnight, may become their most convenient access to basic health care, both for maternal and child health, as well as for common ailments. Households may participate to receive integrated service delivery, along with information about ongoing micro-credit and thrift schemes. Government and non-government functionaries will be expected to function in harmony to ensure integrated service delivery. The panchayat will promote this coordination and exercise effective supervision. (iii) Empowering Women for Improved Health and Nutrition 1. Create an enabling environment for women and children to benefit from products and services disseminated under the reproductive and child health programme. Cluster services for women and children at the same place and time. This promotes positive interactions in health benefits and reduces service delivery costs. 20 2. As a measure to empower women, open more child care centres in rural areas and in urban slums, where a woman worker may leave her children in responsible hands. This will encourage female participation in paid employment, reduce school drop-out rates, particularly for the girl child, and promote school enrolment as well. The aanganwadis provide a partial solution. 3. T o empower women, pursue programmes of social afforestation to facilitate access to fuelwood and fodder. Similarly, pursue drinking water schemes for increasing access to potable water. This will reduce long absences from home, and the need for large numbers of children to perform such tasks. 4. In any reward scheme intended for household levels, priority may be given to energy saving devices such as solar cookers, or provision of sanitation facilities, or extension of telephone lines. This will empower households, in particular women. 5. Improve district, sub-district and panchayat-level health management with coordination and collaboration between district health officer, sub-district health officer and the panchayat for planning and implementation activities. There is need to: v Strengthen the referral network between the district health office, district hospital and the community health centres, the primary health centres and the subcentres in management of obstetric and neo-natal complications. v Strengthen community health centres to provide comprehensive emergency obstetric and neo-natal care. These may function as clinical training centres as well. Strengthen primary health centres to provide essential obstetric and neo-natal care. Strengthen subcentres to provide a comprehensive range of services, with delivery rooms, counseling for contraception, supplies of free contraceptives, ORS and basic medicines, together with facilities for immunisation. v Establish rigorous problem identification mechanisms through maternal and peri-natal audit, from village level upwards. 6. Ensure adequate transportation at village level, subcentre levels, zila parishads, primary health centres and at community health centres. Identifying women at risk is meaningful only if women with complications can reach emergency care in time. 7. Improve the accessibility and quality of maternal and child health services through: v Deployment of community mid-wives and additional health providers at village levels; cluster services for women and children at the same place and time, from village level upwards, e.g. ante-natal and post-partum care, monitoring infant growth, availability of contraceptives and medicine kits; and routinised immunisations at subcentre levels. v Strengthen the capacity of primary health centres to provide basic emergency obstetric and neo-natal health care. v Involve professional agencies in developing and disseminating training modules for standard procedures in the management of obstetric and neo-natal cases. The aim should be to routinise these procedures at all appropriate levels. v Improve supervision by developing guidance and supervision checklists. 21 8. Monitor performance of maternal and child health services at each level by using the maternal and child health local area monitoring system, which includes monitoring the incidence and coverage of ante-natal visits, deliveries assisted by trained health care personnel and post-natal visits, among other indicators. The ANM at the subcentre should be responsible and accountable for registering every pregnancy and child birth in her jurisdiction, and for providing universal ante-natal and post-natal services. 9. Improve technical skills of maternal and child health care providers by: v Strengthening skills of health personnel and health providers through classroom and on-the-job training in the management of obstetric and neo-natal emergencies. This should include training of birth attendants and community midwives at district-level hospitals in life-saving skills, such as management of asphyxia and hypothermia. v Training on integrated management of childhood illnesses for infants (1week - 2 months). 10. Support community activities such as dissemination of IEC material, including leaflets and posters, and promotion of folk jatras, songs and dances to promote healthy mother and healthy baby messages, along with good management practices to ensure safe motherhood, including early recognition of danger signs. 11. Programme development, comprising: v Partnership in family health and nutrition. The aanganwadi worker will identify women and children in the villages who suffer from malnutrition and/or micro-nutritional deficiencies, including iron, vitamin A, and iodine deficiency; provide nutritional supplements and monitor nutritional status. v Convergence, strengthening, and universalisation of the nutritional programmes of the Department of Family Welfare and the ICDS run by the Department of Women and Child Development, ensuring training and timely supply of food supplements and medicines. v Include STD/RTI and HIV/AIDS prevention, screening and management, in maternal and child health services. v Provide quality care in family planning, including information, increased contraceptive choices for both spacing and terminal methods, increase access to good quality and affordable contraceptive supplies and services at diverse delivery points, counseling about the safety, efficacy and possible side effects of each method, and appropriate follow-up. 12. Develop a health package for adolescents. 13. Expand the availability of safe abortion care. Abortion is legal, but there are barriers limiting women's access to safe abortion services. Some operational strategies are: v Community-level education campaigns should target women, household decision makers and adolescents about the availability of safe abortion services and the dangers of unsafe abortion. 22 v Make safe and legal abortion services more attractive to women and household decision makers by (i) increasing geographic spread; (ii) enhancing affordability; (iii) ensuring confidentiality and (iv) providing compassionate abortion care, including post-abortion counseling. v Adopt updated and simple technologies that are safe and easy, e.g. manual vacuum extraction not necessarily dependant upon anaesthesia, or non-surgical techniques which are non-invasive. v Promote collaborative arrangements with private sector health professionals, NGOs and the public sector, to increase the availability and coverage of safe abortion services, including training of mid-level providers. v Eliminate the current cumbersome procedures for registration of abortion clinics. Simplify and facilitate the establishment of additional training centres for safe abortions in the public, private, and NGO sectors. Train these health care providers in provision of clinical services for safe abortions. v Formulate and notify standards for abortion services. Strengthen enforcement mechanisms at district and sub-district levels to ensure that these norms are followed. v Follow norms-based registration of service provision centres, and thereby switch the onus of meticulous observance of standards onto the provider. v Provide competent post-abortion care, including management of complications and identification of other health needs of post-abortion patients, and linking with appropriate services. As part of post-abortion care, physicians may be trained to provide family planning counseling and services such as sterilisation, and reversible modern methods such as IUDs, as well as oral contraceptives and condoms. v Modify syllabus and curricula for medical graduates, as well as for continuing education and in-house learning, to provide for practical training in the newer procedures. v Ensure services for termination of pregnancy at primary health centres and at community health centres. 14. Develop maternity hospitals at sub-district levels and at community health centres to function as FRUs for complicated and life-threatening deliveries. 15. Formulate and enforce standards for clinical services in the public, private, and NGO sectors. 16. Focus on distribution of non-clinical methods of contraception (condoms and oral contraceptive pills) through free supply, social marketing as well as commercial sales. 17. Create a national network consisting of public, private and NGO centres, identified by a common logo, for delivering reproductive and child health services free to any client. The provider will be compensated for the service provided, on the basis of a coupon, duly counter-signed by the beneficiary, and paid for by a system to be devised. The compensation will be identical to providers across all sectors. The end-user will choose the provider of the service. A group of management experts will devise checks and balances to prevent misuse. 23 (iv) Child Health and Survival 1. Support community activities, from village level upwards to monitor early and adequate ante-natal, natal and post-natal care. Focus attention on neo-natal health care and nutrition. 2. Set up a National Technical Committee on neo-natal care, to align programme and project interventions with newly emerging technologies in neo-natal and peri-natal care. 3. Pursue compulsory registration of births in coordination with the ICDS Programme. 4. After the birth of a child, provide counseling and advocacy about contraception, to encourage adoption of a reversible or a terminal method. This will also contribute to the health and well-being of both mother and child. 5. Improve capacities at health centres in basic midwifery services, essential neo-natal care, including the management of sick neo-nates outside the hospital. 6. Sensitise and train health personnel in the integrated management of childhood illnesses. Standard case management of diarrhoea and acute respiratory infections must be provided at subcentres and primary health centres, with appropriate training, and adequate equipment. Besides, training in this sector may be imparted to health care providers at village levels, especially in indigenous systems. 7. Strengthen critical interventions aimed at bringing about reductions in maternal malnutrition, morbidity and mortality, by ensuring availability of supplies and equipment at village levels, and at sub centres. 8. Pursue rigorously the pulse polio campaign to eradicate polio. 9. Ensure 100 percent routine immunisation for all vaccine preventable diseases, in particular tetanus and measles. 10. As a child survival initiative, explore promotional and motivational measures for couples below the poverty line who marry after the legal age of marriage, to have the first child after the mother reaches the age of 21, and adopt a terminal method of contraception after the birth of the second child. 11. Children form a vulnerable group and certain sub-groups merit focused attention and intervention, such as street children and child labourers. Encourage voluntary groups as well as NGOs to formulate and implement special schemes for these groups of children. 12. Explore the feasibility of a national health insurance covering hospitalisation costs for children below 5 years, whose parents have adopted the small family norm, and opted for a terminal method of contraception after the birth of the second child. 24 13. Expand the ICDS to include children between 6-9 years of age, specifically to promote and ensure 100 percent school enrolment, particularly for girls. Promote primary education with the help of aanganwadi workers, and encourage retention in school till age 14. Education promotes awareness, late marriages, small family size and higher child survival rates. 14. Provide vocational training for girls. This will enhance perception of the immediate utility of educating girls, and gradually raise the average age of marriage. It will also increase enrolment and retention of girls at primary school, and likely also at secondary school levels. Involve NGOs, the voluntary sector and the private sector, as necessary, to target employment opportunities. (v) Meeting the Unmet Needs for Family Welfare Services 1. Strengthen, energise and make publicly accountable the cutting edge of health infrastructure at the village, subcentre and primary health centre levels. 2. Address on priority the different unmet needs detailed in Appendix IV, in particular, an increase in rural infrastructure, deployment of sanctioned and appropriately trained health personnel, and provisioning of essential equipment and drugs. 3. Formulate and implement innovative social marketing schemes to provide subsidised products and services in areas where the existing coverage of the public, private and NGO sectors is insufficient in order to increase outreach and coverage. 4. Improve facilities for referral transportation at panchayat, zilla parishad and primary health centre levels. At subcentres, provide ANMs with soft loans for purchase of mopeds, to enhance their mobility. This will increase coverage of ante-natal and post natal check-ups, which, in turn, and will bring about reductions in maternal and infant mortality. 5. Encourage local entrepreneurs at village and block levels to start ambulance services through special loan schemes, with appropriate vehicles to facilitate transportation of persons requiring emergency as well as essential medical attention. 6. Provide special loan schemes and make site allotments at village levels to facilitate the starting of chemist shops for basic medicines and provision for medical first aid. (vi) Under-Served Population Groups (a) Urban Slums 1. Finalise a comprehensive urban health care strategy. 2. Facilitate service delivery centres in urban slums to provide comprehensive basic health, reproductive and child health services by NGOs and private sector organisations, including corporate houses. 25 3. Promote networks of retired government doctors and para-medical and non-medical personnel who may function as health care providers for clinical and non-clinical services on remunerative terms. 4. Strengthen social marketing programmes for non-clinical family planning products and services in urban slums. 5. Initiate specially targeted information, education and communication campaigns for urban slums on family planning, immunization, ante-natal, natal and post-natal check-ups and other reproductive health care services. Integrate aggressive health education programmes with health and medical care programmes , with emphasis on environmental health, personal hygiene and healthy habits, nutrition education and population education. 6. Promote inter-sectoral coordination between departments/municipal bodies dealing with water and sanitation, industry and pollution, housing, transport, education and nutrition, and women and child development, to deal with unplanned and uncoordinated settlements. 7. Streamline the referral systems and linkages between the primary, secondary and tertiary levels of health care in the urban areas. 8. Link the provision of continued facilities to urban slum dwellers with their observance of the small family norm. (b) Tribal Communities, Hill Area Populations and Displaced and Migrant Populations 1. Many tribal communities are dwindling in numbers, and may not need fertility regulation. Instead, they may need information and counseling in respect of infertility. 2. The NGO sector may be encouraged to formulate and implement a system of preventive and curative health care that responds to seasonal variations in the availability of work, income and food for tribal and hill area communities and migrant and displaced populations. To begin with, mobile clinics may provide some degree of regular coverage and outreach. 3. Many tribal communities are dependent upon indigenous systems of medicine which necessitates a regular supply of local flora, fauna and minerals, or of standardised medication derived from these. Husbandry of such local resources and of preparation and distribution of standardised formulations should be encouraged. 4. Health care providers in the public, private and NGOs sectors should be sensitised to adopt a "burden of disease" approach to meet the special needs of tribal and hill area communities. 26 (c) Adolescents 1. Ensure for adolescents access to information, counseling and services, including reproductive health services, that are affordable and accessible. Strengthen primary health centres and subcentres, to provide counseling, both to adolescents and also to newly weds (who may also be adolescents). Emphasise proper spacing of children. 2. Provide for adolescents the package of nutritional services available under the ICDS programme. Comment: Improvements in health status of adolescent girls has an inter-generational impact. It reduces the risk of low birth weight and minimizes neo-natal mortality. Malnutrition is a problem that seriously impairs the health of adolescent and adult women and has its roots in early childhood. The causal linkages between anaemia and low birth weight, prematurity, peri-natal mortality, and maternal mortality has been extensively studied and established. 3. Enforce the Child Marriage Restraint Act, 1976, to reduce the incidence of teenage pregnancies. Preventing the marriage of girls below the legally permissible age of 18 should become a national concern. Comment: It will promote higher retention of girls at schools, and is also likely to encourage their participation in the paid work force. 4. Provide integrated intervention in pockets with unmet needs in the urban slums, remote rural areas, border districts and among tribal populations. (d) Increased Participation of Men in Planned Parenthood 1. Focus attention on men in the information and education campaigns to promote the small family norm, and to raise awareness by emphasising the significant benefits of fewer children, better spacing, better health and nutrition, and better education. 2. Currently, over 97 percent of the sterilisations are tubectomies. Repopularise vasectomies, in particular the no-scalpel vasectomy, as a safe, simple, painless procedure, more convenient and acceptable to men. 3. In the continuing education and training at all levels, there is need to ensure that the no-scalpel vasectomy, and all such emerging techniques and skills are included in the syllabi, together with abundant practical training. Medical graduates, and all those participating in "in-service" continuing education and training, will be equipped to handle this intervention. (vii) Diverse Health Care Providers 1. At district and sub-district levels, maintain block-wise data base of private medical practitioners whose credentials may be certified by the Indian Medical Association (IMA). Explore the possibility of accrediting these private practitioners for a year at a time, and assign to each a 27 satellite population, not exceeding 5,000 (depending upon distances and spread), for whom they may provide reproductive and child health services. The private practitioners would be compensated for the services rendered through designated agencies. Renewal of contracts after one year may be guided by client satisfaction. This will serve as an incentive to expand the coverage and outreach of high quality health care. Appropriate checks and balances will safeguard misuse. 2. Revive the earlier system of the licensed medical practitioners who, after appropriate certification from the IMA, may participate in the provision of clinical services. 3. Involve the non-medical fraternity in counseling and advocacy so as to demystify the national family welfare effort, such as retired defence personnel, retired school teachers and other persons who are active and willing to get involved. 4. Modify the under/post-graduate medical, nursing, and paramedical professional course syllabi and curricula, in consultation with the Medical Council of India, the Councils of ISMH, and the Indian Nursing Council, in order to reflect the concepts and implementation strategies of the reproductive and child health programme and the national population policy. This will also be applied to all in-service training and educational curricula. 5. Ensure the efficient functioning of the First Referral Units i.e. 30 bed hospitals at block levels which provide emergency obstetric and child health care, to bring about reductions in Maternal Mortality Ratio (MMR) and Infant Mortality Rate (IMR). In many states, these FRUs are not operational on account of an acute shortage of specialists i.e. gynaecologist/obstetrician, anaesthetist and pediatrician. Augment the availability of specialists in these three disciplines, by increasing seats in medical institutions, and simultaneously enable and facilitate the acquisition of in-service post-graduate qualifications through the National Board of Medical Examination and open universities like IGNOU in larger numbers. As an incentive, seats will be reserved for those in-service medical graduates who are willing to abide by a bond to serve for 5 years at First Referral Units after completion of the course. States would need to sanction posts of Specialists at the FRUs. Further, these specialists should be provided with clear promotion channels. (viii) (a) Collaboration with and Commitments from the Non-Government Sector 1. There remain innumerable hurdles that inhibit genuine long-term collaboration between the government and non-government sectors. A forum of representatives from government, the non-government organisations and the private sector may identify these hurdles and prepare guidelines that will facilitate and promote collaborative arrangements. 2. Collaboration with and commitments from NGOs to augment advocacy, counseling and clinical services, while accessing village levels. This will require increased clinic outlets as well as mobile clinics. 28 3. Collaboration between the voluntary sector and the NGOs will facilitate dissemination of efficient service delivery to village levels. The guidelines could articulate the role and responsibility of each sector. 4. Encourage the voluntary sector to motivate village-level self-help groups to participate in community activitie. 5. Specific collaboration with the non-government sector in the social marketing of contraceptives to reach village levels will be encouraged. (viii) (b) Collaboration with and Commitments from Industry 1. The corporate sector and industry could, for instance, take on the challenge of strengthening the management information systems in the seven most deficient states, at primary health centre and subcentre levels. Introduce electronic data entry machines to lighten the tedious work load of ANMs and the multi-purpose workers at subcentres and the doctors at the primary health centres, while enabling wider coverage and outreach. 2. Collaborate with non-government sectors in running professionally sound advertisement and marketing campaigns for products and services, targeting all segments of the population, from village level upwards, in other words, strengthen advocacy and IEC, including social marketing of contraceptives. 3. Provide markets to sustain the income-generating activities from village levels upwards. In turn, this will ensure consistent motivation among the community for pursuing health and education-related community activities. 4. Help promote transportation to remote and inaccessible areas up to village levels. This will greatly assist the coverage and outreach of social marketing of products and services. 5. The social responsibility of the corporate sector in industry must, at the very minimum, extend to providing preventive reproductive and child health care for its own employees (if >100 workers are engaged). 6. Create a national network consisting of voluntary, public, private and non-government health centres, identified by a common logo, for delivering reproductive and child health services, free to any client. The provider will be compensated for the service provided, on the basis of a coupon system, duly counter-signed by the beneficiary and paid for by a system that will be fully articulated. The compensation will be identical to providers, across all sectors. The end user exercises choices in the source of service delivery. A committee of management experts will be set up to devise ways of ensuring that this system is not abused. 7. Form a consortium of the voluntary sector, the non-government sector and the private corporate sector to aid government in the provision and outreach of basic reproductive and child health care and basic education. 29 8. In the area of basic education, set up privately run/managed primary schools for children up to age 14-15. Alternately, if the schools are set up/managed by the panchayat, the private corporate sector could provide the mid-day meals, the text -books and/or the uniforms. (ix) Mainstreaming Indian Systems of Medicine and Homeopathy 1. Provide appropriate training and orientation in respect of the RCH programme for the institutionally qualified ISMH medical practitioners (already educated in midwifery, obstetrics and gynaecology over 5-1/2 years), and utilise their services to fill in gaps in manpower at appropriate levels in the health infrastructure, and at subcentres and primary health centres, as necessary. 2. Utilise the ISMH institutions, dispensaries and hospitals for health and population related programmes. 3. Disseminate the tried and tested concepts and practices of the indigenous systems of medicine, together with ISMH medication at village maternity huts and at household levels for ante-natal and post-natal care, besides nurture of the newborn. 4. Utilise the services of ISMH 'barefoot doctors' after appropriate training and orientation towards providing advocacy and counseling for disseminating supplies and equipment, and as depot holders at village levels. (x) Contraceptive Technology and Research on RCH 1. Government will encourage, support and advance the pursuit of medical and social science research on reproductive and child health, in consultation with ICMR and the network of academic and research institutions. 2. The International Institute of Population Sciences and the Population Research Centres will continue to review programme and monitoring indicators to ensure their continued relevance to strategic goals. 3. Government will restructure the Population Research Centres, if necessary. 4. Standards for clinical and non-clinical interventions will be issued and regularly reviewed. 5. A constant review and evaluation of the community needs assessment approach will be pursued to align programme delivery with good management practices and with newly emerging technologies. 6. A committee of international and Indian experts, voluntary and non-government organisations and government may be set up to regularly review and recommend specific incorporation of the advances in contraceptive technology and, in particular, the newly emerging techniques, into programme development. 30 (xi) Providing for the Older Population 1. Sensitize, train and equip rural and urban health centres and hospitals towards providing geriatric health care. 2. Encourage NGOs and voluntary organizations to formulate and strengthen a series of formal and informal avenues that make the elderly economically self-reliant. 3. Tax benefits could be explored as an encouragement for children to look after their aged parents. (xii) Information Education and Communication 1. Converge IEC efforts across the social sectors. The two sectors of Family Welfare and Education have coordinated a mutually supportive IEC strategy. The Zila Saksharta Samitis design and deliver joint IEC campaigns in the local idiom, promoting the cause of literacy as well as family welfare. Optimal use of folk media has served to successfully mobilize local populations. The state of Tamil Nadu made exemplary use of the IEC strategy by spreading the message through every possible media, including public transport, on mile stones on national high ways as well as through advertisement and hoardings on roadsides, along city/rural roads, on billboards, and through processions, films, school dramas, public meetings, local theatre and folk songs. 2. Involve departments of rural development, social welfare, transport, cooperatives, education with special reference to schools, to improve clarity and focus of the IEC effort, and to extend coverage and outreach. Health and population education must be inculcated from the school levels. 3. Fund the nagar palikas, panchayats, NGOs and community organizations for interactive and participatory IEC activities. 4. Demonstration of support by elected leaders, opinion makers, and religious leaders with close involvement in the reproductive and child health programme greatly influences the behaviour and response patterns of individuals and communities. This serves to enthuse communities to be attentive towards the quality and coverage of maternal and child health services, including referral care. Public leaders and film stars could spread widely the messages of the small family norm, female literacy, delayed marriages for women, fewer babies, healthier babies, child immunization and so on. The involvement and enthusiastic participation of elected leaders will ensure dedicated involvement of administrators at district and sub-district levels. Demonstration of strong support to the small family norm, as well as personal example, by political, community, business, professional, and religious leaders, media and film stars, sports personalities and opinion makers, will enhance its acceptance throughout society. 5. Utilise radio and television as the most powerful media for disseminating relevant socio-demographic messages. Government could explore the feasibility of appropriate regulations, and even legislation, if necessary, to mandate the broadcast of social messages during prime time. 31 6. Utilise dairy cooperatives, the public distribution systems, other established networks like the LIC at district and sub-district levels for IEC and for distribution of contraceptives and basic medicines to target infant/childhood diarrhoeas, anaemia and malnutrition among adolescent girls and pregnant mothers. This will widen outreach and coverage. 7. Sensitise the field level functionaries across diverse sectors (education, rural development, forest and environment, women and child development, drinking water mission, cooperatives) to the strategies, goals and objectives of the population stabilisation programmes. 8. Involve civil society for disseminating information, counseling and spreading education about the small family norm, the need for fewer but healthier babies, higher female literacy and later marriages for women. Civil society could also be of assistance in monitoring the availability of contraceptives, vaccines and drugs in rural areas and in urban slums. 32 Appendix II MILESTONES IN THE EVOLUTION OF THE POPULATION POLICY OF INDIA v 1946 Bhore Committee Report v 1952 Launching of Family Planning Programme v 1976 Statement of National Population Policy v 1977 Policy Statement on Family Welfare Programme Both statements were laid on the Table of the House in Parliament, but never discussed or adopted. v 1983 The National Health Policy of 1983 emphasized the need for "securing the small family norm, through voluntary efforts and moving towards the goal of population stabilisation". While adopting the Health Policy, Parliament emphasized the need for a separate National Population Policy. v 1991 The National Development Council appointed a Committee on Population with Shri Karunakaran as Chairman. The Karunakaran Report (Report of the National Development Council (NDC) Committee on Population) endorsed by NDC in 1993 proposed the formulation of a National Population Policy to take a " a long term holistic view of development, population growth and environmental protection" and to "suggest policies and guidelines (for) formulation of programs" and "a monitoring mechanism with short, medium and long term perspectives and goals" (Planning Commission, 1992). It was argued that the earlier policy statements of 1976 and 1977 were placed on the table, however, Parliament never really discussed or adopted them. Specifically, it was recommended that "a National Policy of Population should be formulated by the Government and adopted by Parliament". v 1993 An Expert Group headed by Dr. M.S. Swaminathan was asked to prepare a draft of a national population policy that would be discussed by the Cabinet and then by Parliament. v 1994 Report on a National Population Policy by the Expert Group headed by Dr. Swaminathan. This report was circulated among Members of Parliament, and comments requested from central and state agencies. It was anticipated that a national population policy approved by the National Development Council and the Parliament would help produce a broad political consensus. 33 v 1997 On the 50th anniversary of India's Independence, Prime Minister Gujral promised to announce a National Population Policy in the near future. During 11/ 97 Cabinet approved the draft National Population Policy with the direction that this be placed before Parliament. However, this document could not be placed in either House of Parliament as the respective Houses stood adjourned followed by dissolution of the Lok Sabha. v 1999 Another round of consultations was held during 1998, and another draft National Population Policy was finalised and placed before the Cabinet in March, 1999. Cabinet appointed a Group of Ministers (headed by Dy Chairman, Planning Commission) to examine the draft Policy. The GOM met several times and deliberated over the nuances of the Population Policy. In order to finalise a view about the inclusion/exclusion of incentives and disincentives, the Group of Ministers invited a cross-section of experts from among academia, public health professionals, demographers, social scientists, and women's representatives. The GOM finalised a draft population policy, and placed the same before Cabinet. This was discussed in Cabinet on 19 November, 1999. Several suggestions were made during the deliberations. On that basis, a fresh draft was submitted to Cabinet. 34 Appendix III DEMOGRAPHIC PROFILE India is following the demographic transition pattern of all developing countries from initial levels of “high birth rate - high death rate” to the current intermediate transition stage of “high birth rate - low death rate” which leads to high rates of population growth, before graduating to levels of “low birth rate - low death rate”. 1. Age Composition 1. (i) The age distribution of the population of India is projected to change by 2016, and these changes should determine allocation of resources in policy intervention. The population below 15 years of age (currently 35 percent) is projected to decline to 28 percent by 2016. The population in the age group 15 - 59 years (currently 58 percent) is projected to increase to nearly 64 percent by 2016. The age group of 60 plus years is projected to increase from the current levels of 7 percent to nearly 9 percent by 2016. Table 4 : Age Composition as Percentage of the Total Population2 Year Below Between Between + 60 years 5 years 0-15 >15 -59 years 1991 12.80 37.76 55.58 6.67 2001 10.70 34.33 58.70 6.97 2011 10.10 28.48 63.38 8.14 2016 9.7 27.73 63.33 8.94 2. Inter-State Differences 2. (i) India is a country of striking demographic diversity. Substantial differences are visible between states in the achievement of basic demographic indices. This has led to significant disparity in current population size and the potential to influence population increases during 1996-2016. There are wide inter-state, male-female and rural-urban disparities in outcomes and impacts. These differences stem largely from poverty, illiteracy, and inadequate access to health and family welfare services, which coexist and reinforce each other. In many parts, the widespread health infrastructure is not responsive. 2. (ii) At least 9 states and union territories in India have already achieved replacement levels of fertility. These are ranked in accordance with their total fertility rates. Additionally, in each of the three tables below, the current population of each state/union territory, the ratio of this population to the country population, the infant mortality rate and the contraceptive prevalence rate of the state / union territory is also indicated: 2 Technical Group on Population Projections, Planning Commission. 35 Table 5 : Population Profile of 9 States and Union Territories of India with TFR less than or equal to 2.1 State Population Percent of Total Infant Contraceptive Size (in millions) Total Fertility Mortality Prevalence as on Population Rate Rate Rate 1 March 1999 1997 1998 1999 INDIA 981 .3 3.3 72 44 % Group A (TFR less than or equal to 2.1) Goa 1.5 0.2 1.0@ 23 27.1 Nagaland 1.6 0.2 1.5@ NA 7.8 Delhi 13.4 1.4 1.6@ 36 28.8 Kerala 32.0 3.3 1.8 16 40.5 Pondichery 1.1 0.1 1.8@ 21 56.9 A&N Islands 0.4 0.04 1.9@ 30 39.9 Tamil Nadu 61.3 6.2 2.0 53 50.4 Chandigarh 0.9 0.09 2.1@ 32 35.0 Mizoram 0.9 0.09 NA 23 34.6 @ Three year moving average TFR1995-97 Population Projections by Technical Group on Population Projections,1996 Source: Registrar General of India 2.(iii) There are 11 states and union territories that have a total fertility rate of more than 2.1 but less than 3.0, ranked accordingly : Table 6 : Population Profile of 11 States and Union Territories of India with TFR > 2.1 but < 3 State Population Percent of Total Infant Contraceptive Size (in millions) Total Fertility Mortality Prevalence as on Population Rate Rate Rate 1 March 1999 1997 1998 1999 Group B (TFR > 2.1 and < than 3.0) Manipur 2.21 0.2 2.4@ 25 20.1 Daman & Diu 0.1 0.01 2.5@ 51 30.2 Karnataka 51.4 5.2 2.5 58 55.4 Andhra Pradesh 74.6 7.6 2.5 66 50.3 Himachal Pradesh 6.5 0.7 2.5 64 48.2 Sikkim 0.5 0.06 2.5 52 21.9 West Bengal 78.0 7.9 2.6 53 32.9 Maharashtra 90.1 9.2 2.7 49 50.1 Punjab 23.3 2.4 2.7 54 66.0 Arunachal Pradesh 1.2 0.1 2.8@ 47 14.0 Lakshadweep 0.07 0.01 2.8@ 37 9.1 @ Three year moving average TFR1995-97 Population Projections by Technical Group on Population Projections,1996 Source: Registrar General of India 36 2. (iv) However, there are at least 12 states and union territories that have a total fertility rate of over 3.0. These have been listed below: Table 7 : Population Profile of 12 States and Union Territories of India with TFR greater than or equal to 3 State Population Size Percent of Total Infant Contraceptive (in millions) Total Fertility Mortality Prevalence as on Population Rate Rate Rate 1 March 1999 1997 1998 1999 Group C (>3.0) Orissa 35.5 3.6 3.0 98 39 Gujarat 47.6 4.8 3.0 64 54.5 Assam 25.6 2.6 3.2 78 16.7 Haryana 19.5 2.0 3.4 69 49.7 Dadra & Nagar Haveli 0.2 0.02 3.5@ 61 29.1 Tripura 3.6 0.3 3.9@ 49 25.2 Meghalaya 2.4 0.2 4.8@ 52 4.6 Madhya Pradesh 78.3 8.0 4.0 98 46.5 Rajasthan 52.6 5.4 4.2 83 36.4 Bihar 98.1 10.0 4.4 67 19.7 Uttar Pradesh 166.4 17.0 4.8 85 38.2 Jammu & Kashmir 9.7 1.0 NA 45 15.0 @ Three year moving average TFR1995-97 Population Projections by Technical Group on Population Projections,1996 Source : Registrar General of India 2.(v) The five states of Bihar, Madhya Pradesh, Orissa, Rajasthan and Uttar Pradesh that currently constitute nearly 44 percent of the total population of India, are projected to comprise 48 percent of the total population in 2016. In other words, these states alone will contribute an anticipated 55 percent increase during the period 1996-2016. Demographic outcomes in these states will determine the timing and size of population at which India achieves population stabilisation. 3. Maternal Mortality 3.(i) With 16% of the world's population, India accounts for over 20% of the world's maternal deaths. The maternal mortality ratio, defined as the number of maternal deaths per 100,000 live births, is incredibly high at 408 per 100,000 live births for the country (1997), which is unacceptable when compared to current indices elsewhere in Asia. Table 8 : Maternal Mortality Ratios in Asia3 Sri Lanka China Thailand Pakistan Indonesia India Bangladesh Nepal 30 115 200 340 390 437 850 1500 3 UNFPA, The State of World Population, 1999, 6 Billion : A Time for Choices, 1995 estimates. 37 3. (ii) Within India, the inter-state differentials are a matter of concern. Table 9 : Inter-State Differences within India in Maternal Mortality Ratios4 Kerala Bihar Madhya Pradesh Rajasthan Uttar Pradesh Orissa 87 451 498 607 707 739 4. Infant Mortality 4. It is estimated that about 7 percent of new-born infants perish within a year. Poor maternal health results in low birth weight and premature babies. Infant and childhood diarrhoeal diseases, acute respiratory infections and malnutrition contribute to high infant mortality rates. Additionally, in India, across the board (rural or urban areas), there are more female deaths in the age group of 0-14 than elsewhere5. Although the Infant Mortality Rate (IMR) has decreased from 146 per 1000 births in 1951 to 72 per 1000 births (1997), and the sex differentials are narrowing, again there are wide inter-state differences recorded in 1998, as is clear from Table 4-6. In comparison, we note the infant mortality rates in South Asia and elsewhere: Table 10 : Infant Mortality Rates in Asia6 Sri Lanka Thailand China Indoneisa India Pakistan Bangladesh Nepal 18 29 41 48 72 74 79 83 5. Sex Ratio 5. (i) India shares a distinctive feature of South Asian and Chinese populations as regards the sex ratio, with a century's old deficit of females. The (female to male) sex ratio has been steadily declining. From 1901 to 1991, the sex ratio has declined from 972 to 927. This is largely attributed to the son preference, discrimination against the girl child leading to lower female literacy, female foeticide, higher fertility and higher mortality levels for females, in all age groups up to 45. 4 Registrar General of India. 5 UNICEF, The Progress of Indian States, 1995. India Country Office, New Delhi. 6 UNFPA, The State of the World Population, 1999, 6 Billion : A Time for Choices. 38 Appendix IV UNMET NEEDS & DEFICIENCIES IN CONTRACEPTIVES SERVICES, HEALTH INFRASTRUCTURE, SPECIALISTS AND TRAINED MANPOWER WITH IMPLICATIONS FOR FUNDING The unmet need for contraceptive services is estimated at 28%, necessitating an additionality of approx. Rs. 150 crores (for contraceptives, laproscopes, tubal rings, vaccines and RCH drugs). Health infrastructure is inadequate, with estimated shortages as: v 7,683 subcentres (1991), now estimated at 23,190 subcentres for the projected population in Y ear 2002. Capital cost of one subcentre is Rs. 3 lacs, with a recurring cost of Rs. 0.5 lacs. The Finance Minister in his Budget Speech, 1999-2000, announced a scheme for strengthening rural health infrastructure, to be implemented with responsibility for funding shared between the panchayat, state and central governments. Accordingly, the Department of Family Welfare is formulating a scheme for opening new subcentres where required, providing buildings and equipment to existing subcentres, wherever necessary. v A shortage of 1,513 primary health centres (1991), now estimated at 4,212 PHCs for the population projected in 2002. Capital cost of one PHC is Rs. 24.50 lacs, with a recurring liability of Rs 13 lacs. These expenditures are met by the State Governments under the basic minimum services (BMS) programme. However, the financial position of the State Governments does not enable them to make these investments in health infrastructure. v A shortage of 2,899 community health centres (1991), now estimated at 3,776 CHCs for the projected population for the year 2000. CHCs serve, mostly, as the First Referral Units and are critical for reducing the MMR and IMR, besides serving as operating theatres for family planning services. Capital cost of one CHC is Rs. 80.5 lacs, with a recurring liability of 27 lacs, currently met by State Governments under the BMS programme. v The Department of Family Welfare funds 5,435 Rural FW Centres, some of which are being used as First Referral Units. Others are functioning as block-level PHCs. The estimated additionality towards infrastructure is: (Rs. In crores) A/C 1991 2002 Subcentres Cap 230 695 Rec 38 116 PHCs Cap 370 1032 Rec 196 547 CHCs Cap 2320 3021 Rec 783 1020 TOTAL Capital 2920 4748 Recurring 1017 1683 39 Inadequacies in Trained Manpower v Shortage in manpower is estimated as 27,501 ANMs, 64,860 male muti-purpose workers, and 4,224 LHVs, 5,126 Health Assistants (male), 2,475 medical officers in PHCs, 1,429 surgeons, 1,446 gynaecologists, 1,525 physicians, 1,774 pediatricians, and an overall shortage of 6,635 specialists. v Other health manpower reflects a shortfall of 1,171 radiographers, 6,045 pharmacists, 12,793 Lab Technicians, and 18,851 nuse mid-wives, in the rural primary health care delivery system. The financial requirement to address these unmet needs for trained manpower is approximately Rs. 2,300 crores. v For safe abortion services, no MTP kits have been made available since 1997. However, during the CSSM programme, 1,748 MTP kits were distributed to the FRUs. Most of these are lying unused, on account of shortage of trained manpower. This year an additional 180 MTP sets are being procured. Training v For training, since the population policy emphasises convergence of training requirements as well as decentralisation to sub-district and village levels, the estimated additionality for the present is Rs. 10 crores.
14799	https://stats.stackexchange.com/questions/239637/pnorm-function-in-r	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 pnorm function in R Ask Question Asked Modified 8 years, 11 months ago Viewed 25k times 3 $\begingroup$ How does the pnorm function in R work? I mean how it computes $P(X<10)$ with only the mean and standard deviation given. Does it work same like $z$-score to compute percentages? r probability normal-distribution cumulative-distribution-function Share Improve this question edited Oct 11, 2016 at 19:53 dsaxton 12.3k11 gold badge2828 silver badges4949 bronze badges asked Oct 11, 2016 at 18:05 Khan SaabKhan Saab 22511 gold badge44 silver badges1212 bronze badges $\endgroup$ 1 2 $\begingroup$ Because a normal distribution is fully specified by its mean and standard deviation. $\endgroup$ dsaxton – dsaxton 2016-10-11 18:13:13 +00:00 Commented Oct 11, 2016 at 18:13 Add a comment \| 2 Answers 2 Reset to default 5 $\begingroup$ pnorm calculates cumulative distribution function of normal distribution, i.e. $$ \Pr(X \le x) = F(x) = \frac12\left[1 + \operatorname{erf}\left( \frac{x-\mu}{\sigma\sqrt{2}}\right)\right] $$ where $\mu$ is mean and $\sigma$ is standard deviation. As noted by dsaxton to calculate the probabilities for normally distributed random variable $X$ you need only to know $\mu$ and $\sigma$ parameters and apply the function. If you are not familiar with those concepts you should read carefully the Wikipedia articles I linked above and you should check some introductory handbook on statistics, e.g. All of Statistics by Larry Wasserman. Harvard University provides online lecture introducing probability theory called Statistics 110: Probability by Joe Blitzstein, it's also a good starting point. Share Improve this answer answered Oct 11, 2016 at 18:23 TimTim 144k2727 gold badges275275 silver badges529529 bronze badges $\endgroup$ 1 2 $\begingroup$ Indeed the C code for pnorm has "The main computation evaluates near-minimax approximations derived from those in "Rational Chebyshev approximations for the error function" by W. J. Cody, Math. Comp., 1969, 631-637. This transportable program uses rational functions that theoretically approximate the normal distribution function to at least 18 significant decimal digits. The accuracy achieved depends on the arithmetic system, the compiler, the intrinsic functions, and proper selection of the machine-dependent constants." ... looks like it literally uses an implementation of $\text{erf}$ $\endgroup$ Glen_b – Glen_b 2016-10-12 00:15:06 +00:00 Commented Oct 12, 2016 at 0:15 Add a comment \| 1 $\begingroup$ Function dnorm gives PDF and pnorm gives CDF. For example, if you want to find $P(x<2)$ for standard normal distribution (mean=0 and sd=1), you can use pnorm(2), which is $0.977$. If you want to find $P(1mean=1 and sd=2 you can do pnorm(2,mean=1,sd=2)-pnorm(1,mean=1,sd=2), which is $0.191$ Share Improve this answer answered Oct 11, 2016 at 18:23 HXDHXD 37.7k2727 gold badges152152 silver badges247247 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions r probability normal-distribution cumulative-distribution-function See similar questions with these tags. 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