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3200	http://www.matematicasvisuales.com/english/html/analysis/seriegeom/progregeom.html	\| \| \| --- \| \| \| \| \| Geometric sequences or geometric progressions \| \| \| \| \| \| \| \| \| Geometry Real Analysis Complex Analysis Probability History Help Contact Links Site Map Updates Español \| \| \| \| --- \| \| \| \| \| A sequence is an ordered list of numbers. Some sequences follow a pattern. Each number in a sequence is called a term. If we see the sequence as a function, its domain is the natural numbers. A Geometric sequence (or geometric progression) is a sequence of numbers where each term after the first is given by multiplying the previous one by a fixed non-zero number, a constant, called the common ratio. Any term of a geometric sequence can be expressed by the formula for the general term: When the ratio r is greater than 1 we have an increasing sequence (expontential growth). Even if the ratio is very small the sequence starts increasing slowly but after enough steps the growth becomes bigger and bigger. For example, this is the result after 300 steps if the ratio is 1.01: If the ratio r is positive and less than 1 the sequence is decreasing and the general term tends towards 0. When the ratio r is negative the sequence is alternating. If the ratio r is between -1 and 0 the alternating sequence has a general term that tends towards 0. If the ratio r is less than -1 the alternating sequence in absolute value tends towards infinity (unsigned if we consider the value, due to the alternating sign). We can consider the sum of the n terms of a geometric sequence. We can deduce a formula: In the next application we can play with different cases with a positive common ratio: You can see the behavior if the common ratio is greater than 1 (the sum grows and grows): If the common ratio is less than 1 the sum seems to approach a number: A series is the sum of the infinite terms of a sequence. If a positive r is less than 1 you can sum up these infinite numbers and the result is a number. We can say that the series is convergent (it approaches some limit). If a positive r is greater or equal to 1 then the series do not approaches some number because it becomes bigger and bigger. Then we can say that the series is divergent. In the next application we can play with a general case. Now the common ration can be positive or negative: A divergent alternatig series: A convergent alternatig series: The condition for the convergence of a geometric series with a non-zero common ratio r is: And the formula is: NEXT Sum of a geometric series of ratio 1/4 One intuitive example of how to sum a geometric series. A geometric series of ratio less than 1 is convergent. MORE LINKS Sum of a geometric series of ratio 1/2 The geometric series of ratio 1/2 is convergent. We can represent this series using a rectangle and cut it in half successively. Here we use a rectangle such us all rectangles are similar. Exponentials and Logarithms (1): Exponential Functions We can study several properties of exponential functions, their derivatives and an introduction to the number e. Convergence of Series: Integral test Using a decreasing positive function you can define series. The integral test is a tool to decide if a series converges o diverges. If a series converges, the integral test provide us lower and upper bounds. Integral of powers with natural exponent The integral of power functions was know by Cavalieri from n=1 to n=9. Fermat was able to solve this problem using geometric progressions. Standard Paper Size DIN A There is a standarization of the size of the paper that is called DIN A. Successive paper sizes in the series A1, A2, A3, A4, and so forth, are defined by halving the preceding paper size along the larger dimension. Dilative rotation A Dilative Rotation is a combination of a rotation an a dilatation from the same point. Equiangular spiral In an equiangular spiral the angle between the position vector and the tangent is constant. Multiplying two complex numbers We can see it as a dilatative rotation. The product as a complex plane transformation The multiplication by a complex number is a transformation of the complex plane: dilative rotation. Complex Geometric Sequence From a complex number we can obtain a geometric progression obtaining the powers of natural exponent (multiplying successively) \| \| \| \| Geometry\| Analysis\| Complex Analysis\| Probability\| History\| Help\| Contact\| Links\| Site map\| Updates \|
3201	https://www.chegg.com/homework-help/questions-and-answers/answer-following-physics-questions-use-g-10m-s-2-instead-98m-s-2-answer-questions-leaving--q93599053	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Answer the following physics questions. Use g=10m/s^2 instead of 9.8m/s^2. Answer the questions leaving no space between the number and the unit. For example: 3m/s and not 3 m/s. Remember to give the answers with units. If a calculation results in an irrational number, round it to two decimals or leave it as a fraction. For instance: if you get 1/3, just Answer the following physics questions. Use g=10m/s^2 instead of 9.8m/s^2. Answer the questions leaving no space between the number and the unit. For example: 3m/s and not 3 m/s. Remember to give the answers with units. If a calculation results in an irrational number, round it to two decimals or leave it as a fraction. For instance: if you get 1/3, just leave it like that or type 0.33. By how much does the gravitational force between two objects increase when the distance between their centers is halved?. EXPLAIN. According to the equation for gravitational force, what happens to the force between two bodies if the mass of one of the bodies is tripled?. EXPLAIN. By how much does the gravitational force that the Earth exerts on you change if you climb a palm tree?. EXPLAIN. Why doesn ́t Mercury fall and crash into the Sun? What is the final horizontal velocity (when it reaches the ground) of an object that is thrown with a horizontal velocity of 10m/s?. EXPLAIN. If you throw a ball with an angle, at what part of its trajectory does the ball have minimum speed?, and maximum?. A person throws an object from a 20 meter cliff. How much time does it take for the object to reach the ground if it is thrown horizontally with a velocity of 10m/s?. Earth and the Moon are gravitationally attracted to the Sun, but they don’t crash into the Sun. Why? What is the vertical acceleration that acts on a projectile when there is no air resistance? What is horizontal acceleration that acts on a projectile when there is no air resistance? What would be the path of the Moon if somehow all gravitational forces on it vanished to zero? Upon which is the gravitational force greater: a 1-kg piece of iron or a 1-kg piece of glass? Defend your answer. According to Chegg guidelines , I can solve atleast 4 parts if the question containing more than 4 parts. So I could solved only 4 parts. ( 4 ) Planets are constantly falling on the Sun, but they always fall beyond the curvature of the Sun and so t… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
3202	https://www.chegg.com/homework-help/questions-and-answers/convert-manometer-readings-differential-pressure-data-note-pressure-loss-indicated-incline-q79454714	Solved -Convert the manometer readings to differential \| 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 Engineering Mechanical Engineering Mechanical Engineering questions and answers -Convert the manometer readings to differential pressure data. Note that the pressure loss (∆𝑝), as indicated by the inclined U-tube manometer is: Δ𝑝 = 𝛾∆ℎ = 𝛾∆𝐿 sin (𝜃) where ∆h is the head loss, 𝛾 is the specific weight of the manometer fluid (water), Δ𝐿 is the difference in the upper and lower manometer reading (measured along the incline) and 𝜃 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: -Convert the manometer readings to differential pressure data. Note that the pressure loss (∆𝑝), as indicated by the inclined U-tube manometer is: Δ𝑝 = 𝛾∆ℎ = 𝛾∆𝐿 sin (𝜃) where ∆h is the head loss, 𝛾 is the specific weight of the manometer fluid (water), Δ𝐿 is the difference in the upper and lower manometer reading (measured along the incline) and 𝜃 -Convert the manometer readings to differential pressure data. Note that the pressure loss (∆𝑝), as indicated by the inclined U-tube manometer is: Δ𝑝 = 𝛾∆ℎ = 𝛾∆𝐿 sin (𝜃) where ∆h is the head loss, 𝛾 is the specific weight of the manometer fluid (water), Δ𝐿 is the difference in the upper and lower manometer reading (measured along the incline) and 𝜃 is the angle of the manometer with respect to horizontal. -Plot the pressure loss 𝛥𝑝 (in Pascals) as function of the flow rate 𝑄. Show transcribed image text There’s just one step to solve this.Solution Share Share Share done loading Copy link Step 1 Page-1 Sol:- Applying monometer are between points A and B P A+γ ω h A C+γ ω h e 0=P B ∴P B−P A=γ ω h A C+v ω h C D View the full answer Answer Unlock Previous questionNext question Transcribed image text: Pipe internal diameter: 12.7 mm Distance between pressure tops: 1.57 m Water temperature: 20°C Manometer gauge fluid: water Table 1: Manometer deflection at an inclination angle of 45°. Flow Appearance Flow rate (litres/s) 0 Manometer deflection (mm) 0 N/A 0.02 6 Laminar 0.04 23 0.06 47 0.08 77 0.1 118 0.12 157 Mostly turbulent Fully turbulent Fully turbulent Fully turbulent Fully turbulent Fully turbulent Fully turbulent Fully turbulent Fully turbulent 0.14 199 0.16 265 0.18 316 0.2 396 Table 2: Manometer deflection at an inclination angle of 15°. Flow rate Flow Appearance (litres/min) 0 Manometer deflection (mm) 0 Laminar 0.80 6 Laminar 1.00 8 Laminar 1.20 10 Laminar 1.40 15 1.60 26 1.80 31 2.00 46 Occasional turbulent pockets, mostly laminar Mixed laminar/turbulent Mixed laminar/turbulent Mixed laminar/turbulent Mixed laminar/turbulent Mostly turbulent Fully turbulent Fully turbulent 2.20 47 2.40 60 2.60 69 2.80 84 Not the question you’re looking for? Post any question and get expert help quickly. 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3203	https://api.pageplace.de/preview/DT0400.9781482296372_A35484908/preview-9781482296372_A35484908.pdf	I n t r o d u c t i o n t o Series Editor KENNETH H. ROSEN DISCRETE MATHEMATICS AND ITS APPLICATIONS CODING THEORY Charles J. Colbourn and Jeffrey H. Dinitz, The CRC Handbook of Combinatorial Designs Charalambos A. Charalambides, Enumerative Combinatorics Steven Furino, Ying Miao, and Jianxing Yin, Frames and Resolvable Designs: Uses, Constructions, and Existence Randy Goldberg and Lance Riek, A Practical Handbook of Speech Coders Jacob E. Goodman and Joseph O’Rourke, Handbook of Discrete and Computational Geometry Jonathan Gross and Jay Yellen, Graph Theory and Its Applications Jonathan Gross and Jay Yellen, Handbook of Graph Theory Darrel R. Hankerson, Greg A. Harris, and Peter D. Johnson, Introduction to Information Theory and Data Compression Daryl D. Harms, Miroslav Kraetzl, Charles J. Colbourn, and John S. Devitt, Network Reliability: Experiments with a Symbolic Algebra Environment David M. Jackson and Terry I. Visentin, An Atlas of Smaller Maps in Orientable and Nonorientable Surfaces Richard E. Klima, Ernest Stitzinger, and Neil P. Sigmon, Abstract Algebra Applications with Maple Patrick Knupp and Kambiz Salari, Verification of Computer Codes in Computational Science and Engineering Donald L. Kreher and Douglas R. Stinson, Combinatorial Algorithms: Generation Enumeration and Search Charles C. Lindner and Christopher A. Rodgers, Design Theory Alfred J. Menezes, Paul C. van Oorschot, and Scott A. Vanstone, Handbook of Applied Cryptography Richard A. Mollin, Algebraic Number Theory Richard A. Mollin, Fundamental Number Theory with Applications Richard A. Mollin, An Introduction to Cryptography Richard A. Mollin, Quadratics Series Editor Kenneth H. Rosen, Ph.D. AT&T Laboratories Middletown, New Jersey and DISCRETE MATHEMATICS ITS APPLICATIONS Continued Titles Richard A. Mollin, RSA and Public-Key Cryptography Kenneth H. Rosen, Handbook of Discrete and Combinatorial Mathematics Douglas R. Shier and K.T. Wallenius, Applied Mathematical Modeling: A Multidisciplinary Approach Douglas R. Stinson, Cryptography: Theory and Practice, Second Edition Roberto Togneri and Christopher J. deSilva, Fundamentals of Information Theory and Coding Design Lawrence C. Washington, Elliptic Curves: Number Theory and Cryptography Kun-Mao Chao and Bang Ye Wu, Spanning Trees and Optimization Problems Juergen Bierbrauer, Introduction to Coding Theory This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microﬁlming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Speciﬁc permission must be obtained in writing from CRC Press LLC for such copying. Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identiﬁcation and explanation, without intent to infringe. Visit the CRC Press Web site at www.crcpress.com © 2005 by Chapman & Hall/CRC No claim to original U.S. Government works International Standard Book Number 1-58488-421-5 Library of Congress Card Number 2004049447 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper Library of Congress Cataloging-in-Publication Data Bierbrauer, Juergen. Introduction to coding theory / Juergen Bierbrauer. p. cm. — (Discrete mathematics and its applications) Includes bibliographical references and index. ISBN 1-58488-421-5 (alk. paper) 1.Coding theory. I. Title. II. Series. QA268.B48 2004 003'.54--dc22 2004049447 Preface The theory of error-correcting codes is a branch of discrete mathematics with close ties to other mathematical disciplines, like design theory, combinatorial theory, linear algebra, ﬁnite ﬁelds, rings, Galois geometry, geometric algebra, algebraic curves over ﬁnite ﬁelds and group theory. The best known appli-cation is in the transmission of messages over noisy communication channels. Other ﬁelds of application are to be found in statistics (design of experiments), cryptography (authentication, the design of ciphers) and in many areas of theoretical computer science. In this textbook we present a self-contained introduction to mathematical coding theory and to its major areas of application. High school algebra and some exposition to basic linear algebra are suﬃcient as mathematical back-ground. Part I is designed for use in a one semester undergraduate course. A second semester would start with the theory of cyclic codes. In Part II the emphasis is on cyclic codes, applications of codes and the geometric de-scription of linear codes. The mathematical tools are developed along the way. A brief overview The historical origins of coding theory are in the problem of reliable com-munication over noisy channels. This is a typical problem of the discipline now called Information Theory. Both disciplines, Coding Theory and Information Theory, originated with Claude Shannon’s famous 1948 paper . It contains the celebrated channel coding theorem (see Chapter 9) which states roughly that good long codes are guaranteed to exist, without giving a clue how to construct them. Closely related is the development of Cryptography. Its aim is to ensure reliable communication in the presence of ill-willed opponents. These problems are rather diﬀerent. In the coding the-ory scenario we have to overcome a technical problem (the shortcomings of a communication channel), whereas in Cryptography we have to beat an oppo-nent. Nonetheless the mathematical tools used in these two areas have a large intersection. Historically the development of both disciplines was boosted by the eﬀorts of World War II. Another famous paper of Claude Shannon, from 1949, is perceived as the origin of modern cryptography. vii viii The information-theoretic problem prompted the deﬁnition of a mathemat-ical structure called error-correcting code or simply code. Coding theory can be seen as the study of error-correcting codes, their construction, bounds on their parameters, their implementation and so forth. The most important parameter is the minimum distance. It measures the code’s capability of correcting transmission errors. Progress in coding theory was slow but steady. One important development was the theory of cyclic codes, which is traditionally couched in the language of ring theory. Cyclic codes are particularly useful because they admit a fast decoding algorithm. The theory of cyclic codes is a core topic of Part II. It is developed in Chapter 13, preceded by an introduction to some relevant features of ﬁnite ﬁelds in Chapter 12. Our approach is diﬀerent from the traditional approach. It is based on the trace and the action of the Galois group. Ring theory does not come into play at all. Only the single most famous cyclic code, the binary Golay code, is intro-duced in Part I, along with a closely related structure, the large Witt design (Chapter 7). The ties between coding theory and several areas of pure mathematics have grown stronger all the time. The most important insight goes back to the early 1980s. It is the discovery, by Goppa and Manin, of a close relationship be-tween codes and algebraic curves (in algebraic language function ﬁelds). Algebraic curves are objects of number theory and algebraic geometry, mainstream mathematical disciplines with a long and rich history. The obser-vation by Goppa and Manin makes it possible to use these number-theoretic tools for the construction of codes. The theory of those algebraic-geometric codes (AG-codes) was out of reach for our text. It presupposes an introduc-tion to the theory of algebraic curves with ﬁnite ﬁelds of constants. There is a short section in Chapter 18 which serves as an appetizer. Coding theory and combinatorics are closely connected. As an example, block designs are important objects of modern discrete mathematics. For more information see the CRC Handbook of Combinatorial Designs . We will encounter them repeatedly in the text. A formal deﬁnition is in Chapter 7, where we also derive the large Witt design from the binary Golay code. Other examples of block designs in the text include projective planes, projective and aﬃne geometry over ﬁnite ﬁelds (Chapter 16), the small Witt design, which is derived from the ternary Golay code in Section 16.1, and the Denniston arcs in the same section. Linear codes can be studied from a geometric point of view. From this perspective coding theory can be seen as part of Galois geometry. The basic objects of Galois geometry are aﬃne and projective spaces deﬁned over ﬁnite ﬁelds (see Hirschfeld or the beginning of Chapter 16). Linear codes can be equivalently described as sets of points in projective spaces. In many cases the geometric language is more appropriate than the algebraic approach. In Part I we study 3-dimensional codes from this point of view (Chapter 10). This case is particularly easy to understand as the underlying ix geometrical structures are classical projective planes PG(2, q). The general mechanism is developed and used in Chapter 16. In many cases this leads to a better understanding of the codes. For instance, we use the geometric method to construct the ternary Golay code in Section 16.1. Caps are sets of points in projective or aﬃne geometry no three of which are on a line. They are formally equivalent with linear codes of minimum distance d = 4. It turns out that caps are best understood from a geometric point of view. This is why we study caps in Chapter 16. The case of caps in projective planes and 3-spaces leads to another link with classical algebra. In fact, parabolic and elliptic quadrics yield canonical examples of caps in those dimensions. We include a self-contained introduction to geometric algebra in Section 16.2 which gives a better understanding of those caps. Duality is emphasized throughout the text. The dual of a linear code with minimum distance d is an orthogonal array of strength d −1. Originally orthogonal arrays were deﬁned in the framework of design of experiments, in statistics. The same is true of block designs. The deﬁning properties of orthogonal arrays and of block designs are both uniformity conditions. They look very similar. Orthogonal arrays can be interpreted as families of random variables (functions deﬁned on sample spaces), which satisfy certain statistical independence conditions (see Chapter 6). The strength measures the degree of statistical independence. Such families of random variables are heavily used not only in statistics but also in the theory of algorithms. Whenever we construct a good linear code we also obtain such a statistical object. Typically in coding theory duality is deﬁned with respect to the usual dot product. However each non-degenerate bilinear (or sesquilinear) form de-ﬁnes a notion of duality. An application to the construction of quantum codes in Chapter 17 demands the use of a special bilinear form, the symplec-tic form. This is another motivation for covering the theory of bilinear forms in Chapter 16. Some of the classical bounds, the Singleton bound, the Hamming bound, the Plotkin bound and the Griesmer bound on codes as well as the Bose-Bush bound on orthogonal arrays of strength 2 are derived and used in the text. In fact there is a multitude of bounds, each of which is better than all the others in certain parameter ranges, both for codes (when the minimum distance is the central parameter) and for orthogonal arrays (when the strength is in the center of attention). A general algebraic mechanism for the derivation of bounds is related to orthogonal polynomials (in the case of codes these are the Kravchouk polynomials) and linear programming. There is a general linear programming bound. All the bounds used in the text, with the exception of the Griesmer bound, are special cases of the LP-bound. This interesting topic of LP-bounds is not pursued in the text. It is sketched in a section of Chapter 18. On the level of LP-bounds there is a relation of duality between bounds on codes and bounds on orthogonal arrays. This is another reason why the notion of an orthogonal array should be seen as the dual of x the notion of an error-correcting code, even in the nonlinear case. This leads us to applications of codes. Traditionally coding theorists are bi-ased towards the information-theoretic application that we encounter so often in this text. It still is one of the major applications. We use it as a motivation in the early chapters, discuss syndrome decoding in Chapter 3 and the decod-ing algorithm of BCH-codes based on the Euclidean algorithm in Chapter 13. There is however a plethora of applications of a completely diﬀerent nature. Many of them have surfaced in theoretical computer science, in particular in cryptography. Universal hash families yield a nice paradigmatic example. One version is presented in Part I, Chapter 6, while a more in depth treat-ment is in Chapter 15. This chapter is dedicated to applications alltogether. They range from statistics and cryptography to numerical integration and the theory of algorithms. The plan: Part I Part I forms an elementary introduction to the theory of codes and some typical applications. It assumes only high school mathematics. Some exposi-tion to the basics of linear algebra would be helpful as well. Chapter 1 introduces some basic concepts like bits and bitstrings and de-scribes the basic problem when messages are sent via a noisy channel. The transmission of pictures from space serves as an illustration. The ﬁrst steps towards the algebraization of the problem are taken by introducing the ﬁeld F2 of two elements and giving the bitstrings of length n the structure of a vector space. We encounter the basic idea of error correction and the basic notion Hamming distance. This leads to the formal deﬁnition of a binary code and of a q-ary code for an arbitrary natural number q ≥2. The binary sym-metric channel is the most elementary model which describes how errors are introduced into the message. Basic facts on binomial numbers are reviewed. The sphere-packing bound (or Hamming bound) is our ﬁrst general bound on codes. The football pool problem is a possible application of ternary (q = 3) codes. Chapter 2 introduces the basics of binary linear codes. Key notions are: minimum weight, dimension, generator matrix. After a review of basic facts of linear algebra (applied for vector spaces = codes over F2) we see how block coding works and study the eﬀect on the probability of transmission errors. The dual code is deﬁned with respect to the dot product. The repetition codes and the sum 0 codes are our ﬁrst pair of dual codes. A check matrix of a code is a generator matrix of its dual code. An important family of codes are the binary Hamming codes. Their duals are the binary Simplex codes. The principle of duality shows how to read oﬀthe minimum distance from a xi check matrix. This leads to the notion of a binary orthogonal array (OA). A linear OA has strength t if and only if its dual has minimum weight t + 1. Chapter 3 generalizes linear codes from the binary (q = 2) to the q-ary, where q is an arbitrary prime-power. It is shown how ﬁnite ﬁelds Fq of q elements can be constructed. The deﬁnition of linear q-ary codes is given and the basic facts generalized from the binary to the q-ary: dimension, genera-tor matrix, dual code, check matrix, principle of duality, orthogonal arrays. Basic methods of linear algebra are reviewed (rank, Gauß elimination, deter-minants). Mutually orthogonal latin squares are recognized as special para-metric cases of OA. The MacWilliams formula links the weight distribution of a linear code and its dual. Our proof is probabilistic. This motivates the deﬁnition of probability spaces. The game of SET leads to natural questions concerning ternary (q = 3) linear codes. Syndrome decoding can in principle be used to decode linear codes. A large and important family of linear codes are the Reed-Solomon codes of Chapter 4. They meet the Singleton bound with equality (they are MDS-codes). Lagrange interpolation shows that they form OA of index λ = 1. The dual of an RS-code is an RS-code. Mutually orthogonal latin squares yield non-linear MDS-codes. Covering arrays and their use in software testing are discussed. Chapter 5 introduces some recursive constructions of codes: shortening and puncturing, the parity check bit for binary codes, the residual code and the Griesmer bound, concatenation and the (u, u + v)-construction. Chapter 6 presents our ﬁrst application in computer science. The concept of universal hashing is introduced. ϵ-universal hash classes turn out to be formally equivalent with codes. Chapter 7 is a direct construction of the binary Golay code and the large Witt design. This motivates the deﬁnition of t-designs. Classical projective planes PG(2, q) are introduced. Chapter 8: Meaning and basic properties of the Shannon entropy of a probability space. The binary entropy function. The Jensen inequality as a tool. In Chapter 9 the concept of asymptotic bounds for inﬁnite families of codes with length − →∞is introduced and the asymptotic version of the Singleton bound given. The Plotkin bound on codes of large distance is proved. It implies an asymptotic bound and the Bose-Bush bound on OA of strength 2. Shannon’s channel coding theorem is proved. Basic notions and facts on probability theory (random variable, expectation, variance, Tschebyscheﬀin-equality) are introduced. These are used in the proof. The Gilbert-Varshamov bound is an existence bound on linear codes (certain codes are guaranteed to exist), based on counting arguments. The Justesen codes form an explicit family of codes with asymptotically nontrivial parameters. Chapter 10 is an introduction to the geometric method. 3-dimensional codes are described as multisets of points in the projective plane PG(2, q). An application allows the construction of congestion-free networks. xii The plan: Part II Chapter 12 starts with more basic properties of ﬁnite ﬁelds: Primitive el-ements, ﬁeld extensions, the Frobenius automorphism, the Galois group, the trace from a ﬁeld to a subﬁeld. Trace codes and subﬁeld codes are deﬁned. Delsarte’s theorem describes the dual code. We introduce the Galois closure of a linear code with respect to a subﬁeld, prove the second main theorem and sketch the general strategy to construct cyclic codes. Diﬀerent notions of equivalence of codes are discussed. Chapter 13 is dedicated to the general machinery of cyclic codes. An ex-ample (binary, length 15) is used to illustrate this. This is a subﬁeld code (and a trace code) of a Reed-Solomon code deﬁned over F16. Basic notions of the general construction are cyclotomic cosets, the dimension formula and the BCH-bound on the minimum distance. Parametric examples of cyclic codes are given as well as an application to ﬁngerprinting. The Roos bound and the van Lint-Wilson method allow improvements on the BCH-bound in special situations. Generator matrices and check matrices of cyclic codes are almost canonically determined. BCH-codes are special cyclic codes. Their decoding algorithm is based on the Euclidean algorithm. Constacyclic codes are gener-alizations of cyclic codes. They can however be described within the theory of cyclic codes. This central chapter ends with two families of particularly good quaternary (q = 4) constacyclic codes and a comparison with the traditional ring-theoretic approach to cyclic codes. Chapter 14 complements Chapter 5 by introducing further recursive con-structions of codes. Constructions X and XX can be applied using cyclic codes as ingredients. The covering radius is another basic parameter. It is related to lengthening and has its own applications. We describe an application in steganography and an application in the reduction of switching noise. Chapter 15 discusses various applications. OA are interpreted as families of random variables with certain independence properties and as perfect local randomizers. We introduce linear shift register sequences and their relation to Simplex codes, describe the role of minimum distance and strength in the construction of block ciphers and the use of OA in two-point based sam-pling and chips testing. Further topics include the relation between OA and resilient functions, applications of resilient functions for the wire-tap channel and the generation of random bits, applications of OA in the derandomization of Monte Carlo-algorithms, as well as a more detailed study of universal hash classes, their construction from codes and their applications in cryptography (authentication). Chapter 16 studies the geometric approach to codes. Linear codes are described as multisets of points in projective geometry PG(k−1, q). The main theorem determines the minimum distance in terms of hyperplane intersection sizes. The hexacode, ovals and hyperovals, extended Reed-Solomon codes xiii and the Simplex codes are best understood from this point of view. Codes of dimension 2 and 3 are studied. The ternary Golay code is constructed starting from its parameters [12, 6, 6]3. Barlotti arcs and Denniston arcs as well as the corresponding codes are described. Caps are sets of points no three of which are on a line. They are formally equivalent to linear codes of minimum distance d = 4. We introduce the theory of bilinear forms and quadratic forms. This yields canonical models of large caps in PG(2, q) (parabolic quadrics) and in PG(3, q) (elliptic quadrics). Direct constructions of caps in PG(4, q) and general bounds on caps in arbitrary dimension are derived. Recursive constructions of caps use as ingredients elliptic quadrics or, in the ternary case, the Hill cap in PG(5, 3). Chapter 17 introduces codes, whose alphabet forms a vector space over a ground ﬁeld. They generalize the linear codes. Chen projection is a simple recursive construction. Application to caps yields codes which have been used in computer memory systems. Application to Reed-Solomon codes produces codes which have been used in deep space communication. Another recursive construction simpliﬁes the Bose-Bush construction of OA of strength 2. We develop a theory of cyclic additive codes which generalizes the approach from Chapter 13. Quantum codes are obtained as an application. The sections on tms-nets and on sphere packings in Euclidean spaces (Chap-ter 18) are treated somewhat lighter than the applications in Chapter 15. The theory of tms-nets is to be seen in the context of quasi-Monte Carlo algorithms (related to uniformly distributed point sets in Euclidean space). This appli-cation of coding theory to numerical integration and the pricing of exotic options is rather surprising. The construction of dense sphere packings is a classical problem in Euclidean space with links not only to discrete mathemat-ics but also to algebra and algebraic geometry. A section on list decoding of Reed-Solomon codes has been included as this represents a rather recent de-velopment which uses some basic algebra/geometry in a transparent way. The remaining sections of Chapter 18 have the character of brief survey articles. How to use this text Part I arose from several one semester undergraduate courses on coding theory. Chapters 2 to 6 form the core of such an introductory course. The only section which can (and maybe should) be skipped is Section 3.5 on the MacWilliams transform. The remaining time should be dedicated to one or several of the later chapters of Part I. Chapter 7 would be a natural choice as the binary Golay code probably is the most famous of all codes. Usually I cover Chapter 6 as it is the ﬁrst example of a non-standard application and it can be done in one lecture. xiv Another choice would be to cover Chapter 8 and some of the material from Chapter 9. These chapters form a unit as the entropy function is used in the asymptotic expressions of Chapter 9. The canonical starting point for the second semester of a two semester course is the theory of cyclic codes and their implementation, Chapters 12 and 13, as well as Chapter 14. Some of the later parts of Chapter 13 are optional (the application to ﬁngerprinting, the Roos bound, the van Lint-Wilson bound, the comparison with the traditional approach and Section 13.4 on constacyclic codes). Section 13.3 on the decoding algorithm of BCH-codes may be sacriﬁced as well. Chapter 16 may be considered another core area of coding theory as it gives a better understanding of many important codes. It would be a pity to sacriﬁce the main theorem of the ﬁrst section. Some of the applications in the ﬁrst section ought to be covered. From here on there are several choices. For a thorough introduction to the theory of codes and its links with Galois geometry one might concentrate entirely on Chapter 16. Another possibility is to cut short on Chapter 16 and to cover several ap-plications from Chapter 15 instead. The sections on tms-nets and on sphere packings in Chapter 18 are then a good choice to round oﬀa graduate course. This also has the advantage that the course ends with an introduction to an exceptional object, the Leech lattice. A third strategy is to concentrate on the theory of cyclic codes and their applications. Such a course would end with Chapter 17. The database If you construct a linear code and want to compare it with what is known, there is A. E. Brouwer’s interactive data base of parameters [n, k, d]q of linear codes For q ≤9, n not too large and k ≤n it gives you two numbers: the largest d that is known (the lower bound) and the largest d which may theoretically be possible. xv Textbooks An early classic among the textbooks is Algebraic Coding Theory by E. R. Berlekamp, which presents in particular an excellent introduction to the traditional theory of cyclic codes. The 1977 book The Theory of Error-Correcting Codes by MacWilliams and Sloane is considered something like the bible of traditional coding theory. A more recent introduction is J. H. van Lint’s Introduction to Coding Theory , a relatively short and dense text which helped a great deal in attracting pure mathematicians to the area. H. Stichtenoth’s book Algebraic Function Fields and Codes is a self-contained introduction to the theory of algebraic function ﬁelds in one variable (equivalently: algebraic curves) and to the codes derived from them. Among the undergraduate textbooks we mention Vera Pless, The Theory of Error-Correcting Codes and R. Hill, A ﬁrst course in coding theory . Hill’s book is the ﬁrst of its kind which presents an introduction to the geometric approach to codes. Orthogonal Arrays: Theory and Applications by Hedayat,Sloane and Stufken introduces to orthogonal arrays (dual codes) and their applications to the design of experiments. My thanks go to Yves Edel; Wolfgang Ch. Schmid, Peter Hellekalek and the Institut f¨ ur Mathematik of the University of Salzburg; Fernanda Pambianco, Giorgio Faina and the dipartimento di matematica e informatica of the University of Perugia; Bernd Stellmacher and the Mathematisches Seminar of the University of Kiel; C. L. Chen, Ron Crandall, Ludo Tolhuizen, Albrecht Brandis, Allan Struthers, Michigan Technological University and the National Security Agency. About the author J¨ urgen Bierbrauer received his PhD in mathematics at the University of Mainz (Germany) in 1977, with a dissertation from the theory of ﬁnite groups. His mathematical interests are in algebraic methods of discrete mathematics and its applications. He held a position at the University of Heidelberg (Germany) from 1977 to 1994. Currently he is full professor at Michigan Technological University in Houghton, Michigan. His non-mathematical interests include Romance languages, literature, and the game of Go. xvii To Stella, Daniel and my mother Contents I An elementary introduction to coding 1 1 The concept of coding 3 1.1 Bitstrings and binary operations . . . . . . . . . . . . . . . . 3 1.2 The Hamming distance . . . . . . . . . . . . . . . . . . . . . 7 1.3 Binary codes . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.4 Error-correcting codes in general . . . . . . . . . . . . . . . . 12 1.5 The binary symmetric channel . . . . . . . . . . . . . . . . . 14 1.6 The sphere-packing bound . . . . . . . . . . . . . . . . . . . 18 2 Binary linear codes 23 2.1 The concept of binary linear codes . . . . . . . . . . . . . . . 23 2.2 Block coding . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.3 The eﬀect of coding . . . . . . . . . . . . . . . . . . . . . . . 29 2.4 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.5 Binary Hamming and Simplex codes . . . . . . . . . . . . . . 33 2.6 Principle of duality . . . . . . . . . . . . . . . . . . . . . . . 37 3 General linear codes 41 3.1 Prime ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.2 Finite ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 3.3 Linear codes over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . 47 3.4 Duality and orthogonal arrays . . . . . . . . . . . . . . . . . 50 3.5 Weight distribution . . . . . . . . . . . . . . . . . . . . . . . 58 3.6 The game of SET . . . . . . . . . . . . . . . . . . . . . . . . 63 3.7 Syndrome decoding . . . . . . . . . . . . . . . . . . . . . . . 66 4 Singleton bound and Reed-Solomon codes 71 5 Recursive constructions I 81 5.1 Shortening and puncturing . . . . . . . . . . . . . . . . . . . 81 5.2 Concatenation . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6 Universal hashing 93 7 Designs and the binary Golay code 97 8 Shannon entropy 101 xxi xxii 9 Asymptotic results 111 10 3-dimensional codes, projective planes 123 11 Summary and outlook 129 II Theory and applications of codes 131 12 Subﬁeld codes and trace codes 133 12.1 The trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 12.2 Trace codes and subﬁeld codes . . . . . . . . . . . . . . . . . 138 12.3 Galois closed codes . . . . . . . . . . . . . . . . . . . . . . . 141 12.4 Automorphism groups . . . . . . . . . . . . . . . . . . . . . . 144 13 Cyclic codes 149 13.1 Some primitive cyclic codes of length 15 . . . . . . . . . . . . 149 13.2 Theory of cyclic codes . . . . . . . . . . . . . . . . . . . . . . 152 13.3 Decoding BCH-codes . . . . . . . . . . . . . . . . . . . . . . 167 13.4 Constacyclic codes . . . . . . . . . . . . . . . . . . . . . . . . 178 13.5 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 14 Recursive constructions, covering radius 185 14.1 Construction X . . . . . . . . . . . . . . . . . . . . . . . . . 185 14.2 Covering radius . . . . . . . . . . . . . . . . . . . . . . . . . 193 15 OA in statistics and computer science 201 15.1 OA and independent random variables . . . . . . . . . . . . 201 15.2 Linear shift register sequences . . . . . . . . . . . . . . . . . 204 15.3 Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . 212 15.4 Two-point based sampling . . . . . . . . . . . . . . . . . . . 215 15.5 Resilient functions . . . . . . . . . . . . . . . . . . . . . . . . 217 15.6 Derandomization of algorithms . . . . . . . . . . . . . . . . . 226 15.7 Authentication and universal hashing . . . . . . . . . . . . . 231 16 The geometric description of codes 247 16.1 Linear codes as sets of points . . . . . . . . . . . . . . . . . . 247 16.2 Quadratic forms, bilinear forms and caps . . . . . . . . . . . 273 16.3 Caps: constructions and bounds . . . . . . . . . . . . . . . . 293 17 Additive codes 309 17.1 Basic constructions and applications . . . . . . . . . . . . . . 309 17.2 Additive cyclic codes . . . . . . . . . . . . . . . . . . . . . . 320 17.3 Quantum codes . . . . . . . . . . . . . . . . . . . . . . . . . 326 xxiii 18 The last chapter 331 18.1 The linear-programming bound . . . . . . . . . . . . . . . . 331 18.2 Algebraic-geometric codes . . . . . . . . . . . . . . . . . . . . 336 18.3 List decoding . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 18.4 Expander codes . . . . . . . . . . . . . . . . . . . . . . . . . 344 18.5 tms-nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 18.6 Sphere packings . . . . . . . . . . . . . . . . . . . . . . . . . 350 18.7 Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 18.8 Nonlinear codes . . . . . . . . . . . . . . . . . . . . . . . . . 362 References 371 Part I An elementary introduction to coding 1 Chapter 1 The concept of coding 1.1 Bitstrings and binary operations Basic concepts: bits, bitstrings, transmission of messages, transmitting pic-tures from space, the Morse code, XORing, the ﬁeld F2, the model of message transmission, a ﬁrst idea of error correction. The object of Coding Theory is the transmission of messages over noisy channels. Below is the standard picture visualizing the situation. At ﬁrst we need to understand what the elements of this picture mean: what is a message, a channel, what is noise? Along the way we will encounter more basic notions. In this ﬁrst chapter some of these will be explained. We start with the message. receiver decode errors channel encode transmitter FIGURE 1.1: Information transmission over a noisy channel 3 4 Introduction to Coding Theory If there are 8 possible messages to be sent, say, then we can represent each message as a bitstring of length 3, like 000 or 011 or 110. We will generally assume this has been done and deﬁne a message to be a bitstring. So here are some examples of bitstrings and their lengths: bit string length 000 3 110 3 110011 6 0000011111 10 Transmitting pictures from space Assume we wish to transmit a photograph from outer space, like one of the pictures of Saturn taken by the Voyager spacecrafts in the early 1980s (Viger for Star Trek buﬀs). The picture is divided into 800 × 800 pixels, each pixel is assigned one of 256 = 28 degrees of brightness. The brightness of a pixel is thus represented by a bitstring of length 8 and the total black and white picture consists of 800 × 800 × 8 bits. As the picture really is in color, the same photo is transmitted three times, each time through a diﬀerent color ﬁlter. The full color picture will thus be represented by a bitstring of length 3 × 800 × 800 × 8 = 15, 360, 000. This is our message. The channel is determined by the properties of space between the spacecraft and the receiver on Earth, above all by the Earth’s atmosphere. A certain number of bits will be destroyed. Here we only consider errors of the type that 0 is transmitted and 1 is received or vice versa. The Morse code Another illustration for the claim that every message can be represented by bitstrings is the Morse code, which has been in use for telegraphy since the 1840s. It represents each of the 26 letters A, B, . . . , Z, each digit 0, 1, . . ., 9 as well as the period, the comma and the question mark by a sequence of at most ﬁve dots and dashes. The dot stands for a short signal, the dash for a long signal. Dashes are about three times as long as dots. For example the letter E is represented by a single dot, S is represented by dot-dot-dot and Z is dash-dash-dot-dot. The concept of coding 5 However, the graphical representation is purely conventional. We can rep-resent dash by 1, dot by 0 and obtain a representation of letters and numbers as bitstrings: E=0, S=000, Z=1100, and T=1, Q=1101, V=0001. Back to the general model Assume we wish to send one of 8 possible messages (the bitstrings of length 3), for example message 011. If it should happen along the way (in the channel) that the second bit is ﬂipped (the second coordinate is in error), then 001 will be received. We want to express this situation in mathematical terminology. 1.1 Deﬁnition. F2 = {0, 1} with addition 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0 Here the letter F stands for ﬁeld. It indicates that this tiny little structure of only 2 elements has something in common with the ﬁeld of real numbers and the complex number ﬁeld: each of these structures satisﬁes the same set of axioms. Each of them is a ﬁeld. We will come back to this topic later. The idea behind the addition in F2 is to model the errors in information transmission: a coordinate of a message is in error if and only if 1 is added to the entry in this coordinate. This explains why we must have 1 + 1 = 0 : if a bit is ﬂipped twice then no error occurs, or: if 1 is ﬂipped, then 0 is received. Addition in F2 is also known as XORing, in particular in the computer science literature. Here XOR stands for exclusive or, a logical operation, where 1 stands for true and 0 for false. The relation 1 + 1 = 0 (true or true = false) is what makes it “exclusive”: in order for the result to be true one of the two ingredients has to be true, but not both (the ordinary or operation would have 1 + 1 = 1). Another motivation for our binary addition comes from arithmetic: if we distinguish even and odd integers, the following familiar rules hold: even + odd = odd + even = odd, even + even = odd + odd = even. With even = 0 and odd = 1 these are exactly the rules of binary addition. Addition in F2 describes what happens in each coordinate. Calculation with bitstrings is formalized as follows: 1.2 Deﬁnition. Fn 2 consists of bitstrings x = (x1, x2, . . . , xn) of length n, where xi ∈F2. Addition in Fn 2 is coordinatewise. For example, let x = 001 ∈F3 2 and y = 101 ∈F3 2. Then x + y = 100. With this terminology it is easier to express the relation between messages, errors 6 Introduction to Coding Theory and received messages. Let x = 001100 be the message sent. Assume errors occur in coordinates 2 and 4. We can express this by adding the error vector e = 010100. The received message will then be the sum y = x + e = 011000. Let us do this the other way around: if x = 010101 was sent and y = 010111 was received, then an error occurred in coordinate 5. This means e = 000010. Return to the situation where we send one of 8 messages (the elements of F3 2), for example x = 011. No matter what the error vector e is, the received message y = x + e is again one of the legitimate messages. There is no way for the receiver to suspect that an error occurred, let alone to correct it. So how can we hope to correct errors? Error correction: the ﬁrst idea Here is the easiest of all error-correcting systems: encode each 0 of the message to be sent as a block 000 and analogously each 1 by 111. If the original message is x = 011, the encoded message is now x′ = 000111111. The receiver knows what encoding scheme has been used. He will therefore divide the received message in blocks of length 3. If such a block is 000 or 111 then decoding is obvious: 000 →0, 111 →1. Assume a block of the received message is 101. The receiver knows that at least one transmission error must have happened. It is a basic assumption that a small number of errors is more probable than many errors. The decoding will therefore be 101 →1 (by majority decision). The initial picture begins to make sense now. We have seen that messages can be encoded by the transmitter and decoded by the receiver such that the following holds: if not more than one error occurs during transmission (in the channel), then the error will automatically be corrected. What we have used is known as the repetition code of length 3. 1. We learned how to calculate with bitstrings. 2. Fn 2 consists of the bitstrings of length n. 3. The received message is the binary sum of the sent message and the error vector. 4. The repetition code of length 3 corrects one bit error. The concept of coding 7 Exercises 1.1 1.1.1. Compute the sum of 11001 and 01110 1.1.2. Assume 000000 was sent and two errors occurred. List all possible received messages. 1.1.3. Let x = 1101 be the message to be sent. Encode x using the repetition code of length 3. 1.1.4. Assume the repetition code of length 3 is used and 000110111101 is received. What is the result of decoding? 1.1.5. Why does the Morse code represent letters E and T by strings of length 1, whereas letters like Q,V,Z are represented by longer bitstrings? 1.2 The Hamming distance Basic concepts: The Hamming distance as a metric, the weight of a bitstring. 1.3 Deﬁnition. Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) be bitstrings in Fn 2. The distance (or Hamming distance) between x and y is d(x, y) = number of coordinates i where xi ̸= yi. Here are some examples: d(0000, 1111) = 4, d(00110, 00101) = 2, d(111111, 001100) = 4. Expressed in the context of messages and errors d(x, y) is the minimum number of errors transforming x into y. In fact, consider the second example above: x = 00110 and y = 00101 diﬀer in the last two coordinates: d(x, y) = 2 and x + 00011 = y. Things get even easier when we use the weight. 8 Introduction to Coding Theory 1.4 Deﬁnition. The weight wt(x) of the bitstring x ∈Fn 2 is the number of nonzero coordinates in x. Here are some examples: wt(0000) = 0, wt(1111) = 4, wt(00110) = 2, wt(001101) = 3. The weight of a bitstring is its distance from the all-0 bitstring. If the all-0 bitstring is sent and w errors occur during transmission, then the received message has weight w. If x is sent, e is the error vector and y = x + e is received, then d(x, y) = wt(e). The Hamming distance is also called the Hamming metric. The general notion of a metric is widely used in mathematics. Here is the deﬁnition: 1.5 Deﬁnition. Let X be a set. For every pair x ∈X, y ∈X let a real number d(x, y) be given (the distance from x to y). The function d is called a metric if the following are satisﬁed: • d(x, y) ≥0 for all x, y. • d(y, x) = d(x, y) for all x, y. • d(x, y) = 0 if and only if x = y. • d(x, z) ≤d(x, y) + d(y, z) for all x, y, z. The last requirement is the most important. It is known as the triangle inequality. A famous metric is the Euclidean metric in Euclidean space. If for example x = (x1, x2) and y = (y1, y2) are two points in the plane, then their Euclidean distance is (x1 −y1)2 + (x2 −y2)2. 1.6 Theorem. The Hamming distance is a metric on Fn 2. Most of the properties of Deﬁnition 1.5 are obvious. Only the triangle inequality is a little interesting. This is left as an exercise. 1. The Hamming distance between two bitstrings of same length is the number of coordinates where they diﬀer. 2. d(x, y) is the number of bit errors needed to transform x into y. 3. The Hamming distance is a metric. 4. The weight is the distance from the all-0 vector. 5. d(x, y) = wt(x + y). 6. If bitstring x is sent and y is received, then d(x, y) = wt(e) is the weight of the error vector. The concept of coding 9 Exercises 1.2 1.2.1. Compute d(11001, 01110) and d(0000, 0110). 1.2.2. Find wt(00110) and wt(10111). 1.2.3. List all vectors in F6 2 at distance 3 from 111000. 1.2.4. The alphabet has 26 letters. If we want to represent all possible words of length ≤3 (all letters, pairs of letters and triples of letters) as bitstrings of the same length n, what is the smallest number n such that this is possible? 1.2.5. Prove that the Hamming distance is a metric. 1.2.6. Assume x is sent and y = x + e is received. What can we say about d(x, y) and about wt(e) if not more than 3 errors have occurred? 1.3 Binary codes Basic concepts: Length, minimum distance. The idea of error correction. We saw that no errors can be detected or corrected if all elements of Fn 2 are used as messages. The obvious idea is to use only a certain subset. Such subsets will be called codes. Let us send bitstrings of length 6. Instead of using all elements of F6 2 as (encoded) messages, we use only the following subset: 000000 001011 100110 101101 010101 011110 110011 111000 Such a family of bitstrings of length 6 is also called a binary code of length 6. Its elements are codewords. In our example we have 8 codewords. The most important property of this code is the following: any two diﬀerent codewords are at distance ≥3. We say that 3 is the minimum distance of 10 Introduction to Coding Theory e e e e e e e e e FIGURE 1.2: Non-overlapping balls centered at codewords the code. Please check for yourself that this is true. The parameters of this binary code are then recorded as (6, 8, 3)2 : we have a binary code (indicated by subscript 2), of length 6, consisting of 8 codewords, with minimum distance of 3. The idea of error correction Transmitter and receiver agree on the code to be used. Only codewords will be sent. If only one error occurs in the channel, then the received word will be in a ball of radius 1 around a codeword (in the Hamming metric). Assume the code has been chosen such that any two codewords are at distance at least 3. Then the balls of radius 1 do not overlap: if a bitstring has distance 1 from some codeword, then it has larger distance from any other codeword. In other words, the receiver will decode any vector at distance ≤1 from some codeword as that codeword. In the picture: the whole ball (or call it a disc) of radius 1 is decoded as the center of the ball, or: the received tuple is decoded as the codeword which it resembles most closely. If not more than 1 error occurred, then this error will be corrected. Observe that Figure 1.2 serves only as an illustration. The metric in the Euclidean plane is used to illustrate the situation in a rather diﬀerent metric, the Hamming metric. The concept of coding 11 1.7 Deﬁnition. A binary code (n, M, d)2 is a set of M bitstrings in Fn 2 such that any two diﬀerent elements of the code (codewords) are at distance ≥d. We call n the length and d the minimum distance of the code. A code of minimum distance 5 can correct 2 errors, minimum distance 7 can correct 3 errors and so on. A fundamental problem of coding theory is the following: Given n and d, ﬁnd the largest M such that there is a code (n, M, d)2. 1. A binary code (n, M, d)2 is a collection of M bitstrings of length n such that any two diﬀerent of these codewords are at Hamming distance at least d. 2. A basic problem of coding theory: determine the maximum M such that an (n, M, d)2-code exists. 3. A code can correct e errors provided its minimum distance is d ≥2e + 1. 4. We saw a (6, 8, 3)2−code. Exercises 1.3 1.3.1. If we want to correct 8 bit errors, what would the minimum distance of the code have to be? 1.3.2. Using our code (6, 8, 3)2, decode the following received vectors: 111100, 111011, 000001, 011110. 1.3.3. Does a code (5, 6, 3)2 exist? 12 Introduction to Coding Theory 1.4 Error-correcting codes in general Basic concepts: Basic code parameters. Telegraphy codes as early examples. The notion of a binary code is too narrow, although it is most frequently used in information transmission. Here is the general concept of an (error-correcting) code: 1.8 Deﬁnition. Let A be a ﬁnite set of q elements (the alphabet). A q-ary code C of length n is a family of n-tuples with entries in A : C ⊆An. For example, let q = 3 and A = {0, 1, 2}. Then A4 consists of the 34 tuples of length 4 with entries 0, 1, 2, like for example 0000, 0102, 2221 or 2100. A 3-ary code (also called ternary) of length 4 consists of a collection of such ternary 4-tuples. The Morse code from Section 1.1 really is a ternary code. The reason is that the individual letters need to be seperated. If we represent dot by 0, dash by 1 and a gap between letters as 2, the message SOS will be represented by the ternary word 00021112000. 1.9 Deﬁnition. Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) be elements (strings, vectors, words) in An. The distance (or Hamming distance) between x and y is deﬁned as d(x, y) = number of coordinates i where xi ̸= yi. This is the same as in Deﬁnition 1.3. As before, the minimum distance of a code is the minimum of the distances between diﬀerent codewords, and again the Hamming distance deﬁnes a metric. 1.10 Deﬁnition. A q-ary code (n, M, d)q is a set of M strings in An (where \|A\| = q) such that any two diﬀerent elements of the code (codewords) are at distance ≥d. Call n the length and d the minimum distance of the code. Here are the words of a ternary code (4, 27, 2)3 : The concept of coding 13 0000 1002 2001 0102 1101 2100 0201 1200 2202 0012 1011 2010 0111 1110 2112 0210 1212 2211 0021 1020 2022 0120 1122 2121 0222 1221 2220 As in the binary case, in order to correct e errors we need a code of minimum distance at least 2e+1. Our code (4, 27, 2)3 will not suﬃce to correct one error. For example, if 0000 was sent and 0010 received (only one error occurred), the received vector has distance 1 not only from 0000 but also from 0012 and from 0210 and from 2010. A basic problem of coding theory is the following: given q, n, d, ﬁnd the maximum number M such that a code (n, M, d)q exists. Error detection in telegraphy codes Error detection is a more modest aim than error correction. It is suitable in situations where the channel is very good. In the rare occasions that an error occurs (and is detected), the receiver can then simply ask for retransmission. The alphabet of the telegraphy codes consists of the 26 letters A, B, . . . , Z. The classical commercial codes use ﬁve letter groups as codewords. Each trade had its own elaborate codes. The primary aim of these codes was to save transmission time and thus to save money. As an example take the Acme Code. It saves time to send the codeword BUKSI when Avoid arrest if possible is intended, and AROJD is shorter than Please advertise the birth of twins. It is a little unclear if PYTUO for Collided with an iceberg really achieves much in this respect as such collisions do not happen all the time. In modern terminology this business of representing messages by short strings is called Data Compression or Source Coding. It is not our concern in this book. However, commercial telegraphy codes also took the reliability of message transmission into consideration. A general rule known as the two-letter dif-ferential stipulated that any two codewords had to diﬀer in at least two letters. This means that each commercial code has minimum Hamming dis-tance ≥2, enough to detect single errors. The Acme code also safeguarded against a diﬀerent type of error: no two codewords (necessarily of Hamming distance 2) may result from each other by transposition of two adjacent letters. For instance, if AHXNO is a codeword (it stands for Met with a fatal accident in the Acme code), then HAXNO, AXHNO, AHNXO, AHXON cannot be codewords. This material is from Chapter 22 of D. Kahn’s The Codebreakers . 14 Introduction to Coding Theory 1. A q-ary code (n, M, d)q is a collection of M q-ary n-tuples (the codewords)such that any two diﬀerent codewords are at Hamming distance at least d. 2. A basic problem of coding theory: given q, n, d maximize M such that an (n, M, d)q-code exists. 3. A code can correct e errors provided its minimum distance is d ≥2e + 1. 4. We saw a ternary code (4, 27, 2)3. Exercises 1.4 1.4.1. Find the smallest length n such that an (n, 27, 2)3 exists. 1.4.2. Prove the following: if there is an (n, M, d)q, then there is an (n + 1, M, d)q. 1.4.3. Prove the following: If there is an (n, M, d)q−1, then there is an (n, M, d)q. 1.5 The binary symmetric channel Basic concepts: The BSC, binomial numbers, subsets and paths, the Pascal triangle. So how do we model the noise mentioned in Section 1.1? In the case of the binary alphabet the easiest and most widely used model is the binary symmetric channel (BSC). It assumes that there is a certain ﬁxed probability, a number p, where 0 < p < 1, for a bit transmission error. Clearly a value The concept of coding 15 0 1 0 1 1-p p p 1-p FIGURE 1.3: The BSC p > 1/2 does not make sense (why?) and p = 1/2 would mean that pure noise is received. We can therefore assume p < 1/2, and for all practical purposes p will be rather small. The probability that a sent 0 will be received as 0 is 1 −p. Likewise the probability that a sent 1 will be received as 1 is 1 −p. The probability that a sent 0 is received as a 1 is p, just like the probability that a sent 1 is received as 0. This model of a channel is called symmetric because 0 and 1 play sym-metric roles. It is also called a memoryless channel because the probability of a bit error is independent of the prehistory. Can you think of situations when this model will not be appropriate? In order to be able to do some combinatorial counting based on the BSC we take a look at the binomial numbers. Binomials, subsets and paths The number of bitstrings of length n and weight m is the binomial number n m . This is the number of error patterns that can occur in n bits when the total number of errors is m. For example, the bitstrings of length 4 and weight 2 are 0011, 0110, 0101, 1001, 1010, 1100. Accordingly, 4 2 = 6. 16 Introduction to Coding Theory 1 5 10 10 5 1 4 6 4 1 1 3 3 1 1 2 1 1 1 1 1 FIGURE 1.4: Pascal’s triangle Another interpretation of the binomials uses subsets: identify each coordi-nate of a bitstring of length n with an element of a set, for example with the numbers 1, 2, . . . , n. Each bitstring of length n can then be identiﬁed with a subset of {1, 2, . . ., n}. If the entry in the corresponding coordinate is 1, we include that element in the subset, if the entry is 0 we will not include it. For example, 0011 corresponds to the subset {3, 4}, 1001 to the subset {1, 4}, 1111 to the total set {1, 2, 3, 4} and 0000 to the empty set. We see that the bit-strings of length n and weight m correspond precisely to the subsets of m elements of a ﬁxed set with n elements. A third interpretation of the binomials involves paths in a triangle, see Figure 1.4. Consider paths starting at the top of the triangle, where in each step the choice is between going Southeast or Southwest. We may encode this decision by a string of E and W, for example EEWWE for going at ﬁrst Southeast twice, then Southwest twice and a ﬁnal step in Southeast direction. Denote the top level by level 0. Then our path will end at level 5, at a node labelled 10 in Figure 1.4. Why that label? Our string with entries E and W is a bitstring in disguise. We can write 1 for E and 0 for W, obtaining bitstring 11001. Each path in the triangle is described by a bitstring. Bitstrings of length n end on level n. Two bitstrings end in the same spot if they have the same length and the same weight. This explains our labels: the label of the node on level n The concept of coding 17 and weight m (start with weight 0 on the western end of the level, end with weight n on the eastern end) is the number of paths ending there. This is the number of bitstrings of length n and weight m, in other words the binomial n m . The endpoint of our path is labeled 10 as 5 3 = 10 is the number of paths ending there. The labels 1 on the western border are the numbers n 0 = 1 (there is only one bitstring with all entries 0 of any given length), the labels 1 on the eastern border are the numbers n n = 1 (there is only one bitstring with all entries 1). The triangle is known as the Pascal triangle. It can be used to compute the binomials recursively. In fact, each label not on the border is the sum of the labels directly to its Northeast and to its Northwest, in formulas n m = n −1 m −1 + n −1 m for 0 < m < n. A direct formula to compute the binomials is n m = n(n −1) . . . (n −m + 1) m(m −1) . . . 2 · 1 . As examples, 4 2 = 4 · 3 2 = 6, 5 2 = 5 · 4 2 = 10, 6 3 = 6 · 5 · 4 3 · 2 = 20 and in general n 2 = n(n −1) 2 . Back to the channel Assume a word (a bitstring) of length n is sent through the channel. The probability that no bit error occurs is (1 −p)n. The probability that exactly one error occurs is np(1 −p)n−1 (there are n choices where the error could happen; the probability of a certain error pattern involving just 1 error is p(1 −p)n−1). The probability of a certain error pattern with precisely k errors is pk(1 −p)n−k. The number of such error patterns is the number of possibilities of choosing k elements out of n. This is n k . The probability that precisely k errors occur is therefore n k pk(1 −p)n−k. If we want to compute the probability that at most k errors happen, we have to sum up these probabilities.
3204	https://www.pbslearningmedia.org/resource/ate10.sci.phys.energy.snellslaw/the-index-of-refraction-and-snells-law/	FOR TEACHERS Is WOSU your local station? FOR TEACHERS Student site The Index of Refraction and Snell’s Law Video Grades: 9-12 Collection: Advanced Technological Education In this video from the ICT Center, learn about the index of refraction and Snell's law. Review a simple example of refraction, the speed of light in materials and the formula for calculating the index of refraction. Explore a table of the refractive indices of common media and materials used in fiber optics and semiconductors. Finally, examine Snell's law to understand the relationship between incident and refracted angles. This video is available in both English and Spanish audio, along with corresponding closed captions. Permitted use Stream, Download, Share, and Modify Accessibility Auditory, Visual, Caption Credits You May Also Like 4:11 Interference and Diffraction Video 1:04 Laser Waterfall Video 2:04 Observing Refraction of Light Video ##### Fiber Optics Lesson Plan Explore related topics Science Physical Science Waves and Light Light as Electromagnetic Radiation Engineering & Technology Systems & Technologies Communication and Information Data More from the Advanced Technological Education Collection Collection 117 ##### Planning Your Future Career in Advanced Technology Interactive Lesson 3:02 Engineering Technology: Electronic Maintenance Technician Video 2:46 Process Technology: Process Maintenance Engineer Video 3:08 Biomanufacturing Supervisor Video 3:02 Welding Supervisor Video See the Collection Made Possible Through producer, contributor funder funder Unlock the Power of PBS LearningMedia Create a free account to gain full access to the website Save & Organize Resources See State Standards Manage Classes & Assignments Sync with Google Classroom Create Lessons Customized Dashboard
3205	https://math.stackexchange.com/questions/738973/finding-the-fourth-vertex-of-a-square	Skip to main content Finding the fourth vertex of a square Ask Question Asked Modified 11 years, 4 months ago Viewed 18k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I was given this problem: Without drawing a graph and given the following 3 vertices, find the coordinates of the last vertex of the square: (3,2)(0,5)(−3,2) So first, I found the length of the sides and the diagonal of the square, which are 18−−√ and 6 respectively. By graphing, I know the solution is (0,−1). Then, I assume that since the length between (3,2) and (−3,2) is the diagonal, then the distance between (0,5) and the remaining vertex must be the diagonal too. And since the length of the side is 6, then the distance between the vertex and either (3,2) or (−3,2) must be 6. So: (x−3)2+(y−2)2−−−−−−−−−−−−−−−√=18−−√ (x−0)2+(y−5)2−−−−−−−−−−−−−−−√=6 Which gives (after a bit of cleaning up): x2+y2−10y=11 x2−6x+y2−4y=5 Then, replacing the second expression into the first one: x2−6x+y2−4y=5⇒x2=5+6x−y2+4y 5+6x−y2+4y+y2−10y=11 5+6x+4y−10y=11 6x−6y+6 x−y=1 x=1+y Up to this point, I know I'm not entirely wrong because the expression is true for the actual coordinates of the vertex, because 0=1+(−1) is true. But I wouldn't know how to proceed if I hadn't known the answer beforehand. I need to find both x and y, is there a linear equation I'm missing to find the exact coordinates of the last vertex? Is my process okay or is there a simpler way to do it? algebra-precalculus geometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Apr 4, 2014 at 1:15 user122283 asked Apr 4, 2014 at 1:05 Juan José CastroJuan José Castro 37522 gold badges55 silver badges1515 bronze badges 2 1 I assume you aren't allowed to use logic: (−3,2) and (3,2) are mirrored, therefore (0,5) will be mirrored across their line? – Display Name Commented Apr 4, 2014 at 1:19 The answer is: your process is fine; see my answer below. – M Turgeon Commented Apr 4, 2014 at 1:28 Add a comment \| 3 Answers 3 Reset to default This answer is useful 2 Save this answer. Show activity on this post. You're almost there. What you have found is a relationship between x and y, namely x=1+y. But there are infinitely many pairs satisfying this relation! On the other hand, you also have that x2+y2−10y=11. You can plug in the first expression: (1+y)2+y2−10y=11. Solving this quadratic equation should be easy, but it will give you two roots. On the other hand, only one of the roots can give you the y-coordinate of the fourth vertex. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 4, 2014 at 1:24 M TurgeonM Turgeon 10.8k33 gold badges4343 silver badges8383 bronze badges 1 Yeah! The two roots are 5 and -1, which makes sense since those are the y values for the vertices on that diagonal. Thank you! – Juan José Castro Commented Apr 4, 2014 at 2:21 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Hint: The coordinates are (3,2),(0,5),(−3,2),(x,y). The segment connecting (3,2),(0,5) is parallel to the segment connecting (x,y),(−3,2), and the segment connecting (0,5),(x,y) is perpendicular to the line connecting (3,2),(−3,2). Another extremely simple way is that (3,2) when reflected across the y-axis gives (−3,2), and thus (0,5) when reflected across the x-axis gives (x,y). Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 4, 2014 at 1:19 user122283user122283 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Solution without the calculation: You have three points, A,B,C. Calculate the distances between Every pair of points. You find side length s and diagonal d (d will be larger and only show up as one distance). Let A,C be the two points adjacent to the point you need to find (you know which to pick because they are the only pair d apart). Write down the equation for a circle of radius s centered at A and C. Using the quadratic equation, solve one equation in terms of one of the variables, plug that into the other equation,and solve the resulting quartic. This lets you find both points of intersection (fun fact, these circles can't be tangent). The fourth point of the square is the point of intersection further away from B (or, you can just check which one is a distance d from B. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 4, 2014 at 1:27 Stella BidermanStella Biderman 31.5k66 gold badges4949 silver badges9494 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 algebra-precalculus geometry See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 2 Finding the fourth point of a perfect square (without knowing order of points) Related 4 How to find the other vertices of an equilateral triangle given one vertex and centroid 0 Finding the length from an interior point of a triangle to a vertex given distances to the other two 2 Square in the complex plane given three vertices. 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3206	https://www.meddean.luc.edu/lumen/meded/therapy/pharmii_blockii_13.pdf	Pharmacology/Therapeutics II Block II Lectures 2012‐13 59. Drugs of Abuse, tolerance & Dependence – Dr. Bakowska 60. Pharmacology of Sedative Hypnotic Drugs – Dr. Battaglia 61. Sedative Hypnotic Drugs in Treating Anxiety & Sleep Disorders – Dr. Battaglia 62. Drugs to Treat Rheumatoid Arthritis & Gout – Dr. Clipstone 63. Treatment of Parkinsonism & Dementia – Dr. Lee (To be posted Later) 64. HIV – Dr. Cuevas 65. Anti‐Viral Drugs – Dr. Gallagher 66. Pediatric PsychoPharm‐Treatment of ADHD – Dr. Gutierrez Pharmacology & Thereapeutics Drugs of Abuse, Tolerance and Dependence February 6, 2013 Joanna Bakowska, Ph.D. Drugs of Abuse Date: Feb 6, 2013 Instructor: Joanna C. Bakowska, Ph.D., Assistant Professor, Department of Pharmacology Reading Assignment: Basic & Clinical Pharmacology, B.G. Katzung, 9th Ed. Chapter 32 KEY CONCEPTS AND LEARNING OBJECTIVES 1. To understand the basis underlying the conceptualization of addiction as a disease. 3. To understand the differential diagnostic criteria for drug abuse vs dependence 5. To know the mechanism of action within the central nervous system of major drugs of abuse 6. To know the signs and symptoms of overdosage caused by the major drugs of abuse and appropriate therapeutic interventions 7. To know the signs and symptoms of opioid withdrawal 8. To know the options for pharmacotherapy of opioid abuse/dependence and alcohol dependence Drugs of Abuse 2 Drugs of Abuse 1. Introduction 1.1 Why medical students need to learn to recognize a substance abuse problem regardless of your specialty 1. You will be encountering individuals with drug problems in your practice a. 15.6% of adults (29 million) used illegal drugs in 2007 b. Two-thirds of those who abuse or are dependent visit a primary care physician or emergency department every 6 months 2. Failure to consider substance use as a contributor to the clinical picture will compromise treatment 1.2 Drugs of Abuse A. Psychomotor stimulants – cocaine, amphetamines B. Opiates and Opioids – heroin, morphine, codeine, oxycodone, hydromorphine C. Cannabinoids – marijuana D. Alcohol E. Anti-depressants – barbiturates, benzodiazepines F. Nicotine G. Hallucinogens – LSD, mescaline “club drugs”, MDMA 1.3 DSM-IV-TR Criteria for Abuse (Prevalence 1.4% in recent national sample) A maladaptive pattern of substance use leading to clinically significant impairment or distress, as manifested by one (or more) of the following occurring at any time in the same 12-month period (but symptoms can never meet criteria for dependence, thus dependence and abuse are mutually exclusive) 1. Recurrent use resulting in failure to fulfill major role obligations at work, school or home (missing work, missing school, neglect of children) 2. Recurrent use in situations in which it is physically hazardous (driving, operating a machine) 3. Recurrent substance-related legal problems (DUI) 4. Continued use despite persistent or recurrent social or interpersonal problems caused or exacerbated by effects of the drug (fights with spouse) 1.5 Diagnostic and Statistical Manual (DSM)-IV-TR Criteria for Dependence A maladaptive pattern of substance use, leading to clinically significant impairment or distress, as manifested by three (or more) of the following occurring at any time in the same 12-month period 1. Tolerance 2. Withdrawal (marker of physiological [physical] dependence) 3. Use of larger amounts of substance than intended Drugs of Abuse 3 4. Persistent desire or unsuccessful efforts to cut down or control use 5. Great deal of time spent in obtaining drug, use, or recovery 6. Recreational activities given up or reduced 7. The substance use is continued despite knowledge of physical and psychological problems caused by drugs 1.6 Withdrawal (marker of physiological dependence) 1) Signs and symptoms emerge when use of the drug is stopped 2) Signs and symptoms are reversed when drug is administered again Tolerance a) decreased effect with repeated use of the drug b) a need to use more drug to have the same effect 1.7 Tolerance due to decreased effect with repeated use of the drug (constant amount of the drug) 1.8 Tolerance – increased dose of the drug needed to have the same effect (shift to right in dose-effect curve). 1.9 Drug abuse and drug dependence • Drug abuse is using a drug excessively, or for purposes for which it was not medically intended. • Drug dependence (addiction) is compulsively using a substance, despite its negative and sometimes dangerous effects • Drug dependence results in: a) stereotyped withdrawal syndrome due to discontinuation of a drug b) tolerance 1.10 Important distinctions between abuse and dependence • Abuse and dependence in DSM-IV are mutually exclusive • Abuse does not necessarily lead to dependence and this trajectory varies across drugs 1.11. Mesolimbic dopamine system- a major target of addictive drugs 1. General neurobiological theory of reinforcement (most clearly established for stimulants) a. Ability of a drug to control behavior (be abused) is related to the release of dopamine in the mesolimbic dopaminergic pathway (VTAnucleus accumbensprefrontal cortex), termed the “pleasure center” or “reward center” b. Inhibitory inputs onto dopamine neurons come from GABA-ergic neurons present within the VTA or as a feedback loop from the nucleus accumbens 2. Given evidence of these brain changes, addiction/dependence has come to be considered a disease Drugs of Abuse 4 a. Disease rooted in neuropathology produced by the repeated administration of the drug (pharmacological insult) b. Pathological changes in brain function are in circuits that regulate how a person interprets and behaviorally respond to motivationally relevant stimuli A.1 Psychostimulants: Cocaine and Amphetamines Cocaine 1. Derived from Erythoxylon coca, which is a cultivated plant from South America 2. Isolated in 1855 by Niemann in Germany 3. First human experiments conducted by Freud on himself were published in 1884 4. Regulated by Pure Food and Drug Act of 1906 and Harrison Narcotic Act of 1914 Amphetamines 1. Synthetic phenylethylamine synthesized in 1800s 2. Marked to treat asthma and narcolepsy and later for obesity 3. Used extensively by military during WWI and left over supplies led to epidemic use in several countries. A2. Cocaine 1) used for medical purposes - powerful stimulant of central nervous system (used by Freud to treat depression) - an appetite suppressant (used for obesity) - topical anesthetic (historically used for eye and nasal surgery, currently used for nasal and lacrimal duct surgery) 2) Used for popularization before; 1906 when the Pure Food and Drug Act was passed - coca leaves were included in several wines and ciragettes - Coca leaves were included in original 1886 recipe for Coca-Cola for nearly 20 years A3 and A4. Site of action of Cocaine and Amphetamines in Presynaptic Neurons Cocaine Amph DA DAT Amph DAT VMAT DA DA Amphetamines DA DA DAT Presynaptic terminal Cocaine Amph DA DAT Amph DAT VMAT DA DA Amphetamines DA DA DAT Presynaptic terminal Cocaine Drugs of Abuse 5 Cocaine –inhibits the action of dopamine transporters Amphetamine – inhibits the function of VMAT and impedes the filling of synaptic vesicles, cytoplasmic DA increases – leads to reversal of DAT direction and increase of extracellular DA concentration A5. Acute Effects of Psychostimulants Behavioral and physiological effects (sympathomimetic-mimicking the effects of the sympathetic nervous system) 1. Euphoria, arousal, well-being, increased energy and activity, decreased appetite, self-confidence, reduces feelings of fatigue and boredom, “rush” often described as orgasmic 2. Increases in heart rate and blood pressure, bronchodilation, pupillary dilation A6. Pharmacokinetics of Cocaine 1. Rapidly absorbed into the brain and short-acting. 2. Onset, magnitude of effect (differences in potency), and duration depend upon route of administration (smoked reaches peak in 2 min, injected 1-3 min, “snorting” – 10 min; oral – 30 min) 3. Half-life varies from 40-80 min, requiring repeated administration to maintain blood levels 4. Rapidly metabolized by cholinesterases into benzoylecgnonine (inactive compound) monitored in biological fluids (salive, blood, urine, milk), measured in urine test where; it remains detectable for up to 8 days. 5. Cocaethylene is formed in the body when cocaine is ingested with alcohol; it is pharmacologically active and enhances the effects of cocaine. Cocaethylane has a longer duration action than cocaine itself and is more cardiotoxic than cocaine. 6. Cocaine also blocks voltage-gated membrane sodium ion channels; this action accounts for local anesthetic effect and may contribute to cardiac arrhythmias. A7. Consequences of Long term use of psychostimulants Results either in 1) Sensitization – increased drug response (low-doses and intermittent exposure) 2) Tolerance – decreased drug response Impairment of neurocognitive functions - visuomotor performance, attention Increased risk of infections to viral hepatitis and HIV Physical dependence is controversial Increased risk of developing autoimmune or connective tissue diseases such as lupus, Goodpature’s syndrome, Stevens-Johnson syndrome Drugs of Abuse 6 A8. Overdose signs and symptoms 1. Hyperactivity, agitation, diaphoresis, dilated pupils, tremor, tachycardia, hypertension, hyperpyrexia, stereotypical behavior, chills, nausea/vomiting, weight loss, muscle weakness, tactile hallucination, chest pain, cardiac dysrhythmias, confusion, dyskinesia, seizures, paranoia, coma 2. These can be exacerbated with co-administration of alcohol (formation of cocaethylene) 3. Death can occur secondary to myocardial infarction, cerebrovascular accident, cardiac arrhythmias, seizures or respiratory depression A9. Withdrawal (peaks at 2-4 days) Signs 1. Anxiety, agitation, fatigue, depression, nightmares, headache, sweating, muscle cramps, hunger, craving A10. Detection of Use 1. Look for symptoms noted above 2. Urine tests (2 to 4 days) 3. Other clues: AIDS, hepatitis, track marks, abscesses, bacterial endocarditis, chronic respiratory symptoms A11. Treatment of Cocaine Withdrawal A. Acute withdrawal-symptomatic treatment - Bromocriptine (dopamine agonist) – ameliorates dopamine deficiency state of cocaine withdrawal - Benzodiazepines (lorazepam) - in patients with severe agitation and sleep disturbance B. Treatment of Long-term addiction - No FDA-approved pharmacological therapies - in 2011, the researchers invented vaccine against cocaine by combining a cocaine- like molecule with a part of virus (tests in mice) - Cognitive-Behavioral Therapies – two components 1) Functional analysis – to identify the patient's thoughts, feelings, and circumstances before and after the cocaine use to understand reason for using cocaine 3) Skills Training to help cocaine users to cope with intrapersonal and interpersonal problems B. Opioids (opium, morphine, codeine, heroin, oxycodone) B1. History 1. Opium is derived from extracts of the juice of the opium poppy, Papaver somniferum, and has been used since 3400 BC to relieve suffering, largely pain and asthma 2. Morphine and codeine are derived from opium. 3. Heroin is semi-synthetic opioid synthesized from morphine Drugs of Abuse 7 B2. Mechanisms of action 1. Opioids exert their pharmacodynamic effects through three principal opioid receptors- mu, delta and kappa 2. Opioids cuase disinhibition of mesolimbic dopaminergic system 3. The dependence producing properties of opioids are mediated through the mu receptors B3. Patterns of Abuse 1. Oral, intravenous, subcutaneous (skin popping), smoking, snorting (becoming more prevalent because of fear of AIDS) and intravenous B4. Patterns of Use 1. Heroin’s effects last about 3-5 hrs. 2. Average addict uses 2-4 times/day 2. Tolerance develops which results in a gradually increasing frequency/quantity of use. Physical dependence also develops B5. Sings of Opioid Overdoes 1. Unconsciousness 2. Miosis 3. Hypotension 4. Bradycardia 5. Respiratory depression 6. Pulmonary edema B6. Pharmacokinetics 1. Tolerance to one opioid is usually associated with tolerance to other opioids (cross-tolerance) 1. Heroin is a pro-drug that is rapidly converted into 6-monoacetylmorphine by esterases present in the blood, brain and very tissue 2. 6-monoacetylmorphine is further metabolized to morphine which contributes to the duration of effect of heroin 3. Withdrawal begins 12 hours after last dose, peaks at 1½ - 3 days, and is mostly over by 5-7 days 4. Lingering symptoms can persist for months (“Protracted Abstinence Syndrome”) and are associated with relapse. 5. Withdrawal is profoundly painful and unpleasant though not life-threatening B7. Withdrawal Symptoms 1. Anxiety and dysphoria 2. Craving and drug-seeking 3. Sleep disturbances 4. Nausea, vomiting and diarrhea 5. Lacrimation Drugs of Abuse 8 6. Rhinorrhea 7. Yawning 8. Sweating, chills, gooseflesh (“cold turkey”) 9. Mydriasis (excessive dilation of the pupil) 10. Cramps 11. Hyperpyrexia 12. Involuntary movements (“kicking the habit”) B8. Treatment of Opioid Addiction • Pharmacotherapy • Many individuals are not treated with medications (drug free) – Self-help groups such as Narcotics Anonymous – Inpatient detoxification facilities/residential – Individual therapy rare largely due to sociodemographic characteristics of users – Dependence on prescription opioids presents a new challenge for treatment B9. Goals of Pharmacotherapy • “Cure” of withdrawal or overdose • To improve the holding power of outpatient treatment • To reduce drug craving • To create a “window of opportunity” during which patients can receive psycho-social intervention to decrease the risk of relapse • To serve as short or long-term maintenance agents for patients who can’t function without them, but can lead productive lives with them B10. Treatment approaches of opioid overdose and withdrawal syndrome • Treatment of opioid overdose (bradykardia, respiratory depression, pulmonary edema) - Naloxone (m-opioid receptor antagonist) • Treatment of withdrawal syndrome and maintenance - Methadone - m-opioid receptor agonist - Buprenorphine partial - m-opioid receptor agonist B11. Treatment of Opioid Overdose • Naloxone [Narcan, Nalone] m-opioid antagonists with very high affinity – Fast acting (2 min) but the duration of action (about 45 min) is much shorter than heroin • Therefore, individuals treated for overdose with these antagonists must be kept under observation for the duration of the opioid drug’s effects to determine if additional antagonist treatment is needed B12. Treatment with Methadone • Methadone – agonist to m-opioid receptor with long half-life (15-60h) Drugs of Abuse 9 – Oral administration – Lasts at least 24 hrs • Methadone: Prevents withdrawal symptoms and cravings, has a cross-tolerance with other opioids • Can only be dispensed in federally licensed clinics – Requires almost daily clinic visits even for individuals with long term success B14. Treatment with Buprenorphine • Buprenorphine is a partial m-opioid receptor agonist – has less potential for respiratory depression (hard to overdose) • It has both high affinity for receptors (competes easily) and dissociates slowly (long acting so withdrawal is minimized) • Marketed in a formulation with naloxone (Suboxone) taken sublingually. When misused iv will result in withdrawal symptoms due to the presence of Naloxon (m-opioid receptor antagonist). . • If buprenorphine is initiated prior to the onset of acute withdrawal signs, it may lead to abrupt withdrawal syndrome resulting from displacement of full agonists (e.g. heroin) for the mu-opiod receptor by a buprenorphine C. Cannabinoids: Marijuana C1. History 1. Like other drugs of abuse derived from natural plant products, marijuana has been used for 1000s of years 2. Delta-9-tetrahydrocannabinol (THC) is active constituent 3. There are 400 additional constituents that are found in Cannabis C2. Mechanism of action 1. THC activates a cannabinoid receptor in the VTA to initiate its action, a relatively new scientific finding 2. THD inhibits GABA-ergic interneurons 3. Cannobinoids cause disinhibition of mesolimbic DA system. C3. Acute Effects of Marijuana • Sedation, relaxation • Mood alteration, sense of well-being • Altered perception and time estimation • Impaired judgment, memory, and concentration • Increased heart rate, dry mouth • Increased appetite (“munchies”) – due to drop of sugar levels and also stimulation of cannabinoid receptors in the hypothalamus. • Injection of the conjunctiva (“red eyes”) – due to decreased pressure in the eyes and increased blood pressure Drugs of Abuse 10 C4. Overdose effects of Marijuana Use 1. Panic, delirium, psychosis (often paranoid) 2. Long-term use: Amotivational syndrome, inattention, poor judgment, distractability, impaired social relationships but results from field and laboratory studies are often inconsistent 3. Tolerance occurs but whether or not physical dependence develops in humans remains an open question (but has been demonstrated in mice) 4. Personality changes and cognitive deficits (loss of short-term memory) 5. Many believe it is a “gateway” drug, i.e., its use leads to initiation of use of other drugs of abuse C5. Withdrawal Symptoms - is not life threatening - begin on 1 or 2 day of abstinence, peak day 2-6, resolve within 7-14 days - fatigue (31%) - hypersomia (26%) - psychomotor retardation (25%) - anxiety (19%) - depression (16%) - anorexia - irritability - C6. Differential Diagnosis of Cannabis Intoxication - Intoxication from other addictive substances - Panic disorder - Major depressive disorders - Bipolar I or II disorder - Schizophrenia C7.. Treatment of Marijuana Abuse is Symptomatic • Anxiolytics (benzodiazepines) for anxiety and panic • Antipsychotics for delirium and paranoia • Cognitive behavioral therapy for dependence • No pharmacotherapies but discovery of CB1 receptor and development of an antagonist is promising D. Alcohol • Health care cost of alcohol problems - $186 Billion • Alcohol dependence is a complex disorder in which many factors act together to produce the illness • Approximately 50% of the risk is attributed to genetics • May arise in individuals without family history of alcohol dependence as a result of environmental factors D1. Subtypes of Alcohol Dependence Type A Alcohol Dependence Drugs of Abuse 11 • Late onset (>25 years old) • Few familial alcohol-dependency • Slower disease progression • Milder form of alcohol dependence • Important environmental influence • Minimal criminality Type B Alcohol Dependence • Early onset (25 years old) • Paternal type B alcohol dependence • More severe form of alcohol dependence • Little environmental influence • Frequent criminality • Frequent presence of personality disorder D2. Disease Associated with Chronic Alcohol Use Primary Diseases • Alcohol poisoning • Alcoholic heart disease (cardiomyopathy) • Alcoholic gastritis • Alcoholic liver cirrhosis • Alcoholic nerve disease (polyneuropathy) • Alcoholic psychoses Secondary Diseases • Cancer (lip, mouth, pharynx, esophagus, larynx, liver, stomach) • Diabetes • Gastrointestinal disease • Heart disease (hypertension, stroke) • Liver disease • Pancreatitis (acute, chronic) D3. Effects of Acute Alcohol on Neural Circuits I Dopamine and Opioid Systems • Indirectly increases dopamine levels in the mesocorticolimbic system – Associated with positively reinforcing effects of alcohol • Indirect interaction with opioid receptors results in activation of opioid system – Associated with reinforcing effects via µ-receptors D4. Effects of Acute Alcohol on Neural Circuits I GABA and Glutamine Systems • Increases the effects of GABA, the major inhibitory neurotransmitter in the brain • Inhibits the effects of glutamate, the major excitatory neurotransmitter in the brain Chronic exposure of alcohol leads to a compensatory 1) Reduction in the levels of GABA-gic receptors 2) Up-regulation of NMDA receptor Drugs of Abuse 12 Sudden reduction in chronic alcohol intake results in overactivation of NMDA system D5. Treatment Stages of Alcohol Dependence Stage 1: Identification - obtain history of current and past alcohol use and family history of alcohol problem - use standardized screening tests (e.g., 4-question CAGE) - evaluate patients in terms of the DSM-IV criteria for alcohol abuse and dependence and determine whether patient wants to abstain Stage 2: Detoxification/Withdrawal - Mild or severe withdrawal symptoms mild withdrawal symptoms - agitation, anxiety, insomnia, nausea severe withdrawal symptoms – autonomic hyperactivity, seizures, delirium tremens - relieve immediate symptoms of withdrawal – benzodiazepines – indirect agonist for GABA receptors (reverses effect of alcohol) - Benzodiazepines with long half-life have less chance of recurrent withdrawal. Diazepam has a long half-life Lorazepam has as shorter half-life but is not metabolized by liver – good for patients with cirrhosis Stage 3: Rehabilitation - restructure life without alcohol - relapse prevention – psychotherapy, pharmacotherapy Stage 4: Aftercare - AA meetings, family support D6. Symptoms of Alcohol Withdrawal 1) minor withdrawal (6-36 hrs) - due to CNS hyperactivity – Hyperarousal - mild anxiety, headache, sweating, GI upset, insomnia, nausea 2) Seizures – 6-48 hrs after the last drink 3% of chronic alcoholics can develop tonic-clonic seizures, some develop status epilepticus (seizures without regaining consciousness for 30 min) Can be life threatening 3) alcoholic hallucinations (12-48 hrs after the last drink) visual, auditory, tactile – can last for months 4) delirum tremens (48-96 hrs after the last drink, can last 1-5 days) D7. Delirium Tremens (DT) • Caused by withdrawal from long-term alcohol consumption • 5% of alcohol withdrawal leads to DT (mortality can be up to 35% when untreated) • Caused by withdrawal from benzodiazepines Drugs of Abuse 13 Symptoms - Confusion, disorientation, paranoia, hallucinations – mostly visual and tactile - Uncontrollable tremors of the extremities - Severe autonomic instability (fever, tachycardia, hypertension) - Some patients experience seizures Treatment - Pharmacotherapy is symptomatic and supportive - Benzodiazepines, such as diazepam (Valium), lorazepam (Ativan) - In extreme cases stronger benzodiazepines like temazepam (Restoril) might be used D8. FDA-Approved Pharmacotherapies for Alcohol Dependence Disulfiram – Alcohol Aversion Therapy Naltrexone – Opioid antagonist Acamprosate – restores balance between neuronal excitation and inhibition D9. Disulfiram (Antabuse) – alcohol aversion therapy - Inhibits aldehyde dehydrogenase - When taken with alcohol, increased levels of [acetaldehyde] that leads to nausea, dizziness, headache, hypotension, vomiting - Decreases desire to drink alcohol but does not increase abstinence - Increased risk of hepatotoxicity NOTE: Asian descents who have an ALDH22 genetic variant of the ALDH enzyme that metabolizes acetaldehyde slowly – protects against alcohol dependence D 10. Naltrexone (ReVia) - Opioid antagonist - Blocks release of dopmine from the Nucleus Accumbens - Reduces alcohol cravings - Avoid Naltrexone with Disulfiram – both are potential hepatotoxins - Avoid Naltrexone in patients dependent on opioids- it will precipitate acute withdrawal syndrome D. 11 Acamprosate – (Campral) - Decreases excitatory glutamate neurotransmission and increases GABA-ergic activity - Minor side effects: diarrhea, allergic reactions, irregular heartbeats - Dose adjustment in patients with moderate renal disease (ceratinine clearance 30-50 mL/min) - Contradicted in severe renal disease (ceratinine clearance < 30 mL/min E. Anti-depressants: Benzodiazepines Therapeutic Uses of Benzodiazepines - Benzodiazepines are the most commonly prescribed sedative drugs - Severe anxiety, panic attacks and phobias (because of their anxiolytic properties) Drugs of Abuse 14 - Insomnia - In muscular disorders - effective muscle relaxants - Alcohol withdrawal - Epilepsy – anticonvulsant Benzodiazepines are; indirect agonists of the GABA receptor E1. Benzodiazepine Withdrawal Syndrome • Anxiety, agitation • Increased sensitivity to light and sound • Muscle cramps • Sleep disturbance • Dizziness • Mycolonic jerks Withdrawal Management Treatment with diazepam (Valium) - benzodiazepine with a long-half life - gradually tapering off the drug over a period of months F. Nicotine • Is among the most addictive drugs, accounts for 440,000 deaths yearly • Selective agonist of the nicotinic acetylcholine receptor (nAChR) that is normally activated by acetylcholine • Nicotine acts on nAChR receptors, stimulate dopaminergic neurons in the VTA and increases the release of dopamine in the nucleus accumbens Treatment • Nicotine patches, nasal spray, nicotine lozenge • Varenicline [Chantrix] – non-nicotine medication – partial agonist that binds subunits of nicotine acetylcholine receptors. Because it stimulates the receptors, it relives cravings and withdrawal symptoms during abstinence from smoking - binds to the nAChR receptors with greater affinity than nicotine – thus reduces the pharmacologic reward from cigarette smoking G. Hallucinogens (LSD, mescaline) - Cause change of sensation, illusions, called – mind-bening drugs - The drugs do not induce dependence or addition - Serotonin receptors 5-HT2A in cortex are molecular targets - Treatment: for nonpychotic agitation – anti-anxiety drugs (Diazepam) for severe agitation – use antipsychotic drugs Drugs of Abuse 15 LIST OF DRUGS DISCUSSED IN THIS LECTURE. For more detail see on-line reference www.rxlist.com Generic Name Trade Name Half-life Mechanism of Action Rx Bromocriptine Cycloset 12-14 hrs Dopamine agonist Symptomatic treatment in cocaine withdrawal Lorazepam (is not metabolized by liver) Ativan 10-20 hrs Indirect Agonist of GABA receptor In severe agitation and sleep disturbance during cocaine, alcohol and benzodiazepines withdrawal Diazepam Valium 20–100 hrs (36-200 hrs for main active metabolite desmethyldiazepam) Indirect Agonist of GABA receptor In severe agitation and sleep disturbance during cocaine, alcohol and benzodiazepines withdrawal Naloxone Narcan, Nalone 45 min – 1 hr µ-opioid antagonist Opioid Overdose Methadone Symoron 15-60 hrs µ-opioid agonist Opioid withdrawal syndrome Bupernorphine Subutex 20-60 hrs Partial µ-opioid receptor agonist, has very high affinity Opioid withdrawal syndrome Buprenorphine: Naloxone (4:1) Suboxone Bupernorphine – 20-60 hrs Naloxone – 1 hr Partial µ-opioid receptor agonist in a formulation with Naloxone (m-opioid antagonist) Opioid withdrawal syndrome Disulfiram Antabuse 60-120 hrs Inhibits Aldehyde Dehydrogenase Therapy for Alcohol Dependence Naltrexone ReVia Naltrexone – 4 hrs, is metabolized to 6-beta-naltrexol –13 hrs Opioid Receptor Antagonist Therapy for Alcohol Dependence Acamprosate Campral 20-30 hrs Mechanism unknown Therapy for Alcohol Dependence (renal elimination) Varenicline Chantrix 24 hrs Partial agonist of nicotine acetylcholine receptors Nicotine addiction Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 1 PHARMACOLOGY OF SEDATIVE –HYPNOTIC DRUGS & DRUGS Date: February 13, 2013 – 9:30 am Reading Assignment: Katzung, Basic & Clinical Pharmacology; 12th Edition, Chapter 22, pp. 373-385. KEY CONCEPTS AND LEARNING OBJECTIVES 1. You should understand the structural aspects of the GABAA receptor and the receptor components (i.e. binding sites) mediating the effects of various drugs that can modulate GABAA receptor activity. 2. You should understand the difference between GABAA receptors that contain a BZ1 versus a BZ2 binding sites and the know the effects various sedative–hypnotics will have on these different isoforms of GABAA receptors. 3. You should understand the differences in the pharmacokinetics of the benzodiazepines with respect to the type of metabolism they undergo, the production of active metabolites and the relative overall half-lives (i.e. short, intermediate or long) of the different drugs. 4. You should know the similarities and differences among the benzodiazepines with respect to their relative: (1) time of onset, (2) potency, (3) metabolism and (4) elimination half-lives. 5. You should know the general pharmacodynamic and pharmacokinetc characteristics of the barbiturates and the pharmacological characteristics of this class of drugs that contributed to their limited clinical use relative to the benzodiazepines or other sedative-hypnotics. 6. You should know the similarities, differences and distinguishing characteristics between the different classes of sedative-hypnotics and among the drugs within a given class. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 1 PHARMACOLOGY OF SEDATIVE–HYPNOTIC DRUGS What do we mean by the term Sedative-Hypnotic? Sedative (Anxiolytic) – reduces anxiety and exerts a calming effect. Hypnotic – produces drowsiness and facilitate the onset and maintenance of sleep (involves a greater depression of CNS than sedation). 1. All sedative-hypnotics produce graded dose-dependent depression of CNS function. 2. The magnitude of the depression of CNS function with increasing dose is not the same for all classes of sedative-hypnotic drugs. This is shown below. From: Trevor and Katzung’s Pharmacology, Examination and Board review, 6th ed.page 206 3. All sedative-hypnotic drugs produce their effects by interacting with GABAA receptors and potentiating GABAergic activity at all levels of the neuraxis, from spinal cord to cerebral cortex. Gamma-Amino Butyric Acid (GABA) is a major inhibitory neurotransmitter in the CNS (approximately 30% of synapses in mammalian cerebral cortex are GABAergic) . – Two (2) major classes of GABA receptors have been identified based on function. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 2 1. GABAA Receptors – an ionotropic receptor whose activation increases the opening of chloride channels leading to an inhibitory post-synaptic potential (hyperpolarization). GABAA receptor activation can be modulated by benzodiazepines and other sedative-hypnotic drugs. Musimol is a prototypic agonist and bicuculline the antagonist, can bind to the same site (i.e. the same pharmacophore) that GABA binds to on the receptor to pharmacologically mimic or antagonize, respectively, the effects of GABA. 2. GABAB Receptors - a metabotropic G-protein linked receptor located on: (1) presynaptic terminals controlling the release of GABA (“homoreceptors”) or other neurotransmitters (“heteroreceptors”) via decreases in Ca++ conductance, and (2) postynaptic membranes producing hyperpolarization of the membrane via increases in K+ conductance. (-)-baclofen (Lioresal®) is a selective GABAB agonist that is used clinically as an anti-spastic drug. GABAB receptors are not modulated by benzodiazepines or other sedative-hypnotic drugs. . The COMPLEXITY OF THE GABAA RECEPTOR COMPLEX GABAA receptors are heteropentameric glycoprotein receptors that are formed from the co-assembly of five(5) subunits (420- 450 amino acids) from various polypeptide classes designated as: , , , and , , , & . Each subunit has: a) 4 transmembrane domains with both the amino and carboxy terminus in the extracellular side and b) 2 intracellular loops that provide sites for phosphorylation. Isoforms of these subunits have been identified. There are 6 isoforms of the  subunit (i.e. 1-6), 3 isoforms of  (i.e. 1-3), 3 isoforms of  (i.e 1-3) as well as 3 isoforms of . The most abundant GABAA receptor subtype consists of a generic composition of: Two(2)  subunits, two(2)  subunits, and one(1)  subunit. Given the number of identified subunits/isoforms, thousands of pentameric subunit combinations are theoretically possible. Yet, the majority of GABAA receptors identified are heteropentamer combinations derived from a relatively few number of different subunit isoforms. The majority of GABAA receptors (60%) consist of: 122 while 15-20% consist of 232 with the remaining receptors being represented by various combinations of the isoform subtypes. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 3 The figure below illustrates the organization of the subunits that form the GABAA receptor and the different regions of the receptor that different drugs can target (i.e. bind to and alter function) From: Trevor and Katzung’s Pharmacology, Examination and Board review, 6th ed.,page 205 The binding of GABA (or GABA agonists & antagonists) is to a region located at the  and  interface, providing two sites that can bind GABA. The binding of benzodiazepine (and imidazopyridines) occurs at a “binding site” that resides at the / interface, providing one binding site for BZ1 or BZ2 drugs. Although this “binding sitet” is called a “benzodiazepine receptor”, it is actually an allosteric modulatory site, since binding to these sites can not alter GABAA receptor function in the absence of GABA. The benzodiazepine and imidazopyridine) binding pocket (“receptor”) is largely determined by the type of  subunit that is co-assembled with the 2 subunit (the most abundant of the gamma subunits in the brain). GABAA receptors with either 1,2,3 & 5 exhibit sensitivity to diazepam while those with 4 & 6 are diazepam-insensitive. The imidazopyridine class (e.g. zolpidem) exhibit a greater degree of selectivity for GABAA receptors that have the 1 isoforms with the following degree of selectivity for zolpidem (1 > 2 = 3 >> 5). Based on  isoforms and drug selectivity, 2 types of “benzodiazepine” binding sites have been identified that can modulate GABAA receptor function: BZ1 (omega1) - contains the 1 isoform (Imidazopyridines bind selectively to the BZ1 receptor) BZ2 (omega2) – contains 2,3 or 5 isoforms receptor subtypes Benzodiazepines can bind appreciably to both BZ1 and BZ2 sites (receptors with 1,2,3 or 5). Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 4 The type of alpha subunit may also contribute to the qualitative effects produced by these drugs. Preclinical studies of point mutations (histadine  arginine) in the N-terminal domain of various isoforms of the  subunit suggest that: 1. 1 subunits may mediate the sedative, amnesia and ataxic effects of the benzodiazepines and non-benzodiazepine (BZ1 selective) drugs but not the anxiolytic or muscle relaxing effects. 2. 2 (and possibly 3) subunits may contribute to the anxiolytic and muscle relaxant effects of BZs. 3. 5 may be important in memory impairment and the development of tolerance to the sedative effects of the benzodiazepines. PHARMACODYNAMICS of Drugs that Modulate GABAA Receptor Function 1. Non-Selective “Agonists” – Benzodiazepines that bind to both BZ1 and BZ2 sites and are positive allosteric modulators of GABAA receptors. They DO NOT compete directly with GABA for its binding site (at the / interface) on the GABAA receptor and are ineffective in the absence of GABA (or GABA agonists). 2. Non-Benzodiazepine “Agonists” - The FDA approved imidazopyridines (Zolpidem & Zaleplon) and pyrrolopyrazine drug eszopiclone (Lunestra) that bind to BZ1 sites act as positive allosteric moduators of GABAA receptor function , as they also require GABA to produce their therapeutic effects. 3. Antagonists (flumazenil; Romazicon®) - a competitive antagonist that has a high affinity for the benzodiazepine (BZ1 and BZ2) receptor and blocks the actions of benzodiazepines and imidazopyradine (e.g. zolpidem) & pyrrolopyrazine drugs (e.g.Eszopiclone) but does not anatgonize the actions of other sedative-hyponotics such as the barbiturates, meprobamate or ethanol. When given i.v., flumazenil acts rapidly but has a short t1/2 (0.7-1.3h) due to hepatic clearance. It may precipitate a severe abstinence syndrome in patients physiologically dependent on BDZs. 4. Inverse Agonists- act as negative allosteric modulators of GABAA receptor function (decrease GABAA binding). Inverse agonists (e.g. Beta Carbolines) bind to BZ1 and BZ2 type modulatory sites and can produce anxiety and seizures (via reducing GABA receptor function) as well as block the effects of drugs that bind to BZ1 and BZ2 binding sites. 5. Barbiturates – bind to sites on the GABAA receptor that are distinct from the benzodiazepine binding site and are likely on the  subunits of the receptor. In the presence of GABA, barbiturate binding increases the duration of opening of the chloride channel. At very high doses, barbiturates can directly produce channel opening. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 5 6. Neuroactive steroids - bind to the GABAA receptor at other sites distinct from the BZ binding site to increase the effects of GABA. Some neurosteroids (e.g. alphaxalone) may directly open chloride channels at high concentrations. 7. Ethanol - thought to alter GABAA neurotransmission, since it produces many of the same effects as the BDZs (e.g., anxiolysis) and stimulates Cl- uptake into isolated brain vesicles. THE BENZODIAZEPINES - since being introduced in the 1960s, the BZs remain among the most widely prescribed drugs in the world. This is primarily due to the fact that these drugs have less capacity to produce extreme and potentially fatal CNS depression in comparison with the barbiturates and older sedative-hypnotics. In healthy patients, hypnotic doses of benzodiazepines produce no significant effects on respiration and cardiovascular function. (REMEMBER: “I AM PAM, a LAM, a benzodiazepine”) PHARMACODYNAMICS - Benzodiazepines act as non-selective positive allosteric modulators (“agonists”) of GABA receptor function. Benzodiazepines can bind to BZ1 or BZ2 sites on the GABAA receptor and their effects at BZ1 or BZ2 sites can be antagonized by the BZ antagonist, flumazenil . Mechanism of Action: - the therapeutic efficacy of the benzodiazepines is due to their ability to potentiate GABAA receptor-mediated inhibition. Benzodiazepine binding to BZ1 or BZ2 sites on the GABAA receptor increases the affinity of GABA for its binding site on the receptor and increases the frequency of opening of chloride channels. This results in hyperpolarization and a greater degree of inhibition of neuronal firing. Thus, because they require GABA to affect receptor function, they act as positive allosteric modulators of GABA receptor function, rather than true receptor agonists (e.g. musimol) that can mimic the effects of GABA. In addition to their use as sedative hypnotics, some benzodiazepines have anticonvulsant effects (diazepam, clonazepam & nitrazepam), produce relaxation of skeletal muscles (diazepam) and can be given IV as adjuncts for anesthesia (midazolam & diazepam). PHARMACOKINETICS Absorption & Distribution: Although all benzodiazepines are lipid soluble, the lipophilicity, which can vary over 50–fold, contributes to differences in their rates of absorbtion, onset of action & redistribution. Pregnancy: All sedative-hypnotics cross the placental barrier during pregnancy, and appear in breast milk. Thus, they can reduce neonatal vital functions and may be teratogenic if continued throughout pregnancy Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 6 Bioavailability - is very good; 60-100% depending on the individual drug. Protein binding – moderate, 70% (alprazolam) to high, 99% (diazepam); thus drug interactions with other highly protein bound agents are likely (e.g., between phenytoin and diazepam). Metabolism: Most benzodiazepines undergo metabolism via both microsomal oxidation (Phase I), by the P450 family CYP3A4 and CYP2C19, and conjugation (Phase II) to form glucuronides that are excreted in the urine. Three benzodiazepines are only conjugated and excreted (i.e. undergo only Phase II metabolism). Pharmacokinetic differences largely determine the clinical applications of the Benzodiazepines. Important factors for consideration include: elimination half-life of parent drug and any bioactive intermediates, rapidity of onset of effects and drug potency. Drugs Whose Metabolism includes both Phase I (oxidation) and Phase II (conjugation) The overall elimination half-life of these drugs is often longer than the half life of the parent compound due to formation of one or more bioactive metabolites (Phase I) each of which may be active for an extensive period of time before they are conjugated and excreted (Phase II). Diazepam; Flurazepam (2-Keto derivatives) and Clordiazepoxide: Although chlordiazepoxide is not a 2-keto-benzodiazepine it is converted to desmethyldiazepam. Parent compounds vary in their elimination half-lives (t ½ = 0.1-35 h), but their active metabolites (N-desmethyldiazepam and N-desalkylflurazepam) have half-lives between 50-100 hr. Clonazepam, flunitrazepam (7-Nitro derivatives) 7-nitro substitution appears to enhance anticonvulsant activity. Clonazepam (t ½ = 22-33 h) is reduced to a 7-amino derivative that is inactive centrally but is extensively metabolized (via acetylation, hydroxylation and conjugation) then eliminated. Flunitrazepam (Rohypnol®) is not approved in the U.S.; it is highly abused and is the“date rape” drug. Midazolam , alprazolam and triazolam (Triazolo derivatives): Mean elimination half-lives of these drugs vary (2.5h for midazolam and 11.0 h for alprazolam). They are oxidized to form compounds that are rapidly eliminated. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 7 Drugs Whose Metabolism is Exclusively Phase II Conjugation (L O T ) Lorazepam, Oxazepam and Temazepam (3-hydroxy derivative): These agents are conjugated (Phase II biotransformation only) and thus are not metabolized to bioactive intermediates. Elimination half-lives range between 5-25 h. These are preferred for aged patients and those with impaired hepatic function. Comparative Pharmacokinetics of Various Benzodiazepines Adapted from: B.G. Katzung, Basic and Clinical Pharmacology 9th ed.,page 354 “Side” & Adverse Effects: Frequent: Drowsiness, ataxia, amnesia (don’t remember events during the drugs duration). Occasional: Confusion, paradoxical excitement (in children and the elderly), dizziness. Rare: Paradoxical rage reaction; extrapyramidal symptoms with chlordiazepoxide, allergic reaction, etc. Middle age and elderly patients (50+ years) are especially susceptible to these reactions when given daily high doses of potent benzodiazepines such as triazolam. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 8 Tolerance and Dependence: Pharmacodynamic tolerance seen with chronic use – there is a decrease in the drugs effects requiring higher doses to achieve effects comparable to that before drug exposure (may be due to  in BZ bindig sites. Cross-tolerance can develop to ethanol and other sedative-hypnotics that affect GABAA receptor function. Psychological dependence highly possible; BZ are often abused in conjunction with alcohol. Withdrawal symptoms (anxiety, insomnia) appear upon discontinuation after frequent use. BZ1- Selective Non-Benzodiazepines (Primarily Used to Treat Sleep Disorders) The non-benzodiazepine imidazopyridine drugs zolpidem (Ambien®) and zaleplon (Sonata®) and the newer pyrrolopyrazine drug, eszopiclone (Lunestra®) bind selectively to the BZ1 binding site on the GABAA receptor and act as positive allosteric modulators of GABA receptor function in a fashion similar to the benzodiazepines (that bind to BZ1 & BZ2 sites). In addition, these drugs: - do not produce a dangerous degree of CNS depression (even in overdose) unless taken in combination with other CNS depressants, a combination that can be lethal. - produce effects that can be antagonized by the BZ1 & BZ2 antagonist, flumazenil - may be habit forming with long-term use - lack anxiolytic, anticonvulsant and muscle relaxant efficacy of some of the benzodiazepines - have side effects similar to the benzodiazepines which include: headache and dizziness and somnolence; nausea, vomiting, diarrhea, anterograde amnesia and rebound insomnia (especially at high doses) - can cause “sleep-driving” or “sleep-eating” without any memory of the event - have recently (1/10/13) received an FDA recommendation to lower the dose in females since the data indicate a slower metabolism than in males. 1. Zolpidem (Ambien®) – the first FDA-approved BZ1 selective drug - also available in an extended release form (Intermezzo) - rapidly and completely absorbed from the GI tract, reaching peak plasma levels in 1-2 hours. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 9 - hepatic metabolism via Phase I and Phase II biotransformation; t1/2 of 1.5-3.0 hours. - Half life is prolonged in the elderly and in patients with liver disease. - lacks significant muscle relaxant, anxiolytic and anticonvulsant properties. 2. Zaleplon (Sonata®): resembles zolpidem in its effects. Also rapidly absorbed (peak concentration reached in 1.0 h) from the GI tract, and has a very short t1/2 (1.0 h.) with no active metabolites. Biotransformed hepatically (Type I), thus dosage should be reduced in the elderly and patients with liver disease. Note that the metabolism of zaleplon is inhibited by the OTC H-2 histamine receptor blocker, cimetidine (generic; Tagamet®). Next day effects are less common than after zolpidem or other sedative-hypnotics due to the very short half-life. 3. Eszopiclone (Lunestra) – the S(+) isomer of zopiclone (a pyrrolopyrazine drug already approved for use in Canada) with no structural similarity to zolpidem, zaleplon or the benzodiazepines. - FDA approved (12/04) for treatment of sleep disorders. It does not significantly alter the stages of sleep. - unlike zolpidem and zaleplon, eszopliclone is not restricted in its labeling to short-term use. . Pharmacokinetics: rapid absorbtion from the GI tract; peak plasma concentrations at 1-2 hrs t1/2  6 hrs (slightly longer than the other BZ1 selective drugs). Typical dose is 2-3 mg. The elimination is slower in the elderly and doses should be started at 1 mg. Drug Interactions: Various CYP3A4 inhibitors such as itraconazole (Sporanox) clarithyromycin (Biaxin) and ritonovir (Norvir) can increase serum concentrations and prolong the duration of action. Conversely, Rifampin, a CYP3A4 inducer can decrease serum concentrations and the effectiveness of a given dose of eszopiclone. Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 10 The BARBITURATES: (One of “Mothers Little Helpers” referred to in the 60’s Rolling Stones song). Derivatives of barbituric acid that once enjoyed great popularity to induce and maintain sedation and sleep. However, with the exception of phenobarbital, methohexital and thiopental, their clinical use as sedative-hypnotic agents has waned significantly since the introduction of the benzodiazepines and other drugs. (Remember: “AL”the barbiturate) Mechanisms of Action - Bind to sites on the ionotropic GABAA receptors at sites distinct from the site that binds benzodiazepines and are likely on the  subunits. - They do not exhibit GABAA receptor subtype specificity - At low doses, the binding facilitates the actions of GABA by increasing the duration of opening of the chloride channels. - At high doses, barbituarates can be “GABA-mimetic” and directly activate the opening of the chloride channels in the absence of GABA (i.e no ceiling effect on CNS depression). - Barbiturates also depress the actions of excitatory neurotransmitters and exert non-synaptic membrane effects. Pharmacokinetics The barbiturates are classified primarily based their duration of action, as observed for the benzodiazepines. Short Acting: (hours) - Thiopental, Mexihexital; rapid onset – used for induction of anesthesia, Intermediate Acting (18-48hrs): - Amobarbital (Amytal), secobarbital (Seconal) and pentobarbital (Nembutal) Long acting (4-5 days): - Phenobarbital (Luminol Sodium) Limited Clinical Uses: Treatment of Epilepsy - Phenobarbital Induction of Anesthesia - Thiopental and methohexital . Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 11 Adverse Effects of the Barbiturates 1. They have a low therapeutic index, related to their potency to depress respiration (especially in combination with alcohol). A lethal dose can be < 10x the hypnotic dose. 2. They produce physical dependence; discontinuation of barbiturates after repeated use leads to a withdrawal syndrome which can be life threatening and difficult to treat. 3. Stimulate cytochrome P450 activity and induce hepatic microsomal oxidases. a. Pharmacokinetic tolerance – higher doses of barbiturates may be required over time due to the increase in their own metabolism. b. Cross tolerance - to benzodiazepines or other sedative hypnotics c. Drug interactions – due to increased metabolism of other drugs metabolized by microsomal oxidases. 4. In contrast to the benzodiazepines, the barbiturates can induce anaesthesia. The Older Sedative–Hypnotics 1. Carbamates (Meprobamate) – introduced in 1955 as an antianxiety drug, and remains its only approved use in U.S, although clinical proof for its efficacy as an anxiolytic is lacking.Properties resembling the benzodiazepines. 2. Alcohols – (chloral hydrate) - transformed to trichloroethanol is the pharmacologically active metabolite; 6-10 hour half-life. A Novel Sedative-Hypnotic Ramelteon (Rozerem) – a melotonin receptor agonist that received FDA approval in July 2005 for the treatment of insomnia characterized by difficulty in falling asleep. - no evidence of physical dependence or abuse potential - appears to be well tolerated when administered for long treatment courses Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 12 Mechanism of Action: binds selectively to MT1 and MT1 melatonin receptors to mimic and enhance the actions of the endogenous melatonin that has been associated with the maintenance of circadian sleep rhythms. No measurable affinity for BZ receptors or other sites. Pharmacokinetics - rapid absorbtion – high fat meals delays Tmax and increases AUC ( 30%) - extensive first pass metabolism - short half-life of 1-3 hrs - moderate protein binding (82%) - large Vd (  74 L) - metabolism via CYP 1A2, 2C9, 3A4 - metabolism may be decreased by: Fluvoxamine (Luvox), a broad inhibitor of CYP isozymes, the strong 3A4 inhibitor Ketoconazole (Nizoral) & the strong 2C9 inhibitor Fuconazole (Difulcan) - metabolism may be increased by the strong CYP inducer, Rifampin Side Effects – occurred at rates generally comparable to placebo and included headache, somnolence, fatigue, dizziness, nausea, exacerbated insomnia Caution is advised in using Rameteon in patients with: - liver disease - sleep apnea - depression or suicidal thoughts - individuals over 65; may need to adjust dose Pharmacology & Therapeutics Pharmacology of Sedative-Hypnotic Drugs February 13, 2013 G. Battaglia, Ph.D. 13 RELATIVE LONGEVITY OF ACTION BASED ON THE OVERALL HALF-LIVES OF DRUGS DICUSSED IN THE LECTURES ON SEDATIVE/HYPNOTICS SHORT-ACTING Diphendydramine (Benadryl®) Eszopiclone (Lunestra®) Zolpidem (Ambien®) Zaleplon (Sonata®) Flumazenil (generic, Romazicon) Midazolam (Versed®) Triazolam (Halcion®) Ramelteon (Rozerem®) Thiopental Sodium (Pentathal®) INTERMEDIATE-ACTING Alprazolam (Xanax®) Estazolam (ProSom®) Lorazepam (generic; Ativan®) Oxazepam (Serax®) Temazepam (Restoril®) Hydroxyzine(Atarax®) Meprobamate (Equinil®,Miltown® ) Trazadone(Deseryl) Nefazodone (Serzone®) LONG-ACTING Amitriptyline (generic; Elavil®) Clonazepam (generic, Klonopin®) Chlordiazepoxide (generic, Librium®) Chlorazepate (generic, Tranxene®) Cyclobenzaprine (Flexeril®) Diazepam (Valium®) Flunitrazepam (Rohypnol®) Flurazepam (Dalmane®) Imipramine (Trofanil®) Doxepin (Sinequan®) Phenobarbital (generic, Luminol Sodium®) Pentobarbital (generic, Nembutol Sodium®) Mirtazapine (Remeron®) Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 1 DRUGS TO TREAT SLEEP DISORDERS Date: February 13, 2013 – 10:30 am. Reading Assignment: Katzung, Basic & Clinical Pharmacology; 12th Edition, Chapter 22, pp. 373-385. KEY CONCEPTS AND LEARNING OBJECTIVES 1. You should know the target sites of action of the benzodiazepines and other drugs that can be used to treat sleep disorders 2. You should understand the distinguishing differences in the pharmacokinetics and metabolism of the benzodiazepines and the clinical implications of these respective drug differences. 3. You should understand how the different classes of sedative–hypnotics are used clinically, know the factors to consider in choosing the most appropriate drug for specific clinical situations and/or individuals and understand the potential adverse effects that can be produced by drugs from the respective classes. 4. You should understand the characteristics of benzodiazepines and other sedative-hypnotics that contribute to their different degrees of abuse liability and withdrawal symptoms. 5. You should understand the concepts of psychological and physiological dependence and tolerance and the pharmacokinetic & pharmacodynamic factors that contribute to the expression of these phenomena. 6. . Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 1 DRUGS TO TREAT SLEEP DISORDERS In prescribing drugs for "insomnia", it is essential to first establish the etiology of the disorder (drug dependence, sleep apnea, restless leg syndrome, psychological). If a rational basis for hypnotics can be established, then various factors can be considered in choosing an appropriate hypnotic drug. Overview of Sedative Hypnotics Sleep Disorders are most often treated with (a) benzodiazepines (that bind to both BZ1 and BZ2 sites on the GABAA receptor), (b) drugs that bind selectively to the BZ1 receptors, & (c) drugs that bind to Melotonin receptors. However, other classes of sedating drugs (discussed later and included in Talbe on page 10) can also used to induce sleep. Although a number of different benzodiazepines may be used to treat sleep disorders, the benzodiazepines specifically approved for the treatment of insomnia include: Estazolam (Prosom®), Temazepam (Restoril®), Quazepam (Doral), Flurazepam (Dalmane) Triazelam (Halcion®) [Triazolam (Halcion®) should be avoided, if possible, due to its high abuse potential]. Lorazepam (Ativan®) is also commonly prescribed, but not specifically FDA-approved for treatment of insomnia. The non-benzodiazepine BZ1-selective sedative hypnotics approved for sleep disorders: Zolpidem (Ambien®), Zaleplon (Sonata®) Eszopiclone (Lunestra®) Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 2 And now for something completely different (w/respect to site of action)…… Ramelteon (Rozarem®), an MT1 & MT2 melatonin receptor agonist that has been FDA approved to treat sleep disorders. Important Considerations in Choosing a Benzodiazepine to Treat Sleep Disordrs 1. Rapidity of onset of action 2. Duration of action 3. The half-life of the drug and its route of metabolism 4. the patients age, medical condition and prior drug history It is advised that benzodiazepines be used only for a short-term (1 -3 months) General Pharmacokinetic Considerations in Choosing a Benzodiazepine for Sleep Disorders The half-life of Elimination - drugs with long half-lives (either the parent compound or the active metabolite) will accumulate with prolonged use (as shown in the figure below). Figure from Dr. Lorens 2002 Therapeutics lecture notes which was graciously provided by Dr. David Greenblatt Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 3 Pharmacokinetic Considerations in Choosing a Benzodiazepine in the Elderly Benzodiazepine half-lives are longer in older people because they metabolize the drugs more slowly. Thus, drug concentrations will be greater in the elderly upon daily dosing. Figure from Dr. Lorens 2002 Therapeutics lecture notes which was graciously provided by Dr. David Greenblatt Age-related effects on drug half-lives: - Most likely for drugs that are converted into active metabolites by the liver, drugs such as chlordiazepoxide, diazepam, chlorazepate, quazepam, flurazepam. - Least likely for benzodiazepines that are only conjugated by the liver and have no active metabolites such as: Lorazepam, Oxazepam & Temazepam. Metabolism of the Benzodiazepines (Figure adapted from B.G. Katzung, Basic and Clinical Pharmacology, 9th ed. 2004, page 354) Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 4 Benzodiazepine issues in treating insomnia 1. Rebound insomnia upon stopping medication. This could involve: (a) the reoccurrence of the original symptoms, or (b) symptoms that are the same as original - but greater in intensity. 2. Risk of Abuse, Dependence and Withdrawal – not as great a risk for the BZs as for the barbiturates. The Development of Tolerance and Dependence 1. Psychological Dependence – similar to the behavioral pattern observed with heavy coffee drinkers or cigarette smokers. If more compelling, it can lead to physiologic dependence and tolerance. 2. Tolerance – this is a reduction in the effect of the drug over time that requires higher doses to achieve the original effect. This is analogous to the phenomena of desensitization and is due in part to decreases in the number of benzodiazepine receptors. Tolerance has been observed to the sedating effects of benzodiazepines but not to their anxiolytic or muscle relaxant effects. 3. Physiologic Dependence – a state of response to a drug in which the removal of the drug (“withdrawal”) produces unpleasant symptoms that differ from the original symptoms. The symptoms are usually opposite (compensatory) to the drug’s effects. Common withdrawal symptoms include: sweating, irritability, tachycardia, and abdominal discomfort. Benzodiazepines and other sedative-hypnotics (e.g. barbiturates, alcohol) all can produce: (1) Dependence (2) Tolerance (3) Addiction (4) Withdrawal symptoms The propensity to develop addiction, tolerance and dependence, and more severe withdrawal depend on a number of factors: 1) The rapidity of the time of onset of a drug’s affect. 2) The drug’s potency (can usually be assessed from the clinically prescribed dose). 3) The dose of drug taken. 4) The half-life of the drug. 5) The length of time that the drug has been taken. Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 5 In general; a the severity of withdrawal from sedative-hypnotic drugs will be greater given the factors listed below – (not listed in any order of priority) 1) higher doses 2) longer duration of use 3) higher potency 4) shorter half-life 5) shorter time of onset to produce their effects As the withdrawal from some sedative hypnotic drugs may be fatal, they should never be abruptly discontinued Strategy for Discontinuation: 1. Taper down dose and/or 2. Switch to a longer half-life drug, and preferably one of lower potency and less rapid onset of effect (decreases the reinforcement properties of the drug). Some Drugs that do not produce the characteristics described above: Non-Benzodiazepines Used to Treat Sleep Disorders Immidazopyridines and Pyrrolopyrazines - these drugs do not produce a dangerous degree of CNS depression (even in overdose) unless taken in combination with other CNS depressants, a combination that can be lethal. - may be habit forming with long-term use - do not exhibit significant muscle relaxant, anxiolytic or anticonvulsant effects - may produce anterograde amnesia and rebound insomnia, especially at high doses Zolpidem (Ambien®) – first BZ1-selective drug to receive FDA approval to treat sleep disporders. Approved only for short –term use. Zaleplon (Sonata®): resembles zolpidem in its effects with a slightly shorter half-life (1 hr). Recommendation that dosage be reduced in the elderly and patients with liver disease. The metabolism of zaleplon is inhibited by the OTC H-2 histamine receptor blocker, cimetidine (generic; Tagamet®). - next day effects are less common than after zolpidem due to very short half- life. - similar to zolpidem, its approved only for short-term use. Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 6 Eszopiclone (Lunestra) – the S(+) isomer of zopiclone (a pyrrolopyrazine drug already approved for use in Canada) with no structural similarity to zolpidem, zaleplon or the benzodiazepines. FDA approved (12/04) for treatment of sleep disorders. It does not significantly alter the stages of sleep. - unlike zolpidem & zaleplon, eszopliclone is not restricted in its labeling to short-term use. - its elimination is slower in the elderly and doses should be started at 1 mg instead of the typical dose of 2-3 mg. Drug Interactions: Various CYP3A4 inhibitors (e.g. clarithyromycin -Biaxin) or inducers (e.g. Rifampin) can increase and decrease, respectively, serum concentrations and alter the duration of action of a given dose of eszopiclone. Drugs Used to Treat Sleep Disorders that Target Sites other than The GABAA Receptor 1. A Melatonin Receptor Agonist Ramelteon (Rozerem) – the M1 & M2 melotonin receptor agonist does not appear to have high abuse potential. However, caution should be used in patients with: - liver disease - sleep apnea - depression or suicidal thoughts - individuals over 65; may still use but dose may need to be adjusted 2. The Tricyclic Antidepressants (TCAs) Amitriptyline (Elavil®) -a tricyclic antidepressant (TCA) that is particularly effective in treating sleep disorders associated with or contributing to chronic pain syndromes such as fibromyalgia Doxepin (Sinequan®) & Imipramine (Trofanil®) The TCAs are relatively high affinity antagonists at H-1 histamine receptors, and lower affinity (but comparable) blockade of muscarinic cholinergic and alpha1-adrenergic receptors. Consequently, the TCAs produce several aversive side effects such as: 1) postural (orthostatic) hypotension, 2) cardiotoxicity, and 3) confusion with memory dysfunction, particularly in the elderly. Tricyclic antidepressants SHOULD NOT be prescribed for elderly patients (65+ years) because of their liability for inducing a toxic and confused state. Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 7 3. The Atypical Antidepressants Mirtazapine (Remeron®) is an alpha2-adrenergic receptor antagonist. It blocks presynaptic alpha2 receptors on both noradrenergic and serotonergic nerve terminals leading to an increase in the release of NE and 5-HT. Mirtazapine also blocks H1-histamine, 5-HT2A, 5-HT2C, 5-HT3 serotonin receptors & increases 5-HT at 5-HT1A receptors, inducing anxiolytic & antidepressant effects at longer treatment times At low doses, mirtazapine is highly sedating. Increasing the dose produces less sedation and greater excitation. Mirtazapine does not produce sexual dysfunction, nausea or GI problems. Trazadone (Deseryl) – originally approved as an antidepressant, it is highly sedating and currently is marketed primarily as a hypnotic drug Nefazodone (Serzone®) is a 5-HT2A receptor antagonist and 5-HT reuptake inhibitor. It is mildly sedating and does not interfere with sexual function. Nefazodone is chemically related to the antidepressant drug, trazodone (Desyrel®). . 4. The Antihistamines Cyclobenzaprine (Flexeril) - an H-1 histamine receptor antagonist. Hydroxyzine (Atarax®) – another H-1 histamine receptor antagonist (antihistamine). Diphenhydramine (over the counter Benadryl) – an H-1 histamine receptor antagonist. 5. Nonprescription “sleeping pills”: Many “sleeping pills”, previously available as OTC sleep agents contained the antihistamines pyrilamine or methapyriline, and possibly an analgesic or anticholinergic drug. These were found to produce tolerance and rebound insomnia and to not be more effective than placebo and subsequently removed from the market. Unisom is the only FDA approved OTC sleep aid that is available at this time. Compoz - methapyrilene and pyrilamine Nytol - methapyrilene and salicylamide (salicyate) Sleep-Eze - methapyrilene and scopolamine Sominex - methapyrilene, scopolamine and salicylamide (salicyate) Unisom - contains the antihistamine, doxylamine. Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 8 5. Herbal Preparations: These may be worth trying since there is some evidence that they contain constituents that may be biologically active. Consideration should be given to any prescribed medications, or other OTC medications that may interact with these herbal preparations. Valeriana officinalis (Valerian) - as a standardized 70% ethanol extract (600 mg HS) is a safe hypnotic. Sesquiterpenes are the active compounds that mediate GABA release and the inhibition of GABA breakdown. Its usefulness to treat insomnia for up to 4 weeks has been documented. Valerian does not produce a “next day” hangover or other aversive effects and does not induce any serious drug interactions. Chamomile (Matricaria recutita) - Apigenin is the active indgredient; benzodiazepine agonist Chamomile tea is relaxing when ingested HS. Kava - Kava lactones are the active components that facilitate the binding of GABA; reported to have calming effects. “Passion flower” - Chrysin is the active compound that is a benzodiazepine partial agonist; has been reported to be an effective and safe hypnotic but this view has not been substantiated. Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 9 Generic Name Trade Name Half-life Mech. of Action Characteristics &/or Use BENZODIAZEPINES (Actions at BZ1 and BZ2 receptors) Alprazolam Xanax Intermediate BZ1 & BZ2 agonist, ↑ freq. of Cl- channel opening Panic Disorders, cocaine withdrawal Clonazepam Klonopin Intermediate Same as above Anxiety & sleep disorders, fibromyalgia Clorazepate Tranxene Very long Same as above Inactive parent compound, multiple active metabolites Chlordiazepoxid e Librium Very long Same as above Multiple active metabolites Diazepam Valium Very long due to active metabolites Same as above Rapid onset, long half life, high abuse liability Estazolam ProSom Long Same as above Flunitrazepam Rohypnol Short Same as above Not approved in U.S., date rape drug, high abuse liability Flurazepam Dalmane Very Long Same as above Short acting parent drug but 2 long half life active metabolites that are longer in the elderly Flumazanil Romazicon Short BZ1 &BZ2 ANTAGONIST Can trigger BZ withdrawal, useful in BZ overdose Lorazepam Ativan Intermediate BZ1 & BZ2 agonist, ↑ freq. of Cl- channel opening No bioactive metabolites Phase II only Oxazepam Serax Intermediate Same as above No bioactive metabolites Phase II only Temazepam Restoril Intermediate Same as above No bioactive metabolites Phase II only Midazolam Versed Very Short (2.5hrs) Same as above Strong antereograde amnesia used in operative procedures Triazolam Halcion Very Short (2-3 hrs) Same as above Very rapid onset, high potency, high abuse liability Non-Benzodiazepine Drugs (BZ1 Selective) Zaleplon Ambien Very Short (1 hr) BZ1 selective agonist Sleep aid, rapid absorbtion, no anxiolytic, anticonvulsant & muscle relaxation Zolpidem Sonata Very Short (1.5–3 hrs) BZ1 selective agonist Sleep aid, rapid absorbtion, no anxiolytic, anticonvulsant and muscle relaxant efficacy Eszopiclone Lunestra Short (6 hrs) BZ1 selective agonist Sleep aid, rapid absorbtion, no anxiolytic, anticonvulsant or muscle relaxant efficacy Melatonin Receptor Agonist Ramelteon Rozarem Very Short M1 & M2 melatonin receptor agonist Rapid absorbtion, large Vd, No abuse potential, No BZ receptor binding, metabolism decreased by Fluvoxamine, Ketoconazole & Fluconazole (via CYP inhibition) Pharmacology & Therapeutics Drugs to Treat Sleep Disorders February 13, 2013 G. Battaglia, Ph.D. 10 Generic Name Trade Name Half-life Mech. of Sedation Other Characteristics Barbiturates Pentobarbital Nembutol Intermediate to Long Bind to GABAA &  the duration of Cl- channel opening High abuse liability, low therapeutic index, induces microsomal oxidases, may be GABA-mimetic at high conc. Phenobarbital Luminol Very long Same as above Same as above Thiopental Sodium Pentathal Short Same as above Same as above TCAs Used to Treat Sleep Disorders Amitriptyline Elavil Intermediate Antihistaminergic & anticholinergic TCA, anticholinergic, antiadrenergic effects Doxepin Sinequin Short to Interrnediate Same as Above TCA, anticholinergic, antiadrenergic effects Imipramine Trofanil Interrmediate to long Same as Above TCA, anticholinergic, antiadrenergic effects Other Sedating Drugs Used for Sleep Disorders Cyclobenzaprine Flexeril Intermediate Antihistaminergic Diphendydramine Benadryl Short to intermediate Same as above OTC meds for sleep and allergies Hydroxyzine Atarax Intermediate to long Same as above Mirtazapine Remeron Long (20-40 hrs) Antihistaminergic, 2 alpha rec. antagonist Antiadrenergic & stimulatory at higher doses Nefazodone Serzone Intermediate 5-HT2A antagonist & 5HT uptake inhibitor Liver toxicity, only generic, Trazadone Deseryl Intermediate Same as above Highly sedating Meprobamate Equinil, Miltown Intermediate Likely similar to BZs Anti anxiety (only approved use) & sedative effects Herbs and OTC Meds Camomile Apigenin is a GABA rec. agonist Relaxing effects at bedtime Kava lactones facilitate GABA binding Reported calming effects Valerian Releases GABA & inhibits its breakdown No next day hangover methapyrilene and pyrilamine Compoz Antihistaminergic No longer approved by FDA methapyrilene and salicylamide Nytol Antihistaminergic No longer approved by FDA methapyrilene and scopolamine Sleep-Eze, Sominex Antihistaminergic & anticholinergic No longer approved by FDA Doxylamine Unisom Antihistaminergic Regained FDA approval (2004) for use as OTC sleep aid Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Drugs Used in the treatment of Rheumatoid Arthritis and Gout Date: Friday February 15th, 2012 – 9:30-10:30am Reading Assignment: Pharmacology: Board Review- Trevor, Katzung and Masters, Chapter 36 KEY CONCEPTS AND LEARNING OBJECTIVES 1. Understand the rationale behind the use of Analgesics, NSAIDs and glucocorticoids in the treatment of Osteoarthritis. 2. Understand the relative therapeutic benefit of NSAIDs, Analgesics, Glucocorticoids, DMARDs and Biological Response Modifiers/Biologics in the treatment of Rheumatoid Arthritis. 3. For the frequently used DMARDs class of drugs used in the treatment of RA (methotrexate, hydrochloroquine, sulfasalazine and leflunomide) Know a) Their major indications b) Their mechanism of action c) Their approximate time to effect d) Their major adverse effects e) Their contraindications especially during pregancy 4. Recognize the less frequently used DMARDs such as Azathioprine, D-penicillamine, Gold Salts, Cyclosporin and Cyclophosphamide in the treatment of Rheumatoid Arthritis. No need to learn specifics of these drugs as they are rarely used now in treatment of RA due to their toxicity. 5. Understand the roles of the distinct classes of Biological Response Modifiers/Biologics in the treatment of Rheumatoid Arthritis. 6. For each of the Biological Response Modifiers/Biologics i.e. TNF- blockers: Etanercept, Infliximab and Adalimumab; Abatacept; Rituximab; Anakinra; Tocilizumab and Tofacitinib Know: a) Their major indications b) Their mechanism of action c) Their major adverse effects d) Their major contraindications 7. Understand the pathophysiology of Gout; the role of uric acid in the etiology of the disease; and the typical disease course including hyperuricemia, acute gouty attack, intercritical phase and chronic gout. 8. Understand the rationale for the use of Colchicine and NSAIDs in the treatment of an acute gouty attack 9. Understand the rationale behind the use of drugs used to treat Chronic Gout (e.g. the Uricosuric probenecid; the Xanthine Oxidase inhibitors: Allopurinol and Febuxostat; and the Uric acid degrading enzyme Pegloticase). For each drug Know: a) Their specific indications b) Their mechanism of action c) Their major adverse effects d) Their major contraindications Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Drugs to be covered in this lecture: Particular emphasis should be placed on the drugs that are highlighted in BOLD TEXT. 1. NSAIDs 2. Acetaminophen (Tylenol®/Paracetemol®) 3. Topical Analgesics e.g. Capsaicin 4. Glucocorticoids (Injectable/Oral) 5. Disease-Modifying anti-Rheumatic Drugs (DMARDs) Commonly used DMARDs: Methotrexate (Rheumatrex®) Hydroxychloroquine (Plaquenil®) Sulfasalazine (Azulifidine®) Leflunomide (Arava®) Less frequently used DMARDs: Azathioprine (Imuran®) D-penicillamine (Depen®) Gold salts Cyclosporin A (Sandimmune® & Neural®) Cyclophosphamide (Cytoxan®) 6. Biological-response Modifiers Etanercept (Enbrel®) - TNFa inhibitor Infliximab (Remicade®) - TNFa inhibitor Adalimumab (Humira®) - TNFa inhibitor Anakinira (Kineret®) - IL-1R antagonist Abatacept (Orenica®) –inhibitor of T cell co-stimulation Rituximab (Rituxan®) – anti-B cell agent Tocilizumab (Actemra®) – anti-IL-6R agent Tofacitinib (Xeljanz®) –small molecule inhibitor of immune cytokine receptor signaling 7. Colchicine 8. Uricosuric agents Probenecid 9. Uric Acid Synthesis Inhibitor Allopurinol (Zyloprim®) Febuxostat (Uloric®) 10. Pegloticase - PEG-coupled PORCINE URICASE- degrades Uric acid to soluble byproduct Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. A. Osteoarthritis Overview - Most common joint disease affecting 21 million in the US - Characterized by loss of articular cartilage, bone remodeling and bone hypertrophy - Most commonly affects the weight bearing joints of the hips knees and lower back - Exact cause unknown, but may result from either excessive load on the joints or the presence of abnormal cartilage or bone - Risk factors include age > 50yrs, joint injury, obesity, high bone density, long-term immobilization of the joint, mechanical stress to the joints (e.g. occupational/sports), and genetics Symptoms - Joint soreness after periods of overuse or inactivity - Stiffness after periods of rest that disappear when activity is resumed - Redness, tenderness and swelling of the affected joints - Pain when moving the knee, Pain and swelling of the finger joints - Stiffness and pain in the neck, shoulders, arms or lower back Pathophysiology - Cartilage loses its elasticity and is more easily damaged - Mechanical wear and tear of the cartilage promotes the proliferation of chondrocytes. - Chondrocytes release proteases and pro-inflammatory mediators - Proteases degrade the matrix and result in the formation of abnormal cartilage - Breakdown of the cartilage can cause bones to rub together causing pain - Inflammation of the cartilage (mild c.f. with RA) causes new bone spurs to form, which causes pain and decreases the mobility of the joint Treatment Treatment goals a) Control pain and other symptoms with medication b) Improve functionality and quality of life- weight loss, exercise and physical therapy Medications 1. Analgesics: - For mild to moderate pain- Acetaminophen is the drug of choice for pain relief with minimal side effects (see below). 2. Topical Analgesics: - A topical analgesic such as Capsaicin (derived from hot peppers) can be used together with oral acetaminophen for pain relief. - Capsaicin works by depleting Substance P, which is present in painful joints and is involved in the transmission of pain to the CNS 3. NSAIDs: e.g. Aspirin, Ibuprofen and Naproxen - For patients with moderate to severe pain and signs of inflammation. - However use of NSAIDs may be limited by their side effects. - 4. Injectable glucocorticoids - Glucocorticoids can be injected directly into the joint for fast targeted pain relief as an alternative to patients with mild to moderate pain that do not respond to acetaminophen or NSAID treatment. - ~80% of patients exhibit a therapeutic response - Injections not given more frequently than once every three months Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. B. Rheumatoid Arthritis. Overview - Chronic inflammatory disease of the joints - Affects 1.5% of Americans, 9 million physician visits and >250,000 hospitalizations/year - Systemic autoimmune disease of unknown etiology - Characterized by inflammation and pain in the joints with progressive joint destruction - Additional extra-articular involvement of the skin, cardiovascular system, lungs and muscle - Causes significant disability, deformity and can even precipitate premature death - Although the exact cause is unknown, disease progression clearly involves the immune system with evidence of both T cell and B cell immune response to self-antigen Pathophysiology - Chronic lymphocytic inflammatory infiltration develops in the synovium (joint lining) - Swelling of the synovium causes pain, warmth, stiffness and redness of the joint - Rapid proliferation of synovial fibroblasts and infiltrating leukocytes causes the snovium to thicken and invade the local cartilage and bone - Macrophages and lymphocytes secrete proteolytic enzymes (e.g. collagenase) and inflammatory mediators (e.g. TNF-, IL-6, IL-1 and prostaglandins) causing further inflammation, the activation of bone-resorbing osteoclasts and ultimately the destruction of bone and cartilage. B1 Treatment of Rheumatoid Arthritis B1.1 Treatment goals a) Decrease pain b) Prevent or control joint damage c) Prevent loss of function and preserve quality of life B1.2 Medications used to treat Rheumatoid Arthritis: An overview B1..2A Drugs to treat Acute Joint Pain-Symptomatic Relief Only 1. NSAIDs (e.g. Aspirin, Ibuprofen, Naproxen and Celecoxib) - Drugs of choice for the reduction of inflammation and pain - Patients usually started on an NSAID immediately after diagnosis - For symptomatic relief only! - NSAIDs do not affect disease course - Choice of NSAID determined by efficacy and side effects - If a particular NSAID is ineffective after a 2 week trial an alternative NSAID is warranted 2. Analgesics (e.g. Acetaminophen, Capsaicin or an Opioid analgesic) - For symptomatic pain relief - Can be combined with a NSAID for improved pain relief and anti-inflammatory effect - A topical analgesic such as Capsaicin (derived from hot peppers) can be used together with oral acetaminophen for pain relief. Capsaicin works by depleting Substance P, which is present in painful joints and is involved in the transmission of pain to the CNS 3. Glucocorticoids a) Glucocorticoids exhibit both anti-inflammatory and immunoregulatory activity b) Can be administered orally, intravenously, or by direct injection into the joint c) Useful early in disease while waiting for slow acting DMARDs to work Shown to: Decrease joint tenderness Decrease joint pain Increase grip strength d) Short term/low dose glucocorticoids are seldom associated with serious side effects Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. e) The use of chronic glucocorticoid therapy is controversial- there is some evidence for the inhibition of disease progression, although their use is associated with increased side effects Adverse Effects: Weight gain Hypertension Osteoporosis Hyperglycemia Increased risk of infection f) Withdrawal from long-term glucocorticoid use is difficult, as this can result in increased disease severity due to suppression of the Hypothalamus-Pituitary-Adrenal axis B1.2B Drugs that can acto to prevent or control joint damage 4. Disease modifying anti-Rheumatic Drugs (DMARDS) (e.g. methotrexate) (SEE BELOW) - To slow joint damage and modify the course of the disease 5. Biological Response Modifiers (BRM)/Biologics (e.g. TNF inhibitors etc) (SEE BELOW) - Specific recombinant protein drugs that inhibit the immune responses that contribute towards joint inflammation. - Shown to slow joint damage and modify the course of the disease B2 Disease-Modifying anti-Rheumatic Drugs (DMARDs) Overview a) Unlike NSAIDs and corticosteroids, DMARDs can potentially reduce/prevent joint damage b) DMARDs work to inhibit the overactive immune system that is present in Rheumatoid Arthrisis c) DMARDs should be considered in Rheumatoid Arthritis where the use of NSAIDs/steroids has not prevented ongoing joint pain or other clinical symptoms d) Use of DMARDs should not be delayed beyond 3 months in these patients e) DMARDs are slow acting anti-rheumatic drugs that can take several weeks to many months to show efficacy and are typically taken for long periods ( i.e. months to years) B2.1. Frequently used DMARDs 1. Hydroxychloroquine (e.g. Plaquenil®) a) An anti-malarial drug that is moderately effective for mild rheumatoid arthritis b) Effectiveness only becomes apparent after 3-6 months c) Often combined with other DMARDs e.g. sulfasalazine and methotrexate d) Considered safe for use during pregnancy Mechanism of action: Unclear- thought to inhibit immune responses in a variety of ways - inhibition of TLR signaling in dendritic cells and B cells - inhibition of antigen presentation to CD4+ T cells Adverse Effects - Nausea, epigastric pain, rash and diarrhea - Rare (1/40,000) retinal toxicity has been reported in elderly patients that can result in irreversible visual loss 2. Sulfasalazine (Azulifidine®) a) Decreases signs and symptoms of disease and slows radiographic evidence of joint destruction (effective in up to 50% of patients). b) more toxic than hydroxychloroquine/somewhat less effective than methotrexate c) Sulfasalazine is a combination of 5-aminosalicylic acid covalently linked to sulfapyridine that is cleaved by colonic bacteria to its active components- it is thought that sulfapyridine is responsible for the therapeutic effect d) Effect can be seen in 1-3 months Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. e) Does not appear to be a teratogen- therefore unlike methotrexate it is safe during pregnancy f) Often combined with other DMARDs e.g. hydroxychloroquine g) Generally well tolerated Mechanism of action: Unclear, but thought to interfere with T and B cell immune responses, potentially by inhibiting the activity of the NF-kB transcription factor Adverse Effects: Nausea, headache, anorexia and rash are common (~30% of patients) Rare – Agranulocytosis with Fever/Rash within 2 weeks of treatment - fully reversible following drug discontinuation 3. Methotrexate (Rheumatrex®) a) Drug of choice for treatment of Rheumatoid Arthritis- especially for patients with active disease b) Oral methotrexate in low doses (15-27.5 mg/week) decreases the appearance of new bone erosions and improves the longterm clinical outcome – up to 70% of patients experience some response to the drug c) N.B. the levels of methotrexate used in RA are ~100-1000x lower than used in cancer treatment d) Effects are apparent after 4-6 weeks e) Well tolerated >50% of patients continue taking the drug for >3 yrs Mechanism of action: Unclear, but appears to be independent of its anti-proliferative activity. Thought to be due to an indirect effect on the increased production of adenosine, which is known to exhibit immunosuppressive properties Adverse Effects - Generally well tolerated (>50% of patients continue taking the drug for > 3 yrs) - MTX is 80-90% renally excreted- side effects more common in patients with renal impairment Common side effects: o Dose-dependent hepatotoxicity o Not recommended for those with pre-existing liver disease or consuming alcohol Rarer Side effects: o bone marrow suppression o Acute pneumonitis occurs in 1-2% of patients o Increased risk of lymphoma - Contraindicated during pregnancy (MTX is actually used as an abortifacient) 4. Leflunomide (Arava®) a) As effective as either slufasalazine or methotrexate in decreasing symptoms of disease b) Alternative to those unable to take MTX or for those non-responsive to MTX c) low cost alternative to expensive TNF inhibitors or for those with a preference for oral vs IV medication d) Responses evident in 1-2 months Mechanism of action: Oral pyrimidine synthesis inhibitor that inhibits dihydrotoroate dehydrogenase and therefore blocks the de novo synthesis of uridine, which leads to cell growth arrest in the G1 phase of the cell cycle. Inhibits both T cell proliferation and the production of autoantibodies by B cells Adverse Effects - Diarrhea occurs frequently (~10-15% of patients) - Alopecia, weight gain, rash, increased blood pressure and an increase in liver enzymes can Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. occur - Severe hepatotoxicity (including several fatalities) have been reported in patients taking both Leflunomide and Methotrexate - Contraindicated during pregnancy - Should not be taken concomitantly with rifampin (anti-tuberculosis medication), as it is known to elevate serum leflunomide levels leading to increased risk of toxicity B2.2. Less frequently used DMARDs 1. Azathioprine a) Orally active purine analog that is cytotoxic to inflammatory cells b) Used for patients with refractory Rheumatoid Arthritis and those with systemic involvement, such as rheumatoid vasculitis c) Treatment for 3-6 months is required to be effective d) Drug is not always well tolerated Adverse effects - Nausea, vomiting, abdominal pains, hepatitis, reversible bone marrow suppression and increased risk of lymphoma 2. Gold salts a) Gold compounds (oral/injected intramuscularly) have been used to treat arthritis since the 1960’s b) Can induce a complete remission c) Sometimes used in patients who cannot tolerate methotrexate d) Gold particles are taken up by macrophages and impair their function e) Response requires 3-6 months Adverse effects - stomatitis, dermatitis, proteinuria, thrombocytopenia, leukopenia and bone marrow suppression (RARELY USED NOW DUE TO TOXIC SIDE EFFECTS) 3. Cyclosporin (e.g. Sandimmune®, Neural®) a) Approved for use in Rheumatoid arthritis and retards appearance of bony erosions b) Acts by inhibiting T lymphocyte activation pathways leading to cytokine production c) Maybe useful in patents with refractory arthritis d) However, associated with significant nephrotoxicity, neurotoxicity, hepatotoxicity and increased hypertension e) Toxicity and costs associated with drug level monitoring limit use 4. Cyclophosphamide (Cytoxan®) a) Major metabolite is phosphoramide mustard, which promotes DNA crosslinks and thereby inhibits DNA replication b) Acts to inhibit T and B cell function by 30-40% c) Is useful in the treatment of severe rheumatoid vasculitis d) Long term use is associated with leukopenia, increased risk of infection, cardiotoxicity, alopecia and an increase risk of malignancy, especially bladder cancer. 5. D-penicillamine (Depen®) a) Can be effective in patients with refractory Rheumatoid Arthritis b) However, more toxic than either methotrexate or sulfasalazine- RARELY USED NOW Because of the increased toxicities of these drugs they are now typically only utilized in RA with severe life-threatening extra-articular symptoms such as systemic vasculitis, or in very severe RA that is refractory to other medications. Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. TNF- Activated Endothelial cells Increased leukocyte recruitment Increased joint Inflammation Synoviocyte proliferation Synoviocyte Chondrocyte Osteoclast Differentiation Matrix Metalloproteases + PGE2, IL-6 Cartilage Breakdown Bone Resorption Bone Erosion NAC 2005 B3 Biological Response Modifier/Biologic Drugs Overview Biological response modifiers are recombinant protein drugs that are specifically designed to inhibit either cytokines (e.g. TNF- and IL-1) or cell types (e.g. T cells/B cells) involved in the regulation of the in vivo immune response. a) Drugs that specifically inhibit the action of TNF-: (i) Etanercept (Enbrel®) (ii) Infliximab (Remicade®) (iii) Adalimumab (Humira®) b) Drugs that interfere with the actions of IL-1 - Anakinra (Kineret®) is an IL-1 blocker c) Drugs that inhibit T lymphocyte activation/co-stimulation: - Abatacept (Orenica®) d) Drugs that deplete B lymphocytes: - Rituximab (Rituxan®) e) Drugs that block the actions of IL-6 - Tocilizumab (Actemra®) f) Drugs that inhibit immune cytokine receptor signaling - Tofacitinib (Xeljanz®)) B3.1The critical role played by TNF- in the pathogenesis of Rheumatoid arthritis: An overview a) TNF- is a pivotal cytokine in the regulation of the immune response b) It is synthesized by macrophages, mast cells and activated CD4+ Th1 cells c) It activates macrophages increasing their phagocytic activity and the production of cytotoxic molecules d) It activates the endothelium and promotes the recruitment of leukocytes to site of inflammation e) It promotes the differentiation of bone-resorbing osteoclasts f) It induces the proliferation of synoviocytes and their production of proteases & inflammatory molecules g) It exhibits pyrogenic activity causing fever and acts systemically to cause pain Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Anti-TNF drugs binds soluble TNF- and prevents it from binding to its receptor and inducing a biological response Anti-TNF- drugs NAC 2007 Fully humanized Anti-HuTNF- Monoclonal antibody TNF- Adalimumab (Humira¨ ) Human IgG1 Mouse Anti-HuTNF- antibody variable region TNF- Infliximab (Remicade¨ ) Human IgG1 Fc Human p75 TNF- receptor Extracellular domain TNF- Etanercept (Enbrel¨ ) Mechanism of Action Biological Response TNF- Receptor APC MHC B7 T cell TcR CD28 T cell activation APC MHC B7 T cell T cell activation CD28 Abatacept Human IgG1 Fc Human CTLA-4 Extracellular domain Abatacept Binds with high affinity to B7 molecules B3.2. Anti-TNF- drugs a) Etanercept (Enbrel®) Recombinant protein consisting of two soluble p75 TNF- receptor moieties linked to the Fc portion of a human IgG1 antibody molecule b) Infliximab (Remicade®) Chimeric human/mouse anti-TNF- monoclonal antibody c) Adalimumab (Humira®) Recombinant fully human monoclonal antibody Anti-TNF- drugs: Mechanism of Action. All three anti-TNF drugs work by binding to soluble TNF- and prevent it from interacting with its cognate receptor expressed on the surface of specific cell types. Anti-TNF- drugs: Clinical Use a) Anti-TNF drugs are given either subcutaneously or intravenously and are typically administered weekly/bi-weekly b) Typical time to effect is 1- 4 weeks c) Clinically shown to: (i) reduce joint pain and swelling (ii) decrease the formation of new bone erosions (iii) slow progression of structural joint damage d) ~ 30-60% of patients will exhibit a 20-50% improvement in their symptoms e) Are used as both monotherapy and in combination with methotrexate. When combined with low dose methotrexate the addition of anti-TNF drugs have been shown to significantly prevent disease progression versus the use of methotrexate alone Anti-TNF- drugs: Common Adverse effects. a) Injection site reactions, injection site pain, headache and rash are common, but usually do not require discontinuation of treatment b) Increased risk of opportunistic infections c) Can result in the reactivation of latent tuberculosis and Hepatitis B virus d) Should not be given to patients with either acute or chronic infections e) Treatment should be discontinued if a serious infection or sepsis develop f) May rarely be associated with the exacerbation of pre-existing congestive heart failure and the development of demyelinating diseases such as multiple sclerosis, and the appearance of malignancies, especially lymphoma g) Should not be given to patients with a recent history of malignancy B3.3 Abatacept (Orenica®) a) Is a recombinant protein fusion between the T cell surface molecule CTLA-4 and human IgG1 (CTLA4-Ig) b) It inhibits T cell activation by binding to the CD80/CD86 (B7) family of co-stimulatory ligands expressed on antigen presenting cells, thereby blocking the delivery of co-stimulation signals to the T cell via the CD28 molecule, which is essential for efficient T cell activation c) Slows damage to bone and cartilage and relieves both the symptoms and signs of arthritis d) Effective in patients non-responsive to anti-TNF- drugs Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Biological Response IL-1 Receptor Endogenous IL-1 receptor Antagonist IL-1RA Biological Response IL-1 Receptor IL-1 Biological Response IL-1 Receptor Anakinra prevents IL-1 from binding to its receptor IL-1 Anakinra (recombinant IL-1RA) Adverse effects - Increased risk of serious infections- screen for latent TB and HBV - Not to be given to patients with either acute or chronic infections - Should not be given in combination with a TNF- blocker as this increases infections B3.4 Rituximab (Rituxan®) a) Rituximab is a chimeric humanized antibody that binds to the CD20 molecule expressed on B lymphocytes b) I.V. infusion of Rituximab depletes B cells from the blood. - since B cells are implicated in disease etiology through antigen presentation and formation of autoantibodies their depletion can slow disease progress c) Clinically shown to decrease signs and symptoms of disease and reduce radiographic evidence of disease progression. d) Effects not seen for 3 months, although effects may last 6 months - 2 yrs following a single infusion. e) Effective in patients not responsive to TNF- inhibitors Adverse Effects - Increased infections - Reactivation of latent viruses e.g. CMV, HSV and Hepatitis B&C - Progressive multifocal leukoencephalopathy (PML-RARE) - fatal demyleinating disease associated with reactivation of the JC virus B3.5 Anakinra (Kineret®) a) Genetically engineered recombinant version of an endogenous IL-1 receptor antagonist (IL-1RA) b) Competitively inhibits the pro-inflammatory effects of endogenous IL-1 c) A subcutaneous dose of 150 mg/day gives a modest reduction in pain and swelling, but a significant reduction in new bone erosions (due to the effects of IL-1 on synoviocyte-mediated cartilage degradation). d) Also given in combination with methotrexate Adverse effects - Local injection site reactions are frequent (~40%) and can lead to discontinuation of the drug - Only a small increase in infections - Should not be given to patients with either acute or chronic infections - Complications (neutropenia and severe bacterial infections) occur more frequently when given together with an anti-TNF- drug B3.6 Tocilizumab (Actemra®) a) Chimeric humanized antibody directed against the IL-6 receptor b) Acts as an antagonist of the IL-6 receptor c) Used in patients non-responsive to TNF inhibitors Adverse Effects - Increased risk of BM suppression (Lymphocytopenia, neutropenia, anemia) - Increased risk of serious infections (including TB and HBV) - Hepatotoxicity (routine liver monitoring) - Increased levels of cholesterol - Increased risk of malignancy (especially in setting of immunosuppression) Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. JAK JAK Y Y Biological Response C Cytokine receptor JAK kinase phosphorylates tyrosine residues in receptor cytoplasmic domain JAK JAK Y Y Biological Response C Cytokine receptor Tofacitinib Tofacitinib Inhibits immune cytokine Receptor signaling Tofacitinib Cytokine Cytokine PM PM B3.7 Tofacitinib (Xeljanz®) a) New class of anti-rheumatic drug b) Small molecule inhibitor that inhibits JAK tyrosine kinases involved in immune cell cytokine signaling Adverse Effects - Lymphocytopenia, neutropenia and anemia - Increased risk of serious infections including TB - Lipid abnormalities (increased cholesterol) - Increased liver enzymes B4 Treatment strategy in Rheumatoid Arthritis a) Initial symptomatic treatment for relief of pain and joint inflammation - NSAIDs (e.g. Naproxen, Indomethacin) - Glucocorticoids (e.g. Prednisone) b) DMARDs therapy should be initiated within 3 months of diagnosis - Hydroxychloroquine + Sulfasalazine is used for mild disease - Methotrexate is the drug of choice for active and/or moderate-severe disease c) NSAID are used in the early stages of DMARD therapy to reduce pain while waiting for the clinical effect of DMARDs to “kick-in” d) If Methotrexate is ineffective- other DMARDs (e.g. Leflunomide) and/or Biological Response Modifiers either alone or in combination should be tried. e) Clinical trials have shown that combination therapy with multiple agents is likely to be more effective than monotherapy. e.g. Methotrexate + Hydroxychloroquine and Sulfasalazine Methotrexate + either Etancerpet, Infliximab or Adalimumab C. Gout Overview a) Gout is an extremely painful form of arthritis. b) It is associated with HYPERURICEMIA: high serum levels of uric acid (>7 mg/dL) - uric acid is a poorly soluble end product of purine metabolism - however not all individuals with hyperuricemia will develop gout c) Hyperuricemia can result from either over production of uric acid (10% of patients), or from decreased excretion of uric acid by the kidney (~90% of all patients). d) It affects primarily men in their 30’s and 40’s (frequency 1 in 100) and is associated with obesity, Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Leukocytes Enzymes Urate Crystales Synoviocytes Phagocytosed Urate Crystales PGs IL-1 Inflammation Bone Bone Synoviocytes Bone Bone Increased Serum Uric Acid NAC 2005 hypertension, hyperlipidemia, type 2 diabetes, a diet rich in purines and the excessive consumption of alcohol. - Historically referred to as the “Disease of Kings” Gout: Pathophysiology a) Urate crystals are deposited in the joints b) Synoviocytes phagocytose urate crystals c) Synoviocytes secrete inflammatory mediators- prostaglandins, IL-1 and lysosmal enzymes d) Inflammatory leukocytes are recruited into the joint e) Recruited macrophages phagocytose urate crystals and release additional inflammatory mediators that promote further recruitment of inflammatory cells f) Inflammation causes pain, heat and swelling and damage to the joint C1. Gout Disease Course 1. Initial stage: Asymptomatic hyperuricemia -only a small percentage of patients with hyperuricemia will go onto develop gout 2. Acute Gouty Attack - rapid onset of an intense period of painful swelling in a single joint, most often in the feet (esp. big toe; first metatarsophalangeal joint). - the symptoms of the attack typically resolve within 3-10 days. 3. Intercritical Phase - hyperuricemia without acute symptoms - 10% of patients may never experience another attack 4. Chronic Gout - recurrent attacks of increasing frequency and severity involving additional joints - chronic high levels of urate crystals lead to the formation of TOPHI - a deposition of urate crystals around the synovial joint that can induce an inflammatory response resulting in the destruction of cartilage and the synovial lining C2. Pharmacological strategies for the treatment of gout C2.1. Drugs that relieve the symptoms of the acute gouty attack (A) Colchicine - traditional treatment, - plant alkaloid that prevents tubulin polymerization into microtubules - blocks leukocyte migration and phagocytosis - anti-inflammatory, but no analgesic properties - effective typically only when given during the first 24-48 hrs of the attack - limited by side effects (especially at high doses)- ~80% of patients develop diarrhea/vomiting within 24hrs - overdose can be life threatening due to bone marrow suppression - because of toxicity now generally relegated to a second line agent Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Renal proximal tubule cell LUMEN URAT1 Organic anion (e.g. lactate) URATE URATE [URATE] basolateral apical Renal proximal tubule cell LUMEN Probenecid Sulfinpyrazone Excretion in Urine URAT1 Organic anion (e.g. lactate) URATE URATE URATE basolateral apical Uric acid Reabsorbtion Reduced serum Uric acid levels High serum Uric acid levels Hypoxanthine (more soluble) Xanthine Uric Acid (Insoluble) Xanthine Oxidase Xanthine Oxidase Allopurinol Febuxostat GOUT (B) NSAIDs o commonly used as a replacement for Colchicine o all NSAIDs except Aspirin and Salicylates have been used successfully in the treatment of gout o effective at reducing pain and disability due to attack o COX-2 inhibitors should be used when other NSAIDs are contraindicated because of history of GI bleeds or use of blood thinners C2.2. Drugs that lower plasma uric acid levels by promoting uric acid excretion – URICOSURIC AGENTS Probenecid a) Probenecid are both weak organic acids that inhibit anion transporters in the proximal tubules of the kidney and decrease net reabsorption of uric acid- thereby promoting uric acid excretion b) Indicated in patients that under excrete uric acid c) Should not be given until 2-3 weeks after the initial attack as drugs can actually initiate and/or prolong the symptoms of an acute gouty attack (due to disruption of urate homeostasis)- usually prophylactic NSAID treatment is given at same time to reduce risk of inducing an attack d) Should not be given to patients that naturally produce high levels of uric acid due to increased risk of kidney stones e) To reduce risk of kidney stones urine volume should be maintained at a high level and the urine pH should be kept > pH 6. Contraindicated: Patients with kidney stones and/or renal insufficiency C2.3. Drugs that lower plasma uric acid levels by decreasing uric acid synthesis Allopurinol (Zyloprim®) and Febuxostat (Uloric®) a) Used in the treatment of chronic gout to block production of Uric Acid b) structural analogue of hypoxanthine that inhibits Xanthine Oxidase, an enzyme that catalyses the finall two steps in purine degradation Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. c) particularly useful in patients with: - a high level of endogenous uric acid synthesis, - Recurrent Kidney stones - Renal Impairment - Grossly elevated Uric Acid Levels ie. The presence of TOPHI d) Should not be given during an acute gouty attack as this can actually worsen symptoms An NSAID is usually prophylatically co-administered at the onset of Allpurinol therapy to reduce the chances of precipitating an acute attack of gout Adverse Effects: a) Can induce an acute gouty attack if NSAID prophylaxis not provided b) Rash, leukopenia, thromobocytopenia & fever can occur in 3-5% of patients c) Allopurinol hypersensitivity syndrome (<0.1% of patients) [NOT FEBUXOSTAT] - a rare, but potentially life threatening reaction (25% mortality rate) - most likely to occur in patients with renal insufficiency + diuretic dosage reduction required in presence of renal impairement - symptoms include: Erythematous rash, fever, hepatitis esinophilia and acute renal failure Drug Interactions: 6-mercaptopurine and azathioprine - purine synthesis inhibitors used in immunosuppression and treatment of leukemia - metabolized by xanthine oxidase to inactive metabolites - toxicity is increased in the presence of allopurinol C3. Management of Chronic Gout a) Hyperuricemia by itself does not indicate that treatment is necessary b) Treatment is indicated for: - Patients with multiple gouty attacks - Those that are more susceptible to future attacks e.g. Renal insufficiency - Patients with very high levels of uric acid (>12 mg/dL) Treatment Goal: To reduce serum uric acid levels to <6mg/dL c) Which drug to use is dependent upon whether the patient is either an over producer or an under secretor of uric acid. 24 hr urinary uric acid excretion < 700 mg/dL - Undersecretion - Probenecid 24 hr urinary uric acid excretion >700 mg/dL - Overproduction - Allopurinol d) Allopurinol is specifically indicated for: - Patients with a uric acid kidney stone - Patients with renal insufficiency- as uricosuric agents are not effective - Patients with TOPHI e) Effective therapy will require lifelong treatment C4. Treatment of drug-resistant chronic gout a) New drug Pegloticase (PEG-coupled Porcine Uricase) is an enzyme that degrades insoluble uric acid to more soluble byproduct. Note: Humans lack the Uricase enzyme b) Reserved for patients that have advanced, actively symptomatic gout that is uncontrolled with other uric acid lowering drugs Especially - presence of frequent flares - Presence of tophi - Contraindication to other gout drugs Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Indications MOA Adverse Effects Misc. Hydroxychloroquine Anti-malarial drug Mild RA Inhibits: a) TLR signaling b) Antigen presentation Rare Occular toxicity Safe in pregnancy Often combined with Sulfasalazine Sulfasalazine Pro Drug:Sulfayradine/ 5-aminosalicylic acid -metabolized to active component by colonic bacteria Mild RA Sulfapyridine - active component MOA unknown Inhibits T & B cells probably via NF-kB Hepatotoxicity Agranuocytosis (Rare) Safe in pregnancy Often combined with HCQ Methotrexate Drug of Choice moderate/severe RA MOA in RA different from use in cancer - Increases Adenosine leading to Immunosuppression Hepatotoxicity (common) Pulmonary toxicity BM suppression Risk of lymphoma Contraindicated in Pregnancy and Liver disease Not recommended In Renal impairment (80-90 % renal CL) Leflunomide Alternative to MTX Moderate/severe RA Inhibitor of dihydroorotate dehydrogenase (Uridine synthesis) -(G1 cell cycle arrest) -- inhibits T and B cell Immune responses Hepatotoxicity (esp with MTX) Hypertension (esp NSAIDs) Diarrhea, nausea (~15%) Contraindicated in Pregnancy and Liver disease Traditional DMARDs Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Biologics/Biological Response Modifiers Indications MOA Adverse Effects Misc. TNF inhibitors Etanercept Adlimumab Infliximab Active RA (monotherapy or combine with MTX) Binds to TNF and prevents its interaction with its receptor Risk of infection Reactivation latent TB/HBV Exacerbates CHF Risk Demyelinating disease Risk of malignancy Screen for latent TB and HBV Contraind acute/ chronic infections Abatacept Active RA Alt. to TNF inhibs DO NOT COMBINE with other Biologics CTLA4-IgG fusion protein Binds CD80/CD86 Blocks T cell co-stimulation via CD28 Risk of infection Screen for latent TB and HBV Contraind acute/ chronic infections Rituximab Active RA Alt. to TNF inhibs DO NOT COMBINE with other Biologics Binds to CD20 on B cells IV infusion depletes B cells Risk of infection PML: Progressive Multifocal Leukoencepalopathy - reactivation JC virus Screen for latent TB and HBV Contraind acute/ chronic infections Tocilizumab Active RA Alt. to TNF inhibs DO NOT COMBINE with other Biologics Humanized anti-IL6R mAb Binds IL6R Blocks IL-6 signaling Risk of infection (inc TB) Hepatotoxicity Hypercholesterolemia BM suppression Risk of malignancy (esp Immuno) Screen for latent TB and HBV Contraind acute/ chronic infections Monitor for DemyelinÕ disease Anakinra Active RA Alt. to TNF inhibs DO NOT COMBINE with other Biologics rIL-1RA IL-1R antagonist Blocks IL-1 signaling Risk of infection (inc TB) BM suppression Risk of malignancy Tofacitinib Active RA Alt. to TNF inhibs DO NOT COMBINE with other Biologics Small molecule inhib Inhibits JAK kinases Blocks Immune Cytokine signaling Risk of infection (inc TB) Hepatotoxicity Hypercholestrolemia BM suppression Risk of malignancy (esp Immuno) Screen for latent TB and HBV Contraind acute/ chronic infections Pharmacology & Therapeutics Drugs to treat RA and Gout February 15th, 2013 Neil Clipstone, Ph.D. Drugs to Treat Gout Indications MOA Adverse Effects Misc. NSAIDs Acute Gouty attack Prophylaxis for other gout medications Inhibits COX-2 Anti-inflammatory and Analgesic GI Toxicity Renal Etc etc Colchicine Acute Gouty Attack Prophylaxis for other gout medications Inhibits tubulin Polymerization Blocks leukocyte migration/phagocytosis Narrow therapeutic window Nausea, Diarrhea Vomiting (~80%) Uricosurics Probenecid Chronic Gout Due to decreased Uric acid excretion Inhibits Renal anion transporter Promotes uric acid excretion Contraindicated: Kidney Stones Renal insufficiency Uric acid overproduction Xanthine Oxidase Inhibitors Allopurinol Feboxostat Chronic Gout Due to Uric Acid Overproduction Inhibits Xanthine Oxidase Decreases URIC ACID production Rash Leukopenia Thrombocytopenia Liver enzymes Pegloticase Drug resistant gout Enzymatic degradation of Insoluble URIC ACID Risk of CV events Generation of Anti-drug Abs limits treatment Allopurinol Hypersensitivity Syndrome (25% mortality) (Allopurinol only) Esp. High dose/Renal Failure Dosage reduction required In renal insufficiency Increased toxicity with Azathioprine Anti-inflammatory No analgesic effects Should be given within 24-48 hrs of attack Can cause Kidney stones In high producers of URIC ACID Pharmacology & Therapeutics Parkinsonism & Its Treatment February 19, 2013 J. Lee, M.D., Ph.D. FINAL COPY RECEIVED: 8/28/07 1 TREATMENT OF PARKINSONISM AND DEMENTIA Date: Tues February 19, 2012 – 8:30AM Reading Assignment: Basic & Clinical Pharmacology, 12th ed., - B.G. Katzung, Chapter 28 and Chapter 60 pgs 1053-56. KEY CONCEPTS AND LEARNING OBJECTIVES (Treatment of Parkinson’s Disease) (Section orignially prepared by Dr. B. Wolozin and edited and updated by Dr. J. Lee) 1. Know the presentation of Parkinson’s disease a. Know other movement disorder diseases 2. Know the function of the 4 dopaminergic systems in the brain: a. Be able to identify pathways in figure 1. 3. Know the functional circuitry of the nigrostriatal system. a. The direct and indirect pathway b. Know which neurons have Dopaminergic (D1, D2) and Muscarinic (M) receptors c. Know which neurons have which neurotransmitters in the striatum: Acetylcholine, GABA, Glutamate, Enkephalin, Substance P 4. Know the classes of pharmacotherapy for PD and the timeline for their use 5. Know individual drugs used for treating PD: a. Precursor and peripheral degradation inhibitors (Sinemet®) i. Mechanism ii. Clinical use iii. Side effects (i.e. dystonia, behavioral and fluctuations) b. Dopaminer Receptor Agonists: Bromocriptine (Parlodel®), Ropinirole (Requip®) and Pramipexole (Mirapex®) i. Mechanism, Clinical Use (Example: For rx of On-Off syndrome) ii. Pharmacokinetics: Bromocriptine iii. Side effects: Ropinerole vs. Bromocriptine vs Pramipexole c. DA releaser (Amantatine): Mechanism of action d. Degradation inhibitors (MAO-B and COMT): Mechanisms and clinical uses e. Anticholinergics (Benztropine) i. Mechanism ii. Side effects 6. Know nature/mechanism of alternative treatments for PD (deep brain stimulation etc) as well as those in development. Pharmacology & Therapeutics Parkinsonism & Its Treatment February 19, 2013 J. Lee, M.D., Ph.D. FINAL COPY RECEIVED: 8/28/07 2 Generic Name Trade Name Half Life Mechanism of Action Elimination Benefit Levodopa Dopar 1-3 hrs Dopamine Precursor Decarbox. Replenishes DA Carbidopa Lodosyn Inhibits DOPA Decarbox. Inhibits DA degradation Levo./Carbi. Sinemet 1-3 hrs Combination Decarbox. Increases DA in brain Bromocriptine Parlodel 3 hrs D1/D2 receptor agonist Phase I/II Alternative to Sinemet Pramipexole Mirapex 8 hrs D3 receptor agonist Unchanged-Urine Reduces On/Off Ropinirole Requip 6 hrs D2 receptor agonist CYP450-1A2 Reduces On/Off Selegiline Eldepryl 1-3 hrs MAO-B inhibitor Early/Late treatment Talcapone Tasmar 2-3 hrs COMT inhibitor Glucuronidation Prolongs L-DOPA Amantadine Symmetrel 2-4 hrs DA releaser Unchanged-Urine Early treatment Entacapone Comatan COMT inhibitor Reduces off time Rasagiline Azilect MAO-B inhibit Reduces off time KEY CONCEPTS & LEARNING OBJECTIVES (Treatment of Dementia) 1. Know what are the laboratory and clinical tests for ruling out reversible forms of dementia in the elderly. 2. Know the basic drug treatments for the reversible forms of dementia? 3. Know the current symptomatic and theoretical preventive drug therapies for Alzheimer’s disease. 4. Know what are the theoretical drug therapies for Huntington’s disease and ALS. 5. Know what are the preventive and symptomatic drug treatments for cerebrovascular disease. DRUG LIST Generic Name Trade Name Half-life Mechanism of Action Elimination Benefit Side Effects tacrine Cognex® > 300 h. Acetylcholinest erase Inhibitor CYP450- IA2 Glucuronidation Increase AcH Liver tox. donepezil Aricept® 70 h. Acetylcholinest erase Inhibitor CYP450- 2D6, 3A4 Glucuronidation Increase AcH Diarrhea GI rivastigmine Exelon® 1.5 h. Acetylcholinest erase Inhibitor Rapid cholinesterase mediated Increase AcH Nausea Vomiting GI Pharmacology & Therapeutics Parkinsonism & Its Treatment February 19, 2013 J. Lee, M.D., Ph.D. FINAL COPY RECEIVED: 8/28/07 3 cevimeline (AF102B) Evoxac® 5 h. Muscarinic Agonist CYP450- 2D6, 3A3144 Off label use for AD fluoxetine Prozac® 1-3 days Serotonin reuptake blocker CYP 450 - 2D6 Depression +/- memory GI insomnia indomethacin Indocin® 5 h. Prostaglandin synthesis Primarily renal excretion, unchanged anti-inflammatory GI Ulcers ibuprofen Motrin® 2 h. Related to PGE inhibition Primarily renal excretion, unchanged anti-inflammatory GI Ulcers celocoxib Celebrex® 11 h. Cyclooxygenase 2 Inhibitor CYP 450- 2C9 anti-inflammatory GI Ulcers, less than NSAIDs vitamin E Tetrabenzine Unique E® very long Co - factor very slow, days to weeks Fat soluble anti-oxidant For abnormal movements in Huntingtons Hyper- vitaminosis Less side effects than Reserpine Pharmacology & Therapeutics Parkinsonism & Its Treatment February 19, 2013 J. Lee, M.D., Ph.D. FINAL COPY RECEIVED: 8/28/07 4 Generic Name Trade Name Half-life Mechanism of Action Elimination Benefit Side Effects haloperidol Haldol® 3 weeks D2 receptor blockage NA Decrease agitation Tardive dyskinesia creatine (OTC) Cardiotropin ® NA precursor mitochondrial substrates NA ATP none riluzole Rilutek® 12 h. glutamate release Na channels CYP450 - 1A2 neuronal toxicity liver tox Clofibrate Atromid® 22 h. liver release LDL’s renal CV disease nausea captopril Capoten® 2 h. ACE I inhibitor renal HTN allergic rxn warfarin Coumadin® 2 - 5 days mult. clotting factors CYP450 multiple isoforms strokes bleeding heparin 10 - 20 min. Anti-coagulant liver metabolism strokes blood thinner atorvastatin Lipitor® 14 h. HMG - CoA reductase inhibitor CYP450 - 3A4 CHL AD liver tox muscle weakness Thiamine (OTC) short water soluble co-enzyme pyruvate metabol. renal protects cells from oxidative damage ? vitamin B12 Mega B® short water soluble enzyme co-factor for folate system renal blocks demyelination in spinal cord levothyroxine memantine Synthroid® Ebixa ® Axura® 6 - 7 days substitutes for endogenous T4 NMDA antagonist glucuronidation into bile treat hypo thyroidism Hyper -thyroidism Pharmacology & Therapeutics Therapeutics of HIV infection February 19, 2013 Bruce D. Cuevas, Ph. D. 1 THERAPEUTICS OF HIV INFECTION Date: Monday, February 19, 2013: 10:00AM-12:00PM Reading assignment: Guidelines for the Use of Antiretroviral Agents in HIV-1 Infected Adults and Adolescents. Department of Health and Human Services. Updated as a Living Document on March 27, 2012. Available from the AIDSinfo website. Extensive information concerning treatment options with numerous summary tables. KEY CONCEPTS AND LEARNING OBJECTIVES 1. Understand how distinct antiretroviral agent classes target different phases of the HIV replication cycle. 2. Be able to recognize the diagnostic criteria and therapeutic goals for the treatment of HIV infection. 3. For each of the major classes of antiretroviral medications used in the treatment of HIV infection, you should be able to describe: a) Mechanism of action b) Indications and clinical use c) Onset and duration of action d) Major adverse effects e) Contraindications f) Significant drug interactions (if any) 4. To understand the utility of combination therapy in the treatment of HIV infection, and that co-morbid conditions may require a modified regimen. 5. Be able to recommend an appropriate antiretroviral regimen for a newly diagnosed patient. Drugs to be covered in this lecture: 1. NUCLEOSIDE/NUCLEOTIDE REVERSE TRANSCRIPTASE INHIBITORS Abacavir Didanosine Emtracitabine Lamivudine Stavudine Tenofovir Zidovudine Pharmacology & Therapeutics Therapeutics of HIV infection February 19, 2013 Bruce D. Cuevas, Ph. D. 2 2. NON-NUCLEOTIDE REVERSE TRANSCRIPTASE INHIBITORS Delavirdine Efavirenz Etravirine Nevirapine Rilpivirine 3. PROTEASE INHIBITORS Fosamprenavir Atazanavir Darunavir Indinavir Lopinavir Nelfinavir Ritonavir Saquinavir Tipranavir 4. VIRAL INTEGRASE INHIBITORS Raltegravir Elvitegravir (new drug) 5. FUSION INHIBITORS Enfuvirtide 6. CCR5 ANTAGONISTS Mariviroc Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 1 Therapeutics of HIV infection I. HIV introduction A. HIV is a retrovirus that infects and kills a subset of immune cells leading to reduced immune function 1. HIV is managed as a chronic disease with antiretrovirals and therapy of opportunistic infections. 2. ARV (AntiRetroviral therapy) doesn’t cure or eliminate HIV infection. 3. HIV targets CD4+ or T-helper lymphocytes, destroying the cells. 4. CD4+ depletion leads to severe immuno-deficiency. 5. CD4+ counts below 500 cells/mm3 associated with opportunistic infections. 6. periodic CD4 counts of infected patients are performed to assess: i. immunologic status ii. risk of opportunistic infections iii. need for ARV iv. response to ARV B. Goal of anti-retroviral therapy-reduce viral load and maintain immune function 1. maximal and durable suppression of viral load to reduce the risk of disease progression 2. restoration and/or preservation of immunologic function 3. improvement in quality of life 4. reduction in HIV-related morbidity and mortality 5. prevent vertical transmission of HIV C. laboratory parameters for HIV 1. virologic suppression can be defined as a sustained reductiojn in HIV RNA level below the assay limit of detection (<50 copies/ml) 2. viral load assessment predicts: i. course of disease ii. need for ARV iii. which ARV to utlize iv. clinical response to ARV 3. genotypic resistance testing to determine the resistance mechanism, perform before initiating ARV and after regimen failure D. When to begin ARV 1. The decision to begin ARV is based on an assessment of disease progression risk. 2. Indicators for initiation of ARV include: i. a history of AIDS-defining illness ii. CD4 count <500 cells/mm3 iii. Pregnancy iv. HIV-associated neuropathy v. HBV co-infection when HBV treatment is indicated 3. Check for latest ARV guideline updates on AIDSinfo.nih.gov. E. HIV-2 infection Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 2 1. Endemic to west Africa, should be considered in that population or if patient has contact with that population 2. Generally shows longer asymptomatic stage, lower viral loads and mortality 3. Multispot HIV-1/HIV-2 Rapid test is approved for differentiating HIV-1 from HIV-2, but most serology and viral load tests are unreliable for HIV-2 4. HIV-2 should be considered in patients when serology and/or viral load are negative but CD4 and clinical conditions suggest HIV infection 5. NNRTI not effective against HIV-2. Use clinical improvement and CD4 count improvement to assess response to treatment. II. FDA approved drugs available for treatment of HIV infection- 6 major drug classes A. Nucleoside/nucleotide reverse transcriptase inhibitors (NRTI) 1. General mechanism of action-nucleoside or nucleotide analogs that lack 3’ hydroxyl group enter the cells, are phosphorylated and form sythetic substrates for viral RT. NRTI compete with native nucleotides, and terminate proviral DNA when incorporated 2. Onset and duration- NRTIs generally eliminated from plasma by renal excretion with half-lifes of 1-10 hours, intracellular reservoirs more persistent. One or two doses daily except zalcitabine (every 8h). Only didanosine has food restrictions. Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 3 3. Relative effectiveness at reducing viral load- Modestly effective as monotherapy (not recommended), valuable when used in combination therapy. Resistance develops slowly compared to NNRTIs or PIs. 4. Major adverse effects- Some in this class capable of inhibiting mitochondrial DNA polymerase, toxicities include anemia, myopathy, and pancreatitis and are associated with serious lactic acidosis-hepatic steatosis syndrome. (didanosine > stavudine > zidovudine) 5. Class members currently in use: a. Abacavir i. Indication: activity against HIV-1 ii. Adverse effect: associated with serious hypersensitivity reaction in patients with HLA-B 5701 genotype iii. Contraindicated: in patients with HLA-B 5701 genotype b. Zidovudine (AZT) i. Indications: activity against HIV-1 and HIV-2, used as exposure prophylaxis, OK for children or adults and in pregnancy ii. Major adverse effects: anemia and neutropenia iii. Contraindications: do not co-administer with stavudine (antagonist) iv. Significant drug interactions: cotrimoxazole or ganciclovir (bone marrow toxicity), ribavirin (antagonist) B. Nonnucleotide reverse transcriptase inhibitors (NNRTI) 1. General mechanism of action- noncompetitive inhibitors that bind to reverse transcriptase and induce a corformational change that greatly reduces enzyme activity 2. Onset and duration- rapidly absorbed and metabolized by hepatic CYPs, half-lives range from delavirdine (2-11h) to efavirenz (40-50h) 3. relative effectiveness at reducing viral load- Activity against HIV-1 but not HIV-2. Resistance can develop rapidly, never use a monotherapy. One advantage of NNRTI-based regimen is that PIs can be reserved for later and thus avoid PI adverse effects. The disadvantages are the prevalence of resistant virus, low genetic barrier to resistance. 4. Major adverse effects/drug interactions: all influence CYP activity, so drug interactions common. Cytochrome P450 system and HIV drug metabolism - the P450 cytochromes influence drug metabolism by oxidizing or reducing substrate drugs - P450 substrate drug metabolism is dependent on one or more P450 enzymes (e.g. CYP3A4) - a P450 inhibitor is a drug that inhibits the metabolism of a P450 substrate -a P450 inducer stimulates increased expression of P450 enzymes Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 4 5. FDA-approved class members currently in use: a. Efavirenz i. Indications: preferred as part of initial antiretroviral therapy, the only once daily dose NNRTI ii. Major adverse effects: birth defects, transient CNS effects iii. Contraindications: 1st trimester pregnancy or women planning to conceive iv. Significant drug interactions: CYP3A4 inducer, so reduces concentration of PIs, methadone b. Nevirapine i. Indications: recommended as an alternative to Efavirenz in treatment naïve women with pretreatment CD4<250 cells/mm and men with CD4<400 cells/mm ii. Major adverse effects: severe hepatotoxicity iii. Contraindications: women with pretreatment CD4 >250 cells/mm3 and men with pretreatment CD4 >400 cells/mm3 iv. Significant drug interactions: CYP inducer, reduces concentration of methadone, PIs C. Protease inhibitors (PI) 1. General mechanism of action- specifically and reversibly inhibit the HIV aspartyl protease and thereby block post-translational processing of viral proteins required to produce a mature viral particle 2. Onset and duration- Most PIs have poor bioavailability, absorption of some enhanced by high fat meals, all metabolized by hepatic CYP system 3. Relative effectiveness at reducing viral load: highly effective as part of combination therapy 4. Major adverse effects: associated with metabolic syndrome, lipodystrophy, lipoatrophy of face and limbs, lipemia, nausea, vomiting, diarrhea and paresthesias. All are substrates and inhibitors of CYPs. 5. Ritonavir (RTV) “boosting” of PIs a. ritonavir is a PI that is a potent CYP3A4 inhibitor b. co-administration of low dose ritonavir enhances or “boosts” exposure of other PIs Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 5 c. RTV boosting allows for reduced dose and dosing frequency d. low dose RTV improves tolerance and is effective at CYP3A4 inhibition 6. Class members currently in use: a. Ritonavir i. Indications: active against HIV-1 and 2, potent CYP3A4 inhibitor that is used to boost availability of drugs that are CYP3A4 substrates and to reduce dose and dosing frequency ii. Major adverse effects: not well tolerated at doses required for antiviral activity, but low doses used for boosting are well tolerated, paresthesias iii. Contraindications/significant drug interactions: similar to other PIs ARV and management of dyslipidemia/statin usage - prior to treating cholesterolemia in HIV-infected patients undergoing ARV, consider statin metabolism - should avoid combining CYP3A4 substrate statins with boosted PIs - switching from PIs to NRTIs may alleviate lipodystrophy, hyperlipidemia PI-induced metabolic syndrome Definition: a cluster of metabolic risk factors that tend to occur together that increases chances of developing heart disease, stroke, and/or diabetes and are associated with PI therapy -hyperlipidemia, hypertriglyceridemia, decreased HDL and increased LDL -insulin resistance, hyperinsulinemia -lipodystrophy, central obesity, facial and limb lipoatrophy -increased macrophage CD36 leading to increased cholesterol uptake, atherosclerosis Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 6 D. Viral integrase inhibitors- Raltegravir 1. Mechanism of action: blocks insertion of reverse-transcribed viral DNA into the host DNA 2. Indications: in combination therapy for experienced patients with suppression failure or excess toxicity 3. Onset and duration: twice daily dose following high fat meal 4. Relative effectiveness at reducing viral load: effective when combined with PI and NRTI 5. Major adverse effects: diarrhea, headache, and nausea 6. Contraindications: not for use as monotherapy 7. NEW DRUG: Elvitegravir introduced this year, but only as a pre-packaged combo drug including Tenofovir, Emtricitabine, and Cobicistat (CYP3A4 inhibitor). E. Fusion inhibitors- Enfuvirtide 1. General mechanism of action: peptide inhibitor that binds HIV surface glycoprotein gp41 to block conformation required for membrane fusion with host cell 2. Indications: combination therapy component in experienced patients with viral suppression failure. Injected twice daily 3. Relative effectiveness at reducing viral load: not active against HIV-2, mutation of the HR1 region of gp41 associated with resistance 4. Major adverse effects: Injection site reaction/inflammation is very common, hypersensitivity 5. Contraindications: do not use in patients with known hypersensitivity F. CCR5 antagonists- Maraviroc 1. General mechanism of action: small molecule slowly-reversible antagonist of the CCR5 interaction with gp120, blocks CCR5-tropic HIV-1 entry Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 7 2. Indications: combination therapy component for experienced patients with viral suppression failure, should perform tropism test first 3. Resistance mechanisms: mutation of the CCR5-binding amino acid sequence in HIV gp120, or emergence of CXCR4-tropic virus 4. Major adverse effects: hepatotoxicity and possible hypersensitivity 5. Contraindications: liver dysfunction 6. Significant drug interactions: CYP substrate, concentration altered by CYP inducers and inhibitors III. Combination therapy for HIV infection A. Complicating conditions and considerations: 1. comorbid conditions such as cardiovascular disease, chemical dependency, tuburculosis, renal, liver, or psychiatric disease 2. potential adverse drug effects 3. potential drug interactions with other medications 4. pregnancy 5. results of drug resistance testing 6. HLA-B5701 testing if considering Abacavir 7. gender and pretreatment CD4 count if considering Nevirapine 8. likelihood of patient adherence with the regimen B. Combination ARV options for treatment-naïve patients-recommended starting regimen consists of either 1 NNRTI + 2 NRTI, or 1 PI (preferably boosted with ritonovir) + 2 NRTI. Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 8 Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 9 C. Changing drug regimen due to treatment failure (virologic or immune) 1. Definition of virologic suppression failure: the inability to achieve or maintain suppression of viral replication to levels below the limit of detection (<50 copies/mL) and may manifest as: a. incomplete virologic response to therapy, or b. virologic rebound (after virologic suppression, repeated detection of HIV RNA above the assay limit of detection) 2. Definition of immune failure: the inability to achieve and maintain adequate CD4 T-cell response despite virologic suppression D. Tuberculosis treatment for HIV-infected patients 1. Rifampin-based antimycobacterials are highly effective in treating MTB infection. Rifamycin is a potent inducer of CYP activity, and markedly effects exposure to multiple ARV drugs. 2. Rifampin dramatically reduces exposure to all PIs and multiple NNRTIs and should not be coadministered (except efavirenz). 3. Rifabutin is preferred drug for HIV patients with active MTB, but still interacts with PIs and requires caution. 4. Bottom line is that coadministration of rifabutin and ARV will require monitoring and dose adjustment. Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 10 F. What NOT to use Regimens rationale exception Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 11 What NOT to use (continued) Components rationale exception Example of a test question: A 29 year-old male was previously diagnosed with M. tuberculosis infection and rifampin treatment was initiated. In a follow up exam, lab results reveal the patient to be infected with HIV, and lab results are as follows: CD4+ count = 460 cells/mm3 viral load = 41,000 copies/ml resistance testing = HIV1 Which of the following treatment regimens would be a recommended option, if any? A. efavirenz, tenofovir, and emtricitabine B. Nevirapine, tenofovir, and emtricitabine C. Atazanavir, abacavir, and lamivudine D. Didanosine, abacavir, and lamivudine E. None of the above Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 12 Summary tables: Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 13 Pharmacology & Therapeutics Therapeutics of HIV infections February 19, 2013 B. Cuevas, Ph.D. 14 2/25/2013 1 Mary Lou Gutierrez, MD Child and Adolescent Psychiatry Department of Psychiatry and Behavioral Neuroscience Clinical practice of pediatric psychopharmacology is challenging Over the past 10 years, significant increase in the use of psychotropic medications in the pediatric population. The largest increase was the use of atypical antipsychotics (138.4%) atypical antidepressants (42.8%) Treatment with psychotropic drugs managed a large proportion of children receive these medications from primary care doctors and pediatric specialists OBJECTIVES To be cognizant of the general pharmacological properties of psychotropic medication for the pediatric population To know the clinical indications/usage of psychotropic medications for childhood presentations of psychiatric disorders as recommended by the FDA To know the major warnings and precautions associated with psychotropic medications To appreciate the evidence based medicine treatment of pediatric psychiatric disorders 2/25/2013 2 For more detailed and up-to-date information on FDA regulations Stimulants Methylphenidate Ritalin, Ritalin LA/SA, Metadate CD/ER Concerta Daytrana Dexmethylphenidate Focalin Amphetamine Sulfate Dextroamphetamine (Dexedrine) Mixed Amphetamine Salts (Adderall) Vyanse Selective Norepinephrine Reuptake Inhibitor Atomoxetine (Strattera) NEUROBIOLOGY OF ADHD Conceptualization of etiology Underactivity of the catecholamergic system mediated by dysregulation of the dopaminergic(DA) and noradrenergic(NE) CNS systems 2/25/2013 3 NEUROBIOLOGY OF ADHD Dopamine neurotransmission Pathways Tuberoinfundibular Nigrostriatal Mesolimbic Mesocortical (executive functioning) Functions Neurophysiologically dopamine enhances the signal for attention skills Stimulants block (inhibition of the neurotransmitter reuptake pump) 1-reuptake of norepinephrine (NE) and of 2-dopamine (DA) in the presynaptic neurons which 3-increase their concentrations (DA) in the intrasynaptic space. Atomoxetine (non-stimulant) increases norepinephrine in the synaptic cleft by inhibiting its presynaptic transporter(5). Children who are CYP2D6 “slow metabolizers” may experience atomoxetine’s serum peak concentrations 5-fold greater than fast metabolizers (6). 2/25/2013 4 2/25/2013 5 Stimulants have FDA indications for the treatment of ADHD and narcolepsy (2). MPH and d-MPH are approved for use in patients at least 6 years old Amphetamines in children > 3 years old. Atomoxetine has FDA indication for the treatment of ADHD in individuals at least 6 years old (2). ALGORITHM FOR THE MEDICATION TREATMENT OF ADHD ADAPTED FROM J AM ACAD CHILD ADOLESC PSYCHIATRY 41: 26S-49S, 2002 Monotherapy Methylphenidate or amphetamine Poor response Monotherapy Amphetamine or methylphenidate Monotherapy Atomoxetine or bupropion Poor response Monotherapy with antidepressant not used above Poor response Methylphenidate or amphetamine plus Antidepressant or alpha-agonist Poor response 2/25/2013 6 STIMULUS PRACTICAL CONSIDERATIONS Start with MPH or AMPH If one stimulant not effective, try the other one Short-acting easier to titrate, then may wish to switch to long acting Start with long acting if school dose a problem Titrate to optimum response Example starting dose Methylphenidate 5mg ½ tab BID-TID Atomoxetine, a suggested second line treatment option for ADHD (4). Optimal effective dose of atomoxetine  1.2-1.3 mg/kg/day, administered once daily. The FDA issued a Black Box Warning reporting sudden death in children prescribed stimulants (2), therefore stimulants should not be used in patients with known structural cardiac abnormalities, cardiomyopathy, and serious heart rhythm abnormalities. Precaution: Temporary slowing in growth has also been found in children medicated with stimulants (8). 2/25/2013 7 MPH has strong evidence supporting its use as the first line treatment for ADHD.  Multimodal Treatment Study of Children with Attention-Deficit/Hyperactivity Disorder (MTA) 14-month multisite trial of 4 different treatment strategies for 579 children, aged 7 to 9.9 years (9)  MTA compared the following treatments:  (a) the fixed-dose of MPH titrated to the child’s “best dose”,  (b) intensive behavioral treatment,  (c) combined MPH and intensive behavioral treatment, and  (d) standard community care.  All four groups improved,  But children in the MPH and combination groups improved significantly more  Several studies comparing the safety and efficacy of MPH and Adderall (15, 16, 17). Adderall was reported to exhibit longer lasting therapeutic effects than MPH (15, 16 ), but more stomachaches and sad mood. Swanson et al. (1998) (16) reported that MPH had a faster onset of clinical effect than Adderall (1.8 hours and 3 hours respectively).  Stimulants are available in short (4 hours), intermediate (6-8 hours), and long-acting sustained-release (10-12 hours) formulations (3).  A short-acting medication may be added to a sustained-release medication in the late afternoon to maintain ADHD symptom control in the evening.  It is recommended that stimulants be given seven days per week (20), but patients with decreased appetite, irritability, or sleep problems may discontinue stimulants on weekends and holidays (21).  Weight-based dosing guidelines are available, but effective maintenance dose of stimulants is guided by the clinical response and the side effects. Only a few studies have examined the long-term safety of stimulant use in pediatric populations(12). 2/25/2013 8 Red Light, Green Light Red Light, Green Light (Complex Rules) Simon Says (Mental Flexibility and Self-Control) Row, Row, Row Your Boat (Focus, Memory, Flexibility and Self-Control) "Executive Function" Interactive games require high levels of executive function, test a child’s ability to pay attention, remember rules and exhibit self-control — qualities that predict academic success. Rule of thumb with Pediatric Pharmacology Go low (dosing), Go slow (titration/taper) 2/25/2013 9 REFERENCES ARE LISTED IN BOOK CHAPTER Overview of Pediatric Psychopharmacology. Sigita Plioplys, M.D., Jennifer Kurth, D.O. and Mary Lou Gutierrez, M.D. Anna Ivanenko, M.D., Ph.D., editor: Sleep and Psychiatric Disorders in Children and Adolescents, Informa Healthcare; 1 edition, May 19, 2008. Slides contributed with direct consent from Shire, U.S.
3207	https://www.zhihu.com/question/35853890	如何证明（1+1/n）∧n的单调性？ - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册如何证明（1+1/n）∧n的单调性？关注问题写回答登录/注册微积分如何证明（1+1/n）∧n的单调性？关注者 27 被浏览 152,299 关注问题写回答邀请回答好问题 4 添加评论分享 7 个回答默认排序梦见在校大学生关注 134 人赞同了该回答也许是最简单的一种证明方法，用到了广为人知的平均值定理「又称“算数几何平均不等式”」，可以参考《数学分析》P48 「徐森林版本」。发布于 2017-07-29 09:39 赞同 13425 条评论分享收藏喜欢暮色关注 20 人赞同了该回答 x n+1 x n=(1−1(n+1)2)n.n+2 n+1\frac{x_{n+1}}{x_{n}}=(1-\frac{1}{(n+1)^2})^{n}.\frac{n+2}{n+1} \ 由伯努利不等式（1+α）n≥1+α n(α>−1)（1+\alpha）^{n}\geq1+\alpha n(\alpha>-1) (1−1(n+1)2)n.n+2 n+1≥(1−n(n+1)2).n+2 n+1≥(1−1 n+2）n+2 n+1=1(1-\frac{1}{(n+1)^2})^{n}.\frac{n+2}{n+1} \geq(1-\frac{n}{(n+1)^2}).\frac{n+2}{n+1}\geq(1-\frac{1}{n+2}）\frac{n+2}{n+1}=1 可知， x n x_{n} 是递增数列欧耶编辑于 2021-03-17 07:26 赞同 209 条评论分享收藏喜欢收起 Lunaris 如无必要，勿增实体关注经评论区提醒，以下求解方法用到了自然底数e的结论，如果你是想通过这个问题来证明 (1+\cfrac{1}{n})^n 单调有界的话，那下面的解法属于循环论证。而如果你只是想用现有工具来解决这个问题本身（即分析函数的单调性），那下面的求解方法就没有问题。因此下面的方法展示了一种程序化求解函数单调性问题的普适方法，供大家参考。问题：求函数 f(n)=(1+\cfrac{1}{n})^n ， n\ge 1 的单调性思路：求导函数，根据导函数来判断它的单调性。（没错思路就是这么简单，下面开始实践） f(n) 的底数和指数部分都有自变量，没有对应的求导公式，因此我们需要把指数给打下来，才能方便求导。打指数最好的办法就是对数，因此两边同取对数（之所以敢这么做，是因为 f(x)=\ln x 在整个定义域上单调递增，因此 x_1=x_2 等价于 f(x_1)=f(x_2) ），过程如下： y=(1+\cfrac{1}{n})^n → \ln y=n\ln(1+\cfrac{1}{n}) → y=e^{n\ln(1+\frac{1}{n})} 化成这种形式就可以结合复合函数求导法则来求导了： f'(n)=(1+\cfrac{1}{n})^n \cdot [\ln(1+\cfrac{1}{n})-\cfrac{1}{n+1}] 然后我们要比较 f'(n) 和 0 的关系：显然前面一项一定大于0，因此只需要关注后面一项和 0 的关系。还是用函数的思维体系，我们把后面一项写为函数的形式，即 g(n)=\ln(1+\cfrac{1}{n})-\cfrac{1}{n+1} ， n\ge 1 。求导： g'(n)=-\cfrac{1}{n(n+1)^2}<0 因此 g(n) 在定义域内单调递减，在 n=1 时取得最大值 g(1)=\ln2-\cfrac{1}{2} ，没有最小值。 \begin{aligned} \displaystyle\lim_{n\rightarrow \infty}g(n)&=\displaystyle\lim_{n\rightarrow \infty}\ln(1+\cfrac{1}{n})-\displaystyle\lim_{n\rightarrow \infty}\cfrac{1}{n+1}\ &=0-0\ &=0 \end{aligned} 由此可知 g(n)>0 。因此 f'(n)>0 。因此 f(n) 在定义域内单调递增。结束。展开阅读全文赞同 102 条评论分享收藏喜欢查看剩余 4 条回答写回答下载知乎客户端与世界分享知识、经验和见解相关问题如何判断（1-1/n）^n的单调性？ 2 个回答如何判断f(x)在区间I上的单调性？ 2 个回答是否存在n≥2使得调和数H_n是正整数？ 2 个回答线性变换的值域和核加起来是等于v吗？ 1 个回答如何证明[1-（1/n)]∧n单调递增？ 4 个回答帮助中心知乎隐私保护指引申请开通机构号联系我们举报中心涉未成年举报网络谣言举报涉企侵权举报更多关于知乎下载知乎知乎招聘知乎指南知乎协议更多京 ICP 证 110745 号 · 京 ICP 备 13052560 号 - 1 · 京公网安备 11010802020088 号 · 互联网新闻信息服务许可证：11220250001 · 京网文2674-081 号 · 药品医疗器械网络信息服务备案（京）网药械信息备字（2022）第00334号 · 广播电视节目制作经营许可证:（京）字第06591号 · 互联网宗教信息服务许可证：京（2022）0000078 · 服务热线：400-919-0001 · Investor Relations · © 2025 知乎北京智者天下科技有限公司版权所有 · 违法和不良信息举报：010-82716601 · 举报邮箱：jubao@zhihu.com 想来知乎工作？请发送邮件到 jobs@zhihu.com 登录知乎，问答干货一键收藏打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
3208	https://www.youtube.com/watch?v=tctafLoNIvY	Derivation of power transmitted by wave along string UNSW Physics 19500 subscribers 140 likes Description 18374 views Posted: 8 Mar 2017 This video will show you how to derive the equation for the power transmitted by a wave traveling along a piece of string. 4 comments Transcript: now we're going to consider the power transmitted by a wave traveling along the strength so we're going to start from the formula for kinetic energy I'll show you this derivation in just a minute we're going to end up with the power is equal to 1/2 new where this is the linear density V that's the way to speed Omega squared the angular frequency squared times a squared which is the amplitude squared so let's derive this formula now so we need to start from our expression for kinetic energy we know that kinetic energy is given by well DK that's for an increment of kinetic energy is given by 1/2 DM times u squared now this is for an element of string so this speed here is not the speed of the wave it's the speed that the piece of string is moving at so in order to work out how quickly the piece of string is moving we're going to need to use Y of X and T is equal to a sine KX minus Omega T plus Phi and then we can differentiate this to find out how quickly a piece of string is moving so U is actually equal to dy DT which you can see is equal to in this case minus Omega a cos KX minus Omega T plus v ok so now if we want to find out how much kinetic energy in an incremental stream we've got DK is equal to 1/2 now the mass of a small piece of string here we can give us a mess if it's gotten length DX then DM is going to be DX times the linear density which is mu so we can write this DM as mu DX and now we've got u squared so that Omega squared a squared cos KX minus Omega T by fire now because we're trying to get power which is energy divided by time let's look at how much of this energy goes through in a tiny increment of time so we're going to divide by BP now so it's got the KBT is it could to 1/2 Moo now we'll actually divide this DX by the DT doesn't matter we should turn we divide but you'll see why we choose this one is a minute Omega squared a squared cause lot this should be squared squared KX minus Omega T plus Phi and now the X DT that's how quickly this wave is traveling in the X direction so that is V so that is equal to 1/2 movie Omega squared a squared cos squared KX minus Omega T plus Phi now what we really want to do is work out what the average value of this DK DT so if we want to look at the average value of the DK BP we're going to need to know the average value of cos squared of something so if we it is 1/2 if you want to work out what it is one way to do it is we can integrate from 0 to 2 pi of cos squared theta D theta and then we have to divide by 2 pi possess the range this range 2 pi and when we integrate this thing this thing is pi so this is actually equal to 1/2 so that's one way to show that this cos squared the average value of that is a heart so we have that DK BT average is equal to 1/2 times 1/2 so that's 1/4 movie Omega squared a squared so we've now worked out how much energy is transmitted as kinetic energy however the string has also got to cancel energy and the potential energy we've seen in simple harmonic motion gets converted between while the energy gets converted between kinetic energy and potential energy and on average these two things the same value so that tells us that p is execute decay DT plus Cu DT and we can take the average of both of these and these are the same so this actually turns out to be equal to 1/2 movie Omega squared a squared so we've now shown the relationship that we set out to proof
3209	https://www.exceldemy.com/find-z-critical-value-in-excel/	How to Find Z Critical Value in Excel? - ExcelDemy Free Excel Courses Create Basic Excel Pivot Tables Excel Formulas and Functions Excel Charts and SmartArt Graphics Advanced Excel Training Data Analysis Excel for Beginners Tutorials Excel Basics Excel Functions Excel Formulas Macros & Excel VBA Advanced Excel Excel Charts Data Analysis with Excel Pivot Table in Excel Excel Power Query Excel Solver Excel for Finance Community Templates About Us Home»Excel Formulas»How to Find Z Critical Value in Excel? How to Find Z Critical Value in Excel? Written byAung Shine Last updated: Aug 4, 2024 in this article, we’ll demonstrate 3 different examples of finding the Z Critical Value in Excel. The Z Critical Value plays a significant role in a Hypothesis Test. A Hypothesis Test generates a Test Statistic as the result, but to determine if this result is statistically significant or not, a comparison should be made between the Test Statistic and the Z Critical Value. Introduction to the NORM.S.INV Function The NORM.S.INV function in Excel determines the inverse of the standard normal cumulative distribution. The Mean is Zero and the Standard Deviation is 1 in the cumulative distribution. Syntax =NORM.S.INV(probability) Arguments probability:A probability corresponding to the standard normal cumulative distribution. How to Find the Z Critical Value in Excel: 3 Suitable Examples For hypothesis test results to be considered statistically significant, the absolute value of the test statistic needs to be greater than the Z critical value. To illustrate, we’ll use the below dataset where the Significance Level (α) is 0.05and compute the Z critical value for 3 different test types. Example 1 – Calculate the Z Critical Value in a Left-Tailed Test Let’s find the critical value by applying the Significance Level (α) as a probability in the argument of the NORM.S.INV function. Steps: Select cell C7. Enter the following formula: =NORM.S.INV(C4) next stay CC Settings Off Arabic Chinese English French German Hindi Portuguese Spanish Font Color white Font Opacity 100%Font Size 100%Font Family Arial Text Shadow none Background Color black Background Opacity 50%Window Color black Window Opacity 0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25% 200%175%150%125%100%75%50% Arial Georgia Garamond Courier New Tahoma Times New Roman Trebuchet MS Verdana None Raised Depressed Uniform Drop Shadow White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% Press Enterto return the value. The desired result is returned. Example 2 – Find the Z Critical Value in a Right-Tailed Test As in the case of a left-tailed test, one critical value is also returned as output of a Right-Tailed test. In a right-tailed test case, we need to subtract the Significance Levelfrom 1, so we’ll modify the Significance Level (α) to insert it in the argument of the NORM.S.INV function. STEPS: Select cell C8. Insert the below formula: =NORM.S.INV(1-C4) Return the result by pressing Enter. The Z critical value for the right-tailed test is returned. Read More:How to Find T Critical Value in Excel Example 3 – Determine the Z Critical Value for a Two-Tailed Test ATwo-Tailed testreturns 2 critical values as output. We’ll input the Significance Level (α) as a probability in the argument of the NORM.S.INV function, but we’ll have to edit it first. As it’s two-tailed, we’ll divide the Significance Levelby 2 to get one critical value. Then, to find the second critical value, we’ll subtract the division output from 1. STEPS: Click on cell C9. Enter the formula below: =NORM.S.INV(C4/2) Press Enterto return the result. To find the second critical value, select cell C10. Enter the formula below: =NORM.S.INV(1-C4/2) Press Enter. The two critical values for the two-tailed test are returned. Download Practice Workbook Find Z Critical Value.xlsx Related Article How to Find Critical Value of r in Excel << Go Back to Critical Value in Excel \| Excel for Statistics\|Learn Excel Get FREE Advanced Excel Exercises with Solutions! Save Saved Removed 0 Tags: Critical Value in Excel Aung Shine Aung Shine completed his bachelor’s in Electrical and Electronics Engineering from Bangladesh University of Engineering and Technology. It has been almost 2 years since he joined SOFTEKO and actively working on the ExcelDemy project. Currently he works as a Team Leader where he guides his team members to create technical content. He has published 150+ articles and reviewed 50+ articles. 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3210	https://tutorial.math.lamar.edu/classes/de/wronskian.aspx	Paul's Online Notes Custom Search \| \| \| \| Go To Notes Practice and Assignment problems are not yet written. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here. Show/Hide Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. Sections Fundamental Sets of Solutions Nonhomogeneous Differential Equations Chapters First Order DE's Laplace Transforms Classes Algebra Calculus I Calculus II Calculus III Differential Equations Extras Algebra & Trig Review Common Math Errors Complex Number Primer How To Study Math Cheat Sheets & Tables Misc Contact Me MathJax Help and Configuration Notes Downloads Complete Book Practice Problems Downloads Problems not yet written. Assignment Problems Downloads Problems not yet written. Other Items Get URL's for Download Items Print Page in Current Form (Default) Show all Solutions/Steps and Print Page Hide all Solutions/Steps and Print Page Paul's Online Notes Home / Differential Equations / Second Order DE's / More on the Wronskian Prev. Section Notes Next Section Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width. Section 3.7 : More on the Wronskian In the previous section we introduced the Wronskian to help us determine whether two solutions were a fundamental set of solutions. In this section we will look at another application of the Wronskian as well as an alternate method of computing the Wronskian. Let’s start with the application. We need to introduce a couple of new concepts first. Given two non-zero functions f(x) and g(x) write down the following equation. cf(x)+kg(x)=0(1) Notice that c=0 and k = 0 will make (1) true for all x regardless of the functions that we use. Now, if we can find non-zero constants c and k for which (1) will also be true for all x then we call the two functions linearly dependent. On the other hand if the only two constants for which (1) is true are c = 0 and k = 0 then we call the functions linearly independent. Example 1 Determine if the following sets of functions are linearly dependent or linearly independent. f(x)=9cos(2x)g(x)=2cos2(x)−2sin2(x) f(t)=2t2g(t)=t4 Show All Solutions Hide All Solutions a f(x)=9cos(2x)g(x)=2cos2(x)−2sin2(x) Show Solution We’ll start by writing down (1) for these two functions. c(9cos(2x))+k(2cos2(x)−2sin2(x))=0 We need to determine if we can find non-zero constants c and k that will make this true for all x or if c = 0 and k = 0 are the only constants that will make this true for all x. This is often a fairly difficult process. The process can be simplified with a good intuition for this kind of thing, but that’s hard to come by, especially if you haven’t done many of these kinds of problems. In this case the problem can be simplified by recalling cos2(x)−sin2(x)=cos(2x) Using this fact our equation becomes. 9ccos(2x)+2kcos(2x)=0(9c+2k)cos(2x)=0 With this simplification we can see that this will be zero for any pair of constants c and k that satisfy 9c+2k=0 Among the possible pairs on constants that we could use are the following pairs. c=1,k=−92c=29,k=−1c=−2,k=9c=−76,k=214etc. As we’re sure you can see there are literally thousands of possible pairs and they can be made as “simple” or as “complicated” as you want them to be. So, we’ve managed to find a pair of non-zero constants that will make the equation true for all x and so the two functions are linearly dependent. b f(t)=2t2g(t)=t4 Show Solution As with the last part, we’ll start by writing down (1) for these functions. 2ct2+kt4=0 In this case there isn’t any quick and simple formula to write one of the functions in terms of the other as we did in the first part. So, we’re just going to have to see if we can find constants. We’ll start by noticing that if the original equation is true, then if we differentiate everything we get a new equation that must also be true. In other words, we’ve got the following system of two equations in two unknowns. 2ct2+kt4=04ct+4kt3=0 We can solve this system for c and k and see what we get. We’ll start by solving the second equation for c. c=−kt2 Now, plug this into the first equation. 2(−kt2)t2+kt4=0−kt4=0 Recall that we are after constants that will make this true for all t. The only way that this will ever be zero for all t is if k = 0! So, if k = 0 we must also have c = 0. Therefore, we’ve shown that the only way that 2ct2+kt4=0 will be true for all t is to require that c = 0 and k = 0. The two functions therefore, are linearly independent. As we saw in the previous examples determining whether two functions are linearly independent or linearly dependent can be a fairly involved process. This is where the Wronskian can help. Fact Given two functions f(x) and g(x) that are differentiable on some interval I. If W(f,g)(x0)≠0 for some x0 in I, then f(x) and g(x) are linearly independent on the interval I. If f(x) and g(x) are linearly dependent on I then W(f,g)(x)=0 for all x in the interval I. Be very careful with this fact. It DOES NOT say that if W(f,g)(x)=0 then f(x) and g(x) are linearly dependent! In fact, it is possible for two linearly independent functions to have a zero Wronskian! This fact is used to quickly identify linearly independent functions and functions that are liable to be linearly dependent. Example 2 Verify the fact using the functions from the previous example. f(x)=9cos(2x)g(x)=2cos2(x)−2sin2(x) f(t)=2t2g(t)=t4 Show All Solutions Hide All Solutions a f(x)=9cos(2x)g(x)=2cos2(x)−2sin2(x) Show Solution In this case if we compute the Wronskian of the two functions we should get zero since we have already determined that these functions are linearly dependent. W=∣∣∣9cos(2x)2cos2(x)−2sin2(x)−18sin(2x)−4cos(x)sin(x)−4sin(x)cos(x)∣∣∣=∣∣∣9cos(2x)2cos(2x)−18sin(2x)−2sin(2x)−2sin(2x)∣∣∣=∣∣∣9cos(2x)2cos(2x)−18sin(2x)−4sin(2x)∣∣∣=−36cos(2x)sin(2x)−(−36cos(2x)sin(2x))=0 So, we get zero as we should have. Notice the heavy use of trig formulas to simplify the work! b f(t)=2t2g(t)=t4 Show Solution Here we know that the two functions are linearly independent and so we should get a non-zero Wronskian. W=∣∣∣2t2t44t4t3∣∣∣=8t5−4t5=4t5 The Wronskian is non-zero as we expected provided t≠0. This is not a problem. As long as the Wronskian is not identically zero for all t we are okay. Example 3 Determine if the following functions are linearly dependent or linearly independent. f(t)=costg(t)=sint f(x)=6xg(x)=6x+2 Show All Solutions Hide All Solutions a f(t)=costg(t)=sint Show Solution Now that we have the Wronskian to use here let’s first check that. If its non-zero then we will know that the two functions are linearly independent and if its zero then we can be pretty sure that they are linearly dependent. W=∣∣∣costsint−sintcost∣∣∣=cos2t+sin2t=1≠0 So, by the fact these two functions are linearly independent. Much easier this time around! b f(x)=6xg(x)=6x+2 Show Solution We’ll do the same thing here as we did in the first part. Recall that (ax)′=axlna Now compute the Wronskian. W=∣∣∣6x6x+26xln66x+2ln6∣∣∣=6x6x+2ln6−6x+26xln6=0 Now, this does not say that the two functions are linearly dependent! However, we can guess that they probably are linearly dependent. To prove that they are in fact linearly dependent we’ll need to write down (1) and see if we can find non-zero c and k that will make it true for all x. c6x+k6x+2=0c6x+k6x62=0c6x+36k6x=0(c+36k)6x=0 So, it looks like we could use any constants that satisfy c+36k=0 to make this zero for all x. In particular we could use c=36k=−1c=−36k=1c=9k=−14etc. We have non-zero constants that will make the equation true for all x. Therefore, the functions are linearly dependent. Before proceeding to the next topic in this section let’s talk a little more about linearly independent and linearly dependent functions. Let’s start off by assuming that f(x) and g(x) are linearly dependent. So, that means there are non-zero constants c and k so that cf(x)+kg(x)=0 is true for all x. Now, we can solve this in either of the following two ways. f(x)=−kcg(x)ORg(x)=−ckf(x) Note that this can be done because we know that c and k are non-zero and hence the divisions can be done without worrying about division by zero. So, this means that two linearly dependent functions can be written in such a way that one is nothing more than a constants time the other. Go back and look at both of the sets of linearly dependent functions that we wrote down and you will see that this is true for both of them. Two functions that are linearly independent can’t be written in this manner and so we can’t get from one to the other simply by multiplying by a constant. Next, we don’t want to leave you with the impression that linear independence and linear dependence is only for two functions. We can easily extend the idea to as many functions as we’d like. Let’s suppose that we have n non-zero functions, f1(x), f2(x),…, fn(x). Write down the following equation. c1f1(x)+c2f2(x)+⋯+cnfn(x)=0(2) If we can find constants c1, c2, …, cn with at least two non-zero so that (2) is true for all x then we call the functions linearly dependent. If, on the other hand, the only constants that make (2) true for x are c1=0, c2=0, …, cn=0 then we call the functions linearly independent. Note that unlike the two function case we can have some of the constants be zero and still have the functions be linearly dependent. In this case just what does it mean for the functions to be linearly dependent? Well, let’s suppose that they are. So, this means that we can find constants, with at least two non-zero so that (2) is true for all x. For the sake of argument let’s suppose that c1 is one of the non-zero constants. This means that we can do the following. c1f1(x)+c2f2(x)+⋯+cnfn(x)=0c1f1(x)=−(c2f2(x)+⋯+cnfn(x))f1(x)=−1c1(c2f2(x)+⋯+cnfn(x)) In other words, if the functions are linearly dependent then we can write at least one of them in terms of the other functions. Okay, let’s move on to the other topic of this section. There is an alternate method of computing the Wronskian. The following theorem gives this alternate method. Abel’s Theorem If y1(t) and y2(t) are two solutions to y′′+p(t)y′+q(t)y=0 then the Wronskian of the two solutions is W(y1,y2)(t)=W(y1,y2)(t0)e−∫tt0p(x)dx for some t0. \| \| \| \| Because we don’t know the Wronskian and we don’t know t0 this won’t do us a lot of good apparently. However, we can rewrite this as W(y1,y2)(t)=ce−∫p(t)dt(3) where the original Wronskian sitting in front of the exponential is absorbed into the c and the evaluation of the integral at t0 will put a constant in the exponential that can also be brought out and absorbed into the constant c. If you don’t recall how to do this go back and take a look at the linear, first order differential equation section as we did something similar there. With this rewrite we can compute the Wronskian up to a multiplicative constant, which isn’t too bad. Notice as well that we don’t actually need the two solutions to do this. All we need is the coefficient of the first derivative from the differential equation (provided the coefficient of the second derivative is one of course…). Let’s take a look at a quick example of this. Example 4 Without solving, determine the Wronskian of two solutions to the following differential equation. t4y′′−2t3y′−t8y=0 Show Solution The first thing that we need to do is divide the differential equation by the coefficient of the second derivative as that needs to be a one. This gives us y′′−2ty′−t4y=0 Now, using (3) the Wronskian is W=ce−∫−2tdt=ce2lnt=celnt2=ct2 \| \| \| --- \| \| \| \|
3211	https://labex.io/tutorials/python-how-to-transform-numeric-base-in-python-495792	Learn Projects Pricing Log InJoin For Free How to transform numeric base in Python PythonBeginner How to transform numeric base in Python Practice Now Practice Now Introduction Numeric base transformation is a fundamental skill in Python programming that enables developers to convert numbers between different representation systems. This tutorial explores comprehensive techniques for transforming numeric bases, providing insights into how Python handles various number systems and offers powerful conversion tools for programmers. Numeric Base Basics Understanding Numeric Bases In computer science and mathematics, a numeric base (or radix) represents the number of unique digits used to represent numbers. The most common bases are: \| Base \| Name \| Digits \| Example \| --- --- \| \| 2 \| Binary \| 0-1 \| 1010 \| \| 10 \| Decimal \| 0-9 \| 42 \| \| 16 \| Hexadecimal \| 0-9, A-F \| 2A3F \| graph LR A[Decimal Base 10] --> B[Binary Base 2] A --> C[Hexadecimal Base 16] A --> D[Octal Base 8] Key Concepts Positional Notation In positional notation, each digit's value depends on its position. For example, in decimal 123: 3 is in the ones place 2 is in the tens place 1 is in the hundreds place Common Base Representations Binary (base-2): Used in computer systems, represented by 0 and 1 Decimal (base-10): Standard human counting system Hexadecimal (base-16): Compact representation of binary data Octal (base-8): Less common, but still used in some computing contexts Python Base Representation Decimal to Other Bases Python provides built-in functions for base conversion: ``` Decimal to Binary print(bin(42)) ## Output: 0b101010 Decimal to Hexadecimal print(hex(42)) ## Output: 0x2a Decimal to Octal print(oct(42)) ## Output: 0o52 ``` Base Conversion Techniques Python offers multiple ways to convert between bases, including: Built-in functions String formatting Custom conversion algorithms Why Base Conversion Matters Base conversion is crucial in: Computer networking Digital electronics Cryptography Low-level system programming At LabEx, we believe understanding numeric bases is fundamental to mastering programming skills and developing a deep comprehension of computer systems. Python Conversion Tools Built-in Conversion Methods int() Function The int() function is the primary tool for base conversion in Python: ``` Convert from different bases to decimal binary_num = int('1010', 2) ## Binary to Decimal hex_num = int('2A', 16) ## Hexadecimal to Decimal octal_num = int('52', 8) ## Octal to Decimal print(binary_num) ## Output: 10 print(hex_num) ## Output: 42 print(octal_num) ## Output: 42 ``` Advanced Conversion Techniques Custom Base Conversion Function ``` def convert_base(number, from_base=10, to_base=2): """ Convert numbers between arbitrary bases """ ## Convert to decimal first decimal_num = int(str(number), from_base) ## Convert decimal to target base if to_base == 10: return decimal_num digits = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ' if decimal_num == 0: return 0 result = [] while decimal_num > 0: decimal_num, remainder = divmod(decimal_num, to_base) result.append(digits[remainder]) return ''.join(result[::-1]) Example usage print(convert_base(42, 10, 2)) ## Decimal to Binary print(convert_base(42, 10, 16)) ## Decimal to Hexadecimal ``` Conversion Methods Comparison \| Method \| Pros \| Cons \| --- \| int() \| Simple, built-in \| Limited to standard bases \| \| Custom Function \| Flexible, supports any base \| More complex implementation \| \| format() \| Concise string formatting \| Less intuitive \| String Formatting Techniques ``` Using format() method decimal_num = 42 Binary representation binary_str = f'{decimal_num:b}' print(binary_str) ## Output: 101010 Hexadecimal representation hex_str = f'{decimal_num:x}' print(hex_str) ## Output: 2a Octal representation octal_str = f'{decimal_num:o}' print(octal_str) ## Output: 52 ``` Handling Large Number Conversions ``` Converting large numbers large_number = 1000000 print(f"Large number in binary: {large_number:b}") print(f"Large number in hex: {large_number:x}") ``` graph TD A[Decimal Number] --> B{Conversion Method} B --> \|int()\| C[Standard Base Conversion] B --> \|Custom Function\| D[Flexible Base Conversion] B --> \|format()\| E[String Formatting] Best Practices Choose the right conversion method based on your specific use case Handle potential conversion errors Be aware of performance implications for large numbers At LabEx, we recommend mastering these conversion techniques to enhance your Python programming skills and understand low-level data representations. Real-World Applications Network Address Conversion ``` def ip_to_binary(ip_address): """Convert IP address to binary representation""" return ''.join([bin(int(x)+256)[3:] for x in ip_address.split('.')]) def binary_to_ip(binary_str): """Convert binary string back to IP address""" return '.'.join([str(int(binary_str[i:i+8], 2)) for i in range(0, 32, 8)]) Example usage ip = "192.168.1.1" binary_ip = ip_to_binary(ip) print(f"IP: {ip}") print(f"Binary: {binary_ip}") ``` Cryptography and Encoding ``` def simple_encryption(message, base=16): """Convert message to hexadecimal for basic encoding""" return ''.join([hex(ord(char))[2:].zfill(2) for char in message]) def simple_decryption(encoded_message): """Decode hexadecimal message back to text""" return ''.join([chr(int(encoded_message[i:i+2], 16)) for i in range(0, len(encoded_message), 2)]) Example secret_message = "LabEx" encoded = simple_encryption(secret_message) decoded = simple_decryption(encoded) print(f"Original: {secret_message}") print(f"Encoded: {encoded}") print(f"Decoded: {decoded}") ``` Data Compression Techniques ``` def compress_binary(number): """Demonstrate binary compression technique""" binary = bin(number)[2:] ## Remove '0b' prefix compressed = len(binary).to_bytes((len(binary) + 7) // 8, byteorder='big') return compressed Example of compression large_number = 1024 compressed = compress_binary(large_number) print(f"Original Number: {large_number}") print(f"Compressed Size: {len(compressed)} bytes") ``` Application Domains \| Domain \| Base Conversion Use Case \| --- \| \| Networking \| IP address manipulation \| \| Cybersecurity \| Encoding/encryption \| \| Digital Electronics \| Binary logic operations \| \| Data Science \| Efficient data representation \| graph TD A[Base Conversion] --> B[Networking] A --> C[Cryptography] A --> D[Data Compression] A --> E[System Programming] Performance Considerations ``` import timeit def benchmark_conversion(): """Compare different conversion methods""" ## Decimal to Binary decimal_num = 1000000 ## Method 1: Built-in bin() bin_builtin = timeit.timeit(lambda: bin(decimal_num), number=10000) ## Method 2: Custom conversion def custom_binary(n): return format(n, 'b') bin_custom = timeit.timeit(lambda: custom_binary(decimal_num), number=10000) print(f"Built-in Method: {bin_builtin}") print(f"Custom Method: {bin_custom}") benchmark_conversion() ``` Best Practices for Base Conversion Choose appropriate conversion methods Consider performance implications Handle edge cases and large numbers Validate input before conversion At LabEx, we emphasize the importance of understanding base conversion as a fundamental skill in advanced programming and system-level development. Summary By mastering numeric base transformation in Python, developers can enhance their programming skills and gain deeper understanding of number representation. The techniques covered in this tutorial provide practical solutions for converting between decimal, binary, hexadecimal, and octal systems, empowering programmers to handle diverse numeric conversion scenarios with confidence and efficiency. share topics LinuxDevOpsCybersecurityDatabaseDockerKubernetesRed Hat Enterprise LinuxGitCompTIAShellNmapWiresharkHydraJavaSQLitePostgreSQLMySQLRedisMongoDBGolangC++CJenkinsAnsibleData ScienceMachine LearningPandasNumPyscikit-learnMatplotlibWeb DevelopmentAlibaba CloudHTMLCSSJavaScriptReactKali Linux User Reviews " Работа с git в корне отличается, но идёт гораздо быстрее из-за банального знания терминала благодаря курсу "Быстрый старт с линукс" Всё это позволяет создавать цельное видение работы операционной системы с системой управления версий. И это только начало! 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3212	https://ideagalaxyteacher.com/linear-functions-practice/	Skip to primary navigation Skip to primary sidebar Skip to footer Idea Galaxy Adventures in Junior High and Mathematics 10 Activities to Practice Linear Functions Like a Boss by Rachel Slope and y-intercept make up the crux of the 8th grade math standards. We spend 5 full weeks on learning slope and y-intercept and how to apply them, not to mention all the other time throughout the year getting ready to learn about them and then reviewing and applying them. It still surprises me how complex it all becomes. When you come to think of it, students have to identify slope and the y-intercept in various way. They also have to create them in various forms, and they have to be able to interpret them in many forms as well. It can reach a point of overwhelm for both the teacher and the students. Every year I pledge to myself that I will do better to help students reach the understanding they need to have and every year I get closer to that goal. In this post I’ll share ten activity ideas that get students lots of practice with using equations of linear models to solve problems. For more activity ideas on comparing the different ways functions are shown, check out “10 Activities to Make Comparing Functions Engaging.” My approach this year has taken a slight turn from years past. This time around, students spent a lot more time just understanding slope before we got into anything else. So far, they know slope better than my students in the past. Also, we’ve focused more on what the variables stand for in the slope intercept form. We’ve gone beyond just explaining it or taking notes on slope intercept form. Students review the slope intercept form and its components each day. One by one you can see a little light bulb go off for each of the students. They aren’t all there yet, but each day they move a little bit closer to the goal. I’m going to share some of the strategies that I use or that I want to use in the future to teach students how to write equations and to help them interpret if a story is talking about slope or y-intercept. . The list of activities: Write an Equation Mazes Linear Functions Match and Paste Whiteboard Graphic Organizer FREEBIE Linear Functions Task Cards Writing Equations of a line online practice Write Equations from Situations Introduction to Linear Functions Task Cards Representing Linear Equations Worksheet Set of Worksheets Stop That Creature Let’s dive in These activities have a wide range of uses. They can be used as the first practice that students do. Also, they work well as cyclical review down the road. They are all engaging ways to get students involved in their own learning. Most of the time they lose track of how much work they’re doing because they’re enjoying themselves at the same time. Write an Equation Mazes At some point last year I realized that students just need a lot of practice writing equations from tables, stories, and graphs. One of the best ways they I know to get them this type of practice is through a maze. In these three Write an Equations mazes students solve about 13-15 problems in 10 minutes and they get the repetitions they need. It helps them see the connection between an equation and the other ways the same math problem is expressed in tables, stories, and graphs. We use mazes as our daily bell ringer in my classroom and I find that it’s a perfect warm-up. If you haven’t already joined the Maze of the Month Club, it’s a great way to get an exclusive, free maze sent right to your inbox every month. Click here to get signed up. Sign me up for the Maze of the Month Club. Plus, when you join the MotM Club you get more fun ideas and freebies sent right to your email. Can’t wait to see you there. Linear Relationships Match and Paste One year I noticed that my students in math lab really struggled to see the difference between slope and y-intercept in the real world. They seemed to be guessing every time we read a story problem about it. It sounded something like this, “This number’s the… slope? I mean, y-intercept! Yeah, that’s it, it’s the slo…intercept.” You know what I mean- when students just yell out a random answer. So, I decided that we need to break this down even further. That’s why I created this match and paste activity where all students do is look at statements and then sort them into slope or y-intercept. We do this as a match and paste in our notebook the first time and then students have it as a reference to look back at later on. Then, I have students complete this activity a second time a couple of days later as a quick formative assessment. Also, when completing this sorting activity I have students underline the key works in each statement that let them know if it’s referring to slope or y-intercept. Whiteboard Graphic Organizer This graphic organizer inside a SmartPal has been a huge hit with my students, as well as with my colleagues. The graphic organizer has a graph, a table, and a place for an equation on it. You fill-in part of the graphic organizer and students have to fill-in the rest. It challenges students to think through the different aspects of slope and y-intercept. My kids love it and every students works to try and figure it out. Using this type of response for students gives you the opportunity to see where they struggle. I have each student show me their work after they finish. If they have it right I tell them, and if they have it wrong then they try again. Sometimes I’ll see that a third or half the class is making the same mistake. That quick formative data gives me a chance to address the misconceptions and fix it right on the spot. You can download a free copy of this graphic organizer here. Because of the strong connection between the two topics, I share more ideas about how to use this download, and the graphic organizer itself, in the post “10 Activities to Make Comparing Functions Engaging.” Task Cards I have to admit that I’ve had a hard time finding activities to practice this standard. When I can’t find an activity that fits my students’ needs, then I usually make one. That is how these task cards were born. I needed my students to get more practice, and task cards are such a versatile tool that I like to use them for every topic we study. In this set, the first 8 task cards gets students to practice writing equations from tables, graphs, and stories. The rest of the cards have students identify and write about the different parts of the slope intercept form. You can see an example below: Task cards are so versatile. They can be used as a partner practice activity or you can save them for class warm-ups. Also, you can use them to do whole class practice with student whiteboards. Either way, usually I model the thought process for a couple of task cards before I have students start working on them. Writing Equations of a line online practice I used this online practice for writing equations as an anticipatory set. It took a little longer than I like for anticipatory sets, but it worked great. Every student was working on a different problem at the same time and they wanted to get it right. They had to be careful about their precision and they couldn’t just throw an answer down. It seemed to help the most for the kids that have struggled to see the difference between slope and y-intercept. I think I’ll use it again in a couple of months as a review for this topic. Write Equations from Situations When you click on this download you’ll find a few different activities. There’s an activity called “Match Races with Their Equations” on page 4. They have a series of 10 stories and students write the equations. You have the option of giving the students the answers and letting them choose from the possible answers. That’s not how I use them. Instead, I really want to push my students to be able to write the equations from the stories on their own. This download also has some other practice activities with stories, equations, and graphs. The downside is that there’s no answer key included. You could use these activities as a practice activity, homework, or an exit ticket. Introduction to Linear Functions Task Cards This free set of task cards on Free to Discover’s blog can be used to get students more practice with linear functions. These cards feature problems that get students practice with writing linear functions from tables, finding the missing value on a table, identifying independent and dependent variables, and describing the range. You could use these task cards as warm-up problems or have students complete them with a partner for practice. Pick a Card and Representing Linear Equations Worksheet When I find worksheets like this one, I have a few go to games that I play with them just to add a little bit of excitement. My latest game is called Pick a Card. I show a problem from the worksheet on the doc cam and students work it out on their whiteboards. When everyone is ready I ask 3 to 4 questions about that problem to ask students. Questions might include: What is the slope? What is the y-intercept? Describe what the slope represents. Each student who answered gets to pick a playing card at random. Before they choose, the rest of the students have to decide who they think will get the highest card before the cards are chosen. Then, I have each of the students pick a card. When they reveal what they’ve picked, the person with the highest cards gets 1,000 points, then it’s 500 points for the second highest, and 0 for lowest. The rest of the class gets the same amount of points as the person they chose before cards were picked. It’s silly, but students get a real kick out of it and they’re more engaged when practicing the math. Set of Worksheets By clicking on this link you’ll find 10 worksheets that will help students practice with a variety of skills related to linear functions. Some of the skills include: Finding the slope from a graphed line Finding the slope and y-intercept from a linear equation Graphing lines Working with linear equations If you need some skill building practice, you’ll find a wide variety here. These worksheets aren’t super engaging, but there are some ways to liven them up. One thing that I do with them is use them as an exit ticket and then the next day have students analyze responses using the “My Favorite No” strategy. This means that I find the work of someone who got the question wrong, but shows a great example of a common misconception. The students have to find the mistake and fix it. Error analysis is a great way to have students think about their own math thinking. Stop That Creature This Stop That Creature game from PBS Kids gets students identifying the rule of functions using a table and a “function machine”. Students have to identify the rule and enter it into the machine to stop the clones. There’s a lot of cool graphics. What a fun way for kids to practice figuring out the rule of a linear function. I would use this as a fast finisher activity during my linear functions unit or as a review game later in the year. Try one thing With the topic of understanding linear functions, there are so many skills to practice. Don’t get overwhelmed! Remember that students just have to build on previous understanding. You don’t have to do everything all at once- you can chunk things out. Try one of the activities above and see how your student respond to it. Once you have tried one and had success then you might want to try another one. Good luck in your math teaching journey. Thanks so much for reading. Until next time! Footer Privacy Policy Recent Posts 10 Ways to Help Students Get Dot Plots, Histograms & Box Plots 12 Mean, Median, Mode & Range Activities that Rock 9 Volume of Prisms Activities Your Students Will Love 10 Sure-fire Ways to Practice Area of Triangles & Quadrilaterals 10 Engaging Exponents & Expressions Activities We use cookies on our website to give you the most relevant experience by remembering your preferences and repeat visits. 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3213	https://math.stackexchange.com/questions/505393/how-many-semantically-different-boolean-functions-are-there-for-n-boolean-variab	logic - how many semantically different boolean functions are there for n boolean variables? - 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 how many semantically different boolean functions are there for n boolean variables? Ask Question Asked 11 years, 11 months ago Modified6 months ago Viewed 82k times This question shows research effort; it is useful and clear 36 Save this question. Show activity on this post. In short, this is an assignment question for a course I am taking - the exact wording is this: "Given n Boolean variables, how many 'semantically' different Boolean functions can you construct?" Now, I had a crack at this myself - and got pretty stuck. The question doesnt state how many boolean operators there are (and, or, xor, nand, nor, iff, implies, not) nor does it state whether brackets should be used, i.e. a ^ (b v c) is different from (a ^ b) v c. So, my question for you is - is this question possible given the limited information available? Is it going to be something like n x n x where x is the number of boolean operators. Any direction here would be greatly appreciated. logic boolean Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Mar 17 at 18:53 Tankut Beygu 4,618 2 2 gold badges 12 12 silver badges 25 25 bronze badges asked Sep 26, 2013 at 1:15 Zack NewshamZack Newsham 537 1 1 gold badge 5 5 silver badges 8 8 bronze badges 11 What do you know about constructing Boolean functions from the course? It seems to me that you're on the right track when you mention a function n x n x. How about this: How many different possible functions are there if you have zero variables? Think of the possibilities. How many different possible functions are there if you have one variable? Think of the possibilities. Now how many are there when you have two variables? Think of the possibilities. Now add up all of the possibilities. Does this help?Matt Groff –Matt Groff 2013-09-26 01:20:15 +00:00 Commented Sep 26, 2013 at 1:20 I did try that initially - however, the list got quite large. For 1 variable the answer is obviously 2 (a and ¬a) for 2 variables - its much larger, (a ^ b, a V b, etc) followed by the ¬ on each side, then on both sides. For 3 variables - its even worse, as with the example above a ^ (b v c) is semantically different from (a ^ b) v c Zack Newsham –Zack Newsham 2013-09-26 01:45:33 +00:00 Commented Sep 26, 2013 at 1:45 I'm still wondering about what you've covered. For instance, you may have covered truth tables. The main reason I'm wondering is because of how we, together, can construct Boolean functions. You've got the concept/example already that for one variable, you have either a a or ¬a¬a. Now, the trick seems to be to forget about "what the function is", and instead to concentrate on "what we can get". By this I mean for two variables, we can get a a or ¬a¬a for a a, and b b or ¬b¬b for b b. How many combinations can you get between the two? (Each can take on one of two values)Matt Groff –Matt Groff 2013-09-26 02:20:34 +00:00 Commented Sep 26, 2013 at 2:20 We have covered truth tables, and proof by contradiction and induction. From what you are saying it sounds like "semantically different function" is a unique entry in a truth table, however the truth table for a ^ b looks the same as the truth table for a V b, however each of these is "semantically different", correct?Zack Newsham –Zack Newsham 2013-09-26 02:27:04 +00:00 Commented Sep 26, 2013 at 2:27 5 Yes. For the case of Boolean variables, there are really only 2 2 n 2 2 n combinations. Either a particular combination out of the 2 n 2 n entries in a truth table is true, or it is not. Thus the 2 2 n 2 2 n total combinations.Matt Groff –Matt Groff 2013-09-26 03:18:48 +00:00 Commented Sep 26, 2013 at 3:18 \|Show 6 more comments 9 Answers 9 Sorted by: Reset to default This answer is useful 32 Save this answer. Show activity on this post. This question, in a sense, is a question of combinations. We can start with a single-valued function of Boolean variables. I claim that there are 2 n 2 n combinations of a single-valued function. For instance, if we start with one variable, there are two combinations; namely, a a and ¬a¬a. If we have two variables, there are four combinations. This is because we can have, for a a, either a a or ¬a¬a. Then, for b b, we can have either b b or ¬b¬b. So there are four combinations between these two variables. Similarly, for three variables, there are 2×2×2=2 3 2×2×2=2 3 combinations between these variables. Now, to consider the set of ALL Boolean functions, we have to consider again each of these combinations. We can say that there are 2 combinations 2 combinations different combinations between Boolean variables. This is because, for each combination, it can be true or false. So in the paragraph above, we have stated that there are 2 n 2 n combinations between the variables. Each of these combinations can be true or false for a particular variable assignment. So, again, we get 2 combinations=2(2 n)2 combinations=2(2 n) combinations between them all. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 26, 2013 at 4:15 Matt GroffMatt Groff 5,739 5 5 gold badges 33 33 silver badges 53 53 bronze badges 2 I had read in a textbook that there are in general k k n k k n different functions you can make in a k-valued logic with n variables, but I had never seen a proof. Thank you for writing this, as it has become a reference for my ~2 year long question.user400188 –user400188 2020-09-11 03:39:12 +00:00 Commented Sep 11, 2020 at 3:39 Glad I didn’t have to wait ~ 2 years to understand this question as well. Thank you @Matt Groff. I was vexed by this since last week and it’s consuming me, rethinking the point of whether I should self-study Discrete Math.MikeTheSapien –MikeTheSapien 2024-12-04 01:03:27 +00:00 Commented Dec 4, 2024 at 1:03 Add a comment\| This answer is useful 26 Save this answer. Show activity on this post. Let's reverse-engineer this: In the case of n=2 n=2 there are 2(2 n)=2 4=16 2(2 n)=2 4=16 distinct functions: F 0...F 15 F 0...F 15. Below you can find the resulting truth table. Each column represents the outcome for function F 0...F 15 F 0...F 15 F 0(0,0)=0 F 0(0,0)=0 F 0(0,1)=0 F 0(0,1)=0 .... F 15(1,1)=1 F 15(1,1)=1 ``` A B\| F0 F1 F2 F3 F4 F5 F6 F7 0 0\| 0 0 0 0 0 0 0 0 0 1\| 0 0 0 0 1 1 1 1 1 0\| 0 0 1 1 0 0 1 1 1 1\| 0 1 0 1 0 1 0 1 A B\| F8 F9 F10 F11 F12 F13 F14 F15 0 0\| 1 1 1 1 1 1 1 1 0 1\| 0 0 0 0 1 1 1 1 1 0\| 0 0 1 1 0 0 1 1 1 1\| 0 1 0 1 0 1 0 1 ``` To verify why there are 16 distinct functions, we start with our first function F 0 F 0 which maps any given pair (A,B)(A,B) to F A L S E F A L S E. For the next function F 1 F 1 it is sufficient to differ at only one position in order to be become distinct from F 0 F 0=>F 1(A=1,B=1)=1=>F 1(A=1,B=1)=1. The same is true for F 2 F 2 with regard to F 1 F 1 and so forth. Finally we arrive at F 15 F 15 which becomes distinct from F 14 F 14 by simply mapping all inputs to TRUE. Let's also do the math: How many different ways can you pick k items from n items set with repetition ( k=2 k=2 n=2 n=2 )? n k n k = 2 2=4 2 2=4. There are 4 ways to form a pair (A,B)(A,B) hence F(A,B)F(A,B) yields also 4 outcomes k 1,k 2,k 3,k 4 k 1,k 2,k 3,k 4. How many different ways can you pick k items from n items set with repetition ( k=4 k=4 n=2 n=2 )? n k n k = 2 4=(2(2 2))=16 2 4=(2(2 2))=16. Thus 16 semantically different boolean functions. See the respective function names and symbols below. (Source: function symbol name F0 0 FALSE F1 A ^ B AND F2 A ^ !B A AND NOT B F3 A A F4 !A ^ B NOT A AND B F5 B B F6 A xor B XOR F7 A v B OR F8 A nor B NOR F9 A XNOR B XNOR F10 !B NOT B F11 A v !B A OR NOT B F12 !A NOT A F13 !A v B NOT A OR B F14 A nand B NAND F15 1 TRUE Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 25, 2021 at 20:45 answered Mar 3, 2014 at 21:05 simousimou 371 4 4 silver badges 5 5 bronze badges 1 So you are saying that given all n length binary sequences (whose number is 2 n 2 n) we can get a single output of length m=2 n m=2 n and number of such output is then 2 m=2 2 n 2 m=2 2 n.Ankit Seth –Ankit Seth 2018-09-12 07:33:02 +00:00 Commented Sep 12, 2018 at 7:33 Add a comment\| This answer is useful 7 Save this answer. Show activity on this post. Think about the truth table, say for a concrete n n like n=3 n=3. There are 2 3 2 3 sequences of length 3 3 made up of 0 0's and/or 1 1's. More generally, there are 2 n 2 n sequences of 0 0's and/or 1 1's of length n n. To make a Boolean function, for each of these sequences, we can independently choose the value of our function at the sequence. Thus there are 2(2 n)2(2 n) Boolean functions of n n variables. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 26, 2013 at 1:20 André NicolasAndré Nicolas 515k 47 47 gold badges 584 584 silver badges 1k 1k bronze badges 4 I'm not sure I quite follow, if we choose three variables: a,b and c. How do you get 256 combinations? Also, this doesn't take into account the ways they can be combined (the number of boolean operators)Zack Newsham –Zack Newsham 2013-09-26 01:50:40 +00:00 Commented Sep 26, 2013 at 1:50 1 It is Boolean functions we are counting, not combinations of operators. There are infinitely many ways to express each Boolean function. For a silly example, p 1 p 1, p 1∧p 1 p 1∧p 1, p 1∧p 1∧p 1 p 1∧p 1∧p 1 and so on all give us the same Boolean function. So does \p 1∨p 1\p 1∨p 1, and many many others. And yes, there are 256 256 different truth tables in the case n=3 n=3.André Nicolas –André Nicolas 2013-09-26 02:01:06 +00:00 Commented Sep 26, 2013 at 2:01 Ok, so it sounds like you are saying it is the entries in the truth table we are counting, if p1 V p1 gives us the same function as p1 ^ p1 ^ p1. Does this reduce to a situation where "semantically different" only corresponds to the number of different assignments in a truth table? a ^ b ^ c and a V b V c both will create identical truth tables? I'm still not sure where 256 comes from, surely there are only 001, 010, 011, etc.Zack Newsham –Zack Newsham 2013-09-26 02:31:43 +00:00 Commented Sep 26, 2013 at 2:31 1 Yes, semantically different means different truth table. Because I am lazy, take n=2 n=2. So we have 00 00, 01 01, 10 10, 11 11. Now on the right hand side of the truth table, the truth value of f f at 00 00 could be 0 0 or 1 1. For each of these choices, the truth value of f f at 01 01 could be 0 0 or 1 1. And so on. So 2 4 2 4 choices.André Nicolas –André Nicolas 2013-09-26 02:52:50 +00:00 Commented Sep 26, 2013 at 2:52 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Here let me add my part. Consider we are having two logic variables a and b. These two variables may be 0 or 1. So the total possibilities are ``` a \| b 0 \| 0 = a'b' 0 \| 1 = a'b 1 \| 0 = a b' 1 \| 1 = a b 2_ X 2__ = 4 possibilities 0 or 1 0 or 1 ``` Now its time to find total number of functions for 2 variables!...We found 4 terms for 2 variables in above example, isn't it? So for A Equation or Function may contain any number of 4 terms that we have found above. Like this a'b' + a'b - contain two terms a b' - contain single term a'b' + a'b + a b - contain three terms a'b' + a'b + ab' + a b - contain four terms (maximum possibilities) So the total combination(count) of functions for 2 variables are ``` _2_ X _2_ X _2_ X _2_ = 16 possibilities term or no term term or no term term or no term term or no term - - - - = 0(false) ab - - - = ab - a'b - - = a'b - - ab' - = ab' - - - a'b' = a'b' ab a'b - - = ab + a'b - a'b ab' - = a'b + ab' - - ab' a'b' = ab' + a'b' ab - ab' - = ab + ab' ab - - a'b' = ab + a'b' ..... //some more possible combinations like this, ( to lazy to type it;) ) and finally it would be like ab a'b ab' a'b' = 1 (True) ``` So The combination(count) of functions that can be formed by n variables is 2^2n n = 2 2 ^ 22 = 16 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 11, 2020 at 0:47 answered Jan 6, 2017 at 12:05 leopragileopragi 161 3 3 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I do not know about semantically correct but it is pretty easy to compute even more general case: how many are there functions of k k arguments, each may take n values and function may produce one of m outcomes for every combination. As this blog says, your arguments provide n k n k combinations of values. You can interpret your function as a function of single argument, which takes one of n k n k values. Now, you start by placing all possible functions into one group [all possible functions] and apply the first value of n k n k. The functions of the group will respond with m m various outcomes. This way, you have resolved one group into m m subgroups. [responded with 1][responded with 2] ... [responded with m] On the next step, you apply next value of n k n k to each subgroup. Again, functions in those subgroups will split into m m further subgroups. After two steps, you have resolved all your functions into m 2 m 2 subgroups. After applying all n k n k input values, you have resolved the initial group into m n k m n k subgrops. You have no more tests to apply. You, therefore, consider all functions in the resulting subgroups identical. You have got m n k m n k different functions. Isn't it beatiful? In case the function arguments and values belong to the same type (type is a range of values that variable can take), you have m=n m=n and n n k n n k different functions. Particularly, in case the type is binary, we may have 2 2 k 2 2 k functions. I am sure that all functions are realizable (there is a notion of functional completness), and, thus are semantically correct if that is what you mean. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 16, 2014 at 13:08 ValVal 1,519 1 1 gold badge 15 15 silver badges 27 27 bronze badges 1 I like your answer the best. Others have hinted at this I guess. Let F F be the set of functions from a finite set X X to a finite set Y Y. Let \|A\|\|A\| denote the number of elements in a set A A. In this case we have \|X\|=2 n\|X\|=2 n and \|Y\|=2\|Y\|=2. In general, \|F\|=\|Y\|\|X\|\|F\|=\|Y\|\|X\|, so in this case \|F\|=2 2 n\|F\|=2 2 n. Yes, an old question. Don't care.awsnap –awsnap 2024-04-15 02:34:21 +00:00 Commented Apr 15, 2024 at 2:34 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Consider this: There are n variables which means there are 2 n entries in the truth table. The number of variables determine the number of rows in the truth table A minterm exists for every row in the truth table. This means that there are 2 n minterms. The column of minterms on the far right. Each function can be written as a sum of minterms. The minterm is either in the sum or not in the sum. This means that such a function can be specified with a string of 1's and 0's with the following meaning: if at position i there is a 1 then the i-th minterm is part of the function sum. if there is a 0 instead then that minterm is not part of the sum. Each function can be expressed as a sum of one or more minterms It is clear that having 2 n minterms makes this function defining string 2 n positions long. At each position there can be a 1 or a 0. Hence the total number of different strings is 2 2 n which is the total number of different functions and also the total number of different sums that the minters can be combined into. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 12, 2018 at 10:41 KozmonautKozmonaut 31 1 1 bronze badge Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. This is a classic problem in circuit theory. See One of the difficulties in getting started is deciding how you'll define "semantically equivalent"; then you'll get pretty deep into group theory before you'd be able to describe your answer accurately in a few paragraphs on e.g. StackExchange. Golomb's 1959 paper in IRE Trans Circuit Theory is a good place to start: "The Boolean functions of k variables, f(x1, x2, ..., xk), fall into equivalence classes (or families) when two functions differing only by permutation or complementation of their variables are considered equivalent. The number of such families is easily computed, as illustrated by Slepian [l]. The next step is to discover the invariants of the logic families, and determine to what extent they characterize the individual families. Given the class decomposition, one also wishes to select a "representative assembly", with one delegate from each family. That is, canonical forms for the logics are sought, with every family having its characteristic canonical form..." Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 25, 2018 at 21:51 answered Mar 25, 2018 at 21:44 Clark ThomborsonClark Thomborson 11 3 3 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I know the answers provided are already correct and really nice. However, there is a simple way of going about it which has not been shared, which is the following: A Boolean function: {0,1}n→{0,1}{0,1}n→{0,1} converts any of the possible 2 n 2 n bits onto 0 0 or 1 1. Think of each bit as a cell, and the 1 as placing a ball in a cell. Then, let a demon crafting functions decide where to put the ball or not. Being a binary decision that the demon has to repeat 2 n 2 n times, if follows that it has 2 2 n 2 2 n functions to choose from. The argument is the same for any k≥2 k≥2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Apr 12, 2024 at 17:37 answered Apr 8, 2024 at 15:11 LautaroLautaro 21 3 3 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Adding an answer that might be repetitive, but perhaps will be useful for its straightforwardness. A Boolean function on n n Boolean variables is a map from a set of binary variables to a binary outcome. This means that a Boolean function is, informally, a rule that returns a 1 1 (or TRUE TRUE, etc.) given the outcome of those n n binary variables. If we want to maximize the number of different outcomes, we will have a unique function for every possible outcome of the n n variables. Let’s make this concrete and simple, then generalize. The NAND function for two variables X 1,X 2 X 1,X 2 returns 1 1 iff¬(X 1=1∧X 2=1)¬(X 1=1∧X 2=1); so, of our four possible outcomes for trials of X 1,X 2 X 1,X 2, we have... X 1 0 1 0 1 X 2 0 0 1 1 N A N D(X 1,X 2)1 1 1 0 X 1 X 2 N A N D(X 1,X 2)0 0 1 1 0 1 0 1 1 1 1 0 So, how did we construct the NAND function? We wrote down every possible outcome of a Boolean experiment with two trials. In general, with n n trials which each admit m m outcomes, we have m n m n values possible for the experiment (forgive the statistics language; that is how my brain works). Here, that becomes 2 n 2 n. While you can consider this result as an application of the "permutation with repetition" formula, as another comment points out, this can be understood as a sum of combinations, where we sum the number of ways to have 0,1,2,...n 0,1,2,...n successes (for more on this connection, see my answer here). Finally, we constructed the Boolean function by defining which of those 2 n 2 n patterns of successes were considered successes. In other words, we have a combination of combinations, or a combination where the number of objects available is itself a combination. There are, then, 2 2 n 2 2 n unique combinations of the outcomes of n n trials; since we said a Boolean function is unique if it returns 1 1 for a unique combination of outcomes of the experiment, there are 2 2 n 2 2 n Boolean functions available for n n variables. To see this in even more detail, consult Nisan and Schocken's Elements of Computing Systems, which tallies all 16 16 possible Boolean functions for just two variables. For example, an XOR function is defined by the combination (X 1=1∧X 2=0)∧(X 1=0∧X 2=1)(X 1=1∧X 2=0)∧(X 1=0∧X 2=1). It is tedious to list them all out, but you might try randomly picking a combination and looking up the name of that logic gate. Obviously, this number can get very large, very fast if we add more variables on. To pick the next simplest example, if n=3 n=3, there are 2 3=8 2 3=8 possible outcomes of an experiment with all three binary variables, and then 2 8=256 2 8=256 possible combinations of experiments which we could define to trigger a 1 1, so that many Boolean functions. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 17 at 18:53 Tankut Beygu 4,618 2 2 gold badges 12 12 silver badges 25 25 bronze badges answered Apr 18, 2024 at 20:40 gjmbgjmb 23 4 4 bronze badges Add a comment\| You must log in to answer this question. Protected question. To answer this question, you need to have at least 10 reputation on this site (not counting the association bonus). The reputation requirement helps protect this question from spam and non-answer activity. Start asking to get answers Find the answer to your question by asking. 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3214	https://www.rileychildrens.org/health-info/lower-respiratory-tract-infection	Skip to Main Content Hospital Visitor Guidelines : Riley Children's Health has implemented hospital inpatient visitor restrictions to protect patients and help prevent the spread of respiratory viruses. View full details for Riley Hospital for Children at IU Health and Riley Children's Health at IU Health North. 317.944.5000 Schedule an Appointment Refer a Patient Plan My Visit Pay a Bill Make a Donation Volunteer Find a Career Health Information Find a Doctor Departments & Programs Support Services Contact & Locations Refer a Patient My IU Health Patient Portal Search 317.944.5000 My IU Health Patient Portal Schedule an Appointment Refer a Patient Plan My Visit Pay a Bill Make a Donation Volunteer Find a Career Search Lower Respiratory Tract Infection A lower respiratory tract infection (RTI) occurs when there is an infection of the lungs, specifically in the lower airways. This infection is usually caused by a virus, but it can also be caused by bacteria or other less common organisms. Common lower RTIs in infants and young children include: Flu. The flu (influenza) is a common viral infection that occurs most often during the winter months. It can be more dangerous to your health if you are very young or elderly. Viral Bronchiolitis. Bronchiolitis is an inflammation of the lining of the bronchioles (the very small passages through which air flows to and from the lungs). This condition is very common in infants and caused by several viruses, including respiratory syncytial virus (RSV). Pneumonia. Pneumonia is an infection that causes inflammation of the air sacs in one or both of the lungs. Its symptoms can range from mild to severe enough to require hospitalization. The main symptom of a lower RTI is cough, which can be severe.Your child may have a dry cough or a wet cough. Even if it is a wet cough, he or she may not be able to cough up phlegm/mucus. Other symptoms of a lower RTI include: Fever Tightness in the chest or chest pain Breathing quickly or in an irregular pattern Difficulty catching your breath Wheezing Overall change in well-being (decreased energy, appetite and fluid intake) Diagnosis of Lower Respiratory Tract Infection Doctors at Riley at IU Health may use one or more of the following exams and tests to diagnose a lower RTI: Medical history and physical exam. The doctor will review your child's medical history and conduct a complete physical exam to check for the various signs, symptoms and physical findings of a lower RTI. Nose swab. The doctor may swab your child’s nose to check for the flu or other viruses. Remember that if your child has a viral infection, antibiotics will not be helpful. X-ray. If your child’s doctor suspects pneumonia, he or she might order an X-ray of the chest to check for this condition. Blood test. The doctor may use a blood test to check for signs of inflammation or signs that your child's body is fighting an infection. Bronchoscopy. If your child has a chronic health condition or has had repeated cases of pneumonia, the doctor may plan for a procedure called flexible bronchoscopy. Your child will be sedated and a thin, flexible scope (tube) will be passed through the nose into the lower airway to evaluate the major lung branches and obtain a mucus sample. This sample will help identify the cause of the pneumonia or respiratory tract infection. Treatments Treatments If your child has a lower RTI caused by a virus, the doctor will make the following recommendations: Drink plenty of clear fluids Get plenty of rest Control the fever with acetaminophen or nonsteroidal anti-inflammatory drugs (NSAIDs) such as ibuprofen (as directed by your child's pediatrician based on age and medical history) If your child has a lower RTI caused by bacteria, the doctor will likely prescribe antibiotics to fight the infection. If your child’s condition worsens, he or she may need to be hospitalized. Doctors may administer oxygen therapy or breathing treatments, which can help your child breathe easier. Key Points to Remember Key Points to Remember A lower RTI occurs when there is an infection in the lower airways. Common lower RTIs in infants and young children include the flu, viral bronchiolitis and pneumonia. The main symptoms of a lower RTI are a persistent cough, fever and sometimes difficulty breathing. Treatment for a lower RTI caused by a virus includes drinking plenty of fluids and getting plenty of rest while the virus runs its course. If the infection is caused by bacteria, antibiotics may be prescribed. Support Services & Resources Support Services & Resources Visit the trusted websites below for more information and support for lower RTI. We offer a broad range of supportive services to make life better for families who choose us for their children's care. Learn More About Riley Support Services Centers for Disease Control and Prevention This U.S. government website provides helpful information about the flu. American Lung Association Learn more about pneumonia, including risk factors and causes. Departments Treating This Condition Departments Treating This Condition Ear, Nose & Throat Pulmonology & Respiratory Care Your opinion about this website is important to us. Would you be willing to answer a few questions to help us evaluate and improve our website?
3215	https://mathspace.co/textbooks/syllabuses/Syllabus-1155/topics/Topic-21886/subtopics/Subtopic-279711/	Book a Demo Log inSign upBook a Demo United States of America Grade 6 1.05 Problem solving with GCF and LCM LessonPractice Introduction We have already learned how to find the greatest common factor (GCF) and the least common multiple (LCM) of two or more numbers. We'll now solve real-life problems involving GCF and LCM. Ideas Problem solving with GCF and LCM Problem solving with GCF and LCM There are many applications of GCF and LCM in real-life. Some examples are when we want to group or split things together, to know when an event will repeat, or when two things will happen again at the same time. But, how can we tell if a word problem requires us to use GCF or LCM to solve? When we are asked to find the GCF between two or more numbers, we are being asked what is the biggest number that they can all be divided by that leaves no remainder (or leftovers). When we are asked to find the LCM, we are being asked to find the smallest multiple that is shared by two numbers. We will use GCF for problems that require us to: Split things to smaller groups or sections like in packs, in bags or in boxes Equally share any quantity of items into their largest grouping Arrange something into rows or columns or sets Some keywords that can help us identify if a problem requires finding GCF are: Biggest, greatest, highest or maximum Dividing, sharing, distributing or cutting into pieces We will use LCM to solve problems that ask us to: Find when an event will be repeated or happen again Find the number of pieces to collect multiple items in order to have enough Some keywords that can help us identify if a problem requires finding LCM are: Smallest, least, or minimum Repeated over and over Next Let's practice with some problems below. Examples Example 1 Mario harvested 18 apples and 12 oranges from his orchard. He wants to group the fruits together in baskets to make identical packs to sell in the market. a What is the greatest number of baskets he should use? Worked Solution Create a strategy Analyze the problem by understanding what is asked. Look for keywords to help decide whether finding the GCF or LCM is needed to solve the problem. Apply the idea The problem asks for the greatest number of baskets Mario should use. The key word "greatest" gives us a clue that we will find the GCF. Listing the factors of 18 and 12: 18:1,2,3,6,9,18 12:1,2,3,4,6,12 The greatest common factor is 6. Mario should use 6 baskets. b Determine the number of apples and oranges he should place in each basket. Worked Solution Create a strategy Dividing the number by its GCF determines the number of apples and/or oranges in each basket. Apply the idea \| \| \| \| --- \| Number of apples \| = \| 18÷6 \| \| \| = \| 3 \| \| \| \| \| --- \| Number of oranges \| = \| 12÷6 \| \| \| = \| 2 \| There should be 3 apples and 2 oranges in each basket. Example 2 A party store sells balloons in packs of 10 and balloon sticks in packs of 12. a What is the least number of balloons and balloon sticks Jenny should buy so that there will be one balloon for each balloon stick? Worked Solution Create a strategy Analyze the problem by understanding what is asked. Look for keywords to help decide whether finding the GCF or LCM is needed to solve the problem. Apply the idea We are asked to find the least number of balloons and balloon sticks that Jenny should buy. The keyword is "least" which gives us a clue that we solve the problem by getting the LCM. By using the listing method, we can see that the first seven multiples of 10 and 12 are: 10:10,20,30,40,50,60,70 12:12,24,36,48,60,72,84 The least common multiple of 10 and 12 is 60. Jenny should buy 60 balloons and ballon sticks. b How many packs of balloons and balloon sticks should Jenny buy? Worked Solution Create a strategy Divide 60 by the number of pieces in each pack. Apply the idea \| \| \| \| --- \| Number of packs of balloons \| = \| 60÷10 \| \| \| = \| 6 \| \| \| \| \| --- \| Number of packs of balloon sticks \| = \| 60÷12 \| \| \| = \| 5 \| Jenny should buy 6 packs of balloons and 5 packs of balloon sticks. Idea summary We will use GCF for problems that require us to: Split things to smaller groups or sections like in packs, in bags or in boxes Equally share any quantity of items into their largest grouping Arrange something into rows or columns or sets We will use LCM to solve problems that ask us to: Find when an event will be repeated or happen again Find the number of pieces to collect multiple items in order to have enough Some keywords that can help us identify if a problem requires finding GCF or LCM are: \| GCF \| LCM \| --- \| \| biggest \| smallest \| \| greatest \| least \| \| highest \| minimum \| \| dividing \| repeated \| \| sharing \| next \| \| distributing \| \| \| cut into \| \| Outcomes 6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. What is Mathspace About Mathspace
3216	https://translate.academic.ru/bring+about+a+rapprochement+between/ru/xx/	bring+about+a+rapprochement+between — с русского на все языки academic.ru RU EN DE ES FR Запомнить сайт Словарь на свой сайт Словари и энциклопедии на Академике Найти! Толкования Переводы Книги Игры ⚽ Перевод: с русского на все языки со всех языков на русский Со всех языков на: Русский С русского на: Английский bring+about+a+rapprochement+between ТолкованиеПеревод 1добиться восстановления дружественных отношений General subject:bring about a rapprochement between Универсальный русско-английский словарь>добиться восстановления дружественных отношений См. также в других словарях: rapprochement — n. to bring about a rapprochement between [ræ prɒʃmɑːŋ] to bring about a rapprochement between … Combinatory dictionary Union between Sweden and Norway — United Kingdoms of Sweden and Norway Förenade konungarikena Sverige och Norge De forenede Kongeriger Norge og Sverige Personal union ← … Wikipedia Relations between the Catholic Church and the state — The relations between the Catholic Church and the state have been constantly evolving with various forms of government, some of them controversial in retrospect. In its history it has had to deal with various concepts and systems of governance,… … Wikipedia Hardenberg, Karl August, Fürst von — ▪Prussian statesman Introduction also called (until 1814) Freiherr (baron) von Hardenberg born May 31, 1750, Essenrode, near Gifhorn, Brunswick died Nov. 26, 1822, Genoa Prussian statesman and administrator, who preserved the integrity… … Universalium GOLDZIHER, IGNAZ — (Isaac Judah; 1850–1921), Hungarian scholar, one of the founders of modern Islamic scholarship. Goldziher, born in Szekesfehervar (Stuhlweissenburg), studied Arabic manuscripts at Leyden and Vienna, and traveled in Egypt, Palestine, and Syria… … Encyclopedia of Judaism Bohr , Niels Hendrik David — (1885–1962) Danish physicist Niels Bohr came from a very distinguished scientific family in Copenhagen, Denmark. His father, Christian, was professor of physiology at Copenhagen and his brother Harald was a mathematician of great distinction.… … Scientists Guy XIV de Laval — Infobox Military Person name=Guy XIV de Laval caption= born=28 January 1406 died=2 September 1486 placeofbirth= placeofdeath=Châteaubriant allegiance= branch= serviceyears= rank= unit= commands= battles=Hundred Years War (Battle of La Brossinière … Wikipedia SELBSTEMANZIPATION — SELBSTEMANZIPATION, the first Zionist newspaper in Western Europe, in German. The idea of the newspaper originated in the group centered on the first Zionist student group in Western Europe, kadimah , and its name reflected the admiration felt by … Encyclopedia of Judaism Marguerite-Élie Guadet — (20 July 1758 – 17 June 1794) was a French political figure of the Revolutionary period. Rise to prominence Born in Saint Émilion, Gironde, Aquitaine, he had already gained a reputation as a lawyer in Bordeaux by the time of the Revolution. In … Wikipedia Emmanuel Hiel — (Sint Gillis Dendermonde, 30 May, 1834 Schaarbeek , 27 August, 1899), was a Flemish Dutch poet and prose writer.During his life he held various jobs, from teacher and government official to journalist and bookseller, busily writing all the time… … Wikipedia Sharif Ahmed — Infobox Military Person name= lived= birth date and age\|1964\|07\|25 placeofbirth= Chabila, Somaliacite web \|date=2006 05 17\| url = id=4971\| title = Interview with Head of Somalia s Islamic courts… … Wikipedia 18+ © Академик, 2000-2025 Обратная связь: Техподдержка, Реклама на сайте 👣 Путешествия Экспорт словарей на сайты, сделанные на PHP, Joomla, Drupal, WordPress, MODx. Пометить текст и поделиться Искать во всех словарях Искать в переводах Искать в Интернете Поделиться ссылкой на выделенное Прямая ссылка: Нажмите правой клавишей мыши и выберите «Копировать ссылку» 1
3217	https://files.ele-math.com/articles/mia-16-09.pdf	Mathematical Inequalities & Applications Volume 16, Number 1 (2013), 127–142 doi:10.7153/mia-16-09 CUBIC AND QUARTIC CYCLIC HOMOGENEOUS INEQUALITIES OF THREE VARIABLES TETSUYA ANDO (Communicated by I. Peri´ c) Abstract. We determine the geometric structures of the families of three variables cubic and quartic cyclic homogeneous inequalities of certain classes. These structures are determined by studying some real algebraic surfaces. 1. Introduction Symmetric or cyclic homogeneous polynomial inequalities are one of the most el-ementary inequalities. But they are not studied well. We may dare say that we know only a few even about three variables cyclic homogeneous inequalities. The aim of this article is to present a geometric method in order to deal with the cubic and quartic cyclic homogeneous inequalities in three variables. We sketch the history. Articles on poly-nomial inequalities are very few. One of the most important symmetric homogeneous inequalities is Muirhead’s inequality published in 1902 (), which says that if l1 + l2 + ···+ ln = m1 + m2 + ···+ mn, l1 + l2 + ···+ lk ⩾m1 + m2 + ···+ mk (∀k = 1, 2,..., n −1), then the inequality ∑ σ∈Sn al1 σ(1)al2 σ(2)···aln σ(n) ⩾∑ σ∈Sn am1 σ(1)am2 σ(2)···amn σ(n) holds for any a1 ⩾0,..., an ⩾0. The following Schur’s inequality is also discovered around this age: (ad + bd + cd)+ abc(ad−3+ bd−3 + cd−3) ⩾(ad−1b + bd−1c+ cd−1a)+ (abd−1 + bcd−1 + cad−1) holds for all a ⩾0, b ⩾0, c ⩾0 and integers d ⩾3. It is mystery that no generalization of Schur’s inequality is known yet in the case of more than three variables, except the case of degree three (see p.271 Q4). During about a hundred years, there is no essential development. Recently, Cˆ ırtoaje discovered some important theorems about three variable homogeneous inequality. One of them is as the following: Mathematics subject classiﬁcation (2010): Primary 26D05. Keywords and phrases: Homogeneous inequality. c ⃝ , Zagreb Paper MIA-16-09 127 128 T. ANDO THEOREM. () (1) Let f(x, y, z) be a quartic symmetric homogeneous poly-nomial. Then, f(x, y, z) ⩾0 for any x, y, z ∈R, if and only if f(1,0,0) ⩾0 and f(x,1,1) ⩾0 (∀x ∈R). (2) Let f(x, y, z) be a symmetric homogeneous polynomial with 3 ⩽deg f ⩽5. Then, f(x, y, z) ⩾0 for any x, y, z ⩾0, if and only if f(x,1,0) ⩾0 and f(x,1,1) ⩾0 (∀x ⩾0). He also obtained similar theorem for symmetric homogeneous polynomials with 6 ⩽deg f ⩽8. But we omit it because its statement is long. The following theorem is also fundamental. THEOREM. () Let p, q, r be any real numbers. The cyclic inequality (a4 + b4 + c4)+ r(a2b2 + b2c2 + c2a2)+ (p + q −r −1)abc(a +b+c) ⩾p(a3b + b3c+ c3a)+ q(ab3+ bc3 + ca3) holds for any a, b, c ∈R if and only if 3(1 + r) ⩾p2 + pq + q2. We analyze the above theorem using a convex cone. Let R+:= x ∈R x ⩾0 , Cd := ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ f(a,b,c) f is a cyclic homogeneous polynomial of degree d , such that f(a, b, c) ⩾0 for ∀a, b, c ∈R, and that f(1, 1, 1) = 0. ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ , C + d := ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ f(a,b,c) f is a cyclic homogeneous polynomial of degree d , such that f(a, b, c) ⩾0 for ∀a, b, c ∈R+ , and that f(1, 1, 1) = 0. ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ , Sd := f ∈Cd f is symmetric , S + d := f ∈C + d f is symmetric . Cˆ ırtoaje’s inequality implies that C4 is an ellipsoid cone in R4. We can also determine the structures of S4 and S + 4 , using theorems in . S4 is an elliptic cone in R3. The base of S + 4 is a domain in R2 enclosed by a part of the ellipse and two line segments. These are explained later. Note that Cd = Sd = 0 if d is odd. It is easy to see that C2 = C + 2 = S2 = S + 2 =R+ ·(a2 + b2 + c2 −bc−ca −ab), and these are a half line. In this article, we shall determine the structures of C + 3 , S + 3 and C + 4 . As consequences, S + 3 is a sector on a plane. C + 3 is a cone in R3 whose base is a domain in R2 enclosed by a part of quartic curve and a line segment. The base of CUBIC AND QUARTIC CYCLIC INEQUALITIES 129 C + 4 ⊂R4 is a domain in R3 enclosed by three surfaces, one is a part of the ellipsoid, the others are parts of ruled surfaces. The following inequalities can be proved as a direct corollary of this fact. Note that these are analogues of Schur’s inequality. Let a ⩾0, b ⩾0, c ⩾0, then the following hold: 3 √ 4 3 (a3 + b3 + c3)+ 3 − 3 √ 4 abc ⩾a2b + b2c+ c2a, (1.1) (a3 + b3 + c3)+ 16 √ 2+ 13−1 2 (a2b + b2c+ c2a) ⩾ 16 √ 2+ 13+ 1 2 (ab2 + bc2 + ca2), (1.2) (a4 + b4 + c4)+ 4 4 √ 3 3 −1 abc(a + b + c) ⩾4 4 √ 3 3 (a3b + b2c+ c3a), (1.3) (a4 + b4 + c4)+ α(a3b + b3c+ c3a) ⩾(α + 1)(ab3 + bc3 + ca3), (1.4) here α = 1.37907443362539958016··· is a root of 4α6 + 12α5 −48α4 −116α3 + 24α2 + 84α + 229 = 0. (a4 + b4 + c4)+ β(a3b2 + b2c2 + c2a2) ⩾(β + 1)(a3b + b3c+ c3a), (1.5) here β = 2.18452974131524781307··· is a root of 4β 5 + 19β 4 −32β 3 + 2β 2 −36β −229 = 0. (a4 + b4 + c4)+ γ(a3b + b3c+ c3a) ⩾(γ + 1)(a2b2 + b2c2 + c2a2), (1.6) here γ = 5.07790940231978661368··· is a root of 4γ5 + γ4 −68γ3 −172γ2 −192γ + 144 = 0. These inequalities are located on the boundary of C + d . Note that Schur’s inequality is located on the boundaries of S + d and C + d . We shall explain the outline of our theory. Let Si,j,k(a,b,c) := aib jck + bicjak + cia jck. Take an index set Id so that the set {Si,j,k (i, j, k) ∈Id is a base of the vector space f(a, b, c) f is a cyclic homogeneous polynomial of degree d . Deﬁne the holomorphic map ϕd : P2 R− →PN R (N = #I −1 = ⌈(d +1)(d +2)/6⌉−1) by ϕ(a : b : c) = Si,j,k(a, b, c) (i, j, k) ∈I . Then Xd := ϕd(P2 R) is (a closed domain of) a real projective surface of degree d . Consider Xd in an afﬁne space RN with the origin (1 : 1 : ··· : 1) ∈PN R , and take the convex cone Dd ⊂RN generated by Xd . Then Cd can be identiﬁed with the dual convex cone of Dd . It seems that Xd is the whole part of algebraic surfaces, but this is not true for symmetric polynomials. That is, it is only a closed domain of a surface. Similarly, let P2 +:= (a : b : c) ∈P2 R a ⩾0, b ⩾0, c ⩾0 , X+ d := ϕd(P2 +), and D+ d be the convex cone generated by X+ d . Then C + d can be identiﬁed with the dual convex cone of D+ d . Thus, if we study an algebraic surface Xd or its closed domain X+ d , we can determine the structure of the convex cone Cd or C + d , and we can obtain the sharpest inequalities. 130 T. ANDO It may be possible to determine the structures of C + d , Cd , S + d , Sd for d ⩾5. But the structure of Xd or X+ d is not so simple for d ⩾5. It may also possible to do similar observation for more than three variables inequalities, if we study the structure of higher dimensional projective varieties. Theoretically it will be possible, but the calculation is complicated. The author tried this in vain, and expects the research in the future. 2. Main Theorems We use the same notation as in the section 1, and we denote Si = Si(a,b,c) := ai + bi + ci, Si,j = Si,j(a,b,c) := aib j + bicj + cia j, U = U(a,b,c) := abc, Ti,j = Ti,j(a,b,c) := Si,j(a,b,c)+ S j,i(a,b,c). THEOREM 1. Let fs(a,b,c) := s2S3 −(2s3 −1)S2,1 + (s4 −2s)S1,2 −3(s4 −2s3 + s2 −2s+ 1)U, f∞(a,b,c) := S1,2 −3U. Then, the following hold. (1) The boundary of the convex cone C + 3 is R+ · fs s ∈[0, ∞] ∪ R+ ·f0+ R+ ·f∞ . (2) If f ∈C + 3 , then we can ﬁnd α , β , s ∈R+ such that f = αfs + βf∞. (3) Let f(a,b,c) = S3 + pS2,1 + qS1,2 + rU be a cyclic polynomial such that 3 + 3p + 3q + r = 0. Then, f ∈C + 3 if and only if 4p3 + 4q3 + 27 ⩾p2q2 + 18pq, or “ p ⩾0 and q ⩾0”. (4) S + 3 =R+ ·(T2,1 −6U)+ R+ · S3 + 3U −T2,1). Note that S3 + 3U −T2,1 ⩾0 is Schur’s inequality. THEOREM 2. Let gp,q(a,b,c) := S4 −pS3,1 −qS1,3 + p2 + pq + q2 3 −1 S2,2 + p + q −p2 + pq + q2 3 US1, g∞(a,b,c) = gp,∞(a,b,c) = g∞,q(a,b,c) := S2,2 −US1, hs(a,b,c) := S3,1 + s2S1,3 −2sS2,2 −(s−1)2US1, CUBIC AND QUARTIC CYCLIC INEQUALITIES 131 h∞(a,b,c) := S1,3 −US1, ks,t(a,b,c) := s2S4 −(2s3 −st)S3,1 + (s3t −2s)S1,3 + (s4 −2s2t + 1)S2,2 + s2 −(s−1)2(s2 + st + 1) US1. Then, the following hold. (1) C4 is an ellipsoid cone whose boundary is R+ · gp,q (p, q) ∈R2 ∪ g∞ . (2) If f ∈C4, then we can ﬁnd p, q ∈R, and α , β ∈R+ such that f = αgp,q + βg∞. (See ) (3) The boundary of C + 4 is R+ · gp,q 9(p + q)2 −(p −q)2 ⩾62, p + q ⩾0 ∪ g∞ ∪R+ · ks,t s ⩾0, t ⩾1 ∪ R+ ·k0,1+ R+ · hs s ∈[0, ∞] . (4) If f ∈C + 4 , then we can ﬁnd α , β , t ∈R+ and s ∈[0, ∞] such that f = αhs + βgp(t,s),q(t,s), here p(t,s) := S1(t,s,1) T2,1(t,s,1)−6U(t,s,1)−3(t −s)(s−1)(1 −t) 2(S2,2(t,s,1)−U(t,s,1)S1(t,s,1)) −1, q(t,s) := S1(t,s,1) T2,1(t,s,1)−6U(t,s,1)+ 3(t −s)(s−1)(1 −t) 2(S2,2(t,s,1)−U(t,s,1)S1(t,s,1)) −1. Note that gp,q ⩾0 is Cˆ ırtoaje’s inequality. COROLLARY 3. Use the same notation as Theorem 2, and let gp := gp,p = S4 −pT3,1 + (p2 −1)S2,2 + (2p −p2)US1. Then, the following hold. (1) S4 is an elliptic cone whose boundary is R+ · gp p ∈R ∪{∞} . Thus, if f ∈S4 , then we can ﬁnd α , β , p ∈R+ such that f = αgp +β(S2,2 −US1). (See ) (2) The boundary of S + 4 is R+ · gp p ∈[1, ∞] ∪ R+ ·g1+ R+ ·(T3,1 −2S2,2) ∪ R+ ·g∞+ R+ ·(T3,1 −2S2,2) . Thus, if f ∈S + 4 , then we can ﬁnd α , β ∈R+ and p ∈[0, ∞] such that f = αgp + β(T3,1 −2S2,2). (See ) We shall prove the inequalities (1.1)—(1.6) in the section 1, using above theorems. (1.1) and (1.2) are obtained from fs ⩾0, putting s = 3 √ 2 or s = 1 + √ 2− 2 √ 2−1 2 respectively. (1.3) is obtained from ks,t ⩾0 putting (s, t) = 4 √ 3, 2/ √ 3 . (1.4) is obtained from ks,t ⩾0, eliminating s and t from α = −(2s+t/s), s2 +1/s2 −2t = 0, 1 −(s−1)2(1 +t/s+ 1/s2) = 0. (1.5) and (1.6) can be obtained by the similar way. 132 T. ANDO 3. Proof of Theorem 1 Throughout this paper, we ﬁx the following notation. Pn R:= (real projective space). (x0 : x1 : ··· : xn) the system of homogeneous coordinates of Pn R. P2 +:= (a : b : c) ∈P2 R a ⩾0, b ⩾0, c ⩾0 . It is well known that for any a, b, c ∈R, the inequalities S4 ⩾S3,1, S4 ⩾S2,2 ⩾US1 hold. Moreover, if a, b, c ∈R+ , then S3 ⩾S2,1 ⩾3U , S3,1 ⩾US1, T3,1 ⩾2S2,2 hold. Proof of Theorem 1. (1) (i) Ws shall prove that fs ∈C + 3 for s ⩾0. Since fs(b,a,c) = s4f1/s(a,b,c), we may assume that 0 ⩽a ⩽b ⩽c = 1. Let k := (1 −b)/(1 −a). Since 0 ⩽a ⩽b, we have 0 ⩽k ⩽1. Then we have fs(a,b,c) = fs(a,1 −k(1 −a),1) = (1 −a)2 a(1 −ks)2(k + s2)+ 1 + (1 −k)s2 (1 −k −s)2 ⩾0. Note that fs(0,s,1) = 0. We recommend readers to use computer to check some com-plicated equalities which appear in this article as the above. (ii) We shall observe X+ 3 . Let ϕ3 : P2 R− →P3 R be the holomorphic map deﬁned by ϕ3(a : b : c) := S3(a,b,c) : S2,1(a,b,c) : S1,2(a,b,c) : U(a,b,c) , and let X+ 3 := ϕ3(P2 +), f0(s) := s3 + 1, f1(s) := s2, f2(s) := s, f3(s) := 0, C3 := (f0(s) : f1(s) : f2(s) : f3(s)) ∈X+ 3 s ∈R+ = (ϕ3(0 : s : 1) ∈X+ 3 s ∈R+ . Note that C3 is the boundary of X+ 3 , and C3 is a nodal plane cubic curve whose node is at (1 : 0 : 0 : 0). (1 : 0 : 0 : 0) (1 : 0 : 0 : 1/3) (1 : 1 : 1 : 0) P3 = (1 : 1 : 1 : 1/3) (1 : 1 : 0 : 0) (1 : 0 : 1 : 1/3) C3 X+ 3 CUBIC AND QUARTIC CYCLIC INEQUALITIES 133 The deﬁning equation of X3 := ϕ3(P4 R) is x3 1 + x3 2 + 9x3 3 −6x1x2x3 −x0x1x2 + 3x0x2 3 + x2 0x3 = 0, (3.1) and X3 has a rational double point of the type A1 at P 3 := (1 : 1 : 1 : 1/3). Let V 3 := (x0 : x1 : x2 : x3) ∈P3 R x0 ̸= 0 ∼ =R3, and we choose a system of coordinates (x, y, z) of V 3 as x = (x,y,z) = x1 x0 −1, x2 x0 −1, x3 x0 −1 3 . Note that the coordinate of P 3 is (x, y, z) = (0, 0, 0). Let D+ 3 ⊂R3 be the convex cone in V 3 generated by X+ 3 ⊂V 3 =R3. (iii) We shall show that C + 3 can be identiﬁed with the dual convex cone of D+ 3 , i.e. (D+ 3 )⊥:= f ∈R3 (f·x) ⩾0 for ∀x ∈D+ 3 . . Any three variables cyclic cubic homogeneous polynomial can be written as f(a,b,c) = p0S3 + p1S2,1 + p2S1,2 + p3U, (∃p0,..., p3 ∈R). For this f , we denote Ff(x0,x1,x2,x3) := p0x0 + p1x1 + p2x2 + p3x3, nf := (p1, p2, p3) ∈R3 . Assume that f ∈C + 3 . Then 3p0 + 3p1 + 3p2 + p3 = 0, and p0 = f(1, 0, 0) ⩾0. Let x ∈X+ 3 ⊂D+ 3 , and assume that x corresponds to (x0 : x1 : x2 : x3) = ϕ3(a : b : c) (∃(a : b : c) ∈P2 +). Since f ∈C + 3 , (nf ·x) = Ff (x0,x1,x2,x3) x0 = f(a,b,c) S3(a,b,c) ⩾0. Since X+ 3 generates D+ 3 , (nf · x) ⩾0 for ∀x ∈D+ 3 . Thus nf ∈(D+ 3 )⊥, and C + 3 can be identiﬁed with (D+ 3 )⊥, corresponding f to nf . (iv) We shall show that fs is located on the boundary of C + 3 . Let Fs be the plane in P3 R which tangents to C3 at Qs := ϕ(0 : s : 1) = (f0(s) : f1(s) : f2(s) : f3(s)) and which passes through P 3. The deﬁning equation of Fs is given by. −3 x0 x1 x2 x3 f0(s) f1(s) f2(s) f3(s) d ds f0(s) d ds f1(s) d ds f2(s) d ds f3(s) 1 1 1 1/3 = s2x0 −(2s2 −1)x1 + (s4 −2s)x2 −3(s4 −2s3 + s2 −2s+ 1)x3. 134 T. ANDO This corresponds to fs. By (i), fs ∈C + 3 = (D+ 3 )⊥. Thus fs lies on the boundary of C + 3 . This fact also implies that Qs s ∈[0, ∞] ⊂C3 generates D+ 3 , and D+ 3 = x ∈V 3 (fs ·x) ⩾0 for ∀s ∈[0, ∞] . Let B3 :=R+ · fs s ∈[0, ∞] , and C +b 3 be the boundary of C + 3 . Above observation implies that B3 ⊂C +b 3 . (v) We shall determine C +b 3 −B3, and shall prove (1). Note that Q0 = Q∞is the node of C3, and C3 is smooth at Qs if s ∈(0, ∞). The boundary of B3 is R+ ·f0∪R+ ·f∞. Let B′ 3 :=R+ ·f0+ R+ ·f∞. A point on B′ 3 corresponds to a surface which tangents X+ 3 at Q0 and which passes through P 3. Thus B′ 3 ⊂C +b 3 , and we conclude that C +b 3 = B3 ∪B′ 3. Therefore, we obtain (1). (2) If f lies on the boundary of C + 3 , then (2) is trivial. Assume that f is an interior point of C + 3 . The half line from f∞to f crosses R+ ·B3 at a point β ′fs (∃β ′ > 0, ∃s ∈[0, ∞)). Then, we can write f in the form f = αfs + βf∞. (3) Eliminate s from p = −2s3 −1 s2 , q = s4 −2s s2 , we obtain 27 + 4p3 + 4q2 = p2q2 + 18pq. If observe the graph of this curve, we have the conclusion. Note that since the dual curve of a plane nodal cubic curve is a quartic curve, B3 is generated by a part of a plane quartic curve. (4) Let ψ3 : P2 R− →Z2 :=P2 R be the holomorphic map deﬁned by ψ3(a : b : c) := (S3(a,b,c) : T3(a,b,c) : U(a,b,c)), and let π3 : P3 R− →Z2 =P2 R be the rational map deﬁned by π3(x0 : x1 : x2 : x3) := (x0 : x1 + x2 : x3). Let Y + 3 := ψ3(P2 +) = π3(X+ 3 ) ⊂Z2, and denote y1 := x1 + x2, y2 := x1 −x2, η3(x0,y1,x3) := 4x2 0x3 + 12x0x2 3 + 36x3 3 −6x3y2 1 + y3 1 −x0y2 1. Then, (3.1) can be written as η3(x0,y1,x3)+ y2 2(x0 + 6x3 + 3y1) = 0. Thus, Y + 3 = (x0 : y1 : x3) ∈Z2 η3(x0,y1,x3) ⩽0, x0 ⩾0, y1 ⩾0, x3 ⩾0. . Note that η3(x0,y1,x3) = 0 deﬁnes the cubic curve which has the cusp at (1 : 2 : 1/3), and which has a parameterization (4 −6m2 + 3m3) : (8 −16m+ 12m2−3m3) : (−4 + 8m−5m2+ m3) , (1 ⩽m ⩽2). As the above ﬁgure, Y + 3 ⊂ (x0 : y1 : x3) ∈Z2 y1 −6x3 ⩾0, x0 + 3x3 −y1 ⩾0. . Thus, the boundary of S + 3 is R+ ·(T2,1 −6U)+ R+ · S3 + 3U −T2,1). □ CUBIC AND QUARTIC CYCLIC INEQUALITIES 135 -y1/x0 6 x3/x0 1 2 1/3 π3(P3) Y + 3 4. Proof of Theorem 2 Proof of Theorem 2. (1) We denote Gp,q(x0,x1,x2,x3,x4) := x0 −px1 −qx2 + p2 + pq + q2 3 −1 x3 + p + q −p2 + pq + q2 3 x4. G∞(x0,x1,x2,x3,x4) := x3 −x4. Hs(x0,x1,x2,x3,x4) := x1 + s2x2 −2sx3 −(s−1)2x4. H∞(x0,x1,x2,x3,x4) := x2 −x4, Ks,t(x0,x1,x2,x3,x4) := s2x0 −(2s3 −st)x1 + (s3t −2s)x2 + (s4 −2s2t + 1)x3 + s2 −(s−1)2(s2 + st + 1) x4. (i) We shall show that gp,q ∈C4 for ∀p, ∀q ∈R. As Cˆ ırtoaje() had shown, 3gp,q(a,b,c) = ∑ cyclic (2a2 −b2 −c2 −pab + (p + q)bc−qca)2 ⩾0. (ii) We shall show that hs , ks,t ∈C + 4 for s ⩾0 and t ⩾1. Since S1,3 ⩾US1 for a, b, c ∈R+ , we have hs(a,b,c) = s2(S1,3 −US1)−2s(S2,2 −US1)+ (S3,1 −US1) = (S1,3 −US1) s−S2,2 −US1 S1,3 −US1 2 + US1(S2 −S1,1)2 S1,3 −US1 ⩾0, ks,t(a,b,c) = s2g2s−1/s,2/s−s(a,b,c)+ s(t −1)hs(a,b,c) ∈C + 4 . Note that p(0, s) = 2s−1/s, q(0, s) = 2/s−s, and gp(s,t),q(s,t)(s,t,1) = 0, hs(0,s,1) = 0, ks,t(0,s,1) = 0. (iii) We shall show that ks,t / ∈C + 4 if t < 1, s ⩾0. 136 T. ANDO Since ks,t(b, a, 1) = s4k1/s,t(a, b, 1), we may assume that 0 < s ⩽1. Let t > 0 and p > max{2, 12t/s}. Then, ks,1−t(st/p,s,1) = −s2t2 p4 p2(p −1)(1 −s)+ p3s2(1 −s2)+ p2s5(1 −s) + p3s5 + p(2p −1)s3t + 2ps5t + ps3(1 −s)t2 + (2p −1)s4t2 + p2(p −1)s3 −p(3p −2)s2t −3p2s4t −ps4t . Since t < ps/12 and p/2 < p −1, we have p(3p −2)s2t + 3p2s4t + ps4t = ps2t((3p −2)+ (3p + 1)s2) < p2s3 12 (3p −2)+ (3p + 1) < p3s3 2 < p2(p −1)s3. Thus ks,1−t(st/p, s, 1) < 0. (iv) We shall observe X4. Let ϕ4 : P2 R− →P4 R be the holomorphic map deﬁned by ϕ4(a : b : c) := (S4 : S3,1 : S1,3 : S2,2 : US1), and let X4 := ϕ4(P2 R). It is easy to see the following equalities hold. (S3,1 + S1,3 +US1)2 −(S4 + 2S2,2)(S2,2 + 2US1) = 0 (S3,1 + S1,3 −2US1)2 + 3(S3,1 −S1,3)2 + (S4 −2S2,2 +US1)2 −(S4 −US1)2 = 0 Thus, the deﬁning equations of the quartic surface X4 is (x1 + x2 + x4)2 −(x0 + 2x3)(x3 + 2x4) = 0, (x1 + x2 −2x4)2 + 3(x1 −x2)2 + (x0 −2x3 + x4)2 −(x0 −x4)2 = 0. (4.1) We knows that X4 has a rational double point of the type A1 at P 4 := (1 : 1 : 1 : 1 : 1), from the above equations. C4 X4 X+ 4 P4 = (1 : 1 : 1 : 1 : 1) CUBIC AND QUARTIC CYCLIC INEQUALITIES 137 Let V 4 := (x0 : x1 : x2 : x3 : x4) ∈P4 R x0 ̸= 0 ∼ =R4, and we choose a system of coordinates (x, y, z, w) of V 4 as x = (x,y,z,w) = x1 x0 −1, x2 x0 −1, x3 x0 −1, x4 x0 −1 . By (4.1), the deﬁning equations of X4 ∩V 4 are (x+ y+ w+ 3)2−(2z+ 3)(z+ 2w+ 3) = 0, (x+ y−2w)2 + 3(x−y)2 + (2z−w)2 −w2 = 0. (4.2) Let W 3 = (x, y, z) x, y, z ∈R be the hyperplane deﬁned by w = −1 in V 4, and let ρ : P4 R− →W 3 be the projection from the center P 4. By (4.2), E := ρ X4 −{P 4} is an ellipsoid E = (x,y,z) ∈W 3 (x+ y+ 2)2 + 3(x−y)2 + (2z+ 1)2 = 1. . Let D4 be the convex cone generated by X4 in V 4. The boundary of D4 is the cone whose base is E . By the same argument as (iii) of the proof of Theorem 1, we conclude that C4 can be identiﬁed with the dual convex cone of D4. (v) We shall determine the boundary of C4, and shall prove (1). Let g0(s,t) := S4(s,t,1), g1(s,t) := S3,1(s,t,1), g2(s,t) := S1,3(s,t,1), g3(s,t) := S2,2(s,t,1), g4(s,t) := U(s,t,1)S1(s,t,1), and let Gs,t (resp. G∞) be the hyperplane in P4 R which tangents to X4 at the point ϕ4(s : t : 1) = (g0(s,t) : ··· : g4(s,t)) (resp. (1:0:0:0)) and which passes through P 4. Since x0 x1 x2 x3 x4 g0(s,t) g1(s,t) g2(s,t) g3(s,t) g4(s,t) ∂ ∂sg0(s,t) ∂ ∂sg1(s,t) ∂ ∂sg2(s,t) ∂ ∂sg3(s,t) ∂ ∂sg4(s,t) ∂ ∂t g0(s,t) ∂ ∂t g1(s,t) ∂ ∂t g2(s,t) ∂ ∂t g3(s,t) ∂ ∂t g4(s,t) 1 1 1 1 1 = −S1(s,t,1) S2(s,t,1)−S1,1(s,t,1) 2 S2,2(s,t,1)−U(s,t,1)S1(s,t,1) × Gp(s,t),q(s,t)(x0,x1,x2,x3,x4), the deﬁning equation of Gs,t is given by Gp(s,t),q(s,t) = 0. Note that the range of p(s,t), q(s,t) is R2 ((s, t) ∈R2). When s2 + t2 →∞, deﬁning equation of Gs,t tends to G∞= 0. Thus gp,q and g∞are on the boundary of C4. Let ψ : P2 R− →W 3 be the rational map deﬁned by ψ(a : b : c) = −S4 −S3,1 S4 −US1 , −S4 −S1,3 S4 −US1 , −S4 −S2,2 S4 −US1 . 138 T. ANDO Take any point Q ∈E . Since ψ = ρ ◦ϕ4, we have ψ(P2 R) = E . Thus, there exists s, t ∈R such that ψ(s : t : 1) = Q, or ψ(1 : 0 : 0) = Q. Then Gs,t or G∞tangents to E at Q. Thus, we conclude that B4 :=R+ · gp,q p, q ∈R ∪ g∞ is the boundary of D⊥ 4 = C4. (2) can be obtained by the similar argument as the proof of (2) of Theorem 1. (3) Let X+ 4 := ϕ4(P2 +), E+ := ψ(P2 +), and let D+ 4 be the convex cone generated by X+ 4 . If we obtain the convex closure E+ of E+, we can determine D+ 4 as the convex cone whose base is E+. Let C +b 4 be the boundary of C + 4 . (−1, −1, −1) E+ (vi) We shall determine B4 ∩C +b 4 . Let k1(s) := s3 s4 + 1 −1, k2(s) := s s4 + 1 −1, k3(s) := s2 s4 + 1 −1, Γ := (k1(s),k2(s),k3(s)) ∈W 3 s ∈R+ = ψ(0 : s : 1) ∈W 3 s ∈R+ . Γ is the boundary of E+. Note that Γ has a node at ψ(0 : 0 : 1) = (−1, −1, −1). Since p(s,0), q(s,0) s ∈R+ = p(0,s), q(0,s) s ∈R+ = (p,q) ∈R2 9(p + q)2 −(p −q)2 = 62 and p + q ⩾0 , we know that the plane deﬁned by Gp,q(0, x, y, z, −1) = 0 tangents to E at a point on Γ if and only if 9(p + q)2 −(p −q)2 = 62 and p + q ⩾0. By the above observation, we know that B4 ∩C +b 4 is the following B+ 4 : B+ 4 :=R+ · gp,q 9(p + q)2 −(p −q)2 ⩾62, p + q ⩾0 ∪ g∞ =R+ · R+ · gp(s,t),q(s,t) s, t ∈R+ ∪ g∞ . (vii) We shall determine one of the another parts of C +b 4 . CUBIC AND QUARTIC CYCLIC INEQUALITIES 139 Let Ks (s ∈[0, ∞])be the plane in W 3 which tangents to Γ at ψ(0 : s : 1), and which passes through (−1, −1, −1). The equation of Ks is given by x y z 1 k1(s) k2(s) k3(s) 1 d dsk1(s) d dsk2(s) d dsk3(s) 0 −1 −1 −1 1 = s2 (s4 + 1)2 Hs(0,x,y,z,−1). Thus, we know that hs lies on the boundary of C + 4 , by (ii). Let ℓs := (1 −τ)(−1,−1,−1)+ τ ·ψ(0 : s : 1) 0 ⩽τ ⩽1 be the line segments connecting (−1, −1, −1) and ψ(0 : s : 1), and let E2 := s⩾0 ℓs. The plane deﬁned by Hs = 0 tangents E+ at the line segment ℓs on E2. Thus, the boundary of E+ is E+ ∪E2. The plane deﬁned by Gp(s,t),q(s,t) = 0 tangents E+ at the point ψ(s : t : 1) on E+. Since Ks,t = s2G2s−1/s,2/s−s + s(t −1)Hs = s2Gp(0,s),q(0,s) + s(t −1)Hs, the plane deﬁned by Ks,t = 0 tangents E+ at the point ψ(0 : s : 1) on Γ (s ⩾0, t ⩾1). Thus, ks,t (s ⩾0, t ⩾1) lies on the boundary of C + 4 , and B′ 4 :=R+ · ks,t s ⩾0, t ⩾1 is a part of C +b 4 . (viii) We shall determine C +b 4 −(B+ 4 ∪B′ 4), and shall prove (3). Let B+b 4 , B′b 4 be the boundaries of B+ 4 , B′ 4. Note that we can identify k0,t = k0,1 with k∞,t . As is shown in the above, B+b 4 =R+ · gp(0,s),q(0,s) s ⩾0 =R+ · ks,1 s ⩾0 , B′b 4 =R+ · hs s ∈[0,∞] ∪R+ · ks,1 s ⩾0 ∪(R+ ·h0+ R+ ·k0,1)∪(R+ ·h∞+ R+ ·k0,1). An element of B′′ 4 :=R+ ·k0,1+ R+ · hs s ∈[0, ∞] corresponds to a plane which tangents to E+ at the point (−1, −1, −1). Thus B′′ 4 ⊂C +b 4 . Therefore, C +b 4 = B+ 4 ∪B′ 4 ∪B′′ 4 , and we complete the proof of (3). (4) For s ⩾0, let Ms :=R+ ·hs+ R+ · gp(t,s),q(t,s) t ⩾1 , and M∞:=R+ ·h∞+ R+ ·g∞. By the above observation, we conclude that s∈[0,∞] Ms = C + 4 . Thus, we have (4). □ 140 T. ANDO REMARK 4. The polynomials Hs and Ks,t appear in the deﬁning equation of the hyperplane which tangents to the boundary of X+ 4 . Let l0(s) := s4 + 1, l1(s) := s3, l2(s) := s, l3(s) := s2, l4(s) := 0, C4 := (l0(s) : l1(s) : l2(s) : l4(s)) ∈P4 R s ∈R+ = ϕ4(0 : s : 1) ∈P4 R s ∈R+ . C4 is the boundary of X+ 4 . Let Ls ⊂P4 R be a hyperplane which tangents to C4 at ϕ4(0 : s : 1) (s ⩾0) and which passes through P 4 = (1 : 1 : 1 : 1 : 1). But these conditions do not determine Ls uniquely. Moreover we assume that Ls passes through a point (t0 : t1 : t2 : t3 : t4). Then the deﬁning equation of Ls is x0 x1 x2 x3 x4 l0(s) l1(s) l2(s) l3(s) l4(s) d dsl0(s) d dsl1(s) d dsl2(s) d dsl3(s) d dsl4(s) 1 1 1 1 1 t0 t1 t2 t3 t4 = −s2(t1 + s2t2 −2st3 −(1 −s)2t4)Gp(0,s),q(0,s) + s2t0 + (s−2s3)t1 + (−2s+ s3)t2 + (1 −s2)2t3 −(1 −s−s2−s3 + s4)t4 Hs. Thus the deﬁning equation of Ls can be written as Ks,t(x0, x1, x2, x3, x4) = 0, if we take a suitable t . REMARK 5. When we eliminate s and t from p = −2s3 −st s2 , q = s3t −2s s2 , r = s4 −2s2t + 1 s2 , we obtain p2q2r2 −4p3q3 + 18p3qr + 18pq3r −4p2r3 −4q2r3 −27p4 −27q4 + 16r4 −6p2q2 −80pqr2 + 144p2r + 144q2r −192pq −128r2+ 256 = 0. But the singularity of the surface deﬁned by the above equation is so complicated to state the similar proposition like (3) of Theorem 1. Proof of Corollary 3. We use the same notation as the above proof. (1) Let ψ4 : P2 R− →Z3 :=P3 R be the holomorphic map deﬁned by ψ4(a : b : c) := (S4 : T3,1 : S2,2 : US1), and let π4 : P4 R− →Z3 =P3 R be the rational map deﬁned by π4(x0 : x1 : x2 : x3 : x4) := (x0 : x1 + x2 : x3 : x4). CUBIC AND QUARTIC CYCLIC INEQUALITIES 141 Put Y4 := ψ4(P2 R) = π4(X4). We choose a system of coordinates of π4(V 4) ∼ =R3 as (u,z,w) = x1 + x2 x0 −2, x3 x0 −1, x4 x0 −1 . Let W 2 := π4(W 3) ∼ =R2, and we choose a system of coordinates of W 2 as (u, z). Note that π4(E) = (u,z) ∈W 2 (u + 2)2 + (2z+ 1)2 ⩽1. is an ellipse. π4(D4) is the convex cone in π4(V 4) whose base is π4(E) and whose vertex is (0, 0, 0). Since the line deﬁned by −pu+(p2−1)z = 2p−p2 (resp. z = −1) tangents to the ellipse (u + 2)2 + (2z + 1)2 = 1 at 2p 1 + p2 −2, − p2 1 + p2 (resp. at (−2, −1)), we conclude that gp (resp. g∞) lies on the boundary of S4 = (π4(D4))⊥. It is easy to see that these surround S4. Thus R+ · gp p ∈R ∪{∞} is the boundary of S4, and we have (1). (2) Let ψ2 : P2 R− →W 2 be the rational map deﬁned by ψ2(a : b : c) = −2S4 −T3,1 S4 −US1 , −S4 −S2,2 S4 −US1 , and let C2 := ψ2(0 : s : 1) s ∈[0, ∞] . Since ψ2 = π4 ◦ψ , π4(E+) = ψ2(P2 +). Since the deﬁning equation of C2 is 2(z+ 5/4)2−(u + 2)2 = 1/8, we have that π4(E+) = (u,z) ∈W 2 (u + 2)2 + (2z+ 1)2 ⩽1, 2(z+ 5/4)2 −(u + 2)2 ⩽1/8 . -u 6 z −1 −2 −1/2 −1 π4(E+) Thus, the convex closure of π4(E+) is (u,z) ∈W 2 (u + 2)2 + (2z+ 1)2 ⩽1 and u −2z ⩾0 . Let D′ 4 be the convex cone generated by π4(E+). By the above observation, we con-clude that the boundary of the dual convex cove (D′ 4)⊥∼ = S + 4 is the union of a surface B′ 1 :=R+ · (−p, p2 −1,2p −p2) p ∈[1, ∞] =R+ · gp p ∈[1, ∞] , 142 T. ANDO and two faces B′ 2 :=R+ ·(−1,0,1)+ R+ ·(1,−2,0) =R+ ·g1+ R+ ·(T3,1 −2S2,2), B′′ 2 :=R+ ·(0,1,−1)+ R+ ·(1,−2,0) =R+ ·g∞+ R+ ·(T3,1 −2S2,2). Thus we have (2). □ R E F E R E N C E S T. ANDO, Some homogeneous cyclic inequalities of three variables of degree three and four, Aust. J. Math. Anal. Appl. 7, 2 (2010), Art. 12. V. Cˆ IRTOAJE,Algebraic Inequalities — Old and New Methods, GIL Publishing House (Romania), 2006. V. CIRTOAJE, On the Cyclic Homogeneous Polynomial Inequalities of Degree Four, J. Inequal. Pure and Appl. Math. 10, 3 (2009), Art. 67. V. CIRTOAJE, Necessary and sufﬁcient conditions for symmetric homogeneous polynomial inequalities, Preprint. R. MUIRHEAD, Some methods applicable to identities and inequalities of symmetric algebraic functions of n letters, Proc. Edinb. Math. Soc. 21 (1902), 144–157. (Received May 23, 2011) Tetsuya Ando Department of Mathematics and Informatics Chiba University Chiba 263-8522 JAPAN e-mail: ando@math.s.chiba-u.ac.jp Mathematical Inequalities & Applications www.ele-math.com mia@ele-math.com
3218	https://math.stackexchange.com/questions/2601869/spivak-chapter-14-problem-9	calculus - Spivak Chapter 14 Problem 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 Spivak Chapter 14 Problem 9 Ask Question Asked 7 years, 8 months ago Modified7 years, 8 months ago Viewed 842 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Let f(x)=cos 1 x f(x)=cos⁡1 x for x≠0 x≠0, and f(x)=0 f(x)=0 for x=0 x=0. Is the function F(x)=∫x 0 f F(x)=∫0 x f differentiable at 0? My progress: lim h→0+F(h)h=lim h→0+∫h 0 f h=lim h→0+cos(1 h)1 lim h→0+F(h)h=lim h→0+∫0 h f h=lim h→0+cos⁡(1 h)1 , which does not exist. We know that F′(0)=lim h→0 F(h)h F′(0)=lim h→0 F(h)h . Therefore, the limit above should be equal to 0.0. However, the answer is that F′(x)=0.F′(x)=0. What did I mess up? calculus Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 12, 2018 at 3:29 Rohan Shinde 10.1k 1 1 gold badge 22 22 silver badges 71 71 bronze badges asked Jan 12, 2018 at 3:11 minimariominimario 505 2 2 silver badges 11 11 bronze badges Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. This is a tricky problem and Spivak has given a hint which you have perhaps overlooked. Consider the function g(x)=x 2 sin(1/x),g(0)=0 g(x)=x 2 sin⁡(1/x),g(0)=0 then it can be proved that g′(x)=2 x sin(1/x)−cos(1/x),g′(0)=0 g′(x)=2 x sin⁡(1/x)−cos⁡(1/x),g′(0)=0 so that g′(x)=h(x)−f(x)g′(x)=h(x)−f(x) where f f is the function in your question and h(x)=2 x sin(1/x),h(0)=0 h(x)=2 x sin⁡(1/x),h(0)=0 We have now via Fundamental Theorem of Calculus g(x)=g(x)−g(0)=∫x 0 g′(t)d t=∫x 0 h(t)d t−∫x 0 f(t)d t g(x)=g(x)−g(0)=∫0 x g′(t)d t=∫0 x h(t)d t−∫0 x f(t)d t or F(x)=∫x 0 h(t)d t−g(x)F(x)=∫0 x h(t)d t−g(x) and using Fundamental Theorem of Calculus again we get F′(0)=h(0)−g′(0)=0 F′(0)=h(0)−g′(0)=0 because h h is continuous at 0 0. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 13, 2018 at 2:40 answered Jan 13, 2018 at 2:28 Paramanand Singh♦Paramanand Singh 92.6k 15 15 gold badges 159 159 silver badges 349 349 bronze badges 5 Excellent! OP, you should accept this elegant answer instead of mine (and remember to share hints with the community). @Paramanand Singh, Want to check with you about why h(0)=0. Because you define it that way?Mathemagical –Mathemagical 2018-01-13 02:55:03 +00:00 Commented Jan 13, 2018 at 2:55 @Mathemagical: I have defined it that way to make h h continuous at 0 0. This is crucial in last step when I apply FTC. Also the hint was very cryptic: "Stare at page 165". So you actually need to look in Spivak book page 165 where he discusses about derivative of x 2 sin(1/x)x 2 sin⁡(1/x).Paramanand Singh –Paramanand Singh♦ 2018-01-13 03:39:17 +00:00 Commented Jan 13, 2018 at 3:39 @Mathemagical : h(0)=0 h(0)=0 is also necessary if you want equation g′(x)=h(x)−f(x)g′(x)=h(x)−f(x) to hold for x=0 x=0 but this is secondary and not important. More importantly we need continuity of h h at 0 0.Paramanand Singh –Paramanand Singh♦ 2018-01-13 03:48:29 +00:00 Commented Jan 13, 2018 at 3:48 @Mathemagical: you see what I mean about Spivak Calculus being a little more reasonable? With a hint like this, we might have come up with a nice answer. :)John Hughes –John Hughes 2018-01-13 04:04:29 +00:00 Commented Jan 13, 2018 at 4:04 @JohnHughes yeah. Tried to get a peek on the book now, but it seems no PDF available online. Gotta buy. Parmanand, thanks. Yes, I figured it was for continuity, but glad for your confirmation.Mathemagical –Mathemagical 2018-01-13 04:20:00 +00:00 Commented Jan 13, 2018 at 4:20 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Edits made: Your wanting to use the L’Hospital’s rule is fair, but your conclusion is not valid You do find a zero numerator and denominator. So you are right to try progress by differentiating both. Edit: thanks to what is pointed out in the comments, the derivative is indeed f(h)f(h). However, the limit does not exist. So you cannot conclude anything from L’H rule (see the section “requirement that the limit exist” here for an example). In brief, “L’H computations lead to a no-limit situation” does not imply the limit does not exist. As to why the derivative is zero, an outline: try writing the limit as lim h→0∫h 0 f h=lim M→∞M∫∞M cos y y 2 d y lim h→0∫0 h f h=lim M→∞M∫M∞cos⁡y y 2 d y Now split the integral into intervals of length 2 π 2 π. In each interval, bound the positive part above (for example in [2 N π,(2 N+1)π][2 N π,(2 N+1)π], we have 1 y 2≤1(2 N π)2 1 y 2≤1(2 N π)2 and the negative part from below to show that M times the sum of integrals on these intervals is smaller than a constant times 1/M 1/M. (There is a series sum involved). Reverse the bounds on the positive and negative parts to see that it is bounded below by a constant times −1/M−1/M. As M→∞M→∞, the limit is zero. There might be a more elegant way to show the effect of the cancellation, but that’s all I could work out for now. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 12, 2018 at 14:10 answered Jan 12, 2018 at 3:58 MathemagicalMathemagical 3,652 1 1 gold badge 14 14 silver badges 21 21 bronze badges 9 Actually the derivative of the numerator is correct. The FTC2 requires continuity on the whole interval, not FTC1. If you’re still uncomfortable with that, split the integral into two and take the derivative so the first part will cancel and for the second part you get cos(1/x)user334639 –user334639 2018-01-12 04:02:35 +00:00 Commented Jan 12, 2018 at 4:02 @user334639 FTC1 does require continuity at zero, it is FTC2 that allows it to be continuous just almost everywhere. But your point is right. f(h) is indeed the derivative. The issue is only in drawing conclusions from L’H rule. Having hit a roadblock with it does not mean the limit does not exist.Mathemagical –Mathemagical 2018-01-12 05:51:54 +00:00 Commented Jan 12, 2018 at 5:51 I like this approach (and it's where I was headed as I tried to finish this up this morning), but it leaves me wondering: What was Spivak thinking? Because this comes way before the chapters on sequences/series in the book.John Hughes –John Hughes 2018-01-12 13:17:57 +00:00 Commented Jan 12, 2018 at 13:17 BTW, working out the lower bound on the same interval, I find that on each "period", the integral is bounded by something proportional to 1 M−1 3 1 M−1 3 ... but my algebra could contain errors. At first, I was worried that the positive and negative lobe might cancel in a way that left the order of the difference being 1/M 1/M, and that'd be no use at all.John Hughes –John Hughes 2018-01-12 13:26:08 +00:00 Commented Jan 12, 2018 at 13:26 @JohnHughes I got (taking M=2 π K), the integral’s upper bound to be K 2 π∑∞n=K 2 n+1 n 2(n+1)2 K 2 π∑n=K∞2 n+1 n 2(n+1)2. So over each period, the leading order term was indeed like 1 n 3 1 n 3. This sums to be proportional to 1 K 1 K, so it does the job. As to what Spivak was thinking, I haven’t used this book, but once worked through his “manifolds” book. Usually, the brevity was fine. but sometimes, it drove me up the wall. Things that “immediately follow” didn’t follow at all.Mathemagical –Mathemagical 2018-01-12 14:04:34 +00:00 Commented Jan 12, 2018 at 14:04 \|Show 4 more comments This answer is useful 1 Save this answer. Show activity on this post. Where does your second equality come from? Why is the integral of f(x)f(x) (you've lost a "dx") from 0 0 to h h the same as c o s(1/h)c o s(1/h)? The mean value theorem tells you it's the same as cos(1/c)cos⁡(1/c) for some c c between 0 0 and h h, but there's no guarantee that this c c is actually h h. When you look at cos(1/c)cos⁡(1/c), you realize that this could be any number between −1−1 and 1 1, so that this is not a profitable way to approach the problem. Also, "We know that F′(x)=lim…F′(x)=lim…" is wrong; that should be "We know that F′(0)=…F′(0)=…". And the thing inside the limit should be F(h)−F(0)h F(h)−F(0)h although perhaps you simplified by evaluating F(0)F(0). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 12, 2018 at 3:16 answered Jan 12, 2018 at 3:14 John HughesJohn Hughes 102k 4 4 gold badges 88 88 silver badges 160 160 bronze badges 11 L'Hospital's rule? Since at h=0 h=0, both the top and bottom go to 0. (By the way, my book doesn't usually use dx)minimario –minimario 2018-01-12 03:16:04 +00:00 Commented Jan 12, 2018 at 3:16 1 If your book is Spivak, it writes either ∫b a f∫a b f or ∫b a f(x)d x∫a b f(x)d x, but never a halfway mix of the two.John Hughes –John Hughes 2018-01-12 03:18:16 +00:00 Commented Jan 12, 2018 at 3:18 Alright, fixed that :). Is something wrong with L'H here?minimario –minimario 2018-01-12 03:19:03 +00:00 Commented Jan 12, 2018 at 3:19 For L'Hopital to apply, you need to know that the limit of the top and bottom both exist and are zero. I don't think you know that without a bit more work.John Hughes –John Hughes 2018-01-12 03:19:48 +00:00 Commented Jan 12, 2018 at 3:19 Ah, so my fault comes from assuming that lim h→0+∫h 0 f=0 lim h→0+∫0 h f=0 as h→0 h→0, which doesn't seem to be true. Thanks!minimario –minimario 2018-01-12 03:21:41 +00:00 Commented Jan 12, 2018 at 3:21 \|Show 6 more comments This answer is useful 1 Save this answer. Show activity on this post. Naively one would say you can “apply L’Hopital” because you have 0/0, however L’Hopital is inconclusive if the ratio between the derivatives does not have a limit. In order to show the derivative is zero, you need to check that, because of positive and negative cancellations you have the integral less than c h c h for every c c provided h h is small enough. You can try to show that using some change of variable. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 12, 2018 at 3:48 user334639user334639 1,674 13 13 silver badges 11 11 bronze badges 4 Do you have an example of this: "however L’Hopital is inconclusive if the ratio between the derivatives does not have a limit"minimario –minimario 2018-01-12 03:52:31 +00:00 Commented Jan 12, 2018 at 3:52 1 @minimario The original question is an example. The theorem says the limit of a quotient is the limit of the quotient of the derivatives IF such limit exists. If the limit does not exist, which is the case here, then we are dealing with a perfect case of an inconclusive theorem.user334639 –user334639 2018-01-12 03:57:34 +00:00 Commented Jan 12, 2018 at 3:57 Do you have another example that...I can work with better, for example, with more "useful" functions? ("Useful" being "better-defined", maybe...)minimario –minimario 2018-01-12 04:01:14 +00:00 Commented Jan 12, 2018 at 4:01 I can try, please ask it as a separate question and link it here.user334639 –user334639 2018-01-12 04:05:19 +00:00 Commented Jan 12, 2018 at 4:05 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 See similar questions with these tags. 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3219	https://artofproblemsolving.com/wiki/index.php/Floor_function?srsltid=AfmBOoorw_-J8y2aCpkE505MaajZWXK2A8UoHI3FRh4HSTCATZTQ9oDX	Art of Problem Solving Floor function - 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 Floor function Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Floor function The greatest integer function, also known as the floor function, gives the greatest integer less than or equal to its argument. The floor of is usually denoted by or . The action of this function is the same as "rounding down." On a positive argument, this function is the same as "dropping everything after the decimal point," but this is not true for negative values. Contents 1 Properties 2 Examples 3 Alternate Definition 4 Problems 4.1 Introductory Problems 4.2 Intermediate Problems 4.3 Olympiad Problems 5 See Also Properties for all real . Hermite's Identity: Examples A useful way to use the floor function is to write , where y is an integer and k is the leftover stuff after the decimal point. This can greatly simplify many problems. Alternate Definition Another common definition of the floor function is where is the fractional part of . Problems Introductory Problems Let denote the largest integer not exceeding . For example, , and . How many positive integers satisfy the equation . (2017 PCIMC) Intermediate Problems Find the integer satisfying . Here denotes the greatest integer less than or equal to . (1999-2000 Hong Kong IMO Prelim) What is the units (i.e., rightmost) digit of (1986 Putnam Exam, A-2) How many of the first 1000 positive integers can be expressed in the form , where is a real number, and denotes the greatest integer less than or equal to ? (1985 AIME) Olympiad Problems If is a positive real number, and is a positive integer, prove that where denotes the greatest integer less than or equal to . (1981 USAMO, #5) (Discussion 1) (Discussion 2) Let denote the integer part of , i.e., the greatest integer not exceeding . If is a positive integer, express as a simple function of the sum (1968 IMO, #6) See Also Ceiling function Fractional part Retrieved from " Category: Functions 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.
3220	https://www.scribd.com/document/545316696/Heisler-Diagram	Heisler Diagram (Flat Plate) \| PDF Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 219 views 4 pages Heisler Diagram (Flat Plate) The document contains three Heisler diagrams that plot the dimensionless temperature θ versus the dimensionless radial coordinate X for different geometries - flat plate, cylinder, and spher… Full description Uploaded by Ram Singh AI-enhanced title and description Go to previous items Go to next items Download Save Save Heisler Diagram For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Heisler Diagram For Later You are on page 1/ 4 Search Fullscreen Heisler Diagram (Flat Plate) 0 1 2 3 4 5 10 –3 10 –2 10 –1 10 0 θ c 0.01 0.1 1 10 1 0 2 0 3 0 4 0 5 0 0.01 0.1 1 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 0.01 0.1 F o 0 1 2 3 4 5 10 –1 10 0 θ c 0.01 0.1 1 10 F 0 0 0.5 1 0 0.5 1 X = x/L θ Bi = 1.0 Bi = 10 0.01 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0 F 0 0.4 0 0.5 1 0 0.5 1 F 0 θ c , θ X=0.5 , θ X=1 Bi = 0.1 Bi = 1 Bi = 10 θ c θ X=0.5 θ X=1 adDownload to read ad-free Heisler Diagram (Cylinder) F o 0 1 2 3 4 5 10 –3 10 –2 10 –1 10 0 θ c 0.01 0.1 1 10 1 0 2 0 3 0 4 0 5 0 0.01 0.1 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 0.01 0 1 2 3 4 5 10 –1 10 0 θ c 0.01 0.1 1 10 F 0 0 0.5 1 0 0.5 1 X = x/L θ Bi = 1.0 Bi = 10 0.01 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0 F 0 0.4 0 0.5 1 0 0.5 1 F 0 θ c , θ X=0.5 , θ X=1 Bi = 0.1 Bi = 1 Bi = 10 θ c θ X=0.5 θ X=1 adDownload to read ad-free Heisler Diagram (Sphere) 0 1 2 3 4 5 10 –3 10 –2 10 –1 10 0 θ c 0.01 0.1 1 10 1 0 2 0 3 0 4 0 5 0 0.01 0.1 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 0.01 F o 0 1 2 3 4 5 10 –1 10 0 θ c 0.01 0.1 1 10 F 0 0 0.5 1 0 0.5 1 X = x/L θ Bi = 1.0 Bi = 10 0.01 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0 F 0 0.4 0 0.5 1 0 0.5 1 F 0 θ c , θ X=0.5 , θ X=1 Bi = 0.1 Bi = 1 Bi = 10 θ c θ X=0.5 θ X=1 adDownload to read ad-free Change in Temperature Distribution 0 0.5 1 0 0.5 1 X = x/L θ Bi = 1.0 Bi = 10 0.01 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0 F 0 0.4 0 0.5 1 0 0.5 1 X = x/L θ Bi = 1.0 Bi = 10 0.01 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0 F 0 0.4 0 0.5 1 0 0.5 1 X = x/L θ Bi = 1.0 Bi = 10 0.01 0.1 0.2 0.3 0.5 0.6 0. 0.8 0.1.0 F 0 0.4 F l a t P l a t e C y l i n d e r S p h e r e Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Unit-2 Vector Calculus Notes No ratings yet Unit-2 Vector Calculus Notes 28 pages Techneo Dom Unit - 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3221	https://www.splashlearn.com/math-vocabulary/percent-difference	Parents Explore by Grade Preschool (Age 2-5)KindergartenGrade 1Grade 2Grade 3Grade 4Grade 5 Explore by Subject Math ProgramEnglish Program More Programs Homeschool ProgramSummer ProgramMonthly Mash-up Helpful Links Parenting Blog Success Stories Support Gifting Also available on Educators Teach with Us For Teachers For Schools and Districts Data Protection Addendum Success Stories Lesson Plans Classroom Tools Teacher Blog Help & Support More Programs SpringBoard Summer Learning Our Library All By Grade PreschoolKindergartenGrade 1Grade 2Grade 3Grade 4Grade 5 By Subject MathEnglish By Topic CountingAdditionSubtractionMultiplicationPhonicsAlphabetVowels One stop for learning fun! Games, activities, lessons - it's all here! 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What Is the Percent Difference Formula? Percent Difference vs. Percent Change Solved Examples on Percent Difference Practice Problems on Percent Difference Frequently Asked Questions on Percent Difference What Is the Percent Difference in Math? Percent difference is the ratio of the absolute difference between two values to the average of the two values, calculated as a percentage. Percent Difference $= \frac{Absolute \; Difference}{Average} \times 100$ In simple words, the percent difference is the percentage of the relative difference. The relative difference between two quantities compares the difference between the two values in relation to their average. Relative Difference $= \frac{Absolute \; Difference}{Average}$ The percent difference is frequently used when comparing data sets or estimating the degree of change between two numbers. Note that the “percent difference” and the “percent change” are two different concepts. We will compare them later in the article. Also, do not confuse this formula with the percent decrease or increase formulas. Recommended Games Calculate the Difference Between Amounts of Money Game Play Choose the Correct Difference Game Play Cross Out Fractions to Find the Difference Game Play Cross out Parts to Find the Difference Between Fractions Game Play Cross Out to FInd the Difference Between a Mixed Number and a Fraction Game Play Determine the Correct Difference Game Play Determine the Difference and Match Game Play Determine the Difference Game Play Determine the Difference to Match Game Play Estimate the Difference Game Play More Games What Is the Percent Difference Formula? The percentage difference is the ratio of the difference in their values to their average, multiplied by 100. Percent difference formula for the two values A and B can be given as Percent difference $= \frac{A – B}{ \frac{(A + B)}{2}} \times 100$ Recommended Worksheets More Worksheets How to Find Percent Difference Follow the steps given below to learn how to calculate the percent difference between two values A and B. Step 1: Find the absolute difference. Find the absolute value of the difference between A and B. The order does not matter since we will be taking the absolute value. Absolute Difference $= \|A \;-\; B\|$ Step 2: Find the average of A and B. Simply add the values together and divide the sum by 2. Average $= \frac{(A + B)}{2}$ Step 3: Divide the absolute difference by the average and multiply by 100 in order to calculate the percent difference. Percent Difference $= \frac{Absolute \; Difference}{Average} \times 100$ Percent difference $= \frac{\|A \;-\; B\|}{ \frac{(A + B)}{2}} \times 100$ Percent Difference between Two Numbers The percent difference between the two numbers is expressed as follows: Percent Difference $= \frac{\|A – B\|}{ \frac{(A + B)}{2}} \times 100$ Example: Alex has 10 years of working experience. Paxton has 20 years of working experience. What is the percent difference? Here, $A = 10, B = 20$ Note that these values represent the working experience of Alex and Paxton in years. Can you decide which value is more important here? No, both values hold the same importance! In such cases, we can use the percent difference formula. We choose the average value as the reference point. We take absolute difference to treat both values equally. So, the order does not matter when you subtract. $\|A \;-\; B\| = \|10 \;-\; 20\| = \| \;-\;10\| = 10$ $\frac{(A + B)}{2} = \frac{10 + 20}{2} = 15$ Percent Difference $= \frac{\|A – B\|}{\frac{(A + B)}{2}} \times 100$ Percent Difference $= \frac{10}{15} \times 100$ Percent Difference $≈ 66.67\%$ Percent Difference vs. Percent Change \| Percent Difference \| Percent Change \| --- \| \| Percent Difference $= \frac{\|A – B\|}{\frac{(A + B)}{2}}100$ \| Percent Change $= \frac{(Final \;Value\; -\; Initial \;Value)}{Initial \;Value} \times 100$ \| \| It is calculated by taking the absolute difference between the two values, dividing it by the average of the two values, and multiplying by 100. \| It is calculated by taking the absolute change, dividing it by the initial value, and multiplying by 100. \| \| Reference point is the average of two values. \| Reference point is the initial value. \| \| We ignore the minus sign as we take the absolute value of the difference. \| If the value is positive, it means there is an increase. If the value comes out to be negative, it indicates a decrease. \| Facts about Percent Difference The context of the data being analyzed should be regarded when interpreting percent differences. The percent difference just measures the relative difference between two values and does not reveal the pattern or direction of change. It does not distinguish between positive and negative changes. When there is an old value and a new value, we should use the percentage change formula. When there is an approximate value and an exact value, we should use the percentage error formula. When both the values have equal importance or when it is not possible to determine the reference value, we use the percent difference formula. Conclusion You know the percent difference, definition, concept, formula, etc. Let’s move toward the numerical section of a percent difference, including solved examples and MCQs, for a clearer understanding. Solved Examples on Percent Difference 1. Mr. Dunphy’s garden has 78 rose plants. Mr. Prichet’s garden has 106 rose plants. Find the percent difference. Solution: Firstly, we will calculate the difference between the two numbers. Difference $= 78\;-\;106 = \;-\; 28$ We can remove the negative sign by giving the absolute value of the difference. Absolute Difference $= \|Difference\| = \|\;-\;28\| = 28$ Average $= \frac{78 + 106}{2} = \frac{184}{2} = 92$ Now, find the percentage difference. Percentage Difference $= (\frac{Absolute \;Difference}{Average}) \times 100$ Percentage Difference $= \frac{28}{92} \times 100$ Percentage Difference $≈ 30.43\%$ 2. In a video game tournament, Team A scored 980 and Team B scored 1220 points. What is the percentage difference between the scores? Solution: On calculating the difference between the two scores, we get Difference $=$ Score of Team B – Score of Team A $= 1220\;-\;980 = 240$ Absolute difference $= \| 240 \| = 240$ Average $= \frac{1220 + 980}{2} = 1100$ Now, we can calculate the percentage difference between the teams’ scores. Percentage Difference $= (\frac{Difference}{Average})\times100$ Percentage Difference $= (\frac{240}{1100}) \times 100$ Percentage Difference $≈ 21.81\%$ 3. What is the percentage difference between 7.2 and 8.5? Solution: Here, we have to calculate the difference between two numbers. Difference $= 7.2 \;-\; 8.5 = \;-\; 1.3$ We will remove the negative sign by taking the absolute value of the difference. Absolute Difference $= \|Difference\| = \|\;-\;1.3\| = 1.3$ Now, you can calculate the percentage difference. Average $= \frac{7.2 + 8.5}{2} = 7.85$ Percentage Difference $= (\frac{Absolute \;Difference}{Average}) \times 100$ Percentage Difference $= (\frac{1.3}{7.85}) \times 100$ Percentage Difference $≈ 16.56\%$ Practice Problems on Percent Difference How to Calculate Percent Difference - Definition, Formula, Examples Attend this quiz & Test your knowledge. 1 The population of two cities A and B is 50,000 and 55,000 respectively. What is the percentage difference? $5\%$ $5.8\%$ $920\%$ $9.52\%$ CorrectIncorrect Correct answer is: $9.52\%$Population difference $= 55,000 \;-\; 50,000 = 5,000$ Average $= \frac{(50,000 + 55,000)}{2} = 52,500$ Percent difference $= \frac{5,000}{52,500} \times 100 ≈ 9.52\%$ 2 The weight of an object A is 200 grams. The weight of object B is 180 grams. What is the percent difference? $8\%$ $10\%$ $14\%$ $25\%$ CorrectIncorrect Correct answer is: $10\%$Percent Difference $= \frac{\|Weight \;difference\|}{(Average\; weight)} \times 100$ Percent Difference $= (\frac{20}{190}) \times 100$ Percent Difference $≈ 10.53\%$ 3 The population of town A is 10,000 and that of B is 12,500. What is the percent difference? $22\%$ $25\%$ $45\%$ CorrectIncorrect Correct answer is: $22\%$Difference $= 12,500 \;-\; 10,000 = 2,500$ Average$ = 11,250$ Percent difference $=(\frac{2,500}{11.250}) \times 100 ≈ 22.22\%$ 4 The price of a watch is $\$80$. The price of a clock is $\$58$. Choose the correct percent difference in the price. $25\%$ $22\%$ CorrectIncorrect Correct answer is: $32\%$Difference $= \$80 \;-\; \$58 = \$22$ Average $= 69$ Percent difference $=(\frac{16}{72}) \times 100 ≈ 32\%$. Frequently Asked Questions on Percent Difference Why do we take the average of two values in the percent difference formula? Since there is no way to decide which value is the reference value, we consider the average of two values as the reference value. It helps us treat both the values equally. Why do we ignore the minus sign when calculating the percent difference? Since we focus on the magnitude of the change (regardless if it is positive or negative) and not on the direction of the change, we don’t take the minus sign into account. Since both the values are equally important, the use of the negative sign is not useful. When do we use the percent error formula? When we want to compare the exact value with the estimated value (or measurement), we prefer the percent error formula. It helps us quantify the accuracy or precision of a measurement. What does percent change, percent error, and percent difference tell us? Percent change generally indicates growth or decline, the percent error measures accuracy in the measurement, and the percent difference helps us quantify variation or inconsistency in the two values. RELATED POSTS Converting Fractions into Decimals – Methods, Facts, Examples Math Symbols – Definition with Examples 270 Degree Angle – Construction, in Radians, Examples, FAQs Math Glossary Terms beginning with Z Factor Tree – Definition with Examples Math & ELA \| PreK To Grade 5 Kids see fun. You see real learning outcomes. Make study-time fun with 14,000+ games & activities, 450+ lesson plans, and more—free forever. Parents, Try for FreeTeachers, Use for Free
3222	https://www.youtube.com/watch?v=9x8bf-kl83o	The Binomial Theorem with a Real Exponent Jim Simons 1720 subscribers 21 likes Description 546 views Posted: 10 Nov 2024 The Binomial Theorem give us a power series for (1+x)^r. When r is a natural number it is easy to prove, but when it is a general real number, much less so. A common proof uses fairly advanced calculus (Maclaurin's theorem), but at least in English schools, that is not taught until well after the binomial theorem. In this video I talk through a proof that doesn't use calculus at all. 8 comments Transcript: hello in this video I want to talk about the binomial theorem and in particular the binomial theorem when the exponent is a general real number and I want to prove the result without using calculus so let's remind ourselves what all that means the simplest case of the binomial theorem um generalizes this very familiar statement to Powers here that are a natural number so we can increase the power like this and we can decrease it like that and for me natural numb starts zero so I'm going to include zero so the numbers that appear here are of course Pascal's triangle the binomial coefficients this is four choose two for example and that's all very familiar stuff so we can write that slightly more formally have a general real number and a general natural number and we can write 1 plus X the n as this sum running from K N to the N of n chose k x the K and we can write that out in this form so this is just the formula for n choose K this is the general term um and so those mean exactly the same thing and when we advance the exponent being a real number this time we have a general real number as the power here and it's very much the same at least as far as the algebra is concerned except that this series doesn't stop so that's what we're going to prove so let's go back to the natural number version and let's rewrite this slightly so I'm get to run the sum out to Infinity so in the case of the algebra version um that's fairly easy because when K gets Beyond n gets to n plus one this term is zero this factor is zero so this term is zero and so are all the later terms because they all have this Factor so actually this is the same as that and so this is as well the same as that people get squeamish about having n choose K when K is bigger than n but actually it shouldn't be so how many ways are there of choosing seven objects from a set of three well done there aren't any you can't do it so of course it's zero so of course that means the same as that so now what we do is we Define this function here to be that same thing but with r now as a general real number so starting with this version that is exactly what we had before and it's a straightforward function isn't it nothing very complicated about it except that it goes on forever um now writing it this version now this no longer means the binomial conent it no longer means R choose K you can't two three objects and seven and a half in any particular number of ways that doesn't make sense it just another way of writing this um except that there a way of writing it that reminds us that if R happens to be a natural number then this is um that Chooser thing it is the binomial coefficient so we know that R happens to be a natural number this thing here is 1 plus x to the r and our task is to prove that it is even if R is not a a natural number now um this thing obviously is really a function of both X and R isn't it um but I'm going to think of it as a function of r with X fixed so so far from doing any calculus on X I'm actually going to fix X for the rest of the video so we can imagine X is some particular value it's got to be less than one an absolute value it might be 38 or whatever and it's going to be fixed for the rest of the video so if you want to prove it for another X you'll have to go back to the beginning of the video and do it again but anyway so X is fixed and we think of this as a function of r the first thing to do though of course is we have to prove that this means anything at all in other words this infinite series does converge to something okay so here's our infinite series let's consider the ratio of consecutive terms so here it is this is this um denominator is um this general term and the numerator is the next term so it's very similar it's got one extra Factor on the top of R minus K one extra Factor on the bottom of k + 1 and one extra factor of X almost everything cancels and end up with this as K gets large this tends to minus X doesn't it so if if R is a sensible number like three and a half or minus five or something then by the time K gets to a million this is to all in intense and purposes minus X and so and that's the ratio of um consecutive terms and that ratio is less than one in absolute value so it behaves exactly like a geometric sequence and so it's going to converge now there are several points in this video where I'm not strictly going to prove this thing to um as it were University math standards and this is one of them um so I'm expecting you to believe that it's plausible at least that if this thing this series here ends up looking pretty much exactly like a geometric sequence then it's going to converge um the usual proof that you see of of the binomial theorem actually uses mclen theorem if you know what that is um if you don't it doesn't really matter um it's a piece of relatively Advanced calculus and at least in English schools it isn't introduced until way after this binomial theorem and so you can't use that to prove it and so I suspect it often isn't proved for AEV pupils but actually the mclorin theorem is no let out for this sort of slight hand wavess because the proof that mclorin theorem works is actually a little bit subtle and is again sort of University level math although it's used in a level anyway oh we go so we accept that this is converges the other thing we have to accept is that it's going to converge to a continuous function now that's slightly subtler but let me see how that works so we see that the tale of this um series uh behaves like a um geometric series and so we can chw we can chop the series in half um quite far out to the right in such a way that the tail is absolutely tiny uh like 10us 19 or something um and and it's going to be tiny for all sort of nearby values of R if you like and what's before the tail is just a polinomial so of course it's continuous so the sum of those two things is continuous because if you are change R just a little bit this is a polinomial so it only change a little bit and the Tail's so tiny anyway that that doesn't change much either so again that's not a proper proof but I hope you accept that that means that this thing a it converges something and B what it converges to is a continuous function of r okay so let's think about what this function of art actually looks like so we're going to multiply two versions of it together so it's that times that now there's a third point of which I'm going to playay fast slightly FAS and loose I'm going to assume that we can work out what this is by multiplying it term by term all every one of these by every one of these and adding them all up in any way we choose so that again it's a little bit of um quite subtle um maths to prove that that is valid it is for ser like that it's not valid for all series but for one like this that converges rapidly with this sort of geometric series type ending then then it is valid and by the way it was Newton who first did all this stuff about binomial theorems with um with a non-natural number exponent um and he wouldn't have worried about any of this subtlety um so he would have been quite happy with all these things that we've assumed um because that level of sort of um degree of rigor didn't really enter mathematics until um quite some way after Newton so we're just going to do what do the obvious thing here and I suppose Newton would it would never occur to Newton that this isn't valid so anyway let's do that so if we multiply any one of these by any one of these we're going to get something with a power of X in it and and we're going to collect all the powers of X together so constant terms any one of those is that times that and that's one the x is that times that plus that times that so that's that quadratic term a little bit more complicated so it's 1 time that one times that and that times that so let's collect all that together over two factorial and what have we got here so we got R time R -1 we've got s s - one and what's left over is the cross term here 2 r s over two factorial so that's just RS now that begins to look as though it might be equal to f of x of r+ X after all this is 1 + r plus s x plus r plus s r s - 1/ 2 factorial X2 well spoiler alert it is that let's see if we can establish that so let's look at the third term uh so it's a little bit more complicated here we have 1 S3 um R S2 and so on and again I'm using this s23 thing not because it's not because it's actually s23 but it's just a short hand way of saying this okay so we multiply all out put it every three factorial this one here for example is s cubed - 3 3 s^2 + 2 s and all the rest Works um so these ones are only over two factorial so we have to multiply them by three um collect all the terms um by the power so these are all the cubics all the quadratics and all the linear terms and now look what happens here um so that is R plus s cubed isn't it R CU + 3 R 2 s + 3 r s 2 + S cub and this is - 3 r + S 2 and this is 2 r + S and lo and behold that's compare that with this so it's something cubed minus 3 that thing squar plus two that thing so that's that so that's what we wanted it to be great so now let's look at the Ki term so let's look at the cubic term just again that's what we just proved we can write that this way so that's R2 time s23 and so on that's got to be R plus S3 so the general term is going to look like that we need to prove that so we need to prove this which is know as a juu vond identity um there's something funny about this um this guy here is a 13th 14th century Chinese mathematician whose name is usually spelled Zed hu and so I presume it's um pronounced zuu uh but this theorem always seems to be spelled Chu I don't understand that anyway that's who it is and this is the French guy much more recent right so um that's what we're trying to prove for for K being a general natural number and RNs being real now then the first thing to observe this is easy if RNs happen to be natural numbers because this is R plus s choose K and if you're going to choose K things from a set of R plus a set of s well you've got to choose some RS and some s's so you either choose no RS and all K s's or one R and Kus one s's or so on so that's easy that's definitely true if R it's natural numbers so now we consider this thing here if we take that and subtract that from it now this is polom in RNs so RNs are real variables um so it's a polinomial in the real variables R and S and it's a degree at most K because this is a degree K and that's a degree K and it's zero whenever R and S are natural numbers at least with r and s greater R plus s being greater than equal to K well it sounds as though that Jolly ought to be zero doesn't it just there a z the zero polinomial and it is so let's prove that of course um just that it's got a lot of zeros doesn't mean that it is zero so polinomial are two variables usually have infinitely many roots like this one it's got all those roots but the roots we have here are all these so all the natural number points going on forever in both directions a whole plane covered whole top right corner of the plane covered um with with roots and we need to prove that that means that this thing is in fact zero okay so we start with this polom the first thing we do um is we form a new polinomial by adding K to all of the um all of the parameters so if you like that's Shifting the graph of PK um down and to the left which means it's shifting all the roots down to the left so now it's still a pol polinomial of degree most K but it's now zero whenever R and S are natural numbers okay if you don't like the graph shifting idea um just look at if if R and S are natural numbers then these parameters are natural numbers whose sum is greater than K so it's definitely right okay right so now we consider this thing uh we set s equal to zero and what we get then is a polinomial in R just it's ordinary plal fashion polinomial in R of degree less than equal to K but it's got infinitely many roots all along the x- axis if you like well the r axis well that doesn't happen does it so that is actually zero which means that this thing has no terms of the form just R to the a because they're the ones that would survive if you set s equal to zero and of course in the same way this has no terms in the form s to the anything so all the terms are divisible by RS so let's divide by RS so we get a new polinomial dividing that one by RS now this is a polinomial of degree at most K minus 2 and that's zero whenever RNs well whenever RNs are positive integers anyway we can't say anything about whether it's zero when zero because it's sort of nonsense anyway so but it's positive interest but now we can play that translating trick again move it down to the left just one spot and now it's a polinomial of degree K minus two which is zero whenever R and S are natural numbers remind you of anything same here so do it again round and round and round we go and we end up with a polinomial of degree at most one um which is zero whenever R and F are natural numbers well that's not going to happen is it that's definitely impossible so now what what have we done here um we've proved that when we've gone we've done all these transformations of p and ended up with the zero polinomial but what is that transformation we've done a series of translations and of dividing by RS so we can undo all those Transformations and recover this from our original Zero polinomial by doing a succession of things which are either translations uper to the right actually or multiplying by RS but none of those does anything to turn the zero polinomial into anything else so if you undo all those things you still got the zero polinomial which means this was Zero all along long so we done it right so actually we've now proved that ju vandon theorem and we've now proved that we can take those question marks away that really is true so this times that is that well now we've done all the hard work really so we've proved that and we want to prove the sort of the extended version where if we multiply any number of these fxs together we end up with a single FX evaluated at the sum of all the um parameters of this thing well that's a simple inductive proof isn't it so suppose that almost true let's prove the next one along so the product of n plus one of these things is the product of n of them times the last one by the inductive hypothesis that product is equal to that and by the starting version that times that is equal to that so that's true right now then what good is all that we can now use it to work out what FX actually is so let's suppose M and N are co-prime positive integers no common factor and we'll also assume that m is odd so it's a third or four nths or whatever anyway okay so let's look at FX evaluated at this particular sort of rational number and raise it to the power M well by that functional equation we've established um this is the product of M of these fxs so it's a single FX evaluated the sum of all these which is M that so that's FX of N and that's 1 + x to the N because that's we know what that is when is a natural number harah so now that means this must be the MTH root of that which is that harah now now you see why we have to have n odd because if n were even we we couldn't really quite deduce this it would be sort of plus or minus the N root that's to awkward but any rate so now we've actually proved the binomial theorem where the exponent is a rational number whose denominator is odd well actually we only proved it for a positive rational number haven't we but we can do the negative ones easily because now so if Q is is this R positive rational number then FX of minus Q FX of Q is FX of zero which we know is one and so FX of minus Q is is one over FX to the Q which is therefore on plus xus Q so now we've proved the binomial theorem whenever the EXP oent is a a positive or negative rational number whose denominator is odd but now let's think about a real exponent so these two things here are both continuous functions that definitely is and when we established right at the beginning that is too so if they agree on all rational numbers with all denominators they agree everywhere so they're equal for all so we're done but now I want to go back and think about what a real power a particular an irrational power really is because I'm not sure that that's really addressed terribly well um in the a level syllabus so we we learn what a rational power is so we know that x to the 2/3 for example is the cube root of x squ um but then the syllabus at least doesn't mention anything about defining what a an irrational power is but later on we sort of magically work out we we work with the function e to the X where X is any real number but nobody's ever told anybody what a real what an irrational power is so let's look at that so what is two to the pi well it certainly isn't the product of Pi copies of two is it that's a nonsense so what is it well we can work out roughly what value it ought to have because we can we know how to approximate pi and so we can work and these are rational numbers so we can work out those okay and there are those and those are obviously converging to something so we sort of know what to to the pi ought to be and it's a bit odd isn't it that this looks like an algebraic operation but we're going to have to Define it by taking limits well I'm afraid we do that's the way life is it's rather awkward uh let's do this properly um so suppose we have a we've got a real number and we got a ration irrational R well we can consider rationals either side of R if you like Q just a bit smaller and R just a bit bigger so we know what x of the q and x of the s and all these are less than all those because this is an increasing function but there's no biggest x to the Q because given any Q less than R you can find a bigger one and similarly there's no smallest x to the S on the other hand there can't be two numbers between all those and all those because we can choose a q and s as close together as we like and because this thing is is continuous we can get we can make X q and x the S as close together as we like there's no room for two numbers between all the X the q's and all the X the S there's but there's no biggest than no smallest there's exactly one number between all the X to the q's and all the X to the S's and that's our definition of x to the R so that's sort of fundamentally and deeply different from the way we Define rational powers and that is very interesting it's what makes the real number such an interesting and peculiar Place really that we have this combination of basically algebra multiplying adding and stuff and these limit things um called topology if you like uh so um we can see where two to the pi actually is and we can illustrate that last example here so so all these numbers are two to the something that's less than pi and the blue ones are ones that two to the two to the something bigger than Pi so we got the decimal approximations we've also got this one which actually isn't a terribly good approximation and this one which is a super approximation um and there's just one number between all the red ones and the blue ones and that's two to the pi so going back to here that that's how we Define it but of course it's already well doing that uh but it's got to work so we we need to check that when we've defined it that way this sort of operation works because otherwise it's not going to be a useful definition is it well it is because we can choose a q1 and and an S1 either side of R1 and a Q2 and an S2 either side of r two and so when so we then know that's equal to that because those are rational and that's equal to that and these are all all these are less than all those and in just a own way there isn't room in between these two things for for more than one number um but this one is definitely between that one and that one and this one's definitely between that one and that one so they're both in this Gap but there's no room for the two of them so they got to be the same one so yes that is equal so again this topological proof rather than Al roof so anyway I think that's all very interesting so I hope you enjoyed the video
3223	https://www.dictionary.com/browse/condone	Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips View synonyms for condone condone [kuhn-dohn] verb (used with object) condoned, condoning to disregard or overlook (something illegal, objectionable, or the like). The government condoned the computer hacking among rival corporations. 2. to give tacit approval to. By his silence, he seemed to condone their behavior. 3. to pardon or forgive (an offense); excuse. His employers are willing to condone the exaggerations they uncovered in his résumé. 4. to cause the condonation of; justify the pardoning of (an offense). 5. Law.to forgive or act so as to imply forgiveness of (a violation of the marriage vow). His spouse condoned his infidelity from the early years of their marriage. 1 / Video Player is loading. Duration 0:00 / Current Time 0:00 Loaded: 0% Remaining Time -0:00 This is a modal window. Beginning of dialog window. Escape will cancel and close the window. End of dialog window. Share TOP ARTICLES Powered by AnyClip Privacy Policy Keyboard Shortcuts condone / kənˈdəʊn, ˌkɒndəʊˈneɪʃən / verb to overlook or forgive (an offence) law (esp of a spouse) to pardon or overlook (an offence, usually adultery) Discover More Other Word Forms condonable adjective condoner noun uncondoned adjective uncondoning adjective conˈdonable adjective condonation noun conˈdoner noun Discover More Word History and Origins Origin of condone1 First recorded in 1615–25, but in general currency from its use in the British Divorce Act of 1857 (see condone def. 5 ); from Latin condōnāre “to absolve, grant pardon,” equivalent to con- “with, together” + dōnāre “to give”; con-, donate Discover More Word History and Origins Origin of condone1 C19: from Latin condōnāre to remit a debt, from com- (intensive) + dōnāre to donate Discover More Example Sentences Examples have not been reviewed. Samadi doesn’t condone the leadership of Iran, which his parents fled the country to escape. From Los Angeles Times Prestige has told the BBC that it does not condone any form of unprofessional or threatening conduct and is investigating the phone call and the information provided about previous ownership. From BBC In the ASA's ruling, it said it told Mars "not to condone or encourage irresponsible driving that was likely to breach the legal requirements of the Highway Code in their ads." From BBC “ABC News stands for objectivity and impartiality in its news coverage and does not condone subjective personal attacks on others,” the representative said. From Los Angeles Times "ABC News stands for objectivity and impartiality in its news coverage and does not condone subjective personal attacks on others," they shared in a statement. From Salon Advertisement Discover More Related Words excuse forget forgive ignore wink at Word of the Day roc [rok] Meaning and examples Start each day with the Word of the Day in your inbox! By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies. Quiz Q: Which of the following words means “something extremely light, flimsy, or delicate”? diaphanous gossamer idyllic Take the full quiz.Go to all quizzes Advertisement Advertisement Advertisement condonationcondor We Care About Your Privacy We and our 895 partners store and/or access information on a device, such as unique IDs in cookies to process personal data. You may accept or manage your choices by clicking below or at any time in the privacy policy page. These choices will be signaled to our partners and will not affect browsing data. We and our partners process data to provide: Use precise geolocation data. Actively scan device characteristics for identification. Store and/or access information on a device. Personalised advertising and content, advertising and content measurement, audience research and services development.
3224	https://math.stackexchange.com/questions/149622/finding-out-whether-two-line-segments-intersect-each-other	Skip to main content Finding out whether two line segments intersect each other Ask Question Asked Modified 7 years, 5 months ago Viewed 63k times This question shows research effort; it is useful and clear 14 Save this question. Show activity on this post. I need to know whether or not two line segments intersect. I thought the formula for that is y=mx+b, but I don't think that will work for what I need; at least, I think I need to first know whether the line segments intersect, because that formula just gives the point where the two lines will eventually intersect. For example, say I have two line segments. Line 1 has a starting point of (15,10) and its ending point is (30,17). Line 2 has a starting point of (29,5) and an ending point of (33,14). These shouldn't intersect within those coordinates, but, by using y=mx+b, I find that they approximately intersect at (35,19.5). How can I find out whether these line segments happen to intersect each other? algebra-precalculus analytic-geometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Mar 21, 2018 at 23:06 user168764 asked May 25, 2012 at 12:33 daveMacdaveMac 28511 gold badge44 silver badges99 bronze badges Add a comment \| 5 Answers 5 Reset to default This answer is useful 14 Save this answer. Show activity on this post. The only reason that two lines will not intersect is if they are parallel. Being parallel is the same as having the same slope. So, in your example, line 1 has slope 17−1030−15=715. Line 2 has slope 14−533−29=94. Since these two numbers are not the same, the lines are not parallel, and they intersect somewhere. Now, if what you are considering is only the line segment between the two points, I would first consider the slopes as we just did. If they are parallel, then for sure you know that they do not intersect. If, however, the slopes are different, then they will intersect in one point. You then need to find this point, and see if it happens in bounds given on the x-values. So, when you find that the lines intersect at the point (35,19.5), then this intersection is outside the bounds given, since, e.g., your first line only goes from x=15 to x=30. It never reaches x=35. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Mar 21, 2018 at 23:09 user168764 answered May 25, 2012 at 12:41 ThomasThomas 44.6k1313 gold badges7676 silver badges139139 bronze badges 3 1 sorry for lately asking this question. Can you please explain why you need to test only X coordinate to know the lines segment intersect not both X and Y coordinate. Thanks in advance. – Trying Commented Aug 28, 2013 at 10:16 2 @Trying: Two lines can only intersect in one point. If the x-coordinate to that point is outside the given interval, then the two line segments will not intersect. – Thomas Commented Aug 28, 2013 at 15:11 Depending on how ones calculates the intersection, this algorithm may not work for vertical line segments, since the slope is undefined. – Ceasar Commented Apr 23, 2018 at 20:54 Add a comment \| This answer is useful 12 Save this answer. Show activity on this post. I'm assuming you can already check intersection of lines, as that's already been answered and you seem to understand it fine. The segments are AB and CD. They intersect if their intersection point lies within the darker middle rectangle (i.e. the area in space that they both occupy). In other words, if the intersection point is (x,y), then x must be less than the smallest right-side value (the x-coordinate of AH here), and larger than the smallest left-side value (GB). Here the check x>GBx fails, so they don't intersect. The idea is similar for y values, and it has to pass all 4 tests (two for x and two for y) before you can conclude that they intersect. Share CC BY-SA 3.0 Follow this answer to receive notifications answered May 25, 2012 at 16:12 Robert MastragostinoRobert Mastragostino 16.1k33 gold badges3838 silver badges5656 bronze badges 3 I think this is essentially what Thomas described at the end of his answer. – daveMac Commented May 25, 2012 at 16:24 1 I was mainly trying to clarify that concept, and point out that it could be within the right x values but not count as an intersection point because it's still outside of the right y values, which his answer didn't mention. – Robert Mastragostino Commented May 25, 2012 at 17:02 That is very true about the need to check the y as well. (I think I just took that for granted.) – daveMac Commented May 25, 2012 at 17:34 Add a comment \| This answer is useful 12 Save this answer. Show activity on this post. There are two cases to consider when determining if two line segments AB and CD intersect: (1) The line segments are not co-linear (top three images in the following figure); (2) the line segments are co-linear (bottom two images). The standard y=mx+b is not generally useful as it omits vertical lines. Here it is best to consider the following implicit function h(P) for a line passing through A and B: h(P)=(B−A)×(P−A)=0 where U×V=Ux⋅Vy−Uy⋅Vx. Note that h(P) defines a half space by determining where the point P lies relative to the boundary line through AB: h(P)>0h(P)=0h(P)<0P in positive half-spaceP on the lineP in negative half-space Thus we know if the points C and D straddle the (infinite) line through AB if both h(C) and h(D) are non-zero and have opposite signs. In the general case, we know the line segments AB and CD intersect if the points C and D straddle the line through AB and the points A and B straddle the line through CD. First we must handle the co-linear case where h(C)=0 and h(D)=0. Here we have an intersection iff min(Cx,Dx)≤max(Ax,Bx)∧max(Cx,Dx)≥min(Ax,Bx) and min(Cy,Dy)≤max(Ay,By)∧max(Cy,Dy)≥min(Ay,By) Otherwise, in the general case we use our half space equations. The half space g(x) defined by CD is g(P)=(D−C)×(P−C)=0. We have an intersection where (h(C)⋅h(D)≤0)∧(g(A)⋅g(B)≤0). If you wish to know the actual intersection point you plug the parametric equation L(t) for a line through AB L(t)=(B−A)⋅t+A, −∞<t<∞ into g and solve for t: g(L(t))=(D−C)×(L(t)−C)=0(D−C)×((B−A)⋅t+A−C)=0t=(C−A)×(D−C)(D−C)×(B−A) Plug this value for t back into L(t) and you have your intersection point. Of course the assumes the caveat that (D−C)×(B−A)≠0 where the lines are not parallel. Another great explanation for this can be found here. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Sep 7, 2015 at 17:46 wcochranwcochran 86288 silver badges1616 bronze badges 1 2 This is the most comprehensive and easiest to implement answer I have found. – Cristian Arteaga Commented Oct 30, 2021 at 23:24 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. If two lines have unequal slope they will intersect in a a point. If two lines have equal slope, they are either disjointly parallel and never intersect, or they are the same line. Share CC BY-SA 3.0 Follow this answer to receive notifications answered May 25, 2012 at 12:34 ncmathsadistncmathsadist 50.1k33 gold badges8585 silver badges134134 bronze badges 2 5 As it seems the OP is using "line" to mean "line segment", in the case that the slopes are different, it is also necessary to check whether the intersection occurs in the necessary segments, which is easy enough to do from the coordinates of the intersection point. – mdp Commented May 25, 2012 at 12:39 I updated my question to clarify that I am trying to figure out if the line "segments" intersect. – daveMac Commented May 25, 2012 at 14:55 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. A necessary condition to check the turn it makes. Assume U and V to be the vectors of the segments where U=(ux2−ux1,uy2−uy1), likewise for V. If ((ux1,uy2)−(vx1,vy1))∗V<0, they will intersect. If the scalar product is positive does not intersect. Otherwise, this fast test doesn't apply. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Mar 21, 2018 at 23:13 user168764 answered Nov 26, 2016 at 7:20 user393185user393185 1 0 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 algebra-precalculus analytic-geometry See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Community help needed to clean up goo.gl links (by August 25) Linked 4 Determine whether two line segments intersect, given that one of them is axis-aligned Related 1 I want to find 3 planes that each contain one and only one line from a set 5 given coordinates of beginning and end of two intersecting line segments how do I find coordinates of their intersection? 0 Setting two equations equal to each other 1 if I know a point nearest the zero of a polynomial can I tell which zero it is? 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3225	https://eepower.com/technical-articles/resistance-measurement-basics/	Log In Join Join the Community Register Log In Or sign in with Facebook Google GitHub Linkedin SIGN UP FOR OUR NEWSLETTER Technical Article Understanding Resistance Measurement Tools Join our Engineering Community! Sign-in with: 2 Home Technical Articles Understanding Resistance Measurement Tools Understanding Resistance Measurement Tools Learn about resistance measurement through two methods: the Ammeter/Voltmeter approach and the megohmmeter. The Ammeter/Voltmeter method uses standard electrical instruments, while the megohmmeter is designed specifically for high-resistance measurements. Technical Article Sep 24, 2023 by Amna Ahmad Two primary methods are employed in resistance measurement: the ammeter/voltmeter approach and the specialized megohmmeter. The ammeter/voltmeter method measures resistance by measuring voltage and current across a resistor, though errors can arise based on the connection arrangement. On the other hand, the megohmmeter is specialized for high-resistance measurements. It is available in two versions: the hand-cranked and battery-powered variants, each offering specific advantages for precise resistance assessment in diverse electrical applications. Image used courtesy of Freepik Ammeter/Voltmeter Methods If the voltage across a resistor and the current flowing through it are measured, the resistance value can be calculated by applying Ohm's law. However, an error occurs depending on how the ammeter and voltmeter are connected, and this error may be insignificantly small or large enough to be important. Consider the arrangement shown in Figure 1(a). Because the voltmeter is connected directly across Rx, it measures the actual resistor voltage. However, the ammeter measures the resistor current Ix and the current Iv that flows through the voltmeter. Using the voltage and current measurements, we can calculate a resistance value as follows: [Calculated\,Resistance=\frac{Voltmeter\,Reading}{Ammeter\,Reading}] [R_{X1}=\frac{V_{X}}{I_{X}+I_{V}}] Equation 1. (a) Ammeter measuring Iv + Ix (b) Voltmeter measuring VA + Vx Figure 1. Errors can occur when resistance is measured using an ammeter and voltmeter. To minimize errors, the voltmeter should be connected directly across Rx when Rx is low, and the ammeter should be directly in series with Rx when Rx is high. Image used courtesy of Amna Ahmad The actual resistance of Rx is Rx = Vx/Ix, so the presence of Iv in the equation introduces an error in the result. If Iv is much smaller than Ix, the error may be insignificant. This requires that Ix be a large current, which is the case when Rx is a small resistance. Now consider the arrangement illustrated in Figure 1(b), where Rx and the ammeter are in series, and the voltmeter is connected in parallel with the two of them. In this case, the ammeter measures the actual current through Rx, but the voltmeter measures the voltage across Rx plus the voltage across the ammeter. Using these two measurements, the resistance is calculated as: [R_{X2}=\frac{V_{X}+V_{A}}{I_{X}}] Equation 2. Again, the actual resistance of Rx is Rx = Vx/Ix, and consequently, the presence of VA in Equation 2 produces an error in the result. The error could be insignificant if VA is much smaller than Vx. This requires that Vx be a large voltage, which means that Rx should be a large resistance. The voltmeter should be connected directly across Rx for the greatest accuracy when Rx has a small resistance. For the greatest accuracy when Rx has a large resistance, the ammeter should be connected directly in series with Rx. The correct arrangement can be easily determined by first connecting the ammeter in series with Rx, then observing the ammeter reading with the voltmeter temporarily connected directly across Rx [i.e., connected as in Figure 1(a)]. If the ammeter reading is not noticeably altered when the voltmeter is connected, the readings will give an accurate result. When the ammeter reading is noticeably changed by connecting the voltmeter, the voltmeter should be moved to the other side of the ammeter. Example 1 An ammeter and voltmeter connected as in Figure 1(a) indicate 10 A and 99 V, respectively. When the voltmeter is changed to reconnect the circuit, as in Figure 1(b), the readings become 10 A and 100 V. The ammeter has a resistance of 0.1 Ω. The voltmeter is on its 100 V range and its sensitivity is 20 kΩ/V. Calculate the measured resistance for each case and determine which arrangements give the most accurate result. Solution For the circuit of Figure 1(a) [R_{X}=\frac{V_{X}}{I_{X}+I_{V}}=\frac{99V}{10A}=9.9\Omega] The voltmeter resistance is [R_{V}=\frac{20k\Omega}{V}\times(voltmeter\,range)=\frac{20k\Omega}{V}\times100V=2M\Omega] [I_{V}=\frac{V_{X}}{R_{V}}=\frac{99V}{2M\Omega}=49.5\mu A] So [I_{V}< For the circuit of Figure 1(b) [R_{X}=\frac{V_{X}+V_{A}}{I_{X}}=\frac{100V}{10A}=10\Omega] And [V_{A}=I_{A}R_{A}=10A\times0.1\Omega=1V] In this case, VA is not much smaller than Vx, so the connection shown in Figure 1(a) gives the most accurate result. Example 2 The insulation resistance of an electrical cable is to be measured using a 1000 V source, a 1500 V voltmeter with a sensitivity of 10 k/V, and a 500 A ammeter with a resistance of 100. If the insulation resistance is 5 M, determine the measured resistance using the two methods of connecting the instruments. Solution Connection (a) Ix = 1000 V/5 MΩ = 200 µA Rv = 1500 V × 10 kΩ/V = 15 MΩ Iv = 1000 V/15 MΩ = 66.7 µA Measured R = V/ (Ix + Iv) = 100 V/ (200 µA + 66.7 µA) = 3.75 MΩ Connection (b) Ix = 1000 V/ (5 MΩ + 100 Ω) ≈ 200 µA Measured R = V/Ix = 1000 V/200 µA = 5 MΩ Megohmmeters for Resistance Measurement The megohmmeter, commonly referred to as a megger, measures exceptionally high resistances, such as the insulation resistance in electrical cables. To achieve this, a high-voltage source is necessary to pass a measurable current through these resistances. In essence, a megger functions like an ohmmeter but employs a low-current instrument and a high-voltage source. As depicted in Figure 2(a), the voltage is typically generated using a hand-cranked generator, ranging from 100 V to 2.5 kV. Similar to a low-resistance analog ohmmeter, the megger's scale reflects specific values. When measuring an open circuit, it shows infinity (∞), zero for a short circuit, and half-scale when the unknown resistance matches a standard resistor within the megohmmeter. For other positions on the scale, the deflection is directly proportional to the ratio of the unknown and standard resistors. The instrument's range is adjustable by switching various standard resistor values into the circuit. Megohmmeters operating on battery power are also available, essentially functioning as ohmmeters for very high resistance values. Figure 2(b) illustrates one such instrument. The battery voltage is typically increased to around 1000 V using electronic circuitry to generate a measurable current through the unknown resistance. The measurement is taken by pressing and holding the power button briefly, an arrangement that reduces the current drain on the battery. Hand-cranked megohmmeter Battery-powered megohmmeter Figure 2. A megohmmeter (megger) is essentially an ohmmeter for measuring very high resistances. A high-voltage source and a low-current meter are required. Image used courtesy of Amna Ahmad Resistance Measurement Takeaways Two primary methods for measuring resistance are the ammeter/voltmeter and the specialized megohmmeter approach. The ammeter/voltmeter approach calculates resistance by measuring voltage and current across a resistor, but circuit connections may affect its accuracy. On the other hand, the megohmmeter is specifically designed for high-resistance measurements and is available in two versions: hand-cranked and battery-powered. The hand-cranked megohmmeter operates using a hand-cranked generator to measure high resistances. Conversely, the battery-powered megohmmeter offers convenience and efficiency, enabling accurate measurements while minimizing current drain through a press-and-hold mechanism. Related Content Modern Approach to the Electrification of Consumer Outdoor Power Tools Industry Articles Using Probability Models to Understand Electrical System Reliability Engineering Technical Articles Double Pulse Testing eGuide Industry White Papers A Solution for Accurate High-Voltage Power Measurement Industry Articles Debunking GaN Cost and Reliability Myths Industry Articles Microsoft’s Quantum Computing Tools Expedite Battery Materials Research News AMI 2.0: Grid Management Tools Integrate Artificial Intelligence News Realizing Power-Saving Analog-to-Digital Conversion for Highly Accurate Measurements Technical Articles Learn More About ammeter resistance measurements Megohmmeters voltmeter Comments (2) P PicoPete September 29, 2023 Below figure (a), it’s shown resistance Rx = Vx/(Iv + Ix). That’s incorrect. The current Iv and Ix are separate from each other at least as long as V the voltmeter are in the circuit, where Iv is the current through the Voltmeter V, and the current Ix is the current through the resistor Rx. Current Iv does not flow through Rx. The current flowing through V the voltmeter may be significantly different than the current flowing through Rx. Therefore, Rx = Vx/Ix. So the only way the equation in figure (a) would be correct is if the current Iv = 0, which would presume an ideal voltmeter that uses absolutely no current in its coil/meter movement, which of course is not true in reality. Like. Reply + D Dale Wilson September 30, 2023 Hi PicoPete! You are correct. The equations below Figures 1a and 1b were meant to imply the value for Rx that was calculated in those cases, not the actual Rx value. As you pointed out though, that didn't match with the way the figures were drawn. We have removed the equations below the figures and tried to clarify the equation descriptions to indicate that these are calculated values for Rx, not the actual values for Rx. It is these differences due to measurement errors that the article is trying to highlight. Thanks for being an EE Power reader and bringing this issue to our attention! Please let us know if this is still unclear. Like. Reply) You May Also Like ### Traction Inverter Functional Safety Design with SiC Auxiliary Power Supply Technical Articles ### An Introduction to Wave Propagation in Long-Distance Power Transmission Technical Articles ### Scalable AI Microgrid Makes Hydrogen Fuel Onsite News ### Advanced Load Management Shows How Buildings Can Power EV Charging for All Live Session Welcome Back Or log in with Facebook Google GitHub Linkedin
3226	https://artofproblemsolving.com/wiki/index.php/Diophantine_equation?srsltid=AfmBOoocZOeQkEsd0_aHi5BP362p3AJgcCpQcoEWq0fBDged_BDwLvmR	Art of Problem Solving Diophantine equation - 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 Diophantine equation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Diophantine equation A Diophantine equation is an equation relating integer (or sometimes natural number or whole number) quanitites. Finding the solution or solutions to a Diophantine equation is closely tied to modular arithmetic and number theory. Often, when a Diophantine equation has infinitely many solutions, parametric form is used to express the relation between the variables of the equation. Diophantine equations are named for the ancient Greek/Alexandrian mathematician Diophantus. Contents [hide] 1 Linear Combination 2 Pythagorean Triples 2.1 Method of Pythagoras 2.2 Method of Plato 2.3 Babylonian Method 3 Sum of Fourth Powers 4 Pell Equations 5 Methods of Solving 5.1 Coordinate Plane 5.2 Modular Arithmetic 5.3 Induction 5.4 General Solutions 6 Fermat's Last Theorem 7 Problems 7.1 Introductory 7.2 Intermediate 7.3 Olympiad 8 References 9 See also Linear Combination A Diophantine equation in the form is known as a linear combination. If two relatively prime integers and are written in this form with , the equation will have an infinite number of solutions. More generally, there will always be an infinite number of solutions when . If , then there are no solutions to the equation. To see why, consider the equation . is a divisor of the LHS (also notice that must always be an integer). However, will never be a multiple of , hence, no solutions exist. Now consider the case where . Thus, . If and are relatively prime, then all solutions are obviously in the form for all integers . If they are not, we simply divide them by their greatest common divisor. See also: Bézout's identity. Pythagorean Triples Main article: Pythagorean triple A Pythagorean triple is a set of three integers that satisfy the Pythagorean Theorem, . There are three main methods of finding Pythagorean triples: Method of Pythagoras If is an odd number, then is a Pythagorean triple. Method of Plato If , is a Pythagorean triple. Babylonian Method For any (), we have is a Pythagorean triple. Sum of Fourth Powers An equation of form has no integer solutions, as follows: We assume that the equation does have integer solutions, and consider the solution which minimizes . Let this solution be . If then their GCD must satsify . The solution would then be a solution less than , which contradicts our assumption. Thus, this equation has no integer solutions. If , we then proceed with casework, in . Note that every square, and therefore every fourth power, is either or . The proof of this is fairly simple, and you can show it yourself. Case 1: This would imply , a contradiction. Case 2: This would imply , a contradiction since we assumed . Case 3: , and We also know that squares are either or . Thus, all fourth powers are either or . By similar approach, we show that: , so . This is a contradiction, as implies is odd, and implies is even. QED [Oops, this doesn't work. 21 (or ) are equal to and not even...] Pell Equations Main article: Pell equation A Pell equation is a type of Diophantine equation in the form for natural number. The solutions to the Pell equation when is not a perfect square are connected to the continued fraction expansion of . If is the period of the continued fraction and is the th convergent, all solutions to the Pell equation are in the form for positive integer . Methods of Solving Coordinate Plane Note that any linear combination can be transformed into the linear equation , which is just the slope-intercept equation for a line. The solutions to the diophantine equation correspond to lattice points that lie on the line. For example, consider the equation or . One solution is (0,1). If you graph the line, it's easy to see that the line intersects a lattice point as x and y increase or decrease by the same multiple of and , respectively (wording?). Hence, the solutions to the equation may be written parametrically (if we think of as a "starting point"). Modular Arithmetic Sometimes, modular arithmetic can be used to prove that no solutions to a given Diophantine equation exist. Specifically, if we show that the equation in question is never true mod , for some integer , then we have shown that the equation is false. However, this technique cannot be used to show that solutions to a Diophantine equation do exist. Induction Sometimes, when a few solutions have been found, induction can be used to find a family of solutions. Techniques such as infinite Descent can also show that no solutions to a particular equation exist, or that no solutions outside of a particular family exist. General Solutions It is natural to ask whether there is a general solution for Diophantine equations, i.e., an algorithm that will find the solutions for any given Diophantine equations. This is known as Hilbert's tenth problem. The answer, however, is no. Fermat's Last Theorem Main article: Fermat's Last Theorem is known as Fermat's Last Theorem for the condition . In the 1600s, Fermat, as he was working through a book on Diophantine Equations, wrote a comment in the margins to the effect of "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." Fermat actually made many conjectures and proposed plenty of "theorems," but wasn't one to write down the proofs or much other than scribbled comments. After he died, all his conjectures were re-proven (either false or true) except for Fermat's Last Theorem. After over 350 years of failing to be proven, the theorem was finally proven by Andrew Wiles after he spent over 7 years working on the 200-page proof, and another year fixing an error in the original proof. Problems Introductory Two farmers agree that pigs are worth dollars and that goats are worth dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way? (Source) Intermediate Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product . (Source) Olympiad Determine the maximum value of , where and are integers satisfying and . (Source) Solve in integers the equation . References Proof of Fermat's Last Theorem See also Number Theory Pell equation Retrieved from " Category: Number theory 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. 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3227	https://www.ck12.org/flexi/cbse-math/polynomial-division/what-is-the-quotient-remainder-theorem/	Flexi answers - What is the quotient remainder theorem? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Polynomial Division Question What is the quotient remainder theorem? Flexi Says: The quotient remainder theorem, also known as the division algorithm, is a fundamental theorem in arithmetic that states that if a whole number a is divided by a non-zero whole number b, then there exist whole numbers q and r such that a=b q+r, where r=0 or, r<b. This can also be expressed as: i Where a is the dividend, b is the divisor, q is the quotient and r is the remainder. Using these terms, the division algorithm can be restated as i Let us take any two whole numbers and divide them. i Note that 161=20×8+1 The number here (161) which is to be divided is called the dividend. The number (8) by which the dividend is divided is called the divisor. The number of times (20) the divisor is contained in the dividend is called the quotient. The number (1) which is left over after division is called the remainder. To learn more click here! Analogy / Example Try Asking: How do you divide quadratic equations by polynomials?How do you prove the remainder theorem?How do you do polynomial long division with remainders? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy × Image Attribution Credit: Source: License:
3228	https://math.fandom.com/wiki/Scalar_triple_product	Math Wiki Sign In Don't have an account? Register Sign In Skip to content Math Wiki 1,422 pages in: Linear algebra stubs, Linear algebra, Vector calculus Scalar triple product Sign in to edit History Purge Talk (0) The scalar triple product of three vectors , , and is the dot product of one vector with the cross product of the other two. Technically, it does not produce a scalar but a pseudoscalar, as reversing the direction of one of the vectors will change the sign of the scalar. Properties[] A parallelepiped whose edges are formed by the vectors , , and will have a volume equal to the scalar triple product of the three vectors. If any of the vectors are identical, the scalar triple product will be zero. See also[] Dot product Cross product Vector triple product Parallelepiped This linear algebra-related article contains minimal information concerning its topic. You can help the Mathematics Wikia by adding to it. Category list Community content is available under CC-BY-SA unless otherwise noted.
3229	https://www.organic-chemistry.org/namedreactions/diazotisation.shtm	Organic Chemistry Portal Reactions >> Name Reactions Further Information Literature Related Reactions Azo Coupling Balz-Schiemann Reaction Pschorr Reaction Sandmeyer Reaction Diazotisation The nitrosation of primary aromatic amines with nitrous acid (generated in situ from sodium nitrite and a strong acid, such as hydrochloric acid, sulfuric acid, or HBF4) leads to diazonium salts, which can be isolated if the counterion is non-nucleophilic. Diazonium salts are important intermediates for the preparation of halides (Sandmeyer Reaction, Schiemann Reaction), and azo compounds. Diazonium salts can react as pseudohalide-type electrophiles, and can therefore be used in specific protocols for the Heck Reaction or Suzuki Coupling. The intermediates resulting from the diazotization of primary, aliphatic amines are unstable; they are rapidly converted into carbocations after loss of nitrogen, and yield products derived from substitution, elimination or rearrangement processes. Mechanism of Diazotisation Recent Literature A convenient and general one-step preparation of aromatic and some heterocyclic iodides in good yields includes a sequential diazotization-iodination of aromatic amines with KI, NaNO2, and p-TsOH in acetonitrile at room temperature. E. A. Krasnokutskaya, N. I. Semenischeva, V. D. Filimonov, P. Knochel, Synthesis, 2007, 81-84. Sulfonic Acid Based Cation-Exchange Resin: A Novel Proton Source for One-Pot Diazotization-Iodination of Aromatic Amines in WaterV. D. Filimonov, N. I. Semenischeva, E. A. Krasnokutskaya, A. N. Tretyakov, H. Y. Hwang, K.-W. Chi, Synthesis, 2008, 185-187. Unusually Stable, Versatile, and Pure Arenediazonium Tosylates: Their Preparation, Structures, and Synthetic ApplicabilityV. D. Filimonov, M. Trusova, P. Postnikov, E. A. Krasnokutskaya, Y. M. Lee, H. Y. Hwang, H. Kim, K.-W. Chi, Org. Lett., 2008, 10, 3961-3964. A Simple and Effective Synthesis of Aryl Azides via Arenediazonium TosylatesK. V. Kutonova, M. E. Trusova, P. S. Postnikov, V. D. Filimonov, J. Parello, Synthesis, 2013, 45, 2706-2710. Methanol-Promoted Borylation of Arylamines: A Simple and Green Synthetic Method to Arylboronic Acids and ArylboronatesC.-J. Zhao, D. Xue, Z.-H. Jia, C. Wang, J. Xiao, Synlett, 2014, 25, 1577-1584. Efficient Synthesis of 2-Amino Acid by Homologation of β2-Amino Acids Involving the Reformatsky Reaction and Mannich-Type Imminium Electrophile R. Moumne, S. Lavielle, P. Karoyan, J. Org. Chem., 2006, 71, 3332-3334. Synthesis and properties of bis(hetaryl)azo dyes M. Wang, K. Funabiki, M. Matsui, Dyes and Pigments, 2003, 57, 77-86. Advertise here Main Categories Organic Reactions Org. Chem. Highlights Abstracts Chemicals Job Market Presentations Chemistry Tools Ads & Imprint Popular Subcategories Name Reactions Protecting Groups Organic Synthesis
3230	https://study.com/academy/lesson/how-and-why-to-use-the-general-term-of-an-arithmetic-sequence.html	General Term of an Arithmetic Sequence \| Overview, Formula & Uses - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses Algebra II: High School General Term of an Arithmetic Sequence \| Overview, Formula & Uses Contributors: Stephanie Stanglin, Yuanxin (Amy) Yang Alcocer Author Author: Stephanie Stanglin Show more Instructor Instructor: Yuanxin (Amy) Yang Alcocer Show more Learn the definition of arithmetic sequence and general term of a sequence. Learn the formula for general term of a sequence and see examples. Updated: 11/21/2023 Table of Contents Term of a Sequence What is Arithmetic Sequence? General Term of a Sequence General Term Formula How to Find a Term in an Arithmetic Sequence? Lesson Summary Show FAQ What is the general term of sequence? The general term of a sequence is the ability to find any term in the sequence given the first number, common difference, and term number. The purpose is to be able to find it using a formula without having to count out (using the common difference) to that term number. How do you find the general term of a sequence? To find the general term of a sequence use the general term formula. The general term formula is: subtracting 1 from the desired term number, multiplying that by the common difference, then adding that result to the first term in the sequence. What is arithmetic series? An arithmetic series is the sum of a given number of terms in the associated arithmetic sequence. It is found by finding the first and last terms in the section of the sequence (using the general term formula), adding those up, multiplying that result by the number of terms, then dividing by 2. What is arithmetic sequence example? An arithmetic sequence example is one where there is a common difference between each term, helping to find the next term. An example of this would be {4, 8, 12, 16...} where the common difference is 4 because to get from one term to the next, add 4. Create an account LessonTranscript VideoQuizCourse Click for sound 5:02 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. 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Your next lesson will play in 10 seconds 0:01 An Arithmetic Sequence 1:23 The General Term 2:04 Using the General Term 3:26 Why We Use the General Term 4:08 Lesson Summary QuizCourseView Video Only Save Timeline 62K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Arithmetic Series: Formula & Equation 5:02 ##### Sum of Natural Numbers \| Overview, Formula & Examples 4:58 ##### Consecutive Integers \| Definition, Formula & Examples 5:42 ##### How to Find and Classify an Arithmetic Sequence 9:09 ##### How to Calculate an Arithmetic Series 5:45 ##### How to Evaluate & Write Variable Expressions for Arithmetic Sequences 7:08 ##### Common Difference \| Definition, Formula & Examples 4:01 ##### Arithmetic Sequences \| Definition, Formula & Examples 5:44 ##### Linear Pattern Formula & Examples 4:39 ##### Finding the Sum of Consecutive Numbers 4:10 ##### Recognizing and Solving Mathematical Patterns 4:41 ##### Recurrence Relation \| Definition, Examples & Formula 9:21 ##### Factorial Practice Problems 6:04 ##### Summation \| Definition, Rules & Examples 3:36 ##### Recursive Sequence Formula, Properties & Example 5:44 ##### Sequences in Math \| Overview & Types 5:37 ##### Arithmetic and Geometric Series: Practice Problems 10:59 ##### Infinite Sequence: Definition & Examples 6:39 ##### Finite Sequence: Definition & Examples 6:07 ##### Sequences \| Study.com ACT® Math Test Prep 6:52 ##### Algebra I: High School ##### Math Review for Teachers: Study Guide & Help ##### Common Core Math - Number & Quantity: High School Standards ##### Common Core Math - Algebra: High School Standards ##### Common Core Math - Statistics & Probability: High School Standards ##### Common Core Math - Geometry: High School Standards ##### Common Core Math - Functions: High School Standards ##### Math 105: Precalculus Algebra ##### Algebra II: High School ##### Glencoe Algebra 1: Online Textbook Help ##### McDougal Littell Algebra 2: Online Textbook Help ##### DSST Business Mathematics Study Guide and Test Prep ##### Glencoe Math Connects: Online Textbook Help ##### Glencoe Pre-Algebra: Online Textbook Help ##### Glencoe Math Course: Online Textbook Help ##### Algebra I: High School ##### Math Review for Teachers: Study Guide & Help ##### Common Core Math - Number & Quantity: High School Standards ##### Common Core Math - Algebra: High School Standards ##### Common Core Math - Statistics & Probability: High School Standards Term of a Sequence ------------------ A sequence is a list of numbers that continues on forever and follows a pattern to get from one term to the next. Each number in the sequence is a called a term of the sequence, meaning the number in a certain place in the sequence. For example, in the sequence {2,4,6,8,10...} The pattern is adding 2 to get to the next number. The first term is 2, the second term is 4, and so on. The ellipses at the end signifies that the sequence can go on forever. To unlock this lesson you must be a Study.com memberCreate an account What is Arithmetic Sequence? ---------------------------- An Arithmetic Sequence is a sequence where there is a common difference between one term and the next, which is how the next term is found. In the previous example, the common difference is 2, because that is what is added to one term to get the next term. It is important to note that this common difference can only be addition or subtraction; multiplication and division are geometric sequences. An arithmetic sequence can also be represented on a graph where the term number is the x-value and the term value is the y-value. For example: Graph of the Example To unlock this lesson you must be a Study.com memberCreate an account General Term of a Sequence -------------------------- A general term of a sequence is any term in the sequence that someone wants to find. It is without a number, because there is a way to find any term in the sequence using a formula. Use of General Term In the previous example, it would be easy to find the 6th term, 10th term, or 50th term by continuing to add the common difference (2) to the terms. However, this is a lot of work and not the most efficient way to find a term, especially when the term number gets so large. This is why people use the general term process, to find terms that are much farther in the sequence and time consuming to find. To unlock this lesson you must be a Study.com memberCreate an account General Term Formula -------------------- The general term formula was created to help people find a term quickly and accurately without needing to repeat the common difference many times. The general term formula for an arithmetic sequence is: x n=a+d(n−1) where x n is the value of the nth term, a is the starting number, d is the common difference, and n is the term number. To unlock this lesson you must be a Study.com memberCreate an account How to Find a Term in an Arithmetic Sequence? --------------------------------------------- Finding a term in an arithmetic sequence is done by plugging in numbers for n, a, and d. In the previous example to find the 25th term, plug 25 in for n, 2 in for a because it was the first number in the sequence, and 2 in for d because it is the common difference. x 25=2+2(25−1) This simplifies to x 25=2+2(24)=2+48=50, so the 25th term in the sequence is 50. That can be checked by listing out the first 25 terms in the sequence: {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50}. This matches up with the answer from the formula, but took much longer than using the general term formula. On a graph this sequence would look like Graph of the Arithmetic Sequence The point (25,50) can be seen on the graph, further proving that the formula works and that plotting all these points would take a long time. General Term Arithmetic Sequence Examples Example 1: Given the sequence {3,7,11,15,19...}, find the 13th term. Step 1: Identify n, a, and d n = 13 (the term number) a = 3 (the first number) d = 4 (the common difference, adding 4 to get to the next term) Step 2: Plug in n, a, and d into the general term formula x 13=3+4(13−1) Step 3: Simplify x 13=3+4(13−1)=3+4(12)=3+48=51 So the 13th term of the sequence is 51. Example 2: Given the sequence {20,15,10,5,0...}, find the 24th term. Step 1: Identify n, a, and d n = 24 (the term number) a = 20 (the first number) d = -5 (the common difference, subtracting 5 to get to the next term) Step 2: Plug in n, a, and d into the general term formula x 24=20+−5(24−1) Step 3: Simplify x 24=20+−5(24−1)=20+−5(23)=20−115=−95 So the 24th term of the sequence is -95. Example 3: Given the sequence {2,2.75,3.5,4.25,5...}, find the 100th term. Step 1: Identify n, a, and d n = 100 (the term number) a = 2 (the first number) d = 0.75 (the common difference, adding 0.75 to get to the next term) Step 2: Plug in n, a, and d into the general term formula x 100=2+0.75(100−1) Step 3: Simplify x 100=2+0.75(100−1)=2+0.75(99)=2+74.25=76.25 So the 100th term of the sequence is 76.25. This third example is a perfect case for why the general term formula exists for two reasons: 1) because the term number was so large so it would take a long time to count by hand, and 2) the common difference was a decimal so it is easy to get tripped up and make a mistake when adding decimals. To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- Anarithmetic sequence is a sequence, or list, of numbers that follow a pattern or common difference (adding or subtracting the same number each time to get the next term). To find any general term in the sequence, use the general term formula which is x n=a+d(n−1) where x n is the value of the nth term, a is the starting number, d is the common difference, and n is the term number. This is a much easier and faster way of finding term values in an arithmetic sequence. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript An Arithmetic Sequence We will be talking about the general term of an arithmetic sequence in this video lesson. You will see the formula we use for the general term of any arithmetic sequence and how easy it is to use it. We begin with our arithmetic sequence, which is a string of numbers where each number is the previous number plus a constant. So, if you look at our number line, you will see that it is an arithmetic sequence since each number is the previous number plus 1. Number line This difference between the numbers is called the common difference. We write our sequences inside curly brackets with commas in between the numbers. Another way of thinking about arithmetic sequences is to think of things that grow over time. For example, we can represent the number of leaves a peach tree has each day with an arithmetic sequence. If our peach tree begins with 10 leaves and grows 15 new leaves each day, we can write the arithmetic sequence {10, 25, 40, . . .} to show the number of leaves that our peach tree has each day. This sequence ends when our peach tree stops growing. So, while some arithmetic sequences have an end, like our peach tree sequence, some don't and continue on forever, like our number line. Because our arithmetic sequences have a pattern, we have a formula for the general term of an arithmetic sequence. The General Term This general term is the formula that is used to calculate any number in an arithmetic sequence. General term formula The formula tells us that if we wanted to find a particular number in our sequence, x sub n, we would take our beginning number, a, and add our common difference, d, times n minus 1, which is the location of our desired number minus 1. If we are looking for the 30th number, our n is 30, so our formula begins with x sub 30, and n - 1 equals 29 (30 - 1). Using the General Term Let's look at an example to see how we use this general term. Here's a new sequence: {2, 4, 6, . . .}. We want to find the value of the number that is the 25th place in our sequence. Our n in this case is 25. Before we go further, we first have to check whether this sequence is an arithmetic sequence or not. To do that, we look at the difference between each successive pair of numbers to see if this difference is the same. If it is, then we are looking at an arithmetic sequence. If it isn't, then we can't use our general term formula to find our answer. In our case, we subtract 4 - 2 and also 6 - 4 to see if they equal the same thing. They do; they both equal 2. So, this means the sequence is an arithmetic sequence; it is an arithmetic sequence that jumps by 2 with every number. Now that we know this sequence is an arithmetic sequence, now we label our beginning number and our common difference. Our beginning number is 2, so 2 is a. Our common difference is 2, so our d is also 2. Now we go ahead and fill in these numbers into the formula. We get the 25th number is equal to 2 plus 2 times 24. Evaluating this, we get 2 plus 48 equals 50. So, the 25th number is 50. Why We Use the General Term Why do we use this general term? Let's go back to our peach tree example. You would use this general term if you, as the peach tree owner, wanted to find out how many leaves the peach tree will have on the 67th day of growing new leaves. If you use the formula for the general term, you only need to perform three calculations, one for the n minus 1, another to multiply this n minus 1 with the d, and a final one for adding a. Compare doing just 3 calculations to performing 67 additional problems just to find one number. You can see which one is quicker and easier to do. Lesson Summary Let's review what we've learned now. An arithmetic sequence is a string of numbers where each number is the previous number plus a constant. This constant difference between each pair of successive numbers in our sequence is called the common difference. The general term is the formula that is used to calculate any number in an arithmetic sequence. The formula tells us that if we wanted to find a particular number in our sequence, x sub n, we would take our beginning number, a, and add our common difference, d, times n minus 1, which is the location of our desired number minus 1. We use this formula to find a particular number in our sequence. This is particularly useful if our sequence is long, and we want to find a number that is not at the beginning of our sequence. Learning Outcomes Following this lesson, you'll have the ability to: Define arithmetic sequence and common difference Explain what the general term is and identify a formula for finding it Describe when you would want to use the formula and how to calculate it Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now Algebra II: High School 22 chapters 197 lessons Chapter 1 Algebra II: Real Numbers Types of Numbers & Its Classifications 6:56 min Graphing Rational Numbers on a Number Line \| Chart & Examples 5:02 min Notation for Rational Numbers, Fractions & Decimals 6:16 min The Order of Real Numbers: Inequalities 4:36 min Finding the Absolute Value of a Real Number 3:11 min How to Rationalize the Denominator with a Radical Expression 3:52 min Transcendental vs. Algebraic Numbers \| Overview & Examples 6:23 min Pythagorean Triple \| Definition, List & Examples 3:41 min Chapter 2 Algebra II: Complex and Imaginary Numbers Imaginary Numbers \| Definition, History & Examples 8:40 min How to Add, Subtract and Multiply Complex Numbers 5:59 min Complex Numbers & Conjugates \| Multiplication & Division 6:40 min How to Graph a Complex Number on the Complex Plane 3:28 min How to Solve Quadratics with Complex Numbers as the Solution 5:59 min Chapter 3 Algebra II: Exponents and Exponential Expressions Review How to Use Exponential Notation 2:44 min Scientific Notation \| Definition, Conversion & Examples 6:49 min Simplifying and Solving Exponential Expressions 7:27 min Exponential Expressions & The Order of Operations 4:36 min Multiplying Exponents \| Overview, Methods & Rules 4:07 min Dividing Exponential Expressions 4:43 min The Power of Zero: Simplifying Exponential Expressions 5:11 min Negative Exponents: Writing Powers of Fractions and Decimals 3:55 min Power of Powers: Simplifying Exponential Expressions 3:33 min Chapter 4 Algebra II: Properties of Functions Review Function in Math \| Definition & Examples 7:57 min Transformations: How to Shift Graphs on a Plane 7:12 min How to Add, Subtract, Multiply and Divide Functions 6:43 min Domain & Range of a Function \| Definition, Equation & Examples 8:32 min How to Compose Functions 6:52 min Inverse Functions \| Definition, Methods & Calculation 6:05 min Applying Function Operations Practice Problems 5:17 min Chapter 5 Algebra II: Linear Equations Review Linear Equations \| Definition, Formula & Solution 7:28 min Applying the Distributive Property to Linear Equations 4:18 min Forms of a Linear Equation \| Overview, Graphs & Conversion 6:38 min Abstract Algebraic Examples and Going from a Graph to a Rule 10:37 min Undefined & Zero Slope Graph \| Definition & Examples 4:23 min Parallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 min Parallel & Perpendicular Lines \| Equation, Graph & Examples 6:07 min Linear Equation \| Parts, Writing & Examples 8:58 min Chapter 6 Algebra II: Systems of Linear Equations Review System of Equations in Algebra \| Overview, Methods & Examples 8:39 min How Do I Use a System of Equations? 9:47 min How to Solve a System of Linear Equations in Two Variables 4:43 min How to Solve a Linear System in Three Variables With a Solution 5:01 min Solving System of Equations with 3 Variables \| Steps & Examples 6:04 min Chapter 7 Algebra II: Inequalities Review Inequality Signs in Math \| Symbols, Examples & Variation 7:09 min Graphing Inequalities \| Definition, Rules & Examples 7:59 min Inequality Notation \| Overview & Examples 8:16 min Graphing Inequalities \| Overview & Examples 12:06 min Solve & Graph an Absolute Value Inequality \| Formula & Examples 8:02 min Absolute Value Inequalities \| Definition, Calculation & Examples 9:06 min Translating Math Sentences to Inequalities 5:36 min Chapter 8 Algebra II: Absolute Value Review Absolute Value \| Explanation & Examples 4:42 min Absolute Value Expression \| Evaluation, Simplification & Examples 5:28 min Solving Absolute Value Functions & Equations \| Rules & Examples 5:26 min Absolute Value \| Overview & Practice Problems 7:09 min Absolute Value \| Graph & Transformations 8:14 min Graphing Absolute Value Functions \| Definition & Translation 6:08 min Chapter 9 Algebra II: Graphing and Factoring Quadratic Equations Review Tables & Graphs in the Real World \| Uses & Examples 5:50 min Scatter Plot Graph \| Overview, Uses & Examples 7:17 min Parabola \| Definition & Parabolic Shape Equation 4:36 min Types of Parabolas \| Overview, Graphs & Examples 6:15 min Multiplying Binomials Using FOIL and the Area Method 7:26 min Multiplying Binomials \| Overview, Methods & Examples 5:46 min Factoring Quadratic Equations Using Reverse Foil Method 8:50 min Factoring Quadratic Equations \| Solution & Examples 7:35 min Quadratic Trinomial \| Definition, Factorization & Examples 7:53 min How to Complete the Square \| Method & Examples 8:43 min Completing the Square Practice Problems 7:31 min How to Solve a Quadratic Equation by Factoring 7:53 min Chapter 10 Algebra II: Quadratic Equations Review Quadratic Equation \| Definition, Formula & Examples 5:13 min Solving Quadratics: Assigning the Greatest Common Factor and Multiplication Property of Zero 5:24 min Quadratic Function \| Formula, Equations & Examples 9:20 min How to Solve Quadratics That Are Not in Standard Form 6:14 min Solving Quadratic Inequalities Using Two Binomials 5:36 min Chapter 11 Algebra II: Factoring Factoring in Algebra \| Definition, Equations & Examples 5:32 min Finding the Prime Factorization of a Number \| Meaning & Examples 5:36 min Using Prime Factorizations to Find the Least Common Multiples 7:28 min Equivalent Expressions and Fraction Notation 5:46 min Using Fraction Notation: Addition, Subtraction, Multiplication & Division 6:12 min Factoring Out Variables: Instructions & Examples 6:46 min Combining Numbers and Variables When Factoring 6:35 min Transforming Factoring Into A Division Problem 5:11 min Factoring by Grouping \| Definition, Steps & Examples 7:46 min Rational Root Theorem \| Overview & Examples 7:05 min Chapter 12 Algebra II: Polynomials How to Evaluate a Polynomial in Function Notation 8:22 min Understanding Basic Polynomial Graphs 9:15 min Basic Transformations of Polynomial Graphs 7:37 min Cubic, Quartic & Quintic Equations \| Graphs & Examples 11:14 min Adding, Subtracting & Multiplying Polynomials \| Steps & Examples 6:53 min Pascal's Triangle \| Overview, Formula & Uses 7:26 min Binomial Theorem \| Coefficient Calculation, Formula & Examples 13:35 min Polynomial Long Division \| Overview & Examples 8:05 min Synthetic Division of Polynomials \| Method & Examples 6:51 min Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11 min Factor & Remainder Theorem \| Definition, Formula & Examples 7:00 min Operations with Polynomials in Several Variables 6:09 min Chapter 13 Algebra II: Rational Expressions Review How to Multiply and Divide Rational Expressions 8:07 min Multiplying and Dividing Rational Expressions: Practice Problems 4:40 min Adding & Subtracting Rational Expressions \| Overview & Examples 8:02 min Practice Adding and Subtracting Rational Expressions 9:12 min Rational Equations \| Definition, Formula & Examples 7:58 min Rational Equations: Practice Problems 13:15 min Division and Reciprocals of Rational Expressions 5:09 min Simplifying Complex Rational Expressions \| Steps & Examples 4:37 min Solving Direct Variation \| Equation, Problems & Examples 5:12 min Reflection Rules in Math \| Graph, Formula & Examples 6:07 min Solving Rational Equations and Finding the Least Common Denominator 6:48 min Chapter 14 Algebra II: Graphing and Functions Graphing Basic Functions 8:01 min Compounding Functions and Graphing Functions of Functions 7:47 min Inverse Function \| Graph & Examples 7:31 min Y Square Root of X \| Graph, Domain & Common Mistakes 4:32 min Polynomial Functions: Properties and Factoring 7:45 min Polynomial Functions: Exponentials and Simplifying 7:45 min Exponentials, Logarithms & the Natural Log 8:36 min Slopes of a Line \| Graphs, Formula & Examples 10:05 min Equation of a Line Using Point-Slope Formula 9:27 min Horizontal and Vertical Asymptotes 7:47 min Implicit Function Overview, Formula & Examples 4:30 min Chapter 15 Algebra II: Conic Sections Defining and Graphing Ellipses in Algebra 5:00 min How to Write the Equation of an Ellipse in Standard Form 6:18 min The Circle: Definition, Conic Sections & Distance Formula 3:43 min Hyperbola \| Definition, Equation & Graphs 10:00 min Hyperbola Standard Form \| Definition, Equations & Examples 8:14 min Parabola \| Definition, Formula & Examples 8:33 min Parabola \| Equation, Formula & Examples 8:17 min Conic Sections \| Definition, Equations & Types 6:33 min Chapter 16 Algebra II: Roots and Radical Expressions Review Square Root \| Definition, Formula & Examples 7:05 min Estimating Square Roots \| Overview & Examples 5:10 min Simplifying Square Roots When not a Perfect Square 4:45 min Simplifying Square Root Expressions \| Steps & Examples 7:03 min Radicands and Radical Expressions 4:29 min Evaluating Square Roots of Perfect Squares 5:12 min Factoring Radical Expressions 4:45 min Simplifying Square Roots of Powers in Radical Expressions 3:51 min Multiplying then Simplifying Radical Expressions 3:57 min How to Divide Radicals, Square Roots & Rational Expressions 7:07 min Simplifying Square Roots \| Overview & Examples 4:49 min Rationalizing the Denominator \| Overview & Examples 7:01 min Addition and Subtraction Using Radical Notation 3:08 min How to Multiply Radical Expressions 6:35 min Solving Radical Equations \| Overview & Examples 6:48 min Solving Radical Equations with Two Radical Terms 6:00 min Chapter 17 Algebra II: Exponential and Logarithmic Functions Exponential Function \| Definition, Equation & Examples 7:24 min Exponential Growth & Decay \| Formula, Function & Graphs 8:41 min Logarithms \| Overview, Process & Examples 5:23 min Change of Base Formula \| Logarithms, Examples & Proof 4:54 min Graphing Logarithms \| Overview, Transformations & Examples 6:53 min Evaluating Logarithms \| Properties & Examples 6:45 min Logarithmic Properties \| Product, Power & Quotient Properties 5:11 min Practice Problems for Logarithmic Properties 6:44 min Exponential Equations \| Definition, Solutions & Examples 6:18 min Solving Logarithmic Equations \| Properties & Examples 6:50 min Chapter 18 Algebra II: Sequences & Series Sequences in Math \| Overview, Types & Examples 4:57 min How to Use Factorial Notation: Process and Examples 4:40 min How to Write a Series in Summation Notation \| Overview & Examples 4:16 min Viewing now General Term of an Arithmetic Sequence \| Overview, Formula & Uses 5:01 min Up next Understanding Arithmetic Series in Algebra 6:17 min Watch next lesson Geometric Sequence \| Definition, Formula & Examples 5:14 min Sum of a Geometric Series \| Formula & Examples 4:57 min Sum of Infinite Geometric Series \| Formula, Sequence & Examples 4:41 min Recursive Rule \| Formulas & Examples 5:52 min Using Sigma Notation for the Sum of a Series 4:44 min Mathematical Induction: Uses & Proofs 7:48 min How to Find the Value of an Annuity 4:49 min How to Use the Binomial Theorem to Expand a Binomial 8:43 min Special Sequences and How They Are Generated 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Calculation & Examples 6:31 min Chapter 21 Algebra II: Statistics Measures of Central Tendency \| Definition, Formula & Examples 8:30 min Measures of Dispersion and Skewness 4:12 min Normal Distribution \| Curve, Table & Examples 11:40 min Regression Analysis: Definition & Examples 4:35 min Organizing and Understanding Data with Tables & Schedules 6:33 min Pie Chart vs. Bar Graph \| Overview, Uses & Examples 9:36 min Chapter 22 Algebra II: Trigonometry Transforming sin & cos Graphs \| Graphing sin and cosine Functions 8:39 min Graphing Tangent Functions \| Period, Phase & Amplitude 9:42 min Unit Circle Quadrants \| Converting, Solving & Memorizing 5:15 min Trig Functions using the Unit Circle \| Formula & Examples 5:46 min Special Right Triangles \| Definition, Types & Examples 6:12 min Law of Sines Formula & Examples 6:04 min Law of Cosines \| Definition & Equation 8:16 min Double Angle Formula \| Sin, Cos & Tan 9:44 min Radians to Degree Formula & Examples 7:15 min How to Solve 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3231	https://www.mytutor.co.uk/answers/55676/A-Level/Maths/Prove-the-property-log-a-x-log-a-y-log-a-xy/	+44 (0) 203 773 6024 Prove the property: log_a(x) + log_a(y) = log_a(xy). The derivation of the property starts with the basic representation of logarithms as powers. Lets consider a^(log_a(xy)). Then, a^(log_a(xy)) = xy. However, x = a^(log_a(x)) and y = a^(log_a(y)). Therefore, xy = a^(log_a(y)) a^(log_a(x)) = a^(log_a(x) + log_a(y)). Hence, log_a(x) + log_a(y) = log_a(xy). 5838 Views Need help with Maths? One-to-one online tuition can be a great way to brush up on your Maths knowledge. Have a Free Meeting with one of our hand picked tutors from the UK's top universities Download MyTutor's free revision handbook? This handbook will help you plan your study time, beat procrastination, memorise the info and get your notes in order. 8 study hacks, 3 revision templates, 6 revision techniques, 10 exam and self-care tips. Free weekly group tutorials Join MyTutor Squads for free (and fun) help with Maths, Coding & Study Skills. Related Maths A Level answers Given that y = 4x^3 -1 + 2x^1/2 (where x>0) find dy/dx. The curve C has equation y=3x^3-11x+1/2. The point P has coordinates (1, 3) and lies on C . Find the equation of the tangent to C at P. Solve the equation: log5 (4x+3)−log5 (x−1)=2. What is the amplitude and period of y=3sin(5x)? We're here to help Company Information Popular Requests CLICK CEOP Internet Safety Payment Security Cyber Essentials MyTutor is part of the IXL family of brands: IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Wyzant Trusted tutors for 300 subjects Education.com 35,000 worksheets, games, and lesson plans Vocabulary.com Adaptive learning for English vocabulary Emmersion Fast and accurate language certification Thesaurus.com Essential reference for synonyms and antonyms Dictionary.com Comprehensive resource for word definitions and usage SpanishDictionary.com Spanish-English dictionary, translator, and learning resources FrenchDictionary.com French-English dictionary, translator, and learning Ingles.com Diccionario ingles-espanol, traductor y sitio de apremdizaje ABCya Fun educational games for kids © 2025 by IXL Learning
3232	https://dummit.cos.northeastern.edu/docs/dynamics_3_chaotic_dynamics.pdf	Dynamics, Chaos, and Fractals (part 3): Chaotic Dynamics (by Evan Dummit, 2015, v. 1.00) Contents 3 Chaotic Dynamics 1 3.1 Symbolic Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3.1.1 Orbits of qc(x) = x2 + c for c < −2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3.1.2 Nested Intervals, Itineraries, and Cantor Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3.1.3 Metric Spaces and the Sequence Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.1.4 Equivalent Dynamical Systems: Homeomorphisms and Conjugation . . . . . . . . . . . . . . 9 3.1.5 Equivalence of qc(x) = x2 + c and the Shift Map for c < −2 . . . . . . . . . . . . . . . . . . . 13 3.2 Chaotic Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.2.1 Motivation for Chaos: Properties of the Shift Map on the Sequence Space . . . . . . . . . . . 15 3.2.2 The Formal Denition of Chaos, and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.3 Sarkovskii's Theorem and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.3.1 The Period-3 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.3.2 The Sarkovskii Ordering and Sarkovskii's Theorem . . . . . . . . . . . . . . . . . . . . . . . . 24 3 Chaotic Dynamics In this chapter, our goal is to study chaotic dynamical systems, with an ultimate goal of trying to understand the chaotic behaviors that we saw in our plots of orbit diagrams. In order to do this, we will introduce symbolic dynamics, a powerful tool that will help us understand a number of chaotic systems. Some more technical facts from topology and analysis will be necessary (primarily, some facts about metric spaces, the topology of the real numbers, and homeomorphisms), so we will develop this material to the extent necessary to apply it. We will then provide a denition of a chaotic dynamical system that is amenable to practical use, and prove that a number of simple systems are chaotic. Finally, we nish with a discussion of Sarkovskii's Theorem, a striking and unexpected result that (among other things) implies that any continuous function that possesses a 3-cycle exhibits chaotic behavior. 3.1 Symbolic Dynamics • Let us summarize what we have learned about the quadratic maps qc(x) so far: ◦If c > 1 4, then every orbit of the quadratic map qc(x) tends to ∞, and if c = 1 4 the unique xed point is weakly attracting on the left and weakly repelling on the right. ◦If c < 1 4 then if p+is the larger xed point, then all orbits outside the interval (−p+, p+) tend to ∞. ◦When −2 < c < 1 4, any point lying in the interval (−p+, p+) will have its orbit completely conned to this interval. We can glean some insight about the behavior of qc from the orbit diagram: some values appear to have an attracting cycle, while other values do not. ◦When c < −2, the critical orbit diverges to +∞. • We would now like to understand the dynamics of the maps qc when c < −2 more precisely. 1 ◦To do this, we will ultimately show that the dynamics of qc are modeled by the dynamics of a simple map on a sequence space. We will then switch to analyzing the sequence space, which (it turns out) we can understand completely. ◦Understanding and formalizing the connection between the dynamics on these two spaces will involve some technical results from analysis and topology, which we will develop as needed. 3.1.1 Orbits of qc(x) = x2 + c for c < −2 • It might appear that the dynamics of qc(x) = x2 + c are uninteresting when c < −2, given that the critical orbit diverges to +∞, but this is not at all the case. ◦Explicitly, suppose c < −2 is xed, and let I = [−p+, p+], where p+ = 1 + √1 −4c 2 is the larger xed point of qc(x). ◦Note: All graphics are produced for the case c = −91/36, but the analysis holds in general. • Notice that qc maps I onto an interval strictly containing I (since qc(0) = c does not lie in I), and that the set of points in I whose image also lies in I forms a pair of intervals I0 and I1. Here is a picture: • If a point x ∈I has qc(x) ̸∈I, then the orbit of x necessarily diverges to +∞by our earlier analysis. ◦So, if the orbit of x does not diverge to +∞, it is necessarily the case that qn c (x) ∈I for every integer n ≥1 ◦Equivalently, the orbit of x will not diverge provided that x ∈q−n c (I) for every integer n ≥1. • Therefore, the set of points x whose orbit does not go to ∞is given by the innite intersection Λ = ∞ \ n=1 q−n c (I). A very reasonable question is: what does this set Λ look like? ◦A natural way of trying to understand Λ is to look at each of the terms in the intersection. We already saw that q−1 c (I) is the union of the two intervals I0 and I1. ◦Then q−2 c (I) = q−1 c (I0 ∪I1) = q−1 c (I0)∪q−1 c (I1) by basic properties of the inverse image. Each of q−1 c (I0) and q−1 c (I1) consists of a pair of closed intervals, one in I0 and the other in I1: 2 ◦Notice that q−2 c (I) consists simply of the two intervals I0 and I1, but with a middle portion removed from each: this follows because qc maps I0 bijectively onto the interval I, so the set of points in I0 that q2 c sends into I will be all of I0 except the points in the middle of the interval satisfying q−1 c (x) < −p+. (The same argument holds for I1.) ◦In a similar way we see that q−3 c (I) will be a collection of eight closed intervals obtained by removing a piece from the middle of each of the four intervals in q−2 c (I): ◦We can see (by an easy induction) that q−n c (I) will consist of 2n closed intervals, half of which are in I0 and the other half of which are in I1. Here is a typical picture of the successive inverse images q−n c (I) for 0 ≤n ≤6: • We will show later that whenever c < −2, the sizes of the intervals will shrink to zero as we take n →∞. Assuming for now that all these intervals have size shrinking to 0, a very reasonable question is: how do we know that the set Λ = ∞ \ n=1 q−n c (I) is even nonempty? (After all, the intersection cannot contain any intervals at all.) ◦One way is simply to exhibit some points that lie in the intersection: namely, the set q−k c (p+) for any k ≥1: after k iterations, every point in this set lands at p+, so the orbit certainly always stays in I. ◦Notice that q−k c (p+) is a set containing 2k points, since each point in I has two preimages under qc, so in fact we have exhibited innitely many points in the intersection. Indeed, it is easy to see that q−k c (p+) is the set of endpoints of the intervals in q−k c (I). • It might seem that, as we iteratively remove the middle portion of each interval, as we take the limit we will only be left with the endpoints of the intervals. But in fact, there are many more points in Λ, as we will show. ◦To do so will require a technical result about the topology of R. 3.1.2 Nested Intervals, Itineraries, and Cantor Sets • We can describe the set Λ more precisely using a result known as the nested intervals theorem: • Theorem (Nested Intervals): If {Ji} is a collection of nite closed intervals in R for i ≥1 , and Ji+1 ⊆Ji for each i, then the intersection ∞ \ i=1 Ji is nonempty. Furthermore, if the length of Ji →0 as i →∞, then the intersection consists of a single point. ◦Remark: The assumption that the intervals are nite is necessary, because an innite intersection of innite closed intervals can be empty: ∞ \ i=1 [i, ∞) = ∅. 3 ◦Proof: Let Ji = [ai, bi] where by assumption ai ≤bi for each i. ◦Since Ji+1 ⊆Ji we also have ai ≤ai+1 and bi+1 ≤bi for each i, so the sequence {ai} is increasing and bounded above (by b1). ◦Thus, by the monotone convergence theorem (a bounded monotone sequence of real numbers has a limit), the sequence ai has a limit L as i →∞. Similarly, {bi} is decreasing and bounded below, so the sequence bi has a limit M as i →∞. ◦Then [L, M] is contained in each interval Ji and hence also in the intersection ∞ \ i=1 Ji. But no x < L can be in the intersection: otherwise, it would be contained in every interval Ji and thus each ai would be less than x, contradiction the assumption that the ai →L. Similarly, no x > M is in the intersection. ◦Thus, ∞ \ i=1 Ji = [L, M]. Furthermore, since L ≤M (since M is an upper bound for the sequence {ai} and L is a lower bound for the sequence {bi}), this interval is not empty. ◦For the second part, if the length if Ji →0 as i →∞, then bi −ai →0 as i →∞, so L = M, and thus the intersection is a single point. • Using the nested intervals theorem, we can give a better description of Λ. ◦Given q−n c (I), we saw that q−(n+1) c (I) is obtained by removing a piece from the middle of each of the intervals in q−n c (I). ◦Describing an interval in q−n c (I) is therefore equivalent to recording whether we took the left interval or the right interval each time we applied f −1. ◦We can summarize this information using an n-digit binary string, with 0 meaning left and 1 meaning right, and thus label each of the intervals accordingly: ◦To any innite sequence of binary digits {d0d1d2 · · · }, we can then construct elements of Λ: if we take Jn to be the interval Id0d1d2···dn, then the sequence J1, J2, J3, ... is a nested sequence of closed intervals of length tending to zero (by our earlier proposition), so by the nested interval theorem, the intersection is a single point. ◦Conversely, if x ∈Λ, then x necessarily lies in some subinterval of q−n c (I) for every n ≥1, so by writing down the labels of the sequences, we get an innite sequence of binary digits. ◦An almost equivalent way of dening this sequence is to compute the iterates of x and determine which of I0 and I1 each iterate lands in: • Denition: For x ∈Λ, the itinerary of x is the innite binary string S(x) = {d0d1d2 · · · }, where di = 0 if qn c (x) ∈I0 and dn = 1 if qn c (x) ∈I1. ◦We will return to study this map later, but it serves as our primary motivation for studying sequence spaces. ◦Our ultimate goal is to prove that the itinerary map is a homeomorphism (i.e., a continuous bijection with continuous inverse) when viewed in the appropriate context, and can be used to relate the dynamics of a simple map on the space of binary sequences to the dynamics of the quadratic map qc on R. • Before we discuss sequence spaces, we will briey mention Cantor sets, of which Λ is one example. The most famous Cantor set is likely the Cantor ternary set: • Denition: The Cantor ternary set is the subset ∞ \ n=0 Cn, where C0 = [0, 1] and Cn+1 is obtained by deleting the open middle third of each interval in Cn, for each n ≥0. 4 ◦Thus, C1 = [0, 1 3] ∪[2 3, 1], C2 = [0, 1 9] ∪[2 9, 3 9] ∪[6 9, 7 9] ∪[8 9, 1], and so forth. Here is a picture: ◦Observe that the picture is almost identical to the picture of the set Λ. Indeed, as we will show later, these two sets are homeomorphic (i.e., there exists a continuous function with continuous inverse mapping Λ to the Cantor ternary set). ◦It is possible to prove things about the Cantor ternary set using the nested intervals theorem. However, there is a nicer description of the points in the set using base-3 decimal expansions. • Recall that the base-3 decimal expansion of a (nonnegative) real number has the form n.d1d2d3d4 · · ·3 = n + ∞ X i=1 di 3i , where n is an integer and each di is 0, 1, or 2. ◦Notation: We will generally put a subscript of k when working with decimals in base k, when k ̸= 10, although it should always be clear from the context what base the expansions should be considered in. ◦Such a series always converges by comparison to the geometric series 0.2222 . . .3 = 1. ◦For example, we have 1 3 = 0.13 and 1 4 = 0.010101 . . .3 = 0.013, where the overline (or vinculum) means that the indicated portion repeats indenitely. ◦The real numbers of the form m 3n , for m and n nonnegative integers, have two base-3 expansions: one ending in an innite string of 0s, and another ending in an innite string of 2s. (For example, 1 3 = 0.103 = 0.023.) All other real numbers have a unique ternary expansion. • Proposition: A point α ∈[0, 1] lies in the Cantor ternary set if and only if it has a ternary (base-3) expansion containing only the digits 0 and 2. ◦Proof: Suppose α has a ternary expansion whose rst digit is 1. Then α will be removed at the rst stage of the construction of the Cantor set: the open middle third of [0, 1] consists of all points whose rst base-3 decimal digit is a 1, except for the points 1 3 = 0.1000 · · · = 0.0222 . . . and 2 3 = 0.1222 · · · = 0.2000 . . . . But each of the endpoints has a representation containing no 1s, and they are both preserved since we only remove the open middle third. ◦In exactly the same way, if α has a ternary expansion whose nth digit is 1, then α will be removed at the nth stage of the construction of the Cantor set, unless α happens to be one of the endpoints (which both have a representation that does not contain a 1). ◦Conversely, if α has a representation containing no 1s, then it will never lie in the open middle third of any interval during the construction, so α is in the Cantor ternary set. • As a corollary of the proposition above, we can see that the Cantor ternary set is uncountable: the elements of the Cantor set are the real numbers in [0, 1] having a ternary expansion consisting of only 0s and 2s, and this set is in a bijection with the set of innite binary sequences (namely, by replacing all of the 2s by 1s), which is uncountable. ◦Recall that a set is countable if it can be put in a one-to-one correspondence with some subset of the positive integers, and a set is uncountable otherwise. ◦A typical example of a countable set is the set of rational numbers. ◦The set of innite binary sequences is uncountable, as originally proven by Cantor using his famous diagonal argument. In summary: 5 ∗Suppose by way of contradiction that the set of innite binary sequences were countable. Arrange all of them into an innite array: a1 = d1,1d1,2d1,3 · · · a2 = d2,1d2,2d2,3 · · · a3 = d3,1d3,2d3,3 · · · . . . . . . . . . ∗Now construct the binary sequence x whose ith digit is 1 if di,i = 0 and is 0 if di,i = 1. (These digits run down the diagonal of the array, whence the name of the argument.) ∗Then x cannot be equal to any of the ai, because it diers in at least one place from every element in the list. This is a contradiction, because we assumed all binary sequences were in the array. ◦The set of real numbers, or even the real numbers in the interval [0, 1], is also uncountable, as it can be put into a bijection with the set of innite binary sequences: in essence, we associate each element in the interval with the innite sequence of digits in its base-2 decimal expansion. ∗This is not quite a bijection since some real numbers have two base-2 expansions. However, there are only countably many such numbers, so it is straightforward to x this issue. ∗Explicitly, remove all of the elements with two binary expansions from [0, 1] and place them in a sequence a1, a2, a3, . . . , and also remove all of the elements from the set of binary sequences corresponding to these elements of [0, 1], and place them in a sequence b1, b2, b3, .... ∗Then dene the bijection normally on the elements outside these sequences, and also identify ai with bi. • From our arguments, we can conclude that there is actually a bijection between the points in the Cantor ternary set and the points in [0, 1]. ◦Somehow, the Cantor ternary set is still large, even though the sum of the lengths of the intervals at each stage tends to 0 exponentially rapidly! ◦There are many natural ways to alter the construction to create generalized Cantor sets. Rather than pursuing the topic more now, we will return to discuss Cantor sets when we study fractals. 3.1.3 Metric Spaces and the Sequence Space • Denition: The sequence space on two symbols is the set Σ2 = {(d0d1d2 · · · ) : di = 0 or 1 for each i}. ◦We think of this space as the set of innite binary sequences, although we could equally well use any symbols in place of 0 and 1. ◦Although we will not use it, the more general sequence space on n symbols is the set of innite sequences (d0d1d2 · · · ) where each di lies in the set {0, 1, · · · , n −1}. • The sequence space, so far, has no structure: it is simply a set of sequences. In order to do anything with it, we need to specify some additional structure on the space. ◦We will do this by dening a distance function (or metric) that allows us to measure how far apart elements are. ◦First, however, we will outline some of the basic theory of metric spaces. • Denition: If M is a set, a function d : M × M →R is called a metric, and the pair (M, d) a metric space, if it obeys the following three properties: 1. (Nonnegativity) For any x, y ∈M, d(x, y) ≥0, with d(x, y) = 0 if and only if x = y. 2. (Symmetry) For any x, y ∈M, d(x, y) = d(y, x). 3. (Triangle Inequality) For any x, y, z ∈M, d(x, z) ≤d(x, y) + d(y, z). 6 • The prototypical example of a metric is the Euclidean distance function d(x, y) = \|\|x −y\|\| on Rn. ◦Explicitly, if x = (x1, . . . , xn) and y = (y1, . . . , yn), then d(x, y) = p (x1 −y1)2 + · · · + (xn −yn)2. ◦When n = 1, for example, this is the familiar d(x, y) = \|x −y\| on R. ◦The rst two axioms are trivial, and the third is geometrically obvious, as it is the (actual) triangle inequality: namely, that the shortest distance between x and z is the straight line joining them. ◦The third axiom requires some actual work to prove algebraically, and ultimately, it reduces to the Cauchy-Schwarz inequality x·y ≤\|\|x\|\| \|\|y\|\|, where x·y denotes the dot product. (Recall that \|\|x\|\|2 = x·x.) ∗Explicitly, if we set a = x −y and b = y −z, then we want to show that \|\|a + b\|\| ≤\|\|a\|\| + \|\|b\|\|. ∗We have (a + b) · (a + b) = a · a + 2(a · b) + b · b ≤\|\|a\|\|2 + 2 \|\|a\|\| \|\|b\|\| + \|\|b\|\|2. ∗Taking the square root immediately gives the desired result. ◦For completeness, we give a one-line proof of the Cauchy-Schwarz inequality in Rn: observe that 1 2 n X i=1 n X j=1 (xiyj −xjyi)2 = n X i=1 x2 i n X j=1 y2 j − n X i=1 xiyi !2 = \|\|x\|\|2 \|\|y\|\|2 −(x · y)2 and since the left-hand side is a sum of squares, the right-hand side must be nonnegative. • For posterity we will record the denitions of open and closed sets in a metric space: • Denition: If (M, d) is a metric space, then if x ∈M and r > 0, the open ball Br(x) of radius r centered at x is dened to be the set of points within a distance r of x: namely, Br(x) = {y ∈M : d(x, y) < r}. ◦In the case where X = Rn with the Euclidean metric, the open ball is a literal ball (i.e., the points lying strictly inside the n-dimensional sphere of radius r centered at x). • Denition: A subset U ⊆M is an open set if, for each x ∈U, there is some open ball Bϵ(x) of positive radius ϵ centered at x that is contained in U. A subset C ⊆M is a closed set if its complement M\C is open. ◦Ultimately, one can think of open sets as sets that are the nite or innite union of open balls. ◦There is much that can be said about open and closed sets, even in general metric spaces, but it is not necessary to develop any more point-set topology for our purposes. • There are many other examples of metric spaces, and they are one of the fundamental objects of study in real analysis. ◦For example, if B is the set of all functions that are bounded on the interval [a, b], then d(f, g) = max a≤t≤b \|f(t) −g(t)\| denes a metric on B. ◦If L2 is the set of integrable functions f on the interval [a, b] such that ´ b a [f(t)]2dt is nite, then d(f, g) = h´ b a [f(t) −g(t)]2dt i1/2 denes a metric on L2. ◦More generally, if ⟨·, ·⟩is any inner product on a real vector space V , then the function d(x, y) = p ⟨x −y, x −y⟩is a metric on V . ◦The idea of a metric space provides a way to study general features of convergent sequences in an abstract and general way. In particular, one can apply the results to study spaces of functions, sequences of functions, the dierent types of convergence of sequences of functions in various settings, and so forth. ◦Our present goal is not to develop all of real analysis, so we will stop our discussion here. Instead, we use these examples as motivation for the denition of the metric on the sequence space: • Proposition (Sequence Space Metric): If Σ2 is the sequence space on two symbols, then the distance function d(x, y) = ∞ X i=0 \|xi −yi\| 2i is a metric on Σ2, where x = (x0x1x2 · · · ) and y = (y0y1y2 · · · ). 7 ◦Observe that the series is bounded by the geometric series ∞ X i=0 1 2i = 2, so it always converges. ◦Examples: If x = (111111 · · · ), y = (000000 · · · ), and z = (110110 · · · ), then d(x, y) = 1 + 1 2 + 1 4 + 1 8 + 1 16 + 1 32 = ∞ X i=0 1 2i = 2 d(x, z) = 0 + 0 + 1 4 + 0 + 0 + 1 32 + · · · = ∞ X i=0 1 23i+2 = 4 7 d(y, z) = 1 + 1 2 + 0 + 1 8 + 1 16 + 0 + · · · = ∞ X i=0 1 23i + 1 23i+1 = 12 7 . ◦Proof (of proposition): Clearly, d(x, y) ≥0 and equality occurs only when x = y, since all terms in the series are nonnegative, and they are all zero only when x = y. It is also obvious that d(x, y) = d(y, x) since \|xi −yi\| = \|yi −xi\|. Finally, if x, y, z are in Σ2, then by the usual triangle inequality for real numbers we see \|xi −zi\| ≤\|xi −yi\| + \|yi −zi\| for each i: then summing the appropriate terms yields d(x, z) ≤d(x, y) + d(y, z) as required. • Intuitively, from the denition of the metric, we can see that the earlier terms in an element of Σ2 matter much more than the later terms. (It is quite similar to the behavior of digits in the base-2 decimal expansion of a real number.) • Proposition (Nearby Sequences): If x = (x0x1x2 · · · ) and y = (y0y1y2 · · · ) are elements of Σ2 such that xi = yi for each i ≤n, then d(x, y) ≤2−n. Conversely, if d(x, y) < 2−n, then xi = yi for each i ≤n. ◦In other words: if the early terms of two elements of Σ2 agree, then the distance between them must be small. Conversely, if two elements of Σ2 are close together, then their early terms must agree. ◦Proof: If xi = yi for each i ≤n then d(x, y) = ∞ X i=0 \|xi −yi\| 2i = ∞ X i=n+1 \|xi −yi\| 2i ≤ ∞ X i=n+1 1 2i = 1 2n . Con-versely, suppose that d(x, y) < 2−n. Then each of the terms \|xi −yi\| 2i for i ≤n must be zero, otherwise that term alone would cause the sum to be at least 2−i ≥2−n: so xi = yi for i ≤n. • In order to model the dynamics of the quadratic map qc on the space Λ, we need to introduce the function that plays the analogous role in the sequence space. • Denition: The shift map σ : Σ2 →Σ2 is the map dened by σ(x0x1x2x3 · · · ) = (x1x2x3 · · · ). ◦In other words, σ is the map that deletes the rst term in the sequence, thereby shifting the remaining terms one slot to the left. The kth iterate of σ is equally simple: it deletes the rst k terms. ◦Example: If x = (101010 · · · ) then σ(x) = (010101 · · · ) and σ2(x) = (101010 · · · ) = x, so x is a periodic point of period 2 for σ. ◦Example: If y = (011111 · · · ) then σ(y) = (111111 · · · ) and σ2(y) = (111111 · · · ) = σ(y), so y is an eventually xed point for σ. ◦Notice that σ is a surjective, two-to-one map on Σ2: for any given y ∈Σ2, there are two x satisfying σ(x) = (y0y1y2 · · · ), namely x = (0y0y1y2 · · · ) and (1y0y1y2 · · · ). • It is a simple matter to write down all of the periodic points for σ: they are the periodic sequences s = (s0s1 · · · sn−1s0s1 · · · sn−1 · · · ) = (s0s1 · · · sn−1). ◦Explicitly: it is obvious that any such sequence satises σn(s) = s. Conversely, if σn(s) = s then each block of n terms must repeat in precisely the given manner. ◦Thus, we see immediately that there are 2n sequences of period dividing n for the shift map σ. ◦It is a more dicult problem to determine exactly how many n-cycles there are, since the tally 2n counts all of them n times, and also includes all of the cycles of period dividing n (and most of these are overcounted as well). 8 ◦It is not hard to answer the question for small n simply by writing down all the cycles. For example, there are two 3-cycles given by {(001), (010), (100)} and {(011), (110), (101)}, and there are three 4-cycles given by {(0001), (0010), (0100), (1000)}, {(0011), (0110), (1100), (1001)}, and {(0111), (1110), (1101), (1011)} but this rapidly becomes cumbersome. ◦Ultimately, determining the answer in general requires a technique from number theory known as Möbius inversion. ◦Explicitly, if we dene the Möbius function as µ(n) = ( 0 if n is divisible by the square of any prime (−1)k if n is the product of k distinct primes , where µ(1) = 1, then the number of n-cycles for the shift map σ is equal to 1 n X d\|n 2n/dµ(d), where the sum is taken over all divisors d of n. ◦In any case, the fact remains that it is still quite easy to write down periodic points for σ, in stark contrast to the situation for the other functions we have analyzed. (Note, however, that the set of periodic points is countably innite, so they are still comparatively rare in Σ2.) • A central property of the shift map is that it is continuous: • Proposition: The shift map σ : Σ2 →Σ2 is continuous at every point of Σ2. ◦In general, if f : X →Y is a map from one metric space to another (with associated metrics dx and dy), we say that f is continuous at a point a ∈X if, for any ϵ > 0, there exists a δ > 0 such for any x ∈X satisfying dx(x, a) < δ it is true that dy(f(x), f(a)) < ϵ. ◦If f : X →Y is continuous at every point in X, then (as usual) we will simply say f is a continuous function. ◦In the case where X = Y = R with the usual metric, this reduces down to the familiar denition of a continuous real-valued function: f is continuous at a if, for any ϵ > 0, there exists a δ > 0 such that the statement \|x −a\| < δ implies \|f(x) −a\| < ϵ. ◦Proof: Suppose we are given ϵ > 0 and a point a = (a0a1a2 · · · ). ∗Since ϵ > 0, there exists a positive integer n such that 2−n < ϵ: we then claim that δ = 2−(n+1) will satisfy the denition of continuity. ∗By the nearby sequences proposition, if x = (x0x1x2 · · · ) and d(x, a) < 2−(n+1), then xi = ai for i ≤n + 1. ∗But then σ(x) = (x1x2x3 · · · ) and σ(a) = (a1a2a3 · · · ), so these sequences agree from the 0th term up through the nth term, so again by the nearby sequences proposition, we see that d(σ(x), σ(a)) ≤ 2−n < ϵ, as required. ∗Since a was arbitrary, σ is continuous everywhere. 3.1.4 Equivalent Dynamical Systems: Homeomorphisms and Conjugation • Now that we have developed some basic properties of the shift map σ on the sequence space Σ2, we would like to show that the behavior of σ on Σ2 is the same as the behavior of the quadratic maps qc(x) = x2 + c, for c < −2, on the set Λ of points whose orbits remain bounded under qc. ◦The itinerary map provides a way to relate these two systems: recall that for each x ∈Λ, we dened the itinerary of x as the innite binary string S(x) = {d0d1d2 · · · }, where di = 0 if f n(x) ∈I0 and dn = 1 if f n(x) ∈I1, and I0, I1 were the two intervals in the inverse image of [−p+, p+] under the map qc. ◦However, before showing that the behavior of σ on Σ2 is the same as the behavior of qc on Λ, we need to dene what it means for two dynamical systems to be equivalent. • Our rst step is to dene an equivalence on metric spaces: • Denition: If X and Y are metric spaces, then F : X →Y is a homeomorphism if it is a continuous bijection whose inverse F −1 is also continuous. If there exists a homeomorphism between two spaces, we say they are homeomorphic. 9 ◦Recall that a bijection is a one-to-one (injective) map that is also onto (surjective). ◦Example: The function F(x) = tan(x) mapping X = (−π 2 , π 2 ) to Y = R is a homeomorphism, since tan(x) is clearly a continuous bijection, and its inverse tan−1(x) is also continuous. ◦Example: The function F(t) = (cos(t), sin(t)) mapping X = R modulo 2π to Y = S1 (the unit circle, embedded in the Cartesian plane) is a homeomorphism. It is clearly continuous and (mostly) clearly bijective, and its inverse map is given by G(x, y) =the angle formed by the vector ⟨x, y⟩and the positive x-axis, considered modulo 2π. • Proposition: The relation X is homeomorphic to Y is an equivalence relation on the collection of metric spaces. ◦In other words, if X ∼Y denotes there exists a homeomorphism h : X →Y , then ∼obeys the three properties of an equivalence relation: 1. For any X, X ∼X, 2. For any X and Y , X ∼Y implies Y ∼X, and 3. For any X, Y, Z, X ∼Y and Y ∼Z together imply X ∼Z. ◦Proof: For any metric space X, the identity map h(x) = x is a homeomorphism from X to itself, showing (1). ◦For (2), if h : X →Y is a homeomorphism, then h−1 : Y →X is also a homeomorphism (since the inverse of a bijection is a bijection). ◦Finally, for (3), if h1 : X →Y and h2 : Y →Z are homeomorphisms, then h2 ◦h1 : X →Z is also a homeomorphism, since the composition of continuous functions is continuous and the composition of bijections is a bijection. • Homeomorphisms (in a precise sense) are the most topologically natural kind of equivalence relation of metric spaces, because homeomorphisms preserve the topological structure of a metric space. ◦For example, if h is a homeomorphism and U ⊆X is any subset, then h(U) is open if and only if U is open, and h(U) is closed if and only if U is closed. ◦Similarly, if h : X →Y is a homeomorphism and x, xi ∈X, the statement that lim n→∞xn = x is equivalent to the statement that lim n→∞h(xn) = h(x). • In the dynamical realm, we will think of a homeomorphism h : X →Y as a change of coordinates from the space X to the space Y . We would now like to extend this to dene an equivalence of dynamical systems, but rst we need to formalize the denition of a dynamical system: • Denition: A (discrete) dynamical system is a pair (X, f) where X is a metric space and f : X →X is a function on X. ◦Now, suppose we have a dynamical system (X, f) along with a homeomorphism h : X →Y . We would like to know: what function g : Y →Y corresponds to the function f : X →X once we change coordinates using the homeomorphism h? ◦In other words, given a homeomorphism h : X →Y and a map f : X →X, how can we get a function g : Y →Y ? ◦One way to do this is to observe that the inverse homeomorphism h−1 is a map from Y →X, so to create a map g : Y →Y using f, we can rst apply h−1 to obtain a point in X, then apply f, and then apply h to obtain a new point in Y . ◦This suggests the correct denition is to take g = h ◦f ◦h−1. ◦A (slightly) more sensible way to organize this information is with a commutative diagram, where we draw each of the relevant spaces and maps: X h f / X h Y g / Y 10 ◦The idea is that if we start with some x ∈X, then following either path of arrows down to Y in the lower right should yield the same result. (In other words, the paths commute.) ◦In other words, we want it to be the case that h(f(x)) = g(h(x)) for every x ∈X. ◦Since h is invertible (and a bijection) by setting y = h(x) we can rephrase this as g(y) = h(f(h−1(y))), and this is equivalent to the expression we obtained before. ◦We can see how this works more clearly if we reverse the direction of one of the arrows in the commutative diagram: X f / X h Y h−1 O g / Y Now it is easier to see that g(y) = h(f(h−1(y))) by comparing the two possible paths from the lower-left Y to the lower-right Y . • Denition: Two dynamical systems (X, f) and (Y, g) are conjugate if there exists a homeomorphism h : X →Y such that g = h ◦f ◦h−1. (Or equivalently, if h(f(x)) = g(h(x)) for every x ∈X.) ◦If X = Y , then we will simply say that the maps f and g themselves are conjugate if there is a homeomorphism h : X →X such that g = h ◦f ◦h−1. ◦Remark (for those who like linear algebra): Conjugation arises in linear algebra in the context of a change of basis for a vector space. Namely, if T : V →V is a linear transformation on a nite-dimensional vector space V , then T has a matrix representation that depends on which basis is chosen for V . If the representation is A in one basis and B in another basis, with change of basis matrix M, then A = MBM −1. ◦Remark (for those who like group theory): Conjugation also plays an important role in the study of groups. Explicitly: if f and h are elements of a group G, then the conjugate of f by h is the element g = hfh−1. The conjugation action of an element on a group is a central tool in elementary group theory. • Conjugate dynamical systems behave essentially identically: the conjugating homeomorphism converts dy-namical properties of f on X to properties of g on Y . In particular, the orbit structure of f on X is the same as the orbit structure of g on Y . More explicitly: • Proposition (Conjugate Orbits): Suppose (X, f) is conjugate to (Y, g) via the homeomorphism h. Then {x1, · · · , xn} is an n-cycle for f if and only if {h(x1), . . . , h(xn)} is an n-cycle for g. Furthermore, if the underlying metric space is R in both cases, both sets are n-cycles, and f, g, h are all dierentiable with h′ everywhere nonzero, then the cycles have the same (weakly) attracting/repelling behavior. ◦Proof: Suppose that (X, f) and (Y, g) are conjugate with g = h ◦f ◦h−1. ◦First, by a trivial induction we can see that gk = h◦f k◦h−1 for each k ≥1. (There are k−1 cancellations h−1 ◦h in the resulting expression.) Equivalently, gk ◦h = h ◦f k for each k. ∗In particular, f k(x) = x holds if and only if gk(h(x)) = h(x) holds. Thus, x is periodic under f with period (dividing) k if and only if h(x) is periodic with period (dividing) k for g, for each k ≥1. ∗So we see that {x1, . . . , xn} is an n-cycle for f if and only if {h(x1), . . . , h(xn)} is an n-cycle for g. ◦Now suppose f and g are both dierentiable real-valued functions, that f(x0) = x0, and that h is dierentiable and that h′ ̸= 0. ∗Applying the chain rule to h(f(x)) = g(h(x)) yields h′(f(x0))f ′(x0) = g′(h(x0))h′(x0). ∗Since f(x0) = x0 we obtain h′(x0)f ′(x0) = g′(h(x0))h′(x0), and since h′ ̸= 0 everywhere we may cancel to obtain f ′(x0) = g′(h(x0)). ∗Therefore, by the attracting xed point criterion, we see that the behavior of x0 as a xed point of f is the same as the behavior of h(x0) as a xed point of g. ◦For n-cycles, we can apply the xed-point result to f n (since it is conjugate to gn as we also just showed) to see that the cycle {x1, · · · , xn} has the same behavior as {h(x1), · · · , h(xn)} 11 ◦Finally, suppose that x is a weakly attracting xed point of f. (An analogous argument will cover the case where x is a weakly repelling xed point of f, and these arguments trivially extend to cover the case of n-cycles as well.) ∗Then there is an open interval I = (x −ϵ, x + ϵ) such that every y ∈I has f n(y) →x. ∗Since h is a homeomorphism, h(I) is an open set containing h(x), so (by denition) it contains some open interval J = (h(x) −δ, h(x) + δ) containing h(x). ∗For any b ∈J, since h is a homeomorphism there is some a ∈I such that h(a) = b. ∗Then lim n→∞f n(a) = x so since h is a homeomorphism, we can apply h to see that lim n→∞h(f n(a)) = h(x) = y, and h(f n(a)) = gn(h(a)) = gn(b). ∗Thus, lim n→∞gn(b) = y so y is a weakly attracting xed point of g. • Example: Show that the dynamical systems (R, f) and (R, g) are conjugate, where f(x) = x2 −2x + 2 and g(x) = x2. ◦We want to nd a homeomorphism h(x) such that h(f(x)) = g(h(x)). ◦If we search for linear functions h(x) = ax + b, we must have a(x2 −2x + 2) + b = (ax + b)2, so that ax2 −2ax + (2a + b) = a2x2 + 2abx + b2. Solving this as an identity in x produces a = 1, b = −1. ◦Thus, if h(x) = x −1, it is true that h(f(x)) = (x −1)2 = g(h(x)), so these two maps are indeed conjugate. • Example: Show that any quadratic map p(x) = a1x2 + a2x + a3 with a1 ̸= 0 is conjugate to a quadratic map of the form qc(x) = x2 + c for some value of c. ◦Like in the previous example, we will search for linear functions of the form h(x) = ax + b. ◦We want to ensure that h(p(x)) = qc(h(x)), so we get a(a1x2 +a2x+a3)+b = (ax+b)2 +c, or aa1 = a2, aa2 = 2ab, and aa3 + b = b2 + c. ◦Thus, we can take a = a1 and b = a2/2: then h(p(x)) = qc(h(x)) where c = a1a3 + a2 2 −a2 2 4 . ◦So, every quadratic map is conjugate to one of the maps qc(x) = x2 + c. ◦In particular, all of our previous analysis of the family qc (e.g., our analysis of the orbit diagram) actually extends to the set of all quadratic polynomials. • Example: Show that (R mod 1, D) and (S1, g) are conjugate, where D = ( 2x for 0 ≤x < 1/2 2x −1 for 1/2 ≤x < 1 is the doubling map and g(cos t, sin t) = (cos 2t, sin 2t) is the angle doubling map on the unit circle S1. ◦We claim that the homeomorphism h(x) = (cos(2πx), sin(2πx)) is a conjugation between f and g. ◦To see this, we rst compute h(D(x)): for 0 ≤x < 1 2, this is (cos(4πx), sin(4πx)), and for 1 2 ≤x < 1 this is (cos(4πx−2π), sin(4πx−2π)) = (cos(4πx), sin(4πx)). So in either case, h(D(x)) = (cos(4πx), sin(4πx)). ◦We also compute g(h(x)) = g(cos(2πx), sin(2πx)) = (cos(4πx), sin(4πx)). ◦Thus, since, h(D(x)) = g(h(x)) for all x, the two systems are conjugate. • Example: Show that the dynamical systems (R, f) and (R, g) are not conjugate, where f(x) = x2 −x and g(x) = x3 −3x. ◦Observe that f has two xed points x = 0 and x = 2, while g has three xed points x = 0 and x = ±2. ◦By our proposition above, conjugate systems must have the same orbit structure. Since f and g do not, we conclude that they are not conjugate. 12 3.1.5 Equivalence of qc(x) = x2 + c and the Shift Map for c < −2 • Our main goal is to show that, if c < −2, the dynamical system (Λ, qc(x)) is conjugate to the system (Σ2, σ) via the itinerary map. This will allow us to (fully) understand the dynamics of qc(x) by studying the (much simpler) shift map σ. ◦Recall our earlier notation: if qc(x) = x2 + c, then p+ denotes the larger xed point of qc, I denotes the interval [−p+, p+]. ◦We saw that q−1 c (I) = I0 ∪I1 was the union of two closed intervals, and we dened the set Λ to be the the points in I whose orbits do not blow up to ∞, equal to the innite intersection Λ = ∞ \ n=1 q−n c (I). ◦Finally, we dened the itinerary of x ∈Λ as the binary sequence whose ith digit records which of I0 and I1 the ith iterate of x lands in. • We start by proving our earlier assertion that the set q−n c (I) is a union of 2n intervals whose lengths tend to 0 as n →∞, and also show that our labeling of these intervals was self-consistent: • Theorem (Iterated Inverse Images of I Under qc): Let qc(x) = x2 + c for c < −2 and I = [−p+, p+] for p+ = 1 + √1 −4c 2 . Then q−n c (I) consists of 2n disjoint subintervals of I, each mapped bijectively onto I, and whose lengths tend to zero as n →∞. Furthermore, for each n-digit binary string d0d2 · · · dn−1, if we let Id0···dn−1 = {x ∈I : qi c(x) ∈Idi for each 0 ≤i ≤n −1}, then q−1 c (Id0···dn−1) = I0d0···dn−1 ∪I1d0···dn−1. In particular, the intervals in q−n c (I) are precisely the intervals Id0···dn−1. ◦Proof: We rst show by induction that q−n c (I) consists of 2n disjoint subintervals of I each of which is mapped bijectively onto I. ∗The base case is provided by the observation that q−1 c (I) = I0 ∪I1, where I0 = [−p+, −√−c −p+] and I1 = [√−c −p+, p+], and it is easy to check that qc maps each interval bijectively onto I. ∗For the inductive step, suppose that q−n c (I) consists of 2n disjoint subintervals of I each mapped bijectively onto I. Then q−(n+1) c (I) = q−1 c (q−n c (I)) is the inverse image under qc of these 2n subin-tervals of I0 ∪I1. Since qc is two-to-one and continuous on I, for each closed interval J ∈q−n c (I), we see that q−1 c (J) is the union of two closed intervals J0 ∪J1 (one in I0 and the other in I1) each of which is mapped bijectively onto J by qc. ∗Since qn c maps J bijectively onto I, we conclude that qn+1 c maps each of J0 and J1 bijectively onto I, as required. ◦Next, we show that q−1 c (Id1···dn) = I0d1···dn ∪I1d1···dn, also by induction on n. (The last statement in the theorem then follows by iteratively applying this statement.) ∗For the base case, we already know that q−1 c (I) = I0 ∪I1, and the notation for I0 and I1 is consistent with the given denition. ∗For the inductive step, we rst observe that qc(Id0d1···dn) = Id1···dn for each choice of the digit d0: by denition, Id0d1···dn is the set of points y ∈I such that y ∈Id0, qc(y) ∈Id1, ... , qn c (y) ∈Idn. By applying qc, we see that qc(Id0d1···dn) is the set of points x ∈I such that x ∈Id1, qc(x) ∈Id2, ... , qn−1 c (x) ∈Idn: but this is simply Id1···dn. ∗Furthermore, by the result shown above, q−1 c (Id1···dn) is the union of two closed intervals, one lying in I0 and the other lying in I1, and that each of these intervals is mapped bijectively by qc onto Id1···dn. But this precisely describes the two intervals I0d1···dn and I1d1···dn, so q−1 c (Id1···dn) = I0d1···dn∪I1d1···dn as claimed. ◦Finally, for the statement about lengths, we will prove the result under the additional assumption that c < −(5 + 2 √ 5)/4 = −2.368. (The result is still true for general c < −2 but the proof is more intricate.) ∗For c < −(5 + 2 √ 5)/4, it is a straightforward computation that \|q′ c(x)\| > 1 on all of q−1 c (I). Now let λ be any constant larger than 1 for which \|q′ c(x)\| > λ on all of q−1 c (I) and take J = [a, b] to be any subinterval of q−1 c (I). ∗Then, since qc is monotone on J (since q′ c is never zero), the endpoints of qc(J) are qc(a) and qc(b). 13 ∗Now apply the mean value theorem on [a, b]: there is some constant γ ∈(a, b) for which qc(b) −qc(a) b −a = q′ c(γ), so \|qc(b) −qc(a)\| \|b −a\| = \|q′ c(γ)\| > λ. ∗Equivalently, \|b −a\| ≤λ−1 \|qc(b) −qc(a)\|: thus, the length of J is at most λ−1 times the length of f(J). ∗If we then take J to be any subinterval of q−n c (I), applying this argument n times shows that the length of J is at most λ−n times the length of qn c (J) = I. Since λ > 1, as n →∞this quantity tends to zero, as claimed. • Theorem (Conjugacy of Shift Map and Quadratic Map): For any c < −2, if Λ is the set of points x ∈R whose orbit under qc(x) = x2 + c remains nite, and S : Λ →Σ2 is the itinerary map , then S is a homeomorphism and the dynamical systems (Λ, qc) and (Σ2, σ) are conjugate via S. ◦Most of the hard work has already been done in proving the theorem that describes q−n c (I): we just need to relate that information to the itinerary map. The key observation is that if S(x) = (d0d1d2 · · · dn · · · ), then x ∈Id0d1···dn···. ◦Proof: First, we begin by showing that S(qc(x)) = σ(S(x)) for each x ∈Λ. ∗To see this suppose that x ∈Λ has itinerary S(x) = (d0d1d2d3 · · · ). ∗Then, by denition, qi c(x) ∈Idi, so by reindexing we see that qi c(qc(x)) ∈Idi+1. ∗Thus, the itinerary of qc(x) is (d1d2d3 · · · ) = σ(S(x)), as claimed. ◦Second, we show that S is injective. ∗Suppose that S(x) = S(y) = (d0d1d2d3 · · · ); we wish to show that x = y. ∗By denition, then qi c(x) ∈Idi for each i ≥0. By the denition of the intervals in our earlier theorem about the iterated inverse images of I under qc, we see that x ∈Id0d1···dn for each n ≥0. Similarly, y ∈Id0d1···dn for each n. ∗Suppose \|x −y\| = ϵ > 0. From our theorem, we know that the length of Id0d1···dn tends to 0 as n →∞, so there is some n for which the length is less than ϵ. But this is impossible because an interval of length less than ϵ cannot contain two points of distance ϵ. ∗Thus, \|x −y\| = 0 and so x = y. ◦Third, we show that S is surjective. ∗Given a digit string (d0d1d2 · · · ), we will construct an x such that S(x) = (d0d1d2d3 · · · ). ∗Consider the innite intersection ∞ \ i=0 Id0d1···di: this is an intersection of an innite nested sequence of closed intervals of lengths tending to 0, so the intersection is a single point. ∗Take x to be this intersection point: then since x ∈Id0d1···di, we see that qi c(x) ∈Idi. This holds for each i, so we conclude that S(x) = (d0d1d2d3 · · · ), as claimed. ◦Fourth, we show that S is continuous. ∗Suppose ϵ > 0 and x ∈Λ are given. We need to show there exists δ such that for any y with \|x −y\| < δ, it is true that d(S(x), S(y)) < ϵ. ∗To do this, choose n such that 2−n < ϵ, and let S(x) = (d0d1 · · · dndn+1 · · · ). ∗Now, q−(n+1) c (I) consists of 2n+1 disjoint intervals, each pair of which is separated by a positive distance. Choose δ to be suciently small so that the interval (x −δ, x + δ) does not intersect any interval in q−(n+1) c (I) except for Id0d1···dn: this is possible because x lies in Id0d1···dn, and the endpoints of this interval are a positive distance away from each of the other intervals in q−(n+1) c (I). ∗Then \|x −y\| < δ implies that y ∈Id0d1···dn, so S(y) = (d0d1 · · · dnen+1en+2 · · · ) for some digits ei. ∗By our nearby sequences proposition, we then have d(S(x), S(y)) ≤2−n < ϵ, as required. ◦Finally, we show that S−1 is continuous. ∗Notice that S−1 is the map which maps the sequence (d0d1d2 · · · ) to the single point in the innite intersection ∞ \ i=0 Id0d1···di. 14 ∗Suppose ϵ > 0 and x ∈Σ2 are given. We need to show there exists δ such that for any y with d(x, y) < δ, it is true that S−1(x) −S−1(y) < ϵ. ∗To see this, choose n such that the length of every interval in q−n c (I) is less than ϵ. (This is possible because the lengths of these intervals tend to 0 as n →∞.) ∗We claim that δ = 2−n is sucient: by our nearby sequences proposition, d(x, y) < 2−n implies that the 0th through nth terms of x and y agree: say they are d0d1 · · · dn−1. ∗Thus, by the denition of S, we see that S−1(x) and S−1(y) both lie in Id0d1···dn−1. But now, by assumption, this interval has length less than ϵ: thus, S−1(x) −S−1(y) < ϵ as required. • This theorem is the origin of the name symbolic dynamics: it allows us to use the symbolic sequence space to understand the dynamics of the polynomial map qc. ◦For example, we can write down all of the n-cycles of the shift map on Σ2. The fact that the shift map is conjugate to qc then tells us that each n-cycle of σ gives rise to an n-cycle of qc. ◦So, in particular, for any c < −2, the map qc has exactly two 3-cycles and exactly three 4-cycles. (Try proving this fact analytically!) ◦More generally, there are 2n points of period dividing n, so there is at least 2n−2n−1−2n−2−· · ·−2−1 = 1 point of period exactly n. (We gave an exact formula using Möbius inversion earlier: it is 1 n X d\|n 2n/dµ(d) where µ(d) is the Möbius function.) ◦Using the itinerary map and the results above, we can even describe how to compute these cycles numerically: a point with itinerary (d0d1d2d3 · · · dn · · · ) must lie in the interval Id0d1d2d3···dn. This interval can be found by computing, successively, the intervals J0 = Idn, J1 = Idn−1dn, J2 = Idn−2dn−1dn, ... , Jn = Id0d1···dn, where Jk = Idn−k ∩q−1 c (Jk−1). (Each successive interval is one of the two intervals lying in q−1 c of the previous term.) 3.2 Chaotic Dynamical Systems • In this section we will give Devaney's denition of chaos and show that a number of the systems we have analyzed are chaotic. ◦We will note that there is no universally accepted denition for a chaotic system, but Devaney's denition is one that is frequently used. 3.2.1 Motivation for Chaos: Properties of the Shift Map on the Sequence Space • To motivate the denition of chaos, we will study some of the properties of the shift map σ on the sequence space Σ2. • Our rst observation is that, for any sequence x ∈Σ2, there are periodic points of σ that are arbitrarily close to x. • Proposition: For any x ∈Σ2, there is a sequence xi of periodic points for σ such that lim i→∞xi = x. ◦Proof: We will construct the points explicitly, so suppose x = (d0d1d2 · · · ). ◦Now let xi = (d0d1d2 · · · di). Then since the 0th through ith terms of x and xi agree, by our proposition on nearby sequences we immediately have d(x, xi) ≤2−i. ◦Also, lim i→∞xi = x, since the rst n digits of xn do not change in any subsequent xi. • An equivalent way to state the previous proposition involves the topological notion of denseness: • Denition: If X is a metric space, then we say a subset S is dense in X if for any x ∈X, there is a sequence of elements si ∈S such that lim i→∞si = x. 15 ◦Example: The set of rational numbers Q is a dense subset of R, since every real number can be exhibited as the limit of a sequence of rational numbers. (For example, take the sequence of truncations of its decimal expansion in base 10 or any other base, for that matter.) ◦Example: The set of irrational numbers is also a dense subset of R, since every real number x can be exhibited as the limit of a sequence of irrational numbers. (For example, take any sequence of rational numbers converging to x − √ 2, and add √ 2 to each of them.) ◦Example: The set of periodic points for σ is a dense subset of Σ2, as we just showed. ◦Non-Example: The set {1, 1 2, 1 3, 1 4, 1 5, · · · } is not dense in the interval [0, 1]. For example, there are no points in this set that are within a distance 0.2 of 3 4. Indeed, much more is true: the only limit points of this set (i.e., points that are the limit of some sequence of elements in the set) are 0 and the points already in the set. • Although it may seem like a dense set must be large if it is to be close to every point in the metric space X, this is not necessarily the case. For example, even though Q is a dense subset of R, we can cover Q with a union of open intervals whose total length is arbitrarily small. ◦Explicitly: let ϵ > 0, list the rationals as r1, r2, r3, ...., and take Ii to be the open interval of length 2−iϵ around ri for each i ≥1. Then the innite union ∞ [ i=1 Ii contains Q and is therefore dense in R, but the sum of the lengths of all the intervals is ∞ X i=1 2−iϵ = ϵ, which can be arbitrarily small! ◦However, it is not possible to cover the set of irrational numbers with a union of open intervals having a nite sum of lengths. (The problem is that the set of irrational numbers is uncountable, and an uncountable sum of positive numbers is necessarily innite.) ◦A fuller discussion of this kind of phenomenon belongs to measure theory, which is a discipline of analysis concerned with assigning a measure, or size, to well-behaved subsets of a metric space in a way that is both consistent and analytically useful. • There is another interesting property of σ involving denseness: • Proposition: There exists an element s ∈Σ2 such that the orbit of s under σ is dense in Σ2. ◦Proof: We claim that the element s = ( 0 1 \|{z} length 1 00 01 10 11 \| {z } length 2 000 001 · · · 111 \| {z } length 3 · · · ), constructed by listing all sequences of length 1, then all sequences of length 2, then all sequences of length 3, and so forth, has a dense orbit. ◦To see this, observe that by applying an appropriate shift map σk to s, we can obtain a sequence whose 0th through nth terms form any desired sequence (since any block of length n+1 appears in the expansion of s). ◦Thus, for any x = (d0d1d2 · · · ) and any n, there is an integer an such that the 0th through nth terms of σan(s) are d0d1 · · · dn: then d(x, σan(s)) ≤2−n. ◦In other words: for any x, there is a sequence of terms in the orbit of s that approach x. Since x was arbitrary, this means the orbit of s is dense in Σ2, as claimed. • Related to the existence of a dense orbit is the idea of transitivity: • Denition: A dynamical system (X, f) is transitive if, for every x and y in X and any ϵ > 0 there is a z ∈X whose orbit contains a point within ϵ of x and another point within ϵ of y. ◦A dynamical system (X, f) with a dense orbit is necessarily transitive, since (by denition) the dense orbit contains points within ϵ of any point of X. ◦Thus in particular, (Σ2, σ) is transitive. 16 • The converse of the above statement, that a transitive dynamical system necessarily contains a point with a dense orbit, is also true if X is a compact metric space. (We will not actually use this result, but it is useful as a general fact.) ◦The proof of this statement is not easy: it is essentially nonconstructive and invokes a technical result known as the Baire category theorem. ◦A metric space is compact if any open covering possesses a nite subcover. Explicitly: X is compact if, for any collection of open sets {Ui}i∈I indexed by some set I with the property that X = S i∈I Ui, then there is some nite collection Ui1, ... , Uin such that X = Ui1 ∪· · · ∪Uin. ◦The Heine-Borel theorem states that a subset of Rn is compact if and only if it is closed and totally bounded (i.e., if the set is closed and contained in a ball of some nite radius). Thus, for example, the interval [0, 1] and the Cantor ternary set are both compact. • Example: Show that the doubling map D(x) = ( 2x for 0 ≤x < 1/2 2x −1 for 1/2 ≤x < 1 is transitive on [0, 1). ◦Suppose x and y are elements in [0, 1) and ϵ > 0 is given. We will nd an element z ∈[0, 1) whose orbit comes within ϵ of both x and y. ◦To do this, choose n with 2−n < ϵ, and write x = 0.x1x2 . . . xnxn+1 . . . and y = 0.y1y2 . . . yn . . . , with each expression taken in base 2. ◦Then let z = 0.x1x2 . . . xnxn+1y1y2 . . . yn. ◦Then \|x −z\| ≤ ∞ X i=n+2 \|xi −yi−n−1\| 2i ≤ ∞ X i=n+2 2 2i = 2−n < ϵ. ◦Also, notice that Dn+1(z) = 0.y1y2 . . . yn, so y −Dn+1(z) = ∞ X i=n+1 yi 2i ≤ ∞ X i=n+1 1 2i = 2−n < ϵ. ◦So the orbit of z comes within ϵ of both x and y, so D is transitive. • Non-example: Show that the dynamical system (R, f) where f(x) = x2 is not transitive. ◦We will show that there is no orbit that contains a point within 0.01 of 0.5 and of 0.4. ◦Suppose, by way of contradiction, that the orbit of x ∈R did have this property. ◦Since f(x) −0 x −0 = x < 1 for x ∈(0, 1), every orbit in (0, 1) moves monotonically closer to 0. So in particular, the point within 0.01 of 0.5 must be before the point within 0.01 of 0.4. ◦But if x ∈(0.49, 0.51) then f(x) ∈(0.492, 0.512) = (0.2401, 0.2601), and all subsequent orbits will be even closer to 0. ◦In particular, the orbit of a point within 0.01 of 0.5 cannot contain a point within 0.01 of 0.4. • The last relevant property of the shift map on the sequence space is that it displays sensitive dependence: moving even a small distance away from a starting point will eventually introduce large changes in the orbit. • Denition: A dynamical system (X, f) depends sensitively on initial conditions if there is a β > 0 such that for any x ∈X and any ϵ > 0, there is a y ∈X and an integer k such that d(x, y) < ϵ but d(f k(x), f k(y)) > β. ◦In other words, no matter which point x ∈X we start at, and no matter how close we start to x, there are points within that small distance of x that eventually move a distance at least as large as β from x. (Note that, importantly, the constant β is uniform and does not depend on the starting point x nor on the size of the starting interval ϵ around x.) ◦Roughly speaking: for any x ∈X, there are points near x whose orbits eventually move far away from x. (The orbit does not have to stay far away: it may move back towards x at some later stage.) 17 ◦Sensitive dependence is an extremely troublesome issue when performing numerical computations: if a system exhibits sensitive dependence, then small errors in the computation are likely to magnify as we compute iterates (due to the phenomenon of orbits moving away from one another). As we iterate further, the results can quite easily overwhelm the results of a computation to the point where it becomes completely inaccurate. • Example: Show that the shift map σ is sensitive to initial conditions at every point of Σ2. ◦We will show that the denition is satised with β = 0.5. Let x = (d0d1d2 · · · ) and choose any ϵ > 0. ◦Take any y = (e0e1e2 · · · ) such that y ̸= x and d(x, y) < ϵ. Then since y ̸= x there is some i ≥0 for which xi ̸= yi, as otherwise y and x would be identical. ◦But then we have d(σi(x), σi(y)) = ∞ X j=0 \|xj+i −yj+i\| 2j ≥\|xi −yi\| 20 = 1 > β, meaning that the orbits of x and y eventually move at least a uniform distance β apart, as required. ◦Note that we have actually proved much more than was required: we actually showed that any point near x has an orbit that eventually moves far away from x. • Non-Example: Show that the map f(x) = 1 2x is not sensitive to initial conditions. ◦We easily see that \|f(x) −f(y)\| = 1 2 \|x −y\|, so any two points will be moved strictly closer to one another by f. ◦In particular, if \|x −y\| < ϵ, then \|f n(x) −f n(y)\| < ϵ as well, so the distances between the orbits will never exceed any xed bound. • Some dynamical systems have sensitive dependence only near particular points. • Example: Show that f(x) = x2 has sensitive dependence at x = 1, but not at x = 1/3. ◦We know that any orbit starting in (0, 1) is attracted to the xed point x = 0, while any orbit starting in (1, ∞) will diverge to +∞. Thus, the map exhibits sensitive dependence at x = 1, since any point near 1 will eventually move a distance greater than β = 0.5 away from 1 (since it will approach either 0 or ∞). ◦The map does not exhibit sensitive dependence at x = 1/3, however, because all points x, y ∈(0, 1/2) have the property that \|f(x) −f(y)\| < \|x −y\|: this follows immediately from the mean value theorem and the fact that \|f ′\| < 1 on this interval. Thus, any two orbits that begin within a distance ϵ from one another will never move further apart. • More generally, if a dynamical system has a (weakly) repelling xed point or cycle, then it has sensitive dependence near that repelling xed point. On the other hand, there is no sensitive dependence anywhere inside the immediate attracting basin of a (weakly) attracting xed point or cycle. 3.2.2 The Formal Denition of Chaos, and Examples • We now give Devaney's denition of a chaotic dynamical system: • Denition: A dynamical system (X, f) is chaotic if it (i) has a dense set of periodic points, (ii) it is transitive, and (iii) it depends sensitively on initial conditions. ◦A chaotic system, in general, is computationally intractable (because of the sensitive dependence) and also cannot be broken down into a simpler systems (because of the transitivity, it cannot be split into two systems that can be analyzed separately). Nonetheless, it still contains some amount of regular and predictable behavior (namely, the dense set of periodic points). ◦Example: The shift map σ on Σ2 is chaotic, since we already showed that it has a dense set of periodic points, that it is transitive, and that it depends sensitively on initial conditions. ◦Non-example: The map f(x) = 2x on R is not chaotic. Although it does depend sensitively on initial conditions (at every point in R) it does not have a dense set of periodic points and it is also not transitive. 18 • Example: Show that the doubling map D(x) = ( 2x for 0 ≤x < 1/2 2x −1 for 1/2 ≤x < 1 is chaotic on [0, 1). ◦First, we check that D has a dense set of periodic points. ∗We claim that every rational number x = p q with q odd is a periodic point for D. ∗This follows by observing that D is a bijection from the set 1 q , 2 q , · · · , q −1 q to itself: clearly D maps this set into itself, and it is also surjective because 2k q = D(k q ) and 2k −1 q = D(k + (q −1)/2 q ). ∗But the rational numbers with odd denominator are dense in [0, 1), so D has a dense set of periodic points. ◦Second, we claim that D has a dense orbit, so (in particular) it is transitive. ∗To do this, let α = 0. 0 1 \|{z} length 1 00 01 10 11 \| {z } length 2 000 001 · · · 111 \| {z } length 3 · · · be the base-2 decimal constructed by listing all sequences of length 1, then all sequences of length 2, then all sequences of length 3, and so forth. ∗Note that, in base 2, D(α) is obtained simply by deleting the rst digit of the base-2 decimal expansion of α (i.e., it acts essentially as the shift map). ∗So in particular, for any sequence of digits, there is a shift of α that begins with that sequence of digits. ∗Now let x = 0.d1d2d3 . . . and ϵ > 0. We will show there is some shift of α within ϵ of x. ∗Choose n with 2−n < ϵ. Then there is a positive integer k such that Dk(α) begins as 0.d1d2 . . . dndn+1, so that Dk(α) and x can only dier past the n + 2nd decimal place. ∗Then Dk(α) −x ≤ ∞ X i=n+2 2 2i < 2−n < ϵ, as required. ◦Finally, we show that D has sensitive dependence. ∗We will show that the value β = 1/3 will satisfy the requirements of the denition. ∗First, observe that if a, b are both in [0, 1/2) or [1/2, 1), then \|D(b) −D(a)\| = 2 \|b −a\|. ∗Also, if a ∈[0, 1/2) and b ∈[1/2, 1) then one of \|b −a\| and \|D(b) −D(a)\| is at least 1/3, since if b −a < 1/3 then \|D(b) −D(a)\| = \|1 −2(b −a)\| is larger than 1/3. ∗Therefore, if x, y are any two distinct points, the value of Dk(y) −Dk(x) will double at each stage until the points x and y land in opposite halves of [0, 1), at which point either \|Dn(y) −Dn(x)\| will exceed 1/3 or Dn+1(y) −Dn+1(x) will. ∗Thus, for any two distinct points x and y, their orbits will eventually be a distance of at least 1/3 apart after iterating some number of times. • We will remark that the denition we have given is not the only possible denition of chaos, and there is no general denition of chaos that is universally accepted. ◦In colloquial usage, a system that exhibits sensitive dependence (but not necessarily the other properties) is often called chaotic due to its unpredictability. ◦The example of f(x) = 2x on R, which exhibits sensitive dependence but is nonetheless extremely predictable, suggests quite strongly that we want something stronger than sensitive dependence in order to call a system chaotic. ◦It is generally held that the most important aspects of a chaotic system are sensitive dependence and transitivity. Indeed, Robinson's denition of chaos uses only these two conditions and discards the requirement for a dense set of periodic points. • It has also been shown that, if X is a compact metric space, a transitive system that has a dense set of periodic points necessarily has sensitive dependence too. • Theorem: If X is an innite metric space and f : X →X is a continuous function that has a dense set of periodic points and is transitive, then f also has sensitive dependence on initial conditions. 19 ◦This theorem can be proven using topological arguments similar to the ones we have already given. The details are rather lengthy and not especially enlightening, however, so we will omit them. • If (X, f) is chaotic and conjugate to (Y, g), it seems reasonable to hypothesize that (Y, g) is also chaotic. This turns out to be true if we assume some mild conditions on f and X. ◦Along the way we will prove that properties (i) and (ii) of a chaotic system (namely, having a dense set of periodic points and transitivity) are preserved by conjugation. ◦However: sensitive dependence is not necessarily preserved by conjugation! ◦As an explicit example, observe that the function f(x) = 2x has sensitive dependence on X = (0, ∞), but the function g(x) = x + ln(2) on Y = R does not have sensitive dependence. However, these maps are conjugate, via the homeomorphism h(x) = ln(x). ◦Ultimately, the issue in this example is that the space X = (0, ∞) is not compact. If we assume X is compact, then sensitive dependence is in fact preserved by conjugation. • We will require a proposition about dense sets, which is useful enough that we include it separately: • Proposition (Continuous Image of a Dense Set): If h : X →Y is a continuous surjective map and D is a dense subset of X, then h(D) is a dense subset of Y . ◦Proof: Suppose y ∈Y . We will nd a sequence of points in h(D) with limit y. ◦First, since h is surjective, there exists x ∈X with h(x) = y. ◦Next, by the assumption that D is dense in X, there exists a sequence of elements {xi}i≥1 ∈X with lim i→∞xi = x. ◦Finally, since h is continuous, we can conclude that lim i→∞h(xi) = h(x) = y. So {h(xi)}i≥1 is a sequence of points in h(D) with limit y, as required. • Theorem (Conjugacy and Chaos): If (X, f) is a chaotic dynamical system, h : X →Y is a continuous surjective map with h(f(x)) = g(h(x)) for all x ∈X, g is continuous, and Y is innite, then (Y, g) is also chaotic. ◦Note that this is stronger than merely saying that chaos is preserved by conjugation: chaos is actually preserved by any continuous surjective map that obeys the conjugation property h(f(x)) = g(h(x)). If h has the additional property that there is some n such that h is at most n-to-one, h is called a semi-conjugacy. ◦Proof: We will show each of the parts of the denition of a chaotic system. (Note that most of the individual sub-results do not require all of the hypotheses.) ◦If f has a dense set of periodic points then so does g: ∗By hypothesis, f is chaotic on X so its set S of periodic points is dense in X. ∗Since h is continuous and surjective, by the proposition on the continuous image of a dense set we see that h(S) is dense in Y . ∗Also, h(S) is contained in the set of periodic points for g: by an easy induction, h(f(x)) = g(h(x)) implies h(f n(x)) = gn(h(x)), so if f n(x) = x then gn(h(x)) = h(f n(x)) = h(x). ∗Thus, g has a dense set of periodic points. ◦If f is transitive then g is transitive: ∗Let y1, y2 ∈Y and ϵ > 0. Take B1 to be the open ball of radius ϵ around y1 and B2 to be the open ball of radius ϵ around y2. We want to show there is some point in Y whose orbit under g contains a point in B1 and a point in B2. ∗Since h is continuous and surjective, h−1(B1) is an open set containing x1, so it contains some ball of positive radius r1 around x1. Similarly, h−1(B2) contains some ball of positive radius r2 around x2. ∗By the assumption that f is transitive, there is z ∈X such that the orbit of z contains points within a distance min(r1, r2) of x1 and of x2. 20 ∗Thus, the orbit of z under f contains a point in h−1(B1) and in h−1(B2), so the orbit of h(z) under g contains a point in B1 and a point in B2, as required. ◦Finally, for sensitive dependence we invoke the theorem from earlier: by hypothesis, Y is an innite metric space and g : Y →Y is a continuous function that has a dense set of periodic points and is transitive (as we just showed), so g also has sensitive dependence on initial conditions and is therefore chaotic. • Using this theorem, we can quickly prove that a number of the systems we have already analyzed are chaotic: • Example: For any c < −2, show that qc(x) = x2 + c is chaotic on the set Λ. ◦We already showed that (Σ2, σ) was chaotic and that (Σ2, σ) is conjugate to (Λ, qc). ◦Also, Λ is clearly innite, and σ and qc are both continuous maps. So all of the requirements of the theorem hold, so (Λ, qc) is also chaotic. • Example: Show that the angle doubling map on the circle S1 is chaotic. ◦The angle doubling map, as we saw earlier, is conjugate to the doubling map on R modulo 1, which we already showed (directly from the denition) to be chaotic. ◦Since R modulo 1 is innite and the doubling and angle doubling maps are continuous, the result is immediate. • Example: Show that the map q−2(x) = x2 −2 is chaotic on the interval I = [−2, 2]. ◦We will construct a semi-conjugacy between this map and the angle doubling map D on the circle S1. ◦Explicitly, we claim that h(cos t, sin t) = 2 cos t is a semi-conjugacy between (S1, D) and (I, q−2). ◦To see this, rst note that the map is continuous and surjective from the unit circle to [−2, 2]. ◦Also, h(D(cos t, sin t)) = h(cos 2t, sin 2t) = 2 cos 2t = 4 cos2 t −2 = q−2(2 cos t) = q−2(h(cos t, sin t)). ◦So all of the conditions of the theorem are satised, meaning that q−2(x) = x2 −2 is chaotic on the interval I = [−2, 2] as claimed. ◦To motivate where this semi-conjugacy comes from, notice that the angle doubling map sends (cos t, sin t) 7→ (cos 2t, sin 2t). Now consider only what it does to the x-coordinate: it sends cos t 7→cos 2t = 2 cos2 t −1, or, in other words, it sends the value x to the value 2x2 −1. (And this is a quadratic map.) • As a side-note, we can actually use the semi-conjugacy above to give a simple formula for the periodic points of q−2. ◦Explicitly: our calculations show that for x = 2 cos(t) we have q−2(x) = 2 cos(2t). ◦Then a trivial induction then gives qn −2(x) = 2 cos(2nt). ◦From basic trigonometry, we know that cos(a) = cos(b) is equivalent to a = ±b + 2kπ for some integer k. ◦Applying this shows that 2 cos(t) = qn −2(x) = 2 cos(2nt) is equivalent to 2nt = ±t + 2kπ, so that t = 2kπ 2n ± 1 where k is an integer. ◦Thus, the points of period (dividing) n for q−2 are the values having the form x = 2 cos 2kπ 2n + 1 or x = 2 cos 2kπ 2n + 1 for some integer k. ◦It is not too hard to verify that the full list consists of the 2n−1 values x = 2 cos 2kπ 2n + 1 for 1 ≤k ≤ 2n−1, along with the 2n−1 values x = 2 cos 2kπ 2n −1 for 0 ≤k ≤2n−1 −1. ◦For example, the two 3-cycles can be written explicitly as 2 cos 2π 9 , 2 cos 4π 9 , 2 cos 8π 9 and 2 cos 2π 7 , 2 cos 4π 7 , 2 cos 8π 7 . 21 ◦Using a computer we can factor q3 −2(x) −x q−2(x) −x = (x3 −3x + 1)(x3 + x2 −2x −1), and indeed one can show that the roots of the rst polynomial are 2 cos 2π 9 , 2 cos 4π 9 , 2 cos 8π 9 and the roots of the second polynomial are 2 cos 2π 7 , 2 cos 4π 7 , 2 cos 8π 7 . 3.3 Sarkovskii's Theorem and Applications • In this section we will discuss Sarkovskii's theorem, a result about the structure of periodic points of a continuous function f : R →R that is both strong and surprising for its lack of hypotheses. • We will frequently need to refer to periodic points of exact period n: for shorthand, we will call these period-n points. 3.3.1 The Period-3 Theorem • In 1975, in a paper called Period three implies chaos (which was the rst use of the word chaos to describe a dynamical system), Li and Yorke proved the following theorem: • Theorem (Period-3 Theorem): Let f : I →R be a continuous function dened on an interval I. If f has a period-3 point, then f has a period-k point for any k ≥1. ◦We will remark rst that this result requires f to be continuous: the function f : R →R dened as f(x) = 1 −1 x for x ̸= 0 has the property that f 3(x) = x for all x ̸= 0, 1, so all points have period 3 (and there are no points of any other periods at all). ◦Also, the result is heavily dependent on the underlying metric space being R: it is not even true on the circle, as the map (cos t, sin t) 7→(cos(t + 2π/3), sin(t + 2π/3)), which is simply rotation by 1/3 of a full circle, has every point of period 3, and thus has no points of any other periods. • To prove this theorem, we will begin with a pair of observations: • Lemma 1: If f : I →R is continuous and I and J are closed intervals with I ⊆J and f(I) ⊇J, then f has a xed point in I. ◦Proof: Let I = [a, b]. ◦Since I ⊆J ⊆f(I), there is a point c ∈I such that f(c) ≤a, and there is also a point d ∈I such that b ≤f(d). Since c ∈I we see that f(c) ≤a ≤c and similarly d ≤b ≤f(d). ◦Thus, f(c) −c ≤0 and f(d) −d ≥0. Applying the intermediate value theorem to g(x) = f(x) −x shows that there is some point in [c, d] ⊆I where g is zero: this is a xed point of f. • Lemma 2: If f : I →R is continuous and J ⊆f(I) is a closed and bounded interval, then there exists a closed and bounded interval K ⊆I with f(K) = J. ◦There are more highbrow proofs of this fact using topological properties of compact sets, but we will give a direct proof. ◦Proof: Let J = [a, b]: if a = b the result is obvious so assume a < b. ◦By assumption, there exist p, q ∈I with f(p) = a and f(q) = b. Also suppose that p < q (if p > q then the argument is essentially identical). ◦Let c = max{x : p ≤x ≤q, f(x) = a} be the point closest to q in I such that f(c) = a. (Such a point must exist: by the monotone convergence theorem, any monotone increasing sequence of points xi with f(xi) = a has a limit L, and since f is continuous we obtain f(L) = a.) ◦Now let d = min{x : α ≤x ≤q, f(x) = b} be the point closest to α in I such that f(d) = b. ◦We claim that K = [c, d] has f(K) = J. By construction, f(c) = a and f(d) = b so [a, b] ⊆f(K) by the intermediate value theorem. 22 ◦If there were some w ∈[c, d] with f(w) > b, then the intermediate value theorem would imply the existence of a point e ∈(c, w) such that f(e) = b, which is impossible. Similarly, there cannot be any w ∈[c, d] with f(w) < a. Thus, f(K) = J as required. • Now we can prove the period-3 theorem: ◦Proof: Suppose that f : I →R has a 3-cycle {a, b, c}, where a is less than b and c: then either a < b < c or a < c < b. ◦If a < c < b then notice g(x) = −f(−x) is conjugate to f (via h(x) = −x) and has the 3-cycle {−b, −c, −a} where −b < −c < −a, so it is sucient to treat the case where a < b < c. (The case a < c < b is the mirror image of the case a < b < c.) ◦Let I0 = [a, b] and I1 = [b, c]. Since f(a) = b, f(b) = c, and f(c) = a, we see that f(I0) ⊇I1 and f(I1) ⊇I0 ∪I1 by the intermediate value theorem. ◦First, f must have a xed point in I1 by lemma 1, since f(I1) ⊇I1. ◦Next, we show that f must have a period-2 point in I0. ∗Since f(I0) ⊇I1, by lemma 2 there is an interval B1 in I0 such that f(B1) = I1. ∗Then f 2(B1) = f(I1) ⊇I0 ⊇B1, so since f 2 maps B1 onto itself by lemma 2 we conclude that f 2 must have a xed point y0 in B1. ∗If y0 were a xed point of f, then it would lie in B1 ∩f(B1) = B1 ∩I1 ⊆I0 ∩I1 = {b}, but b has period 3. So y0 must have period exactly 2. ◦Now suppose n > 3. We will construct a period-n point by invoking the interval mapping lemmas to show the existence of a point x0 with f n(x0) = x0 whose rst iterate lies in I0 but whose next n −2 iterates each land in I1. ∗Since f(I1) ⊇I1, by lemma 2 there is a closed subinterval A1 ⊆I1 such that f(A1) = I1. ∗Applying lemma 2 again, we see that there is a closed subinterval A2 ⊆A1 such that f(A2) = A1. ∗We can continue this procedure to construct a sequence of closed intervals An−2 ⊆An−1 ⊆· · · ⊆ A2 ⊆A1 ⊆I1 such that f(Ak) = Ak−1 for each 2 ≤k ≤n −2, and also f(A1) = I1. ∗Also by lemma 2, since f(I0) ⊇I1 ⊆An−2, there is a closed interval An−1 ⊆I0 such that f(An−1) = An−2. ∗Finally, again by lemma 2, since f(I1) ⊇I0 ⊇An−1, there is a closed interval An ⊆I1 such that f(An) = An−1. ∗If we put all of this together, we see that f n(An) = f n−1(An−1) = · · · = f(A1) = I1. ∗But because An ⊆I1, by lemma 1 we conclude that there is a xed point x0 of f n lying in I1. ∗Furthermore, one can check that x0 ∈I0, f(x0) ∈I1, f 2(x0) ∈I0, f 3(x0) ∈I0, ... , f n−1(x0) ∈I0. ∗If x0 had exact period k < n, then we would have f k+1(x0) = f(x0), but f k+1(x0) ∈I0 and f(x0) ∈I1, and the only intersection point of these two intervals is I0 ∩I1 = {b}, so it would necessarily be true that f(x0) = b. But then f 2(x0) would equal c, which is not in I0. This is impossible, so x0 has period exactly n. • Example: Show that the function f : [−1, 1] →[−1, 1] dened by f(x) = ( x + 1 for −1 ≤x ≤0 cos(πx) for 0 < x ≤1 has a period-n point for every n ≥1. ◦First, observe that f is continuous on I, since each of the component functions is continuous and they are equal at the transition point x = 0. ◦The desired result then follows from the period-3 theorem, provided we demonstrate that f has a 3-cycle. ◦A very short amount of experimentation reveals that f(−1) = 0, f(0) = 1, and f(1) = −1, so {−1, 0, 1} is a 3-cycle. Thus, by the period-3 theorem, f has a point of exact period n for every n ≥1. • Example: For any c ≤−7/4, show that the quadratic map qc(x) = x2 + c has a period-n point for each n ≥1. ◦Since qc(x) is continuous on R, by the period-3 theorem it is sucient to show that qc(x) has a period-3 point. 23 ◦To do this, we will show that q3 c(x) has a saddle-node bifurcation at c = −7/4. ◦Some algebra shows that q3 −7/4(x) −x q−7/4(x) −x = 1 64 8x3 + 4x2 −18x −1 2, and the cubic polynomial 8x3 + 4x2 −18x −1 has three real roots x0, x1, and x2 given approximately by −1.747, −0.055, and 1.302). ◦If we rearrange the expression as q3 −7/4(x) = x + (x −x0)2 · 1 64(q−7/4(x) −x)(x −x1)2(x −x2)2 , then it is straightforward to see that q3 −7/4(x0) = x0 and (q3 −7/4)′(x0) = 1, and that (q3 −7/4)′′(x0) ̸= 0. ◦Finally, we can also compute ∂q3 c ∂c c=−7/4 (x) = 1 16 64x6 −304x4 + 364x2 −89 and verify that it is nonzero at each of x0, x1, and x2. ◦So by the saddle-node criterion we conclude that q3 c has a saddle-node bifurcation at x0, x1, and x2 when c = −7/4, and the bifurcation opens in the direction of negative c. ◦It can also be shown using some polynomial algebra that q3 c has no other xed points where the derivative is equal to 1 (which would be the only places these 3-cycles could disappear), so these two 3-cycles will persist for all c ≤−7/4. • The example above does shed some light on the behavior of qc(x) on the interval [−2, −7/4]: it shows that there are innitely many periodic points inside the interval [−p+, p+]. ◦Having innitely many periodic points inside a nite interval does not guarantee they will be dense, but it is at least suggestive of the chaotic behavior we saw experimentally in the orbit diagram of qc(x). ◦Our computations above also help explain the period-3 window near c = −7/4: when the 3-cycle appears, it is attracting for a brief window before undergoing a period-doubling bifurcation. 3.3.2 The Sarkovskii Ordering and Sarkovskii's Theorem • Given the period-3 theorem, it is natural to wonder how far the result extends: for example, does the existence of a 2-cycle, or a 5-cycle, or a 6-cycle, also guarantee the existence of cycles of other orders? • It turns out that, unbeknownst to Li and Yorke, this question had already been answered in a paper of Sarkovskii published (in Russian) more than a decade earlier in 1964. • To state the theorem, we rst need to dene the Sarkovskii ordering of the positive integers: ◦First, we list all of the odd integers starting with 3, followed by 2 times the odd integers starting with 2·3, followed by 22 times the odd integers starting with 22 ·3, and so forth, and nishing with the powers of 2 in descending order. ◦Explicitly, the ordering is 3 ▷5 ▷7 ▷· · · ▷2 · 3 ▷2 · 5 ▷2 · 7 ▷· · · ▷22 · 3 ▷22 · 5 ▷22 · 7 ▷· · · ▷2n ▷2n−1 ▷· · · ▷22 ▷2 ▷1. ◦Thus for example, under the Sarkovskii ordering we have 7 ▷22, 13 ▷2, 20 ▷24, and 14 ▷32. ◦The Sarkovskii ordering is a total ordering on the positive integers: any two distinct integers a, b either satisfy a ▷b or b ▷a. • Theorem (Sarkovskii): Suppose f : I →R is continuous and has a period-k point. If n is any integer with k ▷n, then f also has a period-n point. ◦As an immediate corollary of Sarkovskii's theorem, we see that if f is continuous and has any cycle whose length is not a power of 2, then f in fact has innitely many periodic points each of whose periods is a power of 2. ◦Thus, in particular, if f has only nitely many periodic points, then the periods of each of these points must be a power of 2. 24 ◦In particular, whenever an orbit diagram indicates the existence of an attracting n-cycle when n is not a power of 2, there are actually innitely many other periodic cycles present as well (but we cannot see them, because they are all repelling). ◦The fact that the powers of 2 appear at the end of the Sarkovskii ordering also helps provide an expla-nation for the period-doubling bifurcations we saw in the orbit diagrams: the emergence of an n-cycle with n ̸= 2d at a parameter value λ requires 2d-cycles already to be present for each d ≥1, and the most natural way for these to arise is via a sequence of period-doubling bifurcations of a xed point. • We will not give the full proof of Sarkovskii's theorem, but instead give a detailed outline. The missing pieces are all elementary, in the sense of not requiring anything more than the intermediate value theorem, but the arguments are rather technically involved. ◦Proof (outline): The rst step is to prove a smaller number of more specic cases (which can all be done using arguments similar to the ones we gave in the period-3 theorem): ∗If f has a period-m point with m ≥3, then f has a period-2 point. ∗If f has a period-m point with m ≥3 odd, then f has a period-(m + 2) point. ∗If f has a period-m point with m ≥3 odd, then f has a period-2m point. ∗If f has a period-m point with m ≥3 odd, then f has a period-6 point. ◦Then, by applying these results to f and its iterates in an appropriate way, one can obtain all of the implications in Sarkovskii's theorem. ◦Explicitly, if f has a period-m point with m ≥3 odd, repeatedly applying the second statement shows that f also has points of each odd period larger than k, and also has a point of period 6. ◦If f has a period-2m point with m ≥3 odd, then f 2 has a point of period m. Thus, by the above, f 2 also has a period-k point for each odd k ≥m, so f has a period-k or period-2k point. But by the third statement, having a period-k point implies the existence of a period-2k point, so in either case f has a period-2k point. Also, f 2 has a period-6 point, meaning that f has a period-12 point. ◦Proceeding inductively in a similar way, we can show that if f has a period-2dm point with m ≥3 odd, then f also has a period-2dk point for any odd k ≥m, and also has a period-2d+13 point. This gives all of the implications in Sarkovskii's theorem except for the ones involving points whose period is a power of 2. ◦For these, suppose rst that f has a period-2dm point where m ≥3 is odd. Then for any k, f 2k has a point of period 2d−km if k < d and m if k ≥d, so in any case we see that f 2k has a period-2 point, whence f has a period-2k+1 point. ◦Finally, suppose f has a period-2d point with d ≥2. Then f 2d−2 has a period-4 point, so by the rst statement f 2d−2 has a period-2 point, and therefore f has a period-2d−1 point. Repeatedly applying this result gives all of the remaining implications. • A natural followup question to Sarkovskii's theorem is: are there any implications that are missing? ◦In other words, is it possible that, even if a ▷b in the Sarkovskii ordering, a continuous f : I →R having a period-b point must necessarily also have a period-a point? ◦It turns out that the answer is no! • Theorem (Sarkovskii Converse): For any integer k ≥1, there exists a (bounded) continuous function f : R →R having a period-k point, but no period-n points for any n with n ▷k. There also exists a bounded continuous f having a period-2d point for every d ≥0 but no periodic points of any other period. ◦To prove this theorem, it suces to construct a single example for each of the possible cases: a function with a period-b point but no period-a point for each consecutive pair of terms a ▷b in the Sarkovskii ordering, along with a function having a period-2d point for every d ≥0 but no periodic points of any other period. ◦There are a number of dierent classes of examples that have been used in dierent proofs. 25 ◦We will construct the required examples using the truncated tent maps Th : [0, 1] →[0, 1], which, for a parameter h ∈[0, 1], are dened as Th(x) =      2x for 0 ≤x ≤1/(2h) h for 1/(2h) ≤x ≤1 −1/(2h) 2 −2x for 1 −1/(2h) ≤x ≤1 . ◦Proof: Let T(x) = T1(x), and observe rst (by an easy induction) that T n(x) maps each interval of the form [ k 2n , k + 1 2n ] monotonically onto the interval [0, 1]. Thus, T n has 2n xed points (one in each such interval) and so T has at least one period-n point for each n ≥1. ◦If {x1, x2, . . . , xm} is an m-cycle for T, and a = max(x1, . . . , xm), then because the only points x for which Ta(x) ̸= T(x) are those where T(x) > a, we see that {x1, x2, . . . , xm} remains an m-cycle for Ta, but is not an m-cycle for Tb for any b < a since max(x1, . . . , xm) is no longer in the range of Tb. ◦Now x m, and, for any m-cycle O of T, let hO denote the largest element in that m-cycle. Dene a = min{hO}, where the minimum is taken over all m-cycles of T. We claim that the map Ta has the property that if k ▷m, then Ta has no k-cycle. ∗For example, if m = 3, one can compute that T has the two 3-cycles {2 9, 4 9, 8 9} and {2 7, 4 7, 6 7}, so the corresponding value of a here would be min(8 9, 6 7) = 6 7. ∗First observe that Ta has a unique m-cycle, since if there were a second m-cycle for Ta then either it would have a smaller maximum element (impossible, because by assumption a was minimal) or a larger maximum element (which is also impossible because {x1, . . . , xm} stops being an m-cycle for Tb when b < max(x1, . . . , xm)). ∗Suppose the elements of this unique m-cycle are {z1, . . . , zm}, in increasing order. Then, by con-struction, zm = a, and it is also straightforward to see that z1 = Ta(a). ∗We also observe that the interval [z1, zm] = [Ta(a), a] is invariant under the map Ta: clearly the maximum value is a (which is achieved at the point preceding zm in the cycle C, and is the largest value in the range of Ta) and the minimum value is min(Ta(z1), Ta(a)) = Ta(a), again since the minimum value of Ta on an interval will occur on one of the endpoints. ∗Now, suppose that Ta also has a k-cycle with k ▷m in the Sarkovskii ordering. Then all points of the k-cycle must lie in the interval [T(a), a], since successively iterating Ta eventually maps all values larger than 0 into this interval. Since the k-cycle contains neither of these endpoints (which are both part of the unique m-cycle), the k-cycle is strictly contained in some smaller interval [c, d]. ∗Now, since k ▷m, we may apply Sarkovskii's theorem on the interval [c, d] to obtain the existence of an m-cycle lying in this smaller interval. This is a contradiction, however, because Ta only has one m-cycle, and it contains the points a and T(a) which do not lie in the interval [c, d]. ◦Finally, for the map having a 2d-cycle for each d ≥1 but no other cycles, if we let a2n denote the value a2n = min{hO} over all 2d-cycles O and let L = lim n→∞a2n, then TL contains a 2d-cycle for each d ≥1. It can be shown using an argument similar to the above that TL does not contain any other cycles. • Example: Suppose f : R →R has a period-18 point. List all other n for which f must necessarily have a period-n point. ◦By Sarkovskii's theorem and its converse, the desired list is precisely those values which follow 18 in the Sarkovskii ordering. ◦These are the integers of the form 2m with m ≥11 odd, 2dm with d ≥2 and m odd, and all the powers of 2. ◦We could summarize this list more compactly as the integers 2m with m ≥11 odd and the multiples of 4, along with 1 and 2. Well, you're at the end of my handout. Hope it was helpful. Copyright notice: This material is copyright Evan Dummit, 2015. You may not reproduce or distribute this material without my express permission. 26
3233	https://baike.baidu.com/item/%E9%94%99%E4%BD%8D%E7%9B%B8%E5%87%8F%E6%B3%95/7986491	网页新闻贴吧知道网盘图片视频地图文库资讯采购百科帮助首页秒懂百科特色百科知识专题加入百科百科团队权威合作错位相减法播报讨论上传视频收藏查看我的收藏 0有用+1 0 本词条由《中国科技信息》杂志社参与编辑并审核，经科普中国·科学百科认证。形如An=BnCn，其中{Bn}为等差数列，通项公式为bn=b1+(n-1)d；{Cn}为等比数列，通项公式为cn=c1q^(n-1)；对数列An进行求和，首先列出Sn，记为式（1）；再把所有式子同时乘以等比数列的公比q，即q·Sn，记为式（2）；然后错开一位，将式（1）与式（2）作差，对从而简化对数列An的求和。这种数列求和方法叫做错位相减法。错位相减法是一种常用的数列求和方法。应用于等比数列与等差数列相乘的形式。中文名 : 错位相减法外文名 : Dislocation Subtraction 别名 : 错位法适用领域 : 代数应用学科 : 数学目录 1条件 2举例 3解题应用 ▪典例1： ▪典例2： ▪典例3： ▪典例4： 4公式的推导条件播报编辑如果数列的各项是由一个等差数列和一个等比数列的对应项之积构成的，那么这个数列的前n项和Sn可用此法来求和。举例播报编辑【典例】：求和Sn=1+3x+5x2+7x3+…+(2n-1)·xn-1（x≠0，n∈N）当x=1时，Sn=1+3+5+…+(2n-1)=n2 当x≠1时，Sn=1+3x+5x2+7x3+…+(2n-1)xn-1 ∴xSn=x+3x2+5x3+7x4+…+(2n-1)xn 两式相减得(1-x)Sn=1+2(x+x2+x3+x4+…+xn-1)-(2n-1)xn 化简得解题应用播报编辑错位相减法是数列求和的一种解题方法。在题目的类型中：一般是a前面的系数和a的指数是相等的情况下才可以用。典例1：求和：Sn=a+2a2+3a3+…+nan(a≠0,n∈N) 分析：分a=1，a≠1两种情况求解，当a=1时为等差数列易求；当a≠1时利用错位相减法即可求得。解： (1)当a=1时，； (2)当a≠1时，Sn=a+2a2+3a3+…+nan……① ①×a得，aSn= a2+2a3+3a4+……+nan+1 ……② ①-②得，(1-a)Sn=a+(2-1)a2+(3-2)a3+(4-3)a4……+(n-n+1)an-nan+1 (1-a)Sn=a+a2+a3+a4+……+an-nan+1=a(1-an)/(1-a) -nan+1 ∴ 综上所述，当a=1时，；当a≠1时，．典例2：求和Sn=1+3x+5x2+7x3+……..+(2n-1)·xn-1（x≠0,n∈N）解：当x=1时，Sn=1+3+5+…..+(2n-1)=n2 当x≠1时，Sn=1+3x+5x2+7x3+……..+(2n-1)xn-1 ∴xSn=x+3x2+5x3+7x4……..+(2n-1)xn ∴两式相减得：(1－x)Sn=1+2x[1+x+x2+x3+...+xn-2]-(2n-1)xn 化简得：典例3：求和：解： ① ①两边同时乘以 ② ①-②得：典例4：已知数列{an}中，a1=3，点(an，an+1)在直线y=x+2上。 (1)求数列{an}的通项公式； (2)若bn=an`3n，求数列{bn}的前n项和Tn。解： (1)∵点（an,an+1)在直线y=x+2上 ∴an+1=an+2,即an+1-an=2 ∴数列{an}是以3为首项，以2为公差的等差数列 ∴an=3+2(n-1)=2n+1 (2)∵bn=an·3n ∴bn=(2n+1)·3n ∴Tn=3×3+5×32+7×33+…+(2n-1)·3n-1+(2n+1)·3n ① 3Tn=3×32+5×33+…+(2n-1)·3n+(2n+1)·3n+1 ② 由①-②得 -2Tn=3×3+2(32+33+…+3n)-(2n+1)·3n+1 =9+2×9(1-3n-1)/(1-3)-(2n+1)·3n+1 =-2n·3n+1 ∴Tn=n·3n+1 公式的推导播报编辑以下进行一切通项公式为等差乘等比( )型数列的求和公式推导：已知数列的通项公式为求其前项和因为用上式减下式，得应用等比数列求和公式可得两边均乘得错位相减法并非差比数列的专属求和方法；当｛bn｝和是公比为q(q≠1)的等比数列时，只要｛an｝使得｛cn・bn ｝(其an- an-1= cn-1，n≥2，n∈N)的前n项和能求出来，就可以用错位相减法求｛an・bn ｝的和；用错位相减法求和时，可以在和式两边乘不是公比且不等于1的非零实数。成长任务编辑入门编辑规则本人编辑内容质疑在线客服官方贴吧意见反馈举报不良信息未通过词条申诉投诉侵权信息封禁查询与解封 ©2025 Baidu 使用百度前必读 \| 百科协议 \| 隐私政策 \| 百度百科合作平台 \| 京ICP证030173号
3234	https://bghojogh.github.io/assets/files/engg-6140-w23/ENGG6140_KKT.pdf	KKT Conditions Optimization Techniques (ENGG6140) School of Engineering, University of Guelph, ON, Canada Course Instructor: Benyamin Ghojogh Winter 2023 KKT Conditions 1 / 42 The Lagrangian Function KKT Conditions 2 / 42 Optimization Problem minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. (1) Inequality constraints: yi(x) ≤0 ki(x) ≥0 = ⇒−ki(x) ≤0 = ⇒yi(x) ≤0, where yi(x) := −ki(x) Maximization: maximize x f (x) = ⇒minimize x −f (x) = ⇒minimize x g(x) Examples: for fi(x) : x⊤x + a⊤x + b, x3 1 −4x2 −2x2 3 , tan(x1) −4 sin(x3) −2 for yi(x) ≤0 : x⊤x + a⊤x + b ≤0 for yi(x) ≤0 : x2 1 + 3x2 + 2 ≤0, x3 2 −4x3 −2 ≤0, tan(x3) −4 sin(x1) −2 ≤0, ... for hi(x) = 0 : x⊤x + a⊤x + b = 0 for hi(x) = 0 : x2 1 + 3x2 + 2 = 0, x3 2 −4x3 −2 = 0, tan(x3) −4 sin(x1) −2 = 0, ... KKT Conditions 3 / 42 Lagrangian and Dual Variables minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. Definition (Lagrangian and dual variables) The Lagrangian function for the optimization problem (1) is L : Rd × Rm1 × Rm2 →R, with domain D × Rm1 × Rm2, defined as: L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) = f (x) + λ⊤y(x) + ν⊤h(x), (2) where {λi}m1 i=1 and {νi}m2 i=1 are the Lagrange multipliers, also called the dual variables, corresponding to inequality and equality constraints, respectively. λ := [λ1, . . . , λm1]⊤∈Rm1, ν := [ν1, . . . , νm2]⊤∈Rm2, y(x) := [y1(x), . . . , ym1(x)]⊤∈Rm1, h(x) := [h1(x), . . . , hm2(x)]⊤∈Rm2. Eq. (2) is also called the Lagrange relaxation of the optimization problem (1). KKT Conditions 4 / 42 Sign of Terms in Lagrangian minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) = f (x) + λ⊤y(x) + ν⊤h(x). sometimes, the plus sign behind Pm2 i=1 νihi(x) is replaced with the negative sign. As hi(x) is for equality constraint, its sign is not important in the Lagrangian function. hi(x) = 0 = ⇒−hi(x) = 0, L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) − m2 X i=1 νihi(x) = f (x) + λ⊤y(x) −ν⊤h(x). However, the sign of the term Pm1 i=1 λiyi(x) is important because the sign of inequality constraint is important. KKT Conditions 5 / 42 Interpretation of Lagrangian minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. Our goals: minimize the cost function: minimize x f (x) while satisfying the constraints by penalizing them: yi(x) ≤0 = ⇒minimize x,λi λiyi(x), where λi ≥0. hi(x) = 0 = ⇒minimize x,νi νihi(x), where νi ≥0. let’s combine these using regularized minimization: min x,λ,ν L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) = f (x) + λ⊤y(x) + ν⊤h(x). So, Lagrangian is the relaxation of optimization problem to an unconstrained problem. KKT Conditions 6 / 42 Interpretation of Lagrangian minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. We want to minimize the objective function f (x). We create a cost function consisting of the objective function. The optimization problem has constraints so its constraints should also be satisfied while minimizing the objective function. Therefore, we penalize the cost function if the constraints are not satisfied. For this, we can add the constraints to the objective function as the regularization (or penalty) terms and we minimize the regularized cost. The dual variables λ and ν can be seen as the regularization parameters which weight the penalties compared to the objective function f (x). This regularized cost function is the Lagrangian function or the Lagrangian relaxation of the problem (1). Minimization of the regularized cost function minimizes the function f (x) while trying to satisfy the constraints. min x,λ,ν L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) = f (x) + λ⊤y(x) + ν⊤h(x). KKT Conditions 7 / 42 Lagrange Dual Function Definition (Lagrange dual function) The Lagrange dual function (also called the dual function) g : Rm1 × Rm2 →R is defined as: g(λ, ν) := inf x∈D L(x, λ, ν) = inf x∈D f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) . (3) For convex minimization: the dual function g is a concave function. We will see later that we maximize this concave function in a so-called dual problem. For convex maximization: the dual function g is a convex function. We will see later that we minimize this convex function in a so-called dual problem. KKT Conditions 8 / 42 Primal and Dual Feasibility KKT Conditions 9 / 42 Primal Feasibility minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. Definition (The optimal point and the optimum) The solution of this optimization problem is the optimal point denoted by x∗. The minimum function from this solution, i.e., f ∗:= f (x∗), is called the optimum function of this problem. The optimal point x∗is one of the feasible points which minimizes function f (.) with constraints in problem (1). Hence, the optimal point is a feasible point: yi(x∗) ≤0, ∀i ∈{1, . . . , m1}, (4) hi(x∗) = 0, ∀i ∈{1, . . . , m2}. (5) These are called the primal feasibility. KKT Conditions 10 / 42 Feasibility in dual function minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}. L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) = f (x) + λ⊤y(x) + ν⊤h(x). The optimal point x∗minimizes the Lagrangian function because Lagrangian is the relaxation of optimization problem to an unconstrained problem. On the other hand, according to Eq. (3), g(λ, ν) := inf x∈D L(x, λ, ν), the dual function is the minimum of Lagrangian w.r.t. x. Hence, we can write the dual function as: g(λ, ν) (3) = inf x∈D L(x, λ, ν) = L(x∗, λ, ν). (6) KKT Conditions 11 / 42 Dual Feasibility Lemma (Dual function as a lower bound) For minimization problem, if λ ⪰0, then the dual function is a lower bound for f ∗: g(λ, ν) ≤f ∗. (7) Proof. Let λ ⪰0 which means λi ≥0, ∀i. Consider a feasible e x for problem (1). We have: L(e x, λ, ν) (2) = f (e x) + m1 X i=1 λi \|{z} ≥0 yi(e x) \| {z } ≤0 + m2 X i=1 νi hi(e x) \| {z } =0 ≤f (e x). (8) Therefore, we have: f (e x) (8) ≥L(e x, λ, ν) ≥inf x∈D L(x, λ, ν) (3) = g(λ, ν). Hence, the dual function is a lower bound for the function of all feasible points. As the optimal point x∗is a feasible point, the dual function is a lower bound for f ∗. KKT Conditions 12 / 42 Dual function as a lower bound For minimization problem: g(λ, ν) ≤f ∗ For maximization problem: g(λ, ν) ≥f ∗ KKT Conditions 13 / 42 Nonnegativity of dual variables for inequality constraints From the above lemma, we conclude that for having the dual function as a lower bound for the optimum function, the dual variable {λi}m1 i=1 for inequality constraints (less than or equal to zero) should be non-negative, i.e.: λ ⪰0 or λi ≥0, ∀i ∈{1, . . . , m1}. (9) We assume that the inequality constraints are less than or equal to zero. If some of the inequality constraints are greater than or equal to zero, we convert them to less than or equal to zero by multiplying them to −1. Or, if the inequality constraints are greater than or equal to zero, we should have λi ≤0, ∀i because yi(x) ≥0 = ⇒−yi(x) ≤0. KKT Conditions 14 / 42 The Dual Problem KKT Conditions 15 / 42 Primal and dual problems Definition (Primal and dual problems) We saw that the dual function is a lower bound for the optimum function: g(λ, ν) ≤f ∗. We want to find the best lower bound so we maximize g(λ, ν) w.r.t. the dual variables λ, ν. Eq. (9) says that the dual variables for inequalities must be nonnegative: λ ⪰0 Hence, we have the following optimization: maximize λ,ν g(λ, ν) subject to λ ⪰0. (10) The problem (10) is called the Lagrange dual optimization problem for problem (1). The problem (1) is also referred to as the primal optimization problem. The variable of problem (1), i.e. x, is called the primal variable while the variables of problem (10), i.e. λ and ν, are called the dual variables. Let the solutions of the dual problem be denoted by λ∗and ν∗. We denote: g∗:= g(λ∗, ν∗) = sup λ,ν g. KKT Conditions 16 / 42 Weak and strong duality Definition (Weak and strong duality) For all convex and nonconvex minimization problems, the optimum dual problem is a lower bound for the optimum function: g∗≤f ∗ i.e., g(λ∗, ν∗) ≤f (x∗). (11) This is called the weak duality. For some optimization problems, we have strong duality which is when the optimum dual problem is equal to the optimum function: g∗= f ∗ i.e., g(λ∗, ν∗) = f (x∗). (12) The strong duality usually holds for convex optimization problems. KKT Conditions 17 / 42 Weak and strong duality Corollary Eqs. (11) and (12) show that, in a minimization problem, the optimum dual function always provides a lower-bound for the optimum primal function: g∗≤f ∗. (13) KKT Conditions 18 / 42 Weak and strong duality in iterative optimization If optimization is iterative, the solution is updated iteratively until convergence. The series of primal optimal and dual optimal converge to the optimal solution and the dual optimal, respectively. in convex problem: {x(0), x(1), x(2), . . . } →x∗, {ν(0), ν(1), ν(2), . . . } →ν∗, {λ(0), λ(1), λ(2), . . . } →λ∗, f (x(0)) ≥f (x(1)) ≥f (x(2)) ≥· · · ≥f (x∗), g(λ(0), ν(0)) ≤g(λ(1), ν(1)) ≤· · · ≤g(λ∗, ν∗). g(λ(k), ν(k)) ≤f (x(k)), ∀k. (14) KKT Conditions 19 / 42 Slater’s condition Lemma (Slater’s condition ) For a convex optimization problem in the form: minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, Ax = b, we have strong duality if it is strictly feasible, i.e.: ∃x ∈int(D) : yi(x) < 0, ∀i ∈{1, . . . , m1}, Ax = b. (15) In other words, for at least one point in the interior of domain (not on the boundary of domain), all the inequality constraints hold strictly. This is called the Slater’s condition. KKT Conditions 20 / 42 Example for Slater’s condition Example: minimize x=[x1,x2]⊤ x2 1 + x2 2 + 1 subject to x1 + x2 ≥1, x1 −x2 = 2. The constraints: x1 + x2 ≥1 = ⇒−x1 −x2 + 1 ≤0, x1 −x2 = 2 = ⇒x1 −x2 −2 = 0. So, the problem is: minimize x=[x1,x2]⊤ x2 1 + x2 2 + 1 subject to −x1 −x2 + 1 ≤0, x1 −x2 −2 = 0. KKT Conditions 21 / 42 Example for Slater’s condition minimize x=[x1,x2]⊤ x2 1 + x2 2 + 1 subject to −x1 −x2 + 1 ≤0, x1 −x2 −2 = 0. Can we find at least one [x1, x2]⊤to satisfy the following? −x1 −x2 + 1 ≤0, x1 −x2 −2 = 0. We have: x1 −x2 −2 = 0 = ⇒x1 = x2 + 2 = ⇒−(x2 + 2) −x2 + 1 < 0 = ⇒x2 > −0.5 We can take x2 = 0, then: x1 = x2 + 2 = 0 + 2 = 2 So, we can find x1 = 2, x2 = 0 which satisfies the Slater’s condition. Therefore, strong duality holds for this problem. KKT Conditions 22 / 42 Stationarity Condition KKT Conditions 23 / 42 Stationarity condition L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x). As was explained before, the Lagrangian function can be interpreted as a regularized cost function to be minimized. Hence: minimize x L(x, λ, ν). (16) We can find its minimum by setting its derivative w.r.t. x, denoted by ∇xL, to zero: ∇xL(x, λ, ν) = 0 = ⇒∇xf (x) + m1 X i=1 λi∇xyi(x) + m2 X i=1 νi∇xhi(x) = 0. (17) This equation is called the stationarity condition because this shows that the gradient of Lagrangian w.r.t. x should vanish to zero (note that a stationary point of a function is a point where the derivative of function is zero). This derivative holds for all dual variables and not just for the optimal dual variables. KKT Conditions 24 / 42 Complementary Slackness KKT Conditions 25 / 42 Complementary slackness Assume that the problem has strong duality with x∗, λ∗, and ν∗. According to Eq. (12), g(λ∗, ν∗) = f (x∗), and Eq. (3), g(λ, ν) := infx∈D L(x, λ, ν), we have: f (x∗) (12) = g(λ∗, ν∗) (3) = inf x∈D f (x) + m1 X i=1 λ∗ i yi(x) + m2 X i=1 ν∗ i hi(x) (a) = f (x∗) + m1 X i=1 λ∗ i yi(x∗) + m2 X i=1 ν∗ i hi(x∗) (b) = f (x∗) + m1 X i=1 λ∗ i yi(x∗) (c) ≤f (x∗), (18) where (a) is because x∗is the primal optimal solution for problem (1) and it minimizes the Lagrangian, (b) is because x∗is a feasible point and satisfies hi(x∗) = 0, and (c) is because λ∗ i ≥0 according to Eq. (9) and the feasible x∗satisfies yi(x∗) ≤0, so we have: λ∗ i yi(x∗) ≤0, ∀i ∈{1, . . . , m1}. (19) KKT Conditions 26 / 42 Complementary slackness We obtained Eqs. (18) and (19): f (x∗) = f (x∗) + m1 X i=1 λ∗ i yi(x∗) ≤f (x∗), λ∗ i yi(x∗) ≤0, ∀i ∈{1, . . . , m1}. Therefore: f (x∗) = f (x∗) + m1 X i=1 λ∗ i yi(x∗) ≤f (x∗) = ⇒ m1 X i=1 λ∗ i yi(x∗) = 0 (19) = ⇒λ∗ i yi(x∗) = 0, ∀i. Therefore, the multiplication of every optimal dual variable λ∗ i with yi(.) of optimal primal solution x∗must be zero. This is called the complementary slackness: λ∗ i yi(x∗) = 0, ∀i ∈{1, . . . , m1}. (20) KKT Conditions 27 / 42 Complementary slackness The complementary slackness: λ∗ i yi(x∗) = 0, ∀i ∈{1, . . . , m1}. These conditions can be restated as: λ∗ i > 0 = ⇒yi(x∗) = 0, (21) yi(x∗) < 0 = ⇒λ∗ i = 0, (22) which means that, for an inequality constraint, ▶If the dual optimal is nonzero, its inequality function of the primal optimal must be zero. ▶If the inequality function of the primal optimal is nonzero, its dual optimal must be zero. KKT Conditions 28 / 42 KKT Conditions KKT Conditions 29 / 42 Karush-Kuhn-Tucker (KKT) conditions In previous slides, we derived the primal feasibility, dual feasibility, stationarity condition, and complementary slackness. These four conditions are called the Karush-Kuhn-Tucker (KKT) conditions [2, 3]. The primal optimal variable x∗and the dual optimal variables λ∗= [λ∗ 1, . . . , λ∗ m1]⊤, ν∗= [ν∗ 1 , . . . , ν∗ m2]⊤must satisfy the KKT conditions. KKT Conditions 30 / 42 Karush-Kuhn-Tucker (KKT) conditions We summarize the KKT conditions in the following: 1 Stationarity condition: ∇xL(x, λ, ν) =∇xf (x) + m1 X i=1 λi∇xyi(x) + m2 X i=1 νi∇xhi(x) = 0. (23) 2 Primal feasibility: yi(x∗) ≤0, ∀i ∈{1, . . . , m1}, (24) hi(x∗) = 0, ∀i ∈{1, . . . , m2}. (25) 3 Dual feasibility: λ ⪰0 or λi ≥0, ∀i ∈{1, . . . , m1}. (26) 4 Complementary slackness: λ∗ i yi(x∗) = 0, ∀i ∈{1, . . . , m1}. (27) As listed above, KKT conditions impose constraints on the optimal dual variables of inequality constraints because the sign of inequalities is important. KKT Conditions 31 / 42 Karush-Kuhn-Tucker (KKT) conditions Recall the dual problem (10): maximize λ,ν g(λ, ν) subject to λ ⪰0. The constraint in this problem is already satisfied by the dual feasibility in the KKT conditions. Hence, we can ignore the constraint of the dual problem: maximize λ,ν g(λ, ν), (28) which should give us λ∗, ν∗, and g∗= g(λ∗, ν∗). This is an unconstrained optimization problem and for solving it, we should set the derivative of g(λ, ν) w.r.t. λ and ν to zero: ∇λg(λ, ν) = 0 (6) = ⇒∇λL(x∗, λ, ν) = 0. (29) ∇νg(λ, ν) = 0 (6) = ⇒∇νL(x∗, λ, ν) = 0. (30) Setting the derivatives of Lagrangian w.r.t. dual variables always gives back the corresponding constraints in the primal optimization problem. KKT Conditions 32 / 42 Karush-Kuhn-Tucker (KKT) conditions Eqs. (23), (29), and (30) state that the primal and dual residuals must be zero: ∇xL(x, λ, ν) = ∇xf (x) + m1 X i=1 λi∇xyi(x) + m2 X i=1 νi∇xhi(x) = 0, ∇λL(x∗, λ, ν) = 0, ∇νL(x∗, λ, ν) = 0. Eqs. (3) and (28) can be summarized into the following max-min optimization problem: sup λ,ν g(λ, ν) (3) = sup λ,ν inf x L(x, λ, ν) = L(x∗, λ∗, ν∗). (31) KKT Conditions 33 / 42 Where the KKT name came from? The reason for the name KKT is as follows . In 1952, Kuhn and Tucker published an important paper proposing the conditions . However, later it was found out that there is a master’s thesis by Karush, in 1939, at the University of Chicago, Illinois . That thesis had also proposed the conditions; however, researchers including Kuhn and Tucker were not aware of that thesis. Therefore, these conditions were named after all three of them. KKT Conditions 34 / 42 Method of Lagrange Multipliers KKT Conditions 35 / 42 Method of Lagrange multipliers We can solve the optimization problem (1): minimize x f (x) subject to yi(x) ≤0, i ∈{1, . . . , m1}, hi(x) = 0, i ∈{1, . . . , m2}, using duality and KKT conditions. This technique is also called the method of Lagrange multipliers. For this, we should do the following steps: 1 We write the Lagrangian as Eq. (2): L(x, λ, ν) := f (x) + m1 X i=1 λiyi(x) + m2 X i=1 νihi(x) = f (x) + λ⊤y(x) + ν⊤h(x). 2 We consider the dual function defined in Eq. (3) and we solve it: x† := arg min x L(x, λ, ν). (32) It is an unconstrained problem and according to Eqs. (3) and (23), we solve this problem by taking the derivative of Lagrangian w.r.t. x and setting it to zero, i.e., ∇xL(x, λ, ν) set = 0. This gives us the dual function: g(λ, ν) = L(x†, λ, ν). (33) KKT Conditions 36 / 42 Method of Lagrange multipliers 3 We consider the dual problem, defined in Eq. (10) which is simplified to Eq. (28) because of Eq. (26). This gives us the optimal dual variables λ∗and ν∗: λ∗, ν∗:= arg max λ,ν g(λ, ν). (34) It is an unconstrained problem and according to Eqs. (29) and (30), we solve this problem by taking the derivative of dual function w.r.t. λ and ν and setting them to zero, i.e., ∇λg(λ, ν) set = 0 and ∇νg(λ, ν) set = 0. The optimum dual value is obtained as: g∗= max λ,ν g(λ, ν) = g(λ∗, ν∗). (35) 4 We put the optimal dual variables λ∗and ν∗in Eq. (23) to find the optimal primal variable: x∗:= arg min x L(x, λ∗, ν∗). (36) It is an unconstrained problem and we solve this problem by taking the derivative of Lagrangian at optimal dual variables w.r.t. x and setting it to zero, i.e., ∇xL(x, λ∗, ν∗) set = 0. The optimum primal value is obtained as: f ∗= f (x∗). (37) KKT Conditions 37 / 42 Example for Method of Lagrange multipliers Example: minimize x=[x1,x2]⊤ x2 1 + x2 2 + 1 subject to x1 + x2 ≥1, x1 −x2 = 2. The constraints: x1 + x2 ≥1 = ⇒−x1 −x2 + 1 ≤0, x1 −x2 = 2 = ⇒x1 −x2 −2 = 0. The Lagrangian is: L(x, λ, ν) = x2 1 + x2 2 + 1 + λ(−x1 −x2 + 1) + ν(x1 −x2 −2). ∇xL = ∇x1L ∇x2L = 2x1 −λ + ν 2x2 −λ −ν = 0 0 . Therefore: 2x1 −λ + ν = 0 2x2 −λ −ν = 0 = ⇒x† 1 = λ −ν 2 , x† 2 = λ + ν 2 . KKT Conditions 38 / 42 Example for Method of Lagrange multipliers L(x†, λ, ν) = (x† 1 )2 + (x† 2 )2 + 1 + λ(−x† 1 −x† 2 + 1) + ν(x† 1 −x† 2 −2) = ( λ −ν 2 )2 + ( λ + ν 2 )2 + 1 + λ(−λ −ν 2 −λ + ν 2 + 1) + ν( λ −ν 2 −λ + ν 2 −2) = −1 2 λ2 −1 2 ν2 + λ −2ν + 1 Hence: g(λ, ν) = L(x†, λ, ν) = −1 2 λ2 −1 2 ν2 + λ −2ν + 1. We have: ∇λg(λ, ν) = −λ + 1 = 0 ∇νg(λ, ν) = −ν −2 = 0 = ⇒λ∗= 1, ν∗= −2. KKT Conditions 39 / 42 Example for Method of Lagrange multipliers We found λ∗= 1, ν∗= −2. Therefore: L(x, λ∗, ν∗) = x2 1 + x2 2 + 1 + λ∗(−x1 −x2 + 1) + ν∗(x1 −x2 −2) = x2 1 + x2 2 + 1 + (1)(−x1 −x2 + 1) + (−2)(x1 −x2 −2) = x2 1 + x2 2 −3x1 + x2 + 6. We have: ∇xL(x, λ∗, ν∗) = ∇x1L(x, λ∗, ν∗) ∇x2L(x, λ∗, ν∗) = 2x1 −3 2x2 + 1 = 0 0 . 2x1 −3 = 0 2x2 + 1 = 0 = ⇒x∗ 1 = 3 2 , x∗ 2 = −1 2 . KKT Conditions 40 / 42 Example for Method of Lagrange multipliers In summary, we have: x∗ 1 = 3 2 , x∗ 2 = −1 2 , λ∗= 1, ν∗= −2. f ∗= f (x∗) = 3(x∗ 1 )2 + 2(x∗ 2 )2 + 1 = 3( 3 2 )2 + 2(−1 2 )2 + 1 = 3.5 g∗= g(λ∗, ν∗) = L(x†, λ∗, ν∗) = −1 2 (λ∗)2 −1 2 (ν∗)2 + λ∗−2ν∗+ 1 = −1 2 (1)2 −1 2 (−2)2 + 1 −2(−2) + 1 = 3.5 It is expected to have f ∗= g∗, as we have strong duality (we saw in previous slides that this problem satisfies Slater’s condition; see the example we provided for Slater’s condition). KKT Conditions 41 / 42 References M. Slater, “Lagrange multipliers revisited,” tech. rep., Cowles Commission Discussion Paper: Mathematics 403, Yale University, 1950. W. Karush, “Minima of functions of several variables with inequalities as side constraints,” Master’s thesis, Department of Mathematics, University of Chicago, Chicago, Illinois, 1939. H. W. Kuhn and A. W. Tucker, “Nonlinear programming,” in Berkeley Symposium on Mathematical Statistics and Probability, pp. 481–492, Berkeley: University of California Press, 1951. T. H. Kjeldsen, “A contextualized historical analysis of the Kuhn–Tucker theorem in nonlinear programming: the impact of world war II,” Historia mathematica, vol. 27, no. 4, pp. 331–361, 2000. KKT Conditions 42 / 42
3235	https://www.calculateme.com/math/percent-change/90-percent-less-than-90	Calculate 90% Less than 90 calculateme Math Contact Us Calculate 90% Less than 90 What is a 90 percent decrease from 90? Reduce 90 by 90%, which is the same as decreasing by nine tenths (9/10). % Calculate 90% less than 90 is 9 What is a 90% Discount on $90? If something costs $90 and the price is reduced by 90%, the new price would be $9.00. The new price is $81.00 less than the old price. Nearby Results \| 90% less than \| Result \| --- \| \| 90.0 \| 9 \| \| 90.1 \| 9.01 \| \| 90.2 \| 9.02 \| \| 90.3 \| 9.03 \| \| 90.4 \| 9.04 \| \| 90.5 \| 9.05 \| \| 90.6 \| 9.06 \| \| 90.7 \| 9.07 \| \| 90.8 \| 9.08 \| \| 90.9 \| 9.09 \| \| 91.0 \| 9.1 \| \| 91.1 \| 9.11 \| \| 91.2 \| 9.12 \| \| 91.3 \| 9.13 \| \| 91.4 \| 9.14 \| \| 91.5 \| 9.15 \| \| 91.6 \| 9.16 \| \| 91.7 \| 9.17 \| \| 91.8 \| 9.18 \| \| 91.9 \| 9.19 \| \| 92.0 \| 9.2 \| \| 92.1 \| 9.21 \| \| 92.2 \| 9.22 \| \| 92.3 \| 9.23 \| \| 92.4 \| 9.24 \| \| 92.5 \| 9.25 \| \| 92.6 \| 9.26 \| \| 92.7 \| 9.27 \| \| 92.8 \| 9.28 \| \| 92.9 \| 9.29 \| \| 93.0 \| 9.3 \| \| 93.1 \| 9.31 \| \| 93.2 \| 9.32 \| \| 93.3 \| 9.33 \| \| 90% less than \| Result \| --- \| \| 93.4 \| 9.34 \| \| 93.5 \| 9.35 \| \| 93.6 \| 9.36 \| \| 93.7 \| 9.37 \| \| 93.8 \| 9.38 \| \| 93.9 \| 9.39 \| \| 94.0 \| 9.4 \| \| 94.1 \| 9.41 \| \| 94.2 \| 9.42 \| \| 94.3 \| 9.43 \| \| 94.4 \| 9.44 \| \| 94.5 \| 9.45 \| \| 94.6 \| 9.46 \| \| 94.7 \| 9.47 \| \| 94.8 \| 9.48 \| \| 94.9 \| 9.49 \| \| 95.0 \| 9.5 \| \| 95.1 \| 9.51 \| \| 95.2 \| 9.52 \| \| 95.3 \| 9.53 \| \| 95.4 \| 9.54 \| \| 95.5 \| 9.55 \| \| 95.6 \| 9.56 \| \| 95.7 \| 9.57 \| \| 95.8 \| 9.58 \| \| 95.9 \| 9.59 \| \| 96.0 \| 9.6 \| \| 96.1 \| 9.61 \| \| 96.2 \| 9.62 \| \| 96.3 \| 9.63 \| \| 96.4 \| 9.64 \| \| 96.5 \| 9.65 \| \| 96.6 \| 9.66 \| \| 96.7 \| 9.67 \| \| 90% less than \| Result \| --- \| \| 96.8 \| 9.68 \| \| 96.9 \| 9.69 \| \| 97.0 \| 9.7 \| \| 97.1 \| 9.71 \| \| 97.2 \| 9.72 \| \| 97.3 \| 9.73 \| \| 97.4 \| 9.74 \| \| 97.5 \| 9.75 \| \| 97.6 \| 9.76 \| \| 97.7 \| 9.77 \| \| 97.8 \| 9.78 \| \| 97.9 \| 9.79 \| \| 98.0 \| 9.8 \| \| 98.1 \| 9.81 \| \| 98.2 \| 9.82 \| \| 98.3 \| 9.83 \| \| 98.4 \| 9.84 \| \| 98.5 \| 9.85 \| \| 98.6 \| 9.86 \| \| 98.7 \| 9.87 \| \| 98.8 \| 9.88 \| \| 98.9 \| 9.89 \| \| 99.0 \| 9.9 \| \| 99.1 \| 9.91 \| \| 99.2 \| 9.92 \| \| 99.3 \| 9.93 \| \| 99.4 \| 9.94 \| \| 99.5 \| 9.95 \| \| 99.6 \| 9.96 \| \| 99.7 \| 9.97 \| \| 99.8 \| 9.98 \| \| 99.9 \| 9.99 \| © H Brothers Inc, 2007–2025 Privacy Policy \| Contact Us Search CalculateMe × Go
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Modified 25-01-2023 Sulphur Dioxide: Preparation, Structure, Properties, and Uses Sulphur Dioxide:Do you know what a burning matchstick smells like? It smells like sulphur dioxide. It’s a major contributor to greenhouse gas emissions and a key ingredient in the production of the ‘king of chemicals.’ When we start our car in the morning, we come into contact with sulphur dioxide. We will go over the formula, structure, preparation, properties, and applications of sulphur dioxide in this blog. Sulphur Dioxide Sulphur dioxide is created as a waste gas when sulphur-containing materials, such as metal, are heated, such as when smelting metal or burning coal or oil. It’s a major pollutant that’s currently considered a hazard to the environment. Structure of Sulphur Dioxide Sulphur dioxide has a bent structure with polar covalent bonds between the sulphur atom and two oxygen atoms. The chemical formula Sulphur dioxide is ({\rm{S}}{{\rm{O}}_2}). Sulphur in sulphur dioxide is ({\rm{s}}{{\rm{p}}^2})hybridised. Ideally, the bond angle should be ({120^{\rm{o}}}) but the ({\rm{O}} – {\rm{S}} – {\rm{O}}) bond angle is found to be ({119^{\rm{o}}}). This happens due to the lone pair of electrons on the Sulphur atom. The lone pair-bond pair repulsion decreases the bond angle from ({120^{\rm{o}}})to ({119^{\rm{o}}}). Study About Carbon Chemical Formula Here Each oxygen atom in the Sulphur dioxide molecule is attached to the Sulphur atom by a sigma ((\sigma ))and ((\pi ))bond. The (\sigma )bonds between Sulphur and Oxygen atom is formed by ({\rm{s}}{{\rm{p}}^2}{\rm{ – p}})overlap while one of the (\pi )bonds is formed by the ({\rm{p\pi – p\pi }})overlap and other from ({\rm{p}}\pi – {\rm{d}}\pi )overlap. The electronic configuration of ({\rm{S}}) is (1{{\rm{s}}^2}2{{\rm{s}}^2}2{{\rm{p}}^6}3{{\rm{s}}^2}3{{\rm{p}}^4}). During the formation of ({\rm{S}}{{\rm{O}}_2}), one electron from the (3{\rm{p}})orbital goes to the (3{\rm{d}})orbital, and ({\rm{S}}) undergoes ({\rm{s}}{{\rm{p}}^2})hybridisation. Two of these orbitals form sigma bonds with two oxygen atoms, and the third contains a lone pair. ({\rm{p}})-orbital and ({\rm{d}})-orbital have an unpaired electron each. One of these electrons forms({\rm{p}}\pi – {\rm{ p}}\pi )bond with one oxygen atom, and the other forms({\rm{p}}\pi – {\rm{ d}}\pi )bond with the other oxygen. This is the reason ({\rm{S}}{{\rm{O}}_2})has a bent structure with a bond length of (143\,{\rm{pm}}). However, both the ({\rm{S}} – {\rm{O}}) bonds are identical due to resonance. Preparation of Sulphur Dioxide When sulphur burns in the air, sulphur dioxide is formed along with traces of sulphur trioxide (\left( {6 – 8\% } \right)). The equation is shown below: ({\rm{S}}({\rm{s}}) + {{\rm{O}}_2}({\rm{g}}) \to {\rm{S}}{{\rm{O}}_2}({\rm{g}})) Laboratory Preparation of Sulphur Dioxide Sulphur dioxide is prepared in the laboratory by the action of dilute sulphuric acid on sulphites. Sodium sulfate and water are also the byproducts of this reaction. We can write the chemical reaction as: ({\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_3}({\rm{s}}) + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}({\rm{aq}}) \to {\rm{S}}{{\rm{O}}_2}({\rm{g}}) + {\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}})) Sulphur dioxide is also prepared by gently heating copper turnings in a flask with concentrated sulfuric acid. We can write the chemical reaction as, ({\rm{Cu}}({\rm{s}}) + 2{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}({\rm{aq}}) \to {\rm{S}}{{\rm{O}}_2}({\rm{g}}) + {\rm{CuS}}{{\rm{O}}_4}({\rm{aq}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}})) Industrial Preparation of Sulphur Dioxide Commercially, sulphur dioxide is produced as a byproduct of the roasting of sulphide ores such as Iron Pyrites or Zinc blende. (4{\rm{Fe}}{{\rm{S}}_2}({\rm{s}}) + 11{{\rm{O}}_2}({\rm{aq}}) \to 2{\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}({\rm{s}}) + 8{\rm{S}}{{\rm{O}}_2}({\rm{g}})) Physical Properties of Sulphur Dioxide Sulphur dioxide is a colourless, toxic gas with a suffocating and pungent odour. It is heavier than air and readily soluble in water. One volume of water can dissolve about eight volumes of the gas at (0\;^\circ {\rm{C}}). Under the pressure of (2{\rm{ atm}}) and at room temperature, it can be easily liquified. Its boiling point is (263\, {\rm{ K}}), and itszing point is (197.8\, {\rm{ K}}). It is a non-flammable gas that acts as a solvent for phosphorus, sulphur, iodine, etc. Chemical Properties of Sulphur Dioxide The chemical properties of Sulphur Dioxide are explained below 1. Acidic Character of Sulphur Dioxide When Sulphur dioxide dissolves in water, sulphurous acid is formed. Its aqueous solution is acidic that turns blue litmus red. Hence, ({\rm{S}}{{\rm{O}}_2})is considered an anhydride of sulphurous acid. ({\rm{S}}{{\rm{O}}_2}({\rm{g}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_3}({\rm{aq}})) Because of its acidic character: (a) Sulphur dioxide reacts readily with sodium hydroxide solution to form sodium sulphite solution, which reacts with more sulphur dioxide to form sodium hydrogen sulphite. (2{\rm{NaOH}}({\rm{aq}}) + {\rm{S}}{{\rm{O}}_2}({\rm{g}}) \to {\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_3}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}})) ({\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_3}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) + {\rm{S}}{{\rm{O}}_2}({\rm{g}}) \to 2{\rm{NaHS}}{{\rm{O}}_3}({\rm{aq}})) (b) When Sulphur dioxide gas is bubbled through lime water, the solution turns milky. On passing excess sulphur dioxide gas, the milkiness disappears due to the formation of calcium bisulphite. ({\rm{Ca}}{({\rm{OH}})_2}({\rm{aq}}) + {\rm{S}}{{\rm{O}}_2}({\rm{g}}) \to {\rm{CaS}}{{\rm{O}}_3}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}})) ({\rm{CaS}}{{\rm{O}}_3}({\rm{aq}}) + {{\rm{H}}_2}O({\rm{l}}) + {\rm{S}}{{\rm{O}}_2}({\rm{g}}) \to {\rm{Ca}}{\left( {{\rm{HS}}{{\rm{O}}_3}} \right)_2}({\rm{aq}})) 2. Combination with Oxygen Sulphur dioxide reacts with oxygen on heating to form sulphur trioxide. The above reaction is slow and reversible. Therefore, the reaction is carried out in the presence of some catalysts such as platinised asbestos, vanadium pentoxide, etc. 3. Reaction with Halogens Sulphur dioxide combines directly with the halogens (fluorine, chlorine, and bromine) to form sulphuryl fluoride, chloride, and bromide. For example: ({\rm{S}}{{\rm{O}}_2})combines with chlorine in the presence of charcoal to give sulphuryl chloride. The charcoal acts as a catalyst for the reaction. ({\rm{S}}{{\rm{O}}_2}({\rm{g}}) + {\rm{C}}{{\rm{l}}_2}({\rm{g}}) \to {\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}({\rm{l}})) 4. As an Oxidising Agent ({\rm{S}}{{\rm{O}}_2})oxidises powerful reducing agents like ({{\rm{H}}_2}\;{\rm{S}},\,{\rm{HI}},\,{\rm{Mg}},\,{\rm{Fe}}), etc., and itself gets reduced to sulphur or sulphide. For example: ({\rm{S}}{{\rm{O}}_2}({\rm{g}}) + 2{{\rm{H}}_2}\;{\rm{S}}({\rm{g}}) \to 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 3\;{\rm{S}}) 5. ({\rm{S}}{{\rm{O}}_2})as a Reducing Agent Aqueous ({\rm{S}}{{\rm{O}}_2}) shows reducing character and itself gets oxidised to ({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}). It reduces acidified orange potassium dichromate solution to light green chromium sulphate. ({{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + 3{\rm{S}}{{\rm{O}}_2} \to {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + {\rm{C}}{{\rm{r}}_2}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_3} + {{\rm{H}}_2}{\rm{O}}) 2. It reduces acidified potassium permanganate solution and pink colour of ({\rm{KMn}}{{\rm{O}}_4}) is discharged. (2{\rm{KMn}}{{\rm{O}}_4} + 5{\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + 2{\rm{MnS}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}) 3. It reduces halogens (\left( {{\rm{C}}{{\rm{l}}_2},\,{\rm{B}}{{\rm{r}}_2},\,{{\rm{l}}_2}} \right)) in aqueous solution. To respective halides. Eg. ({\rm{C}}{{\rm{l}}_2} + {\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to 2{\rm{HCl}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}) 4. It reduces ferric salt to ferrous salt, and colour changes from yellow to light green. (2{\rm{FeC}}{{\rm{l}}_3} + {\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to 2{\rm{FeC}}{{\rm{l}}_2} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + 2{\rm{HCl}}) 5. It reduces potassium iodate (\left( {{\rm{KI}}{{\rm{O}}_3}} \right)) solution to iodine. (2{\rm{KI}}{{\rm{O}}_3} + 5{\rm{S}}{{\rm{O}}_2} + 4{{\rm{H}}_2}{\rm{O}} \to {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + {{\rm{I}}_2} + 4{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}) 6. As a Bleaching Agent ({\rm{S}}{{\rm{O}}_2})acts as a bleaching agent in the presence of moisture. It can bleach coloured wool, silk, flower, hair, etc. The formation of nascent hydrogen accounts for the bleaching action of sulphur dioxide, which reduces colouring substance to colourless reduced product. ({\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + 2[{\rm{H}}]) ({\rm{Colouring}}\,{\rm{matter}} + [{\rm{H}}] \to {\rm{Colourless}}\,{\rm{matter}}) The bleaching action of ({\rm{S}}{{\rm{O}}_2})is temporary. The bleached colourless compound will gradually regain its original colour standing in the air due to oxidation by air. Difference Between Sulphur Dioxide and Chlorine Sulphur dioxide ({\rm{S}}{{\rm{O}}_2})Chlorine ({\rm{C}}{{\rm{l}}_2}) 1. Sulphur dioxide reacts with water to yield nascent hydrogen, which bleaches coloured substances to the colourless product by reduction. ({\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + 2[{\rm{H}}])1. Chlorine reacts with water to yield nascent oxygen, which bleaches coloured substances to the colourless product by oxidation. ({\rm{C}}{{\rm{l}}_2} + {{\rm{H}}_2}{\rm{O}} \to 2{\rm{HCl}} + [{\rm{O}}]) 2. Bleaching by ({\rm{S}}{{\rm{O}}_2}) may sometimes be due to the formation of colourless addition products 2. No such colourless addition product is formed with chlorine. 3. Bleaching by ({\rm{S}}{{\rm{O}}_2}) is reversible, i.e., a temporary process.3. Bleaching by ({\rm{C}}{{\rm{l}}_2}) is irreversible, i.e., a permanent process. 4. ({\rm{S}}{{\rm{O}}_2}) is a mild bleaching agent.4. ({\rm{C}}{{\rm{l}}_2}) is a strong bleaching agent. 5. It is used to bleach delicate articles such as wool, silk, cane sugar, etc.5. It is used to bleach wood, textiles, paper, etc. Test of Sulphur Dioxide The reduction of acidified potassium permanganate solution is a convenient test to detect the presence of this gas. The pink colour of ({\rm{KMn}}{{\rm{O}}_4}) is discharged, which confirms the presence of sulphur dioxide gas. (2{\rm{KMn}}{{\rm{O}}_4} + 5{\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + 2{\rm{MnS}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}) Uses for Sulfur Dioxide Sulfur dioxide is the primary ingredient for the most commonly used chemicals across the globe, i.e., sulfuric acid. It is the precursor to the synthesis of many sulphites such as sodium hydrogen sulphite, calcium hydrogen sulphite, etc. These sulphites are used in the preservation of jams, pickles, and jellies. Sulfur dioxide is a widely-used pest control product and can be used to disinfect materials, such as wood or straw. It acts as a fumigating agent to eliminate insects and other pests. It is used for refining petroleum and sugar and for bleaching delicate articles such as wool. Sulphur dioxide is also used as antichlor, i.e., for removing excess chlorine from bleached articles. Liquid ({\rm{S}}{{\rm{O}}_2})is used as a solvent to dissolve a number of organic and inorganic chemicals. It is also used as a refrigerant. Summary Sulphur dioxide is an important greenhouse gas and a major chemical ingredient used to synthesise sulphuric acid. Similar to chlorine, sulphur dioxide also acts as a bleaching agent. In this article, we learnt the formula, structure, preparation, and properties of sulphur dioxide. We also learned some of its uses and how sulphur dioxide’s bleaching action is different from that of chlorine. FAQs on Sulfur Dioxide _Q.1. Is sulfur dioxide a compound? Ans:_ Yes, sulphur dioxide is an inorganic compound. It has a bent structure with polar covalent bonds between the sulphur atom and two oxygen atoms. _Q.2. What is the confirmatory test to detect the presence of Sulphur dioxide gas? Ans:_ The reduction of acidified potassium permanganate solution is a convenient test to detect the presence of this gas. The pink colour of ({\rm{KMn}}{{\rm{O}}_4}) is discharged, which confirms the presence of sulphur dioxide gas. (2{\rm{KMn}}{{\rm{O}}_4} + 5{\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + 2{\rm{MnS}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}) _Q.3. How does Sulphur dioxide form acid rain? Ans:_ Acid rain is caused by a chemical reaction that takes place when compounds like sulfur dioxide and nitrogen oxides are released into the air. These compounds react with water, oxygen, and other chemicals to form sulfuric and nitric acids.These acids fall to the ground as acid rain. _Q.4. Is sulfur dioxide harmful to humans? Ans:_ Sulphur dioxide is a major greenhouse gas. It causes wheezing, shortness of breath and chest tightness, and other respiratory-related problems. _Q.5. Does sulfur dioxide contribute to global warming? Ans:_ Sulfur dioxide, along with carbon emissions, contributes to the formation of aerosols. These compounds directly or indirectly affect the warming and cooling of the earth’s atmosphere. Beyond sulfur dioxide’s environmental impact, this gas also is detrimental to human health. Study Oxides of Sulphur Here We hope this article on Sulphur Oxide is helpful to you. If you have any questions related to this page, reach us through the comment box below and we will get back to you as soon as possible. Related Articles Ellipse: Definition, Properties, Applications, Equation, Formulas ----------------------------------------------------------------- AcademicEllipse: Do you know the orbit of planets, moon, comets, and other heavenly bodies are elliptical? Mathematics defines an ellipse as a plane curve surrounding... 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3237	https://groups.google.com/g/xataface/c/-yKbkAePvOk	Wildcard Searches Groups Groups Conversations All groups and messages Send feedback to Google Help Training Sign in Groups Groups Xataface Conversations About Privacy•Terms     Wildcard Searches 20,558 views Skip to first unread message  Alan Dobkin unread, Apr 29, 2015, 6:42:01 PM 4/29/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to xata...@googlegroups.com Is it possible to do a wildcard or partial-word search using the built-in search action? Partial words are currently not found using the default search box. I also tried the URL conventions syntax, as well as , %, and ?, but none of those seem to work. Thanks, Alan Steve Hannah unread, Apr 30, 2015, 1:16:19 AM 4/30/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to xata...@googlegroups.com Not with the "-search" parameter. However you can perform partial searches on field searches. From the search instructions in the old advanced find form (I need to add these to the new one... and I need to add this to the URL conventions page on the wiki. To perform wildcard searches, prefix your query with a tilde ("~"), then include one of the supported wildcard characters in your query. Supported Wildcards: %: Matches any string of length 0 or more characters. _: Matches a single character. Examples: "~J%y" will match only if the field value begins with "J" and ends with "y". E.g. It will match both "January" and "Joy", but not "Joyous" or "AJoy" "~B%" will match any rows where the field value begins with "B" "~B_b" will match any rows beginning and ending with "b" and is 3 characters long. This would match "bib" and "bob" but not "boob"   -- You received this message because you are subscribed to the Google Groups "Xataface" group. Visit this group at To view this discussion on the web visit -- Steve Hannah Web Lite Solutions Corp. Halsted Bernard unread, Jul 24, 2015, 6:07:19 PM 7/24/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to Xataface, st...@weblite.ca Hi, Steve: Is it possible to use a wildcard search to return all records that have no value in a field? Kind regards, Halsted  Steve Hannah unread, Jul 24, 2015, 6:11:51 PM 7/24/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to xata...@googlegroups.com Yes. Just search for "=". This translates to an exact match for nothing. The opposite search would be for "<", i.e. finds everything greater than nothing. Steve   To view this discussion on the web visit  Halsted Bernard unread, Jul 24, 2015, 6:19:11 PM 7/24/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to Xataface, st...@weblite.ca That worked. Thanks, Steve.  Steve Hannah unread, Jul 24, 2015, 6:20:53 PM 7/24/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to Halsted Bernard, Xataface Sorry... second example should have been ">" (not "<", because ">" is greater than and "<" is less than) Steve  Halsted Bernard unread, Jul 24, 2015, 6:25:02 PM 7/24/15    Reply to author Sign in to reply to author Forward Sign in to forward Delete You do not have permission to delete messages in this group Copy link Report message Show original message Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message to Xataface, st...@weblite.ca Thanks for the clarification.  Reply all Reply to author Forward  0 new messages  Search Clear search Close search Google apps Main menu
3238	https://openstax.org/books/precalculus/pages/9-key-concepts	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Precalculus Key Concepts PrecalculusKey Concepts Search for key terms or text. Key Concepts ## 9.1 Systems of Linear Equations: Two Variables A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 1. Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution. One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 2. Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 3. A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 4. It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example 5, Example 6, and Example 7. Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 8. The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 9. Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 10 and Example 11. ## 9.2 Systems of Linear Equations: Three Variables A solution set is an ordered triple that represents the intersection of three planes in space. See Example 1. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. See Example 2. Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example 3. A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example 4. Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See Example 5. Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line. ## 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the parabola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example 1. There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the circle; (3) two solutions, the line intersects the circle in two points. See Example 2. There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle: (1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See Example 3. An inequality is graphed in much the same way as an equation, except for > or <, we draw a dashed line and shade the region containing the solution set. See Example 4. Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See Example 5. ## 9.4 Partial Fractions Decompose by writing the partial fractions as Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See Example 1. The decomposition of with repeated linear factors must account for the factors of the denominator in increasing powers. See Example 2. The decomposition of with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in See Example 3. In the decomposition of where has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers asSee Example 4. ## 9.5 Matrices and Matrix Operations A matrix is a rectangular array of numbers. Entries are arranged in rows and columns. The dimensions of a matrix refer to the number of rows and the number of columns. A matrix has three rows and two columns. See Example 1. We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See Example 2, Example 3, Example 4, and Example 5. Scalar multiplication involves multiplying each entry in a matrix by a constant. See Example 6. Scalar multiplication is often required before addition or subtraction can occur. See Example 7. Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second. The product of two matrices, and is obtained by multiplying each entry in row 1 of by each entry in column 1 of then multiply each entry of row 1 of by each entry in columns 2 of and so on. See Example 8 and Example 9. Many real-world problems can often be solved using matrices. See Example 10. We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See Example 11. ## 9.6 Solving Systems with Gaussian Elimination An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example 1. A matrix augmented with the constant column can be represented as the original system of equations. See Example 2. Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows. We can use Gaussian elimination to solve a system of equations. See Example 3, Example 4, and Example 5. Row operations are performed on matrices to obtain row-echelon form. See Example 6. To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions. See Example 7 and Example 8. A calculator can be used to solve systems of equations using matrices. See Example 9. Many real-world problems can be solved using augmented matrices. See Example 10 and Example 11. ## 9.7 Solving Systems with Inverses An identity matrix has the property See Example 1. An invertible matrix has the property See Example 2. Use matrix multiplication and the identity to find the inverse of a matrix. See Example 3. The multiplicative inverse can be found using a formula. See Example 4. Another method of finding the inverse is by augmenting with the identity. See Example 5. We can augment a matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See Example 6. Write the system of equations as and multiply both sides by the inverse of See Example 7 and Example 8. We can also use a calculator to solve a system of equations with matrix inverses. See Example 9. ## 9.8 Solving Systems with Cramer's Rule The determinant for is See Example 1. Cramer’s Rule replaces a variable column with the constant column. Solutions are See Example 2. To find the determinant of a 3×3 matrix, augment with the first two columns. Add the three diagonal entries (upper left to lower right) and subtract the three diagonal entries (lower left to upper right). See Example 3. To solve a system of three equations in three variables using Cramer’s Rule, replace a variable column with the constant column for each desired solution: See Example 4. Cramer’s Rule is also useful for finding the solution of a system of equations with no solution or infinite solutions. See Example 5 and Example 6. Certain properties of determinants are useful for solving problems. For example: If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. When two rows are interchanged, the determinant changes sign. If either two rows or two columns are identical, the determinant equals zero. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. The determinant of an inverse matrix is the reciprocal of the determinant of the matrix If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. See Example 7 and Example 8. 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: Jay Abramson Publisher/website: OpenStax Book title: Precalculus Publication date: Oct 23, 2014 Location: Houston, Texas Book URL: Section URL: © Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
3239	https://sk.sagepub.com/ency/edvol/download/sage-encyclopedia-of-business-ethics-and-society-2e/chpt/deductive-reasoning.pdf	No internet connection. All search filters on the page have been cleared. Your search has been saved. Entry Reader's guide Entries A-Z Subject index Deductive Reasoning [Page 868]Deductive reasoning lies at the heart of human existence. Science, math, psychology, and philosophy are just a few of the fields that study the topic formally, but every person on earth is a student of the subject. Most people use some form of it almost every day of their lives. The ancient Greeks started writing about it more than two millennia ago, and the rest of the world has been studying it ever since. Deductive reasoning is a process that moves from general statements to specific conclusions. It starts with premises that are believed to be true. If such premises are indeed true and the steps taken to reach a conclusion follow the rules of logic, the conclusion is said to be valid. If the argument is valid, the information in the conclusion statement essentially exists in the premises that begin the argument. Deductive reasoning is nonampliative: It does not add to what is already known. A valid deductive argument is also necessarily truth preserving: If it starts with true statements, it ends with a true conclusion. Such arguments are also considered erosion proof: Adding new true premises and additional logical steps to an argument do not impair validity. Finally, deductive logic is all or nothing: An argument is either valid or invalid, there are no possibilities in between. The following is an example of a deductive argument: All border collies are dogs. Lothar is a border collie. Lothar is a dog. The first two statements are taken as true. They are the premises. The third statement is the conclusion, but it adds no new information to the world. All the information it provides is contained in the first two premises. It is nonampliative and necessarily truth preserving. It will remain so even if we add an additional true premise such as “All dogs are mammals.” The argument is therefore erosion proof. Finally, there is no question about the extent of the validity of the conclusion. Its validity is unambiguous and maintains the all-or-nothing characteristic. Brief History Over the course of his life (384–322 BCE), Aristotle wrote six works that would later be arranged into the Organon (the term organon means “instrument”). In the third work, Prior Analytics, he defines a deduction (or syllogism) as a “discourse in which, certain things being stated, something other than what is stated follows of necessity from their being so.” Aristotle’s fourth work in the Organon, Posterior Analytics, defines induction as “exhibiting the universal as implicit in the clearly known particular.” Inductive reasoning follows a path in the opposite direction of deduction, from a collection of specific data to a general rule. Deductive reasoning is not common in empirical scientific research, which utilizes inductive reasoning. Aristotle was the first to separate and name the two types of logical reasoning. Pure mathematics, theoretical science, and logic use deductive processes, while all empirically based sciences use inductive processes. [Page 869]Around the year 300 BCE, in Alexandria, Egypt, the Greek mathematician Euclid completed his 13-volume work Elements of Geometry. Much of the content had been written about previously by several different authors, but he compiled all of it into a single work. In Elements, Euclid distinguishes between three types of statements: (1) definitions, statements defining concepts that are inherently accepted to be true, but the concepts may or may not exist; (2) postulates, subject-specific statements that are strongly believed to be true but have not been proven; and (3) common notions, broad general statements that are strongly believed to be true but have not been proven. The use of the premises in Elements coupled with the logical steps Aristotle put forth in Prior Analytics became the building blocks for all future analytical geometry. ... Decision-Making Models Deep Ecology Sign in to access this content Get a 30 day FREE TRIAL Read next More like this Sage Recommends We found other relevant content for you on other Sage platforms. Have you created a personal profile? 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3240	https://mathoverflow.net/questions/406508/maximizing-the-area-of-a-region-involving-triangles	Skip to main content Maximizing the area of a region involving triangles Ask Question Asked Modified 3 years, 10 months ago Viewed 434 times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. I thought of a question while making up an exercise sheet for high school students, and posted it on MathStackExchange but did not receive an answer (the original post is here), so I thought perhaps MO would be a more suitable place to attract answers. Let T be an equilateral triangle of unit area, with vertices A1,A2,A3. Place triangles T1,T2,T3 each of unit area such that the centroid Gi of Ti is equal to Ai for i=1,2,3. What is the maximum possible value of the area of the region T1∩T2∩T3, and what configuration and shape of T1,T2,T3 achieves this maximum? mg.metric-geometry euclidean-geometry plane-geometry triangles Share CC BY-SA 4.0 Improve this question Follow this question to receive notifications edited Oct 22, 2021 at 5:29 Martin Sleziak 4,77344 gold badges3838 silver badges4242 bronze badges asked Oct 18, 2021 at 18:05 Stanley Yao Xiao♦Stanley Yao Xiao 29.6k77 gold badges5656 silver badges163163 bronze badges 1 1 Is it clear that the supremum is actually achieved? – Daniel Asimov Commented Oct 19, 2021 at 1:06 Add a comment \| 3 Answers 3 Reset to default This answer is useful 7 Save this answer. Show activity on this post. I find a larger area than @MattF in a non-symmetric solution. A numerical approximation to that solution is: ``` , , ] ``` This can then be turned in an exact solution by first viewing those decimals numbers as rational numbers an then rescaling each triangle to have exactly volume 1 and translating such that they exactly have the desired center. Doing all these calculations in A can be done, but the solutions have large minimial polynomials, too large to fit in the margin at least... Then one can calculate the area of the intersection of these three triangles and gets 0.248595187378.... As given above they approximate the required conditions to at least 10 decimal digits, so even without relying on exact calculations in A, I'm pretty confident that this larger area is not just a numerical artifact. Here's an image of the arrangement: I found this solution trying a out random triangles. Of course this is not the optimal solution, I just wanted to point out that it might be a good idea to look into solution not assuming D3 symmetry, but perhaps rather into solution that only have the Z/2 symmetry of the original triangle, because the solution I found looks like it is close to a local maximum where the green and the orange triangle are mirror images of each other and the purple one is different. Update: Instead of doing random exploration, I used some gradient descent numerical optimization to look for more local optima. As suspected the solution with Z/2-symmetry comes up. So even without imposing symmetry the optimization leads to a symmetric solution of the form ((a,b),(−a,b),(0,c)),((d,e),(f,g),(h,i)),((−d,e),(−f,g),(−h,i)) Choosing approximately (a,b,c,d,e,f,g,h,i)=(−0.23003,−0.571673,3.775494,2.701848,−0.885711,−0.229117,−0.55433,−0.19322,0.123974) gives an area of approximately 0.2517.... This it what it looks like: I also didn't prescribe that the green and yellow triangles should have sides overlapping the sides of the purple triangle, but the optimization led to that too. Looking at the numerical evindence, I would expect that the global optimum is close to the above solution. It might be possible to get exact coordinates for the local optimum. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 23, 2021 at 21:51 user44143 answered Oct 23, 2021 at 11:53 Moritz FirschingMoritz Firsching 10.9k33 gold badges6464 silver badges8989 bronze badges 2 1 Can you find a maximum assuming that the green and orange triangles are symmetric, and that the purple triangle has one side overlapping with each of the green, orange and gray? – user44143 Commented Oct 23, 2021 at 13:01 2 Yeah, I did some more numerical optimization, see the updated answer. Having the purple triangle to overlap the grey triangle leads to sub-optimal solutions, so I dropped that constraint. (Also I dropped all constraints and with the right starting point found solutions with approximately this symmetry optimizing over the entire configuration space sitting in dimension 18. ) – Moritz Firsching Commented Oct 23, 2021 at 20:45 Add a comment \| This answer is useful 5 Save this answer. Show activity on this post. I find an area of .245463 for the intersection when each Ti has a vertex at distance 3.15835 from the center of T. This seems to be maximal among the symmetric options. The gray triangle in the diagram is T, the black triangles are the Ti (with their specified far vertices outside the area of the diagram), and the blue nonagon is the desired intersection. If the distance is (1+2b)R, where R is the circumradius of T, we can write the area exactly as 4(2b−b2−34b2−1−8b2−4b−212b2−1) which is easy to maximize numerically. For a quicker approximate answer, we can assume that the area of intersection is a circle, tangent to the two long sides of T1 at (r,θ) in polar coordinates. Then we can solve for r,R,θ satisfying 3R23–√/4r(secθ−2)/3r2(1+secθ)2cotθ=1, for circumradius of T;=R, for centroid of T1;=1, for area of T1 and the area and distance quoted above are πr2≃.225 and rsecθ≃3.17. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 22, 2021 at 7:39 answered Oct 19, 2021 at 15:09 user44143user44143 3 Certainly not the picture I was expecting! – Stanley Yao Xiao ♦ Commented Oct 21, 2021 at 12:16 Very nice! MattF, while working on this, did you get the sense that "the symmetric options" lead to the largest area? All rotational moves around your configuration are area-downward? Maybe your edit showing the explicit area answers this? Is your configuration at least a local max? – Joseph O'Rourke Commented Oct 22, 2021 at 20:56 1 @JosephO’Rourke, I assumed D3 symmetry, and the result is already mildly asymmetric in that the nonagon has no Z/9Z symmetry. It would be hard to establish anything definitive about solutions with no symmetry, since the overall configuration space of three triangles (18 coordinates) with 3 specified areas and 6 specified centroid coordinates has 9 degrees of freedom. – user44143 Commented Oct 23, 2021 at 0:15 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. too long for a comment: The exact numbers in @Matt's answers are all algebraic and their minimal polynomials are: For the area: root of 81x9−5184x8+60012x7+1080072x6−30787658x5+308408248x4−1618286324x3+4501458360x2−5533131519x+1109800008 near 0.24546347734508587544309378249... Distance from center: root of 617673396283947x36−51122005539352848x32−976537068350570496x28−6031070320116105216x24+83285306175293227008x20−71803381414983892992x16+47645844627271974912x12−12502304826882785280x8+600386125323829248x4−4503599627370496 near 3.1583535536268193745169822798... The corresponding value of b: root of 2304x9−2304x8−1536x7+1920x6−1120x5−512x4+128x3+40x2+5x+1 near 1.2998723034626336283906933307... None of these numbers have a radical expression. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 22, 2021 at 15:24 Peter Taylor 7,55111 gold badge2424 silver badges3333 bronze badges answered Oct 22, 2021 at 13:18 Moritz FirschingMoritz Firsching 10.9k33 gold badges6464 silver badges8989 bronze badges 1 I think this answer can be deleted now…complicated exact expressions for a suboptimal solution are not that interesting! – user44143 Commented Oct 23, 2021 at 21:47 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 mg.metric-geometry euclidean-geometry plane-geometry triangles See similar questions with these tags. Related 2 Maximum possible number of similar three-colored triangles 6 Problem on distances in a polygon Growing a chain of unit-area triangles: Fills the plane? 6 On covering convex 2D regions with rectangles 3 Least area and least perimeter triangles that contain a convex planar region - how different can they be? On 3 centers of triangles Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
3241	https://oercommons.org/courseware/lesson/619/overview	Preview Please log in to save materials. Log in Report Details Standards Resource Library Subject: : Numbers and Operations Material Type: : Lesson Plan Level: : Middle School Grade: : 6 Provider: : Pearson Tags: : - 6th Grade Mathematics - Absolute Values - Negative Numbers Log in to add tags to this item. License: : Creative Commons Attribution Non-Commercial Language: : English Media Formats: : Text/HTML Show More Show Less Education Standards 1 2 3 4 5 ... 6 7 8 9 10 11 12 13 14 MCCRS.Math.Content.6.NS.C.5 Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, debits/credits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation. MCCRS.Math.Content.6.NS.C.6 Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. MCCRS.Math.Content.6.NS.C.6c Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane. MCCRS.Math.Content.6.NS.C.7 Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Understand ordering and absolute value of rational numbers. MCCRS.Math.Content.6.NS.C.7a Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret -3 > -7 as a statement that -3 is located to the right of -7 on a number line oriented from left to right. MCCRS.Math.Content.6.NS.C.7c Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of -30 dollars, write \|-30\| = 30 to describe the size of the debt in dollars. MCCRS.Math.Content.6.NS.C.7d Maryland College and Career Ready Math Standards Grade 6 Learning Domain: The Number System Standard: Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than -30 dollars represents a debt greater than 30 dollars. CCSS.Math.Content.6.NS.C.5 Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, debits/credits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation. CCSS.Math.Content.6.NS.C.6 Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. CCSS.Math.Content.6.NS.C.6c Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane. CCSS.Math.Content.6.NS.C.7 Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Understand ordering and absolute value of rational numbers. CCSS.Math.Content.6.NS.C.7a Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right. CCSS.Math.Content.6.NS.C.7c Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of –30 dollars, write \|–30\| = 30 to describe the size of the debt in dollars. CCSS.Math.Content.6.NS.C.7d Common Core State Standards Math Grade 6 Cluster: Apply and extend previous understandings of numbers to the system of rational numbers Standard: Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than –30 dollars represents a debt greater than 30 dollars. MP4 Mathematical Practice 3 Download Math, Grade 6 Rational Numbers Getting Started Rational Numbers Fractions and Decimals Ratios Expressions Equations and Inequalities Rate Putting Math to Work Distributions and Variability Surface Area and Volume Using Negative and Absolute Numbers Above and Below Sea Level Opposite of a Number Using Negative and Absolute Numbers Possible or Impossible? Inequalities Assess and Revise Gallery Problems Coordinate Plane Drawing Figures on the Coordinate Plane Reflections Peer Review and Revise Gallery Problems Using Negative and Absolute Numbers Overview Students answer questions about low temperatures recorded in Barrow, Alaska, to understand when to use negative numbers and when to use the absolute values of numbers. Key Concepts The absolute value of a number is its distance from 0 on a number line. The absolute value of a number n is written \|n\| and is read as “the absolute value of n.” A number and the opposite of the number always have the same absolute value. As shown in the diagram, \|3\| = 3 and \|−3\| = 3. In general, taking the opposite of n changes the sign of n. For example, the opposite of 3 is –3. In general, taking the absolute value of n gives a number, \|n\|, that is always positive unless n = 0. For example, \|3\| = 3 and \|−3\| = 3. The absolute value of 0 is 0, which is neither positive nor negative: \|0\| = 0. Goals and Learning Objectives Understand when to talk about a number as negative and when to talk about the absolute value of a number. Locate the absolute value of a and the absolute value of b on a number line that shows the location of a and b in different places in relation to 0. Construct and Critique Arguments Mathematical Practices in Action Mathematical Practice 3: Construct viable arguments and critique the reasoning of others. Direct students’ attention to the definition of absolute value. Then watch the video. Have students listen to the dialogue between Jan and Carlos. This video shows students engaged in Mathematical Practice 3: Construct viable arguments and critique the reasoning of others. After students have watched the video, discuss Carlos and Jan’s conversation. Point out that their conversation is an example of constructing viable arguments and critiquing the reasoning of others, which is an important mathematical practice. Have students talk about each discussion question. Emphasize that through making arguments and listening to the arguments of others, they can advance their understanding of mathematics. As students work in math class, they will be expected to explain their thinking and to defend it using objects, drawings, diagrams, or other models. They will learn to listen to and critique the reasoning of others. Tell students that talking about math will help them understand the math better. SWD: Students with disabilities may have difficulty with cooperative learning tasks. Support students who have trouble resolving conflicts or disagreements by coaching the students to recognize the importance of considering alternate perspectives. Also, provide suggestions as to how students can navigate differences of opinion in learning situations. ELL: Oral explanations can be hard to follow when they are in a language that the students don’t fully comprehend. Be sure that your pace is adequate, and, if needed at any point, pose questions to make sure students are following what you are trying to convey. Opening Construct and Critique Arguments The absolute value of a number is the distance a number is from 0 on a number line. The symbol \|a\| is used to indicate the absolute value of a. Watch the video. What did Carlos think absolute value was at the beginning of the video? What did Jan suggest doing to help them think about absolute value? What argument did Carlos use to say that the absolute value of 12 cannot be –12? What model did Jan and Carlos use to help them understand absolute value? What would Carlos have said the absolute value of 8 is at the beginning of the video? What would Carlos have said the absolute value of 8 is at the end of the video? MP4 Mathematical Practice 3 Download Math Mission Lesson Guide Discuss the Math Mission. Students will understand when to use the absolute value of a number. Opening Understand when to use the absolute value of a number. Temperatures in Barrow, Alaska Lesson Guide Have students work independently on each problem and then discuss their answers with their partners. ELL: Some of the words in the questions and prompts can be somewhat difficult for ELLs to follow. If necessary, rephrase using words you know students can understand to allow ELLs to fully participate and to have a fair chance to answer the questions. Interventions Students have difficulty getting started. What information are you given? How can you find the coldest temperature? Will you represent the coldest temperature using a negative number or an absolute value? Explain your thinking. What are you trying to find when you are asked how far below 0°F the coldest temperature is? Will you represent the distance below 0 using a negative number or an absolute value? Explain your thinking. Student has an incorrect solution. [common error] Student represents distance using a negative number, not the absolute value of a number. Explain what this problem is asking. Does your answer to this problem make sense? Explain why or why not. Can a distance be a negative number? Explain your thinking. Student has a solution. Explain your strategy for solving each problem. Why did you use a negative number in your answer to the first problem? Why did you use the absolute value to find the answer to the second problem? Could you have used a number line to help you solve the problems? Mathematical Practices Mathematical Practice 3: Construct viable arguments and critique the reasoning of others. Listen as students explain their answers to their partners. Identify students who begin to use Mathematical Practice 3 so that you can discuss their conversations during Ways of Thinking. Do students’ explanations make sense? Are they defending their answers using a number line? Are they asking questions when their partner’s explanation is unclear? Mathematical Practice 4: Model with mathematics. Watch for students who make a number line to model the temperatures. Choose one or two students to share their strategy during Ways of Thinking. Answers The coldest temperature is –23.6°F. The coldest temperature is 23.6 degrees below 0°F. The difference between the lowest and highest temperatures recorded in the table is 57.2 degrees. Possible explanation: The highest temperature is 33.6°F, so I know it is 33.6 degrees above 0°F. The lowest temperature is –23.6°F, so I know it is 23.6 degrees below 0°F. So, I added together the two amounts: 33.6 + 23.6 = 57.2. Possible answer: When measuring a temperature in relation to 0, you use the negative number. When measuring a change or a difference in the number of degrees, you use the absolute value of a number. Work Time Temperatures in Barrow, Alaska The table shows some low temperatures recorded in Barrow, Alaska, during a 12-month period. Use the table to answer these questions. What is the coldest temperature? How far below 0°F is the coldest temperature? What is the difference between the lowest and highest temperatures, in degrees? Explain how you know. When did you use a negative number and when did you use the absolute value of a number? Ask yourself: Represent the temperatures on a number line. Where is the coldest temperature? Can you represent distance using a negative number or using an absolute value? How many degrees above 0°F is the highest temperature? How many degrees below 0°F is the lowest temperature? How many degrees are between the highest and lowest temperatures? Prepare a Presentation Lesson Guide Check that students understand when to use a negative number and when to use the absolute value. Preparing for Ways of Thinking As students are working, note if there are students who cannot distinguish between a negative number and an absolute value. Note any misconceptions and clarify them during Ways of Thinking. Challenge Problem Possible Answers Problem: The temperature dropped by 12 degrees on Saturday. If the original temperature was 4°F, what was the new temperature? Answer: The new temperature was –8°F. Problem: On the number line, how many units are between the points –3 and –7? Answer: There are 4 units between –3 and –7. Work Time Prepare a Presentation Explain when you use a negative number and when you use the absolute value. Demonstrate how to take the absolute value of a number. Challenge Problem Create a story problem that has a negative answer. Create a story problem that includes a negative value but has a positive answer. Make Connections Lesson Guide Have students share their work on the problems and then give their presentations. Ask questions such as the following when reviewing the temperature problems: How did you solve the problems? What do you think of [Name]’s and [Name]’s methods? How are they similar? How are they different? Whose method makes more sense to you? Why? Did anyone use a number line to help them? Show us what you did. What elements of Mathematical Practice 3 did you use when discussing the problems with your partner? [common error] Student shows an answer of –23.6 degrees for the second problem. Help students understand that this problem is asking for a distance—how far below 0°F—and that a distance is always positive. In the discussion of the last problem, elicit that in the first problem, students use a negative sign to give the temperature. In the second problem, students look at the distance from 0°F, so the answer is 23.6 degrees. A negative sign is not used for distance. In the third problem, students find the distance from –23.6°F to 33.6°F. Since students are finding the distance, the answer is positive. Have students who did the Challenge Problems share their work. Have the class solve some of the problems to see if the answers are the kind of answers the writers wanted. ELL: Be sure that all students (including ELLs) participate in the presentation exercise, and monitor that ELLs do not shy away from this activity, as it is important that they share aloud so that they can hear their own voice and get used to talking in front of large groups. When critiquing students whose proficiency in English is low, focus on what the student is trying to convey and not the grammar mistakes. If you are not sure you understand, ask the student to repeat in different ways. If other students in the class understand (and you don’t), allow them to help you. Performance Task Ways of Thinking: Make Connections Take notes about how to determine the absolute value of a number and when to use the absolute value of a number. Hint: As students present, ask questions such as: How can a number line be used to solve the problem? In what types of situations is it best to use the absolute value of a number? In what types of situations is it best to use a negative number? Absolute Value A Possible Summary In situations involving distance, you can give the distance as the absolute value of some number. The absolute value of a number is its distance from 0 on a number line. Additional Discussion Points The absolute value of a number n is written \|n\| and is read as “the absolute value of n.” A number and the opposite of the number always have the same absolute value. Taking the opposite of n changes the sign of n. For example, the opposite of 3 is –3. Taking the absolute value of n makes n positive, except when n = 0, since 0 is neither positive nor negative. For example, \|3\| = 3 and \|−3\| = 3. In some situations, it is best to talk about a negative number as a negative number, and in other situations, it is best to talk about its absolute value. Formative Assessment Summary of the Math: Absolute Value Write a summary of what you learned about absolute value. Check your summary. Do you include a definition of absolute value? Do you explain how to find the absolute value of a number on a number line? Do you talk about when to use the absolute value? Reflect On Your Work Lesson Guide Have each student write a brief reflection before the end of class. Review the reflections to find out what students know about the absolute value of a number and the opposite of a number. ELL: Before students write their reflection, allow some additional time for ELLs to discuss their ideas with a partner to help them organize their thoughts. Allow ELLs who share the same language of origin to discuss in this language if they wish and to use a dictionary (or dictionaries). Work Time Reflection Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful. When I look at the absolute value of a number and the opposite of a number, I see these connections … `
3242	https://www.quora.com/How-do-you-use-the-discriminant-B-2-4AC-to-decide-whether-the-following-equation-are-ellipse-hyperbola-or-parabolal	Something went wrong. Wait a moment and try again. Conic Sections Quadratic Formula Hyperbolas (geometry) Ellipses (geometry) Basic Algebra Solving Quadratic Equatio... Math Geometry 5 How do you use the discriminant B^2-4AC to decide whether the following equation are ellipse, hyperbola, or parabolal? Enrico Gregorio Associate professor in Algebra · Author has 18.3K answers and 15.9M answer views · 2y The general equation for a conic is Ax2+Bxy+Cy2+Dx+Ey+F=0 An ellipse has no (real) point at infinity, which means that the (homogeneous) equation obtained by removing the terms of lower degree than 2 has no nontrivial real solutions: Ax2+Bxy+Cy2=0 has no real nontrivial solution (that is with either x≠0 or y≠0) if and only if B2−4AC<0. In the case B2−4AC=0 the conic is tangent to the line at infinity, so it’s a parabola. In the case B2−4AC>0 the conic intersects the line at infinity in two distinct (real) points, so it’s a hyperbola. However. a conic with B2−4AC<0 need not be an ellipse: The general equation for a conic is Ax2+Bxy+Cy2+Dx+Ey+F=0 An ellipse has no (real) point at infinity, which means that the (homogeneous) equation obtained by removing the terms of lower degree than 2 has no nontrivial real solutions: Ax2+Bxy+Cy2=0 has no real nontrivial solution (that is with either x≠0 or y≠0) if and only if B2−4AC<0. In the case B2−4AC=0 the conic is tangent to the line at infinity, so it’s a parabola. In the case B2−4AC>0 the conic intersects the line at infinity in two distinct (real) points, so it’s a hyperbola. However. a conic with B2−4AC<0 need not be an ellipse: it may be a conic with no real points, such as x2+y2+1=0 (which is nondegenerate). The above is only valid if you know that the conic is nondegenerate, that is det⎡⎢⎣AB/2D/2B/2CE/2D/2E/2F⎤⎥⎦≠0 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Robert Paxson BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views · 2y Since no equation is given, let Ax2+Bxy+Cy2+Dx+Ey+F=0 Then, if the discriminant, Δ=B2−4AC, is: Δ<0→ ellipse or its degenerate cases. If B=0 and A=C, then this is a circle or its degenerate cases. Δ=0→ parabola or its degenerate cases. Δ>0→ hyperbola or its degenerate cases. If A+C=0, then this is a rectangular hyperbola or its degenerate cases. Subinoy Paul Mathematics Enthusiast · Author has 1.4K answers and 1M answer views · 5y Related How do you identify if an equation is a parabola, a hyperbola, or an ellipse? Let us consider the general quadratic equation ax2+by2+2hxy+2gx+2fy+c=0 The following list summarises all cases. h2<ab⟹Ellipse h2=ab⟹Parabola h2>ab⟹Hyperbola a=b,h=0⟹Circle a+b=0⟹Rectangular Hyperbola Δ=0⟹Two straight lines, real or imaginary Δ=0,h2=ab⟹Two parallel straight lines Where, Δ=abc+2fgh−af2−bg2−ch2=0 Very much daunting eh! Dagim Gizachew 6y Related What is the geometrical proof of hyperbola standard equation which c^2=a^2+b^2? Create a circle by making c a radius. Create a rectangle passing through the two vertices by using 2a and 2b as length and width respectively. then , OF and OB are equal because both are radii of the same circle. so you proved it. OB^2=a^2+b^2 …..from Pythagoras theorem OB=OF OF^2=a^2+b^2 …..substituting OB by OF OF=c…..by definition finally c^2=a^2+b^2….Proved!! Create a circle by making c a radius. Create a rectangle passing through the two vertices by using 2a and 2b as length and width respectively. then , OF and OB are equal because both are radii of the same circle. so you proved it. OB^2=a^2+b^2 …..from Pythagoras theorem OB=OF OF^2=a^2+b^2 …..substituting OB by OF OF=c…..by definition finally c^2=a^2+b^2….Proved!! Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. Related questions How do you find the equation of a hyperbola, given the focus, the directrix equation, and the asymptote angle (analytic geometry, math)? What is the equation for an ellipse? How are ellipses used in real life? Why is there no single equation for an ellipse or hyperbola? How and where to use the formula b+-√(b^2) (4ac) /2a? Where can ellipses and hyperbolas be found in real life? Janet Heberling Lives in San Francisco, CA (2022–present) · Author has 21.5K answers and 9.4M answer views · Updated 5y Related How do you identify if an equation is a parabola, a hyperbola, or an ellipse? If there is only one squared variable, it is a parabola. If there are no squared variables, it is a line. Put the equation into standard form. If there are two squared variables and they have different signs, it is a hyperbola. If the squared variables have the same sign, it is an ellipse. If the squared variables have the same sign and same coefficients, it is a circle. John Sahr PhD, Cornell University, 1990 · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.9K answers and 2.9M answer views · 5y Related How do you identify if an equation is a parabola, a hyperbola, or an ellipse? Suppose that you have a relationship between x and y: ax2+bx+cy2+dy+e=0 If either a or c = 0 (but not both), then you’ve got a parabola. if ac>0 then you’ve got an ellipse if ac<0 then you’ve got a hyperbola. if both a=0 and b=0 then you’ve got a straight line. Sponsored by CDW Corporation How do you balance new objectives and overall efficiency? CDW helps assess, design, orchestrate and manage your hybrid infrastructure while you stay productive. B.L. Srivastava Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 7.6K answers and 8.1M answer views · 5y Related How do you identify if an equation is a parabola, a hyperbola, or an ellipse? General equation of a conic is given as below; f(x,y)=ax^2 + 2hxy + by^2 + 2gx + 2fy +c Let D = (abc + 2fgh -af^2 - bg^2 -ch^2)(called discriminant of f) . If (1) D = 0 then f represent a pair of st. lines (2) If D in not zero, then (i) h^2 < ab ==> f represents an ellipse (ii) h^2 > ab ==> f represents a hyperbola (iii) h^2 = ab ==>f represent a parabola. (iv) a = b and h = 0 ==> f represents a circle. Robert Paxson BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views · 1y Related Given the equation \ (4x^2 + 9y^2 - 24x + 54y - 76 = 0), determine whether the conic is an ellipse, hyperbola, or parabola. If it is an elliptical or hyperbolic, what is its standard form equation? 4x2+9y2−24x+54y−76=0 4(x2−6x)+9(y2+6y)=76 4(x2−6x+9)+9(y2+6y+9)=76+36+81 4(x−3)2+9(y+3)2=193 (x−3)21934+(y+3)21939=1 This is an ellipse that looks like this: 4x2+9y2−24x+54y−76=0 4(x2−6x)+9(y2+6y)=76 4(x2−6x+9)+9(y2+6y+9)=76+36+81 4(x−3)2+9(y+3)2=193 (x−3)21934+(y+3)21939=1 This is an ellipse that looks like this: Sponsored by CDW Corporation Need AI intuition without compromising compute power? Explore AMD Ryzen™ and Windows 11-powered x86 PCs from CDW to accelerate modern business objectives. Related questions What is the pedal equation of the curve hyperbola x 2 a 2 − y 2 b 2 = 1 ? How do you find the standard form of the equation of a hyperbola when given only the vertices and the center? How do I find the $b$ of a hyperbola equation? What is the canonical equation of the hyperbola passing through the point M (24;5) and whose asymptotes are y =5/12x? How do I show that the ellipse x 2 a 2 + y 2 b 2 = 1 and the hyperbola x 2 A 2 − y 2 B 2 = 1 are orthogonal trajectories if A 2 < a 2 and a 2 − b 2 = A 2 + B 2 (so the ellipse and hyperbola have the same foci)? Saba Hemati Knows Persian · Author has 1.8K answers and 1.6M answer views · 6y Related Where does the 4 in this discriminant formula come from (D = b^2 – 4ac)? ax^2 + bx + c = 0 a(x^2 + b/a x) + c = 0 a[(x + b/(2a))^2 - b^2/(4a^2)] + c = 0 a (x + b/(2a))^2 - b^2/(4a) + c = 0 a (x + b/(2a))^2 = b^2/(4a) - c (x + b/(2a))^2 = (1/a)(b^2/(4a) - c) (x + b/(2a))^2 = b^2/(4a^2) - c/a x + b/(2a) = ± √[b^2/(4a^2) - c/a] x + b/(2a) = ± √[b^2/(4a^2) - 4ac/(4a^2)] x + b/(2a) = ± √(b^2 - 4ac) / (2a) x = - b/(2a) ± √(b^2 - 4ac) / (2a) x = (-b ± √(b^2 - 4ac)) / (2a) The 4 is bold above. Dean Rubine Author of: It's not just π; all of trig is wrong! · Upvoted by Terry Moore , M.Sc. Mathematics, University of Southampton (1968) · Author has 10.6K answers and 23.6M answer views · 1y Related A parabola with equation y = ax^2 + bx + c has vertex (h,k). How many of the six quantities a,b, c, h,k and the discriminant =b^2 - 4ac can be negative at the same time? All of them. We want the vertex in the third quadrant, and the negative discriminant means it doesn’t meet the x axis. The negative coefficients jibe with that nicely, how about y=−(x+1)2−1 or y=−x2−2x−2 Discriminant −4, vertex (−1,−1). All of them. We want the vertex in the third quadrant, and the negative discriminant means it doesn’t meet the x axis. The negative coefficients jibe with that nicely, how about y=−(x+1)2−1 or y=−x2−2x−2 Discriminant −4, vertex (−1,−1). Sudhansu Bhushan Mishra A curious teen desperate to get better · Author has 616 answers and 1.8M answer views · 8y Related What is the method to know whether an equation will make a parabola, hyperbola, ellipse, or a straight line? If it's a straight line, then there are going to be no square terms in the simplified equation. For other things there's this general equation: ax2+by2+2gx+2fy+2hxy+C For a circle, you won't have that 2hxy term, and the coefficients of both x and y would have the same value. For a parabola, you will find that h2=ab. You 'll also find that only one square term exists. For a hyperbola, h2>ab For an ellipse, h2<ab EDIT: Two more cases If there is no other term apart from abc+2fgh−af2−bg2−ch2=0 then it’s a pair of straight lines. A rectangular hyperbola is a hyperbola with If it's a straight line, then there are going to be no square terms in the simplified equation. For other things there's this general equation: ax2+by2+2gx+2fy+2hxy+C For a circle, you won't have that 2hxy term, and the coefficients of both x and y would have the same value. For a parabola, you will find that h2=ab. You 'll also find that only one square term exists. For a hyperbola, h2>ab For an ellipse, h2<ab EDIT: Two more cases If there is no other term apart from abc+2fgh−af2−bg2−ch2=0 then it’s a pair of straight lines. A rectangular hyperbola is a hyperbola with eccentricity of √2 Adrian Giles Studied Physics & Mathematics (Graduated 1985) · Author has 3.5K answers and 801K answer views · 1y Related What is the similarity between the equation of a parabola and the equation of an ellipse or hyperbola? When a conic section is cast in polar form with the origin at a focus of an ellipse the equation is: r(θ)=l/(1+e cosθ) where l is the latus rectum. l=a(1-e²) 0≤e<1 ellipse (e=0 circle) e=1 parabola e>1 hyperbola So the similarity is that the same equation yields all these conics with just one parameter change. The parameter e is termed the eccentricity. These plots from Desmos demonstrate this behaviour: The e values are 0.9 (red ellipse), 1 (blue parabola), 1.5 (black hyperbola). The latus rectum is 5. The vertical scale is partly obscured but the plots are done with unit aspect ratio. When a conic section is cast in polar form with the origin at a focus of an ellipse the equation is: r(θ)=l/(1+e cosθ) where l is the latus rectum. l=a(1-e²) For: 0≤e<1 ellipse (e=0 circle) e=1 parabola e>1 hyperbola So the similarity is that the same equation yields all these conics with just one parameter change. The parameter e is termed the eccentricity. These plots from Desmos demonstrate this behaviour: The e values are 0.9 (red ellipse), 1 (blue parabola), 1.5 (black hyperbola). The latus rectum is 5. The vertical scale is partly obscured but the plots are done with unit aspect ratio. Eduard Wikarta Studied at Meridian Junior College · 6y Related Where does the 4 in this discriminant formula come from (D = b^2 – 4ac)? Starting off with the standard representation for a polynomial of degree 2: ax2+bx+c=0 x2+bax+ca=0,a≠0 x2+2⋅b2ax+ca=0 x2+2⋅b2ax+(b2a)2−(b2a)2+ca=0 (x2+2⋅b2ax+(b2a)2)−(b2a)2+ca=0 (x+b2a)2−(b2a)2+ca=0 (x+b2a)2=(b2a)2−ca x+b2a=±\sq Starting off with the standard representation for a polynomial of degree 2: ax2+bx+c=0 x2+bax+ca=0,a≠0 x2+2⋅b2ax+ca=0 x2+2⋅b2ax+(b2a)2−(b2a)2+ca=0 (x2+2⋅b2ax+(b2a)2)−(b2a)2+ca=0 (x+b2a)2−(b2a)2+ca=0 (x+b2a)2=(b2a)2−ca x+b2a=±√(b2a)2−ca x+b2a=±√b24a2−4ac4a2 x+b2a=±√14a2(b2−4ac) x=−b2a±12a√b2−4ac x=−b2a±√b2−4ac2a x=−b±√b2−4ac2a The ‘4’ comes in when we need to bring out the 14a2 within the square root in the 5th last step. Related questions How do you find the equation of a hyperbola, given the focus, the directrix equation, and the asymptote angle (analytic geometry, math)? What is the equation for an ellipse? How are ellipses used in real life? Why is there no single equation for an ellipse or hyperbola? How and where to use the formula b+-√(b^2) (4ac) /2a? Where can ellipses and hyperbolas be found in real life? What is the pedal equation of the curve hyperbola x 2 a 2 − y 2 b 2 = 1 ? How do you find the standard form of the equation of a hyperbola when given only the vertices and the center? How do I find the $b$ of a hyperbola equation? What is the canonical equation of the hyperbola passing through the point M (24;5) and whose asymptotes are y =5/12x? How do I show that the ellipse x 2 a 2 + y 2 b 2 = 1 and the hyperbola x 2 A 2 − y 2 B 2 = 1 are orthogonal trajectories if A 2 < a 2 and a 2 − b 2 = A 2 + B 2 (so the ellipse and hyperbola have the same foci)? How do you calculate a parabolic equation? What are the methods for finding the equations of ellipses and hyperbolas? How can you determine whether it is a parabola, ellipse, or hyperbola? Why is the discriminant value equal to b^2-4ac? What is the relationship between an ellipse and a hyperbola? Is there any connection between an ellipse and a parabola? Do we consider the signs from the equation for (a+b) ^2 and discriminant formula which is b^2 - 4ac? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
3243	https://www.softschools.com/math/trigonometry/the_sine_function_in_right_triangles/	Pre-K Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade Middle School High School Phonics Fun Games Math Math Games Math Worksheets Algebra Language Arts Science Social Studies Literature Languages Themes Quizzes Timelines Login Home > Math > Trigonometry > The Sine Function in Right Triangles The Sine Function in Right Triangles Sine is a trigonometric ratio comparing two sides of a right triangle. Sine is usually shortened to sin but is pronounced sine. This function can be used to determine the length of a side of a triangle when given at least one side of the triangle and one of the acute angles. Quick Review: the three main trig ratios are sine, cosine and tangent. They can be memorized using SOH CAH TAH What does this mean? It means that sine is the ratio of the opposite side divided by the hypotenuse. Sinθ = Let's look at an example at how sine can be used to find the length of the opposite side. To find x write an equation using the sine ratio and then solve for x Sine 30° = use the multiplication property to isolate x (15) sin 30 = (15)You will need to use a calculator to find the value of sin 30° (15) (.5) = x make sure your calculator is in degree mode by verifying that Sin 30 = .5 7.5 = x the side length represented by x has a length of 7.5 Now let's look at how Sine can be used to find the length of the hypotenuse. To find x write an equation using the sine ratio and then solve for x Sine 20° = use the multiplication property to eliminate the fraction (x) sin 20° = (x)You will need to use a calculator to find the value of sin 20°. Round to 4 decimal places Make sure your calculator is in degree mode by verifying that (x) (.3420) = 10 Sin 20 .3420 Divide both sides by .3420 to isolate x x = 29.2398 Round your answer to the nearest tenth x = 29.2 the length of the hypotenuse is 29.2 The trig ratios have many real world and practical applications in fields such as aviation, architecture, surveying. Using the trigonometric ratios, such as sine, allows for the measurement of things that cannot be determined using typical measurement tools. \| \| \| Related Links: Math Trigonometry The Tangent Function in Right Triangles Graphing the Cosine Function \| To link to this The Sine Function in Right Triangles page, copy the following code to your site: More Topics Handwriting Spanish Facts Examples Formulas Difference Between Inventions Literature Flashcards 2020 Calendar Online Calculators Multiplication Educational Videos Quizzes Flashcards Coloring Pages Links Link us Contact Us Privacy policy Terms of Use Search © 2005-2020 Softschools.com
3244	https://web.mit.edu/16.unified/www/FALL/fluids/Lectures/f03.pdf	Fluids – Lecture 3 Notes 1. 2-D Aerodynamic Forces and Moments 2. Center of Pressure 3. Nondimensional Coeﬃcients Reading: Anderson 1.5 – 1.6 Aerodynamics Forces and Moments Surface force distribution The ﬂuid ﬂowing about a body exerts a local force/area (or stress) ⃗ f on each point of the body. Its normal and tangential components are the pressure p and the shear stress τ. ( magnitude greatly exaggerated) V f f τ R M R L D N A α α resultant force, and moment about ref. point alternative components of resultant force p local pressure and shear stress components force/area distribution on airfoil τ r ds r The ﬁgure above greatly exaggerates the magnitude of the τ stress component just to make it visible. In typical aerodynamic situations, the pressure p (or even the relative pressure p −p∞) is typically greater than τ by at least two orders of magnitude, and so ⃗ f is very nearly perpendicular to the surface. But the small τ often signiﬁcantly contributes to drag, so it cannot be neglected entirely. The stress distribution ⃗ f integrated over the surface produces a resultant force ⃗ R, and also a moment M about some chosen moment-reference point. In 2-D cases, the sign convention for M is positive nose up, as shown in the ﬁgure. Force components The resultant force ⃗ R has perpendicular components along any chosen axes. These axes are arbitrary, but two particular choices are most useful in practice. 1 Freestream Axes: The ⃗ R components are the drag D and the lift L, parallel and perpendic-ular to ⃗ V∞. Body Axes: The ⃗ R components are the axial force A and normal force N, parallel and perpendicular to the airfoil chord line. If one set of components is computed, the other set can then be obtained by a simple axis transformation using the angle of attack α. Speciﬁcally, L and D are obtained from N and A as follows. L = N cos α −A sin α D = N sin α + A cos α Force and moment calculation A cylindrical wing section of chord c and span b has force components A and N, and mo-ment M. In 2-D it’s more convenient to work with the unit-span quantities, with the span dimension divided out. A′ ≡A/b N′ ≡N/b M′ ≡M/b V α u u p (s ) p (s ) l l θ θ u u τ (s ) x τl (s ) l y θ u s u ds sl l ds u τ u p u ds b c On the upper surface, the unit-span force components acting on an elemental area of width dsu are dN′ u = (−pu cos θ −τu sin θ) dsu dA′ u = (−pu sin θ + τu cos θ) dsu And on the lower surface they are dN′ ℓ = (pℓcos θ −τℓsin θ) dsℓ dA′ ℓ = (pℓsin θ + τℓcos θ) dsℓ Integration from the leading edge to the trailing edge points produces the total unit-span forces. N′ = Z TE LE dN′ u + Z TE LE dN′ ℓ A′ = Z TE LE dA′ u + Z TE LE dA′ ℓ 2 The moment about the origin (leading edge in this case) is the integral of these forces, weighted by their moment arms x and y, with appropriate signs. M′ LE = Z TE LE −x dN′ u + Z TE LE −x dN′ ℓ+ Z TE LE y dA′ u + Z TE LE y dA′ ℓ From the geometry, we have ds cos θ = dx ds sin θ = −dy = −dy dx dx which allows all the above integrals to be performed in x, using the upper and lower shapes of the airfoil yu(x) and yℓ(x). Anderson 1.5 has the complete expressions. Simpliﬁcations In practice, the shear stress τ has negligible contributions to the lift and moment, giving the following simpliﬁed forms. L′ = cos α Z c 0 (pℓ−pu) dx + sin α Z c 0 pℓ dyℓ dx −pu dyu dx ! dx M′ LE = Z c 0 " pu x + dyu dx yu ! −pℓ x + dyℓ dx yℓ !# dx A somewhat less accurate but still common simpliﬁcation is to neglect the sin α term in L′, and the dy/dx terms in M′. L′ ≃ Z c 0 (pℓ−pu) dx M′ LE ≃ Z c 0 −(pℓ−pu) x dx The shear stress τ cannot be neglected when computing the drag D′ on streamline bodies such as airfoils. This is because for such bodies the integrated contributions of p toward D′ tend to mostly cancel, leaving the small contribution of τ quite signiﬁcant. Center of Pressure Deﬁnition The value of the moment M′ depends on the choice of reference point. Using the simpliﬁed form of the MLE integral, the moment Mref for an arbitrary reference point xref is M′ ref = Z c 0 −(pℓ−pu) (x −xref) dx = M′ LE + L′xref This can be positive, zero, or negative, depending on where xref is chosen, as illustrated in the ﬁgure. At one particular reference location xcp, called the center of pressure, the moment is deﬁned to be zero. M′ cp = M′ LE + L′xcp ≡0 xcp = −M′ LE/L′ 3 M < 0 M = 0 or or xcp l p −p u L L L M > 0 The center of pressure asymptotes to +∞or −∞as the lift tends to zero. This awkward situation can easily occur in practice, so the center of pressure is rarely used in aerodynamics work. For reasons which will become apparent when airfoil theory is studied, it is advantageous to deﬁne the “standard” location for the moment reference point of an airfoil to be at its quarter-chord location, or xref = c/4. The corresponding standard moment is usually written without any subscripts. M′ c/4 ≡M′ = Z c 0 −(pℓ−pu) (x −c/4) dx Aerodynamic Conventions As implied above, the aerodynamicist has the option of picking any reference point for the moment. The lift and the moment then represent the integrated pl−pu distribution. Consider two possible representations: 1. A resultant lift L′ acts at the center of pressure x = xcp. The moment about this point is zero by deﬁnition: M′ cp = 0. The xcp location moves with angle of attack in a complicated manner. 2. A resultant lift L′ acts at the ﬁxed quarter-chord point x = c/4. The moment about this point is in general nonzero: M′ c/4 ̸= 0. The ﬁgure shows how the L′, M′, and xcp change with angle of attack for a typical cambered airfoil. Note that with representation 1, the xcp location moves oﬀthe airfoil and tends to +∞as L′ approaches zero. Fixing the moment reference point, as in representation 2, is a simpler and preferable approach. Choosing the quarter-chord location for this is especially attractive, since M′ then shows little or no dependence on the angle of attack. This surprising fact will come from a more detailed airfoil analysis later in the course. 4 l p −p u xcp xcp xcp c/4 c/4 c/4 α = 5 α = 0 α = −5 L’ (M’=0) L’ M’ 1. 2. Nondimensional Coeﬃcients The forces and moment depend on a large number of geometric and ﬂow parameters. It is often advantageous to work with nondimensionalized forces and moment, for which most of these parameter dependencies are scaled out. For this purpose we deﬁne the following reference parameters: Reference area: S Reference length: ℓ Dynamic pressure: q∞= 1 2ρV 2 ∞ The choices for S and ℓare arbitrary, and depend on the type of body involved. For aircraft, traditional choices are the wing area for S, and the wing chord or wing span for ℓ. The nondimensional force and moment coeﬃcients are then deﬁned as follows: Lift coeﬃcient: CL ≡ L q∞S Drag coeﬃcient: CD ≡ D q∞S Moment coeﬃcient: CM ≡ M q∞Sℓ For 2-D bodies such as airfoils, the appropriate reference area/span is simply the chord c, and the reference length is the chord as well. The local coeﬃcients are then deﬁned as follows. Local Lift coeﬃcient: cℓ≡ L′ q∞c Local Drag coeﬃcient: cd ≡ D′ q∞c Local Moment coeﬃcient: cm ≡ M′ q∞c2 These local coeﬃcients are deﬁned for each spanwise location on a wing, and may vary across the span. In contrast, the CL, CD, CM are single numbers which apply to the whole wing. 5
3245	https://otvet.mail.ru/question/236744876	Решите, пожалуйста, задачу. - man99_1 \| Ответы Mail Призвать силу поиска... Создать пост Войти Главная Моя лента Пространства В ТОПЕ НА ОТВЕТАХ Шутки за 500 Мир и его люди Настоящее хобби Пикантно о любви Бабочки в животе Смотреть все Всё об Ответах man99_1 1г Сборная Домашка +4 Решите, пожалуйста, задачу. В магазине проходит акция «Каждый третий товар — бесплатно». При печати чека покупки выстраиваются по убы- ванию цены, и все товары с номерами, кратными трем, выдают- ся бесплатно. Петя и Маша выбрали товаров в сумме на 30 000 рублей и оплатили их двумя чеками. Благодаря акции каждый из них сэкономил по 1000 руб. Какое максимальное количество денег могли бы они сэкономить, если бы оплачивали покупки сообща одним чеком? знания#деньги#рубль#акция#товар#цены 2 2 1 По дате По рейтингу iu_iu_158 Гуру 1г очевидно, что нельзя сэкономить больше 10 000₽ (30 0001/3) рассмотрим пример, когда Петя и Маша сэкономили бы 10 000₽, если бы оплачивали товары одним чеком: Пусть Петя купил два товара стоимостью 9 000₽ и один товар стоимостью 1000₽ А Маша один товар за 9000₽ и два за 1000 ₽. Все условия задачи выполнены, каждый из них при покупке в два чека сэкономит 1000₽, а при покупке в один чек они сэкономят 10 000₽. пример чека общего,товары идут по убывающей: 9000р 9000р 9000р-третий товар бесплатно 1000р 1000р 1000р-третий товар бесплатно экономия 10000р 0 0 Ответить
3246	https://mathematica.stackexchange.com/questions/238673/behaviour-of-coefficient-for-negative-exponents-in-rational-expressions	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 Behaviour of Coefficient[] for negative exponents in rational expressions Ask Question Asked Modified 3 years, 11 months ago Viewed 188 times 5 $\begingroup$ I've observed a puzzling behaviour of Coefficient[] for negative exponents and I'm wondering if I was always just relying on undefined behaviour or if there is a bug in recent versions of Mathematica. So far, if I had an expression of the form expr = a/x + b/(1+x) + c and ran Coefficient[expr,x,-1] I've always gotten a as the answer, which made a lot of sense to me. I've tried this with a number of different versions of Mathematica (all on Linux if that matters) that I had access to and the behaviour described above is true for 8.0.0, 8.0.1, 8.0.4, 9.0.0, 9.0.1, 10.0.0, 10.0.1, 10.4.0, 10.4.1, 11.0.0 and 11.0.1. With 11.1.0, 11.1.1, 11.3.0 and 12.0.0 I got the answer a + b/(1+x) which I find a bit weird, but maybe I can come up with a rationale behind this. Finally, with 12.1.1 and 12.2.0 I get a/(1+x) + b/(1+x) + c/(1+x) which makes absolutely no sense at all to me. My question is: Is this a bug in the newer versions of Mathematica or was the answer of 8.0.0 to 11.0.1 always just undefined behaviour? And is there a workaround, which would allow me to extract the coefficient a from expressions like the one above (i.e. after partial fractioning a rational function, take only the term that is multiplied by x^k with k a negative integer)? If this is indeed undefined behaviour, shouldn't Mathematica issue a warning or something like that in this case? functions coefficients Share Improve this question edited Jan 25, 2021 at 14:52 arndarnd asked Jan 23, 2021 at 13:13 arndarnd 72533 silver badges1212 bronze badges $\endgroup$ 5 $\begingroup$ I think MMA does not consider a Laurent series to be a polynomial. At least the help does not mention any negative exponents. However, you could determine the smallest negative exponent and multiply the Laurent series with the corresponding power of x to transform it into a polynomial that MMA can digest. $\endgroup$ Daniel Huber – Daniel Huber 2021-01-23 14:17:02 +00:00 Commented Jan 23, 2021 at 14:17 7 $\begingroup$ Looks like a bug. Please report it (among other reasons, so I don't forget to have a look at it). $\endgroup$ Daniel Lichtblau – Daniel Lichtblau 2021-01-23 16:04:27 +00:00 Commented Jan 23, 2021 at 16:04 1 $\begingroup$ Thank you for the feedback. I've now reproduced this behaviour also with Mathematica 12.2.0 (question edited accordingly). I'll write to Wolfram to report this as a potential bug. If there are any updates, I'll try to add the information also here. $\endgroup$ arnd – arnd 2021-01-25 14:57:46 +00:00 Commented Jan 25, 2021 at 14:57 2 $\begingroup$ Please do report the TS tracking number. I will want to follow up in-house. $\endgroup$ Daniel Lichtblau – Daniel Lichtblau 2021-01-26 15:32:40 +00:00 Commented Jan 26, 2021 at 15:32 $\begingroup$ TS tracking no longer needed; I received the bug report. Thanks for raising the issue. $\endgroup$ Daniel Lichtblau – Daniel Lichtblau 2021-01-26 19:07:59 +00:00 Commented Jan 26, 2021 at 19:07 Add a comment \| 1 Answer 1 Reset to default 1 $\begingroup$ I agree with what @DanielHuber said in the comment, since PolynomialQ[a/x + b/(1 + x) + c, x] gives False, and Coefficient is claimed to work for polynomials: Coefficient[expr,form] gives the coefficient of form in the polynomial expr. Coefficient[expr,form,n] gives the coefficient of form^n in expr. It may thus be an undefined behavior for non-polynomials. As a workaround you can define that: coefficient[expr_, form_, n_Integer : 1] := Total@Cases[expr, e_ /; FreeQ[e/(form^n), x] :> e/(form^n)] so that in v12.2 coefficient[expr, x, -1] gives a coefficient[expr, 1 + x, -1] gives b coefficient[expr, x, 0] gives c coefficient[a/x + b x/(1 + x) + c, x, 0] gives c coefficient[Apart[a/x + b x/(1 + x) + c], x, 0] gives b + c since $\frac{x}{1+x}=1-\frac{1}{1+x}$ Share Improve this answer edited Jan 26, 2021 at 16:33 answered Jan 26, 2021 at 16:10 SnzFor16MinSnzFor16Min 2,2501010 silver badges2121 bronze badges $\endgroup$ 0 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functions coefficients See similar questions with these tags. 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3247	https://www.albert.io/blog/is-it-a-geometric-sequence-when-its-an-exponential-function/	Is it a Geometric Sequence when it's an Exponential Function? \| Albert Blog & Resources Skip to content Subjects Classes Pricing Schedule a Demo Resources Help Center Albert Academy Try Albert School Success Free Resources Subjects Classes Pricing Schedule a Demo Resources Help Center Albert Academy Try Albert School Success Free Resources ➜ AP® Precalculus RESOURCES PRACTICE FREE RESPONSE ASSESSMENTS OVERVIEW RESOURCES PRACTICE FREE RESPONSE ASSESSMENTS OVERVIEW Is it a Geometric Sequence when it’s an Exponential Function? The Albert Team Last Updated On: January 22, 2025 What We Review Toggle - [x] Introduction Understanding Sequences and Functions What is a Geometric Sequence? What is an Exponential Function? Comparing Geometric Sequences and Exponential Functions Understanding Arithmetic Sequences and Linear Functions Summary and Key Takeaways Quick Reference Chart Conclusion Need help preparing for your AP® Precalculus exam? Introduction Sequences and functions form the backbone of many mathematical concepts taught in high school. Is it a geometric sequence when it’s an exponential function? It is essential to understand them to grasp more complex topics and to approach section 2.2 Change in Linear and Exponential Functions. Sequences, be they arithmetic or geometric, and functions, like linear and exponential, each have distinct characteristics. However, they can often seem similar. Clarifying these differences is important for students studying mathematics, especially in AP® Precalculus. This guide will explore what makes geometric sequences and exponential functions unique and how they relate to each other. Start practicing AP® Precalculus on Albert now! Understanding Sequences and Functions Sequences: More Than Just a List Ordered lists of numbers follow a specific pattern to create sequences. Arithmetic sequences progress by adding a constant number to each term, while geometric sequences multiply each term by a constant ratio. Functions: Dependable Relationships Functions depend on inputs to produce outputs. Linear functions increase or decrease steadily, while exponential functions grow or decay at increasing rates. Both sequences and functions can describe periodic behaviors but do so with different approaches. What is a Geometric Sequence? Characteristics of a Geometric Sequence A geometric sequence is defined by its first term, g 0 g_0 g 0, and a common ratio, r r r. Each term is the previous term multiplied by r r r. The formula below represents this relationship: g n=g 0⋅r n g_n = g_0 \cdot r^n g n=g 0⋅r n Example: Exploring a Geometric Sequence Consider a geometric sequence with g 0=2 g_0 = 2 g 0=2 and r=3 r = 3 r=3. The task is to find the first five terms: (g 0=2)( g_0 = 2 )(g 0=2) (g 1=2⋅3=6)( g_1 = 2 \cdot 3 = 6 )(g 1=2⋅3=6) (g 2=6⋅3=18)( g_2 = 6 \cdot 3 = 18 )(g 2=6⋅3=1 8) (g 3=18⋅3=54)( g_3 = 18 \cdot 3 = 54 )(g 3=1 8⋅3=5 4) (g 4=54⋅3=162)( g_4 = 54 \cdot 3 = 162 )(g 4=5 4⋅3=1 6 2) Result: The first five terms are 2, 6, 18, 54, and 162. What is an Exponential Function? Characteristics of an Exponential Function An exponential function grows (or decays) by multiplying a base value by a constant base raised to the power of x x x. Expressed as: f(x)=a⋅b x f(x) = a \cdot b^x f(x)=a⋅b x Ready to boost your AP® scores? Explore our plans and pricing here! Example: Evaluating an Exponential Function Evaluate the function f(x)=1⋅2 x f(x) = 1 \cdot 2^x f(x)=1⋅2 x for (x=0,1,2,3,4)( x = 0, 1, 2, 3, 4 )(x=0,1,2,3,4): (f(0)=1⋅2 0=1)( f(0) = 1 \cdot 2^0 = 1 )(f(0)=1⋅2 0=1) (f(1)=1⋅2 1=2)( f(1) = 1 \cdot 2^1 = 2 )(f(1)=1⋅2 1=2) (f(2)=1⋅2 2=4)( f(2) = 1 \cdot 2^2 = 4 )(f(2)=1⋅2 2=4) (f(3)=1⋅2 3=8)( f(3) = 1 \cdot 2^3 = 8 )(f(3)=1⋅2 3=8) (f(4)=1⋅2 4=16)( f(4) = 1 \cdot 2^4 = 16 )(f(4)=1⋅2 4=1 6) Result: The values are 1, 2, 4, 8, and 16. Comparing Geometric Sequences and Exponential Functions Similarities Both geometric sequences and exponential functions start with an initial value and change by a consistent ratio or base. This method of repeated multiplication is a common thread. Differences Discrete vs. Continuous: Geometric sequences have discrete terms (specific points). On the other hand, exponential functions can continuously take any real number input. Different Formulations: Sequences use g n=g 0⋅r n g_n = g_0 \cdot r^n g n=g 0⋅r n, whereas exponential functions use f(x)=a⋅b x f(x) = a \cdot b^x f(x)=a⋅b x. Example: Contrasting Differences Analyze the geometric sequence g n=3⋅2 n g_n = 3 \cdot 2^n g n=3⋅2 n versus the exponential function f(x)=3⋅2 x f(x) = 3 \cdot 2^x f(x)=3⋅2 x: Both begin with 3 and grow by a factor of 2. Geometric: computes g 3=3⋅2 3=24 g_3 = 3 \cdot 2^3 = 24 g 3=3⋅2 3=2 4. Exponential: continuous function f(3)=24 f(3) = 24 f(3)=2 4. Understanding Arithmetic Sequences and Linear Functions J Hokkanen,CC BY-SA 3.0, via Wikimedia Commons Linear Functions: Consistent Increases Linear functions are characterized by their constant rate of change, expressed as f(x)=b+m x f(x) = b + mx f(x)=b+m x, where (m)( m )(m) is the slope. Example: Calculating a Linear Function Given f(x)=2+3 x f(x) = 2 + 3x f(x)=2+3 x, find f(0) f(0) f(0), f(1) f(1) f(1), and f(2) f(2) f(2): (f(0)=2+3⋅0=2)( f(0) = 2 + 3 \cdot 0 = 2 )(f(0)=2+3⋅0=2) (f(1)=2+3⋅1=5)( f(1) = 2 + 3 \cdot 1 = 5 )(f(1)=2+3⋅1=5) (f(2)=2+3⋅2=8)( f(2) = 2 + 3 \cdot 2 = 8 )(f(2)=2+3⋅2=8) Result: Outputs are 2, 5, and 8, each increasing by 3. Connecting Linear Functions to Arithmetic Sequences Arithmetic sequences and linear functions are closely related because both involve a constant rate of change. In an arithmetic sequence, each term increases or decreases by a fixed value, called the common difference. This mirrors the behavior of a linear function, where the slope (m m m) represents the constant rate of change. Example: Relating Arithmetic Sequences to Linear Functions The arithmetic sequence 2,5,8,…2, 5, 8, \dots 2,5,8,… can be expressed as a linear function: f(x)=2+3 x f(x) = 2 + 3x f(x)=2+3 x, where: 2 2 2 is the first term, representing the y y y-intercept (b b b), and 3 3 3 is the common difference, representing the slope (m m m). Using the formula for the n n n-th term of an arithmetic sequence, a n=a 1+(n−1)d a_n = a_1 + (n - 1)d a n=a 1+(n−1)d, we see it aligns perfectly with the linear function format: a n=b+m x a_n = b + mx a n=b+m x. Thus, we can model every arithmetic sequence as a linear function, allowing us to use algebraic tools to analyze and extend patterns in the sequence. Summary and Key Takeaways Understanding the fine lines between geometric sequences and exponential functions helps in distinguishing mathematical relationships. They share their use of multiplication but differ in application—geometric sequences are discrete, while exponential functions are continuous. Nevertheless, recognizing these intricacies aids students in mastering sequences and functions in precalculus. Be sure to also review the differences between linear vs. exponential functions. Quick Reference Chart VocabularyDefinition Arithmetic Sequence A sequence where each term is derived by adding a constant. Geometric Sequence A sequence where each term is derived by multiplying by a constant ratio. Linear Function A function that changes at a constant rate; can be expressed as f(x)=b+m x f(x) = b + mx f(x)=b+m x. Exponential Function A function with output changing by a constant multiplicative factor; expressed as f(x)=a b x f(x) = ab^x f(x)=a b x. Conclusion Students are encouraged to observe sequences and functions in real-life scenarios to enhance their understanding. They should be able to answer the question, “Is it a geometric sequence when it’s exponential functions?” Continued practice with exercises focused on transforming between arithmetic and geometric contexts is beneficial. This will solidify mathematical concepts, boosting confidence in tackling precalculus problems. Sharpen Your Skills for AP® Precalculus Are you preparing for the AP® Precalculus exam? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now! Need help preparing for your AP® Precalculusexam? Albert has hundreds of AP® Precalculus practice questions, free responses, and an AP® Precalculus practice test to try out. Start practicing AP® Precalculus on Albert now! Interested in a school license? 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3248	https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-geometry/cc-8th-angles-between-lines/v/angles-formed-by-parallel-lines-and-transversals	Angles, parallel lines, & transversals (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Skip to lesson content 8th grade math Course: 8th grade math>Unit 5 Lesson 1: Angles between intersecting lines Angles, parallel lines, & transversals Parallel & perpendicular lines Missing angles with a transversal Angle relationships with parallel lines Measures of angles formed by a transversal Equation practice with angle addition Equation practice with angles Math> 8th grade math> Geometry> Angles between intersecting lines © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Angles, parallel lines, & transversals Google Classroom Microsoft Teams About About this video Transcript Parallel lines are lines in the same plane that go in the same direction and never intersect. When a third line, called a transversal, crosses these parallel lines, it creates angles. Some angles are equal, like vertical angles (opposite angles) and corresponding angles (same position at each intersection).Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Shannon 13 years ago Posted 13 years ago. Direct link to Shannon's post “What does the >> and the ...” more What does the >> and the > mean? Answer Button navigates to signup page •12 comments Comment on Shannon's post “What does the >> and the ...” (103 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Goezonme 6 years ago Posted 6 years ago. Direct link to Goezonme's post “They are symbols that tel...” more They are symbols that tell you these lines are parellel.For example,> is not parellel to >>. 22 comments Comment on Goezonme's post “They are symbols that tel...” (65 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Apoorv 4 years ago Posted 4 years ago. Direct link to Apoorv's post “What is the meaning of in...” more What is the meaning of intercept? Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Angelica Chen 4 years ago Posted 4 years ago. Direct link to Angelica Chen's post “The intercept of somethin...” more The intercept of something is a place where something else crosses it. In math, it is the place where a line crosses the corresponding axis. Comment Button navigates to signup page (17 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... H 2 years ago Posted 2 years ago. Direct link to H's post “whats a plane i need an a...” more whats a plane i need an answer what are votes for Answer Button navigates to signup page •Comment Button navigates to signup page (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer ghost184155 2 years ago Posted 2 years ago. Direct link to ghost184155's post “In mathematics, a "plane"...” more In mathematics, a "plane" typically refers to a two-dimensional flat surface that extends infinitely in all directions. It is often represented as a geometric object with no thickness. A plane can be defined by a point and a normal vector, or it can be described by a linear equation in three-dimensional space. The equation of a plane in three-dimensional Cartesian coordinates (x, y, z) can be written in the form Ax + By + Cz = D, where A, B, and C are constants not all equal to zero, and (A, B, C) is a normal vector to the plane. The point (x₀, y₀, z₀) through which the plane passes is often used to uniquely define a specific plane. Planes are fundamental objects in geometry and are used in various branches of mathematics, including linear algebra, calculus, and analytical geometry. They play a crucial role in understanding the geometry of three-dimensional space. Comment Button navigates to signup page (12 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Vedanshi 2 years ago Posted 2 years ago. Direct link to Vedanshi's post “When at 4:17 he says that...” more When at 4:17 he says that there is no proof for this, wouldn't it be easier to just get a protractor and measure it to prove that both those angles are the same degree? Answer Button navigates to signup page •2 comments Comment on Vedanshi's post “When at 4:17 he says that...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer H 2 years ago Posted 2 years ago. Direct link to H's post “ya but hes just teathing ...” more ya but hes just teathing in theory Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... BrenaH 8 months ago Posted 8 months ago. Direct link to BrenaH's post “I'm going to cry WHEN WI...” more I'm going to cry WHEN WILL I EVER USE THIS Answer Button navigates to signup page •4 comments Comment on BrenaH's post “I'm going to cry WHEN WI...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kellan Zhang 4 months ago Posted 4 months ago. Direct link to Kellan Zhang's post “i mean.... if you are gon...” more i mean.... if you are gonna take a field like an engineer or designing/constructing stuff, this could come in handy. from my perspective, i don't think learning about stuff is not for the sake of learning stuff. but rather learning it to help you. i mean, if you find this area not exactly helpful of what you'll use in the future, then.... you might as well change to smth else. but you never know. plus it won't really hurt to learn something new :D 2 comments Comment on Kellan Zhang's post “i mean.... if you are gon...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Zareea 3 years ago Posted 3 years ago. Direct link to Zareea's post “Threw out the whole video...” more Threw out the whole video I didn't understand any of this Answer Button navigates to signup page •2 comments Comment on Zareea's post “Threw out the whole video...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer sunchoi2011 3 months ago Posted 3 months ago. Direct link to sunchoi2011's post “Try coming back to this a...” more Try coming back to this after going back a few videos/exersizes and see if you understand it then. 1 comment Comment on sunchoi2011's post “Try coming back to this a...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Kenneth Morillo 3 years ago Posted 3 years ago. Direct link to Kenneth Morillo's post “what does >- mean?” more what does >- mean? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer David Severin 3 years ago Posted 3 years ago. Direct link to David Severin's post “I assume you mean the gre...” more I assume you mean the greater than or equal to sign, this tells you that the endpoint of the ray counts as part of the solution. On a number line, it is represented by a solid dot (as opposed to a open circle with just >), and on a coordinate plane it is a solid line (as opposed to a dashed line with just a >). Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Renesmee 2 years ago Posted 2 years ago. Direct link to Renesmee's post “is this going to help in ...” more is this going to help in life or is this just math Answer Button navigates to signup page •1 comment Comment on Renesmee's post “is this going to help in ...” (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Zoe Baruch 2 years ago Posted 2 years ago. Direct link to Zoe Baruch's post “It depends on what your j...” more It depends on what your job is, there are definitely some jobs that you wouldn’t use this but there are others that you would. 3 comments Comment on Zoe Baruch's post “It depends on what your j...” (11 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... ParvBhardwaj 9 years ago Posted 9 years ago. Direct link to ParvBhardwaj's post “Hi friends :-) I have a q...” more Hi friends :-) I have a question for you- 1> at 6:27 "Sal" proved us that alternate interior angles are equal , right? so, here is my question ? we know that :- b=c,f=g now b and c are vertical angles and we also know that b = f (corresponding angle) so why can't we say that c = f ,right? and why not alternate exterior angles? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer fr33d0g 9 years ago Posted 9 years ago. Direct link to fr33d0g's post “Ok, so you said b=c,f=g. ...” more Ok, so you said b=c,f=g. Which says b=c and f=g. If you look back at Sals blackboard on the video there is no comma, it simply says b=c=f=g which means they are ALL equal, so yes , c does equal f and c equals g, and b equals g etc. etc.. As for alternate exterior angles, the same rule applies. a=d=e=h. They are all equal. 2 comments Comment on fr33d0g's post “Ok, so you said b=c,f=g. ...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... YoraanZ 3 years ago Posted 3 years ago. Direct link to YoraanZ's post “I think the fact that you...” more I think the fact that you said there is no proof for the two angles are equivalent was kinda funny. But still thank you so much for posting these videos! Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer David Severin 3 years ago Posted 3 years ago. Direct link to David Severin's post “In geometry, we have axio...” more In geometry, we have axioms/postulates, and we have proofs. Anything that is obvious is assumed to be true. Sal states that it is obvious, so it does not need to go through the proof process. Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript Let's say we have two lines over here. Let's call this line right over here line AB. So A and B both sit on this line. And let's say we have this other line over here. We'll call this line CD. So it goes through point C and it goes through point D. And it just keeps on going forever. And let's say that these lines both sit on the same plane. And in this case, the plane is our screen, or this little piece of paper that we're looking at right over here. And they never intersect. So they're on the same plane, but they never intersect each other. If those two things are true, and when they're not the same line, they never intersect and they can be on the same plane, then we say that these lines are parallel. They're moving in the same general direction, in fact, the exact same general direction. If we were looking at it from an algebraic point of view, we would say that they have the same slope, but they have different y-intercepts. They involve different points. If we drew our coordinate axes here, they would intersect that at a different point, but they would have the same exact slope. And what I want to do is think about how angles relate to parallel lines. So right over here, we have these two parallel lines. We can say that line AB is parallel to line CD. Sometimes you'll see it specified on geometric drawings like this. They'll put a little arrow here to show that these two lines are parallel. And if you've already used the single arrow, they might put a double arrow to show that this line is parallel to that line right over there. Now with that out of the way, what I want to do is draw a line that intersects both of these parallel lines. So here's a line that intersects both of them. Let me draw a little bit neater than that. So let me draw that line right over there. Well, actually, I'll do some points over here. Well, I'll just call that line l. And this line that intersects both of these parallel lines, we call that a transversal. This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write this to show that these two are equal and these two are equal right over here. Now the other thing we know is we could do the exact same exercise up here, that these two are going to be equal to each other and these two are going to be equal to each other. They're all vertical angles. What's interesting here is thinking about the relationship between that angle right over there, and this angle right up over here. And if you just look at it, it is actually obvious what that relationship is-- that they are going to be the same exact angle, that if you put a protractor here and measured it, you would get the exact same measure up here. And if I drew parallel lines-- maybe I'll draw it straight left and right, it might be a little bit more obvious. So if I assume that these two lines are parallel, and I have a transversal here, what I'm saying is that this angle is going to be the exact same measure as that angle there. And to visualize that, just imagine tilting this line. And as you take different-- so it looks like it's the case over there. If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle is going to have the same measure as this angle. And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here. So I'm going to use lowercase letters for the angles themselves. So let's call this lowercase a, lowercase b, lowercase c. So lowercase c for the angle, lowercase d, and then let me call this e, f, g, h. So we know from vertical angles that b is equal to c. But we also know that b is equal to f because they are corresponding angles. And that f is equal to g. So vertical angles are equivalent, corresponding angles are equivalent, and so we also know, obviously, that b is equal to g. And so we say that alternate interior angles are equivalent. So you see that they're kind of on the interior of the intersection. They're between the two lines, but they're on all opposite sides of the transversal. Now you don't have to know that fancy word, alternate interior angles, you really just have to deduce what we just saw over here. Know that vertical angles are going to be equal and corresponding angles are going to be equal. And you see it with the other ones, too. We know that a is going to be equal to d, which is going to be equal to h, which is going to be equal to e. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. 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3249	https://en.wikipedia.org/wiki/50_Cent_Is_the_Future	50 Cent Is the Future - 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 Background 2 Track listing 3 Charts 4 References 5 External links 50 Cent Is the Future [x] 7 languages Français 한국어 Italiano Polski Português Русский Українська 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 2002 mixtape by G-Unit \| 50 Cent Is the Future \| \| \| \| Mixtape by G-Unit \| \| Released \| June 1,2002(2002-06-01) \| \| Recorded \| 2001 in Canada \| \| Genre \| Gangsta rap hardcore hip hop \| \| Length \| 49:37 \| \| Label \| Street Dance Thurd World Muzic \| \| 50 Cent chronology \| \| Guess Who's Back? (2002)50 Cent Is the Future (2002)No Mercy, No Fear (2002) \| Professional ratings\| Review scores \| \| Source \| Rating \| \| RapReviews \| 7/10 \| \| Spin \| \| \| XXL \| XL (4/5) \| 50 Cent Is the Future is the second mixtape by American rapper 50 Cent and first one by his rap group G-Unit. It was released on June 1, 2002 via Street Dance/Thurd World Muzic. The lone guest appearance is provided by UTP, which marks the first collaboration between the group and future member Young Buck. In September 2004, the album was charted at number 59 in Switzerland and at number 65 in the United Kingdom. In 2006, the album was ranked at #7 on the 'Top Ten Mixtape List Of All Time' by XXL, and at #22 on 'The 50 Best Rap Mixtapes of the Millennium' by Pitchfork. Background [edit] The mixtape was recorded in Canada in 2001 after 50 Cent was dropped from Columbia Records and blacklisted from the recording industry due to his controversial song "Ghetto Qu'ran (Forgive Me)", leaving his debut studio album Power of the Dollar unreleased. He then traveled to Canada to record the mixtape due to being unable to find a studio in the United States that would allow him to record. The project mostly revisits material by Mobb Deep and features Southern hip hop group UTP represented by Skip, Young Buck and Juvenile. After 50 Cent Is the Future, he recorded his 2002 compilation mixtape Guess Who's Back?, which "G-Unit That's What's Up" is included in. Track listing [edit] \| No. \| Title \| Writer(s) \| Original song and artist(s) \| Length \| --- --- \| 1. \| "U Should Be Here" \| G-Unit \| "Be Here" by Raphael Saadiq and D'Angelo \| 3:33 \| \| 2. \| "Bump Dat Street Mix" \| 50 Cent and Tony Yayo \| "Bump Dat" by Mobb Deep \| 3:06 \| \| 3. \| "The Banks Workout" \| Lloyd Banks and 50 Cent \| "Lyrical Exercise" by Jay-Z \| 4:01 \| \| 4. \| "Whoo Kid Kay Slay Shit!" \| 50 Cent \| "Crawlin'" by Mobb Deep \| 2:33 \| \| 5. \| "50 Cent Just Fucking Around" \| 50 Cent \| "What You Want" by Ma$e and Total "Biggie" by Junior M.A.F.I.A. \| 2:12 \| \| 6. \| "G-Unit Soldiers" \| G-Unit \| "Losin' Weight" by Cam'ron and Prodigy \| 3:06 \| \| 7. \| "Got Me a Bottle" \| 50 Cent and Lloyd Banks \| "Got Me a Model" by R.L. and Erick Sermon \| 2:52 \| \| 8. \| "Tony Yayo Explosion" \| 50 Cent and Tony Yayo \| "Eye for a Eye (Your Beef Is Mines)" by Mobb Deep, Nas and Raekwon \| 2:45 \| \| 9. \| "Clue/50" \| 50 Cent \| \| 1:30 \| \| 10. \| "A Lil Bit of Everything U.T.P." \| G-Unit and UTP \| "Little Bit" by Juvenile, 50 Cent, Lloyd Banks, Skip, Tony Yayo, Wacko, Young Buck \| 4:09 \| \| 11. \| "Cut Master C Shit"" \| 50 Cent \| "No Nuts, No Glory" by the Geto Boys \| 2:56 \| \| 12. \| "Call Me" \| 50 Cent and Tony Yayo \| "Call Me" by Tweet \| 3:03 \| \| 13. \| "50/Banks" \| 50 Cent and Lloyd Banks \| "They Ain't Ready" by Jadakiss, Bubba Sparxxx and Timbaland \| 2:45 \| \| 14. \| "Surrounded by Hoes" \| 50 Cent \| "Round & Round" by Jonell \| 2:10 \| \| 15. \| "G-Unit That's What's Up!" \| G-Unit \| "Y'all Been Warned" by Wu-Tang Clan \| 4:13 \| \| 16. \| "Bad News" \| G-Unit \| "Feeling Good" by Nina Simone \| 4:33 \| \| Total length: \| 49:27 \| Charts [edit] \| Chart (2004) \| Peak position \| --- \| \| Swiss Albums (Schweizer Hitparade) \| 59 \| \| UK Albums (OCC) \| 65 \| References [edit] ^Juon, Steve 'Flash' (May 8, 2007). "50 Cent:: 50 Cent Is the Future:: Third World Music/BCD Music Group". RapReviews. Retrieved December 19, 2020. ^Aaron, Charles (July 2007). "Discography: 50 Cent". Spin. New York. p.84. Retrieved March 4, 2024 – via Google Books. ^DJ Rhude (August 2002). "Official Bootlegs: 50 Cent – 50 Cent is the Future: G-Unit". Critical Breakdown. XXL. No.39. p.139. ^"Top Ten Mixtape List Of All Time - XXL". XXL. April 17, 2006. Retrieved December 19, 2020. ^Drake, David (June 29, 2016). "The 50 Best Rap Mixtapes of the Millennium - Page 3". Pitchfork. Retrieved December 19, 2020. ^Matthews, Adam (May 24, 2000). "SOHH Exclusive: 50 Cent Shot In New York \| Daily Hip-Hop News \| SOHH.com /". SOHH. Archived from the original on May 10, 2007. Retrieved December 19, 2020 – via Wayback Machine. ^"Swisscharts.com – G-Unit feat. 50 Cent – 50 Cent Is The Future". Hung Medien. Retrieved December 19, 2020. ^"Official Albums Chart Top 100". Official Charts Company. Retrieved December 19, 2020. External links [edit] 50 Cent Is The Future at Discogs (list of releases) \| show v t e 50 Cent \| \| Albums discography Singles discography Videography Awards and nominations G-Unit \| \| Studio albums \| Get Rich or Die Tryin' The Massacre Curtis Before I Self Destruct Animal Ambition \| Unreleased \| Power of the Dollar Street King Immortal \| \| \| Compilations \| Best of 50 Cent \| \| Soundtracks \| Get Rich or Die Tryin' OST \| \| Mixtapes \| Guess Who's Back? 50 Cent Is the Future No Mercy, No Fear God's Plan War Angel LP Forever King The Big 10 The Lost Tape 5 (Murder by Numbers) The Kanan Tape \| \| Concert tours \| The Invitation Tour The Final Lap Tour \| \| Video games \| Bulletproof Blood on the Sand \| \| Television \| 50 Cent: The Money and the Power 50 Central \| \| Films \| Films directed Before I Self Destruct Films written Gun All Things Fall Apart \| \| Related articles \| Energy Brands Cheetah Vision SMS Audio Street King The 50th Law G-Unit Records G-Note Records G-Unit Books G-Unit Clothing Company G-Unit Films and Television Inc. G-Unity Foundation Shady/Aftermath The Re-Up Shady XV Jam Master Jay Eminem Dr. Dre Teairra Marí \| \| Category \| \| show v t e G-Unit \| \| 50 Cent Tony Yayo Lloyd Banks Young Buck Kidd Kidd Uncle Murda The Game \| \| Studio albums \| Beg for Mercy T.O.S. (Terminate on Sight) \| \| EPs \| The Beauty of Independence The Beast Is G Unit \| \| Singles \| "Stunt 101" "Poppin' Them Thangs" "Wanna Get to Know You" "Smile" "I Like the Way She Do It" "Rider Pt. 2" "Close to Me" \| \| Promo singles \| "Wanna Lick" \| \| Related articles \| Discography Records Films Books Clothing Company Foundation DJ Whoo Kid 50 Cent: Bulletproof (G-Unit Edition) \| \| Authority control databases \| MusicBrainz release group \| Retrieved from " Categories: G-Unit albums 50 Cent mixtape albums 2002 mixtape albums Hidden categories: Use mdy dates from June 2025 Articles with short description Short description is different from Wikidata Articles with hAudio microformats Album articles lacking alt text for covers Articles with music ratings that need to be turned into prose Album chart usages for Switzerland Album chart usages for UK2 This page was last edited on 4 June 2025, at 04:29(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents 50 Cent Is the Future 7 languagesAdd topic
3250	https://physics.stackexchange.com/questions/829176/work-energy-theorem-problem-involving-springs	homework and exercises - Work-Energy Theorem problem involving springs - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Work-Energy Theorem problem involving springs Ask Question Asked 1 year ago Modified12 months ago Viewed 338 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. The figure below is a top view of blocks A and B of mass m m each connected with ideal massless spring with a spring constant k k. Both blocks lie on frictionless ground and are imparted horizontal velocity v v as shown when spring is unstretched. Find the maximum stretch of the spring. The problem is I can't understand what to consider as the system. When I consider the whole thing as the system I get the right answer. But the issue arises when I attempt to apply work energy theorem on the blocks independently. Here's what I did. Taking only block A as the system. 1/2 k x 2=1/4 m v 2 1/2 k x 2=1/4 m v 2 (taking horizontal velocity only) This gives me the wrong answer. Spring potential energy depends on deformation in the spring. So for each of the blocks it should be 1/2 k x 2 1/2 k x 2, right? And as far as K.E. is concerned, the y component wont change. The change comes in the x component which is why I only considered that. The thing is I've solved other questions with a similar set up (I might add those later to elaborate my doubt) in which I was able to get to the answer even when I used the theorem individually on each block. I would really appreciate if somebody could help me out. My basics are kind of weak. homework-and-exercises newtonian-mechanics energy-conservation work spring Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Sep 28, 2024 at 16:32 entropyentropy asked Sep 26, 2024 at 19:23 entropyentropy 158 9 9 bronze badges 3 1 Voting to reopen, as this is a perfectly clear question with a straight-forward solution, and the questioner's error almost certainly arises from treating a spring stretched from both ends as two back-to-back springs with the same spring constant.gandalf61 –gandalf61 2024-09-27 09:19:45 +00:00 Commented Sep 27, 2024 at 9:19 @gandalf6 Thanks much for the response. This clarifies my doubt really well. I now got my mistake. I remember learning that in case of springs for series and parallel we use the formula similar to that used for capacitors.entropy –entropy 2024-09-28 16:31:35 +00:00 Commented Sep 28, 2024 at 16:31 @StevanV.Saban Now that the question has been reopened I have deleted my comment and replaced it with an answer.gandalf61 –gandalf61 2024-09-29 08:50:35 +00:00 Commented Sep 29, 2024 at 8:50 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Here's what I did. Taking only block A as the system. 1/2kx2=1/4mv2 (taking horizontal velocity only) That's incorrect. For a single block, K E b l o c k=1 2 W s p r i n g K E b l o c k=1 2 W s p r i n g or 1 2 m v 2 x=1 4 k x 2 m a x 1 2 m v x 2=1 4 k x m a x 2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 29, 2024 at 16:30 answered Sep 27, 2024 at 1:40 user334569 user334569 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Given an ideal spring, frictionless surface, and no net external forces acting on the two block/spring system, the total temporary loss of the component of kinetic energy of both blocks (not just one) in the horizontal direction, will equal the increase elastic potential of the spring at maximum extension. The two blocks will then oscillate back and forth compressing and extending the spring while moving at a constant vertical velocity of v y=v sin 45 0 v y=v sin⁡45 0. Hope this helps. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Sep 27, 2024 at 1:31 Bob DBob D 83.3k 7 7 gold badges 61 61 silver badges 160 160 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Where I think you are going wrong when considering each block separately is assuming that a single spring with spring constant k k stretched from both ends is equivalent to two back-to-back springs each with spring constant k k. This is not correct. The two back-to-back springs are each stretched by half the amount of the original spring, so to produce the same force as the original spring they must each have spring constant 2 k 2 k. The energy stored in the two springs each with extension 1 2 x 1 2 x is then 2(1 2(2 k)(1 2 x)2)=1 2 k x 2 2(1 2(2 k)(1 2 x)2)=1 2 k x 2 as required. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Sep 29, 2024 at 8:48 gandalf61gandalf61 64.9k 8 8 gold badges 84 84 silver badges 191 191 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. A key part of the question is the 45∘45∘ angle of velocity v v to the spring, so only v cos 45∘≈0.707 v v cos⁡45∘≈0.707 v, should be considered for the spring extension on each block. (What is option no. 1?) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 27, 2024 at 16:36 answered Sep 26, 2024 at 20:20 narratorjaynarratorjay 131 4 4 bronze badges 2 v is the velocity not the force, no?Bob D –Bob D 2024-09-26 21:11:07 +00:00 Commented Sep 26, 2024 at 21:11 Sorry I forgot to include the options. vroot(m/k) was what I meant.entropy –entropy 2024-09-28 16:33:41 +00:00 Commented Sep 28, 2024 at 16:33 Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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3251	https://www.scribd.com/document/831400929/The-Study-of-Poetry-by-Matthew-Arnold	Matthew Arnold's Poetry Critique \| PDF \| Poetry \| Truth Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 100%(1)100% found this document useful (1 vote) 497 views 5 pages Matthew Arnold's Poetry Critique Matthew Arnold's essay "The Study of Poetry" emphasizes the importance of poetry in providing moral and intellectual clarity amidst the cultural challenges of the Victorian era. He critiques… Full description Uploaded by 22116045 AI-enhanced title and description Go to previous items Go to next items Download Save Save The Study of Poetry by -Matthew Arnold For Later Share 100%100% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save The Study of Poetry by -Matthew Arnold For Later You are on page 1/ 5 Search Fullscreen Matthew Arnold, one of the most influential poets and critics of the 19th century, is renowned for his deep engagement with the cultural and intellectual concerns of his time. His essay "The Study of Poetry" reflects his broader vision of literature’s role in society and his commitment to the idea that poetry should not only entertain but also instruct, elevate, and offer moral and intellectual clarity. Arnold was deeply concerned with the state of English culture during the Victorian era, particularly with the tension between the rising tide of industrialization, scientific progress, and the decline of religious and spiritual authority. He felt that poetry, with its ability to capture universal truths and emotional depth, could provide a counterbalance to this increasingly fragmented and materialistic world. By comparing contemporary poets to the classics — like Homer, Shakespeare, and Dante — he seeks to establish a high standard for poetry that excels mere personal or fashionable trends. His focus on "high seriousness" and the "touchstone method" is not just an aesthetic principle, but a call for poetry that is intellectually rigorous, sincere, and capable of offering profound i nsights into the human condition. "The Study of Poetry" serves as both a critical reflection on the role of poetry in society and a model for how to approach and appreciate literature. It urges us to seek out poetry that is not only technically proficient but also emotionally and intellectually fulfilling — poetry that can endure the test of time. Arnold’s work continues to inspire readers and critics, encouraging us to consider not just what poetry says, but how it says it, and how it reflects the deeper truths of human experience. It explores the purpose, importance, and standards of poetry, asserting that poetry has a unique role in interpreting life and supporting human experience as other sources of meaning — like dogma, science, and traditional religion — become increasingly uncertain. The speaker argues that poetry offers an "immense" future because it engages directly with ideas, not mere facts, which are changeable and unreliable as foundations for meaning. Poetry’s focus on the deeper truth in ideas allows it to become a lasting source of consolation, inspiration, and understanding. To fulfill this high destiny, poetry must maintain high standards of excellence and avoid “charlatanism”— a tendency to confuse what is true and profound with what is shallow and pretentious. Poetry, being an art form and a mode of thought, must remain inviolable and free from deceit, for the stakes are high: only the best, truest poetry can provide the needed “criticism of life” and help humanity confront its challenges with clarity and resilience. The author cautions against two common fallacies in judging poetry: the “historic” and “personal” estimates. The historic estimate overemphasizes a poet’s work because of its place in literary history, while the personal estimate overvalues poetry due to personal preferences or experiences. Both can obscure our ability to appreciate poetry purely on its inherent quality. Only by focusing on a “real estimate”— an appreciation of the best, most excellent poetry based on its intrinsic value — can we fully experience the depth, strength, and joy that poetry offers. It discusses the tendency of literary historians to focus excessively on the origins, development, and past reputations of poets and their works, sometimes to the detriment of understanding and appreciating the true classics in poetry. The author warns against overvaluing works simply because they are historically significant or well-known, especially if their intrinsic quality does not match their reputation. For example, French critics began re-examining their neglected early poetry and found it more authentic than the courtly "classical" poetry of the seventeenth century, which some criticize as overly polished and lacking genuine poetic spirit. However, the author argues that while understanding historical context is useful, it should not overshadow our ability to recognize and enjoy genuine poetic excellence. The author insists on a distinction: if a poet is a genuine classic — meaning their work is of the highest quality — our primary goal should be to deeply enjoy and appreciate this excellence, rather than fixate on historical details or reputations. This “real estimate” focuses on the quality and emotional impact of the work itself, not just its historical or personal significance. While it’s valuable to understand the context and challenges faced by a classic poet, this knowledge is only worthwhile if it enhances our enjoyment and understanding of their work’s true merit. The author also warns against overemphasis on detailed background study (like the “historic origins” of a work) because it can distract from actually experiencing and appreciating the poetry itself. This can be compared to schoolboys who spend so much time on detailed grammatical study of Greek and Latin that they hardly get to enjoy reading the literature. Similarly, those studying a poet’s hi storical context may become more interested in lesser-known works due to the effort involved, leading them to overrate these works. In essence, the author urges readers to focus on experiencing and appreciating the inherent excellence of truly great poetry. While understanding historical relationships is part of literary study, this understanding should serve to deepen our enjoyment of the finest works. The true goal of studying poetry should be to repeatedly return to the core principle of seeking and savoring the excellence of classic poetry. adDownload to read ad-free It elaborates on two main ideas: The potential flaws in how we judge poetry and the best ways to identify truly excellent poetry. The author, Matthew Arnold, critiques how different perspectives (or “estimates”) can distort our understanding and appreciation of poetry. He mentions two specific estimates: the “historic estimate” and the “personal estimate.” The Historic and Personal Estimates: The historic estimate can cause us to overvalue ancient poetry due to its historical significance rather than its quality. Because of its cultural or historical status, we might find ourselves tempted to compare it to works of universally recognized greatness, such as comparing Chanson de Roland to Homer’s Iliad. Arnold illustrate s this with a French critic who praises Chanson de Roland as a work of epic grandeur. Arnold, however, disagrees, pointing out that while Chanson de Roland has value, it lacks the supreme poetic quality of Homer. The personal estimate, on the other hand, skews judgment for contemporary or modern poets with whom we may feel a personal connection. This closeness can lead to an exaggerated view of their significance or quality, potentially clouding a fair assessment. Real (or True) Estimate: Arnold argues that instead of letting historic or personal biases affect our views, we should adopt a “real estimate” of poetry, grounded in its intrinsic quality. To recognize this quality, Arnold suggests keeping in mind lines or passages from universally acknowledged masters — poets such as Homer, Dante, Shakespeare, and Milton . These “touchstones” of poetry serve as benchmarks or standards for identifying the true beauty and excellence of other works. By keeping passages of high poetic quality in mind, readers can develop a refined sense of what makes poetry truly great, allowing them to better evaluate and appreciate other works without bias. Characteristics of High-Quality Poetry: According to Arnold, great poetry has an unmistakable “mark and accent” of beauty, worth, and power. However, he refrains from providing an abstract definition of these qualities, suggesting that attempting to define them would only obscure their true nature. Instead, he argues that these qualities are best experienced directly in the words of master poets . The “substance” (content) and “style” (form) of the poetry itself reveal its excellence, and a genuine reader, by feeling the power of these high-quality passages, will better understand and appreciate poetry across different works. In summary, Arnold advocates for a direct, experience- based approach to evaluating poetry, guided by established “touchstone” examples from the best poets, which can help readers maintain clear and sound judgments about poetic quality. This approach encourages readers to develop an instinctual sense of poetic greatness, which is more reliable than theoretical analysis or personal or historical biases. Matthew Arnold discusses the qualities that give the best poetry its special character. He uses Aristotle’s concept that poetry, compared to history, reaches a "higher truth and seriousness," which elevates it beyond mere factual record. According to Arnold, the greatness of poetry lies in the combination of two aspects: Substance and Matter: Great poetry contains profound truth and seriousness, engaging deeply with significant aspects of human experience. Style and Manner: High poetic quality is also evident in the diction (word choice) and movement (rhythm and flow) of the verse. These stylistic qualities give poetry its unique accent and power. Arnold argues that these two aspects are interdependent : -The depth of truth and seriousness in the content ( substance and matter ) is matched by excellence in expression (style and manner). A lack in one is mirrored by a lack in the other. This synergy between substance and style is what makes poetry memorable and effective. -To illustrate this, Arnold examines the historical significance of early French romance poetry from the twelfth and thirteenth centuries, which played a dominant role in medieval European literature and influenced emerging Italian literature, notably in writers like Dante and Petrarch. He acknowledges the historical importance of this poetry but suggests that, by modern standards, it lacks the poetic excellence he describes. Thus, he warns against overvaluing poetry solely because of its historical context or cultural prominence. Ultimately, Arnold believes that poetry's greatness should be measured by its truth, seriousness, and stylistic power, rather than by its historical or cultural impact alone. He encourages readers to apply these criteria directly and personally when evaluating poetry, allowing the inherent quality of the work to guide their judgments. adDownload to read ad-free Matthew evaluates the significance and limitations of Geoffrey Chaucer's poetry, emphasizing its stylistic beauty but highlighting its absence of what he considers essential in the greatest poetry: “high seriousness.” Arnold points out that Chaucer, though inspired by French and Italian traditions, achieved a unique, lasting appeal. While French romance poets like Christian of Troyes were highly regarded in their own time, their impact has diminished. Chaucer, however, remains a "source of joy and strength," with a timeless quality that continues to resonate and will likely grow in popularity. This enduring quality of Chaucer’s work means it transcends the “historic estimate,” the idea of valuing works solely for their historical or cultural significance. Chaucer’s poetry is valued f or its inherent artistic merit. Arnold attributes Chaucer’s superiority to his "large, free, simple, clear yet kindly view of human life." Unlike the French romance poets, who lack a mature understanding of humanity, Chaucer presents a thoughtful and empathetic perspective on the world. He references the Prologue to The Canterbury Tales as an example, which Dryden famously praised as “God’s plenty” for its vibrant representation of diverse human characters. Arnold believes that Chaucer’s portrayal of life is expansive, makin g his poet ry more substantial and true. Arnold lauds Chaucer’s language for its “divine liquidness of diction” and “ fluidity of movement .” By “liquid diction,” Arnold refers to the harmonious, flowing quality of Chaucer’s words, while “fluid movement” denotes the smooth, rhythmic cadence of his lines. The appeal of Chaucer’s style lies not merely in “smooth numbers” or rhymes, as Dr. Johnson mistakenly believed. Arnold argues that Chaucer’s rhythmic beauty is foundational for English poetry, setting a standard followed by poets like Spenser, Shakespeare, Milton, and Keats. To illustrate Chaucer’s stylistic charm, Arnold contrasts an excerpt from The Prioress’ Tale with Wordsworth’s modernization of it. Wordsworth’s version, though faithful in meaning, lacks the “delicate and evanescent charm” of Chaucer’s original language. Arnold acknowledges that Chaucer’s fluidity is partly due to his use of certain liberties with language, like turning words such as “neck” into two syllables. However, Arnold insists that Chaucer’s fluidity is not dependent on these liberties, as other poets with similar freedoms did not achieve Chaucer’s level of fluidity. This fluidity, Arnold asserts, is primarily the result of Chaucer’s natural poetic talent rather than his use of linguistic l icense. Even poets who lack such freedoms, like Shakespeare and Keats, achieve similar fluidity due to their talent. Arnold believes that Chaucer, despite his strengths, is not one of the “great classics.” His poetry is lovely and wise, but it lacks what A ristotle described as spoudaiotes, or “high seriousness”— a quality Arnold deems essential for the greatest poetry. He compares Chaucer to Dante, Homer, and Shakespeare, whose works embody a profound seriousness and “criticism of life” that provides readers with a foundation upon which they can rest their spirits. This quality, Arnold argues, meets the increasing demands modern readers place on poetry to deliver moral or spiritual depth. As a final comparison, Arnold introduces François Villon, a French poet known for his troubled life and work characterized by moments of remarkable poignancy and seriousness, such as in La Belle Heaulmière. Even though Villon’s work occasionally captures this seriousness, it is not sustained across his entire body of work. Arnold implies that the greatness of poets like Dante or Homer lies in their ability to maintain this seriousness consistently, rather than in occasional moments. Chaucer’s poetry, while enjoyable and humane, lacks this sustained high seriousness, which, for Arnold, limits his standing among the highest ranks of poets. In summary, Arnold’s commentary suggests that Chaucer's poetry, though influential and cherished for its human insight, charm, and stylistic fluidity, lacks the “high seriousness” that elevates poetry to a truly timeless and universal level. He acknowledges Chaucer’s immense contributions to English poetry but suggests that to enter the ranks of poets like Dante, Chaucer would need to impart a deeper, more sustained seriousness and philosophical weight in his work. This quality, Arnold believes, allows readers to find lasting strength and repose, qualities he feels are essential to the world’s greatest poetry. Arnold uses the example of Chaucer, Dryden, Pope, and others to discuss what constitutes "true" poetry, suggesting that classic poetry requires a "high seriousness" that reflects a meaningful critique of life. He praises Chaucer for his originality and poetic truth but acknowledges that Chaucer lacks the gravitas found in great classics. Arnold then shifts to the poetry of Dryden and Pope, analyzing how their works align with the values of their era — the "age of prose and reason." He suggests that while Dryden and Pope contributed significantly to English letters, their poetry serves the rational and orderly prose style of the Restoration and the 18th century rather than the elevated, enduring qualities of poetic classics. Arnold describes the constraints of their time, where religious fervor and the rise of formal prose shaped a li terature that valued precision and balance, albeit at the expense of imaginative and poetic freedom. Therefore, he concludes that Dryden and Pope are classics of English prose rather than poetry. In contrast, Arnold holds Gray in high regard, recognizing him as a poet who, though not as prolific, encapsulates the spirit of classic poetry. Gray’s engagement with Greek poetry and his adoption of its perspectives allowed him to approach the poetic ideal Arnold values, though Gray's works are limited in volume and scope. Finally, Arnold addresses Robert Burns, particularly in the context of his Scottish poetry. He acknowledges that while Burns’s work adDownload to read ad-free is celebrated in Scotland, his themes — centered on "Scotch drink, Scotch religion, and Scotch manners" — might limit his universal appeal, especially outside Scotland. Burns's true brilliance, Arnold suggests, is found in his ability to transcend his local themes and offer a universal "criticism of life." Yet Arnold warns that the cultural specificity of Burns's work might lead to a "personal estimate" that obscures a true, objective evaluation of his place in the poetic canon. Matthew Arnold's essay "The Study of Poetry," where he discusses the qualities that make poetry timeless and universally resonant, using the works of Robert Burns as a focal point to illustrate his ideas. Arnold is critical of the view that Burns is a poet of the highest order. He acknowledges Burns’s skill, vitality, and unique voice, but he contrasts Burns’s work with the works of the "great classics," claim ing that Burns lacks what he calls "high seriousness." Arnold defines this "high seriousness" as a quality born of absolute sincerity and depth of feeling that elevates the poet's criticism of life. Burns, he argues, while often touching upon profound themes, sometimes lapses into bravado or lacks the emotional depth that would make his work resonate at the same level as Dante or Shakespeare. Burns's pieces that aim for moral themes, according to Arnold, often feel like preaching rather than poetry emerging from the soul. Arnold ultimately appreciates Burns’s strength in his lively and ironic perspective on life, comparing him favorably to Chaucer in his vivid portrayals of human nature. In particular, Arnold finds "The Jolly Beggars" and poems like "Tam o' Shanter" to be triumphant poetic expressions, even if they lack the weighty solemnity of classic works. Despite his criticism, Arnold concludes that Burns’s work is poetically sound at its core, especially in its lighter, witty, and more genuine moments, offering a wholesome counterpoint to poets like Shelley, whose work he sees as overly abstract and distant from practical life. Arnold's approach, often referred to as the "touchstone" method, advocates measuring poetry against the enduring works of classic poets, to determine if it truly touches on universal truths with sincerity. This, for Arnold, helps distinguish the truly great works from those that might appeal to personal taste but lack the lasting value of the classics. Poetry is Immense: Arnold suggests that poetry holds a vast and enduring significance because it is more than just entertainment or clever wordplay; it captures and expresses the deepest truths of human existence, transcending time and place. In this sense, poetry is "immense" because it addresses universal concerns, tapping into a deep emotional resonance that is timeless. Great poetry connects not only with the individual but with the collective human experience across ages. Religion has Materialized: This likely refers to the way religious or spiritual matters have been reduced to simplistic, material concerns in society, no longer conveying the deep, transcendent truths they once did. Arnold suggests that religion, when it becomes an institution or a system of rigid rules, loses its spiritua l force and becomes something more "material" or less meaningful. In the same way, poetry, when disconnected from true emotion or sincerity, can become superficial or formulaic. Charlatanism: In the context of Arnold’s essay, charlatanism refers to the pretensions of poets or writers who claim to offer profound insight or artistic mastery but lack true substance. These individuals are skilled in surface-level techniques or gimmicks (much like a charlatan in any other field) but fail to provide the deeper, genuine qualities that characterize great poetry. Arnold criticizes poetry that lacks true emotion, depth, and sincerity — essentially, poetry that is "pretend" or disingenuous. Who is the Classic: A "classic," in Arnold’s view, is a work of literature (or a poet) whose qualities transcend the limitations of its time and culture. These works remain relevant and resonant across generations because they embody fundamental truths of the human experience. For Arnold, Dante, Shakespeare, and Homer are examples of "classics." They provide a lasting touchstone for all subsequent poetry, offering a model of what great poetry should be — timeless, sincere, and deeply rooted in truth. Touchstone Method: The "touchstone" method is Arnold's critical approach for evaluating poetry. It involves comparing a poet’s work to the standards set by the greatest classic poets. In this approach, the greatness of a work is determined by how it holds up against these timeless examples of poetic excellence. The method emphasizes high seriousness, sincerity, and the ability to convey universal truths, making it a reliable test for the value of any poetry. Substance, Matter, Style: Arnold argues that great poetry consists of both substance (the depth of the ideas or themes it addresses) and style (the way those ideas are expressed). The substance of a poem refers to its themes — its insights into life, nature, and human emotions. 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3252	https://www.mayocliniclabs.com/test-catalog/overview/608083	Test Id : HBEL1 Hemoglobin Electrophoresis Evaluation, Blood Overview Specimen Clinical & Interpretive Performance Fees & Codes Setup & Updates Useful For Suggests clinical disorders or settings where the test may be helpful Diagnosis and classification of hemoglobin disorders, including thalassemias and hemoglobin variants Profile Information A profile is a group of laboratory tests that are ordered and performed together under a single Mayo Test ID. Profile information lists the test performed, inclusive of the test fee, when a profile is ordered and includes reporting names and individual availability. \| Test Id \| Reporting Name \| Available Separately \| Always Performed \| --- --- \| \| HBELI \| Hb Electrophoresis Interpretation \| No \| Yes \| \| HGBCE \| Hb Variant, A2 and F Quantitation,B \| Yes \| Yes \| \| HPLC \| HPLC Hb Variant, B \| No \| Yes \| Reflex Tests Lists tests that may or may not be performed, at an additional charge, depending on the result and interpretation of the initial tests. \| Test Id \| Reporting Name \| Available Separately \| Always Performed \| --- --- \| \| HPFH \| Hb F Distribution, B \| No \| No \| \| MASS \| Hb Variant by Mass Spec, B \| No \| No \| \| SDEX \| Sickle Solubility, B \| Yes \| No \| \| IEF \| Isoelectric Focusing, B \| No \| No \| \| UNHB \| Hb Stability, B \| No \| No \| \| WASQR \| Alpha Globin Gene Sequencing, B \| Yes, (Order WASEQ) \| No \| \| WBSQR \| Beta Globin Gene Sequencing, B \| Yes, (Order WBSEQ) \| No \| \| WGSQR \| Gamma Globin Full Gene Sequencing \| Yes, (Order WGSEQ) \| No \| \| HBEL0 \| Hb Electrophoresis Summary Interp \| No \| No \| \| WBGDR \| Beta Globin Gene Cluster, Del/Dup,B \| Yes, (Order WBGDD) \| No \| \| WAGDR \| Alpha Globin Clustr Locus Del/Dup,B \| Yes, (Order AGDD) \| No \| Testing Algorithm Delineates situations when tests are added to the initial order. This includes reflex and additional tests. This evaluation will always include hemoglobin (Hb) A2 and HbF and hemoglobin electrophoresis utilizing capillary electrophoresis and cation exchange high-performance liquid chromatography methods. Reflex testing, performed at additional charge, may include any or all of the following to identify rare hemoglobin variants present: sickle solubility (hemoglobin S screen); hemoglobin heat and isopropanol stability studies (unstable hemoglobin); isoelectric focusing, intact globin chain mass spectrometry (hemoglobin variant by mass spectrometry); HbF distribution by flow cytometry; DNA Sanger sequencing assays for: 1) beta-chain variants and the most common beta thalassemias (beta-globin gene sequencing), 2) alpha-chain variants and less common nondeletional alpha thalassemias (alpha-globin gene sequencing), or 3) gamma-chain variants and nondeletional hereditary persistence of fetal hemoglobin (HPFH) (gamma-globin full gene sequencing); multiplex ligation-dependent probe amplification assays for: 1) large deletional alpha thalassemias and alpha-gene duplications (alpha-globin gene analysis), or 2) beta-globin gene cluster locus large deletions and duplications, including large deletional HPFH, delta-beta thalassemia, gamma-delta-beta thalassemia, epsilon-gamma-delta-beta thalassemia and large deletional beta or delta thalassemia (beta-globin cluster locus deletion/duplication). If test results in the profile are abnormal, results may be reviewed by a hematopathology consultant, and a summary interpretation provided. One or more of the following molecular tests may be reflexed: -WAGDR / Alpha Globin Cluster Locus Deletion/Duplication, Blood -WASQR / Alpha-Globin Gene Sequencing, Blood -WBSQR / Beta-Globin Gene Sequencing, Blood -WBGDR / Beta-Globin Gene Cluster Deletion/Duplication, Blood -WGSQR / Gamma-Globin Full Gene Sequencing, Varies For cases with molecular testing added, a preliminary interpretation will be reported that discusses the protein test results. After all test results are finalized, an additional consultative interpretation that summarizes all testing and incorporates subsequent genetic results will be provided. Method Name A short description of the method used to perform the test HBELI, HBEL0: Medical Interpretation HGBCE: Capillary Electrophoresis HPLC: Cation Exchange/High-Performance Liquid Chromatography (HPLC) IEF: Isoelectric Focusing MASS: Mass Spectrometry (MS) HPFH: Flow Cytometry UNHB: Isopropanol and Heat Stability NY State Available Indicates the status of NY State approval and if the test is orderable for NY State clients. Yes Reporting Name Lists a shorter or abbreviated version of the Published Name for a test Hb Electrophoresis Evaluation Aliases Lists additional common names for a test, as an aid in searching A2 Hemoglobin Alpha Globin Variant Alpha Thalassemia Barts Hemoglobin Beta Globin Variant Beta Thalassemia H Disease Hemoglobin A2 Hemoglobin Cascade Hemoglobin Electrophoresis Cascade Level 1 Hemoglobin Molecular studies Hemoglobin Variant HGB (Hemoglobin) Electrophoresis Isoelectric Focusing Capillary electrophoresis HPLC High performance liquid chromatography Mass Spectrometry Microcytosis Sickling Test Thalassemia Sickle Prep FTA2 HB Testing Algorithm Delineates situations when tests are added to the initial order. This includes reflex and additional tests. This evaluation will always include hemoglobin (Hb) A2 and HbF and hemoglobin electrophoresis utilizing capillary electrophoresis and cation exchange high-performance liquid chromatography methods. Reflex testing, performed at additional charge, may include any or all of the following to identify rare hemoglobin variants present: sickle solubility (hemoglobin S screen); hemoglobin heat and isopropanol stability studies (unstable hemoglobin); isoelectric focusing, intact globin chain mass spectrometry (hemoglobin variant by mass spectrometry); HbF distribution by flow cytometry; DNA Sanger sequencing assays for: 1) beta-chain variants and the most common beta thalassemias (beta-globin gene sequencing), 2) alpha-chain variants and less common nondeletional alpha thalassemias (alpha-globin gene sequencing), or 3) gamma-chain variants and nondeletional hereditary persistence of fetal hemoglobin (HPFH) (gamma-globin full gene sequencing); multiplex ligation-dependent probe amplification assays for: 1) large deletional alpha thalassemias and alpha-gene duplications (alpha-globin gene analysis), or 2) beta-globin gene cluster locus large deletions and duplications, including large deletional HPFH, delta-beta thalassemia, gamma-delta-beta thalassemia, epsilon-gamma-delta-beta thalassemia and large deletional beta or delta thalassemia (beta-globin cluster locus deletion/duplication). If test results in the profile are abnormal, results may be reviewed by a hematopathology consultant, and a summary interpretation provided. One or more of the following molecular tests may be reflexed: -WAGDR / Alpha Globin Cluster Locus Deletion/Duplication, Blood -WASQR / Alpha-Globin Gene Sequencing, Blood -WBSQR / Beta-Globin Gene Sequencing, Blood -WBGDR / Beta-Globin Gene Cluster Deletion/Duplication, Blood -WGSQR / Gamma-Globin Full Gene Sequencing, Varies For cases with molecular testing added, a preliminary interpretation will be reported that discusses the protein test results. After all test results are finalized, an additional consultative interpretation that summarizes all testing and incorporates subsequent genetic results will be provided. Specimen Type Describes the specimen type validated for testing Whole Blood EDTA Ordering Guidance Multiple hematology evaluations are available. For information on testing that is performed with each evaluation, see Benign Hematology Evaluation Comparison. Necessary Information At minimum, include recent transfusion information and most recent complete blood cell count results. Metabolic Hematology Patient Information (T810) is strongly recommended. Testing may proceed without this information, however if the information requested is received, any pertinent reported clinical features and data will drive the focus of the evaluation and be considered in the interpretation. The laboratory has extensive experience in hemoglobin variant identification and many cases can be confidently classified without molecular testing. However, molecular confirmation is always available, subject to sufficient sample quantity (eg, multiplex ligation-dependent probe amplification testing requires at least 2 mL of sample in addition to protein testing requirements). If no molecular testing or specific molecular tests are desired, utilize the appropriate check boxes on the form. If the form or other communication is not received, the reviewing hematopathologist will select appropriate tests to sufficiently explain the protein findings, which may or may not include molecular testing. Specimen Required Defines the optimal specimen required to perform the test and the preferred volume to complete testing Container/Tube: Preferred: Lavender top (EDTA) Acceptable: Yellow top (ACD solution B) Specimen Volume: 10 mL Collection Instructions: Send whole blood specimen in original tube. Do not aliquot. Special Instructions Library of PDFs including pertinent information and forms related to the test Informed Consent for Genetic Testing Metabolic Hematology Patient Information Benign Hematology Evaluation Comparison Informed Consent for Genetic Testing (Spanish) Forms New York Clients-Informed consent is required. Document on the request form or electronic order that a copy is on file. The following documents are available: -Informed Consent for Genetic Testing (T576) -Informed Consent for Genetic Testing-Spanish (T826) Metabolic Hematology Patient Information (T810) If not ordering electronically, complete, print, and send a Benign Hematology Test Request (T755) with the specimen Specimen Minimum Volume Defines the amount of sample necessary to provide a clinically relevant result as determined by the testing laboratory. The minimum volume is sufficient for one attempt at testing. 1 mL (this volume will limit reflex testing possibilities); 3 mL if multiplex ligation-dependent probe amplification is needed Reject Due To Identifies specimen types and conditions that may cause the specimen to be rejected All specimens will be evaluated at Mayo Clinic Laboratories for test suitability. Specimen Stability Information Provides a description of the temperatures required to transport a specimen to the performing laboratory, alternate acceptable temperatures are also included \| Specimen Type \| Temperature \| Time \| Special Container \| --- --- \| \| Whole Blood EDTA \| Refrigerated \| 7 days \| \| Useful For Suggests clinical disorders or settings where the test may be helpful Diagnosis and classification of hemoglobin disorders, including thalassemias and hemoglobin variants Testing Algorithm Delineates situations when tests are added to the initial order. This includes reflex and additional tests. This evaluation will always include hemoglobin (Hb) A2 and HbF and hemoglobin electrophoresis utilizing capillary electrophoresis and cation exchange high-performance liquid chromatography methods. Reflex testing, performed at additional charge, may include any or all of the following to identify rare hemoglobin variants present: sickle solubility (hemoglobin S screen); hemoglobin heat and isopropanol stability studies (unstable hemoglobin); isoelectric focusing, intact globin chain mass spectrometry (hemoglobin variant by mass spectrometry); HbF distribution by flow cytometry; DNA Sanger sequencing assays for: 1) beta-chain variants and the most common beta thalassemias (beta-globin gene sequencing), 2) alpha-chain variants and less common nondeletional alpha thalassemias (alpha-globin gene sequencing), or 3) gamma-chain variants and nondeletional hereditary persistence of fetal hemoglobin (HPFH) (gamma-globin full gene sequencing); multiplex ligation-dependent probe amplification assays for: 1) large deletional alpha thalassemias and alpha-gene duplications (alpha-globin gene analysis), or 2) beta-globin gene cluster locus large deletions and duplications, including large deletional HPFH, delta-beta thalassemia, gamma-delta-beta thalassemia, epsilon-gamma-delta-beta thalassemia and large deletional beta or delta thalassemia (beta-globin cluster locus deletion/duplication). If test results in the profile are abnormal, results may be reviewed by a hematopathology consultant, and a summary interpretation provided. One or more of the following molecular tests may be reflexed: -WAGDR / Alpha Globin Cluster Locus Deletion/Duplication, Blood -WASQR / Alpha-Globin Gene Sequencing, Blood -WBSQR / Beta-Globin Gene Sequencing, Blood -WBGDR / Beta-Globin Gene Cluster Deletion/Duplication, Blood -WGSQR / Gamma-Globin Full Gene Sequencing, Varies For cases with molecular testing added, a preliminary interpretation will be reported that discusses the protein test results. After all test results are finalized, an additional consultative interpretation that summarizes all testing and incorporates subsequent genetic results will be provided. Clinical Information Discusses physiology, pathophysiology, and general clinical aspects, as they relate to a laboratory test A large number of variants of hemoglobin (Hb) have been recognized. Although many do not result in clinical or hematologic effects, clinical symptoms that can be associated with Hb disorders include microcytosis, sickling disorders, hemolysis, erythrocytosis/polycythemia, cyanosis/hypoxia, anemia (chronic, compensated, or episodic), and increased methemoglobin or sulfhemoglobin results (M-hemoglobins). For many common Hb variants (eg, HbS, HbC, HbD and HbE, among many others), protein studies will be sufficient for definitive identification. However, some Hb conditions may be difficult to identify by protein methods alone and may require molecular methods for confirmation. Hb disorders commonly occur as compound disorders (2 or more genetic variants) that can have complex interactions and variable phenotypes. In these situations, molecular testing may be necessary for accurate classification. It is important to note that although powerful as an adjunct for a complete and accurate diagnosis, molecular methods without protein data can give incomplete and possibly misleading information due to limitations of the methods. Accurate classification of hemoglobin disorders and interpretation of genetic data requires the incorporation of protein analysis results. This profile is well-suited for the classification of hemoglobin disorders. Mayo Clinic Laboratories receives specimens from a wide geographic area and nearly one-half of all specimens tested exhibit abnormalities. The most common abnormality is an increase in HbA2 to about 4% to 8%, which indicates beta-thalassemia minor when present in the correct clinical context. A wide variety of other hemoglobinopathies are also frequently encountered. Ranked in order of relative frequency, these are: Hb S (sickle cell disease and trait), C, E, Lepore, G-Philadelphia, HbH disease, D-Los Angeles, Koln, Constant Spring, O-Arab. Other variants associated with hemolysis, erythrocytosis/polycythemia, microcytosis, cyanosis/hypoxia are routinely identified; however, some will not be detected by routine screening methods and require communication of clinical findings to prompt indicated reflex testing options. Alpha-thalassemia genetic variants are very common in the United States, occurring in approximately 30% of African Americans and accounting for the frequent occurrence of microcytosis in persons of this ethnic group. Some alpha-thalassemia conditions (eg, HbH, Barts) can be identified in the hemoglobin electrophoresis protocol, although Hb Constant Spring may or may not be evident by protein methods alone dependent upon the percentage present. It is important to note, alpha thalassemias that are from only 1 or 2 alpha-globin gene deletions are not recognized by protein studies alone and alpha-gene deletion and duplication testing is required. Reference Values Describes reference intervals and additional information for interpretation of test results. May include intervals based on age and sex when appropriate. Intervals are Mayo-derived, unless otherwise designated. If an interpretive report is provided, the reference value field will state this. Hemoglobin Electrophoresis Interpretation Definitive results and an interpretative report will be provided. Hemoglobin Variant, A2 and F Quantitation HEMOGLOBIN A 0-30 days: 5.9-77.2% 1-2 months: 7.9-92.4% 3-5 months: 54.7-97.1% 6-8 months: 80.0-98.0% 9-12 months: 86.2-98.0% 13-17 months: 88.8-98.0% 18-23 months: 90.4-98.0% or =24 months: 95.8-98.0% HEMOGLOBIN A2 0-30 days: 0.0-2.1% 1-2 months: 0.0-2.6% 3-5 months: 1.3-3.1% or =6 months: 2.0-3.3% HEMOGLOBIN F 0-30 days: 22.8-92.0% 1-2 months: 7.6-89.8% 3-5 months: 1.6-42.2% 6-8 months: 0.0-16.7% 9-12 months: 0.0-10.5% 13-17 months: 0.0-7.9% 18-23 months: 0.0-6.3% or =24 months: 0.0-0.9% VARIANT 1 0.0 VARIANT 2 0.0 VARIANT 3 0.0 Interpretation Provides information to assist in interpretation of the test results The hemoglobin fractions, including hemoglobin variants are identified and quantitated. An interpretive report that summarizes all testing, including the significance of the findings, is issued. Cautions Discusses conditions that may cause diagnostic confusion, including improper specimen collection and handling, inappropriate test selection, and interfering substances Some hemoglobin disorders and variants are not detected by the screening methods including, common alpha-thalassemia conditions and require further reflex testing to identify. If a family history of a known hemoglobin disorder, prior therapy for a hemoglobin disorder, or otherwise unexplained lifelong/familial symptoms, such as hemolysis, microcytosis, erythrocytosis/polycythemia, cyanosis, or hypoxia are present, this should be clearly communicated to the laboratory so appropriate reflex testing can be added, see Metabolic Hematology Patient Information. Recent transfusion may mask protein results including hemoglobin electrophoresis, hereditary persistence of hemoglobin F by flow cytometry, stability studies, and sickle solubility studies depending on percentage of transfused cells present. Some hemoglobin variants can originate from the donor blood product and not from the tested recipient. These are typically found in low percentage. If the patient has undergone a bone marrow transplant, the results may show atypical results and should be interpreted in the context of clinical information. Some therapies cause artefactual effects in protein studies, including hydroxyurea and decitabine (increased hemoglobin F levels), voxelotor (artefactual peaks) and gene therapy (alternate protein detection, beta T87Q, by mass spectrometry). Clear communication of prior therapy is strongly recommended. Clinical Reference Recommendations for in-depth reading of a clinical nature Hoyer JD, Hoffman DR. The thalassemia and hemoglobinopathy syndromes. In: McClatchey KD, Amin HM, Curry JL, eds. Clinical Laboratory Medicine. 2nd ed. Lippincott Williams and Wilkins; 2002:866-895 Oliveira JL. Diagnostic strategies in hemoglobinopathy testing, the role of a reference laboratory in the USA. Thalassemia Reports. 2018;8(1):7476. doi:10.4081/thal.2018.7476 Brancaleonai V, Di Pierro E, Motta I, Cappellini MD. Laboratory diagnosis of thalassemia. Int J Lab Hematol. 2016;38 Suppl 1:32-40. doi:10.1111/ijlh.12527 Hartveld CI. State of the art and new developments in molecular diagnostics for hemoglobinopathies in multiethnic societies. Int J Lab Haematol. 2014;36(1):1-12. doi:10.1111/ijlh.12108 Riou J, Szuberski J, Godart C, et al. Precision of CAPILLARYS 2 for the detection of hemoglobin variants based on their migration positions. Am J Clin Pathol. 2018;149(2):172-180. doi:10.1093/ajcp/aqx148 Method Description Describes how the test is performed and provides a method-specific reference Hemoglobin Electrophoresis: The CAPILLARYS System is an automated system that uses capillary electrophoresis to separate charged molecules by their electrophoretic mobility in an alkaline buffer. Separation occurs according to the electrolyte pH and electro-osmotic flow. A sample dilution with hemolyzing solution is injected by aspiration. A high-voltage protein separation occurs with direct detection of the hemoglobin-protein fractions at 415 nm, which is specific to hemoglobin. The resulting electrophoregram peaks are evaluated for pattern abnormalities and are quantified as a percentage of the total hemoglobin present. Examples of position of commonly found hemoglobin fractions are, from cathode to anode: HbA2', C, A2/O-Arab, E, S, D, G-Philadelphia, F, A, Hope, Barts, J, N-Baltimore and H.(Louahabi A, Philippe M, Lali S, Wallemacq P, Maisin D. Evaluation of a new Sebia kit for analysis of hemoglobin fractions and variants on the Capillarys system. Clin Chem Lab Med. 2006;44:340-345; instruction manual CAPI 3 HEMOGLOBIN(E) Phoresis VS >9.15. Sebia; 12/2020) High-Performance Liquid Chromatography: Hemolysate of whole blood is injected into an analysis stream passing through a cation exchange column using high-performance liquid chromatography. A preprogrammed gradient controls the elution buffer mixture that also passes through the analytical cartridge. The ionic strength of the elution buffer is raised by increasing the percentage of a second buffer. As the ionic strength of the buffer increases the more strongly retained hemoglobins elute from the cartridge. Absorbance changes are detected by a dual-wavelength filter photometer. Changes in absorbance are displayed as a chromatogram of absorbance versus time.(Huismann TH, Scroeder WA, Brodie AN, Mayson SM, Jakway J. Microchromotography of hemoglobins. III. A simplified procedure for the determination of hemoglobin A2. J Lab Clin Med. 1975;86:700-702; Ou CN, Buffone GJ, Reimer GL, Alpert AJ. High-performance liquid chromatography of human hemoglobins on a new cation exchanger. J Chromatogr. 1983;266:197-205; Szuberski J, Oliveira JL, Hoyer JD. A comprehensive analysis of hemoglobin variants by high-performance liquid chromatography [HPLC]. Int J Lab Hematol. 2012;34(6):594-604; instruction manual: Bio-Rad Variant II Beta-thalassemia Short Program Instructions for Use, L70203705. Bio-Rad Laboratories, Inc; 11/2011) PDF Report Indicates whether the report includes an additional document with charts, images or other enriched information No Day(s) Performed Outlines the days the test is performed. This field reflects the day that the sample must be in the testing laboratory to begin the testing process and includes any specimen preparation and processing time before the test is performed. Some tests are listed as continuously performed, which means that assays are performed multiple times during the day. Monday through Thursday Report Available The interval of time (receipt of sample at Mayo Clinic Laboratories to results available) taking into account standard setup days and weekends. The first day is the time that it typically takes for a result to be available. The last day is the time it might take, accounting for any necessary repeated testing. 2 to 25 days Specimen Retention Time Outlines the length of time after testing that a specimen is kept in the laboratory before it is discarded Whole blood: 7 days; Abnormal samples: 14 days Performing Laboratory Location Indicates the location of the laboratory that performs the test Mayo Clinic Laboratories - Rochester Main Campus CLIA Number: 24D0404292 Fees : Several factors determine the fee charged to perform a test. Contact your U.S. or International Regional Manager for information about establishing a fee schedule or to learn more about resources to optimize test selection. Authorized users can sign in to Test Prices for detailed fee information. Clients without access to Test Prices can contact Customer Service 24 hours a day, seven days a week. Prospective clients should contact their account representative. For assistance, contact Customer Service. Test Classification Provides information regarding the medical device classification for laboratory test kits and reagents. Tests may be classified as cleared or approved by the US Food and Drug Administration (FDA) and used per manufacturer instructions, or as products that do not undergo full FDA review and approval, and are then labeled as an Analyte Specific Reagent (ASR) product. This test has been modified from the manufacturer's instructions. Its performance characteristics were determined by Mayo Clinic in a manner consistent with CLIA requirements. This test has not been cleared or approved by the US Food and Drug Administration. CPT Code Information Provides guidance in determining the appropriate Current Procedural Terminology (CPT) code(s) information for each test or profile. The listed CPT codes reflect Mayo Clinic Laboratories interpretation of CPT coding requirements. It is the responsibility of each laboratory to determine correct CPT codes to use for billing. CPT codes are provided by the performing laboratory. 82664 (if appropriate) 83068 (if appropriate) 83789 (if appropriate) 88184 (if appropriate) 83020-26 (if appropriate) LOINC® Information Provides guidance in determining the Logical Observation Identifiers Names and Codes (LOINC) values for the order and results codes of this test. LOINC values are provided by the performing laboratory. \| Test Id \| Test Order Name \| Order LOINC Value \| --- \| HBEL1 \| Hb Electrophoresis Evaluation \| 94538-6 \| \| Result Id \| Test Result Name \| Result LOINC Value Applies only to results expressed in units of measure originally reported by the performing laboratory. These values do not apply to results that are converted to other units of measure. \| --- \| 41927 \| Hb A \| 20572-4 \| \| 41928 \| Hb F \| 32682-7 \| \| 41929 \| Hb A2 \| 4552-6 \| \| 41930 \| Variant 1 \| 24469-9 \| \| 41931 \| Variant 2 \| 24469-9 \| \| 41932 \| Variant 3 \| 24469-9 \| \| 41933 \| HGBCE Interpretation \| 78748-1 \| \| 65615 \| HPLC Hb Variant, B \| No LOINC Needed \| \| 608088 \| Hb Electrophoresis Interpretation \| 49316-3 \| \| 609421 \| Hb Electrophoresis Interp Cancel \| No LOINC Needed \|
3253	https://www.practiceproblems.org/problem/Discrete_Math/Recurrences_and_Generating_Functions/Solve_Recurrence_Relation_with_Initial_Conditions	Solve Recurrence Relation with Initial Conditions - Discrete Math Recurrences and Generating Functions \| PracticeProblems.org Skip to Content PracticeProblems.org PracticeProblems.org Toggle Menu Search Problems...⌘K Solve Recurrence Relation with Initial Conditions Home \| Discrete Math \| Recurrences and Generating Functions \| Solve Recurrence Relation with Initial Conditions Recurrences and Generating FunctionsMediumVideo Solve the recurrence relation a n=3 a n−1−2 a n−2 a_n = 3a_{n-1} - 2a_{n-2}a n=3 a n−1−2 a n−2 with initial conditions a 0=1 a_0 = 1 a 0=1 and a 1=3 a_1 = 3 a 1=3. Solving recurrence relations is a fundamental task in discrete mathematics, often encountered in algorithm analysis and combinatorics. Recurrence relations express each element of a sequence using the preceding elements, which can be integral in modeling problem states or sequential steps in computational problems. Understanding how to solve them is crucial for computer science students, especially for designing and analyzing algorithms. For this problem, the relation is linear with constant coefficients, a common type of recurrence where methods like characteristic equations are applied. Initial conditions present a specific solution path, allowing us to determine particular constants for the relation's general solution. This problem introduces students to analyzing linear homogeneous recurrence relations, a foundational skill for more complex recurrence types like non-homogeneous or those involving variable coefficients. After grasping basic recurrence structures and solutions, learners can expand this knowledge to derive closed-form solutions, facilitating more efficient computations than iterative methods. This understanding prepares you not only for theoretical applications but also for practical problem-solving scenarios often encountered in areas like dynamic programming, a key algorithm design paradigm in computer science. Posted by Gregory 2 months ago Related Problems Ratio of Consecutive Terms in Fibonacci Sequence Given a sequence generated by the rule x n=x n−1+x n−2 x_n = x_{n-1} + x_{n-2}x n=x n−1+x n−2, determine the ratio of consecutive terms as it approaches a limit, and prove that this ratio is the golden ratio 1+5 2\frac{1 + \sqrt{5}}{2}2 1+5 or its negative inverse. Recurrences and Generating FunctionsMediumVideo Generating Functions for Candy Combination Problem Using generating functions, determine how many ways there are to combine 10 candies when the candies are red, blue, and green with the conditions: even number of red candies, more than six blue candies, and less than three green candies. Recurrences and Generating FunctionsMediumVideo Recurrence Relation for Geometric Sequence Define the recurrence relation for a geometric sequence as a n=a n−1×2 a_n = a_{n-1} \times 2 a n=a n−1×2 for n≥1 n \geq 1 n≥1, starting with a 0=3 a_0 = 3 a 0=3. Recurrences and Generating FunctionsEasyVideo Second Order Linear Homogeneous Recurrence Relation with Initial Conditions Solve the second-order linear homogeneous recurrence relation a n=5 a n−1−6 a n−2 a_n = 5a_{n-1} - 6a_{n-2}a n=5 a n−1−6 a n−2 with initial conditions. Recurrences and Generating FunctionsMediumVideo Home • Privacy • Terms of Service © 2025 PracticeProblems.org
3254	https://es.scribd.com/document/352988410/Greenwood-Earnshaw-Chemistry-of-the-Elements	Greenwood Earnshaw Chemistry of The Elements \| PDF \| Natural Sciences \| Nature Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 2K views 4 pages Greenwood Earnshaw Chemistry of The Elements The document discusses the book "Chemistry of the Elements" written by N.N. Greenwood and A. Earnshaw. The book provides a descriptive approach to discussing the chemistry of the elements an… Full description Uploaded by emre AI-enhanced title and description Go to previous items Go to next items Download Save Save Greenwood Earnshaw Chemistry of the Elements For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Greenwood Earnshaw Chemistry of the Elements For Later You are on page 1/ 4 Search Fullscreen Greenwood Earnshaw Chemistry Of The Elements Greenwood Earnshaw Chemistry Of The Elements DOWNLOAD NOW CHEMISTRY OF THE ELEMENTS, SECOND EDITION: N. N. GREENWOOD ... Mon, 08 Dec 1997 2 3:57:00 GMT chemistry of the elements, ... [n. n. greenwood, a. earnshaw] ... approach based on descriptive chemistry. the chemistry of the elements is still discussed within ... 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GREENWOOD AND EARNSHAW, CHEMISTRY OF THE ELEMENTS \| VIPER Tue, 13 Jun 2017 06:33:00 GMT greenwood and earnshaw, chemistry of the elements ... an advanced (graduate) textbook that covers the reactivity of the elements. a great descriptive chemistry resource. CHEMISTRY OF ELEMENTS - SCRIBD Wed, 10 Oct 2012 23:57:00 GMT chemistry of elements. ... greenwood a.earnshaw origin of the elements. ... various theories of the origin of the chemical elements are all very recent and the ... CHEMISTRY OF THE ELEMENTS: N. N. GREENWOOD, A. EARNSHAW ... Mon, 06 Mar 2017 17:35:00 GMT chemistry of the elements: n. n. greenwood, ... a. earnshaw (author) 4.8 out of 5 ... approach based on descriptive chemistry. the chemistry of the elements is still ... CHEMISTRY OF THE ELEMENTS EBOOK: N. N. GREENWOOD, A ... Thu, 18 May 2017 11:15:00 GMT chemistry of the elements ebook: n. n. greenwood, a. earnshaw: amazon: kindle store amazon try prime kindle ...chemistry of the elements ... N. N. GREENWOOD, A. 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GREENWOOD ...CHEMISTRY OF THE ELEMENTS - GBVCHEMISTRY OF THE ELEMENTS BY GREENWOOD, N. N., EARNSHAW, A ...CHAPTER 8: CHEMISTRY OF THE MAIN GROUP ELEMENTSCHEMISTRY OF ELEMENTS 2ND ED. - N. N. GREENWOOD & A. EARNSHAWCHEMISTRY OF THE ELEMENTS BY N. N. GREENWOOD; A. EARNSHAW ...CHEMISTRY OF THE ELEMENTS BY GREENWOOD - ABEBOOKSCHEMISTRY OF THE ELEMENTS (BOOK, 2006) [WORLDCAT]CHEMISTRY OF THE ELEMENTS (BOOK, 1984) [WORLDCAT]sitemap index Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like I Want To Eat Your Pancreas (2018) by Yoru Sumino 100% (6) I Want To Eat Your Pancreas (2018) by Yoru Sumino 232 pages Chemistry of The Elements Second Edition by N. N. Greenwood 0% (2) Chemistry of The Elements Second Edition by N. N. 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3255	https://taylorandfrancis.com/knowledge/Medicine_and_healthcare/Anatomy/Alveolar_process/	Alveolar process – Knowledge and References – Taylor & Francis Skip to content Regions Africa China Japan MENA Contact Careers Search Who we serveExpand submenu Academia Authors Journal authors Book authors Faculty/instructors Librarians Journal editors Industry & government Practitioners R&D professionals Authors Government Healthcare professionals Marketing & brand managers Partners Societies Booksellers Funders KnowledgeExpand submenu Medicine & healthcare Education Business & management Psychology Environmental sciences Bioscience Computer & information sciences Mathematics Engineering Civil Electrical Industrial & manufacturing Mechanical Chemical engineering Automotive engineering Energy and oil Materials science research Explore our research portfolio Corporate solutionsExpand submenu Pharmaceutical Finance Materials sciences Technology Law Automotive engineering Chemical engineering Energy and oil Defense NewsExpand submenu Newsroom Insights, analysis, and blogs Events AboutExpand submenu Company Our brands Executive leadership team Our history Careers Our policies Open research Corporate responsibility Artificial intelligence (AI) Regions Africa China Japan MENA Contact Careers Search Publish with us I want to publish a book in a journal To find out how to publish or submit your book proposal: Go to Author Services > To find a journal or submit your article to a journal: Go to our journal suggester > Alveolar process The alveolar process is a bony ridge that contains the teeth and forms the dental arch. It functions as a supporting tissue for the teeth, and there is a definite interplay between teeth and alveolar bone in maintaining dental health. The mandibular alveolar process is a specific part of the alveolar process located in the lower jaw.From: Clinical Head and Neck Anatomy for Surgeons , Fundamentals of Craniofacial Growth more Related Topics Alveoli Dental implants Jaw Maxillary Oral hygiene Periodontal disease Bone Current Research Clinical Trials (United States) Clinical Trials (Europe) Clinical Trials (Australia/New Zealand) Clinical Trials (India) About this page The research on this page is brought to you by Taylor & Francis Knowledge Centers. This collection is automatically generated from our most recent books and journals on this topic. Home Knowledge Medicine and Healthcare Anatomy Alveolar process Explore chapters and articles related to this topic Introduction to Oral and Craniofacial Tissue Engineering View Chapter Purchase Book Published in Vincenzo Guarino, Marco Antonio Alvarez-Pérez, Current Advances in Oral and Craniofacial Tissue Engineering, 2020 María Verónica Cuevas González, Eduardo Villarreal-Ramírez, Adriana Pérez-Soria, Pedro Alberto López Reynoso, Vincenzo Guarino, Marco Antonio Alvarez-Pérez Root Cementum (RC) is a mineralized tissue that surrounds the superficial root of the tooth; their function is to support the tooth through the PDL and alveolar bone (Yamamoto et al. 2016). Alveolar Bone (AB) is another mineralized tissue and is associated with the formation of membranous bone of both mandibular and maxillary tissues during the development of the first dentition, two components form this kind of bone, the first belong to the alveolar process, which in turn is composed by the cortical and cancellous bone tissue, the last one stores Haversian systems required for maintenance and remodeling of the bone; the second component is the alveolar bone itself which corresponds to the bone portion that covers the dental surface and serves as a union site to the Sharpey fibers from PDL (Chu et al. 2014). Periodontal ligament (PDL) is formed by collagen fibers which could be classified according to their localization of the fibers onto the alveolar crest, oblique, transseptal, horizontal, inter-radicular or apical (Maheaswari et al. 2015). The union of these fibers to the soft tissue provides a natural coupling of the roots of the tooth in the alveolus: the union of the PDL to the RC or the AB facilitates the transfer of loads of the teeth towards the bone, because the bone-cement/PDL-binding sites contain areas between 10–15 μm rich in biochemical gradients, which are known as enthesis sites that facilitate cell-cell interactions and communications (Lee et al. 2015). Experimental Stomatology View Chapter Purchase Book Published in Samuel Dreizen, Barnet M. Levy, Handbook of Experimental Stomatology, 2020 Samuel Dreizen, Barnet M. Levy Both endochondral and periosteal bone formation were inhibited by the deficiency of tryptophan. The alveolar process was affected to a greater degree than the long bones. The mandibular condyle of the deficient animals was smaller in size and had arrested chondrogenesis and osteogenesis, leading to a flattening of the condylar head. Both the radicular and interseptal areas of the alveolar bone were markedly affected. The resulting osteoporosis was characterized by an almost complete disappearance of spongy bone trabeculae. There were localized areas of fragmentation and loss of periodontal attachment fibers in the interradicular bone. Inhibition and retardation of endochondral and periosteal bone formation were the direct result of reduced protein synthesis. This was evidenced by a reduction in width of the calcifying cartilage. All pathologic changes underwent reversal upon repletion with tryptophan. Diseases of the Masticatory Complex View Chapter Purchase Book Published in Lars Granath, William D. McHugh, Systematized Prevention of Oral Disease: Theory and Practice, 2019 Gunnar E. Carlsson, Bengt Ingervall, George A. Zarb Bone resorption in the edentulous alveolar process is inevitable and gradually leads to functional problems in many complete denture wearers.3 The best prevention for this situation is, of course, to avoid extraction of the last few teeth. There are a series of therapeutic alternatives, which all are better than conventional complete dentures. They vary from simple overdentures to sophisticated partial denture constructions which utilize the remaining teeth as abutments for different types of retention.55 Peri-implant bone resorption risk of anterior maxilla narrow single implants: a finite-element analysis View Article Journal Information Published in Biomaterial Investigations in Dentistry, 2022 Ivan Onone Gialain, Leonardo Folmer Rodrigues da Silva, Marlene Kasumi Gantier Takano, Rafael Yagüe Ballester, Marina Guimarães Roscoe, Josete Barbosa Cruz Meira The current study was designed to simulate a unique geometry of an anterior maxillary alveolar process for all implant sizes. Therefore, as the implant diameter decreased, the buccal bone thickness increased (Table 1). Except for this aspect, the computational models enabled a complete control of research variables, which is an important advantage over clinical studies since the effect of the study variable is not adulterated by the effect of uncontrolled ones. However, the observed peri-implant bone resorption risk indexes are valid for a similar clinical scenario, with a favorable anatomy condition, in which bone augmentation is not necessary. In the clinical decision-making process, the dentist needs to evaluate other variables that influence the peri-implant strain and stress, such as intensity of masticatory forces , occlusal pattern, presence of parafunction, implant position and inclination , bone quality , crown-to-implant ratio, the distance between the implant neck plan and the load application point, and the heterogeneity of peri-implant bone properties. Parameter identification for the simulation of the periodontal ligament during the initial phase of orthodontic tooth movement View Article Journal Information Published in Computer Methods in Biomechanics and Biomedical Engineering, 2021 Albert Heinrich Kaiser, Ludger Keilig, Reinhard Klein, Christoph Bourauel The periodontium is a compound of several tissues that support the teeth. It includes the gingiva, the cementum, the periodontal ligament and the alveolar bone proper. According to Hand and Frank (2015), ‘The periodontal ligament attaches the tooth root to alveolar bone, and it serves to absorb and resist the forces of occlusion on the tooth. It consists of collagenous fiber bundles… Interstitial areas containing loose connective tissue, blood vessels, and nerves are present between the fiber bundles in the periodontal ligament. These interstitial areas are continuous with openings through the alveolar bone (Volkmann’s canals) to the marrow spaces of the alveolar process.’ Alveolar bone remodeling after tooth extraction in irradiated mandible: An experimental study with canine model View Article Journal Information Published in Ultrastructural Pathology, 2018 Venni Heinonen, Timo J. Ruotsalainen, Lauri Paavola, Jopi J. Mikkonen, Pekka Asikainen, Arto P. Koistinen, Arja M. Kullaa Bone remodeling in the alveolar process is important for understanding tooth loss associated with periodontal diseases, osteonecrosis of the mandible and bone resorption after tooth extraction. Currently there are many studies focusing on the side-effects of RT and their treatment.13 Volumetric follow-up of bone fill during healing of an extraction socket has been presented in the literature, and it has been demonstrated that the bone healing after tooth extraction in irradiated head and neck cancer patients is delayed.14 However, the role of RT in bone healing after tooth extraction is not completely understood. Scroll to Top Explore Our company Our leadership team Careers Contact us Taylor & Francis Online Routledge.com Our policies Privacy policy Cookies Terms and conditions Accessibility Modern slavery Anti-bribery and corruption Corporate responsibility Gender pay gap report Information for Book Authors Journal Authors Librarians Editors Societies Booksellers Press and media Support Customer support Rights and permissions Royalty recipients Follow us Taylor & Francis Informa Routledge © 2025 Informa plc. 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If you need exact metric measurements, we recommend referencing data sheets and specifications for the products in question. You can also contact us for measurement assistance or for any other questions - simply follow the link at the button below. Related Resources Can you Convert AWG to the Metric System? Solid Conductor AWG Dimensions Chart Amperage Conversion Chart Metric to AWG Conversion Table \| Metric Size (mm2 cross-sectional area) \| AWG Wire Gauge Size \| --- \| \| 0.05 \| 30 \| \| 0.08 \| 28 \| \| 0.14 \| 26 \| \| 0.25 \| 24 \| \| 0.34 \| 22 \| \| 0.38 \| 21 \| \| 0.5 \| 20 \| \| 0.75 \| 19 \| \| 1 \| 18 \| \| 1.5 \| 16 \| \| 2.5 \| 14 \| \| 4 \| 12 \| \| 6 \| 10 \| \| 10 \| 8 \| \| 16 \| 6 \| \| 25 \| 4 \| \| 35 \| 2, 1 \| \| 50 \| 1/0 \| \| 55 \| 1/0 \| \| 70 \| 2/0, 3/0 \| \| 95 \| 4/0 \| \| 120 \| 250MCM \| \| 150 \| 300MCM \| \| 185 \| 350MCM,400MCM \| \| 240 \| 450MCM, 500MCM \| \| 300 \| 600MCM \| \| 400 \| 750MCM, 800MCM \| \| 500 \| 1000MCM \| AWG to Metric Conversion \| AWG Wire Gauge Size \| Metric Size (mm2 cross-sectional area) \| --- \| \| 30 \| 0.05 \| \| 28 \| 0.08 \| \| 26 \| 0.14 \| \| 24 \| 0.25 \| \| 22 \| 0.34 \| \| 21 \| 0.38 \| \| 20 \| 0.5 \| \| 19 \| 0.75 \| \| 18 \| 1 \| \| 16 \| 1.5 \| \| 14 \| 2.5 \| \| 12 \| 4 \| \| 10 \| 6 \| \| 8 \| 10 \| \| 6 \| 16 \| \| 4 \| 25 \| \| 2 \| 35 \| \| 1 \| 50 \| \| 1/0 \| 50 \| \| 2/0 \| 70 \| \| 3/0 \| 95 \| \| 4/0 \| 120 \| \| 250MCM \| 120 \| \| 300MCM \| 150 \| \| 350MCM \| 185 \| \| 400MCM \| 185 \| \| 450MCM \| 240 \| \| 500MCM \| 240 \| \| 600MCM \| 300 \| \| 750MCM \| 400 \| \| 1000MCM \| 500 \| Related Resources Can you Convert AWG to the Metric System? 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3257	https://www.uptodate.com/contents/kidney-disease-in-systemic-sclerosis-scleroderma-including-scleroderma-renal-crisis	Kidney disease in systemic sclerosis (scleroderma), including scleroderma renal crisis - UpToDate Subscribe Sign in English Deutsch Español 日本語 Português Why UpToDate? Product Editorial Subscription Options How can UpToDate help you?Select the option that best describes you Medical Professional Resident, Fellow, or Student Hospital or Institution Group Practice Subscribe Kidney disease in systemic sclerosis (scleroderma), including scleroderma renal crisis View Topic Share Font Size Small Normal Large Bookmark Rate Feedback Tools Formulary drug information for this topic No drug references linked in this topic. Find in topic Formulary Print Share Feedback Font Size Small Normal Large Outline SUMMARY AND RECOMMENDATIONS INTRODUCTION PREVALENCE SCLERODERMA RENAL CRISIS Risk factors Clinical presentation Pathology Diagnosis - Initial evaluation - Establishing the diagnosis - Differential diagnosis Surveillance Prevention Treatment - Goals of therapy - ACE inhibitors in all patients Initial captopril regimen Long-term therapy - Other antihypertensive agents - Monitoring response to therapy - Experimental therapies Prognosis Kidney transplantation - Minimum wait time for transplantation - Graft and patient outcomes - Recurrent disease OTHER TYPES OF KIDNEY DISEASE SOCIETY GUIDELINE LINKS SUMMARY AND RECOMMENDATIONS REFERENCES GRAPHICS Tables - Criteria for acute kidney injury Pictures - Schistocytes on peripheral smear - Scleroderma glomerulus light - HUS vascular I - HUS vascular II - HUS vascular III - HUS IF IV RELATED TOPICS Clinical manifestations and diagnosis of systemic sclerosis (scleroderma) in adults Clinical spectrum of antineutrophil cytoplasmic autoantibodies Cyclosporine and tacrolimus nephrotoxicity Diagnosis of immune TTP Diagnostic approach to adult patients with subacute kidney injury in an outpatient setting Diagnostic approach to suspected TTP, HUS, or other thrombotic microangiopathy (TMA) Kidney effects of ACE inhibitors and ARBs in hypertension Kidney transplantation in adults: Chronic allograft nephropathy Kidney transplantation in adults: Clinical features and diagnosis of acute kidney allograft rejection Major side effects of angiotensin-converting enzyme inhibitors and angiotensin II receptor blockers Moderate to severe hypertensive retinopathy and hypertensive encephalopathy in adults Pathogenesis of systemic sclerosis (scleroderma) Society guideline links: Chronic kidney disease in adults Treatment and prevention of hyperkalemia in adults Kidney disease in systemic sclerosis (scleroderma), including scleroderma renal crisis Authors:John Varga, MDAndrew Z Fenves, MDSection Editors:Richard J Glassock, MD, MACPGary C Curhan, MD, ScDDeputy Editor:Eric N Taylor, MD, MSc, FASN Contributor Disclosures All topics are updated as new evidence becomes available and our peer review process is complete. Literature review current through:Aug 2025. This topic last updated:Oct 14, 2024. Please read the Disclaimerat the end of this page. INTRODUCTION— The pathologic hallmarks of systemic sclerosis (SSc; scleroderma) are diffuse fibrosis (uncontrolled accumulation of collagen) and widespread sclerosis (thickening and narrowing) of small- and medium-sized vessels. (See "Pathogenesis of systemic sclerosis (scleroderma)".) SSc may affect only the skin and subjacent tissues but is generally associated with systemic involvement. (See "Clinical manifestations and diagnosis of systemic sclerosis (scleroderma) in adults", section on 'Clinical features'.) The most serious kidney manifestation of SSc is scleroderma renal crisis (SRC), which occurs in a minority of patients. A review of SSc-associated kidney disease will be presented here, with the emphasis on SRC. Extrarenal manifestations of SSc are discussed separately. (See "Clinical manifestations and diagnosis of systemic sclerosis (scleroderma) in adults", section on 'Major organ involvement'.) PREVALENCE— Autopsy studies reveal kidney pathology in approximately 60 percent of patients with SSc . In addition, as many as 50 percent of patients with SSc have markers of kidney disease, as manifested by mild proteinuria, elevated serum creatinine concentration, and/or hypertension [2,3]. However, many of these findings can be attributed to causes other than SSc. The most serious kidney manifestation of SSc is SRC. This life-threatening complication occurs in as many as 5 to 20 percent of patients with diffuse cutaneous SSc [3-10]. SRC occurs in only 1 to 2 percent of patients with limited cutaneous SSc . However, since limited cutaneous SSc is the more common form of the disease, a substantial proportion of patients with SRC have limited cutaneous SSc [8,9]. Antineutrophil cytoplasmic antibody (ANCA)-related glomerulonephritis develops rarely, in less than 1 percent of patients with SSc [11,12]. (See 'Other types of kidney disease' below.) SCLERODERMA RENAL CRISIS— SRC is a life-threatening complication of SSc. Prior to the widespread use of angiotensin-converting enzyme (ACE) inhibitors, almost all patients with SSc who had significant kidney involvement died within one year [13,14]. Risk factors—A number of risk factors for SRC have been identified: ●Diffuse skin involvement–The most important risk factor for SRC is diffuse skin involvement, particularly if it is rapidly progressive [5,7-9,15]. A study of 826 patients with diffuse cutaneous SSc found that rapid progression of skin thickness was an independent predictor of early SRC (odds ratio [OR] 2.05, 95% CI 1.10-3.85) . The presence of palpable tendon friction rubs, common in patients with SSc who have diffuse skin involvement, is associated with a greater than twofold increase in the risk for SRC . SRC occurs infrequently in patients with limited cutaneous SSc and only rarely in patients with SSc who lack skin involvement (ie, systemic sclerosis sine scleroderma) . ●Glucocorticoid use– The use of glucocorticoids, particularly in high doses, is associated with the development of SRC [8,9,15,19-22]. In a case-control study of 110 patients with SSc, moderate- to high-dose glucocorticoid therapy (≥15 mg/day of prednisone or equivalent) in the preceding six months was associated with a markedly increased risk of SRC (OR 4.37, 95% CI 2.03-9.43) . Other studies have found similar results, with approximately 60 percent of patients with SRC having had prior recent exposure to glucocorticoids [8,9,23]. Higher doses of glucocorticoids may cause salt and volume retention, the initiation or worsening of hypertension, and the triggering of SRC in a subset of these patients. Glucocorticoids may also stimulate increased expression of endothelin-1 receptors in the kidney. ●Serum autoantibodies – The presence or absence of certain serum autoantibodies appears to predict the risk for SRC: •Autoantibodies directed against ribonucleic acid (RNA) polymerase III are associated with a higher risk of SRC [8,24-26]. In a study of patients with SSc, anti-RNA polymerase III autoantibodies were detected in the serum of 59 percent of 96 patients who developed SRC compared with only 12 percent of 735 patients who did not develop SRC . Another analysis of 1029 patients with SSc reported anti-RNA polymerase III autoantibodies in 52 percent of those who developed SRC . •By contrast, anticentromere antibodies are associated with lower risk of SRC . Anticentromere autoantibodies were observed in 1.8 percent of patients who developed SRC versus 29 percent in patients who did not. ●Cyclosporine – There is anecdotal evidence that cyclosporine, a renal vasoconstrictor, may accelerate kidney disease in patients with SSc. In one report, acute kidney injury (AKI) developed in three of eight patients treated with cyclosporine . However, a causal relation to the drug, rather than to SSc itself, could not be determined. (See "Cyclosporine and tacrolimus nephrotoxicity".) ●Other risk factors – Additional factors that may identify patients with SSc who are at increased risk for SRC include contractures at the large joints, new-onset anemia, and new cardiac events such as heart failure or pericardial effusion [5,15,24,28]. Clinical presentation—SRC is a relatively early complication of SSc that almost invariably occurs within the first five years after the onset of the disease.In a series of 110 cases of SRC, SRC occurred at a median duration of 7.5 months from the first non-Raynaud clinical manifestation of SSc . In some cases, SRC may even be the initial manifestation of SSc [29,30]. As described above, rapidly advancing skin induration is a major risk factor for SRC. However, SRC can occur in the absence of warning signs. (See 'Risk factors' above.) SRC is characterized by the following features [4,5,7,23]: ●Abrupt onset of moderate to marked hypertension, sometimes accompanied by manifestations of malignant hypertension such as hypertensive retinopathy (hemorrhages and exudates) and hypertensive encephalopathy [23,31]. In a review of 145 patients with SRC, 85 percent had new diastolic hypertension, with a mean peak blood pressure of 178/102 mmHg . The acute blood pressure elevation in SRC is typically associated with an increase in plasma renin activity. In approximately 10 percent of patients, SRC occurs in the absence of hypertension . However, some of these patients have blood pressures that are higher than their baseline values (eg, 130/80 mmHg in a young, thin woman whose baseline value was 100/60 mmHg). ●Acute kidney injury. ●A normal urine sediment. Glomerulonephritis is not a feature of SRC, and, accordingly, microscopic hematuria or cellular casts are uncommon. Heavy proteinuria is not a typical finding in SRC. Other findings that may be associated with SRC reflect both the underlying vasculopathy and the marked hypertension. Laboratory parameters indicative of a thrombotic microangiopathy may be present, such as thrombocytopenia, anemia, elevated lactate dehydrogenase (LDH), low haptoglobin, and schistocytes on peripheral blood smear (picture 1). Additional clinical manifestations of malignant hypertension may include heart failure, pericardial effusion, headache, and seizures. Pathology—The primary histopathologic changes in the kidney are localized in the small arcuate and interlobular arteries and the glomeruli [5,32]. The characteristic finding is intimal proliferation and thickening that leads to narrowing and obliteration of the vascular lumen, with concentric "onion-skin" hypertrophy. These histopathologic findings are similar to SSc-associated vascular lesions found in other organs (picture 2A-E). SRC is a thrombotic microangiopathy similar to malignant hypertension (formerly called malignant nephrosclerosis), thrombotic thrombocytopenic purpura (TTP), hemolytic uremic syndrome (HUS), radiation nephritis, chronic kidney transplant rejection, and the antiphospholipid antibody syndrome. Because of the nonspecific kidney histologic findings common to all these entities, a kidney biopsy does not definitively establish the diagnosis of SRC. Diagnosis Initial evaluation—The diagnosis of SRC should be considered in all patients with SSc who present with acute kidney injury, particularly if the patient has a concomitant sudden increase in blood pressure. Because acute kidney injury in patients with SSc is not always due to SRC, patients should be evaluated thoroughly for other causes of acute kidney injury using the same approach as that for patients without SSc. This is discussed in more detail elsewhere. (See "Diagnostic approach to adult patients with subacute kidney injury in an outpatient setting".) In addition, patients suspected of SRC should be evaluated for the presence of a thrombotic microangiopathy with a complete blood count, a peripheral blood smear, and measurement of serum LDH and haptoglobin. (See "Diagnostic approach to suspected TTP, HUS, or other thrombotic microangiopathy (TMA)".) Establishing the diagnosis—SRC is a clinical diagnosis based upon the presence of characteristic features in patients with SSc. Kidney biopsy is generally not required to diagnose SRC, but sometimes may be helpful by ruling out other diagnoses and/or by providing useful prognostic information. (See 'Pathology' above.) Although there is no generally accepted or validated definition of SRC, diagnostic and classification criteria have been proposed [33,34]. We take the following approach to the diagnosis of SRC: ●In patients with SSc who have any high-risk features (ie, acute kidney injury occurring within five years of SSc onset, autoantibodies to RNA polymerase III, or moderate- to high-dose glucocorticoid therapy [≥15 mg/day of prednisone or equivalent] in the preceding six months), we make the diagnosis of SRC if the patient meets the following two criteria: •Acute increase in blood pressure defined as any of the following: new onset of elevated systolic blood pressure ≥140 mmHg or diastolic blood pressure ≥90 mmHg in a patient with previously normal blood pressure, an increase in systolic blood pressure of ≥30 mmHg above baseline, or an increase in diastolic blood pressure ≥20 mmHg above baseline. Blood pressure should be checked at least twice, with measurements separated by at least five minutes. •Acute kidney injury, as defined by Kidney Disease Improving Global Outcomes (KDIGO) guidelines (table 1), not explained by other causes. ●In patients with SSc who do not have any high-risk features, we make the diagnosis of SRC if the patient meets the two criteria above plus at least one of the additional supportive findings below: •Evidence of microangiopathic hemolytic anemia and thrombocytopenia, as suggested by the following: new or worsening anemia not due to other causes, the presence of schistocytes on a peripheral blood smear, thrombocytopenia, elevated LDH, and/or low haptoglobin. •Evidence of target organ dysfunction, such as: -Hypertensive retinopathy (see "Moderate to severe hypertensive retinopathy and hypertensive encephalopathy in adults", section on 'Clinical manifestations and diagnosis') -Hypertensive encephalopathy (see "Moderate to severe hypertensive retinopathy and hypertensive encephalopathy in adults", section on 'Clinical manifestations and diagnosis') -Acute heart failure •Characteristic changes on kidney biopsy (see 'Pathology' above) ●Rarely, patients with SRC may present without characteristic increases in blood pressure . In patients with SSc who do not meet the blood pressure criteria above, we make the diagnosis of SRC only if the patient has a high-risk feature, acute kidney injury not explained by other causes, and at least one of the following additional findings: evidence of microangiopathic hemolytic anemia and thrombocytopenia, and/or characteristic changes on kidney biopsy. Differential diagnosis—SRC must be distinguished from other forms of thrombotic microangiopathy, particularly TTP and HUS. Patients with TTP/HUS characteristically present with thrombocytopenia, purpura, and prominent microangiopathic hemolysis and may have rapidly progressive kidney failure. The clinical manifestations may mimic SRC, but patients with TTP/HUS have no clinical or serologic signs of SSc. Distinguishing TTP/HUS from SRC at presentation may present a considerable and urgent diagnostic challenge. The diagnosis of TTP/HUS is sometimes suggested by the presence of a potential inciting event, such as diarrhea in children or certain forms of chemotherapy in adults. In addition, TTP, but not SRC, is characterized by an absence or marked reduction (ie, <10 percent) in the serum activity of the metalloprotease ADAMTS13. Kidney biopsy does not distinguish between these disorders. (See "Diagnostic approach to suspected TTP, HUS, or other thrombotic microangiopathy (TMA)" and "Diagnosis of immune TTP".) Surveillance—SRC is a life-threatening complication of SSc. Patients with the diffuse cutaneous form of SSc, autoantibodies to RNA polymerase III, and early-stage disease are at greatest risk for SRC, particularly if skin involvement is advancing and/or tendon friction rubs are present. Close monitoring is most important during the first four to five years of SSc since SRC most frequently occurs during this phase of the disease. (See 'Clinical presentation' above.) The primary rationale for monitoring for SRC is that treatment is more likely to be effective when initiated early, before irreversible kidney injury has occurred. However, no studies to date have evaluated the effectiveness of monitoring patients with SSc for the development of SRC in terms of reducing morbidity and mortality. We use the following surveillance regimen: ●Blood pressure should be measured on a regular basis in all patients with SSc. For patients at high risk for SRC (such as those with early-stage diffuse cutaneous disease, rapidly progressive cutaneous involvement with tendon friction rubs, or the presence of autoantibodies to RNA polymerase III), we advocate daily home blood pressure measurements and, for others, twice-weekly measurements. In patients with SSc, a persistent rise of 15 mmHg in the systolic blood pressure or a 10 mmHg rise in the diastolic blood pressure raises concern and the need for further evaluation, including determination of the serum creatinine. In patients with SSc taking antihypertensive medications, sustained hypertension >140/90 mmHg, which is not readily corrected with dose adjustment and dietary salt restriction, would trigger further evaluation . ●We measure serum creatinine at every clinic visit (eg, every three months) in all patients with SSc. An increase in the serum creatinine from baseline may be a warning sign of impending SRC. Prevention—There are no prospective studies that demonstrate that the avoidance and/or administration of any agent lowers the incidence or severity of SRC. ●Avoidance of glucocorticoids – As previously mentioned, the use of high-dose glucocorticoids is associated with a marked increase in the risk of SRC (see 'Risk factors' above). When glucocorticoid use is unavoidable in patients with SSc (eg, when there is concomitant myositis), we limit the dose of prednisone to <15 mg/day and use it for the shortest possible period. ●No role for ACE inhibitors, ARBs, or calcium channel blockers – We do not use ACE inhibitors, angiotensin receptor blockers (ARBs), or calcium channel blockers (CCBs) for prevention of SRC. The crucial role of ACE inhibitors in the treatment of SRC is discussed below. (See 'ACE inhibitors in all patients' below.) Observational studies suggest the counterintuitive possibility that ACE inhibitors may increase the risk of developing SRC. In a cohort of over 14,000 patients with SSc who were followed for approximately five years, the use of ACE inhibitors was associated with an increased risk of developing SRC (hazard ratio [HR] 2.6, 95% CI 1.7-4.0) . By contrast, the use of ARBs or CCBs was not associated with the development of SRC. In another study of 87 patients with SRC, prior exposure to ACE inhibitors (ie, before the diagnosis of SRC) was associated with an increased risk of death (HR 2.4, 95% CI 1.02-5.75) . It is possible these data indicate that ACE inhibitors delay recognition and/or aggressive treatment of serious kidney complications in patients with SSc. There are no trial data evaluating the impact of ACE inhibition specifically on the risk of SRC. However, one trial compared the effect of quinapril versus placebo on the progression of cutaneous and visceral disease among approximately 200 patients with SSc . Mean baseline creatinine in both groups was 0.9 mg/dL. At a follow-up of approximately three years, compared with placebo, there was a relative increase in creatinine in the quinapril group (OR 3.2, 95% CI 0.9 to 5.5). However, the episodes of increase in creatinine were not diagnosed as SRC. The use of CCBs in SSc was associated with a lower risk of SRC (HR 0.09, 95% CI 0.04 to 0.24) in one observational study of over 400 patients . However, the overall number of SRC events was low. In addition, this association has not been replicated in other studies [21,35,38]. ●Choice of blood pressure medications – Among patients who are already on an ACE inhibitor for treatment of hypertension, our approach varies depending upon their risk for SRC. We replace the ACE inhibitor with another antihypertensive agent only among patients at high risk (ie, those with early-stage diffuse cutaneous disease or the presence of RNA polymerase III antibodies). We do not modify antihypertensive therapy among patients previously on an ARB or a CCB, regardless of their risk for SRC. (See 'Risk factors' above.) Treatment—As a life-threatening condition, SRC is a medical emergency that requires hospital admission for prompt and aggressive treatment. Without intervention, SRC can progress to end-stage kidney disease (ESKD) over a period of one to two months, with death usually occurring within one year . Successful treatment of SRC is dependent upon its initiation before irreversible kidney damage has occurred. Goals of therapy—The mainstay of therapy in SRC is effective and prompt blood pressure control. The principal goal is to return the patient to their previous baseline blood pressure within 72 hours. The optimal antihypertensive agent is an ACE inhibitor. If this blood pressure goal cannot be achieved with ACE inhibitor therapy alone, other antihypertensive agents should be added. (See 'ACE inhibitors in all patients' below and 'Other antihypertensive agents' below.) For patients with SRC and hypertension who present with evidence of central nervous system (CNS) involvement, additional goals for immediate blood pressure lowering are the same as for any patient with hypertensive retinopathy and/or encephalopathy. (See "Moderate to severe hypertensive retinopathy and hypertensive encephalopathy in adults", section on 'Goal of therapy'.) Since the development of hypertension is usually acute in SRC, rapid blood pressure reduction to baseline does not usually carry the risks seen with rapid blood pressure lowering in patients with longstanding hypertension. ACE inhibitors in all patients—For all patients with SRC, we recommend blood pressure control with an angiotensin-converting enzyme (ACE) inhibitor rather than other antihypertensive agents. We use captopril as initial therapy rather than other ACE inhibitors because of greater clinical experience with this agent and because its rapid onset (peak effect at 60 to 90 minutes) and short duration of action permit rapid dose escalation. Other ACE inhibitors (such as enalapril or ramipril) may provide comparable benefit although data are limited. For patients who are unable to take oral medications, we administer a single dose of intravenous (IV) enalaprilat, followed by oral captopril administered through a nasogastric tube. We avoid additional doses of IV enalaprilat due to its longer duration of action. Initial dosing of captopril is discussed below. (See 'Initial captopril regimen' below.) For patients who are unable to take an ACE inhibitor (eg, due to allergy), an ARB is an alternative option. However, it is not known whether ARBs are as effective as ACE inhibitors, since these agents have not been adequately evaluated in the treatment of SRC. Observational data have shown that ACE inhibitors are associated with greater antihypertensive efficacy, better preservation of kidney function, and lower mortality in patients with SRC [8,31,39-41]. Prior to the widespread use of ACE inhibitors, almost no patients with SRC had recovery of kidney function, and almost all died within one year [13,14,21,39]. The benefits of ACE inhibitor therapy are illustrated by the following studies: ●In a prospective cohort study of 108 patients with SRC conducted before and after the availability of ACE inhibitors, treatment with an ACE inhibitor (mostly captopril) was associated with a significantly higher rate of recovery of kidney function . Among patients who survived dialysis for more than three months, 11 of 20 patients who continued ACE inhibitor therapy were able to discontinue dialysis after 3 to 15 months compared with 0 of 15 patients who were not treated with ACE inhibitors (55 versus 0 percent). One-year survival for SRC was 76 percent in patients treated with ACE inhibitors compared with 15 percent in patients treated with other drugs. ●In a retrospective case series of 110 patients with SRC, all but two of whom were treated with ACE inhibitors, 24 of the 72 patients (33 percent) who started dialysis successfully discontinued it . One- and five-year survival rates were 82 and 59 percent, respectively. ●A prospective cohort study of 145 patients with SRC who were continuously treated with ACE inhibitors and followed for 5 to 10 years reported that 55 patients (38 percent) did not require dialysis, and 34 patients (23 percent) who started dialysis successfully discontinued it after 2 to 18 months (mean 8 months) . Initial captopril regimen—Our initial approach to treatment varies with the patient's blood pressure and whether the patient has SRC alone or in combination with CNS involvement (ie, hypertensive encephalopathy and/or retinopathy). ●Hypertension without CNS involvement– For patients with SRC and hypertension who lack evidence of CNS involvement, we begin captopril at a dose of 6.25 to 12.5 mg. We progressively escalate the dose in 12.5 to 25 mg increments at four- to eight-hour intervals until the goal blood pressure is reached (see 'Goals of therapy' above). The maximum captopril dose is 300 to 450 mg/day, usually divided in three daily doses. ●Hypertension with CNS involvement – For patients with SRC and hypertension who have evidence of central nervous system involvement (eg, encephalopathy, papilledema), we administer the same captopril dose escalation regimen as above and, for acute blood pressure control, add a very short-acting parenteral antihypertensive drug such as nitroprusside (see 'Goals of therapy' above). To allow for rapid up-titration of captopril, parenteral antihypertensive agents should be discontinued as soon as possible. We avoid the use of intravenous labetalol (and other beta blockers) in SRC because of the theoretical risk of worsening vasospasm. ●Normotension– For patients with SRC who are normotensive, we begin captopril at a dose of 6.25 mg and, if tolerated, increase the dose to 12.5 mg at the second dose. Further dose escalation should be pursued as needed to lower the patient's blood pressure to baseline and must be done carefully to prevent the development of hypotension. (See 'Goals of therapy' above.) Long-term therapy—We continue ACE inhibitor therapy indefinitely, even if no longer needed for blood pressure control and even if only low doses are tolerated. For long-term therapy, longer acting ACE inhibitors (such as enalapril or ramipril) may be used rather than captopril to improve patient adherence. We do not stop ACE inhibitors in patients who progress to dialysis, since an appreciable proportion will recover sufficient kidney function to discontinue kidney replacement therapy [8,31,42]. The benefits of indefinite ACE inhibitor therapy are unclear. There are no studies examining ACE inhibitor withdrawal in patients with a history of SRC. Other antihypertensive agents—For patients without sufficient blood pressure control despite maximum dose of an ACE inhibitor, we add a dihydropyridine calcium channel blocker such as amlodipine. In addition to their blood pressure lowering efficacy, dihydropyridine calcium channel blockers are first-line pharmacologic therapy for Raynaud phenomenon. Other antihypertensive drugs that can be added, if necessary, include diuretics and alpha blockers. Beta blockers are usuallyavoided in patients with SSc because of the theoretical risk of worsening vasospasm. We avoid combination therapy with ARBs and ACE inhibitors for SRC since multiple studies of patients with other diseases have suggested that patients treated with both an ACE inhibitor and ARB are at higher risk for adverse events compared with those treated with only one agent. (See "Major side effects of angiotensin-converting enzyme inhibitors and angiotensin II receptor blockers", section on 'Increased adverse effects'.) However, there are no studies that specifically address the role of second-line blood pressure therapy in patients with SRC. Monitoring response to therapy—In addition to close monitoring of blood pressure, we assess kidney function, electrolytes, and hemolysis parameters during therapy. Our approach is as follows: ●We measure the serum creatinine concentration daily after initiating an ACE inhibitor. Patients starting therapy commonly show an initial, transient rise in serum creatinine concentration because of the ACE inhibitor-induced fall in efferent arteriolar resistance and intraglomerular pressure (see "Kidney effects of ACE inhibitors and ARBs in hypertension", section on 'Renovascular hypertension'). However, because of the importance of ACE inhibitor therapy for survival and eventual kidney recovery, we do not stop ACE inhibitors even for a progressive rise of serum creatinine. ●We measure serum electrolytes daily after initiating an ACE inhibitor. Patients with acute kidney injury who are treated with an ACE inhibitor are at risk of hyperkalemia. Despite aggressive medical therapy to control hyperkalemia, some patients may require dialysis initiation to tolerate ongoing therapy with an ACE inhibitor or ACE inhibitor dose escalation. (See "Major side effects of angiotensin-converting enzyme inhibitors and angiotensin II receptor blockers", section on 'Hyperkalemia' and "Treatment and prevention of hyperkalemia in adults".) ●After the initial assessment of hemoglobin, platelet count, serum levels of LDH and haptoglobin, and the peripheral blood smear, we recheck these laboratories daily until the hemoglobin stabilizes and the other parameters normalize. Because the degree of microangiopathic hemolysis often reflects the activity of the disease process, laboratory evidence of ongoing hemolysis is an indication to intensify ACE inhibitor therapy, regardless of blood pressure control. Experimental therapies—There are anecdotal reports of the use of other therapies in patients with SRC, but we do not use any of these agents as initial therapy: ●In a single-center retrospective chart review of 20 patients with SRC, 10 with associated microangiopathic hemolytic anemia were treated with plasma exchange therapy in combination with ACE inhibitors . Patients treated with the combination therapy had a better kidney survival rate at one year (80 percent) compared with those who were treated with ACE inhibitors only (45 percent). ●The nonselective endothelin-1 receptor antagonist bosentan was studied in six patients with SRC . When used in combination with ACE inhibitors for six months, bosentan was associated with a trend toward improved blood pressure control and preserved kidney function. ●Intravenous prostacyclin (iloprost or epoprostenol), which is believed to help the microvascular lesion without precipitating hypotension, has been administered for SRC based upon anecdotal observations of benefit. ●Limited evidence has been presented for the use of the complement inhibitor eculizumab as a rescue therapy in SRC [45,46]. However, there are no controlled studies examining the safety and efficacy of complement inhibitors in patients with SRC. Prognosis—Despite the improved prognosis associated with long-term ACE inhibitor therapy, outcomes in SRC remain poor. As an example, approximately 20 to 50 percent of patients with SRC will require dialysis [8,21,31]. The prognosis of SRC may be worse for patients with the following characteristics: ●End-stage kidney disease – Mortality is higher for patients who progress to ESKD. In a prospective cohort study of 145 patients with SRC treated with ACE inhibitors, patients who did not require dialysis or required only temporary dialysis had long-term outcomes similar to patients with diffuse SSc who did not have SRC, with mortality rates of 10 percent at five years and 15 to 20 percent at eight years . By contrast, the mortality rates in patients who required permanent dialysis were approximately 60 percent at five years and 75 percent at eight years. Survival on dialysis in patients with SRC is worse than in other forms of ESKD, as illustrated in a study from the United States Renal Data System that included 364,000 patients with ESKD on maintenance dialysis, 820 of whom had SSc . Two-year survival was significantly lower in patients with SSc (49 percent versus 64 percent in all other patients), even though the patients with SSc were younger. Similar findings have been reported in a study of 342 patients with SSc in Europe . Since vascular access is particularly troublesome in patients with SSc, complications with vascular access may have contributed to the higher mortality. ●Normotensive SRC – Patients with normotensive SRC have worse kidney outcomes and higher mortality than patients with SRC who are hypertensive [19,23]. Worse outcomes in normotensive renal crisis have been attributed to several factors, including delayed recognition of SRC and the possibility that normotensive SRC might indicate cardiac failure in these patients. ●Specific findings on kidney biopsy – Greater numbers of thrombosed vessels, the severity of ischemic glomerular collapse, and peritubular C4 staining observed on kidney biopsy may correlate with poor outcomes . Kidney transplantation—There is limited experience with kidney transplantation in patients with SRC, in part because transplantation is sometimes precluded by the severity of the extrarenal manifestations of SSc [49,50]. The United Network for Organ Sharing (UNOS) database included 260 transplants performed between 1987 and 2004 for the kidney diagnosis of SSc . Minimum wait time for transplantation—We generally do not perform kidney transplantation for at least six months after the initiation of dialysis. Recovery of kidney function can occur in SRC, permitting discontinuation of dialysis in many patients. Since the improvement in kidney function can continue for up to 18 months, decisions regarding kidney transplantation do not have to be made during or immediately following an episode of SRC. Graft and patient outcomes—Kidney allograft survival in patients with SRC is reduced compared with that of transplant recipients with ESKD due to other kidney disorders; however, this difference has diminished as more successful immunosuppression protocols have been utilized [28,42,51-53]. Reported graft survival rates in the United States from the UNOS registry were 68 to 79 percent at one year, 60 to 70 percent at three years, and 57 percent at five years [28,52]. However, these data included some patients who were treated in the pre-cyclosporine era and therefore likely had more episodes of rejection and worse outcomes. A European database of patients transplanted from 2002 to 2013 reported much better one- and five-year graft survival rates (90 and 72 percent, respectively), likely reflecting the use of more effective immunosuppressive strategies . Despite these relatively low allograft survival rates, patient survival after kidney transplantation is superior to that of patients with SRC on dialysis who remain on the waitlist. This was shown in a study of 258 patients with SRC who were listed for kidney transplantation between the years 1985 to 2002 . The one- and three-year patient survival with transplantation was 90 and 80 percent, respectively, compared with 81 and 55 percent in those who remained on the waitlist. Recurrent disease—The incidence of recurrent SRC in the transplanted kidney is difficult to ascertain with certainty because the primary kidney histopathologic changes associated with SRC (mucoid intimal thickening of the interlobular arteries and fibrinoid necrosis in the glomeruli) may be difficult to differentiate from acute or chronic kidney allograft rejection. (See "Kidney transplantation in adults: Clinical features and diagnosis of acute kidney allograft rejection" and "Kidney transplantation in adults: Chronic allograft nephropathy".) However, recurrence rates in transplanted kidneys are low (fewer than 5 percent) [28,52,53]. As an example, in a review from the UNOS database of 260 transplants for SRC performed between 1987 and 2004, recurrent disease developed in five (1.9 percent) . Most recurrences occurred within the first one to two years after transplantation, with many occurring within a few months. Early loss of native kidney function due to SRC appears to be a risk factor for recurrence in the transplanted kidney . In addition, recurrent SRC may be preceded by one or more clinical markers that are predictive of severe SSc, such as progressive skin thickening, new-onset anemia, and cardiac complications such as pericardial effusion or congestive heart failure . The optimal immunosuppressive regimen to avoid recurrent SRC in patients who undergo kidney transplantation is unknown . Transplant centers have successfully used calcineurin inhibitors for maintenance immunosuppression and high-dose glucocorticoids for induction therapy in patients with a history of SRC . However, because of the potential association between cyclosporine use and risk of SRC, it is possible that noncalcineurin inhibitor-based immunosuppressive regimens (eg, belatacept) may be superior for allograft patients with SSc. Because glucocorticoids are an established risk factor for SRC, we avoid their use at high doses (ie, >20 mg prednisone per day) in transplant patients with SSc whenever feasible. (See 'Risk factors' above.) We continue treatment with ACE inhibitors indefinitely after transplantation. Recurrent SRC in the transplant may follow a similar course as the primary disease. The treatment of recurrent SRC is the same as that for disease in the native kidney. (See 'Treatment' above.) OTHER TYPES OF KIDNEY DISEASE— Kidney disease in patients with SSc is not always due to SRC. This is particularly true in patients with few risk factors for SRC. (See 'Risk factors' above.) Acute kidney injury in SSc may reflect prerenal disease associated with heart failure, pulmonary hypertension, nonsteroidal antiinflammatory drugs, diuretics, or hypovolemia due to gastrointestinal involvement and/or malabsorption . Rarely, patients with SSc present with antineutrophil cytoplasmic antibody (ANCA)-related crescentic glomerulonephritis [55,56]. The diagnosis is confirmed by kidney biopsy showing a pauci-immune crescentic glomerulonephritis. (See "Clinical spectrum of antineutrophil cytoplasmic autoantibodies", section on 'Disease associations'.) SOCIETY GUIDELINE LINKS— Links to society and government-sponsored guidelines from selected countries and regions around the world are provided separately. (See "Society guideline links: Chronic kidney disease in adults".) SUMMARY AND RECOMMENDATIONS ●Prevalence – As many as 50 percent of patients with systemic sclerosis (SSc; scleroderma) have mild proteinuria, elevated serum creatinine concentration, and/or hypertension. However, many of these findings can be attributed to causes other than SSc. Scleroderma renal crisis (SRC) occurs in 5 to 20 percent of patients with diffuse cutaneous SSc and in 1 to 2 percent of patients with limited cutaneous SSc. (See 'Prevalence' above.) ●Scleroderma renal crisis– SRC is a life-threatening complication of SSc that almost always occurs within the first five years of disease onset. •Risk factors–The most important risk factor for SRC is diffuse skin involvement, particularly if it is rapidly progressive. Other risk factors include glucocorticoid use and autoantibodies directed against RNA polymerase III. (See 'Risk factors' above.) •Clinical presentation – SRC is characterized by acute kidney injury (AKI), abrupt onset of moderate to marked hypertension, and a normal urine sediment without heavy proteinuria. Evidence of a microangiopathic hemolytic anemia may be present (picture 1). Although approximately 10 percent of patients with SRC are normotensive, some of these patients have a substantial increase in blood pressure from baseline. (See 'Clinical presentation' above.) •Diagnosis – SRC is a clinical diagnosis based upon characteristic features in patients with SSc and the exclusion of other causes of AKI. Kidney biopsy is generally not required. Our approach to diagnosis incorporates the presence or absence of specific high-risk features, blood pressure criteria, and, as necessary, additional supportive findings. SRC must be distinguished from other forms of thrombotic microangiopathy, particularly thrombotic thrombocytopenic purpura and hemolytic-uremic syndrome. (See 'Diagnosis' above.) •Surveillance and prevention – We advocate monitoring for SRC with regular assessments of home blood pressure and serum creatinine. We avoid glucocorticoid use in patients with SSc whenever possible. We do not use angiotensin-converting enzyme (ACE) inhibitors, angiotensin receptor blockers (ARBs), or calcium channel blockers (CCBs) to prevent SRC. Among patients previously taking an ACE inhibitor for hypertension, we replace it with an alternative agent only if they are at high risk for SRC. (See 'Surveillance' above and 'Prevention' above.) •Treatment – SRC is a medical emergency that requires hospital admission. We use the treatment approach below: -Goals of therapy – The principal goal is to return the patient to their previous baseline blood pressure within 72 hours. For patients with SRC and hypertension who present with evidence of central nervous system (CNS) involvement, additional goals for immediate blood pressure lowering are the same as for any patient with hypertensive retinopathy and/or encephalopathy. (See 'Goals of therapy' above.) -ACE inhibitors – For all patients with SRC, we recommend blood pressure control with an ACE inhibitor rather than other antihypertensive agents (Grade 1B). We use captopril because its rapid onset and short duration of action permit rapid dose escalation. For patients who are unable to take an ACE inhibitor, an ARB is an alternative option. Initial dosing and duration of ACE inhibitor therapy are discussed above. (See 'ACE inhibitors in all patients' above.) -Additional antihypertensive agents – For patients without sufficient blood pressure control despite maximum dose of an ACE inhibitor, we add a dihydropyridine calcium channel blocker such as amlodipine. Beta blockers are usually avoided in patients with SSc because of the theoretical risk of worsening vasospasm. (See 'Other antihypertensive agents' above.) -Monitoring response to therapy – We follow kidney function, electrolytes, and laboratory markers of microangiopathic hemolysis during treatment. Because of the importance of ACE inhibitor therapy for survival and eventual kidney recovery, we do not stop ACE inhibitors even for a progressive rise of serum creatinine. (See 'Monitoring response to therapy' above.) •Kidney transplantation – Kidney transplantation is an option for patients with SRC who progress to end-stage kidney disease. We do not perform kidney transplantation for at least six months after the initiation of dialysis, given the chance of recovery of kidney function. (See 'Kidney transplantation' above.) REFERENCES 1. D'Angelo WA, Fries JF, Masi AT, Shulman LE. Pathologic observations in systemic sclerosis (scleroderma). 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Scleroderma at end stage renal disease in the United States: patient characteristics and survival. J Nephrol 2002; 15:236. 48. Batal I, Domsic RT, Shafer A, et al. Renal biopsy findings predicting outcome in scleroderma renal crisis. Hum Pathol 2009; 40:332. 49. Paul M, Bear RA, Sugar L. Renal transplantation in scleroderma. J Rheumatol 1984; 11:406. 50. Merino GE, Sutherland DE, Kjellstrand CM, et al. Renal transplantation for progressive systemic sclerosis with renal failure: case report and review of previous experience. Am J Surg 1977; 133:745. 51. Bleyer AJ, Donaldson LA, McIntosh M, Adams PL. Relationship between underlying renal disease and renal transplantation outcome. Am J Kidney Dis 2001; 37:1152. 52. Gibney EM, Parikh CR, Jani A, et al. Kidney transplantation for systemic sclerosis improves survival and may modulate disease activity. Am J Transplant 2004; 4:2027. 53. Chang YJ, Spiera H. Renal transplantation in scleroderma. Medicine (Baltimore) 1999; 78:382. 54. Bertrand D, Dehay J, Ott J, et al. Kidney transplantation in patients with systemic sclerosis: a nationwide multicentre study. Transpl Int 2017; 30:256. 55. Arnaud L, Huart A, Plaisier E, et al. ANCA-related crescentic glomerulonephritis in systemic sclerosis: revisiting the "normotensive scleroderma renal crisis". Clin Nephrol 2007; 68:165. 56. Kamen DL, Wigley FM, Brown AN. Antineutrophil cytoplasmic antibody-positive crescentic glomerulonephritis in scleroderma--a different kind of renal crisis. J Rheumatol 2006; 33:1886. Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. It is not meant to be comprehensive and should be used as a tool to help the user understand and/or assess potential diagnostic and treatment options. It does NOT include all information about conditions, treatments, medications, side effects, or risks that may apply to a specific patient. 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Topic 7183 Version 41.0 Topic Feedback Tables Criteria for acute kidney injury Criteria for acute kidney injury Pictures Peripheral smear in microangiopathic hemolytic anemia showing presence of schistocytesLight microscopy of glomerulus in scleroderma renal crisisLight microscopy showing thrombotic microangiopathy with subintimal fibrin deposition in an interlobular arteryLight microscopy showing thrombotic microangiopathy with mucoid intimal thickening of a muscular renal arteryLight microscopy showing thrombotic microangiopathy with onion-skin thickening of a muscular renal arteryImmunofluorescence microscopy showing fibrin deposition in thrombotic microangiopathy Peripheral smear in microangiopathic hemolytic anemia showing presence of schistocytesLight microscopy of glomerulus in scleroderma renal crisisLight microscopy showing thrombotic microangiopathy with subintimal fibrin deposition in an interlobular arteryLight microscopy showing thrombotic microangiopathy with mucoid intimal thickening of a muscular renal arteryLight microscopy showing thrombotic microangiopathy with onion-skin thickening of a muscular renal arteryImmunofluorescence microscopy showing fibrin deposition in thrombotic microangiopathy Company About Us Editorial Policy Testimonials Wolters Kluwer Careers Support Contact Us Help & Training Citing Our Content News & Events What's New Clinical Podcasts Press Announcements In the News Events Resources UpToDate Sign-in CME/CE/CPD Mobile Apps Webinars EHR Integration Health Industry Podcasts Follow Us Sign up today to receive the latest news and updates from UpToDate. 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3258	https://www.geeksforgeeks.org/maths/pascals-triangle-formula/	Pascal's Triangle Formula - 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 Number System and Arithmetic Algebra Set Theory Probability Statistics Geometry Calculus Logarithms Mensuration Matrices Trigonometry Mathematics Sign In ▲ Open In App Pascal's Triangle Formula Last Updated : 23 Jul, 2025 Comments Improve Suggest changes 8 Likes Like Report Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The triangle starts with a "1" at the top, and each subsequent row contains the coefficients of the binomial expansion. Here is an example of Pascal's Triangle: The formula to calculate any element in Pascal's Triangle is based on binomial coefficients, denoted as: (n k)=n!k!(n−k)!\dbinom{n}{k} = \dfrac{n!}{k!(n - k)!}(k n)=k!(n−k)!n! Pascal's triangle is a beautiful concept of probability developed by the famous mathematician Blaise Pascal, which is used to find coefficients in the expansion of any binomial expression. It is a method to know the binomial coefficients of terms of binomial expression (x + y)n, where n can be any positive integer and x, and y are real numbers. This triangle is used in different types ofprobabilityconditions. Here each row represents the coefficient of expansion of (x + y)n. (x + y)0 = 1 (x + y)1 = 1x + 1y (x + y)2 = 1x 2 + 2xy + 1y 2 (x + y)3 = 1x 3 + 3x 2 y + 3xy 2 + 1y 3 (x + y)4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3+1y 4 (x + y)5 = 1x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5xy 4 + 1y 5 Here the power of y in any expansion of (x + y)n represents the column of Pascal's Triangle. n represents the row of Pascal's triangle. Row and column are 0 indexed in Pascal's Triangle. Pascal's Triangle Construction It's quite simple to make a Pascal's Triangle. Start by creating the top row (the 0th row) with just the number 1. In the following rows, each new number in Pascal's Triangle is the sum of the two numbers directly above it. For example, to find the number in row 4, column 2: Add the numbers from row 3, column 1, and row 3, column 2. So, the number in row 4, column 2 is 1 + 2 = 3. Pascal's Triangle Construction Properties of Pascal's Triangle Each number in Pascal's Triangle is the sum of two numbers above it. Numbers in a row are symmetric. Each number represents a binomial coefficient. Numbers on the left and right sides of the triangle are always 1. nth row contains (n+1) numbers in it. Read more: Important Facts about Pascal's Triangle Pascal's Triangle Other Formulas The Pascal's Triangle formula to find the element in the n-th row and k-th column of the triangle is: (p q)=(p−1 q−1)+(p−1 q)\binom{p}{q} = \binom{p-1}{q-1} + \binom{p-1}{q}(q p)=(q−1 p−1)+(q p−1) Here, 0 ≤ q ≤ p, p is a non-negative number Alternatively, the formula to find the number in the nnn-th row and r-th column is given by: (p q)=p!q!(p−q)!\binom{p}{q} = \frac{p!}{q!(p-q)!}(q p)=q!(p−q)!p! And another recursive expression is: (p q)=(p q−1)+(p−1 q−1)\binom{p}{q} = \binom{p}{q-1} + \binom{p-1}{q-1}(q p)=(q−1 p)+(q−1 p−1) Pascal's Triangle Binomial Expansion As we already know tascal's triangle defines the binomial coefficients of terms of binomial expression (x + y)n, So the expansion of (x + y)n is: (x + y)n = a 0 x n + a 1 x n-1 + ......a n-1 xy n-1 + a n y n Practice Questions on Pascal's Triangle Question 1: Find the coefficient of the term x2yin the expansion of(x + y)3. Solution: Method 1: We look at the row 3rd row of Pascal's Triangle because n is 3 and 1st column of the Pascal's Triangle because power of y is 1 in the term x 2y. So the coefficient is 3. Method 2: We simply apply n C r where n = 3, r = 1. So coefficient of x 2y in the expansion of(x + y)3 is 3 C 1 = 3 Question 2: Find the coefficient of the term x2y2in the expansion of(4x + 3y)4. Solution: Method 1: We look at the row 4th row of Pascal's Triangle because n is 4 and 2nd column of the Pascal's Triangle because power of y is 2 in the term x 2 y 2. So number in Pascal's Triangle is 6. But we see that coefficient of x is 4 and y is 3 now since power of x is 2 and y is 2 in the term x 2 y 2 so pascal Triangle number will be multiplied by 4 2 and 3 2 to find the coefficient. Coefficient = 6 x 4 2 x 3 2 = 864 Method 2: We simply apply n C r where n = 4, r = 2. So Pascal Triangle number of term x 2 y 2 in the expansion of(4x +3y)4 is 4 C 2 = 6. But we see that coefficient of x is 4 and y is 3 now since power of x is 2 and y is 2 in the term x 2 y 2 so pascal Triangle number will be multiplied by 4 2 and 3 2 to find the coefficient. Coefficient =6 x 4 2 x 3 2 = 864 Question 3: Write the 6th row of the Pascal's Triangle Solution: 6th row can be written as : 6 C 0 6 C 1 6 C 2 6 C 3 6 C 4 6 C 5 6 C 6 1, 6, 15, 20, 15, 6, 1 Question 4: Find the coefficient of the term x4in the expansion of(2x + y)4. Solution: Method 1: We look at the row 4th row of Pascal's Triangle because n is 4 and 0th column of the Pascal's Triangle because power of y is 0 in the term x 4. So number in Pascal's Triangle is 1. But we see that coefficient of x is 2 and y is 0 now since power of x is 4 and y is 0 in the term x 4 so Pascal Triangle's number will be multiplied by 2 4 and 1 0 to find the coefficient. Coefficient =1 x 2 4 x 1 0= 16 Method 2: We simply apply n C r where n = 4, r = 0. So Pascal Triangle number of term x 4 in the expansion of(2x + y)4 is 4 C 0 = 1. But we see that coefficient of x is 2 and y is 0 now since power of x is 4 and y is 0 in the term x 4 so Pascal Triangle's number will be multiplied by 2 4 and 1 0 to find the coefficient. Coefficient = 1 Question 5: Find the coefficient of the term xy2in the expansion of(2x + y)3. Solution: Method 1: We look at the row 3rd row of Pascal's Triangle because n is 3 and 2nd column of the Pascal's Triangle because power of y is 2 in the term xy 2. So number in Pascal's Triangle is 3. But we see that coefficient of x is 2 and y is 1 now since power of x is 2 and y is 1 in the term xy 2 so Pascal Triangle's number will be multiplied by 2 1 and 1 2 to find the coefficient. Coefficient =3 x 2 1 x 1 2 = 6 Method 2: We simply apply n C r where n = 3, r = 2. So Pascal Triangle number of term xy 2 in the expansion of(2x + y)3 is 3 C 2 = 3. But we see that coefficient of x is 2 and y is 1 now since power of x is 2 and y is 1 in the term xy^2 so Pascal Triangle's number will be multiplied by 2 1 and 1 2 to find the coefficient. 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3259	https://greenbugenergy.com/get-educated-knowledge/power-versus-energy	Power versus energy - GreenBug Energy - micro hydro Skip to content ▼ home Get Inspired Dare Mighty Things. Project Case Studies Micro Hydro Sites Multimedia Get Educated Learn more about waterpower. knowledge skills guides issues Take Action Get started today. Purchase a micro hydro system Lease Your Waterpower Site Recommend us Be an advocate Get Your Site Assessed Events Invest Connect with others Learn from or share with others connect on Facebook Share what you're doing to help the cause Attend Events In The News The latest news and updates. Glossary explaining terms, expressions and unusual words FAQ Questions, answers, action. Document Library All the documents available to download from our website are right here in one place. about us Elevator Pitch Get the essential information – quickly Core values GreenBug Energy’s work is governed by its’ core values. 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Power is measured in watts (W) or kilowatts (kW). 1 Watt (W)= joule per second 1 Kilowatt (KW) = 1000 watts 1 Magawatt (MW) = 1000 kilowatts 1 Horsepower = 745.699872 watts When you’re discussing waterpower and say “Power” you have to know what “Power” you’re measuring or talking about because there are 3 spots to measure power. The hydraulic power – or power in the water. The mechanical power – the turbine changes the hydraulic power to mechanical power at the output shaft of the turbine. The mechanical power will be less than the hydraulic power because of efficiency losses due to friction etc. The electrical power – the generator changes or converts the mechanical power in the rotating turbine shaft into electrical power. As power is converted from one form to another some power is lost at each stage. [dropcap3]1[/dropcap3]Hydraulic Power Calculation To calculate hydraulic power you need to measure Head (meters) and Flow (liters per second). Hydraulic Power = Head(meters) x Flow(lps) x 9.81 Hydraulic Power = Watts or Hydraulic Power = Heat (meters) x Flow (cubic meters per second) x 9.81 Hydraulic Power = kilowatts The greater the head and flow the greater the power. [titled_box title=”Where did this formula come from?”] The energy released by a falling body of water = weight x distance And weight is the downward force the water exerts which = mass x acceleration due to gravity. And mass = density x volume Therefore put those all together and you get: Hydraulic Power = Joules per second = Watts = Density x Volume x acceleration due to gravity x distance. And density of water = 1000 kg per cubic meter. And volume of water = Q = cubic meters (m 3) (use cubic meters, not liters per second because density of water is being measured in terms of cubic meters) And acceleration due to gravity = 9.81 meters per second squared (m⁄s 2) If you assemble all this together in the formula again Hydraulic Power = Watts = 1000 (kg⁄m 3) x Q (m3) x 9.81 (m⁄s 2) x H (m) Watts = 9810 QH kilowatts = 9.81QH, where Q is cubic meters per second and H is meters of head.[/titled_box] Note: Head referred to above is the vertical distance between upper and lower water levels (and there are some rules for proper measurement of this distance). However, the turbine may not be installed at the current upper water level height. If it’s installed lower then the useful head is less. This can be referred to as the “Net Head” or “Useful Head”. Therefore we have to recalculate the hydraulic power available to the turbine using useful head. Net Hydraulic Power Available to Turbine= Useful Head (meters) x Flow (lps) x 9.81 = Watts [dropcap3]2[/dropcap3]Mechanical Power Mechanical Power = Net Hydraulic Power (watts) x Turbine Efficiency (%) Mechanical Power = Watts [dropcap3]3[/dropcap3]Electrical Power Electrical Power = Mechanical Power (watts) x Generator Efficiency (%) Electrical Power = Watts Energy is the unit of power used over a period of time. Energy = Power x period for which power is used i.e. kWh= power in kW x hours for which the power is used or produced. i.e. 1 kWhs = 1 kw of power used or produced for 1 hour. Privacy Policy\|Security Statement\|Legal Information\|Site Map © 2020 GreenbugEnergy.com Original text Rate this translation Your feedback will be used to help improve Google Translate
3260	https://www.youtube.com/watch?v=ahUOM_ieB14	Magnetic Field of Rotating Disk 1.avi Viken Kiledjian 7850 subscribers 231 likes Description 27516 views Posted: 5 Apr 2011 This video illustrates how to derive the equation of the magnetic field of a rotating disk and check what its behavior is near and far from the disk. 19 comments Transcript: we're going to calculate the magnetic field of a rotating disc so let's say the disc is a solid charged disc rotating like this and I want to calculate the magnetic field uh along the X the axis of the disc a distance D away from the center of the disk so what we're going to do is we're going to use a result that we have proven earlier the magnetic field of a current carrying wire with a current I and a radius of R so if it's just one wire like that and has radius R the magnetic field along the along its Central axis we had gotten this result um B was equal to mu0 i r 2 over 2 R 2 + x² to the 3 power so so we had seen that that equation actually makes sense because when we set the x equal to zero we get um magnetic field is equal to mu0 I over when x = 0 we have R 2 to R Cub you get mu0 I over 2 R which we know is the magnetic field of a coil right and if it has n turns it's n mu0 i over 2 R so the equation makes sense so now we're going to use that as the basis of a solid disc so let's say the solid disc has a charge Q rotating at a rate of Omega Omega is the radians per second so uh what we could do here is say okay the solid disc is made up of a bunch of uh circular charge elements DQ right which are rotating uh per given time with a Time t with a period T so the current they kind of basically create a current because what is current it's charge divided by time right so the current di created by that one element there DQ divided by the period T that that's equal to the DI so uh from uh physics we also know that uh Omega is equal to 2 pi F which equals 2 piun / period therefore the period equals 2 pi/ Omega so therefore T di is equal to uh DQ / 2 pi/ Omega therefore di I equals Omega DQ over 2 pi so that creates its own little current and the magnetic field of that current will be DB so I can go to this equation here and we can call this DB and then we can call this di right and then we can say uh DB is equal to d r 2 so DB is equal to mu0 and then di is going to equal my Omega DQ R 2 over uh 2 pi so two you already have the two here 2 2 pi that's going to be 4 pi and then you have r^ 2 + x^2 to the 3 okay the next thing that we do is going to be uh to say okay DQ is equal to Sigma da the surface charge density so now I'm going to do it with if the surface charge density is uniform if the surface charge density is not not uniform then you can do techniques that I've shown earlier with uh potential and electric field of the discs where the surface charge density is nonuniform so uh you can always put that function there for whatever the surface charge density is so let's assume it's uniform now so what's my da da is 2 pi r Dr uh it's the circumference of the element times the thickness of the element so you have Sigma is 2 pi RDR that goes into here DQ is equal to Sigma 2i R Dr R okay now I already have an r s okay so that's going to make this R cubed right because it was 2 pi RDR and then I already have an R 2 then on the bottom I have 4 piun R 2 + X2 to the 3es now I want to integrate this from what from zero to some Big R zero to some Big R the whatever the radius of the disc is my X here is the same thing as what I meant here distance D I'm not integrating along the X because I'm trying to find the magnetic field at a certain point I'm integrating along the RS right so my X is just a constant it's just D okay I mean I could keep it as X probably appears better um okay so now let U equal R 2 + x^2 du = 2 R Dr r r r^ 2 is = to u - x^2 okay so magnetic field here a lot of things are constant here 2 pi cancels the 4 Pi gives 2 mu0 Omega Sigma mu 0 Omega Sigma / 2 cancels out then you have integral 0 to R and then you have here Rd R right right RDR well I can split this up into r s and RDR right so it was R Cub I I'm splitting it up RDR is equal to d/2 okay this half came from came out already so this is D over2 and then I have an extra R 2 which is equal to u- X2 okay and then on the bottom I have what this whole thing is u u to the 3es and then the limits of the integral change so when little r is equal to z u is equal to X2 when little r is equal to Big R uh U is equal to r^ 2 + X2 okay so this is actually a doable integral because magnetic field mu0 Omega Sigma over 4 and then integral of uh this one is going to be U to the half uh U to the 2 halves right it's a one so it's going to be U to the2 minus X2 U to the -3 d uh du equation here for the magnetic field of a disc and then the integral of U the2 is U half over half U the half over half and then integral of u32 is you do the negative half over negative half okay and then the limits of the integral are X2 to R 2 + X2 okay so I could add this these two cancels this makes it a two mu0 Omega Sigma / 2 then you could put the R 2 + X2 square root plus and then you have X2 and then this goes to the bottom square root of r s + X2 then you put this in minus Square < TK of X2 which is x minus this time you have another x s here over and then you're putting the x s on the bottom right this is uh this is uh U to the half on the bottom so when the r you put R2 + x s here you went to the bottom then you have an extra X squ then you did sub subtract X2 when here give you X then you have subtract X2 this one already is x s so you keep it there and then X2 goes into the bottom X2 and then square root of X2 is just X so that's actually going to be what it looks at first as if it's going to cancel but it's really not if you keep track of the signs it's not going to cancel it's going to give you -2X okay so that's probably as simple as you can get it D is equal to Mu Omega now the sigma is the total charge of the disc divided by the surface area so Q over PK R 2 squ < TK of r² + X2 + X2 / < TK R 2 + X2 - 2x okay
3261	https://us.sofatutor.com/math/videos/distance-rate-time-different-directions	Try sofatutor for 30 Days Do you already have an account? Log in: Try sofatutor for 30 Days Discover why over 1.6 MILLION students choose sofatutor! Math Algebra 1 Equations Distance - Rate - Time – Different Directions Distance - Rate - Time – Different Directions Watch videos Start exercises Show worksheets Improve your grades now while having fun Unlock this video in just a few steps, and benefit from all sofatutor content Videos for all grades and subjects that explain school material in a short and concise way. Boost your confidence in class by studying before tests and mock tests with our fun exercises. Learn on the go with worksheets to print out – combined with the accompanying videos, these worksheets create a complete learning unit. 24-hour help provided by teachers who are always there to assist when you need it. Try it for 30 Days Cancel at any time Rating Ø 4.5 / 4 ratings You must be logged in to be able to give a rating. Wow, thank you!Please rate us on Google too! We look forward to it! Go to Google The authors Susan S. Basics on the topic Distance - Rate - Time – Different Directions A boat departs from London and travels at a certain speed, and another boat leaves from Amsterdam. The two boats travel on the same path but in the opposite direction… Does this problem make you seasick? Don’t worry, it’s a distance-rate-time problem with travel from different directions. Use the distance-rate-time formula to solve this kind of problem. The DRT triangle can help you remember the three formulas used to solve problems of how far, how fast, and how long. Distance is equal to rate times time, rate is equal to distance divided by time, and time is equal to distance divided by rate. Which formula you should use depends on which quantity is unknown. Distance, rate, and time problems when travel is from different directions won't be complicated as long as you stay organized, set up a table and enter all the information you know, and use variables to represent the unknown values. Be careful when using different units of time. You cannot calculate minutes and hours in the same problem, so before calculating the distance, rate, or time, convert minutes to hours or hours to minutes. To learn more about how to calculate the distance, rate, or time of travel from different directions, watch this video and see how Johnny Crash uses the DRT formulas to solve his problems. Create equations that describe numbers or relationships. CCSS.MATH.CONTENT.HSA.CED.A.1 Transcript Distance - Rate - Time – Different Directions Tim Carcrashian, true to his name, works for a vehicle crash testing company. His job is to set up the slow-motion-camera used to document car crashes. The set-up of the camera is very expensive and time-consuming. To do the job just right, Tim needs to know exactly when and where the cars will crash. How can we help him? The distance-rate-time pyramid This is a distance-rate-time problem - with cars coming from different directions. For the test crash, Car 1 and Car 2 will drive towards each other. Car 1 will pass Point A going 50 miles per hour, and Car 2 will pass Point B going 70 miles per hour. The distance from Point A to Point B is 1.2 miles. Exactly where and when will the cars crash? Let’s figure out how to solve this problem. Distance is equal to rate times time. This pyramid can help us remember the formula. A shorter, more common, way to say this d = rt. Write equations Ok, let's get started...It's a good idea to use a table to organize the data. First, let's fill in the values with what we already know, the rate for each car. We don’t know the value for time yet, just that it's the same for both cars, so for now, we can represent this with the variable, 't'. We can write two equations to describe what's going on here: d = 50t. And...d = 70t. Let's write that information into the table. Since we know the total distance the two cars will a travel together is 1.2 miles, we can set up an equation and solve for the unknown time. 1.2 = 50t + 70t. Evaluating this we get: t = 1/100 of an hour. Let's fill in the table... Now, we can use the value we just calculated for the time to determine the actual distances both cars will drive before crashing and fill in the table. d = 50(0.01), which is equal to 0.5d = 70(0.01), giving us 0.7 The distance-rate-time formula This is just great! Tim is so happy! Using the distance-rate-time formula, coming from different directions, now, Tim knows Car 1 will drive exactly 1/2 of a mile and Car 2 will drive 7/10 of a mile, then crash, bang, boom! As a bonus, Tim also knows exactly when this will happen. The cars will crash after 36 seconds, which is equal to 1/100 of an hour. He sets up the Slow-Motion-Camera Unit just right. He’s so good at his job, maybe he'll get a promotion and a big raise! Oh no. One week later, Tim learns the test track will be under construction in order to improve the testing conditions. During construction, the right side of the track will be shortened. Instead of the track having a distance of 1/2 of a mile on the left side and 7/10 of a mile on the right side... Calculate a new rate The right side will be only 4/10 of a mile long. Tim doesn't want to change the set-up of the camera. What can he do so the cars will still crash at the same place and time? He can't change the distance or the time, but he can change the speed of Car 2. Let's calculate the new rate using the distance-rate time formula. As before, we have d = rt. To isolate the rate, divide distance by time. r = 0.4 / 0.01. The rate is 40 miles per hour. So, if Car 2 travels on the track at a rate of 40 miles per hour for 4/10 of a mile, after 1/100 of an hour, the two cars will crash in just the right spot. Change the distance Another week goes by... Tim gets an email... The track improvements are all done, and the new plan is to crash test one of the cars driving at a very high speed. Car 2 will be tested driving 90 miles per hour. Tim doesn't have time to change the camera. What should he do? He can't change the speed or the time, but he can change the distance Car 2 will drive. If Car 2 drives 90 miles per hour for 1/100 of an hour, use the distance-rate-time formula to calculate the new distance it must drive so the camera can capture the crash. Let's reuse our formula from before: distance is equal to rate times time. 90(0.01) = 0.9 miles. So Tim can get the perfect camera shot, Car 2 must drive 9/10 of a mile at 90 miles per hour for 1/100 of an hour . Summary Let's summarize: The distance-rate-time pyramid is a handy graphic organizer to help you remember and understand how to solve these kind of problems. t = d/r, r = d/t, d = rt. After all his hard work, Tim can finally watch the crash test. Oh no, that's not test Car 2...That looks a lot like Tim's boss' car... Distance - Rate - Time – Different Directions exercise Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Distance - Rate - Time – Different Directions. Determine when and where the car will crash. Hints Fill in the table: $\begin{array}{\|l\|c\|c\|c\|} \hline &r&t&d\ \hline \text{car 1}&50&t&?\ \hline \text{car 2}&70&t&?\ \hline \end{array}$ Don't forget the mnemonic device. Determine the unknown quantity, and use the appropriate equation: $d=r\times t$ $r=\frac dt$ $t=\frac dr$ The sum of the distances is equal to $1.2$ (miles). Solution The red car travels at $50$ mph, while the blue car travels in the opposite direction at $70$ mph. This is a distance-rate-time-problem with different directions. Let's start with a table: $\begin{array}{\|l\|c\|c\|c\|} \hline &r&t&d\ \hline \text{car 1}&50&t&50t\ \hline \text{car 2}&70&t&70t\ \hline \end{array}$ Use the equation $d=r\times t$. Since the car cannot travel a distance greater than $1.2$ miles, we write the equation: $\begin{array}{rclcl} 50t+70t & = & 1.2 &&\|\text{Combine Like Terms} \ 120t&=&1.2 \ \color{#669900}{\div120} && \color{#669900}{\div120} &&\|\text{Opposite Operations} \ t &=& 0.01 \end{array}$ $t=0.01$ is the time in hours. If we want to know how many seconds this is, we can calculate the conversion. There are $60$ minutes in an hour and $60$ seconds for every minute. Multiply... there are $3,600$ seconds in one hour. $\begin{array}{rcll} \dfrac{60 ~\text{min.}}{1 ~\text{hr.}} \times \dfrac{60 ~\text{sec.}}{1 ~\text{min.}} &=& \dfrac{3,600 ~\text{sec.}}{1 ~\text{hr.}} \end{array}$ $~$ $\dfrac{3,600 ~\text{ sec.}}{1 ~\text{ hr.}}\times 0.01 ~\text{ hr.} = 36 ~\text{ sec.}$ Now we know that the crash will happen after $36$ seconds. To find out where the crash will occur, we use the equation $d=r\times t$. Plugging in the known values gives us: $d = \dfrac{50 ~\text{mi.}} {\text{hr.}} \times 0.01 ~\text{hr.} = 0.5~\text{mi.}$ for the red car $d = \dfrac{70 ~\text{mi.}} {\text{hr.}} \times 0.01 ~\text{hr.} = 0.7~\text{mi.}$ for the blue car To document the crash, Johnny has to set up the camera $0.5$ miles far away from Point A. ### Calculate the rate at which the car travels. Hints Keep this triangle in mind - it's magic. If it takes $0.5$ hours to travel a distance of $30$ miles, the rate is $60$ mph. Solution You can remember the formulas by using the Distance Rate Time Triangle. Here we know Time: $t=0.01$ hours Distance: $0.4$ milesWe need to know the rate of the blue car, so we choose the formula in the middle $r=\frac dt$. By plugging in the known values, we get: $r=\frac{0.4~\text{mi.}}{0.01~\text{hr.}}=40 ~\text{mph.}$ ### Evaluate the time, if the rate changes. Hints First gather all given information: $r=80$ $t=0.01$ Decide which formula to use. If you already know the rate and the time, use $d=r\times t$ to find the distance. For example: The given information: $r=90$ $t=0.01$To calculate the distance, we can calculate: $d = rt$ $d=90\times 0.01=0.9$ miles Solution The speed of the blue car is increased to $80$ mph. Johnny's a bit lazy and doesn't want to change the camera position. He can't change the rate or the time, but he can change the distance: To figure this out, use the equation: $d=r\times t$. Given: Rate: $80$ mph Time: $0.01$ hrs.$d=80\times 0.01=0.8$ So the blue car must be a distance of $0.8$ miles from the crash point. The distance between the two cars at the start is $0.5+0.8=1.3$ miles. ### Find out when and where the two brothers will meet. Hints This is a distance, rate, and time problem with different directions. Use the formula $d=r\times t$. Keep in mind that the sum of the distances traveled is $3$ miles. Use the opposite operations: Multiplication ($\times~\longleftrightarrow~\div$) Division ($\div~\longleftrightarrow~\times$) Addition ($+~\longleftrightarrow~-$) Subtraction ($-~\longleftrightarrow~+$) You have to calculate the time first, and then you can calculate the distance. Solution Eugene is riding his bike at a rate of $10$ mph, while Shane is running at a rate of $2$ mph. Here we have a distance rate time problem involving different directions. First let's consider the distance. We know that $d=r\times t$, and the time $t$ is unknown. $d=10t$ for Eugene $d=2t$ for ShaneBut, we do know the total distance, so we can write the equation: $10t+2t=3$ Combine the like terms: $12t=3$. Finally, divide by $12$: $t=\frac3{12}=\frac14=0.25$ hours. So the brothers will meet after $0.25$ hours or $15$ minutes. To determine the distance each boy will travel, we use the formula $d=r\times t$ once again: Eugene: $d=10\times 0.25=2.5$ miles Shane: $d=2\times 0.25=0.5$ milesIn summary: We know that the brothers will meet after 15 minutes. Eugene starts $2.5$ miles away from the meeting point. Shane starts $0.5$ miles away from the meeting point. Summarize the distance, rate, and time problem. Hints Use the magic triangle: $d$ over $r$ as well as $d$ over $t$ $r$ and $t$ on the same level Remember what the variables stand for: Distance Rate Time Solution To solve problems, use the Distance Rate Time triangle. Distance Rate TimeStarting at the top, if we want to calculate distance, $d$, we can see that $r$ and $t$ are on the same level. So our equation is modified: $d=r\times t$. To calculate the rate, $r$, we look at the other two variables, $d$ and $t$. $d$ stands over $t$ in the triangle, so we have $r=\frac dt$. To find time, $t$, we notice that $d$ is over $r$, so our final equation is: $t=\frac dr$. Determine when Jim has to leave his home to make it to his appointments on time. Hints First determine the time, $t$, Jim needs to travel each distance. You should use the following equation: $t=\frac dr$ Next, calculate the time in hours. Convert to minutes if necessary. Subtract your answer from the time Jim plans to arrive at each destination. Remember, there are $60$ minutes in an hour. Solution What do we know? The distance from Jim's house to each destination Jim's walking rate is $2$ mphTo calculate the time it takes Jim to travel to each of these destinations, we use the formula $t=\frac dr$. School Distance = $0.8$ miles $t=\frac{0.8}{2}=0.4$ hours. Multiply by 60: $t=0.4\times 60=24$ minutes. Subtract $24$ minutes from the arrival time: $8:00-0:24=7:36$ a.m. To get to school on time, Jim has to leave his house by $7:36$ a.m. Grandma's House Distance = $1$ mile $t=\frac{1}{2}=0.5$ hours. Again, multiply by $60$: $t=0.5\times 60=30$ minutes. Subtracting $30$ minutes from the arrival time gives us: $10:00-0:30=9:30$ a.m. Jim has to leave by $9:30$ a.m. to get to his grandma's house by $10:00$ a.m. Jack Distance = $3$ miles $t=\frac{3}{2}=1.5$ hours or, $1$ hour and $30$ minutes. Subtract this time from the arrival time (We used the $24$-hour clock for convenience.): $14:00-1:30=12:30$, or 12:30 pm. Jim has to leave by 12:30 p.m. to meet Jack at $2$ p.m. Cinema Distance = $4$ miles Oh, what a day. Jim deserves a little bit of time to kick back and relax. To get to the movies on time, he has to start $t=\frac42=2$ hours before the film starts playing. So he should leave his house at $6$ p.m. He walks $2$ hours to watch a movie!? I hope it's a good movie! 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3262	https://par.nsf.gov/servlets/purl/10485532	Vol.:(0123456789) Int. J. Res. Undergrad. Math. Ed. 1 3 Students’ Thinking about the Structure of Constructive Existence Proofs Kristen Vroom 1 · Tenchita Alzaga Elizondo 2 Accepted: 22 March 2022 © The Author(s), under exclusive licence to Springer Nature Switzerland AG 2022 Abstract Undergraduate students are expected to produce and comprehend constructive existence proofs; yet, these proofs are notoriously difficult for students. This study investigates students’ thinking about these proofs by asking students to validate two arguments for the existence of a mathematical object. The first argument featured a common structural error while the second was a valid argument of the claim. We found that the students often considered the logical structures of the arguments when validating them. They provided reasons for their evaluations, including why they thought the structure of the first argument functioned to prove the claim and why they thought the structure of the second argument did not function to prove the claim. We discuss how these reasons provide insights into why constructive existence proofs might be challenging for students. We end the paper with implica-tions for the teaching and learning of constructive existence proofs and their proof frameworks. Keywords Constructive existence proofs · Existence claims · Proof framework · Explanatory proof · Proof validation · Proof Introduction Proofs and proof-activity are central components of mathematics students’ under-graduate courses and as such, have received wide attention from mathematics education researchers. Scholars often differentiate students’ activity when work-ing on the formal-rhetorical part of a proof from the problem-centered part (e.g., Kristen Vroom vroomk@oregonstate.edu Tenchita Alzaga Elizondo halzaga@pdx.edu 1Oregon State University, Corvalis, OR, USA 2Portland State University, Portland, OR, USA Int. J. Res. Undergrad. Math. Ed. 1 3 Fukawa-Connelly, 2012; McKee et al., 2010; Selden & Selden, 2010, 2013). Cre-ating the formal-rhetorical part includes unpacking and using the logical structure of the associated claim to create or make sense of the logical structure of an argu-ment, the result of this process is a proof framework (see Fig. 1 for an example of a proof with its framework in bold). Selden and Selden (2013) explain that while proof frameworks can be difficult for students with little experience with proof, they generally do not require a deep understanding of the concepts involved. Yet, that is not to say that important mathematical work is absent from creating the formal-rhetorical part of proof. We see proof frameworks as one of the many con-ventional systems that the larger mathematics community uses to communicate mathematical meanings, thus providing a rich context for discussions about lin-guistic choices that are used to discern these meanings (Kontorovich & Zazkis, 2017; Larsen et al., 2022; Vroom, 2022). This study investigates students’ think-ing about the structure of proofs, specifically for proofs that argue the existence of a mathematical object. Undergraduate students are expected to produce and comprehend existence proofs, or proofs that argue the existence of a mathematical object. Claiming the existence of a mathematical object is prevalent in undergraduates’ proof-activity as it is common in both theorems (e.g., Mean Value Theorem, Cauchy’s Theorem in group theory) and defining properties (e.g., 𝜀 − 𝛿 definition of continuity, the identity element of a group). Regardless of its prevalence, scholars have recog-nized that existence proofs are notoriously difficult for undergraduate students, especially since their prior experiences in secondary mathematics rarely neces-sitate arguing the existence of an object (De Guzmán et al., 1998). In our classroom experiences, as well as discussions with other experienced instructors of proof-based courses, we have noticed that when attempting to prove the existence of a mathematical object, students often produce an argument with a flawed framework. Their argument starts with a desired property and then contin-ues solving for the object. For instance, in Fig. 2 a student produced such an argu-ment to show that given a group G and two elements a, b ∈ G , there exists x ∈ G such that ax = b . The argument deduced from the property ax = b that x = a−1 ⋅ b .Rather than proving the existence of the desired mathematical object, the argu-ment shows that if ax = b , then x = a−1 ⋅ b . Given the perceived prevalence of this framework error and our view that students can and will engage in meaning making, we wondered what mathematical thinking students engaged in as they produced or endorsed an argument of this nature. In this study, we explored students’ thinking about an argument for the exist-ence of a mathematical object that featured an error in the framework, as well as students’ thinking about a valid argument of the same existence claim. In doing Fig. 1Example of valid exist-ence proof with framework in bold 1 3 Int. J. Res. Undergrad. Math. Ed. so, we gain a better understanding of the mathematical work that students engage in when working on the formal-rhetorical part of a proof. Theoretical Grounding In what follows we first describe what we mean by proof, proof-activity, and con-structive existence proofs. Then, we situate our work in relevant research literature. Proof and Proof Activity We draw on Stylianides’ (2007) characterization of proof in school mathematics: “Proof is a mathematical argument , a connected sequence of assertions for or against a mathematical claim, with the following characteristics: 1. It uses statements accepted by the classroom community ( set of accepted state-ments ) that are true and available without further justification; 2. It employs forms of reasoning ( modes of argumentation ) that are valid and known to, or within the conceptual reach of, the classroom community; and 3. It is communicated with forms of expression ( modes of argument representation )that are appropriate and known to, or within the conceptual reach of, the class-room community” (p. 291). We emphasize two points about this definition. First, what constitutes a proof is dependent on the audience’s accepted statements, modes of argumentation, and rep-resentation. For example, an argument that is viewed as a proof by a mathematician may be rejected by a student because it uses a form of reasoning unfamiliar to the student. Throughout the paper, we specify when we consider the students’ evalu-ation of an argument. When we classify an argument as valid (or invalid), we are Fig. 2Example of student argument with logical error with typed proof below Int. J. Res. Undergrad. Math. Ed. 1 3 referring to our view of whether the larger mathematics community would accept (or reject) the mathematical argument as a proof of a given statement based on for-mal logic. Second, Stylianides’ (2007) definition indirectly suggests that an argument that meets the three characteristics establishes the truth of the statement. To make this explicit, we further clarify that the associated mathematical claim plays a role in the second characteristic: when evaluating whether a mathematical argument is a proof, one should consider whether the argument uses acceptable modes of argumentation that are for (or against) a known mathematical claim. We see proof frameworks as a way to determine this. For instance, in the student argument above (Fig. 2), the stu-dents’ proof framework assumes ax = b to deduce x = a−1 ⋅ b, which is inconsistent with the associated claim and thus does not establish its truth. While we acknowl-edge that a given argument could prove a different claim than what is given, when we say that an argument was (or was not) viewed as a proof, we mean that the argu-ment was (or was not) viewed as a proof of the given claim. Scholars have distinguished between proofs that only convince and proofs that simul-taneously convince and explain (i.e., explanatory proofs ) (e.g., Bartlo, 2013; Lockwood et al., 2020; Weber, 2010). In this study, we leverage two aspects of explanatory proofs that are consistent with Lockwood et al. (2020) characterization. First, a proof that is explanatory to a mathematician may not give the same insight to a student, and thus, explanatory proofs can be audience-dependent. Because of this, a prover who wanted to provide insight into why a statement is true would consider their audience. Second, the explanatory nature of a proof is closely tied to the activity of constructing the proof; a reader often finds a proof to be explanatory if it gives them insight into the informal reasoning that was used to create it. Thus, a prover who wanted to produce an explana-tory proof for a particular reader might include an appropriate amount of their problem-solving that led them to determine why the statement was true. The research literature on students’ proof-activity has historically focused on construction, comprehension, or validation. Constructing a proof is the mental and physical actions of producing a proof, comprehending a proof is making meaning of the given text as one reads a proof, and validating a proof is checking whether a mathematical argument proves a given mathematical claim. As with many scholars (e.g., Melhuish et al., 2021; Selden & Selden, 2003, 2017), we see these activities as inherently connected. We see comprehension and validation as mutually dependent activities; one cannot validate a proof without comprehending it nor can one com-prehend a proof without understanding its validity. Selden and Selden (2003, 2017) explain that students’ proof construction and validation are necessarily interrelated: “…one constructs a proof with an eye toward ultimately validating it and may often validate parts of it during the construction process. In fact, the final part of a proof construction is likely to be a validation of that proof” (Selden & Selden, 2003, p. 6). Because of this, we argue that we can gain insight into why students construct an argument like the one in Fig. 2 by investigating their validation of such an argument. One common aspect among these proof activities that is particularly relevant to this study is attention to the proof framework. While comprehending proofs, students can coordinate the statement being proven, the proof technique, and the necessary assump-tions and conclusions (Mejía-Ramos et al., 2012; Weber, 2015). While validating proofs, 1 3 Int. J. Res. Undergrad. Math. Ed. students can evaluate whether the structure works to prove the claim (Selden & Selden, 2003). When constructing a proof, Selden et al. (2018) recommend students create a proof framework to organize the logical structure of their argument. The student intro-duces the claim’s assumptions by introducing the appropriate variables, leaves space to later fill in the middle of the argument, and ends with the statement’s conclusion. The prover can add another layer of their framework by “unpacking” the conclusion. Selden and Selden (2013) note that with enough practice, identifying an appropriate proof framework can be a relatively procedural task involving little to no problem-solving. Once their framework is complete, they can begin problem solving how to fill in the middle of the argument. This problem-solving can sometimes include starting by fictitiously assuming the conclusion to deduce what is known to be true based on the givens of the statement. Then, the prover can reverse the argument to start with the givens and end with the sought-after conclusion. This strategy is consistent with what is called the analysis-synthesis method (Lakatos, 1978). In this study, we asked students to engage in proof validation of two arguments: one is a valid argument while the other is invalid because it uses an invalid proof framework (similar to the argument in Fig. 2). We investigate the students’ valida-tion activity to not only understand students’ reasons for the evaluations of the argu-ments, but also to better understand how students might think about the arguments and why students might construct a proof with this structural error. Proofs of Existence Claims Brown (2017) explained that existence proofs take two forms: constructive or non-constructive. A constructive existence proof “either explicitly produces the desired result or provides an algorithm for its production” (p. 468–469). Whereas a noncon-structive proof “establishes the theorem as a consequence of previous theorems or as a logical necessity” (p. 469). Brown provided two proofs of the claim that there exist irrational numbers a and b such that a b is rational. The constructive proof introduced the irrational numbers a = √2 and b = log 2 9 and then demonstrated that the desired property, a b is rational, was met. The nonconstructive existence proof relied on the logical necessity that either (1) a = b = √2 or (2) a = √2 √2 and b = √2 must sat-isfy the desired property since √2 √2 is either rational or irrational. In Brown’s (2017) study, she investigated students’ conviction of the two proofs. Despite prior literature supporting the conjecture that students would be more convinced by the constructive argument (Harel & Sowder, 1998; Leron, 1985), Brown found that the students in her study tended to deem the constructive proof as less convincing, providing some evidence for students’ skepticism for construc-tive existence proofs. Further, the students often gave unexpected reasons for their selection. For instance, more than half of the students that viewed the constructive proof as more convincing did not mention its constructive nature and some students seemed to interpret the non-constructive proof as a constructive one. Thus, Brown highlighted that students’ interpretations of a proof might differ from the standard Int. J. Res. Undergrad. Math. Ed. 1 3 mathematical interpretation. This suggests that there are potential methodological issues in work that draws on preconceived notions about what is communicated with a given proof to understand how the students interpret it. Research that aims to understand students’ interpretations should use approaches that bring to light the arguments from the students’ point of view. We argue that another aspect of constructive existence proofs may impact how students comprehend and validate them. Published existence proofs, and likely those provided by instructors, vary in the extent to which the process of identifying the desired object (or a procedure for generating it) are communicated. On one end of this spectrum, a prover offers an object as a candidate early in the proof without details of how the prover found it and then shows that the object fits the desired property (e.g., the proof in Fig. 1). On the other end, the prover details the construc-tion process. The proof in Fig. 3 shows that given any 𝜀 > 0, there exists a math-ematical object, denoted 𝛿 , that fits a desired property (i.e., if 0 < x − 2 < 𝛿 then g(x) − 4 < 𝜀 ). This proof shares with the reader more information about how the prover went about selecting 𝛿 = min{ 1, 𝜀 ∕5}.The prover must strategically decide an adequate amount of information to share with the reader in such a way that the argument is logically connected with the asso-ciated claim. For example, presumably the prover of the argument in Fig. 1 selected x = a−1 ∗ b by first solving the desired equation a ∗ x = b for x . However, the prover tactically left this information out since proving if a ∗ x = b, then x = a−1 ∗ b was not the goal of the argument. In contrast, the prover of the proof in Fig. 3 was quite strategic by including their problem-solving work without a logical flaw. The amount of detail that the prover shares with a reader might also impact how stu-dents comprehend and validate the proof. The framework of the proof in Fig. 1 is straightforward in that one could match the framework to the statement; however, such a concise framework contains a great deal of precise meanings that students may not yet be able to unpack. The proof in Fig. 3 describes more informal reason-ing for how the prover found the proposed value for 𝛿 ; however, a consequence of the lengthier description is that the framework is more hidden. Fig. 3An example of a more explanatory constructive exist-ence proof (borrowed from Abbot, 2015, p. 117) 1 3 Int. J. Res. Undergrad. Math. Ed. Insights from the Literature While the research literature on students’ thinking about proof and proving in gen-eral is mounting, there are only a handful of studies that explicitly discuss students’ thinking about existence proofs (Brown, 2017; Samper et al., 2016; Schaub, 2021). In what follows, we review the literature on students’ proving activity to gain insight into how students might think about constructive existence proofs. There are several studies that have investigated students’ attention (or lack thereof) to the structure of arguments in general. These studies show that students might view an invalid argument with a flaw in the framework as a proof because they ignore the logical flaw or doubt the structure of the argument matters (Selden & Selden, 2003; Weber, 2009, 2010). Weber (2009, 2010) interviewed 28 mathemat-ics majors asking them to read various arguments presented as potential proofs, or purported proofs. One of which featured a logical flaw by assuming “ n is divisible by 3” to prove that “If n2 is divisible by 3, then n is divisible by 3” (i.e., assumed the conclusion). Similar to Selden and Selden’s (2003) findings which also used this argument, Weber found that more than half of the students viewed the argument as a proof. Twelve of these students gave the argument the highest marks in terms of their personal conviction and its rigor despite the flaw in the proof framework. Weber explained that many of the students did not check whether the argument used appropriate assumptions and conclusions. This is consistent with other findings that students tend to focus on “surface features” such as computational aspects of an argument rather than the logical structure (Inglis & Alcock, 2012; Selden & Selden, 2003). Interestingly, Weber (2009) also found that some of the students who viewed the argument as a proof attended to the structure by identifying that the argument started with what they wanted to prove but, as one student put it, “I don’t know if the formatting matters” (p. 15). Thus, when students are considering proofs of existence claims, the structure of the argument may not impact their validation. Studies have also identified that students have difficulties with the mathematical language used in proofs (Moore, 1994; Selden & Selden, 1995), and specifically, that students may not understand the nuanced ways that mathematical objects are introduced and used (Lew & Mejía-Ramos, 2019). In the context of existence proofs, Samper et al. (2016) argue that teachers need to deliberately support the students in understanding the strategy for proving existence claims in geometry. This included supporting the students to understand what object to introduce, how they could introduce the object, and why it was guaranteed. This is because in their experience, students tended to introduce random objects and attempt to force the desired proper-ties on them. Given that constructive existence proofs often introduce a particular object as a candidate early in the argument, these studies suggest that students might not fully understand the subtle way that such an argument establishes the existence of the desired object and may need intentional support to gain this understanding. Of particular relevance to our study, Schaub (2021) presented a classroom epi-sode of an inquiry-oriented Real Analysis class constructing a proof of an existence claim. During the episode, the class discussed leaving out the algebraic work that led the students to find a desired mathematical object from their final proof while proving a particular sequence converged by employing the 𝜀 − N definition. In the Int. J. Res. Undergrad. Math. Ed. 1 3 case that a sequence converges, the definition claims the existence of a mathematical object, denoted N : for every 𝜀 > 0, there exists a positive integer N such that for all n ≥ N , a n − L < 𝜀 . During the episode, the instructor transformed the class board work for finding an N into a proof, which included “Let N > ⌈3∕𝜀 − 1∕2⌉” imme-diately after introducing an arbitrary positive epsilon. After the instructor wrote the full proof on the board, she acknowledged that it seemed off putting because it was unclear why the prover chose the particular N , explaining: “…[The proof] has this weird like ‘where did you get that?’ and the answer is we got that by working out the problem backwards. Assuming what we wanted to prove, finding what N should be, and then sticking it in and proving that it works. And this step is super important because this is a proof in the right order. Start with epsilon, find big N , and show that beyond that point, we are always close, and close means within epsilon, of our limit” (Schaub, 2021, p. 168). A student then expressed some discomfort with this strategy, explaining, “I’ve always had a problem with this way of writing a proof” adding, “because we do things, we find the answer the only way we can, and then we pretend like we didn’t do it that way” (Schaub, 2021, p. 168). This student’s comment suggests that stu-dents might have issues with constructive existence proofs that are not transparent in how the prover found the mathematical object. The reviewed literature suggests that constructive existence proofs are non-trivial for students. Students’ difficulties may be related to the proof framework, how the object is introduced and used in the proof, and lack of transparency for how the prover went about finding the mathematical object. In the current study, we inves-tigate: how do students think about arguments for the existence of a mathematical object? To answer this question, we created models of the students’ thinking as they evaluated arguments for the existence of a mathematical object. That is, we used observable data (students’ utterances, gestures) to describe our understanding of the students’ thinking. This understanding adds to the current research literature by identifying why constructive existence proofs might be challenging for students. In doing so, we view our study as important steps towards better supporting students to engage in constructing, comprehending, and validating constructive existence proofs. Methods Interviews This study is part of a larger design research project that is developing inquiry-oriented Introduction to Proof curriculum and instructor support materials ( Advancing Students’ Proof Practices in Mathematics through Inquiry, Reinvention, and Engagement project, NSF DUE #1916490). The course is designed so that the instructors can use materials in the context of real analysis (via exploring the Intermediate Value Theorem, IVT) and/or group theory (via exploring Symmetry groups). The course was intended for stu-dents who had not yet taken any undergraduate proof-based courses. During the course, students reinvented key concepts (e.g., sequence convergence or the definition of group) 1 3 Int. J. Res. Undergrad. Math. Ed. and engaged in various related proof construction tasks. Of particular relevance, the students were sometimes tasked to prove various existence claims, during which the instructors often mentioned the problem-solving needed to find the desired object prior to writing the proof. At this point in the design project, the curriculum did not include tasks designed to explicitly support students in learning about how constructive exist-ence proofs can be a valid proof of an existence claim, or why an argument that starts with the desired property is not a valid proof. Students exiting the courses were invited to complete a survey which asked them to interpret a collection of mathematical statements as well as describe their experi-ences in the course, views about mathematics, and demographic information. At the end of the survey, students were invited to participate in an additional interview. The data for this study comes from 16 semi-structured interviews (not including pilot interviews) from students at two universities enrolled in three different classes with different instructors. At one university two instructors implemented both group theory and real analysis materials during a 10-week term whereas the instructor at the other university implemented real analysis materials during a 15-week semester. The interview tasks were either in the context of groups or functions depending on the course context. See Table 1 for an overview of the data. There were two interview components that corresponded to two distinct research goals. The first component followed-up with the students regarding their experiences in the course. The goal for this component was to better understand these experi-ences in order to inform later course implementations. The second component is the focus of this report and was designed to engage students in proof validation tasks. The goal was to understand how the students thought about the purported proofs. We then used our interpretations of their thinking to design additional instructional tasks for the course that would target students’ proof comprehension, validation, and construction of existence claims. The full interview typically lasted an hour with the second component lasting between 30 and 40 min. We were both present for each of the interviews. The interviews were facilitated remotely via Zoom and a shared Google Doc and were video-recorded capturing the students’ gestures and typed work. When we refer to individual students, we use a code to indicate their proof validation for each argument and the context of the interview tasks. For instance, the label “NPG-7” represented that the participant did not view the first argument as a proof (represented by “N”), viewed the second argu-ment as a proof (represented by “P”), the arguments were in the group context (rep-resented by “G”), and they were the seventh interviewee (represented by “7”). We used “F” to represent the tasks in the function context and “U” to represent the one instance in which a student was ultimately undecided about the validity. Interview Tasks There were three main parts of the second component of the interview: we asked students to (1) (re)interpret a given existence statement, (2) describe their sense making and validation of the Invalid Argument, and then (3) describe their sense Int. J. Res. Undergrad. Math. Ed. 1 3 making and validation of the Valid Argument. See Figs. 4 and 5 for the statements and the Invalid and Valid Arguments. The students did not see the labels “Invalid Argument” or “Valid Argument” during the interview. Additionally, the students did not initially see the underlined text in the arguments (we will elaborate on this methodological choice later). The students interviewed with the group context had previously discussed the statement in detail and approaches for proving it during the class. In the function context, the students had previously conjectured various func-tion properties that would guarantee the existence of a real root (e.g., IVT). Prior to the interviews, the students had not engaged with tasks that asked them to evaluate the two arguments given in Figs. 4 or 5. We constructed the Invalid Arguments in Figs. 4 and 5 based on our experiences with students. It was relatively common during the interviews for students to com-ment that the argument was how they would construct a proof of the given state-ment. For instance, before PPG-11 saw the Invalid Argument, he explained that he would start proving the given statement by “left multiplying both sides of this equa-tion [a ∗ x = b] by a−1 and then it would just be x = a−1 ∗ b”. When he was given the Invalid Argument, he confirmed that this was the approach that he explained earlier, saying “That’s what I was getting at, yeah.” While we acknowledge that in Table 1Overview of Data Term/Semester School Course Content Interview Context Number of Participants Summer 2020 Term Community College AGroup Theory Group 3 (Pilot Data) Fall 2020 Term University A Group Theory +Real Analysis Group 6Winter 2021 Term University A Group Theory +Real Analysis Group 5Spring 2021 Semester University B Real Analysis Function 5 Fig. 4Group version of the statements and purported proofs 1 3 Int. J. Res. Undergrad. Math. Ed. both contexts the solved-for object must fit the desired criteria, we view this argu-ment as invalid since this is not always the case for existence claims of this form. We constructed the Valid Arguments to show two parts of the desired property: 1) the mathematical object is a solution to the desired equation (e.g., a ∗ x = b) and 2) the mathematical object is a member of the desired set (e.g., x ∈ G ). The first part is accomplished through the string of equalities at the end of the proof while the second is accomplished by the underlined text after the mathematical object is introduced. We conducted three pilot interviews which supported refinements to the inter-view protocol. We found one strategy to be particularly effective in eliciting stu-dents’ thinking about the arguments: during the second and third part of the inter-view we periodically added, deleted, or adapted lines of the arguments, asking students whether the change altered the meaning of the argument and if the change was necessary. We found this to be a useful interviewing technique in two ways. First, by asking about the necessity of specific lines, we focused students’ attention beyond the surface features and on the inner-workings of the argument in relation to the statement. Second, it supported us to see the arguments from the students’ point of view: the students’ responses included their view of how each line func-tioned to convey the prover’s argument, and thus, centering the student’s view of the prover’s argument. We added this interviewing technique to the protocol in two ways. First, we added the underlined text in Figs. 4 and 5 after the students had dis-cussed their initial thinking about the purported proof. We decided to focus on these lines because they were part of the proof frameworks. We decided to add the lines after their initial thinking to allow the students opportunities to identify potential disconnects between the proofs and the statements without our prompting. Second, we added individualized follow-up questions of this nature depending on the discus-sion. This methodological choice is consistent with Vroom’s (2022) technique that was rooted in social semiotic theory (Halliday & Matthiessen, 2013) that surfaced students’ views of meaningful grammatical choices. Additionally, we view this approach as following Brown’s (2017) recommendation that research that engages students in comparing proofs should seek to understand student activity from the Fig. 5Function version of the statements and purported proofs Int. J. Res. Undergrad. Math. Ed. 1 3 eyes of the student as opposed to interpreting it through an observer’s predetermined criteria. Part 1: Interpret the Statement The task-based interview began with the student and researchers opening a shared Google Doc that displayed the statement. The students that were interviewed using the group context were previously asked to determine if the statement was true or false and explain why on the exit survey. The students were prompted to re-read the statement and their response and then asked whether they agreed with their earlier response. Students interviewed using the function context were not surveyed about their interpretation of the statement. In this case, the students were given the state-ment and then asked about the meaning, whether the statement was true or false, and why. In either case, we continued to ask follow-up questions until we were satisfied with our understanding of the students’ interpretations. At this point in the inter-views, all of the interviewees indicated that the given statement was true. Part 2: Make Sense of the Invalid Argument Next, we pasted the Invalid Argument (without the underlined text) into the docu-ment and presented it as a student’s proof of the statement. After we gave the par-ticipant time to read the argument, we asked them to discuss their thinking about the purported proof following up with whether or not they thought it was a valid proof and why. We then added the underlined text to the argument and prompted the student to explain whether or not the new text was needed and whether it changed the meaning of the argument. We continued to follow up with the student regarding what they saw as the prover’s argument. Part 3: Make Sense of the Valid Argument Then, we pasted the Valid Argument into the document explaining that it was a dif-ferent student’s proof of the statement. Again, we allowed participants time to read the argument and asked the same follow-up questions as we did with the Invalid Argument. We then added the two underlined texts one at a time to allow discussion for each. Throughout this discussion we elicited how the participant considered the first prover’s argument as different or similar to the second prover’s argument. Addi-tionally, we asked how the students saw the statement as connected or disconnected to each of the purported proofs. Data Analysis Our data analysis process was consistent with a thematic analysis (Braun & Clarke, 2006). Together we engaged in a cyclic process examining each interview tran-script and corresponding video data for evidence of the students’ thinking about the 1 3 Int. J. Res. Undergrad. Math. Ed. existence statement and two arguments. To do so, we focused on the following guid-ing questions: • How did the student describe their interpretation of the statement? – Did the student’s description of the statement change over the course of the interview? (If yes, how so, and why?) • How did the student describe their initial thinking about the Valid/Invalid Argu-ments? – Did the student indicate it was a proof or not? Why? – How did the student describe the logical argument? – Did the student change the way they described their thinking about the Valid/ Invalid Argument over the course of the interview? (If yes, how so, and why?) • How did the student discuss the two arguments in comparison to each other? For each student, we discussed our answers to these guiding questions until we could come to an agreement on how we would interpret their thinking. We docu-mented our answers to the questions with relevant quotes and our shared-interpretation of the quotes in an analytic memo. By answering these questions, we developed initial descriptions of our understanding of each of the student’s thinking about the two argu-ments and why they did or did not view them as proofs. After creating each analytic memo, we compared our understanding of the student’s thinking to the previous. Dur-ing this comparison, we generated and refined a list of the ways in which we under-stood how the participants thought about the arguments and how this related to their views of the arguments as proofs or not. Results In this section, we share our interpretations of the students’ ways of thinking about the mathematical arguments, and how, if at all, this could explain the students’ view of the arguments as proofs. We summarize our reasons for why the students in our study viewed the arguments as proofs (or not) at the end of each subsection. Table 2 offers an overview of the students’ assessments of the arguments by providing the frequency of each of the proof validation combinations. The Invalid Argument Shows How to Find the Mathematical Object The majority of the participants (N = 12) indicated that the Invalid Argument showed how to find the mathematical object. For instance, PNG-3 explained that it described “the process of finding that element [ x ] for which that [a ∗ x = b] is true”. The students whose thinking did not fit in this category either did not elabo-rate on the prover’s logical argument after they pointed to an error (NPG-8, NPG-9, Int. J. Res. Undergrad. Math. Ed. 1 3 NPG-10) or else explained that the prover showed that if there exists an x in G such that a ∗ x = b, then x must be in the form a−1 ∗ b (NPG-1). Some of the students who saw the Invalid Argument as showing how to find the mathematical object viewed it as a proof (N = 7), while others did not (N = 5). One salient reason that these students claimed it was a proof was that it elaborated why the existence claim was true by showing the algebraic work to construct the claimed object. For instance, when PPG-11 compared the two arguments, he explained that he had a preference for the Invalid Argument since it showed how to find the object, saying: “I prefer to see the steps, right. Like, I think I prefer to see like the thought process happen in real time…” It was common for these students to mention that the Invalid Argument was an appropriate proof for more novice readers. For instance, PPF-16 explained, “I could see where somebody at a lower math level would appre-ciate having that”. We see these students as thinking the Invalid Argument was an explanatory proof for less experienced students as it provided insights about how to construct the claimed object. Some of the students who viewed the Invalid Argument as a proof since it showed how to find the object explained why starting with the desired equation was not a flaw in the logic. PPG-11 explained: “No, they don’t start with the thing that they’re trying to prove. They’re trying to prove that x needs to be an element in G. Right. That’s what we’re trying to prove, that there exists an x in G such that the statement is true.” While this student seemed to view the property a ∗ x = b as a way to describe a desired characteristic of x, the student focused on showing that such a group element existed. PPF-16 explained that it was appropriate to start by “assuming that there exists a real number t that makes the function equal to zero” because “we know that whatever number we’re going to solve t to be is always going to make that [ f (t)]zero”. To this student, the prover could show that there existed a real number t such that f (t) = 0 by solving the desired equation for t because the value that they solve for must be the value that makes f (t) = 0 true 1.There were other students who deemed the Invalid Argument as a proof who pointed to the logical necessity of showing how to find the mathematical object. Table 2Overview of students’ validations of the two arguments aThis count does not include PUG-4 was undecided about the valid-ity of the Valid Argument Student Validations of Invalid Argument Student Validations of Valid Argument Frequency aNot a proof Proof 7Proof Proof 4Proof Not a proof 2Not a proof Not a proof 21If the reader interpreted the implicit warrants in the proof to be bi-directional, then they would feel confident that the solved for object would have to be a solution. 1 3 Int. J. Res. Undergrad. Math. Ed. PUG-4 explained that she preferred the Invalid Argument because it made sense to her and she could see the statement connecting to the argument. She said: “I think it makes more sense mathematically. But also in context of the problem that we’re supposed to show ‘for every a there exists an x…’, you know? So, I like that it gets, you know, it starts here, this is the equation that we’re given. This is what we have and then here’s the x that exists and that’s like proof. That’s why I like it, because it’s kind of in the order of the proof...” We interpret her comment to mean that she not only could follow the algebraic steps, but she saw it as fitting the logical structure of the given statement. PNG-6 further articulated why from her perspective the Invalid Argument was a proof. When asked how she saw the argument connected to the statement she explained that the prover showed that “no matter what the a is, we’re able to figure out an x,that will equal b when you do that [operation].” To these students, an appropriate approach to show that one could find a desired mathematical object was to show how to find it. Four of the 5 students who did not view showing how to find the object as a proof explained that it assumes there exists the desired object when it solves the desired equation for the value. Some of these students commented on the utility of the alge-braic work even though it should not be part of the proof. For instance, NPG-5 said: “That [‘Then a ∗ x = b�] assumes the conclusion and then works from that. Which is like one of those good strategies you use to write a proof, but you do it separately and then try and write it the other way.”. Interestingly, the fifth student who did not see the Invalid Argument as a proof viewed the argument as a necessary part of the proof, but alone was incomplete. To NNF-15, the Invalid Argument showed why x = − b∕m and that it was a real number but “it has not proven that f (x) = 0. It is just stating ‘hey, this is what we’re trying to solve for’…”. In summary, we saw most of our students as identifying the prover of the Invalid Argument as showing the process of finding the desired mathematical object. We viewed the students in our study as thinking about the argument in at least one of the following ways: • The argument was a proof since it showed the existence of the desired object by explaining why such an object fit the desired property, • The argument was a proof since it fits the logical structure of the statement, • The argument was not a proof since it assumed the existence of the desired object, or • The argument was not a proof since it had only solved for the desired object but also needed to show the desired property was true. The Valid Argument Shows an Instance of the Mathematical Object All the students in our study (N = 16) were able to identify that the Valid Argument introduced an instance of the mathematical object. For instance, when asked to walk through the prover’s logic in the group theory context, NPG-8 explained “they’re Int. J. Res. Undergrad. Math. Ed. 1 3 just choosing an x, which is essentially what the problem is asking for. It’s just say-ing, just find, just find one of them”. While all students saw the Valid Argument as identifying a particular instance of the mathematical object, some students did not view it as a proof (N = 5) 2. We iden-tified two reasons that these students had this view. First, four students explained that by starting with an instance of the mathematical statement, the prover incor-rectly assumed the existence of the object instead of deducing its existence. For instance, when discussing why the Valid Argument may not be a proof, PUG-4 stated: “because it starts with the x - that’s a little confusing to be like ‘oh, for every a, an x exists’ but that’s what they’re saying right here, this is the x that exists”. For this student, like the other three, the prover could not start with x = a−1 ∗ b because by doing so they assume that the object exists. The fifth student who did not view the argument as a proof had a different reason. Similar to his reasoning for why the Invalid Argument was not a proof, NNF-15 saw the Valid Argument as only provid-ing half of the necessary argument, saying “you would also have to prove why x is equal to −b∕m, you can’t just state that”. Unlike the other four students, NNF-15 viewed the Valid Argument as a necessary part of the proof since the prover showed the mathematical object met the desired equation (i.e., f (x) = 0). To NNF-15, you would need the two arguments together to make a proof. Among the 11 students who thought the Valid Argument was a proof, one salient reasoning was that it was logical but had a jump in explanation. For example, “The second [argument], just like tosses that up there. So, they maybe did some side work or maybe were just able to see it. But, uh, but they kind of leave a lot of their reasoning off the page” (PPG-11). While these students also acknowledged that the prover omitted relevant reason-ing from the reader by introducing an object up front, we see these students as think-ing that the proof only needed to convince rather than explain to the reader how to find the object. There were several students who explained that the Valid Argument was a proof since it only needed to show the existence of one element that met the desired prop-erty. NPF-12 explained: “So, since we only have to prove there exist one number we don’t have to prove for all real numbers, we just have to find that one real number for which the equation is equal to zero, so I guess, we could say that, like to let it be equal to −b∕m and then use that to show that the equation is zero.” Several of these students (among others) did not see the line that the mathemati-cal object was an element of the claimed set (i.e., “By closure, x is in G since a−1 is in G and b is in G” or “since m ≠ 0, t is a real number) as being necessary for the argument. For instance, NPG-9 explained: “I don’t think it’s necessary, but I think it’s helpful. [...] Explaining why x is in there like, to me, it was just kind of intuitive because you define it in terms of things that are already in G…” 2 This count includes PUG-4 who offered a reason why the Valid Argument was not a proof, but did not ultimately commit to her decision. 1 3 Int. J. Res. Undergrad. Math. Ed. NPG-9 as well as others saw this line as simply adding further explanation and clarity for the reader rather than a necessary part of the conclusion. In summary, we saw all the students as thinking the Valid Argument introduced a specific mathematical object. We viewed the students in our study as thinking about the argument in at least one of the following ways: • The argument was not a proof because it assumed the existence of the object instead of deducing its existence, • The argument was not a proof because it only showed that the object met the desired property but not how they found the object, • The argument was a proof since it showed the existence of the mathematical object but had a jump in explanation when the mathematical object was intro-duced, • The argument was a proof since it only needed to show the existence of one ele-ment that met the desired property, or • The argument was a proof since it showed the existence of the mathematical object even without arguing the mathematical object was in the desired set. The Arguments are Structured in Different Ways Most of the students in our study identified the different logical structures of the arguments (N = 15). Many of these students described them as “opposite” of each other or explained how the first deduced information from the desired equation whereas the second concluded the desired equation. For instance, when comparing the two arguments PNG-3 stated that “this one [the Invalid Argument] shows how they got to that x and this one [the Valid Argument] just like here’s the x and you get b when you multiply by a”. The one student’s thinking that did not fit in this cat-egory saw both arguments as proofs that were structured in the same way. Some students indicated that the structural differences mattered to them, that is, only one of the frameworks structured the argument to prove the claim. This depended on the students’ view of the goal of the argument. These students dis-cussed two different goals. First, some students thought that a proof should con-clude with the mathematical object instead of assuming it at the start - these students tended to be the ones who thought a) the Invalid Argument was a proof because it explained how to find the mathematical object and b) the Valid Argument was not a proof because it assumed the existence of the object instead of deducing its exist-ence. PNG-3 explained: “Well, we’re trying to conclude that there exists an x. Okay yeah, so I don’t think proof two [the Valid Argument] is [a proof], because that is the conclusion, that there exists an x such that a ∗ x = b. Whereas in this one [the Invalid Argument] we’re definitely looking for that x, in the first one. Because I think that the conclu-sion should be that there exists an x in G such that a ∗ x = b, and our givens are that a is in G and that b is in G, essentially and that’s it.” She later confirmed that to her the Valid Argument assumed the conclusion by introducing the element x = a−1 ∗ b at the beginning of the argument. Second, there Int. J. Res. Undergrad. Math. Ed. 1 3 were some students who thought that the argument should conclude that the desired property held for an instance of the mathematical object. These students were the ones that a) did not think the Invalid Argument was a proof since it assumed the existence of the desired object and b) thought the Valid Argument was a proof since it only needed to show the existence of one element that met the desired property. Other students indicated that the structural differences did not matter to them. These students either explained that neither argument effectively proved the claim (N = 2) or else viewed both arguments as proofs (N = 3). The students who thought neither were a proof thought so because neither had acceptable or complete frame-works. NNG-2 identified that both arguments had issues with how they started. She discussed that a proof of the existence statement needed to deduce the mathematical object rather than assume it, leading to a clear rejection of the Valid Argument due to it starting “with our conclusion and not with our beginning”. She preferred the Invalid Argument saying it was “the proof in the correct direction” but explained that it needed one fix. She explained: “I wouldn’t just say ‘then’ I would say, like I think that’s more of like a ‘consider’ or ‘if’. [...] Yeah if a ∗ x = b (emphasis added).” This suggests she also viewed the Invalid Argument as making the error by start-ing with an ineffective assumption. To her, the Invalid Argument no longer assumed unknown knowledge when she changed the beginning of the argument to a condi-tional statement by adding “if”. NNF-15 was the other student who viewed neither as a proof. As we have previously discussed, he thought the two arguments were structurally different in that they accomplished different, but necessary parts. To him, the frameworks were incomplete on their own. The other three students who did not think the structural differences mattered viewed both arguments as proofs. These students acknowledged that one could con-struct different proofs to show the existence of the mathematical object. They saw a) the Invalid Argument as a proof since it showed the existence of the desired object by explaining why such an object fit the desired property and b) the Valid Argument as a proof with a jump in explanation when the mathematical object is introduced. We interpret this as them seeing one proof as being more explanatory to certain readers, while another may only convince. For these students, the first proof pro-vided additional information that the second proof left out. While both arguments were valid, the intended reader differed. For instance, PPF-13 explained: “So, this one [the Invalid Argument] feels like it’s more like explain it like your five and the other one [the Valid Argument] is more like assume they already know what it means to go from there and that they’re both valid, but they’re both different ways of doing it.” To PPF-13, and the other two students whose thinking fit in this category, the two arguments accomplished the task of finding the object, they just did so in different ways. As PPG-11 stated: “It’s just the construction of it [the two arguments] that’s different […] They [the provers] both understand that x needs to be equal to a−1 ∗ b whatever a is.” Unlike these students, a fourth student, PPF-14, who viewed both arguments as proofs, thought that they followed a parallel structure. Like the three 1 3 Int. J. Res. Undergrad. Math. Ed. students in this category, PPF-14 also saw the explanatory power of the first argu-ment, but saw it as an expanded version of the second argument. He explained: “They start the same and then this one [the Valid Argument] jumps straight to choose t = − b∕m. So, it’s assuming [...] that hey you were able to complete these steps on your own, essentially. [The Invalid Argument], it’s showing you, step by step. And this one’s [the Valid Argument] just assuming - okay, we start with A [pointing to the first line] we go to B [pointing to the “Choose” line] we hit C [point-ing to the last line with the string of equalities] and there’s our answer. This one [the Invalid Argument] is - we started A [pointing to the first line] here’s how we get to B [highlighting the algebraic work that leads to t = − b∕m] here’s an example of it and then there’s C [pointing to the last line], which is also our answer so it’s just giving those extra steps to somebody.” In summary, we saw that all but one of the students in our study identified that the arguments were structured in different ways. We viewed these students as thinking about the arguments in at least one of the following ways: • Only the Invalid Argument had an effective framework since it was the only one that concluded the mathematical object instead of assuming it at the start, • Only the Valid Argument had an effective framework since it was the only one that concluded the desired property held for an instance of the mathematical object, • Neither of the arguments functioned as proofs since neither had acceptable or complete frameworks, • Both of the arguments functioned as proofs since the Invalid Argument was more explanatory to certain readers while the Valid Argument only convinced. Conclusion and Discussion Students’ Thinking About Constructive Existence Proofs Returning to our research question, our study found that students thought about arguments for the existence of a mathematical object in several different ways. Some students identified that an argument that started with the desired property and then solved for the object was not a proof, while other students offered reasons why to them such an argument proved the claim. Additionally, there were also students who identified that an argument that introduced the object as a candidate early on and then showed that it fit the desired criteria was a proof. Yet, there were other stu-dents who offered reasons why to them such an argument was not a proof. Despite the different ways that the students validated the arguments, their reasons tended to attend to the logical structure of the arguments. This is in contrast to prior studies that have documented that students ignore the structure of proofs or think that they do not matter (Selden & Selden, 2003; Weber, 2009, 2010). Of particular interest, the students provided reasons for why they thought the structure of the Invalid Argu-ment functioned to prove the claim and why they thought the structure of the Valid Int. J. Res. Undergrad. Math. Ed. 1 3 Argument did not function to prove the claim. This offers insights into why con-structive existence proofs might be challenging for students. In reference to the Invalid Argument, we found two reasons for why some stu-dents in our study saw starting with the desired property (e.g., a ∗ x = b) as accept-able. First, some of the students viewed the Invalid Argument’s framework as fitting the logical structure of the existence statement. For them, the statement required that one show how to find the desired object and thus needed to start with the prop-erty and end with the object itself. Second, other students viewed it as a way to add transparency when showing the existence of the desired object by explaining how one would find it. We saw these students as valuing the explanatory nature of Invalid Argument while acknowledging it was not the only way to prove the exist-ence of the object. These ways of reasoning can explain why students may produce or endorse an argument that starts with a desired property and follows with solving for the object: students may think that (a) the structure of the proof should logically show how to find the mathematical object or (b) a reader should gain insight about how the mathematical object was derived. Additionally, we found there were students who did not view the Valid Argument as a proof of the existence claim. In particular, there were students who saw the line that introduced the object (e.g., Choose x = a−1 ∗ b) as assuming the existence of the mathematical object. In other words, the argument did not prove the existence of the object since it assumed it from the start. Such a view explains why students may not think that proofs of existence claims can first introduce the object as a candidate: students may think that such an introduction is logically invalid as it assumes the conclusion. Implications and Future Directions Our findings motivate additional areas for research. First, we echo Brown’s (2017) recommendation: there is a need for research on student’s conceptions of proof and proof-activity from the students’ point of view. By prioritizing the students’ point of view, we were able to describe important mathematical thinking from the students that might have gone unnoticed if we conducted our analysis using preconceived notions about what is communicated with the given proofs. Potential important insights can be gained from future research that takes such an asset lens on students’ thinking about proof and proof-activity focusing on what students can do rather than a deficit lens focusing on what students cannot do. From our work, we see two potential avenues to further study students’ think-ing about existence proofs. Our findings suggest that students might find value in arguments that give insight into the informal reasoning used to construct it. More research is needed to understand how students think about explanatory proofs. In particular, a limitation of our study is that we did not ask students to validate a valid and explanatory proof. It could be insightful to better understand how students think about valid explanatory proofs in addition to invalid explanatory proofs like the ones in our study. An extension of our study could be to ask students to validate four cat-egories of arguments: 1) a valid explanatory proof, 2) a valid non-explanatory proof, 1 3 Int. J. Res. Undergrad. Math. Ed. 3) an invalid explanatory proof, and 4) an invalid non-explanatory proof. To see how students’ views of explanatory proofs extends beyond the context of constructive existence proofs, future research might study students’ views of proofs of claims that typically employ additional proof techniques. It might also be insightful to further investigate students thinking about the labels used in existence proofs, or proofs in general. For instance, one might investigate the influence on the choice of the letter x in the statement and the Valid Argument on the students’ view that the argument assumed the conclusion by presenting students with the Valid Argument without the using x (e.g., “Consider a−1 ∗ b” instead of “Choose x = a−1 ∗ b”). Importantly, there is also a need to find ways to base instruction on student think-ing to support them in learning about proof, including supporting students in learn-ing about proof frameworks. As our findings show, there is important mathematical work in the formal-rhetorical part of a proof, and we argue instructors should sup-port students new to proof to engage in it. We see that discussions that elicit the possible functions of the frameworks as deeply mathematical in the sense that such a discussion would unpack nuanced ways to convey specific mathematical ideas. We see such instruction far more productive than a procedural approach to proof frame-works that simply asks novice students to practice matching the given statement’s assumptions and conclusions to the structure of the argument without any discussion related to the meaningful choices of the framework. Engaging students in comparing the two arguments in this study seemed to be a productive activity for eliciting possible student interpretations of the prover’s argu-ment and how that argument did or did not prove the given statement. We suspect such an activity would lead to productive classroom discussions that could support students in better understanding constructive existence proofs. We emphasize sev-eral points of discussion that our findings suggest would be particularly advanta-geous. First, while finding the mathematical object is a critical step in constructing the proof, it is not the goal of the argument. Instructors should take care to support their students in recognizing that starting from a desired property can be a produc-tive problem-solving strategy to find a claimed object but also why this work may not show-up in a proof, or if it does, how the prover must carefully do so. This is related to our second point: we see that there is a need to support students in under-standing how one might construct a valid explanatory constructive existence proof. That is, if students have this desire, then instructors should support them in add-ing insight into how they found the object while preserving the connection between the logical structures of the associated claim and their argument. Third, instructors could act as a broker between the classroom community and the larger mathematics community (Rasmussen et al., 2009; Vroom, 2020) to support students in under-standing how introducing an object as a candidate early in the proof does not assume the conclusion and how the argument establishes the existence of the desired object. We see our study as important steps in supporting students in engaging in proof-activity related to existence claims by providing different ways in which students might think about constructive existence proofs and their frameworks. There is important mathematical work in understanding the formal-rhetorical part of a proof that students can, and should, engage in, including how a proof framework func-tions to prove the claim. We hope that future work will investigate the affordances of Int. J. Res. Undergrad. Math. Ed. 1 3 instruction that goes beyond procedural frameworks in supporting students in con-structing, comprehending, and validating proofs of existence claims. Acknowledgements We would like to thank Dr. Sean Larsen for his feedback and the anonymous review-ers for their support in refining this work. Funding This work is part of the Advancing Students’ Proof Practices in Mathematics through Inquiry, Reinvention, and Engagement project (NSF DUE #1916490). The opinions expressed do not necessarily reflect the views of the NSF. Declarations Ethics Approval The authors report the study was approved by an Internal Review Board. Consent to Participate The authors report that all participants gave informed consent. Consent for Publication All authors give consent for the publication of this manuscript. Conflicts of Interest We report that there is no conflict of interest. References Abbott, S. (2015). Understanding analysis . Springer. Bartlo, J. R. (2013). Why ask why: an exploration of the role of proof in the mathematics classroom (Doc-toral dissertation, Portland State University). Braun, V., & Clarke, V. (2006). Using thematic analysis in psychology. Qualitative Research in Psychol-ogy, 3(2), 77–101. Brown, S. A. (2017). Who’s there? A study of students’ reasoning about a proof of existence. Interna-tional Journal of Research in Undergraduate Mathematics Education, 3(3), 466–495. De Guzmán, M., Hodgson, B. R., Robert, A., & Villani, V. (1998). Difficulties in the passage from sec-ondary to tertiary education. In Proceedings of the international Congress of Mathematicians (vol. 3, pp. 747–762). 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3263	https://www.youtube.com/watch?v=guEE6QujHOE	Absolute Value Inequalities – How to Solve TabletClass Math 880000 subscribers 110 likes Description 3737 views Posted: 28 Jun 2022 TabletClass Math: How to solve absolute value inequalities. For more math help to include math lessons, practice problems and math tutorials check out my full math help program at Math Notes: Pre-Algebra Notes: Algebra Notes: Geometry Notes: Algebra 2 / Trig Notes: 10 comments Transcript: okay let's talk about how to solve absolute value inequalities and here is one example problem that we're going to be doing in a second if you are uh any sort of math student that's taking algebra or any level of algebra that would include things like Algebra 1 Algebra 2 college algebra Intermediate Algebra there's so many different types of uh flavors of algebra you're going to have to understand absolute value and the respective subtopics of uh absolute value so those would include obviously inequalities absolute value equations how to graph absolute value functions how to evaluate absolute value just with from a basic numeric uh standpoint like what's the absolute value ofg -3 and then of course uh how what's the definition of absolute value so I've done additional videos on all this but in this particular video we're just going to focus in on dealing with absolute value inequalities but again uh you're going to want to know how to solve absolute value equations and all those other things I just mentioned here in a second now I bring that up because a lot of students will study and learn how to solve absolute value equations and then when they look at an absolute value inequality they'll do the same techniques or same uh procedure uh to solve an absolute value inequality they'll be using the absolute value equation procedure and you know this is a very common uh mistake uh that I see uh with math students so you don't want to confuse the two because they you know are completely different but we're going to get into how to deal with absolute value uh inequalities in just one second actually if you know how to uh do this problem go ahead and do this real quick and put your answer in the comment section now we do want to um uh have a graph to this the respective graph as well you won't obviously be able to put that into the comment section but you can put in an inequality uh statement uh that represents the solution to this inequality so if you think you can do this stuff go ahead and put your answer into the comment section but we're going to get into this in just one second but first let me quickly introduce myself my name is John I'm the founder of tablet class math I'm also a middle and high school math teacher I'm going to leave a link to my math he program in the description of this video but if you are frustrated in mathematics maybe uh you know you've math has just not been your your topic and you just have a lot of anxiety about it maybe you don't feel like you're getting enough instruction at school or maybe you don't connect with your teachers teaching style well listen I've been teaching math for decades and I really pride myself in breaking all these concept ceps down into really bite-sized super clear and understandable kind of Concepts so anyone can be successful in math so if you're at the middle school high school even college level in terms of mathematics I can help you succeed in your respective math courses now if you're studying for any test that has a math section on it I'm talking about things like the GED uh sat act GRE GMAT asab accuplacer clip exam teacher certification exam nursing school entrance exam I go on and on and on but you get the idea can help you prepare and pass those exams if you homeschool have an outstanding very comprehensive homeschool math program that you might want to check out and if you don't have any math notes Don't Panic I'm going to leave links to my math notes um in the description of this video but I'm going to tell you right now if you want a great grade in uh math in your math course you got to have great math notes this is uh non-negotiable okay so start improving your notes and you'll see magic start happening all right so let's get into absolute value inequalities again I'm going to be doing this problem here uh in a second but let's get uh into a more basic problem first okay all right so here um basically uh this is the setup to do an absolute value inequality now let's just notice this is a less than okay so we have u a variety of different absolute value um uh inequality symbols we can have along with the absolute value so you can have the less than less than equal to greater than and greater than uh equal to so basically the procedure is effectively the same but you got to be very very careful when you're doing this now I will say as a prerequisite you need to to understand how to solve compound inequalities okay so if you're not familiar with solving linear inequalities or compound inequalities then you going you're going to want to go back and review that but you'll probably be able to to uh pick up how to solve these if you uh forgot as I go through the steps okay so in this basic problem here okay of course we have the absolute value function and an inequality uh symbol less than okay so we have this is less than nine so what is the setup well we can't do anything um uh with this uh setup right as it is right now well well actually we can okay but I guess what I'm getting at is in order to solve an absolute value inequality we have to have only the absolute value function on the left hand side and a number on the right hand side just like this so this is already set up nicely uh for us right now so uh didn't mean to confuse you but what I'm trying to say not every problem is going to be set up all nice and beautiful like this but we're going to start with a nice easy problem all right so notice I have my less than symbol so what you're going to do is you're going to take what's inside the absolute value function and in this case it's x - 5 we're going to put this in the middle okay then we're going to take this number whatever number this is okay we're going to put this over here and we're going to put the opposite or the negative of that number over here and we're going to keep the same inequality symbol uh x - 5 we're going to surround that by the a less than here and a less than here so let's make sure you understand this because this is how we're going to do all these prompts so let me actually take this uh from a different point of view so the uh the first step is you're going to take whatever's in the um inside the absolute value function we're going to put that in the middle and then we're going to surround that by this inequality symbol it's less than here and then we're going to put a less than there and then we're going to drop this number whatever it is we're going to put it right there and then we're going to take the negative of that number and put it right there okay so as long as you understand what I just showed you right here okay you're going to be able to solve any absolute value inequality now you can't do this setup and uh until and unless you have only your absolute value function on the left hand side any a number on the right hand side and not every problem is going to be like that you're going to run into more challenging absolute value problems something like this uh the absolute value 2 uh absolute value 2/3 x minus let's say 5 and then we might have minus one let me uh move this down here I just want to show you an example of something that's going to require more steps greater than equal to let's say 10 okay so here this isn't this absolute value portion right here is not isolated all by itself on the left hand side so we have some cleanup to do in order to get it to this step but um again I want to start off with what we do once uh we have an absolute value uh inequality written where it's just absolute value function and number okay so again uh these PRS are a little bit more challenging I'm going to give you some suggestions on how to continue to practice this but let's get uh into this now all right so if you understand the setup okay what we want to do is kind of like solve for x in the middle right this is a compound inequality here so uh to get X by itself I have x - 5 so I'm going to add five to everything okay again this is how you solve a compound inequality so I have -9 + five is what that's four okay and this is going to be less than uh X all right the fives go away is less than 14 okay so that's another problem right there but let me go and scoot this down give ourselves a little bit more well actually can't do that let me scoot it up isn't it really cool what I could do with this computer all right so anyways so this is uh the solution to this absolute value inequality okay it's all X's greater than four and less than 14 and then we can graphically show this by U making a little number line we'll put our two numbers here four and 14 and the way you want to do this is just put two circles here now because this is less than and not less than or equal to these are going to remain Open Circles so all numbers okay let's just read this all X's that are greater than four are this way but at the same time less than 14 so this is uh this would be the respective graph uh solution uh to this absolute value inequality okay so you can write it this way uh but you also need to understand that you can graph and you'll be required to graph um your Solutions as well again if you don't understand what I'm doing here you need to go back and review inequalities working with linear inequalities and compound inequalities okay so if this makes sense uh well you pretty much got the main idea but let's take a look at a couple more examples so we can uh make sure you have this uh down okay so here and we have our absolute value function all by itself on this side so it's good there and a number on our other side of the inequality so we're ready to set this up so the first thing is we're going to take what's inside remember you dro the absolute value function when you do the setup it's just what's inside the absolute value function in this case it's 2x - 3 I'm going to put that in the middle I'm going to keep my less than here I'm going to surround that by the same sign whatever it is I'm just going to write it twice okay don't flip it around keep that as less than and then here is going to be uh the number itself and then this is going to be the negative of that number of course this will be seven and -7 okay so this is the confusing part for a lot of students is this setup but once you have this setup now we just need to go ahead and get this x solve for this X in the middle I.E solve this compound inequality all right so let let's go ahead and take a look at I'm going to do that so I have 2x - 3 so I'm going to add three to everything again if you're confused about this you need to review compound inequalities that's a kind of prerequisite uh before you get into this so I have -7 + 3 that gives me -4 uh 2x - 3 + 3 that gives me down to 2x so this is like kind of solving for an equation with x in the middle um I'm trying to get X in the middle I.E that's what I'm trying to do so it's not in equation but you're basically taking similar steps as if you were solving an equation so I have 7 plus 3 that is 10 so to get X by itself I'm going to go ahead and divide everything by two okay and when I do that I get4 ID 2 is uh4 ID 2 is -2 I get X and then here I get five and this is the solution okay so what am I going to do well let's go ahead and graph this because again this is very important what this is saying this is a compound uh inequality I want all X's uh all numbers that are greater than -2 but at the same time less than5 so I put -2 and five on my number line okay remember -2 is over here five is over here and all numbers uh in between those um -2 and five are the solution okay not including -2 and five okay so we'll see this uh we'll see another example of less than or equal to or greater than equal to here in a second where you got to shade these circles in but these uh circles are specifically left open okay all right so hopefully this makes sense to you and now let's take on a little bit more interesting problem okay so uh this problem we have our absolute value portion here but we have uh the absolute value portion of this problem is right here then we have a number and now this is a greater than okay so this is a couple uh this is like you know adding up or adding a new twist to this problem so what do we need to do first well the first thing is we want to get the absolute value function by itself on the left hand side and a number on itself on right hand side so whatever basic algebra you need to do to get that uh to look like this that's what you have to do okay so in this case I'm just going to add one to both sides of the equation or S both sides of this absolute value inequality see I have to be careful you know because you know you're doing things that you do when you're solving equations but this is not an equation okay you do take similar steps so here I have um absolute value of 9 - x is greater than or equal to 15 okay so what's the setup well the setup is not going to change okay it's going to be the same um concept we're going to take what's inside the absolute value function 9 - x we're going to put that in the middle all right then we're going to put this 15 down here and then we're going to put the this negative of this number which is going to be5 right here again keeping this is greater than or equal to right uh uh we're being consistent with that we're kind of surrounding that 9 - x by this greater than or equal to just like this now I have the work uh here okay I'm going to show you the full solution you kind of see what I'm doing but uh if you should really pause the video and see if you can finish this prom out because that's a good uh this would be a good little check to see if you really understand compound inequalities because you may not understand them as well as you think you do okay which is going to be a problem when you're dealing with absolute value inequality you got to again understand compound qualities so what i' like you to do is to finish this prom out and then show me the graph as well okay so uh hopefully you've done that if you are not ready go and pause the video but I'm going to carry on here and finish this problem up okay so how do I get uh X by itself well I have nine I have 9 Min - x so let's subtract N9 uh from everything here on all sides of the inequality so that's going to get uh my uh x a little bit closer to be by itself so when I do this I'm going to get -15 +9 is -4 greater than or equal tox okay because the nines go away but I'm still left with this negative X so that's in the middle and I have 15 minus 9 which of course is 6 but this is uh Negative X not a positive X so we have to have a positive 1X okay so how do I deal with this well this this is -1x so to get a positive X I need to divide everything by1 now what happens when you divide by a negative value or multiply by negative value on both sides of an inequality okay what happens is the um inequality symbols flip so if you're dealing with a greater than it's going to go to a less than and in this case we have a greater than or equal to so it's going to go to a less than or equal to so this is a very very little subtle detail that if you don't do you're going to get the problem exactly wrong okay so notice here I'm switching the inequality symbols okay again a prerequisite to do absolute value inequalities is to know how to do uh linear and compound inequalities okay so now we have X is uh let's just read this okay we're we're talking about um all X's that are greater than equal to 24 at the same time we want all X's that are less than or equal to --6 so what's the graph of this particular in uh statement inequality statement going to look like okay well remember -6 is to the left of 24 and you really got to think about this okay so here's -6 and here's 24 just because the 24 is over here and the6 is over here doesn't mean anything in terms of the graph let's put these values where they're um at on a number line okay all right so let's read this again let's start over here this part of this inequality statement says all X's that are greater than or equal to 24 where are those X's at well they're going to be this way these are all the X's that are greater than now equal to we have to fill in that Circle okay so this is very very important that means that 24 is in fact the that number itself is part of the solution set okay all right so now that we have that let's go ahead and read this all X's that are less than or equal to -6 are going to be in what direction this direction right here and of course it's going to include -6 so we got to fill that in as well okay so let's just go to call it a wrap with these um couple PRS here there are more challenging absolute value uh inequality proms so I'm going to get suggest a couple things to you one i' I'm certain uh that I've done additional absolute value uh videos in my pre algebra and algebra playlist on my YouTube channel because I have over a thousand plus uh videos on my channel so I may I have a ton of content so you can check out uh some more of my absolute value uh proms there but I would highly suggest if you are really looking to master this that that you check out one of my um algebra courses like my algebra 1 Algebra 2 course but if you know all of this and you got all this right then I must go ahead and give you an awesome happy face let's go ahead and write it out and let's do the Mohawk today I'll give you a 1984 Mohawk I think the Mohawks were around for like I don't know late 7s 78 and I would say by 80 88 they were kind of fading out and that was a good thing but they were pretty cool when they were around but they just took too much hairspray so if you're my age in the 50s you know exactly what I'm talking about we used a ton of hairspray but I'm pretty sure our hair was kind of flammable probably not a good thing but anyways let me give you an A+ any 100% nice work okay but um here's a deal if you didn't understand this you need to uh to learn this okay and probably a couple of uh areas that may be troubling you is your knowledge of compound inequalities and inequalities in general so make sure you got that down before you start tackling absolute value inequalities and then don't confuse what you do with an absolute value equation with an absolute value inequalities that's why note taking and focus is so critical there's just so much little details that you have to uh really you know be paying attention to in order to get these problems right but you can get them right you just got to break all these down in there small little component pieces and hope this video uh hopefully this video helped you out as a quick review or maybe this is an introduction to absolute value inequalities but you know you need to carry forward and practice more challenging problems but anyways uh don't forget to smash that like button that definitely helps me out and again uh if you're new to my YouTube channel uh please consider subscribing I've been on YouTube for 10 plus series have over a th000 plus math videos basic math all the way to um calculus so again you know my whole you know effort what I'm very passionate about is try to really take a relaxing approach to learn mathematics okay we don't need to make it any more um involved than it already is so let's just relax and learn it one little step at a time so if you like my content please uh you know take advantage of all the videos I've posted but my best math help will always be within my math help program okay so with that being said I definitely wish you all the best in your mathematics Adventures thank you for your time and have a great day
3264	https://pubmed.ncbi.nlm.nih.gov/36600633/	Investigation and monitoring of heavy metal poisoning - 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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Investigation and monitoring of heavy metal poisoning Nicola L Barlow1,Sally M Bradberry2 Affiliations Expand Affiliations 1 Clinical Biochemistry, Black Country Pathology Services, West Bromwich, UK nicola.barlow1@nhs.net. 2 West Midlands Poisons Unit, City Hospital, Birmingham, UK. PMID: 36600633 DOI: 10.1136/jcp-2021-207793 Item in Clipboard Review Investigation and monitoring of heavy metal poisoning Nicola L Barlow et al. J Clin Pathol.2023 Feb. Show details Display options Display options Format J Clin Pathol Actions Search in PubMed Search in NLM Catalog Add to Search . 2023 Feb;76(2):82-97. doi: 10.1136/jcp-2021-207793. Epub 2022 Dec 12. Authors Nicola L Barlow1,Sally M Bradberry2 Affiliations 1 Clinical Biochemistry, Black Country Pathology Services, West Bromwich, UK nicola.barlow1@nhs.net. 2 West Midlands Poisons Unit, City Hospital, Birmingham, UK. PMID: 36600633 DOI: 10.1136/jcp-2021-207793 Item in Clipboard Full text links Cite Display options Display options Format Abstract Historically, heavy metal measurement and interpretation has been a highly specialised area performed only in a handful of centres within the UK. However, recent years have seen a move to more local testing due to the repatriation of referred work into pathology networks and the increased availability of inductively coupled plasma mass spectrometry technology. While management of significant poisoning is still overseen by tertiary care poisoning specialists, management of milder cases may be undertaken locally.Non-specialist clinical scientists and clinicians need to know when heavy metal testing is appropriate, which samples are required (and any specific requirements around collection) and how to interpret and act on the results.This Best Practice article provides guidance on the investigation and monitoring of the toxic elements most frequently encountered in general medical practice; lead, mercury and arsenic. It is intended as a reference guide for the non-specialist and as a comprehensive summary for clinical toxicologists and clinical scientists. Keywords: biochemistry; chemistry, clinical; toxicology. © Author(s) (or their employer(s)) 2023. No commercial re-use. See rights and permissions. Published by BMJ. PubMed Disclaimer Conflict of interest statement Competing interests: SMB is a member of the UKHSA Lead Exposure and Public Health Intervention and Surveillance (LEPHIS) Steering Group. Similar articles The relative impact of toxic heavy metals (THMs) (arsenic (As), cadmium (Cd), chromium (Cr)(VI), mercury (Hg), and lead (Pb)) on the total environment: an overview.Rahman Z, Singh VP.Rahman Z, et al.Environ Monit Assess. 2019 Jun 8;191(7):419. doi: 10.1007/s10661-019-7528-7.Environ Monit Assess. 2019.PMID: 31177337 Review. [Intoxications with trace metal elements: A new concept of the metal profile].Moutaouakkil Y, Mounir R, El Cadi MA, Lamsaouri J, Bousliman Y, ElJaoudi R.Moutaouakkil Y, et al.Ann Biol Clin (Paris). 2024 Aug 30;82(3):254-265. doi: 10.1684/abc.2024.1898.Ann Biol Clin (Paris). 2024.PMID: 39150146 Review.French. Heavy metal content of ayurvedic herbal medicine products.Saper RB, Kales SN, Paquin J, Burns MJ, Eisenberg DM, Davis RB, Phillips RS.Saper RB, et al.JAMA. 2004 Dec 15;292(23):2868-73. doi: 10.1001/jama.292.23.2868.JAMA. 2004.PMID: 15598918 Environment and Health: Heavy Metal Toxicity.Ratnapradipa D.Ratnapradipa D.FP Essent. 2024 Oct;545:13-18.FP Essent. 2024.PMID: 39412504 Review. Failure of chelator-provoked urine testing results to predict heavy metal toxicity in a prospective cohort of patients referred for medical toxicology evaluation.Weiss ST, Campleman S, Wax P, McGill W, Brent J; Toxicology Investigators Consortium.Weiss ST, et al.Clin Toxicol (Phila). 2022 Feb;60(2):191-196. doi: 10.1080/15563650.2021.1941626. Epub 2021 Jun 29.Clin Toxicol (Phila). 2022.PMID: 34184587 See all similar articles Cited by Associations between plasma and urinary heavy metal concentrations and the risk of prostate cancer.Coradduzza D, Sanna A, Di Lorenzo B, Congiargiu A, Marra S, Cossu M, Tedde A, De Miglio MR, Zinellu A, Mangoni AA, Cogoni AA, Madonia M, Carru C, Medici S.Coradduzza D, et al.Sci Rep. 2025 Apr 24;15(1):14274. doi: 10.1038/s41598-025-97682-0.Sci Rep. 2025.PMID: 40274938 Free PMC article. Pollution Risk Prediction for Cadmium in Soil from an Abandoned Mine Based on Random Forest Model.Cao J, Guo Z, Lv Y, Xu M, Huang C, Liang H.Cao J, et al.Int J Environ Res Public Health. 2023 Mar 14;20(6):5097. doi: 10.3390/ijerph20065097.Int J Environ Res Public Health. 2023.PMID: 36982005 Free PMC article. 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3265	https://www.quora.com/How-can-we-efficiently-determine-if-a-large-number-is-a-semiprime-and-what-methods-work-best-for-this	How can we efficiently determine if a large number is a semiprime, and what methods work best for this? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Factorization Methods Semiprime Large Numbers Algorithm Complexity Prime Number Theory Composite Numbers Optimal Algorithms Prime Factorization 5 How can we efficiently determine if a large number is a semiprime, and what methods work best for this? All related (32) Sort Recommended SH SS Studied Semiprime Numbers (Graduated 1966) · Author has 1.3K answers and 98.4K answer views ·May 10 Semiprime number is a number which is a product of two prime numbers. In short it is a COMPOSITE number. If such number is made up of two large prime numbers so that difference between them is comparatively small then we have two different methods to calculate the prime numbers Fist method is use of formula K^2 - n = h^2 . This method highly reduces our calculation. Even if the two prime numbers differ by 1500 , we can within notime calculate the two prime numbers whose product is given.. This method is useful for pen and paper work . The second method is deterministic method. It is highly accurate. But Continue Reading Semiprime number is a number which is a product of two prime numbers. In short it is a COMPOSITE number. If such number is made up of two large prime numbers so that difference between them is comparatively small then we have two different methods to calculate the prime numbers Fist method is use of formula K^2 - n = h^2 . This method highly reduces our calculation. Even if the two prime numbers differ by 1500 , we can within notime calculate the two prime numbers whose product is given.. This method is useful for pen and paper work . The second method is deterministic method. It is highly accurate. But this method increases our calculation. If we prepare an algorithm then it becomes easy for us to find out correct values of the two prime numbers. Let's take one example to discuss first method. EXAMPLE . Product of the two prime numbers is 3646891949 . We have to find out values of P And Q Here given number has dR 05 . Sqrt (PQ) = 60389.501 So values of K = 60390 , 60391, ….. But we can reject a few values from this above given SEQUENCE . Which numbers are likely to be rejected depends upon dR and last digit of given number . Here it is 9 . dR = 05 . So we select k = 60390. 60393. 60396 ….. Let's verify each term of the above given SEQUENCE We find that 6393 is correct value. Sqrt ( 60393×60393 - PQ ) = 650 .00 Hence difference between two prime = 1300 . P = 60393 +650 = 61043 . Q = 60393 - 650 = 59743 . THANKS . ///////////////////////////////////////////////////////////////////////////////////////////// Upvote · Sponsored by State Bank of India Start your Home Loan journey with SBI. Get interest subsidy up to ₹1.80 lakh on Home Loans up to Rs. 25 lakh and turn your dream into reality! Learn More 99 82 Related questions More answers below How can we determine if one number is always greater than another? How can large groups efficiently coordinate airport transportation in Wisconsin? What methods can be used to quickly calculate large numbers without the use of pen and paper or a calculator? What are the efficient methods of doing mathematical calculations with large numbers? Is it possible to determine which of two large numbers is smaller without calculating them? Joe Zbiciak Developed practical algorithms actually used in production. · Upvoted by Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) and Shirish H. Phatak , M.Phil. Computer Science, Rutgers University (2000) · Author has 6K answers and 58.5M answer views ·5y Related Suppose you're given a (relatively small) number p. Is there any efficient method of telling if p is prime or not? Just how small are we talking here? Under 100? Under 1000? Under 100,000? Under 10,000,000? If it’s small enough, you can just look it up in a static lookup table, or perhaps a static bitmap. For numbers under 10,000,000, I could construct a nice bitmap that fits in about 325KiB that would give you very fast O(1)O(1) lookup. You can reject a lot of numbers as “not prime” by taking a suitable modulo, and looking that up. For example, if you compute p mod 30 p mod 30, 22 of the 30 possible results indicate that the number is composite. You eliminate over 2/3ds of the results right there by rejecting a Continue Reading Just how small are we talking here? Under 100? Under 1000? Under 100,000? Under 10,000,000? If it’s small enough, you can just look it up in a static lookup table, or perhaps a static bitmap. For numbers under 10,000,000, I could construct a nice bitmap that fits in about 325KiB that would give you very fast O(1)O(1) lookup. You can reject a lot of numbers as “not prime” by taking a suitable modulo, and looking that up. For example, if you compute p mod 30 p mod 30, 22 of the 30 possible results indicate that the number is composite. You eliminate over 2/3ds of the results right there by rejecting all multiples of 2, 3, and 5. (Note that 2⋅3⋅5=30 2⋅3⋅5=30.) You could extend it further. For example, you could extend the above technique to include 7 and 11, and take a modulo relative to 2310. That improves your ratio to nearly 4/5ths of values get rejected up front. For what remains, though, if you want a precise result, you’ll have to determine whether it has any divisors. If you reject the lookup table/bitmap techniques, then you’re left with a trial-division loop. You might still use some lookup tables to avoid “useless” trial divisions. For example, if you build an appropriate modulo-30 residue table, you can limit your trial divisions to values that don’t run afoul of the mod-30 table (after first testing your input against the mod-30 table, of course). Also, you can stop your trial division loop at √p p. Part of me is curious whether there’s an interesting and compact set of logic gates that would take an N-bit number and return a single true/false value as to whether that N-bit value is prime. I have a database of optimal tree-like sequences for boolean functions up to 5 variables, so I could use that for N = 5. But, that only gets me as far as the range 0..31 (or 0..62 if I trivially reject even values that aren’t exactly 2). Upvote · 999 103 9 8 Raymond Beck Former Former Army Sergeant · Author has 10.2K answers and 1.2M answer views ·May 11 many primes are twin primes like 11 and 13, a difference of 2 semi primes are the product of 2 primes take the nearest integer square root for example 11x13 = 143 144 = 12^2 take 12 plus and minus 1 to get 11 and 13 which are prime, so 143 is semi prime 5x3 =15 sqr16=4, 4+/-1= 3 and 5 which are twin primes so 15 is semi prime Upvote · Jack Smith B.A. in Physics&Mathematics, University of Illinois · Author has 4.2K answers and 11.4M answer views ·Aug 4 Related Why do numbers like 91 often seem prime at first glance, and what tricks can help quickly determine if a number is prime or not? Because the prime factorization 91 = 7 × 13 isn’t very easy to find. The most common prime factors are 2, 3 and 5. The divisibility by them is also very easy. For 2 and 5 you look at the last digit. For 3, you only need to sum the digits to see if it’s divisible by 3. Multiples of 7 up to 70 are found on the multiplication table. Multiples of 11 less than 100 are also easy. If the two digits are the same, the number is divisible by 11. Otherwise, it’s not divisible by 11. This leaves 91 as the only composite number under 100 not divisible by 2, 3, 5 or 11 that is not on the multiplication table. T Continue Reading Because the prime factorization 91 = 7 × 13 isn’t very easy to find. The most common prime factors are 2, 3 and 5. The divisibility by them is also very easy. For 2 and 5 you look at the last digit. For 3, you only need to sum the digits to see if it’s divisible by 3. Multiples of 7 up to 70 are found on the multiplication table. Multiples of 11 less than 100 are also easy. If the two digits are the same, the number is divisible by 11. Otherwise, it’s not divisible by 11. This leaves 91 as the only composite number under 100 not divisible by 2, 3, 5 or 11 that is not on the multiplication table. That is also why some people will erroneously believe it is prime. Finding the prime factorization of 91 isn’t that difficult. There are three ways of doing it: 91 = 70 + 21. As the sum of two numbers divisible by 7, 91 must be divisible by 7. (This is the easiest way to memorize) 91 = 130 - 39. This can be a bit difficult to find. But once you get there, you can see that 130 and 39 are both divisible by 13. 130 is straightforward. You know 39 is divisible by 3 so you can just perform division and see that 39 = 3 × 13. 91 as the difference of two numbers divisible by 13 must as a result be divisible by 13 too. 91 = 100 - 9. As the difference of squares it can be factored. And because 100 and 9 are squares of two non-consecutive numbers, neither factor is 1 so 91 isn’t prime. Upvote · 9 3 9 1 Sponsored by Match2Pay Crypto Payment Solution for Your Business. Start today to streamline your payment processes and elevate your business. Learn More 9 4 Related questions More answers below What is an efficient method to improve my mental calculation? How should I study math? What methods did ancient Indian mathematicians use to calculate large numbers quickly? What is a way to determine if one number is greater than another without using arithmetic operators? Which semiprime number from, 1387, 54953 and 15989 satisfies the condition that (P - q) = 212? (solve this without splitting up the given semiprimes) Gaku Sato Formalist. · Author has 14.4K answers and 86.9M answer views ·Apr 14 Related How can it be determined if very large numbers are prime numbers, without using a computer? Is prime, not is probably prime? Okay, so we take n boxes. For efficiency, we take a big square frame that can easily hold all the boxes, and organize the boxes filling rows from the top down. If there are empty rows, we move boxes from the right column to the bottom then shrink the frame by 1 unit and repeat until the second to last row is full. Now we close the first (top left) box. You can also close all the boxes on the right edge (unless there's only two boxes, in which case 2 is prime). Then we set up a grid. The first point goes to the first open box to the right and the other points keep t Continue Reading Is prime, not is probably prime? Okay, so we take n boxes. For efficiency, we take a big square frame that can easily hold all the boxes, and organize the boxes filling rows from the top down. If there are empty rows, we move boxes from the right column to the bottom then shrink the frame by 1 unit and repeat until the second to last row is full. Now we close the first (top left) box. You can also close all the boxes on the right edge (unless there's only two boxes, in which case 2 is prime). Then we set up a grid. The first point goes to the first open box to the right and the other points keep the interval to the right, keeping count and looping to the next rows as necessary. If a grid point is over a box, then close it. (The grid will repeat after at most a number of rows equal to the size of the interval, so we gain efficiency by repeating the grid.) If the last box ever closes, n is composite. If the last box is still open when the first row is completely closed, n is prime (again keeping the case of 2 boxes in mind). Of course we're taking ceil(sqrt(n)) and seiving. We don't need to keep smaller primes (except n=2), and we can do everything else mechanically and use repetition to speed things up. Wait, here's a simpler (doesn't require counting) but less efficient one. Take n boxes and put them all in a row. Take the rightmost column and move those boxes filling in an incomplete row under the existing rows from left to right. If the new row would be longer than the higher rows, then start a new row with the excess. If the shape becomes taller than it is wide, then n is prime. If the new row is ever the same length as the above rows after competing the column move, stop. n is composite. Repeat (unless you've stopped). Stepping through possible divisors but not skipping composite. My kids went to Montessori, so it's all about the maths manipulatives. Upvote · 9 1 Eleftherios Argyropoulos B.S. in Mathematics&Physics, Northeastern University (Graduated 2002) · Author has 2K answers and 2.5M answer views ·3y Related Is the Mersenne number 2 1277−1 2 1277−1 a semiprime? First of all, we see that 1277 1277 is prime. This makes 2 1277−1 2 1277−1 a candidate for being Mersenne prime, but unfortunately it is not. Using Lucas - Lehmer test with the help of a computer, we can easily verify that the previous Mersenne prime is 2 607−1 2 607−1, while the next Mersenne prime after 2 1277−1 2 1277−1 is just: 2 1279−1 2 1279−1 Now, we will show that every Mersenne number is square free. Let us assume that a Mersenne number M n=2 n−1 M n=2 n−1 is perfect square. Then, since every Mersenne is odd, the square root of it p p, must be odd let say p=2 k+1 p=2 k+1. Then, we will have: M n=p 2…(1)M n=p 2…(1) (2 k+1)2=4 k 2+4 k+1=>p 2≡(2 k+1)2=4 k 2+4 k+1=>p 2≡ Continue Reading First of all, we see that 1277 1277 is prime. This makes 2 1277−1 2 1277−1 a candidate for being Mersenne prime, but unfortunately it is not. Using Lucas - Lehmer test with the help of a computer, we can easily verify that the previous Mersenne prime is 2 607−1 2 607−1, while the next Mersenne prime after 2 1277−1 2 1277−1 is just: 2 1279−1 2 1279−1 Now, we will show that every Mersenne number is square free. Let us assume that a Mersenne number M n=2 n−1 M n=2 n−1 is perfect square. Then, since every Mersenne is odd, the square root of it p p, must be odd let say p=2 k+1 p=2 k+1. Then, we will have: M n=p 2…(1)M n=p 2…(1) (2 k+1)2=4 k 2+4 k+1=>p 2≡1(m o d 4)=>M n≡1(m o d 4)…(2)(2 k+1)2=4 k 2+4 k+1=>p 2≡1(m o d 4)=>M n≡1(m o d 4)…(2) Moreover, we see that: M n=2 n−1=>M n≡3(m o d 4)…(3)M n=2 n−1=>M n≡3(m o d 4)…(3) Hence, since (2)(2) and (3)(3) are incompatible, we have our proof by contradiction. Hence, by the above proof, if a Mersenne number is semiprime, then it must have two distinct prime factors p p and q q. Now, in order to examine if 2 1277−1 2 1277−1 is semiprime or not, we will use theorem 11 from In theorem 11, we see that in order for M n M n to be semiprime, since: n=p 1=1277 n=p 1=1277 then for s=t=1 s=t=1, there must exist distinct positive integers k=l 1 k=l 1 and m=l 2 m=l 2, such that: [2(1277)k+1][2(1277)m+1]=2 1277−1[2(1277)k+1][2(1277)m+1]=2 1277−1 By giving this information to wolfram alpha, we get the result: Solve (2554 k+1)(2554 m+1)=(2 1277−1)(2554 k+1)(2554 m+1)=(2 1277−1) over the integers Result: k=1018787411380618547565117494847698271203529977218745576088401757087275651886585777017642858431120221695616837701084671866815149397839023581454571907368570943904084720966055241363858980977053988887329622214926567268176749916557212105223424176237104570451482858781437042730855186474966288632167661696448590627160162705867318033153723129280829877533090335722022993413735386702790051755 k=1018787411380618547565117494847698271203529977218745576088401757087275651886585777017642858431120221695616837701084671866815149397839023581454571907368570943904084720966055241363858980977053988887329622214926567268176749916557212105223424176237104570451482858781437042730855186474966288632167661696448590627160162705867318033153723129280829877533090335722022993413735386702790051755 and m=0 m=0 and of course: m=1018787411380618547565117494847698271203529977218745576088401757087275651886585777017642858431120221695616837701084671866815149397839023581454571907368570943904084720966055241363858980977053988887329622214926567268176749916557212105223424176237104570451482858781437042730855186474966288632167661696448590627160162705867318033153723129280829877533090335722022993413735386702790051755 m=1018787411380618547565117494847698271203529977218745576088401757087275651886585777017642858431120221695616837701084671866815149397839023581454571907368570943904084720966055241363858980977053988887329622214926567268176749916557212105223424176237104570451482858781437042730855186474966288632167661696448590627160162705867318033153723129280829877533090335722022993413735386702790051755 and k=0 k=0 which wolfram alpha omitted. Hence, if wolfram alpha is correct, then, no, 2 1277–1 2 1277–1 is not semiprime. I tested wolfram alpha for the case of n=59 n=59, for which we have: M 59=2 59−1=(179951)(3203431780337)M 59=2 59−1=(179951)(3203431780337) which means that M 59 M 59 is semiprime. Here is what wolfram alpha gave: Solve (118 m+1)(118 k+1)=2 59−1(118 m+1)(118 k+1)=2 59−1 over the integers Result: k=0 k=0 and m=4885260612740877 m=4885260612740877 k=1525 k=1525 and m=27147726952 m=27147726952 k=27147726952 k=27147726952 and m=1525 m=1525 k=4885260612740877 k=4885260612740877 and m=0 m=0 which is in complete agreement with the validity of theorem 11. Upvote · 9 2 Sponsored by RedHat Customize AI for your needs, with simpler model alignment tools. Your AI needs context, not common knowledge. Learn More 9 9 Mark Gritter recreational mathematician · Upvoted by Tim Farage , Former Professor of Mathematics · Author has 5.7K answers and 11.8M answer views ·Apr 14 Related How can it be determined if very large numbers are prime numbers, without using a computer? There are simple primality tests (or rather compositeness tests) that are feasible to do by hand, though it will naturally be pretty error-prone. The Fermat primality test says: Pick some integer a a that is coprime with n n. (If you can’t find one, then n n is not prime.) Compute a n−1 mod n a n−1 mod n. If the result is not 1, then n n is not prime. There are a few numbers which are composite but fail that test for any coprime a a. They are called Carmichael numbers. However, for hand computation this is still probably the best bet. We can pick a base that makes things easy for us — like a=2 a=2. The odds that we will Continue Reading There are simple primality tests (or rather compositeness tests) that are feasible to do by hand, though it will naturally be pretty error-prone. The Fermat primality test says: Pick some integer a a that is coprime with n n. (If you can’t find one, then n n is not prime.) Compute a n−1 mod n a n−1 mod n. If the result is not 1, then n n is not prime. There are a few numbers which are composite but fail that test for any coprime a a. They are called Carmichael numbers. However, for hand computation this is still probably the best bet. We can pick a base that makes things easy for us — like a=2 a=2. The odds that we will get 1 even though the number is composite is probably much less than that we will make an error in computation. The earliest example is n=341, but there are only 21,853 “pseudoprimes” for a=2 a=2 less than 25,000,000,000. The next problem is that computing a very large exponent by hand is difficult. We might be tempted to work in base 2: Write a zero at the end of the current number (to multiply by 2) Subtract the binary representation of n n, if we would not go negative This would be pretty easy but we’d have to do it n−1 n−1 times, so millions or billions of steps for a large number. Instead, computers use “repeated squaring”: x 2 n=(x n)2 x 2 n=(x n)2 x 2 n+1=x×(x n)2 x 2 n+1=x×(x n)2 So, you will have to have a way to hand-calculate squares, and then do long division to find the remainder modulo n n. People did do large computations before computers, and even before calculators, but you might want to break up the computation into chunks of more manageable size where you can use a hand calculator. We can do this via a formula like: (10 k a+b)2=10 2 k a 2+2⋅10 k a b+b 2(10 k a+b)2=10 2 k a 2+2⋅10 k a b+b 2 That lets us turn one big square into two smaller squares and a smaller multiplication. For example, if you have an 8-digit calculator, you can compute a 8-digit square by computing three 4-digit multiplications, all of which will fit in your calculator. You can also use the Karatsuba algorithm to aid in hand multiplication. All this would be quite tedious; there’s a reason we developed computational aids! Footnotes Primality test - Wikipedia Upvote · 9 7 9 1 João Ferreira Works at Faculdade De Ciências Da Universidade De Lisboa ·9y Related How do I get the modulo of a very large number efficiently? You say you have the number in base 8. I'm assuming this means you have a string with characters from 0 to 7. There are several ways you can do it. First, convert the number back to base 10, then divide by 13 and keep the remainder of the division. This is the best (quickest and most direct) approach, provided you have a way to convert to base 10. You say you can store the number in a Java BigInteger, so I assume you do have a function to convert bases. There is another way, which is less direct but which I think would be slightly quicker. Modulo 13 in base 10 is the same as modulo 13 in any b Continue Reading You say you have the number in base 8. I'm assuming this means you have a string with characters from 0 to 7. There are several ways you can do it. First, convert the number back to base 10, then divide by 13 and keep the remainder of the division. This is the best (quickest and most direct) approach, provided you have a way to convert to base 10. You say you can store the number in a Java BigInteger, so I assume you do have a function to convert bases. There is another way, which is less direct but which I think would be slightly quicker. Modulo 13 in base 10 is the same as modulo 13 in any base (including 8). So let's directly compute de modulo in this base. 8^0 = 1 (mod 13) 8^1 = 8 (mod 13) 8^2 = -1 (mod 13) 8^3 = -8 (mod 13) 8^4 = 1 (mod 13) 8^5 = 8 (mod 13) 8^6 = -1 (mod 13) 8^7 = -8 (mod 13) 8^8 = 1 (mod 13) (if my calculations are correct) Since your string representation can be converted to an expression of the type a_0 8^0 + a_1 8^1 + ... where each a_i is the character i (counting from the last one), you can then get the last character and multiply it by 1, the the next to last and multiply it by 8, then the next and multiply it by -1, etc, and finally sum these values. Upvote · 9 2 Jan M Savage Taught CS at Tsinghua University (2006-2016) · Author has 901 answers and 10.5M answer views ·May 8 Related What is the fastest method to check if two numbers are coprime? Two numbers are coprime if they share no prime numbers in their factorizations. 8 (2³) and 15 (3 × 5) are coprime. 9 (3²) and 15 (3 × 5) are not coprime (They share 3). But you asked about the fastest way. Then it’s Two numbers are coprime (relatively prime) if their greatest common divisor is 1. Fastest way? Raku got this. There are two ways to write it (and you don’t even need to import any library like you would in some languages): The first one is safer, and tells a ‘story’—hence the verbosity adds to the signal. PS: The theme was designed by myself (the IDE is Comma). Continue Reading Two numbers are coprime if they share no prime numbers in their factorizations. 8 (2³) and 15 (3 × 5) are coprime. 9 (3²) and 15 (3 × 5) are not coprime (They share 3). But you asked about the fastest way. Then it’s Two numbers are coprime (relatively prime) if their greatest common divisor is 1. Fastest way? Raku got this. There are two ways to write it (and you don’t even need to import any library like you would in some languages): The first one is safer, and tells a ‘story’—hence the verbosity adds to the signal. PS: The theme was designed by myself (the IDE is Comma). Upvote · 9 7 9 1 Darrell Tangman BS in Mathematics, Michigan State University (Graduated 1969) · Author has 64 answers and 23.1K answer views ·Aug 7 Related Is there an efficient way to determine if the large numbers from this method are prime without extensive calculation? This may be a matter of me not understanding how Quora works, but I don’t see any context that provides a meaning for “this method”. If this question is associated with another question or with an answer to another question, it seems to me that it would make sense to quote enough of the context for this question to be meaningful on its own. If there is a way to find the context for a question like this on Quora, I would appreciate it if someone would point it out to me. Upvote · Related questions How can we determine if one number is always greater than another? How can large groups efficiently coordinate airport transportation in Wisconsin? What methods can be used to quickly calculate large numbers without the use of pen and paper or a calculator? What are the efficient methods of doing mathematical calculations with large numbers? Is it possible to determine which of two large numbers is smaller without calculating them? What is an efficient method to improve my mental calculation? How should I study math? What methods did ancient Indian mathematicians use to calculate large numbers quickly? What is a way to determine if one number is greater than another without using arithmetic operators? Which semiprime number from, 1387, 54953 and 15989 satisfies the condition that (P - q) = 212? (solve this without splitting up the given semiprimes) What are the different methods by which we can split up a given semiprime number 1619833 into two prime numbers? How can you determine if a number is currently in use? What learning methods work best for INTP mathematicians? What is the most efficient method for calculating large prime numbers mentally? What formula would be used for this process? What is the most efficient way to calculate a large number of decimals for Pi? Related questions How can we determine if one number is always greater than another? How can large groups efficiently coordinate airport transportation in Wisconsin? What methods can be used to quickly calculate large numbers without the use of pen and paper or a calculator? What are the efficient methods of doing mathematical calculations with large numbers? Is it possible to determine which of two large numbers is smaller without calculating them? What is an efficient method to improve my mental calculation? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
3266	https://www.acs.org/content/dam/acsorg/msc/downloads/chapter-2/lesson-1/ch2-l1-ngss.pdf	www.acs.org/middleschoolchemistry 1 ©2023 American Chemical Society The Next Generation Science Standards (NGSS) CHAPTER 2, LESSON 1 – HEAT, TEMPERATURE, AND CONDUCTION MS-PS1-4. Develop a model that predicts and describes changes in particle motion, temperature, and state of a pure substance when thermal energy is added or removed. DISCIPLINARY CORE IDEAS PS3.A: Definitions of Energy • Temperature is a measure of the average kinetic energy of particles of matter. The relationship between the temperature and the total energy of a system depends on the types, states, and amounts of matter present. (MS-PS3-3), (MS-PS3-4) • The term “heat” as used in everyday language refers to both thermal energy (the motion of atoms or molecules within a substance) and the transfer of that thermal energy. In science, heat is used only for this second meaning; it refers to the energy transferred due to the temperature difference between two objects. (MS-PS1-4) Students place heated metal washers in room-temperature water and measure the temperature change of the water and the washers. Students begin to develop the understanding that temperature is related to the motion of atoms and molecules and that heat is transferred from a substance at a higher temperature to a substance at a lower temperature. SCIENCE AND ENGINEERING PRACTICES Developing and Using Models • Develop a model to describe unobservable mechanisms. (MS-PS3-2) Engaging in Argument from Evidence Students investigate the question: Does the temperature of an object change when it is placed in hot water? Along with heating and taking temperature readings of metal washers and water, students see a molecular model animation depicting conduction on the molecular level. Students use and further develop this molecular model and apply it to evidence they have observed to explain their observations on the molecular level and to answer the question to investigate. www.acs.org/middleschoolchemistry 2 ©2023 American Chemical Society CROSSCUTTING CONCEPTS Cause and Effect • Cause and effect relationships may be used to predict phenomena in natural or designed systems. (MS-PS1-4) Scale, Proportion, and Quantity • Time, space, and energy phenomena can be observed at various scales using models to study systems that are too large or too small. (MS-PS1-1) Students see and apply the cause and effect relationship between heating and cooling and the motion and arrangement of molecules. Students use a molecular-level explanation to explain the macroscopic effect of the observed change in temperature.
3267	https://www.meegle.com/en_us/topics/algorithm/route-optimization-algorithms	Route Optimization Algorithms Explore diverse perspectives on algorithms with structured content covering design, optimization, applications, and future trends across industries. In today’s fast-paced world, where logistics and transportation are the backbone of industries ranging from e-commerce to healthcare, route optimization algorithms have emerged as a game-changer. These algorithms are not just about finding the shortest path from point A to point B; they are about maximizing efficiency, reducing costs, and improving customer satisfaction. Whether you're managing a fleet of delivery vehicles, planning emergency response routes, or optimizing supply chain logistics, understanding and implementing route optimization algorithms can significantly impact your operations. This article delves deep into the intricacies of route optimization algorithms, exploring their components, benefits, challenges, best practices, and future trends. By the end, you'll have actionable insights to leverage these algorithms effectively in your professional endeavors. Implement [Algorithm] solutions to optimize workflows and enhance cross-team collaboration instantly. Understanding the basics of route optimization algorithms Understanding the basics of route optimization algorithms What is Route Optimization? Route optimization refers to the process of determining the most efficient path for vehicles, goods, or individuals to travel between multiple locations. It involves minimizing travel time, distance, or cost while considering constraints such as traffic conditions, delivery windows, vehicle capacities, and road restrictions. Route optimization algorithms are mathematical models or computational methods designed to solve these complex routing problems efficiently. These algorithms are widely used in industries like logistics, transportation, and field services, where timely and cost-effective delivery is critical. They rely on advanced techniques such as graph theory, machine learning, and heuristic methods to process large datasets and generate optimal routes. Key Components of Route Optimization Algorithms Input Data: Accurate and comprehensive data is the foundation of route optimization. This includes: Objective Function: The goal of the algorithm, such as minimizing travel distance, reducing fuel consumption, or maximizing delivery efficiency. Constraints: Real-world limitations that must be factored into the optimization process, such as: Algorithm Type: Different algorithms are used based on the complexity of the problem: Output: The optimized route plan, which includes the sequence of stops, estimated travel times, and other relevant metrics. Benefits of implementing route optimization algorithms Benefits of implementing route optimization algorithms Efficiency Gains with Route Optimization Algorithms Implementing route optimization algorithms can lead to significant efficiency improvements across various domains. Key benefits include: Reduced Operational Costs: By minimizing travel distances and fuel consumption, businesses can lower transportation expenses. Improved Delivery Times: Optimized routes ensure timely deliveries, enhancing customer satisfaction and loyalty. Enhanced Resource Utilization: Algorithms help allocate vehicles and drivers more effectively, reducing idle time and maximizing productivity. Environmental Impact: Shorter routes and reduced fuel usage contribute to lower carbon emissions, supporting sustainability goals. Scalability: Route optimization algorithms can handle complex scenarios involving hundreds or thousands of locations, making them suitable for businesses of all sizes. Real-World Applications of Route Optimization Algorithms E-commerce Logistics: Companies like Amazon and FedEx use route optimization to ensure fast and cost-effective deliveries. Emergency Services: Ambulances and fire trucks rely on optimized routes to reach destinations quickly during emergencies. Field Service Management: Technicians and service providers use route optimization to plan their daily schedules efficiently. Public Transportation: Transit authorities optimize bus and train routes to improve service reliability and reduce operational costs. Supply Chain Management: Manufacturers and distributors use route optimization to streamline the movement of goods between warehouses and retail outlets. Related: Click here to utilize our free project management templates! Challenges in route optimization algorithm development Challenges in route optimization algorithm development Common Pitfalls in Route Optimization Design Data Quality Issues: Inaccurate or incomplete data can lead to suboptimal routes and inefficiencies. Complex Constraints: Real-world routing problems often involve multiple constraints, making it challenging to find feasible solutions. Scalability: As the number of locations increases, the computational complexity of the problem grows exponentially. Dynamic Changes: Traffic conditions, weather, and other factors can change rapidly, requiring real-time adjustments to routes. Integration Challenges: Integrating route optimization algorithms with existing systems and workflows can be difficult. Overcoming Route Optimization Limitations Invest in High-Quality Data: Use GPS tracking, IoT devices, and other technologies to collect accurate and real-time data. Leverage Hybrid Algorithms: Combine heuristic and metaheuristic approaches to balance accuracy and computational efficiency. Implement Real-Time Updates: Use dynamic routing algorithms that can adapt to changing conditions. Focus on Scalability: Employ cloud-based solutions and parallel computing to handle large-scale routing problems. Collaborate with Experts: Work with data scientists and algorithm developers to design tailored solutions for your specific needs. Best practices for route optimization algorithm implementation Best practices for route optimization algorithm implementation Tools for Enhancing Route Optimization Software Solutions: Platforms like Route4Me, OptimoRoute, and Google Maps API offer robust route optimization capabilities. Machine Learning Models: Use predictive analytics to forecast traffic patterns and delivery times. IoT Integration: Connect vehicles and devices to gather real-time data for dynamic routing. Cloud Computing: Leverage cloud-based systems for scalability and faster processing. Visualization Tools: Use mapping software to visualize routes and identify potential bottlenecks. Case Studies of Successful Route Optimization Implementation Amazon’s Last-Mile Delivery: Amazon uses advanced route optimization algorithms to ensure fast and efficient deliveries, even in densely populated urban areas. UPS’s ORION System: UPS developed the On-Road Integrated Optimization and Navigation (ORION) system, which saves millions of miles annually by optimizing delivery routes. DHL’s Smart Logistics: DHL employs route optimization algorithms to streamline global supply chain operations, reducing costs and improving delivery times. Related: Click here to utilize our free project management templates! Future trends in route optimization algorithms Future trends in route optimization algorithms Emerging Technologies Impacting Route Optimization Artificial Intelligence: AI-powered algorithms can predict traffic patterns and optimize routes in real-time. Autonomous Vehicles: Self-driving cars and drones rely on route optimization for navigation and delivery. Blockchain: Blockchain technology can enhance transparency and security in logistics operations. 5G Connectivity: Faster data transmission enables real-time updates and dynamic routing. Sustainability Initiatives: Algorithms are being designed to prioritize eco-friendly routes and reduce carbon footprints. Predictions for Route Optimization Evolution Increased Automation: Algorithms will become more autonomous, requiring minimal human intervention. Integration with Smart Cities: Route optimization will play a key role in urban planning and traffic management. Personalized Routing: Algorithms will offer tailored solutions based on individual preferences and constraints. Global Expansion: Route optimization will become more accessible to small businesses and developing regions. Enhanced Collaboration: Cross-industry partnerships will drive innovation and adoption of advanced algorithms. Examples of route optimization algorithms in action Examples of route optimization algorithms in action Example 1: Optimizing Delivery Routes for E-commerce An e-commerce company uses route optimization algorithms to plan delivery routes for its fleet of vehicles. By analyzing customer locations, delivery time windows, and traffic patterns, the algorithm generates efficient routes that minimize travel time and fuel consumption. This results in faster deliveries, lower costs, and improved customer satisfaction. Example 2: Emergency Response Routing A city’s emergency services department employs route optimization algorithms to plan routes for ambulances and fire trucks. The algorithm considers factors like traffic congestion, road closures, and proximity to hospitals or fire stations. This ensures that emergency responders can reach their destinations quickly, potentially saving lives. Example 3: Field Service Scheduling A telecommunications company uses route optimization algorithms to schedule daily routes for its technicians. The algorithm accounts for factors like job priority, technician skill levels, and travel distances. This helps the company complete more service calls in less time, improving efficiency and customer satisfaction. Related: Click here to utilize our free project management templates! Step-by-step guide to implementing route optimization algorithms Step-by-step guide to implementing route optimization algorithms Define Objectives: Determine the specific goals of your route optimization efforts, such as reducing costs or improving delivery times. Gather Data: Collect accurate and comprehensive data on locations, constraints, and other relevant factors. Choose an Algorithm: Select the most suitable algorithm based on the complexity of your routing problem. Integrate Tools: Use software platforms and APIs to implement the algorithm and visualize routes. Test and Validate: Run simulations to ensure the algorithm generates optimal routes under various scenarios. Monitor and Adjust: Continuously monitor performance and make adjustments based on real-time data. Tips for do's and don'ts Tips for do's and don'ts \| Do's \| Don'ts \| --- \| \| Use high-quality data for accurate results. \| Ignore real-time updates and dynamic changes. \| \| Test algorithms under different scenarios. \| Overlook scalability when handling large datasets. \| \| Invest in advanced tools and technologies. \| Rely solely on manual route planning. \| \| Collaborate with experts for tailored solutions. \| Neglect integration with existing systems. \| \| Prioritize customer satisfaction in route planning. \| Focus only on cost reduction without considering other factors. \| Related: Click here to utilize our free project management templates! Faqs about route optimization algorithms Faqs about route optimization algorithms What industries benefit most from route optimization algorithms? Industries such as logistics, transportation, e-commerce, healthcare, and field services benefit significantly from route optimization algorithms due to their reliance on efficient routing for operations. How can beginners start with route optimization algorithms? Beginners can start by learning the basics of graph theory and algorithm design, exploring software tools like Route4Me or OptimoRoute, and experimenting with simple routing problems. What are the top tools for route optimization algorithms? Popular tools include Route4Me, OptimoRoute, Google Maps API, and custom-built solutions using programming languages like Python and R. How does route optimization impact scalability? Route optimization algorithms enable businesses to handle complex routing problems involving large datasets, making them scalable for operations of any size. Are there ethical concerns with route optimization algorithms? Ethical concerns include data privacy issues, potential biases in algorithm design, and environmental impacts of routing decisions. Addressing these concerns requires transparency and responsible algorithm development. Implement [Algorithm] solutions to optimize workflows and enhance cross-team collaboration instantly. Navigate Project Success with Meegle Pay less to get more today. Explore More in Algorithm Navigate Project Success with Meegle, Today Limited time offers are available. Pay less to get more today.
3268	https://math.answers.com/math-and-arithmetic/What_is_general_solution_to_a_trigonometric_equation	What is general solution to a trigonometric equation? - Answers Create 0 Log in Subjects>Math>Math & Arithmetic What is general solution to a trigonometric equation? Anonymous ∙ 10 y ago Updated: 6/26/2025 The general solution to a trigonometric equation provides all possible angles that satisfy the equation. For example, for equations involving sine or cosine, the general solutions can often be expressed in the form ( x = n \cdot 2\pi + \theta ) or ( x = n \cdot 2\pi - \theta ) for sine, or ( x = n \cdot 2\pi + \theta ) for cosine, where ( n ) is any integer and ( \theta ) is a specific angle solution. This reflects the periodic nature of trigonometric functions, allowing for infinitely many solutions based on the periodic intervals. AnswerBot ∙ 3 mo ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Math & Arithmetic ### What is the global solution of an ordinary differential equation? The global solution of an ordinary differential equation (ODE) is a solution of which there are no extensions; i.e. you can't add a solution to the global solution to make it more general, the global solution is as general as it gets. ### What is the general solution of a differential equation? It is the solution of a differential equation without there being any restrictions on the variables (No boundary conditions are given). Presence of arbitrary constants indicates a general solution, the number of arbitrary constants depending on the order of the differential equation. ### What algebra 2 problem equals 24? What topics are included in "Algebra 2" may vary depending on the specific textbook. But in general, if you want an equation that has a certain solution, in this case 24, you can start with the equation:x = 24 Then you can do several operation on this equation, always doing the same on both sides, such as: Add or subtract the same number on both sides Multiply or divide both sides by the same number Square both sides Apply functions, such as trigonometric functions, inverses trigonometric functions, exponential functions, etc. In general, you can do this repeatedly. ### When a variable appears in the denominator of one or more term of an equation? It can. And does, for example, in the hyperbolic trigonometric functions. It can make the solution harder but there is no law that says that solutions must be easy! ### How can you tell when an equation in one variable has infinitely many solutions or no solutions? There is no simple method. The answer depends partly on the variable's domain. For example, 2x = 3 has no solution is x must be an integer, or y^2 = -9 has no solution if y must be a real number but if it can be a complex number, it has 2 solutions. Related Questions Trending Questions What are the 8 composite numbers between 31 and 41?What it means to divide the numerator and the denominator of a ratio by the same factor?Is 108 a even number?What is 3 and 4 in a fraction percent in a decimal?How many times does 22 go into 79?What does 4 sixteeths equal?What is 10 x 10 to the 6th power?what is your problem?Is 282109272212719 least or greatest?What do you take after trigonometry?In an experiment to see if the comfort of chairs in a classroom makes a difference in how long students will work students will work on math problems one group was in standard chairs and one group w?How many 24x24 tiles to cover 720sq ft?How do you find average speed on a graph?Where is the oil filtre for soarer 96?What is the mixed number for 73.477?Can have two real numbers if the quotient is not real number?What is 14 over 12 in lowest form?What does po stand for in tpr graph?What is 55 percent your of 80?What will maximize the amount of interest you earn? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
3269	https://palmer.wellesley.edu/~aschultz/w18/math223/homework/w18_223_hwk06_solns.pdf	Math 223, Spring 2018 Assignment 6 – Solutions Due: March 9, 2018 Part A (1) Find the least nonnegative solution to the simultaneous linear congruences below. Be sure to state the modulus for which solutions are deﬁned. (a) x ≡2 (mod 7) x ≡3 (mod 6) x ≡4 (mod 5) Solution. The Chinese remainder theorem tells us that since 7 and 6 and 5 are pairwise relatively prime, there is a unique solution to this equation modulo 7 · 6 · 5 = 210. To ﬁnd it, we ﬁrst ﬁnd the unique solution to the ﬁrst two congruences modulo 42. Note that the ﬁrst congruence tells us that x must take the form 2 + 7k for some k ∈Z. Plugging this expression for x into the second congruence gives 2 + 7k ≡3 (mod 6). Reducing modulo 6 and simplifying shows that this is the same as saying k ≡1 (mod 6). Hence if we take k = 1 then the ﬁrst two congruences have a common solution given by x = 2 + 7 · 1 = 9. (Again, the Chinese remainder theorem says this solution is unique modulo 42.) Hence our system is equivalent to the system x ≡9 (mod 42) x ≡4 (mod 5). To solve this, note the ﬁrst congruence gives us x = 9+42ℓfor some ℓ∈Z. Plugging this expression for x into the last congruence gives 9 + 42ℓ≡4 (mod 5). Reducing modulo 5 and rearranging tells us that 2ℓ≡ (mod 5). We have that ℓ= 0 is an obvious solution, and so plugging this into x gives x = 9 + 42 · 0 = 9. Hence x = 9 is a solution to all three equations. As we noted above, this solution is unique modulo 210 by the Chinese remainder theorem. □ (b) (⋆) x ≡17 (mod 31) x ≡1 (mod 3) x ≡3 (mod 4) x ≡7 (mod 11) Solution. We’ll proceed as we did in the last problem. We have that the solution to this simultaneous congruence exists and is unique modulo 4092. We’ll ﬁnd our solution by ﬁrst ﬁnding a common solution to the ﬁrst two congruences, and another solution to the last two congruences. We’ll then combine these together. To solve the ﬁrst two congruences, note that the ﬁrst tells us that x = 17 + 31k for some k ∈Z. Plugging this into the second congruence gives 17 + 31k = 1 (mod 3). Reducing modulo 3 and rearranging gives us the congruence k ≡2 (mod 3). Hence we have x = 17+31·2 = 79 is a solution to the ﬁrst congruence. The Chinese remainder theorem tells us this solution is unique modulo 93. For the latter pair of congruences, note that the third congruence tells us x = 3 + 4ℓfor some ℓ∈Z. When we plug this into the fourth congruence we get 3 + 4ℓ= 7 (mod 11). Rearranging gives us 4ℓ≡4 (mod 11), and ℓ= 1 is the obvious solution. So the second pair of congruences has the solution 3 + 4 · 1 = 7. The Chinese remainder theorem says the solution is unique modulo 44. Page 1 of ?? Math 223, Spring 2018 Assignment 6 – Solutions Due: March 9, 2018 Hence our original system is equivalent to the system x ≡79 (mod 93) x ≡7 (mod 44). Note that the Chinese remainder theorem says we have a unique solution modulo 93 · 44 = 4092, and so our only job is to ﬁnd it. The ﬁrst congruence gives us x = 79 + 93z for some z ∈Z, and substituting this into the next congruence gives 79 + 93z = 7 (mod 44). Rearranging and reducing modulo 44 then yields the congruence 5z ≡16 (mod 44). We could solve this using our methods for ﬁnding solutions to linear congruences, but instead we observe that 16 ≡60 (mod 44), and so our equation is 5z ≡60 (mod 44). This has the obvious solution of z = 12, and so our desired solution is x = 79 + 93 · 12 = 1195. □ (2) Find all incongruent solutions to the simultaneous linear congruences 3x ≡2 (mod 4) 4x ≡1 (mod 5) 6x ≡3 (mod 9) Solution. First, observe that the form of these linear congruences is not precisely the form that we use when doing Chinese remainder theorem problems, and so this theorem doesn’t tell us much about the solvability of these congruences (or the uniqueness of solutions). So we’ll start by replacing each of these congruences with a congruence of the form x ≡ai (mod ni), so that we can use our Chinese remainder theorem technology. The congruence 3x ≡2 (mod 4) has a unique solution modulo 4 by the big theorem on linear congruences (since gcd(3, 4) = 1), and it’s easy enough to see that x = 2 is a solution. So the ﬁrst congruence is equivalent to x ≡2 (mod 4). The congruence 4x ≡1 (mod 5) also has a unique solution by the big theorem on linear congruences (since gcd(4, 5) = 1), and again it’s relatively easy to see that x = 4 is a solution. So this congruence is equivalent to x ≡4 (mod 5). Finally note that since 3 divides each term in the last congruence, we can use the ”cancellation in congruence” result to say that 6x ≡3 (mod 9) if and only if 2x ≡1 (mod 3). This congruence has a unique solution (again, since gcd(2, 3) = 1), and x = 2 is an obvious solution. So this congruence is equivalent to x ≡2 (mod 3). In total, this means our original system of congruences is equivalent to the system x ≡2 (mod 4) x ≡4 (mod 5) x ≡2 (mod 3). This congruence has a unique solution modulo 60 by the Chinese remainder theorem. Furthermore, since 14 ≡2 (mod 4) and 14 ≡4 (mod 5) and 14 ≡2 (mod 3), we can see that 14 is a solution. Hence this system of congruences is equivalent to x ≡14 (mod 60). □ Page 2 of ?? Math 223, Spring 2018 Assignment 6 – Solutions Due: March 9, 2018 (3) Suppose that p and q are distinct primes with p −1 \| q −1. Prove that if gcd(a, pq) = 1, then aq−1 ≡1 (mod pq). Solution. First, note that from Homework 3 problem A1, since gcd(a, pq) = 1 we have gcd(a, p) = 1 and gcd(a, q) = 1. By Fermat’s little theorem we get aq−1 ≡1 (mod q) directly. Furthermore since p −1 \| q −1 we have (p −1)k = q −1 for some k ∈Z. Since Fermat’s little theorem gives ap−1 ≡1 (mod p), we therefore have aq−1 = ap−1k ≡1k ≡1 (mod p). The above show that aq−1 is a solution to the simultaneous system x ≡1 (mod q) x ≡1 (mod p). But the Chinese remainder theorem says there is only one solution modulo pq to this system, and by inspection we see that x = 1 is also a solution. Hence the solution we’ve already veriﬁed must be equivalent to 1 modulo pq: aq−1 ≡1 (mod pq). □ Part B (1) Prove that if p is an odd prime number, then 2(p −3)! ≡−1 (mod p). [Note: for those who haven’t seen factorials in a while, it might be useful to know that 0! is deﬁned to be 1.] Solution. By Wilson’s theorem we know that (p −1)! ≡−1 (mod p). Now since p ≥3 we can write (p −1)! = (p −1)(p −2)(p −3)!. Observe that p −1 ≡−1 (mod p) and similarly p −2 ≡−2 (mod p). Hence (p −1)(p −2) ≡2 (mod p). Substituting this into our expression above gives −1 ≡(p −1)! ≡(p −1)(p −2)(p −3)! ≡(−1)(−2)(p −3)! ≡2(p −3)! (mod p). □ (2) (⋆) Fifteen toddlers are given a giant bucket of toys to play with. Their parents want to divide the toys evenly among them, but when attempt to do so they ﬁnd that 2 toys are left over. One toddler then gets sick and has to leave the playdate. When another attempt is made to divide the toys evenly among the remaining toddlers, it happens that there is 1 toy left over. This makes 3 of the toddlers throw a tantrum, and in response their parents decide to remove them from the playdate. When another attempt is made to distribute the toys evenly among the remaining toddlers, it succeeds. What’s the minimum number of toys that could be in the bucket? Solution. The problem describes the following word problem, where T is the number of toys T ≡2 (mod 15) T ≡1 (mod 14) T ≡0 (mod 11). We can solve this system using the skills we have built up to solve Chinese remainder theorem problems. For instance, we know from the Chinese remainder theorem that there is a unique solution modulo 15 · 14 · 11 = 2310. Our job is to simply ﬁnd such a solution. First we ﬁnd a value of T that solve the ﬁrst two congruences. We know that a value of T that solves the ﬁrst congruence must take the form T = 2 + 15k for some k ∈Z. Plugging this into the second congruence then gives 2 + 15k ≡1 (mod 14). If we rearrange and reduce modulo 14 we get k ≡−1 (mod 14). So the value T = 2 −15 = −13 solves the ﬁrst congruence. [Naturally this doesn’t translate well to the problem we’re interested in, since T should be a positive number in order to make sense in the context of our problem. But for now we just want to ﬁnd a solution to this system; we can then ﬁnd the least positive solution later. Page 3 of ?? Math 223, Spring 2018 Assignment 6 – Solutions Due: March 9, 2018 Our system is now reduced to the simultaneous congruence T ≡−13 (mod 210) T ≡0 (mod 11). A solution to the ﬁrst congruence takes the form T = −13+210s for some s ∈Z, and plugging this into the next congruence gives −13 + 210s ≡0 (mod 11). Rearranging and reducing coeﬃcients modulo 11 then gives s ≡2 (mod 11). Hence a solution to the system is given by T = −13 + 2 · 210 = 407. Note that since the solution to the original system is unique modulo 2310, the number T of toys must be congruence to 407 modulo 2310. But 407 is the least positive number congruent to 407 modulo 2310, so it must be the number of toys □ (3) Determine the remainder of 3253905 + 546625 upon division by 13. Solution. It turns out that 13 · 25 = 325 and 13 · 42 ≡546, so both 325 ≡0 (mod 13) and 546 ≡0 (mod 13). Hence we get 3253905 + 546625 ≡03905 + 0625 ≡0 (mod 13). Note: My intention wasn’t for this problem to collapse quite so much. I’ll confess that I chose these numbers at random, and so it was pure coincidence that both bases were congruent to 0 modulo 13. It’s worth knowing how to do this problem in the situation that the bases are not congruent to 0 in the given modulus, however, so you should practice some problems like that. See the next problem for examples of computations in this vein. □ (4) Prove that 645 is a 2-psuedoprime. Solution. Our job is to show that 2645 ≡2 (mod 645). We could attempt to do this directly, but it would require a signiﬁcant amount of computation. Instead, we’ll try to to compute the value of 2645 modulo each of its prime factors, and then “glue” these computations together via the Chinese remainder theorem. For this reason, it will be useful to know that 645 = 3 · 5 · 43. First, note that since gcd(2, 3) = 1, we have 22 ≡23−1 ≡1 (mod 3) by Fermat’s little theorem (though, honestly, we don’t need to break out some fancy theorem to check this fact). Since 645 = 322 · 2 + 1 we therefore have 2645 ≡ 22322 21 ≡13222 ≡2 (mod 3). In a similar way, since gcd(2, 5) = 1, Fermat’s little theorem gives 24 ≡25−1 ≡1 (mod 5). Since 645 = 161 · 4 + 1 it then follows that 2645 ≡ 24161 21 ≡11612 ≡2 (mod 5). Finally, since gcd(2, 43) = 1, Fermat’s little theorem gives 242 ≡243−1 ≡1 (mod 43). Note that 645 = 15 · 42 + 15, and so we get 2645 ≡ 24215 215 ≡115215 ≡215 (mod 43). Now one can compute 21 ≡2 22 ≡4 24 ≡16 28 ≡256 ≡41 ≡−2, with all computations performed modulo 43. Hence we get 2645 ≡215 ≡28 · 24 · 22 · 2 ≡(−2)(16)(4)(2) ≡−162 ≡−(−2) (mod 43) where in the last step we used the fact above that 16·16 ≡28 ≡−2 (mod 43) from our last computation. Page 4 of ?? Math 223, Spring 2018 Assignment 6 – Solutions Due: March 9, 2018 In all, we see that 2645 is a solution to the simultaneous congruence x ≡2 (mod 3) x ≡2 (mod 5) x ≡2 (mod 43). The Chinese remainder theorem tells us that there is only one solution to this modulo 3 · 5 · 43 = 645, and it’s obvious that x = 2 is such a solution. Hence our computed solution of 2645 and the obvious solution of 2 must be the same modulo 645. Whence 2645 ≡2 (mod 645), as desired. □ Page 5 of ??
3270	https://pmc.ncbi.nlm.nih.gov/articles/PMC6164648/	p-Value Histograms: Inference and Diagnostics - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Service Alert: Planned Maintenance beginning July 25th Most services will be unavailable for 24+ hours starting 9 PM EDT. Learn more about the maintenance. 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Learn more: PMC Disclaimer \| PMC Copyright Notice High Throughput . 2018 Aug 31;7(3):23. doi: 10.3390/ht7030023 Search in PMC Search in PubMed View in NLM Catalog Add to search p-Value Histograms: Inference and Diagnostics Patrick Breheny Patrick Breheny 1 Department of Biostatistics, University of Iowa, Iowa City, IA 52242, USA; patrick-breheny@uiowa.edu Find articles by Patrick Breheny 1, Arnold Stromberg Arnold Stromberg 2 Department of Statistics, University of Kentucky, Lexington, KY 40508, USA; stromberg@uky.edu Find articles by Arnold Stromberg 2, Joshua Lambert Joshua Lambert 2 Department of Statistics, University of Kentucky, Lexington, KY 40508, USA; stromberg@uky.edu Find articles by Joshua Lambert 2, Author information Article notes Copyright and License information 1 Department of Biostatistics, University of Iowa, Iowa City, IA 52242, USA; patrick-breheny@uiowa.edu 2 Department of Statistics, University of Kentucky, Lexington, KY 40508, USA; stromberg@uky.edu Correspondence: joshua.lambert@uky.edu; Tel.: +1-859-257-6115 Received 2018 Jul 30; Accepted 2018 Aug 30; Collection date 2018 Sep. © 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC6164648 PMID: 30200313 Abstract It is increasingly common for experiments in biology and medicine to involve large numbers of hypothesis tests. A natural graphical method for visualizing these tests is to construct a histogram from the p-values of these tests. In this article, we examine the shapes, both regular and irregular, that these histograms can take on, as well as present simple inferential procedures that help to interpret the shapes in terms of diagnosing potential problems with the experiment. We examine potential causes of these problems in detail, and discuss potential remedies. Throughout, examples of irregular-looking p-value histograms are provided and based on case studies involving real biological experiments. Keywords:p-value, histograms, inference, diagnostics 1. Introduction Since the advent of high-throughput technology, it has become common for experiments in biology and medicine to involve large numbers of hypothesis tests. A natural graphical method for visualizing the body of these tests is to take the p-values from these tests and construct a histogram. If all null hypotheses are true, these p-values follow a uniform distribution, which corresponds to a flat-looking histogram. Figure 1 illustrates an idealized version of this histogram in which 10,000 p-values have been drawn from a uniform random number generator. Figure 1. Open in a new tab Simulated p-values from an idealized setting in which all null hypotheses are true. Of course, one typically hopes that some of these null hypotheses are incorrect, and that there is an overabundance of low p-values. For example, in Rogier et al. , 201 genes were tested for differential expression upon exposure to azoxymethane (AOM) and dextran sulfate sodium (DSS) in neonatal mice using nanostring hybridization. Two sample t-tests were applied to each gene; the resulting p-values are displayed in Figure 2. In the histogram, the p-values appear to be relatively uniform except for the clear overabundance of very low p-values. Figure 2. Open in a new tab Rogier et al. : Overabundance of low p-values in a sufficiently powered experiment. There has been a tremendous amount of work in the past two decades, in particular involving false discovery rates , extending multiple comparison procedures to large-scale simultaneous inference questions such as these. Naturally, the vast majority of this work has focused on questions involving individual hypotheses. Our focus here, however, concerns what the p-value histogram says about the experiment as a whole. Some examples help to illustrate what we mean by this. Figure 3 displays another set of p-values, also from Rogier et al. , this time involving mice which were not injected with AOM. In the experiment, not a single hypothesis could be rejected at the 10% false discovery rate level. However, as we can see in the figure, the p-values clearly do not seem to be uniformly distributed. There is an apparent overabundance of low p-values, suggesting the existence of genes in mice that genuinely responded to DSS. However, the experiment is not sufficiently powered to detect them after making corrections for multiple testing. Figure 3. Open in a new tab Rogier et al. : Overabundance of low p-values in an insufficiently powered experiment. Lastly, Figure 4 presents the p-values of an experiment by Fischl et al. where paired t-tests were used to compare dependent samples. From the histogram, it appears as though something has gone wrong: there is an abundance not of low p-values but of p-values near 0.3. In summary, we have encountered four examples: no interesting features to detect (Figure 1); interesting features easily detected (Figure 2); interesting features present but unable to be detected (Figure 3); and finally, a problematic experiment (Figure 4). We discuss these cases in greater detail below and provide diagnostics for distinguishing between them. Throughout this manuscript, we use the generic term “feature” to refer to the quantity being measured in a high-throughput experiment; in our examples, the features are gene expression levels, but all of the ideas in the article are equally applicable to any high-throughput measurement such as metabolite or protein concentrations. Figure 4. Open in a new tab Fischl et al. : Overabundance of p-values near . Possible problematic experiment. All histograms were produced in R (www.r-project.org). The code that was used to produce them can be found in the Appendix A. 2. Methods 2.1. Background This section provides a brief background on hypothesis testing with multiple comparisons, and defines several terms that will be used throughout the paper. For a p-value to be valid, a p-value must satisfy under the null. For the sake of simplicity, we mostly assume we are working with proper p-values, in which under the null hypothesis and are therefore uniformly distributed. The above properties mean that, if we declare hypotheses to be false if , the proportion of time we incorrectly reject a truly null hypothesis will be no greater than ; this is known as the type I error rate. When multiple hypotheses are being tested, however, there may be many false positives (type I errors) even if the type I error rate is low. Thus, more strict significance criteria are typically required when multiple comparisons are being made. The most widespread alternative to the type I error rate for high-throughput hypothesis testing is the false discovery rate (FDR). Rather than limit the fraction of null hypotheses that can be rejected, the false discovery rate limits the fraction of rejected tests that come from null hypotheses. It is unknown, of course, whether a given test is null, thus, in practice, FDR methods instead control the expected value of this fraction. The most widely used procedure for FDR control was developed by Benjamini and Hochberg , which is what we use throughout in this manuscript, although we note that there is an extensive amount of literature on the subject with many alternative approaches. If the hypothesis being tested is not null, then the above framework places no restrictions on what its distribution may look like. In practice, however, we would like the p-values for these cases to be as small as possible, and design the experiment in order to maximize the probability of obtaining as many low p-values as possible (i.e., the power). Thus, we expect the results of a high-throughput experiment to resemble a combination of a uniform distribution (from the null hypotheses) and a distribution with an overabundance of low p-values (from the non-null hypotheses). For the purposes of this paper, we define a “regular” p-value histogram to be one that reflects such a combination. In other words, a regular p-value histogram is one that is either uniform (Figure 1) or slopes downward left-to-right (as in Figure 2 and Figure 3). We define an “irregular” histogram to be one that has any other shape (for example, Figure 4). 2.2. Higher Criticism For the data presented in Figure 3, not a single null hypothesis could be rejected at a false discovery rate of 5%. However, it seems clear from looking at the histogram that something is going on and that more low p-values are present than one would expect by chance alone. This claim can be tested using quantiles of the binomial distribution. Let b denote the bin width of the histogram, m denote the number of hypotheses being tested, X denote the number of p-values in the bin closest to zero, and denote the -level quantile of the cumulative distribution function (CDF) of the binomial distribution with size m and probability p. Then, under the global null hypothesis for all j, the probability that X exceeds is only 5%. Note that, arguably, the 0.975 quantile could be used instead, as it would be consistent with the standard of always applying two-sided tests, although it would seem a one-sided test makes more sense here. Returning to our example from Rogier et al. in Figure 3, and , so . Figure 5 superimposes this threshold upon our earlier histogram. As the figure illustrates, the fact that 27 p-values fall below 0.05 provides significant evidence to reject the global null hypothesis, even though we cannot specifically reject any individual null hypothesis. Figure 5. Open in a new tab Rogier et al. : Higher criticism threshold superimposed on Figure 3 indicating some genes may genuinely respond to the DSS. This is not a new idea in statistics, and dates back at least to John Tukey, who referred to this question as the “higher criticism” (HC) . Tukey proposed the following test statistic, based on a normal approximation to the binomial: where x is the number of p-values that fall below 0.05. One may then reject the global null at a 5% significance level if HC > 1.645. This leads to a very similar threshold as the above method for large numbers of tests (for example, with tests, the binomial threshold is 62 and the Tukey threshold is 63). We prefer the more accurate binomial threshold for our purposes here, but note that Tukey’s closed-form approach has advantages for theoretical study and has received renewed attention in the past decade in the field of high-dimensional data analysis [5,6,7,8,9]. In particular, one limitation of the procedure we describe above, and of p-value histograms in general, is that they depend on an arbitrarily chosen bin width (in this paper, we use 0.05 throughout). To address this shortcoming, Donoho and Jin considered an adaptive variant of higher criticism that involves varying the HC threshold to detect where the HC statistic is maximized. Similarly, it is widely recommended to vary the bin width when using histograms to visualize a distribution. With p-value histograms, this tends to be particularly relevant at the far left end of the histogram (i.e., in detecting patterns involving very low p-values). However, we have found that a bin width of 0.05 is in general a good place to start. Thus, what should be made of situations such as that in Figure 5? Obviously, the main point of these sorts of experiments is to assess the veracity of individual hypotheses, and in that sense an experiment giving rise to Figure 5 must be viewed as unsuccessful. However, the higher criticism here implies that there is something to find—this experiment failed to find it, but another experiment, perhaps carried out with an improved experimental design or additional observations, might be successful. This is in contrast to the conclusion one would reach after looking at the histogram in Figure 1, which suggests that there is little hope in conducting another experiment investigating the same biological question, as there is simply nothing to find. 2.3. Quality Control The same basic idea can be used to test for departures from uniformity anywhere between 0 and 1, not necessarily only among low p-values. It is straightforward to extend the approach from Section 2.2 to this case using a Bonferroni correction. With a binwidth of 0.05, this amounts to checking 20 bins, and therefore using a corrected significance threshold of 0.05/20 = 0.0025, or equivalently, a frequency threshold of . For the data from the study by Fischl et al. in Figure 4, 23,332 and 0.05, so the frequency threshold is 1261. In Figure 6, this threshold is superimposed on the original histogram. Figure 6. Open in a new tab Fischl et al. : Higher criticism threshold superimposed on Figure 4 indicates a possible problem in quality control. As another example of an experiment whose p-value histogram displays a strange departure from uniformity, Figure 7 presents the p-values of an unpublished NanoString gene expression experiment conducted in 2012 by Dr. Luke Bradley at the University of Kentucky. These p-values were extracted from a two-way interaction effect in a three-way ANOVA model for each gene. Figure 7. Open in a new tab Bradley experiment: Indication of systematic deviation from theoretical null distributions. This procedure and the bound we have described are useful as a test of quality control, which we define broadly to mean any unexpected departure of the test from its distribution under the null hypothesis (notably, not including an abundance of low p-values, which would be expected). Here, it indicates that the excess of p-values around 0.3 in Figure 6 and the excess of p-values around 1 in Figure 7 are not due merely to random chance, but that some systematic deviation from the theoretical null distributions of the test statistics has occurred. In contrast, Figure 8 presents results from an experiment by Matthews and Bridges , in which steers were assigned randomly to graze either in a pasture that contained high levels of ergot alkaloids () or one that did not (). The p-values come from a two-sample t-test of gene expression levels in the liver of the two groups of steers, as measured by NanoString. Although there is something of an abundance of p-values near 0.6, this excess is well within the bounds of random variation. Figure 8. Open in a new tab Matthews and Bridges : Abundance of p-values near , although higher criticism bound suggests that this is likely due to chance. One drawback of the proposed quality control test is that it depends on an arbitrary bin size and is only looking for unimodal departures from uniformity. An alternative approach free of bin-width dependence would be to use a Kolmogorov–Smirnoff (KS) test of the p-values against a uniform reference distribution. In addition, the KS test is more powerful at detecting multimodal departures from uniformity. The main appeal of the quality control test we describe here is that it provides a helpful visual cue with which to interpret the p-value histogram and diagnose potential problems. 2.4. Causes of Anomalous p-Value Histograms In this section, we explore some of the potential causes of the anomalous p-value histograms we have shown above. A related discussion is given by Brad Efron in Section 5 of Efron and Chapter 6.4 of Efron ; we hope to add to Efron’s remarks by providing specific instances of these violations to illustrate the connection between the cause and the resulting shape of the p-value histogram. In Section 2.4.1 and Section 2.4.2, we simulate features belonging to two groups and use a two-sample t-test to test the null hypothesis that the means of the two groups are the same. 2.4.1. Low Power Here, we simulate observations in each of two groups from the standard normal distribution. For 80% of the features, there is no difference in the means. For the remaining 20%, the difference in means was drawn from a Uniform (−2, 2) distribution. The p-value histogram and accompanying higher criticism threshold are shown in Figure 9 (left). Figure 9. Open in a new tab (Left) Simulated data with low power; (Middle) same data in(Left), showing contributions from null and non-null genes; and (Right) data simulated under same conditions as (Left), but with adequate power. With , there is insufficient evidence to reject any of the individual null hypotheses, even at a liberal FDR cutoff of 30%. Nevertheless, the higher criticism threshold indicates that some of the features are non-null. Figure 9 (middle) shows a decomposition of the p-value histogram, revealing the contributions from the null and non-null features. As one might imagine from the shape of the histogram, the rise on the left side results from the fact that most of the non-null features have low p-values. However, this is not true for all of the non-null features. With insufficient power, many of the non-null features turn out to have moderate, or even large p-values and can be found throughout all bins of the histogram. Obtaining these results is likely to be disappointing, since no significant features could be detected, but the p-value histogram and higher criticism indicate reasons for optimism. Although the initial experiment was unable to distinguish null and non-null features, there are indeed interesting features to be discovered, and a second, more adequately powered experiment may be successful at finding them. To illustrate this, we simulated data under the same settings as above, but with a sample size of in each group. In marked contrast to the previous results, we can now safely reject 504 null hypotheses at the 5% FDR level. These results are displayed in Figure 9 (right), and show much clearer separation between null and non-null features. 2.4.2. Incorrect Distributional Assumptions In Figure 10, we simulate observations in each of the two groups from the exponential distribution with rate 1, and then apply a two-sample t-test for each feature. Thus, in this example, all 10,000 features satisfy the null hypothesis. The derivation of p-values from the t-test assumes normally distributed data; here, that assumption is highly inaccurate, the exponential distribution being both highly skewed and having considerably thicker tails than the normal distribution. Figure 10. Open in a new tab Simulated data: A t-test was applied, even though the data come from an exponential distribution. Problems with distributional assumptions can be alleviated by choosing more robust, nonparametric methods. For example, replacing the t-test in the above example with a Wilcoxon rank sum test produces an appropriate, uniform-looking histogram. In addition, distributional problems are alleviated as n increases due to the central limit theorem. Increasing n to 30 in each group for this setting also yields a flat, uniform-looking histogram essentially indistinguishable from Figure 1. 2.4.3. Correlation among Features Perhaps the most common cause of irregular-looking histograms, however, is correlation among features. With respect to p-value histograms, correlation among the features being tested does not necessarily alter the shape of the histogram: marginally, each p-value still follows a uniform distribution under the null. However, it does mean that there is a greater chance of seeing an irregular deviation from uniformity in the p-value histogram. For example, imagine a bundle of highly correlated features. Due to the correlation, these features will have similar p-values. Where the bundle lies is uniformly distributed, but, wherever it lands, a “bump” will appear in the histogram. The higher criticism and quality control bounds in Section 2.2 and Section 2.3 are based on the assumption that the features being tested are mutually independent of each other. The primary practical consequence of correlation among features is that that the QC bound given in Section 2.3 is too low, leading one to conclude that an error has occurred when the irregular shape may simply be explained by correlation among the features. Fortunately, given an adequate sample size, it is possible to assess the impact of correlation among features using permutation approaches. The idea underlying the permutation approach is simple. Let denote the matrix of feature values (here, gene expression data), with each row of denoting an experimental unit consisting of m features. By permuting the rows of , we accomplish two things. First, we eliminate any association between and any other variables or group memberships that we are testing for. Second, by permuting entire rows of intact, we preserve any correlation among the rows that is present in the data. Thus, by carrying out the original test on random permutations of , we obtain p-values from the null distribution but without assuming independence among features. We repeated the test for the two-way interaction in the Bradley data seen in Figure 7 for 1000 random permutations of the expression data. For each permutation, we made a p-value histogram and recorded the count in the most highly populated bin. Figure 11 plots the histogram of the original p-values with two lines superimposed. One is the original quality control metric from Section 2.3 which assumes independence among the hypothesis tests, the other is the 95th percentile of the maximum counts from the permutation histograms. Figure 11. Open in a new tab Bradley experiment: Permutation vs. independence approaches when the correlation between genes is high. The difference between the lines is striking. In this experiment, the correlation between genes is quite high (root-mean-square correlation among the 536 genes selected for the NanoString experiment was 0.75). As a result, the spike of p-values near 0.9 observed in the data could easily have arisen simply from the correlation among genes. In fact, given the correlation among features, the irregular-looking histogram of Figure 7 is not particularly unusual at all, a point clearly communicated by the large gap between the p-value histogram and the “Permutation” line in Figure 11. Correlation among features also affects the higher criticism threshold of Section 2.2, although not as much as for quality control thresholds. The same permutation approach can be applied to obtain correlation-adjusted higher criticism thresholds, although in this case we would examine the 95th percentile of the counts for the first bin rather than the maximum count. For the Rogier et al. data of Figure 5, the higher criticism bound assuming independence was 15, while the higher criticism bound obtained from the permutation approach was 19.4. This is far less dramatic than the difference in Figure 11 because, while correlation leads to bumps in the p-value histogram, those bumps are not systematically located in the lowest bin. Unfortunately, there are limitations to the permutation approach. One is that it can be computer-intensive if p is large or if the tests themselves are time-consuming to perform. The other issue is that permutation approaches cannot be applied to very small samples. For example, we cannot use a permutation approach to investigate the Fischl et al. data in Figure 6, which involves a one-sample t-test with only three pairs of subjects. Although the idea can be extended to paired data (by randomly assigning signs to the differences rather than permuting rows), in this case, there are only four distinct random assignments that can be made, and hence four different null histograms to serve as a reference for comparison, which is not sufficient for estimating a 95th percentile. This is a fundamental limitation with applying permutation approaches to small samples, although the number of available permutations rapidly increases with sample size. For example, in a two-sample study with in each group, only 10 distinct permutations are available; however, with in each group, the number of permutations increases to 92,378. For both reasons (small sample sizes and computational burden), it is desirable to develop an analytic method for estimating higher criticism and quality control thresholds that account for correlation among features. Such a development is beyond the scope of this manuscript, but we re-examine this issue in the discussion. 2.5. Remedies When faced with an irregular p-value histogram, what action should a researcher take? In this section, we describe possible remedies. One potential remedy is to increase the sample size by collecting more data. This is most clearly indicated in situations such as in Figure 3, where there is a clear indication that non-null features are present, but unable to be reliably distinguished from noise. The higher criticism threshold is potentially a very useful tool to guide this decision in terms of whether the additional cost of collecting more data is likely to bear fruit or not. Alternatively, irregular p-value histograms may serve as an indication that the assumptions being made in the statistical analysis are not being met (see Section 2.4.2) and that one should consider an alternative approach—for example, a Wilcoxon rank sum test instead of a two-sample t-test. It is worth noting that higher sample sizes are beneficial here as well. Not only do larger sample sizes increase the robustness of many statistical tests, they also allow one to fit less restrictive statistical models. Lastly, we note that irregular p-value histograms may also indicate that the experimental design should be revised. Although to some extent correlation among features is an unavoidable biological fact, it is also the case that careful experimental designs (randomization, blocking, balance, etc.) reduce this correlation and the potential for confounding factors to induce correlation in an experiment. An element of design particularly relevant to expression and other sorts of “-omic” data is the issue of normalization. Proper normalization procedures substantially reduce correlations in this sort of data . However, while normalization procedures are well-developed for long-standing technologies such as microarray data , this is often not the case for more recent technologies such as NanoString; normalization for RNA-Seq data has greatly advanced over the past decade, although developments continue to be made [15,16]. 3. Discussion 3.1. Alternative Approaches A common alternative to the p-value histogram is the QQ (quantile–quantile) plot, which plots the observed quantiles of the p-values against the expected theoretical quantiles of, in this case, a uniform distribution. The general idea of the plot is that if there is agreement between the observed and theoretical quantiles, the points will fall on the identity line. The biggest difference between QQ plots and histograms is that QQ plots do not involve binning and can therefore reveal more information about individual tests than a histogram, especially at the extremes of the distribution. For example, suppose that we conducted 1000 tests, three of which had a very low p-value. The three significant results would be easily seen in a QQ plot, but could be obscured in the histogram. For the most part, however, QQ plots and p-value histograms convey similar information regarding power and systematic discrepancies between assumptions and test results. The issues that we have described here could also have been detected by inspecting QQ plots; conversely, except for the example given above, anything one could learn from a QQ plot, one can also see in the p-value histogram. Which approach is better largely comes down to familiarity. An analyst familiar with interpreting QQ plots may see little benefit in p-value histograms, but many people, especially outside statistics, find QQ plots foreign and abstract, and are much more familiar with histograms. Much has been written about how to interpret QQ plots, but relatively little on p-value histograms. Another alternative worth mentioning is to construct a histogram of test statistics rather than p-values. In this case, one would compare the resulting histogram to, e.g., a t distribution rather than the uniform. The primary advantage of plotting test statistics is that one can separately visualize each tail. For example, there may be a large number of overexpressed genes in an experiment but no underexpressed genes; such an asymmetry would not be apparent from a p-value histogram (assuming the p-values are from a two-tailed test). The primary disadvantage of plotting test statistics is that it is harder to see differences in the tails such as a subtle but consistent excess of low p-values. Examples of these three plots, each applied to the data in Figure 9, are shown side-by-side for comparison in Figure 12. Figure 12. Open in a new tab (Left) p-value histogram; (Middle) test statistics histogram; (Right) quantile-quantile plot. 3.2. Summary In this article, we have taken a closer look at p-value histograms with respect to two questions of vital practical importance: Higher criticism: Is there a significant excess of low p-values? In other words, is there any evidence of a systematic biological response in the experiment? Quality control: Has something gone wrong in this experiment? We present straightforward analytic diagnostics to address these questions, as well as a permutation-based approach capable of accounting for correlation among features. As Figure 11 demonstrates, correlation among features is an important issue as it has the potential to dramatically affect p-value histograms. Our derivation of higher criticism bounds in Section 2.2 and quality control bounds in Section 2.3 assumes that the p-values are proper (see Section 2.1). Many common tests, however, especially those involving discrete outcomes, are valid but not proper. For these conservative tests, the higher criticism derivation still holds, although similar to the tests themselves, the threshold will be conservative. However, for the quality control bound, this issue causes a problem, as a bump in the histogram could be the result of the conservative nature of the test and not an actual problem with the experiment. The quality control bounds derived in Section 2.3 are not likely to be useful for such tests, although the permutation approach may still be used. An additional factor that can distort p-value histograms, but which is not discussed in Section 2.4, is the effect of correlation among sampling units, possibly brought on by unmeasured confounding variables. The effect of correlation among samples (as opposed to correlation among features) is to broaden the null distribution. If this correlation is not accounted for, it will lead to an inflation of test statistics and a failure to preserve the proper size of the test, rejecting the null hypothesis too often. This is obviously an important issue, although p-value histograms are of little help in diagnosing this issue, since, when this issue is present, the histogram appears similar to “ideal” results, with a clear excess of small p-values. Finally, as noted in Section 2.4.3, it is desirable to develop an analytic method capable of computing higher criticism and quality control thresholds without the need for a permutation approach. Such a method, however, would need to both estimate and account for all pairwise correlations among the features. This is potentially a very large number, especially for genome-wide expression studies. These statistical challenges are not necessarily insurmountable, but they do fall beyond the intended scope of this article. Despite these limitations, it is our hope that the tools and examples presented in this article will be useful to researchers engaged in testing of high-throughput biological data, particularly since advice on “troubleshooting” such experiments is difficult to find in the scientific literature as problematic and underpowered studies often go unpublished. Acknowledgments We would like to thank the two anonymous reviewers, who provided many helpful remarks that led to considerable improvement of this manuscript. Appendix A The histograms can easily be reproduced in R (www.r-project.org) with the following code, which assumes that a vector p of p-values has already been calculated: b <- 0.05 hist(p, breaks=seq(0, 1, b), col=“gray”, border=“white”) Higher criticism: abline(h=qbinom(0.95, length(p), b), col=“red”) Quality control: abline(h=qbinom(1 − b0.05, length(p), b), col=“blue”) Author Contributions Conceptualization, P.B., A.S., and J.L.; Methodology, P.B., and A.S.; Software, P.B.; Validation, P.B., A.S., and J.L.; Formal Analysis, P.B., and A.S.; Investigation, P.B., A.S., and J.L.; Resources, P.B., and A.S.; Data Curation, P.B. and A.S.; Writing—Original Draft Preparation, P.B.; Writing—Review and Editing, P.B., A.S., and J.L.; Visualization, P.B.; Supervision, P.B., A.S., and J.L.; Project Administration, P.B., A.S., and J.L. Funding This research received no external funding. Conflicts of Interest The authors declare no conflict of interest. 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Introduction 2. Methods 3. Discussion Acknowledgments Appendix A Author Contributions Funding Conflicts of Interest 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
3271	https://www.youtube.com/watch?v=r2cF4hC1zWI	How to calculate resultant vector magnitude and direction using Law of Cosines and Law of Sines boyfriendnibluefairy 2180 subscribers 72 likes Description 7036 views Posted: 11 Nov 2020 In this tutorial, you will learn how to calculate the resultant or vector sum of two vectors by applying law of sines and law of cosines. 8 comments Transcript: consider vector a that represents 5 newtons of force with the direction of 30 degrees north of east and vector b that represents 10 newtons of force with direction 30 degrees east of north of course we could get the resultant vectors or the combination of a and b by using the component method and we'll end up with this the length of r is or the magnitude of r is 14.55 newtons and its direction is 50.10 degrees north of east but let's try using the law of cosines and law of sines to determine the resultant vector of vectors a and b consider these two parallel lines and if you try to intersect it with a straight line if this is theta then this is also theta in other words if for example this is 60 degrees then this is also 60 degrees these two angles here are actually called alternate interior angles again when you have two parallel lines and they are intersected by a straight line then alternate interior angles are equal so to use law of cosines and law of sines to obtain the resultant vector of vectors a and b let's try to visualize vector a plus vector b and its resultant vector so consider this x and y axis let's suppose this is vector a which is 5 newtons and its angle is 30 degrees again this is vector a from the arrow head of vector a we put the tail end of vector v so for example there's a coordinate axis here we attach vector b so again this is 10 newtons this 30 degrees east of north is equivalent to obviously 60 degrees north of east the resultant vector is from the tail end of vector a to the arrow head of vector b this is the resultant vector that we are after now to get the length or magnitude of the resultant vector we use the law of cosines let me redraw this triangle here to get a clearer picture so again this is vector a and vector b and we are after for this resultant vector let's assign the angle here that is opposite of the length a with alpha and let's designate the angle here opposite of length b as beta and opposite of the length are here as gamma based on this triangle the law of cosines tells us that the length r is equal to a squared plus b squared minus 2 a b cosine gamma apparently the gamma here can be calculated using this corner of the triangle here so remember that this is 30 degrees apparently this is 90 degrees so this one is based on our argument earlier if this is 30 degrees then this is also 30 degrees they are alternate interior angles basically this tells us that gamma equals 30 degrees plus 90 degrees plus 30 degrees so we can calculate the length of r remember this is 150 degrees [Music] and this is the same as what we have calculated earlier using the component method now to get the direction or angle of the resultant vector let's go back to this figure this is the resultant vector then its direction from the x-axis is let's symbolize it with theta so obviously remember that this corner here is beta and this one is 30 degrees so basically this tells us that theta equals again 30 degrees plus this angle here which is beta but how do we get beta so this is where the law of sines comes in so we get better by using the law of sines again based on this figure sine alpha divided by the opposite length of that angle is equal to sine beta divided by b equals sine gamma divided by c now we can actually use just one equation like for example sine alpha divided by a equals sine beta divided by b or sine beta divided by b equals sine gamma divided by c but since our target variable is beta and we already calculated gamma the value of gamma is 150 remember and c here is actually sorry this is r we already calculated it and it's equal to 14.55 newton so we already have a value for this and we already have a value for b this is given so we will be using this equation to get beta so that once we get beta we just plug it here then we get the direction of the resultant vector so using this equation the direction is the same as what we have obtained using the component method earlier please subscribe to my youtube channel thank you for watching
3272	https://physics.stackexchange.com/questions/321779/finite-difference-method-the-correct-formula	computational physics - Finite difference method: The correct formula - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finite difference method: The correct formula Ask Question Asked 8 years, 6 months ago Modified8 years, 6 months ago Viewed 382 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Suppose I have a uniform 1D grid with spacing Δ x Δ x and want to solve for example the Schrödinger equation on this grid. What is the correct approximation for the second order derivative? Is it d 2 ψ d x 2=1 2 Δ x 2(ψ i+1+ψ i−1−2 ψ i)d 2 ψ d x 2=1 2 Δ x 2(ψ i+1+ψ i−1−2 ψ i) or is it d 2 ψ d x 2=1 Δ x 2(ψ i+1+ψ i−1−2 ψ i)d 2 ψ d x 2=1 Δ x 2(ψ i+1+ψ i−1−2 ψ i) It would seem that I find both in literature. I personally think that the first is the correct one since it agrees with a second order Taylor expansion. On the other hand if I try to insert a plane wave of the form ψ k(x i)=exp(i k x i)ψ k(x i)=exp(i k x i) I obtain only the correct dispersion relation (by Taylor expansion of the cosine) for the second formula. If the last confuses you, see computational-physics differentiation Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Apr 3, 2017 at 4:26 Qmechanic♦ 222k 52 52 gold badges 636 636 silver badges 2.6k 2.6k bronze badges asked Mar 27, 2017 at 19:13 user13514user13514 681 6 6 silver badges 10 10 bronze badges 2 2 The second one is correct.lemon –lemon 2017-03-27 19:17:23 +00:00 Commented Mar 27, 2017 at 19:17 2 Like lemon said, the second one is correct. It is most likely that you've seen the first one because they have included the 1/2 in front of the Laplacian (from the 1/2m part).ConfusedStudent –ConfusedStudent 2017-03-27 19:35:18 +00:00 Commented Mar 27, 2017 at 19:35 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. The second equation is correct. As you suggested, with ψ i+1=ψ i+ψ′i Δ x+1 2 ψ′′i Δ x 2+1 6 ψ′′′i Δ x 3+O(Δ x 4)ψ i+1=ψ i+ψ i′Δ x+1 2 ψ i″Δ x 2+1 6 ψ i‴Δ x 3+O(Δ x 4) and ψ i−1=ψ i−ψ′i Δ x+1 2 ψ′′i Δ x 2−1 6 ψ′′′i Δ x 3+O(Δ x 4),ψ i−1=ψ i−ψ i′Δ x+1 2 ψ i″Δ x 2−1 6 ψ i‴Δ x 3+O(Δ x 4), we have 1 Δ x 2(ψ i+1+ψ i−1−2 ψ i)=ψ′′i Δ x 2+O(Δ x 4)Δ x 2=ψ′′i+O(Δ x 2)1 Δ x 2(ψ i+1+ψ i−1−2 ψ i)=ψ i″Δ x 2+O(Δ x 4)Δ x 2=ψ i″+O(Δ x 2) To leading order, the Hamiltonian acting on ψ ψ is then (H^ψ)i==−ℏ 2 2 m ψ′′i+U i ψ i−ℏ 2 2 m Δ x 2(ψ i+1+ψ i−1−2 ψ i)+U i ψ i(H^ψ)i=−ℏ 2 2 m ψ i″+U i ψ i=−ℏ 2 2 m Δ x 2(ψ i+1+ψ i−1−2 ψ i)+U i ψ i As @KaneBilliot said, maybe some references have included that 2 in the denominator with the expression for ψ′′i ψ i″? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Mar 27, 2017 at 19:45 answered Mar 27, 2017 at 19:33 Eric AngleEric Angle 3,086 21 21 silver badges 16 16 bronze badges 1 So if I want to solve the SE for a non-uniform grid what would the appropriate formula for the second order derivative be? Should I have 1/2 in front of x i+1−x i−1 x i+1−x i−1?user13514 –user13514 2017-03-27 20:47:26 +00:00 Commented Mar 27, 2017 at 20:47 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Your question is essentially asking the finite different coefficients for a particular derivative order and choice of sample points. With this tool you can see that your second choice is correct. In general these can be found by applying a Taylor series to each term and working out coefficients that fit. However, in practice this is simply an algorithmic and tedious exercise and would recommend using a table or the tool I linked. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Mar 27, 2017 at 19:33 JamalSJamalS 19.7k 6 6 gold badges 65 65 silver badges 109 109 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. You can easily derive the correct expression as follows. We can formally write down the Taylor expansion of a function as: exp(h D)f(x)=f(x+h)exp⁡(h D)f(x)=f(x+h) where D D is the differential operator. Then we can find many possible ways to express the derivative in terms of finite differences by writing the r.h.s. in terms of finite difference operators. E.g. we can write: f(x+h)=(1+Δ)f(x)f(x+h)=(1+Δ)f(x) where Δ Δ acts on f f as: Δ f(x)=f(x+h)−f(x)Δ f(x)=f(x+h)−f(x) We can thus write: exp(h D)f(x)=(1+Δ)f(x)exp⁡(h D)f(x)=(1+Δ)f(x) This allows you to formally express the differential operator in terms of the finite difference operator. While this will yield a formally correct expression, it will be in terms of forward differences and will thus yield an asymmetric expression. What we want here is a symmetric expression, but this obtained in just the same say, you just consider the symmetric finite difference expression: [exp(h D)−exp(−h D)]f(x)=f(x+h)−f(x−h)[exp⁡(h D)−exp⁡(−h D)]f(x)=f(x+h)−f(x−h) We can write this as: sinh(h D)f(x)=Δ s f(x)sinh⁡(h D)f(x)=Δ s f(x) where Δ s Δ s is the average between the forward and backward finite difference. We can thus formally write: D=1 h arcsinh Δ s=1 h[Δ s−Δ 3 s 6+3 Δ 5 s 40−5 Δ 7 s 112+⋯]D=1 h arcsinh⁡Δ s=1 h[Δ s−Δ s 3 6+3 Δ s 5 40−5 Δ s 7 112+⋯] The second derivative operator is easily expressed in terms of finite differences by squaring the series, we have: D 2=1 h 2[Δ 2 s−Δ 4 s 3+8 Δ 6 s 45−4 Δ 8 s 35+⋯]D 2=1 h 2[Δ s 2−Δ s 4 3+8 Δ s 6 45−4 Δ s 8 35+⋯] Using only the first term of this series yields the desired expression: Δ 2 s f(x)=1 2 Δ s[f(x+h)−f(x−h)]=1 4[f(x+2 h)−2 f(x)+f(x−2 h)]Δ s 2 f(x)=1 2 Δ s[f(x+h)−f(x−h)]=1 4[f(x+2 h)−2 f(x)+f(x−2 h)] Then the smallest stepsize h h you can take on your grid is h=Δ x 2 h=Δ x 2, because the smallest step that appears in this formula is 2 h 2 h. This then yields the approximation: D 2 f(x)≈f(x+Δ x)−2 f(x)+f(x−Δ x)Δ x 2 D 2 f(x)≈f(x+Δ x)−2 f(x)+f(x−Δ x)Δ x 2 To easily find expressions for higher order terms, it's convenient to introduce the shift operator E E that acts like: E f(x)=f(x+h)E f(x)=f(x+h) Then we have: Δ s=E−E−1 2 Δ s=E−E−1 2 So, the calculation of Δ 2 s Δ s 2 given above is nothing more than the binomial expansion of the square of the r.h.s. These methods involving formal manipulations of differential operators and finite difference operators are also useful for purely theoretical computations, albeit they'll then yield formal expressions that will then lack a rigorous mathematical derivation. E.g. suppose that you want to compute an integral from zero to infinity of some function: ∫∞0 x s−1 f(x)d x∫0∞x s−1 f(x)d x and the series expansion coefficients of the integrand are known, we have: f(x)=∑n=0∞(−1)n c n n!x n f(x)=∑n=0∞(−1)n c n n!x n We can then write c n=E n c 0 c n=E n c 0, where E E is again the shift operator. We thus have: f(x)=∑n=0∞(−1)n E n n!x n c 0=exp(−E x)c 0 f(x)=∑n=0∞(−1)n E n n!x n c 0=exp⁡(−E x)c 0 The integral can then be formally computed as: ∫∞0 x s−1 f(x)d x=Γ(s)E−s c 0=Γ(s)c−s∫0∞x s−1 f(x)d x=Γ(s)E−s c 0=Γ(s)c−s Here one assumes one is allowed to analytically continue the series expansion coefficients in some way, this can be made more rigorous, see here for details. But what should be clear is that using suitably defined operators you can get to results must faster, basically by cutting through the mathematical red tape. 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3273	https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.pdf	23 11 Article 07.1.4 Journal of Integer Sequences, Vol. 10 (2007), 2 3 6 1 47 Complementary Equations Clark Kimberling Department of Mathematics University of Evansville 1800 Lincoln Avenue Evansville, IN 47722 USA ck6@evansville.edu Abstract Increasing sequences a() and b() that partition the sequence of positive integers are called complementary sequences, and equations that explicitly involve both a() and b() are called complementary equations. This article surveys several families of such equations, including b(n) = a(jn) ± r, b(n) = a(jn) + kn, b(n) = f(a(n)), and b(n) = a(b(n −1)) + qn + r. 1 Introduction Under the assumption that sequences a and b partition the sequence N = (1, 2, 3, . . .) of positive integers, the designation complementary equations applies to equations such as b(n) = a(a(n)) + 1 in much the same way that the designations functional equations, diﬀerential equations, and Diophantine equations apply elsewhere. Indeed, complementary equations can be regarded as a class of Diophantine equations. Various pairs of complementary sequences, such as Beatty sequences ([1, 21]), have been widely studied, as evidenced by many entries in the Online Encyclopedia of Integer Se-quences. In particular, complementary sequences have been discussed extensively by Fraenkel in connection with Beatty sequences, spectra of numbers, and combinatorial games; see [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]. See also [16, 18]. However, in the literature the recog-nition of a single equation involving both sequences tends to occur only parenthetically. The purpose of this paper is to recognize classes of such equations explicitly. 1 Throughout, the symbols a and b denote strictly increasing complementary sequences; i.e., every number in N is a(n) or b(n) for some n in N, and no term of a is also a term of b. An ordinary complementary equation (OCE) is deﬁned by the form b(n) = f(b a(n),b b(n), n), (1) where b a(n) = (a (1), a (2) , . . . a(p(n))), a(p(n)) < b(n), b b(n) = (b (1), b (2) , . . . b(q(n))), b(q(n)) < b(n). That is, the nth term of the complement, b, of a, is determined as a function, f, of n and terms of a and b that are previously deﬁned. (It is common and convenient to use time-suggestive descriptors such as previously, but we note that inductive deﬁnitions do not, in fact, depend on time.) In some cases, an OCE (1) forces a(1) = 1, but, as in Example 3 below, this need not be the case. By decreeing an initial value, either a(1) = 1 or else b(1) = 1, the OCE must then have a unique solution b, or equivalently, a. An equation involving both a sequence and its complement and which cannot be rep-resented in the form (1) is a partial complementary equation (PCE). Typically, a PCE determines only a part of a solution; which is to say that there may be many solutions, as in Example 2 just below. Example 1. The OCE b(n) = a(a(n)) + 1 has unique solution a(n) = ⌊nτ⌋, where τ = (1 + √ 5)/2. The complement is given by b(n) = ⌊nτ⌋+ n, which is also the unique solution of the OCE b(n) = a(n) + n, an equation discussed more generally in section 4. Example 2. Every solution of the equation b(n) = b(n −1)a(n + 1) (2) has a(1) = 1. However, the number 2 could be either a(2) or b(1), so that (2) is a PCE. Example 3. The OCE b(n) = a(n −1) + b(n −1) (3) for n ≥2, with initial condition b(1) = 1, has as solution the Hofstadter sequence A005228: b = (1, 3, 7, 12, 18, 26, 35, 45, . . .) with complement a = A030124. Example 4. Bode, Harborth, and Kimberling discuss the equation b(n) = a(n −1) + a(n −2) with prescribed initial terms a(1) and a(2). 2 Example 5. Another PCE is a(b(n)) −b(a(n)) = 1. Solutions include b(n) = 2n and b(n) = ⌊nτ⌋. In the sequel, we shall solve four types of OCEs: b(n) = a(jn) ± r, b(n) = a(jn) + kn, b(n) = f(a(n)), and the PCE b(n) = a(b(n −1)) + qn + r. 2 The step sequence of an OCE Suppose an OCE (1) is given. Because a and b are complementary, when jointly ranked they form the sequence N. The joint ranking has the form a(1), . . . , a(u1), b (1) , . . . , b(v1), a(u1 + 1), . . . , a(u2), b (v1 + 1) , . . . , b(v2), a(u2 + 1), . . . , a(u3), b (v2 + 1) , . . . , b(v3), . . . , where the numbers ui and vi are nonnegative integers. Note that 1 = a(1) < b(1), if u1 > 0; 1 = b(1) < a(1), if u1 = 0. Each n ≥v1 + 1 has a unique representation n = vi−1 + r, where 1 ≤r ≤vi −vi−1, where i −1 = max{m : n ≥vm}. Deﬁne the step sequence s = (s(2), s(3), . . .) by s(n) = ui −ui−1 + 1, if r = 1; 1, if r > 1. Then b is clearly given by b(n) = u1 + 1, if n = 1; b(n −1) + s(n), if n ≥2. (3) In many cases, we shall see, every b(n) is immediately preceded and followed by a term of a, so that vi = i for all i ≥1, and s(n) = un −un−1 + 1 for n ≥2. In the sequel, we shall concentrate on sequences b of this kind. 3 3 The equations b(n) = a(jn) ± r Consider the equation b(n) = a(jn) + r, (4) where 1 ≤r ≤j. In order to solve this OCE, we ﬁnd inductively that a(n) =                  n, if 1 ≤n < j + r; n + 1, if j + r ≤n < 2j + r; n + 2, if 2j + r ≤n < 3j + r; . . . . . . n + q, if qj + r ≤n < (q + 1)j + r. . . . . . . For example, we move from a(n) = n to a(n) = n + 1 when n = j + r in order to make room for b(1) = j + r. The inequality for which a(n) = n + q is equivalent to n −j −r j < q ≤n −r j , so that q = n −r j and a(n) = n + n −r j . Thus, for n = 1, we have a(jn) = j. Replacing n by jn gives a(jn) = jn + jn −r j = jn + n −1. To conclude, we have a(jn) + r = jn + n −1 + r for all n ≥1, so that the solution of (4) is given by b(n) = (j + 1)n + r −1. The same method applies to the OCE b(n) = a(jn) −r, where 1 ≤r ≤j −1, giving the solution b(n) = (j + 1)n −r. 4 The equation b(n) = a(jn) + kn Suppose r and s are positive irrational numbers satisfying Beatty’s equation : 1 r + 1 s = 1. (5) 4 Then the sequences a and b given by a(n) = ⌊nr⌋and b(n) = ⌊ns⌋are a pair of complemen-tary sequences known as Beatty sequences (, , ). The OCE b(n) = a(n) + kn, (6) where k is a positive integer, occurs in Stolarsky where it is solved by means of shift operators, related to morphisms and continued fractions (, ) and also closely related to the step sequences of section 2. Sequences satisfying (6) were also studied by Fraenkel . The solution of (6) is given by the Beatty sequences a(n) = ⌊rn⌋, b(n) = ⌊sn⌋, (7) where r = 1 + √ k2 + 4 −k 2 and s = 1 + √ k2 + 4 + k 2 . (8) We wish to generalize Stolarsky’s result to certain equations of the form b(n) = a(jn) + kn; (9) speciﬁcally, we seek positive integers j and k for which the solution is a pair of Beatty sequences. Write r = m + √p j , (10) where m and p are rational numbers and √p is irrational. Equation (5) then leads to s = j√p + p + jm −m2 p −(m −j)2 . (11) The desired equations ⌊sn⌋= ⌊jrn⌋+ kn are equivalent to sn −δn = jrn −εn + kn, (12) where the fractional parts δn and εn satisfy 0 < δn = sn −⌊sn⌋< 1 and 0 < εn = jrn −⌊jrn⌋< 1 for all n. Dividing both sides of (12) by n and taking the limit as n →∞gives s = jr + k. Thus the coeﬃcient j/(p−(m−j)2) of √p on the right side of (11) must equal the coeﬃcient of √p in jr, which, by (10) is 1, so that j = p −(m −j)2, which implies j = 2m −1 ± p 4(p −m) + 1 2 . 5 In order that j be an integer, p 4(p −m) + 1 must be an odd integer: p 4(p −m) + 1 = 2q −1, so that p = q2 −q + m. Substituting into (11) and simplifying gives s = q + √p. Thus, for given m and q for which q2 −q + m is a nonsquare (below, we shall show that it is always a nonsquare), we put j = m + q −1, k = q −m, r = m + p q2 −q + m j , s = q + p q2 −q + m, and have the solution (7) of the equation (9). Instead of starting with m and q, we can start with j and k to produce q = j + k + 1 2 , m = j −k + 1 2 , √p = p (j + k + 1)2 −4k 2 . It is this latter representation of p that we now use to show that √p is irrational for all positive integers j and k. It suﬃces to show that (j + k + 1)2 −4k is a nonsquare. Let M = j + k + 1, and note that for k = 0 and k = M −1 we have M 2 −4k taking the values M 2 and (M −2)2, respectively. There is only one square between those numbers, namely (M −1)2. Therefore, if M 2 −4k is a square for some k satisﬁng 1 ≤k ≤M −2, then that value of k must satisfy M 2 −4k = (M −1)2. (13) However, (13) implies 4k = 2M −1, a number that is both even and odd. As there is no such number, (j + k + 1)2 −4k is not a square for any positive integers j and k. Examples using Beatty-pair solutions of (9) are now easy to write out, as suggested by a table: j 1 1 2 1 2 3 1 2 3 4 k 1 2 1 3 2 1 4 3 2 1 p 5/4 2 3 13/4 17/4 21/4 5 6 2 8 In connection with heap games, Fraenkel considers the extension of (6) to the OCE b(n) = ja(n) + kn, 6 where j and k are positive integers. For small values of j and k, solutions a and b include the pairs (A045671, A045672), (A045681, A045682), and (A045749, A045750), and (A045774, A045775). Example 6. Taking j = 1 and leaving k arbitrary in (9) gives (8). Example 7. Taking j = k gives the OCE b(n) = a(jn) + jn with solution (7) using r = 1 + p 4j2 + 1 2j and r = 2j + 1 + p 4j2 + 1 2 . 5 The OCE of a dispersion, b(n) = f(a(n)) In this section, rather than starting with an OCE, we start with a certain kind of array consisting of all the positive integers, and we derive an OCE from it. Suppose f and g are strictly increasing complementary sequences and that g(1) = 1. The dispersion, D(f) = {d(i, j)}i,j≥1 of f is deﬁned as the array having ﬁrst column given by d(i, 1) = g(i) and subsequent columns given inductively by d(i, j) = f(d(i, j −1)). We shall see next that the general dispersion D(f) is naturally associated with the OCE b(n) = f(a(n)), (14) and that the dispersion provides a solution to this equation. Note ﬁrst that no member of column 1 of D(f) is an image of f, so that the terms of column 1 belong to the sequence a. Next, every member m of column 2 is of the form f(j) where j is in a, so that m is in b. Consequently, each m′ in column 3 satisﬁes m′ = f(m) for some m in sequence b. Therefore, every term of column 3 is in a; otherwise, if m′ were in b, then m would be in a, a contradiction. This shows that every term of column 3 is in a. Continuing inductively, we conclude that the terms of the odd numbered columns of D(f) are the terms of a, so that a is the ordered union of all the odd numbered columns. Likewise, b is the ordered union of all the even numbered columns. In D(f), every positive integer occurs exactly once (see for a proof), so that every positive integer is in a or b, which conﬁrms that these are complementary sequences. Example 8. Let f(n) = 2n and g(n) = 2n −1. The associated OCE is b(n) = 2a(n). The northwest corner of the dispersion D(f) is 1 2 4 8 16 32 · · · 3 6 12 24 48 96 5 10 20 40 80 160 7 14 28 56 112 224 . . . ... 7 so that a is the ordered sequence of numbers (2i + 1)22j, i ≥0, j ≥0, this being A003159, with complement b = A036554, described as the numbers whose binary representation ends in an odd number of zeros. As suggested by Example 8, the OCE b(n) = ka(n), for k ≥2, has solution a the ordered union of the numbers (ki + r)k2j, 1 ≤r ≤k −1, i ≥0, j ≥0, with b the ordered union of the numbers (ki + r)k2j+1. Example 9. Let f(n) = 2n + 1 for n ≥1, let g(1) = 1 and g(n) = 2n for n ≥2. The associated OCE is b(n) = 2a(n) + 1. The northwest corner of the dispersion D(f) is 1 3 7 15 31 63 · · · 2 5 11 23 47 95 4 9 19 39 79 159 6 13 27 55 111 223 . . . ... so that a is the ordered sequence of numbers 22j+1 −1 and (2i + 1)22j −1, i ≥1, j ≥0, and b is the ordered sequence of numbers 22j+2 −1 and (2i + 1)22j+1 −1, i ≥1, j ≥0. It is natural to ask what OCEs are associated with well-known dispersions. In the next examples, we answer this question for the Wythoﬀarray, the Wythoﬀdiﬀerence array, the Stolarsky array, and the inverse Wythoﬀarray. Example 10. Column 1 of the Wythoﬀarray W = {w(i, j)} is given by w(i, 1) = ⌊⌊iτ⌋τ⌋, and the ordered complement of column 1, written as an increasing sequence, is given by f(n) = ⌊(n + 1)τ⌋−1. The associated OCE is therefore b(n) = ⌊(a(n) + 1)τ⌋−1. 8 Its solution a is the ordered union of odd numbered columns of W, so that a is simply the lower Wythoﬀsequence, A000201. The complement, b, is the rest of W, which as an increasing sequence is the upper Wythoﬀsequence, A001950. Initial terms are given by a = (1, 3, 4, 6, 8, 9, 11, 12, . . .), b = (2, 5, 7, 10, 13, 15, 18, 20, . . .). Fraenkel and Kimberling discuss Example 10 in greater detail. Example 11. The Wythoﬀdiﬀerence array, D = {d(i, j)}, given by A080164, is the dispersion of the upper Wythoﬀsequence, which, when written in increasing order, is given by f(n) = ⌊(τ + 1)n⌋. The associated OCE is therefore b(n) = ⌊(τ + 1)a(n)⌋. Its solution from columns of D is given by initial terms as follows: a = (1, 3, 4, 5, 6, 8, 9, 11, 12, . . .), b = (2, 7, 10, 13, 15, 20, . . .). Example 12. The inverse Wythoﬀarray, X = {x(i, j)}, has ﬁrst column given by x(i, 1) = s(n) = 1, if i = 1; ⌊iτ⌋−1, if i > 1 and ordered complement of column 1 given by f(n) = ⌊(n + 1)τ⌋+ n. (The deﬁnition of the inverse of a dispersion is given in ). The associated OCE is b(n) = ⌊(a(n) + 1)τ⌋+ a(n). Its solution from columns of X is given by initial terms as follows: a = (1, 2, 3, 5, 7, 8, 10, 11, 12, 13, . . .), b = (4, 6, 9, 14, 19, 22, 27, 30, 33, 35, . . .). 6 The form b(n) = a(b(n −1)) + qn + r We begin with the OCE b(n) = a(b(n −1)) + 1. (15) If a(1) = 1, the solution is b(n) = n2 + n 2 + 1, whereas if b(1) = 1, the solution is b(n) = n2 + n 2 . (16) 9 We shall prove the latter, starting with a lemma closely related to (3). Lemma. Suppose a and b satisfy (15) and the initial condition b(1) = 1. Then a(m + 1) = a(m) + 2, if m = b(k) for some k; a(m) + 1, otherwise. Proof. Because b(1) = 1, we have a(1) ≥2, so that b(2) ≥3, by (15). As a ﬁrst step in an induction proof, we therefore have b(2) −b(1) ≥2. Assume for arbitrary k ≥2 that b(k) −b(k −1) ≥2. Then, using (15), b(k + 1) −b(k) = a(b(k)) −a(b(k −1)) ≥b(k) −b(k −1) ≥2. Thus, for every k ≥2, the numbers b(k)−1 and b(k)+1 are terms of the sequence a. As every positive integer is in exactly one of the sequences a and b, we have a (m + 1) = a (m) + 2 if a(m) is in b and a (m + 1) = a (m) + 1 otherwise. ■ Now, we shall prove that (15), with the intial condition b(1) = 1, implies (16). By (15), b(2) = a(b(1)) + 1 = a(1) + 1 ≥3, so that 2 cannot be b(2) and must therefore be a(1). Then b(2) = a(b(1))+1 = a(1)+1 = 3. As a two-part induction hypothesis, assume for arbitrary k ≥1 these two equations: b(k) = k(k + 1) 2 , (17) a(b(k)) = b(k) + k. (18) Then b(k + 1) = a(b(k)) + 1 = b(k) + k + 1 by (18) = (k + 1)(k + 2) 2 by (17), and it remains to be proved that a(b(k + 1)) = b(k + 1) + k + 1. (19) We have a(b(k) + 1) = a(b(k)) + 2 by the lemma = b(k) + k + 2 by (18). Also by the lemma, a(b(k) + 2) = a(b(k) + 1) + 1 a(b(k) + 3) = a(b(k) + 2) + 1 = a(b(k) + 1) + 2 . . . a(b(k) + k) = a(b(k) + k −1) + 1 = a(b(k) + 1) + k −1 a(b(k) + k + 1) = a(b(k) + k) + 1 = a(b(k) + 1) + k. (20) 10 Now, a(b(k) + 1) = a(b(k)) + 2 by the lemma = b(k + 1) + 1 by (15), so that a(b(k) + 1) + k = b(k + 1) + k + 1, which by (20) implies a(b(k) + k + 1) = b(k + 1) + k + 1, and the desired (19) now follows from the fact, already proved, that b(k) + k + 1 = b(k + 1). (Note that (18) is a PCE; one solution is given by (17); another, by b(n) = 3n −2.) Similar inductive proofs can be given for various OCEs, including the following: Equation: b(n) = a(b(n −1)) + r, where r ≥1 Initial value: b(1) = 1 Solution: b(n) = (r −1)(n −1) + n(n + 1) 2 . Equation : b(n) = a(b(n −1)) + r, where r ≥1 Initial value : a(1) = 1 Solution : b(n) = rn + n2 −n + 2 2 . Equation : b(n) = a(b(n −1)) + qn, where q ≥1 Initial value : b(1) = 1 Solution : b(n) = q(n2 + n + 2) + n2 −n + 2 2 . Equation : b(n) = a(b(n −1)) + qn, where q ≥1 Initial value : a(1) = 1 Solution : b(n) = q(n2 + n) + n2 −n + 2 2 . Equation : b(n) = a(b(n −1)) + qn + r, where q ≥1, r ≥1 Initial value : b(1) = 1 Solution : b(n) = q(n2 + n + 2) + n2 + (r + 1)n −2r + 2 2 . Equation : b(n) = a(b(n −1)) + qn + r, where q ≥1, r ≥1 Initial value : a(1) = 1 Solution : b(n) = q(n2 + n) + n2 + (2r −1)n + 2 2 . 11 To summarize, if q ≥0 and r ≥0 and q and r are not both 0, and if an initial value, either a(1) = 1 or b(1) = 1 is assumed, then the equation b(n) = a(b(n −1)) + qn + r holds for a unique second-degree polynomial in n. Several special cases are tabulated here, along with three examples in which r < 0. b(n) −a(b(n −1) = Initial Solution Name 1 a(1) = 1 (n2 + n + 2)/2 A000124, Hogben’s c. p. nos. 2 a(1) = 1 (n2 + 3n + 2)/2 A000217, triangle numbers n b(1) = 1 n2 A000290, square numbers 2n + 2 a(1) = 1 (3n2 + 5n + 2)/2 A000326, pentagonal numbers 3n + 2 a(1) = 1 2n2+3n + 1 A000384, hexagonal numbers 2n + 1 a(1) = 1 2n2+2n + 1 A001844, centered square nos. n + 1 b(1) = 1 n2+n −1 A028387 n −1 b(1) = 1 n2−n + 1 A002061, central polygonal nos. 3n −1 a(1) = 1 2n2+1 A058331 3n −2 b(1) = 1 2n2−1 A000384, hexagonal numbers References J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge University Press, 2003. S. Beatty, Problem 3173, Amer. Math. Monthly 33 (1926) 159; 34 (1927) 159. J.-P. Bode, H. Harborth, and C. Kimberling, Complementary Fibonacci sequences, forthcoming. A. S. Fraenkel, The bracket function and complementary sets of integers, Canadian J. Math. 21 (1969), 6–27. A. S. Fraenkel, Complementing and exactly covering sequences, J. Combinatorial Theory Ser. A 14 (1973), 8–20. A. S. Fraenkel, A Characterization of exactly covering congruences, Discrete Math. 4 (1973), 359–366. A. S. Fraenkel and I. Borosh, A generalization of Wythoﬀ’s game, J. Combinatorial Theory Ser. A 5 (1973), 175–191. A. S. Fraenkel, Further characterizations and properties of exactly covering congruences, Discrete Math. 12 (1975), 93–100. A. S. Fraenkel, Complementary systems of integers, Amer. Math. Monthly 84 (1977), 114–115. 12 A. S. Fraenkel, How to beat your Wythoﬀgames’ opponents on three fronts, Amer. Math. Monthly 89 (1982), 353–361. A. S. Fraenkel, Wythoﬀgames, continued fractions, cedar trees and Fibonacci searches, Theoret. Comput. Sci. 29 (1984), 49–73. A. S. Fraenkel and C. Kimberling, Generalized Wythoﬀarrays, shuﬄes and intersper-sions, Discrete Math. 126 (1994), 137–149. A. S. Fraenkel, Heap games, numeration systems and sequences, Ann. Combinatorics 2 (1998), 197–210. A. S. Fraenkel and D. Krieger, The structure of complementary sets of integers: a 3-shift theorem, Int. J. Pure Appl. Math. 10 (2004), 1–49. R. K. Guy, Max and Mex Sequences, Unsolved Problems in Number Theory, 2nd ed., Springer-Verlag, New York (1994), 227–228, Problem 27. A. Holshouser and H. Reiter, A generalization of Beatty’s theorem, Southwest J. Pure Appl. Math. (2001), 24–29. C. Kimberling, Interspersions and dispersions, Proc. Amer. Math. Soc. 117 (1993), 313–321. A. McD. Mercer, Generalized Beatty sequences, Internat. J. Math. Sci. 1 (1978), 525– 528. N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. N. J. A. Sloane, Classic Sequences In The On-Line Encyclopedia of Integer Sequences: (Part 1:) T K. B. Stolarsky, Beatty sequences, continued fractions, and certain shift operators, Canadian Math. Bull. 19 (1976) 473–482. 2000 Mathematics Subject Classiﬁcation: Primary 11B37. Keywords: Beatty sequence, complementary equation, complementary sequences, disper-sion, inverse, polygonal numbers, Stolarsky array, Wythoﬀarray, Wythoﬀdiﬀerence array, Wythoﬀsequences. (Concerned with sequences A000124 A000201 A000217 A000290 A000326 A000384 A001844 A001950 A002061 A003159 A005228 A028387 A036554 A045671 A045672 A045681 A045749 A045750 A045774 A045775 A058331 and A080164.) Received May 16 2006; Revised versions received July 26 2006; October 11 2006. Published in Journal of Integer Sequences, December 30 2006. Return to Journal of Integer Sequences home page. 13
3274	https://www.ausetute.com.au/heatneutral.html	Enthalpy of Neutralisation or Heat of Neutralization Chemistry Tutorial Key Concepts ⚛ Neutralisation, or neutralization, is the name given to the reaction that occurs between an Arrhenius acid and an Arrhenius base. (1) H+(aq) + OH-(aq) → H2O(l) ⚛ When an acid is added to an aqueous solution of base, the temperature of the solution increases. Or, if a base is added to an aqueous solution of an acid, the temperature of the solution increases. ⚛ Energy (heat) is produced when an acid reacts with a base in a neutralisation reaction. · Neutralisation reactions are exothermic. · ΔH for a neutralisation reaction is negative. ⚛ Molar enthalpy of neutralisation (molar heat of neutralization) is the energy liberated per mole of water formed during a neutralisation reaction. · ΔneutH (or ΔHneut) is the symbol given to the molar enthalpy of neutralisation. (2) · ΔneutH is usually given in units of kJ mol-1 (kJ of energy released per mole of water produced) ⚛ Enthalpy of neutralisation can be determined in the school laboratory using a styrofoam™ cup solution calorimter(3). Calculating the molar enthalpy of neutralisation from experimental results is a 3 step process: Step 1: Calculate the heat evolved: q = m × Cg × ΔT m = total mass of reaction mixture Cg = specific heat capacity of solution ΔT = change in temperature of solution Step 2: Calculate the enthalpy change for the reaction: ΔH = −q Step 3: Calculate the molar enthalpy of neutralisation: ΔneutH = ΔH ÷ n(H2O(l)) ⚛ Molar enthalpy of neutralisation for reactions between dilute aqueous solutions of strong acid and strong base is always the same(4), that is, ΔneutH = -55.90 kJ mol -1 because no bonds need to be broken, and because making the H-O bonds in H2O releases energy (breaking bonds is an endothermic process, making bonds is an exothermic process) ⚛ Less than 55.90 kJ mol-1 of energy is released when: (a) a weak acid neutralises a strong base (b) a strong acid neutralises a weak base (c) a weak acid neutralises a weak base because some of the energy is consumed in the process of breaking weak acid bonds and/or weak base bonds. Please do not block ads on this website. No ads = no money for us = no free stuff for you! Polstyrene Cup Calorimeter for Enthalpy of Neutralisation Experiment Expanded polystyrene (polystyrene foam or styrofoam™) cups are often used as take-away coffee cups because the expanded polystyrene is a good insulator, that is, your coffee stays hot but you don't burn your fingers holding the cup! This same insulating property can be exploited to make a reasonably good calorimeter (a device used to measure energy, or heat, change during a chemical reaction). A schematic diagram of a simple polystyrene foam cup calorimeter is shown below: A known amount of a reactant, such as a dilute aqueous solution of a base, is placed in the polystyrene cup (insulated vessel in the diagram). The following data for this solution are recorded: concentration: c1 = ? mol L-1 volume = V1 = ? mL A small hole is placed in the polystyrene lid to allow a thermometer to be pushed through. The fit must be snug enough to hold the thermometer in place, suspended off the bottom of the cup and immersed in the reactant. It is assumed no heat will be lost through the lid or the hole in the lid. The intial temperature of the reactant is measured and recorded. initial temperature = Ti = ? °C A known amount of the second reactant, for example a dilute solution of acid, at the same temperature is added to the solution in the cup. The following data for this second solution are recorded: concentration: c2 = ? mol L-1 volume = V2 = ? mL The thermometer is also used to stir the solution while the reaction is taking place. The temperature of the solution in the cup will rise. The maximum temperature reached is recorded as the final temperature. final temperature = Tf = ? °C For the calculation of heat of neutralization (enthalpy of neutralisation) we need to determine: (i) total mass, m, of the solution in the cup First assume additivity of volumes so that total volume of solution, Vf, is the sum of the volume of the two reactants: Vf = V1 + V2 Then assume that, because all of the solutions are dilute aqueous solutions, the density of each solution and hence the density of the final solution, d, is the same as water which we will assume is 1.00 g mL-1 density = mass (g) ÷ volume (mL) Substitute the density of water into the equation: 1.00 (g/mL) = mass (g) ÷ Vf (mL) Multiplying both sides of the equation by Vf (mL) Vf ~~(mL)~~ × 1.00 (g/~~mL~~) = ~~Vf (mL)~~ × mass (g) ÷ ~~Vf (mL)~~ Vf × 1.00 = mass (g) = m g (ii) specific heat capacity, Cg, of the solution Assume that, because all of the solutions are dilute aqueous solutions, the specific heat capacity of each solution and hence the specific heat capacity of the final solution, Cg, is the same as water which we will assume is 4.18 J g-1 °C-1 Cg = 4.18 J g-1 °C-1 (iii) change in temperature, ΔT, as a result of the neutralisation reaction: ΔT = (Tf − Ti) °C Now we can calculate the heat released by this neutralisation reaction, q, q = m × Cg × ΔT In order to determine the molar enthalpy of neutralisation (molar heat of neutralization), we need to determine how many moles of water, n(H2O(l)), have been formed as a result of the reaction: H+(aq) + OH-(aq) → H2O(l) So, using the stoichiometric ratio (mole ratio), we can see that: n(H+(aq)) = n(OH-(aq)) = n(H2O(l)) Now we can calculate the energy released per mole of water, or the molar enthaply of neutralisation (molar heat of neutralization), ΔneutH Remember, the reaction is exothermic so the sign of ΔH will be negative! ΔneutH = −q ÷ n(H2O(l)) Do you know this? Join AUS-e-TUTE! Play the game now! ↪ Back to top Enthalpy of Neutralisation: Strong Monoprotic Acid + Strong Monobasic Base HCl(aq) is a strong monoprotic acid, it completely dissociates (ionises) in water to produce hydrogen ions (H+(aq)) and chloride ions (Cl-(aq)): HCl(aq) → H+(aq) + Cl-(aq) n(HCl(aq)) : n(H+(aq)) is 1:1 (that is, monoprotic) NaOH(aq) is a strong monobasic base, it completely dissociates (ionises) in water to produce sodium ions (Na+(aq)) and hydroxide ions (OH-(aq)): NaOH(aq) → Na+(aq) + OH-(aq) n(NaOH(aq)) : n(OH-(aq)) is 1:1 (that is, monobasic) A neutralization reaction occurs when HCl(aq) is added to NaOH(aq) HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) and heat energy is given off (the reaction is said to be exothermic) In an experiment to determine the molar enthalpy of neutralisation, 50.0 mL of 1.00 mol L-1 NaOH(aq) is placed in the styrofoam cup. The temperature of the NaOH(aq) is recorded. 1.00 mol L-1 HCl(aq) at the same temperature is added 10.0 mL at a time. The reaction mixture is stirred between each addition. The maximum temperature the solution reached is then recorded. The results of the experiment are shown in the table below: \| total volume HCl(aq) added (mL) \| temperature incalorimeter (°C) \| --- \| \| 0 \| 18.0 \| \| 10 \| 20.2 \| \| 20 \| 21.8 \| \| 30 \| 22.9 \| \| 40 \| 23.8 \| \| 50 \| 24.6 \| \| 60 \| 24.0 \| \| 70 \| 23.6 \| and the results have been plotted on the graph shown below: \| \| \| --- \| \| temperature (°C) \| HCl(aq) added to NaOH(aq)total volume of HCl(aq) added (mL) \| Initially, the temperature of the reaction mixture in the calorimeter (styrofoam™ cup) increases as HCl(aq) is added. Energy (heat) is being produced by the reaction. The reaction is exothermic. Maximum temperature reached is 24.6°C when 50.0 mL of HCl(aq) had been added. When 50.0 mL of the acid has been added, all the base has been neutralised. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) moles HCl(aq) added = moles of NaOH(aq) present in the calorimeter Adding more acid doesn't increase the temperature in the calorimeter any further(5). We can calculate the molar enthalpy of neutralisation (molar heat of neutralization) for the reaction if we assume that the: (i) density of each dilute aqueous solution is the same as water, 1 g mL-1 at 25°C so, the mass of solution in grams = volume of solution in mL (ii) heat capacity of each solution is the same as for water, Cg = 4.18 J°C-1g-1 Calculating the molar enthalpy of neutralisation using the data from the experiment: Step 1: Extract the data needed to calculate the molar heat of neutralisation for this reaction: V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL V(HCl) = volume of HCl(aq) added to achieve neutralisation = 50.0 mL c(NaOH) = concentration of NaOH(aq) = 1.00 mol L-1 c(HCl) = concentration of HCl(aq) = 1.00 mol L-1 Ti = initial temperature of solutions before additions = 18.0°C Tf = final temperature of solution at neutralisation = 24.6°C d = density of solutions = 1 g mL-1 (assumed) Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed) q = heat liberated during neutralisation reaction = ? J Step 2: Check the units for consistency and convert if necessary: Convert volume of solutions (mL) to mass (g): density × volume = mass since density = 1 g mL-1: 1 × volume (mL) = mass (g) mass(NaOH) = 50.0 g mass(HCl) = 50.0 g Step 3: Calculate the heat produced during the neutralisation reaction: heat produced = total mass × specific heat capacity × change in temperature q = mtotal × Cg × ΔT mtotal = mass(NaOH) + mass(HCl) = 50.0 + 50.0 = 100.0 g Cg = 4.18 J°C-1g-1 ΔT = Tf - Ti = 24.6 - 18.0 = 6.6°C q = 100.0 ~~g~~ × 4.18 J ~~g-1 °C-1~~ × 6.6 ~~°C~~ = 2758.8 J Step 4: Calculate the moles of water produced: OH-(aq) + H+(aq) → H2O(l) 1 mol OH-(aq) + 1 mol H+(aq) → 1 mol H2O moles(H2O) = moles(OH-(aq)) moles(OH-(aq)) = concentration (mol L-1) × volume (L) = 1.0 mol ~~L-1~~ × 50.0 ~~mL~~/1000 ~~mL~~/~~L~~ = 0.050 mol moles of water produced = 0.050 mol Step 5: Calculate the heat liberated per mole of water produced, ΔneutH : ΔneutH will be negative because the reaction is exothermic ΔneutH = heat liberated per mole of water = -1 × q ÷ moles of water ΔneutH = -1 × 2758.8 J ÷ 0.050 mol = -55176.0 J mol-1 We can convert J to kJ by dividing by 1000: ΔneutH = -55176.0 ~~J~~ mol-1 ÷ 1000 ~~J~~/kJ = 55.2 kJ mol-1 Do you understand this? Join AUS-e-TUTE! Take the test now! ↪ Back to top Enthalpy of Neutralisation: Strong Diprotic Acid and Strong Monobasic Base The experiment described above is repeated using 50.0 mL of 1.00 mol L-1 sodium hydroxide, a strong monobasic base, and 1.00 mol L-1 sulfuric acid, a strong diprotic acid, instead of 1.00 mol L-1 hydrochloric acid, a strong monoprotic acid. When plotted on a graph as shown below, the second experiment's results look different when compared to the first experiment's results: \| \| \| --- \| \| temperature (°C) \| Acid + NaOHtotal volume of acid added (mL) \| Initially, the temperature of the reaction mixture in both experiments increases as acid is added. Energy (heat) is being produced by the reaction. The reaction is exothermic. Maximum temperature reached for the reaction with H2SO4(aq) is higher than the maximum temperature reached for the reaction with HCl(aq). Volume of H2SO(aq) added to reach the maximum temperature is less than the volume of HCl(aq) needed to reach maximum temperature. For the reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) we found that ΔneutH = -55.2 kJ mol-1 (per mole of water formed) So, H+(aq) + OH- → H2O(l) ΔneutH = -55.2 kJ mol-1 For the reaction H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) 2H+(aq) + 2OH-(aq) → 2H2O(l) so, H+(aq) + OH-(aq) → H2O(l) We predict that ΔneutH = -55.2 kJ mol-1 We can use the results of the second experiment to calculate the value for the molar enthalpy of neutralisation (ΔneutH), and see if they agree with our prediction. Calculate the molar heat of neutralisation for the reaction from the results of the experiment shown in the graph above: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Step 1: Extract the data needed to calculate the molar heat of neutralisation for this reaction: V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL V(H2SO4) = volume of H2SO4(aq) added to achieve neutralisation = 25.0 mL c(NaOH) = concentration of NaOH(aq) = 1.00 mol L-1 c(H2SO4) = concentration of H2SO4(aq) = 1.00 mol L-1 Ti = initial temperature of solutions before additions = 18.0°C Tf = final temperature of solution at neutralisation = 26.9°C d = density of solutions = 1 g mL-1 (assumed) Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed) q = heat liberated during neutralisation reaction = ? J Step 2: Check the units for consistency and convert if necessary: Convert volume of solutions (mL) to mass (g): density × volume = mass since density = 1 g mL-1: 1 × volume (mL) = mass (g) mass(NaOH) = 50.0 g mass(H2SO4) = 25.0 g Step 3: Calculate the heat produced during the neutralisation reaction: heat produced = total mass × specific heat capacity × change in temperature q = mtotal × Cg × ΔT mtotal = mass(NaOH) + mass(H2SO4() = 50.0 + 25.0 = 75.0 g Cg = 4.18 J°C-1g-1 ΔT = Tf - Ti = 26.9 - 18.0 = 8.9°C q = 75.0 ~~g~~ × 4.18 J ~~g-1 °C~~ × 8.9 ~~°C~~ = 2790.2 J Step 4: Calculate the moles of water produced: OH-(aq) + H+(aq) → H2O(l) 1 mol OH-(aq) + 1 mol H+(aq) → 1 mol H2O moles(H2O) = moles(OH-(aq)) moles(OH-(aq)) = concentration (mol L-1) × volume (L) = 1.0 × 50.0/1000 = 0.050 mol moles of water produced = 0.050 mol Step 5: Calculate the heat liberated per mole of water produced, ΔneutH : ΔneutH will be negative because the reaction is exothermic ΔneutH = heat liberated per mole of water = -1 × q ÷ moles of water ΔneutH = -1 × 2790.2 J ÷ 0.050 mol = -55803 J mol-1 (of water produced) We can divide J by 1000 to convert this enthalpy change to kJ per mole: ΔneutH = -55803 ~~J~~ mol-1 ÷ 1000 ~~J~~/kJ = -55.8 kJ mol-1 (of water produced) Now compare the molar enthalpy of neutralisation (molar heat of neutralization) for: \| \| \| --- \| \| HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) \| ΔneutH = -55.2 kJ mol-1 (of water) \| \| H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) \| ΔneutH = -55.8 kJ mol-1 (of water) \| There is close agreement between the two values for molar heat of neutralisation, so we can generalise and say that the molar enthalpy of neutralisation for the reaction between a strong acid and a strong base is a constant. Can you apply this? Join AUS-e-TUTE! Take the exam now! ↪ Back to top Molar Enthalpy of Neutralisation: Weak Acid + Strong Base The experiment described above is repeated using 50.0 mL of 1.00 mol L-1 sodium hydroxide, a strong monobasic base, and 1.00 mol L-1 aqueous hydrogren cyanide (HCN(aq), hydrocyanic acid or prussic acid), a weak monoprotic acid (Ka ≈ 6 × 10-10), instead of 1.00 mol L-1 hydrochloric acid, a strong monoprotic acid. The results of these experiments are shown in the graph below: \| \| \| --- \| \| temperature (°C) \| Acid + NaOHtotal volume of acid added (mL) \| Initially, the temperature of the reaction mixture in both experiments increases as acid is added. Energy (heat) is being produced by the reaction. The reaction is exothermic. Maximum temperature reached for the reaction with HCN(aq) is much less than the maximum temperature reached for the reaction with HCl(aq). Volume of HCN(aq) added to reach the maximum temperature is the same as the volume of HCl(aq) needed to reach maximum temperature (both volumes are 50.0 mL). For the reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) we found that ΔneutH = -55.2 kJ mol-1 (per mole of water formed) So, H+(aq) + OH- → H2O(l) ΔneutH = -55.2 kJ mol-1 For the reaction HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l) Will ΔneutH for this reaction be the same (-55.2 kJ mol-1) ? We can use the results of the HCN experiment recorded in the graph above to calculate the value for the molar enthalpy of neutralisation (ΔneutH), and see. Step 1: Extract the data needed to calculate the molar heat of neutralisation for this reaction: V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL V(HCN) = volume of HCN(aq) added to achieve neutralisation = 50.0 mL c(NaOH) = concentration of NaOH(aq) = 1.00 mol L-1 c(HCN) = concentration of HCN(aq) = 1.00 mol L-1 Ti = initial temperature of solutions before additions = 18.0°C Tf = final temperature of solution at neutralisation = 19.2°C d = density of solutions = 1 g mL-1 (assumed) Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed) q = heat liberated during neutralisation reaction = ? J Step 2: Check the units for consistency and convert if necessary: Convert volume of solutions (mL) to mass (g): density × volume = mass since density = 1 g mL-1 (assumed): 1 × volume (mL) = mass (g) mass(NaOH) = 50.0 g mass(HCN) = 50.0 g Step 3: Calculate the heat produced during the neutralisation reaction: heat produced = total mass × specific heat capacity × change in temperature q = mtotal × Cg × ΔT mtotal = mass(NaOH) + mass(HCN) = 50.0 + 50.0 = 100.0 g Cg = 4.18 J°C-1g-1 ΔT = Tf - Ti = 19.2 - 18.0 = 1.2°C q = 100.0 ~~g~~ × 4.18 J ~~g-1 °C-1~~ × 1.2 ~~°C~~ = 501.6 J Step 4: Calculate the moles of water produced: NaOH(aq) + HCN(aq) → NaCN(aq) + H2O(l) moles(H2O) = moles(NaOH) moles(NaOH(aq)) = concentration (mol L-1) × volume (L) = 1.0 mol ~~L-1~~ × 50.0 ~~mL~~/1000 ~~mL~~/~~L~~ = 0.050 mol moles of water produced = 0.050 mol Step 5: Calculate the heat liberated per mole of water produced, ΔneutH : ΔneutH will be negative because the reaction is exothermic ΔneutH = heat liberated per mole of water = -1 × q ÷ moles of water ΔneutH = -1 × 501.6 ÷ 0.050 = -10032 J mol-1 We can convert J to kJ by dividing by 1000: ΔneutH = -10032 ~~J~~ mol-1 ÷ 1000 ~~J~~/kJ = -10.0 kJ mol-1 Compare the molar heat of neutralisation for the neuralisation of NaOH(aq) by both HCl(aq) and HCN(aq): \| \| \| --- \| \| HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) \| ΔneutH = -55.2 kJ mol-1 (of water) \| \| HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l) \| ΔneutH = -10.0 kJ mol-1 (of water) \| The heat released per mole of water for the HCN(aq) reaction is much less than the heat released per mole of water for the HCl(aq) reaction. The difference in molar enthalpy of neutralisation is due to the type of reaction taking place: (i) Strong Acid + Strong Base Reaction: ⚛ Strong base, NaOH, fully dissociates in water. The reacting species is OH-(aq) ⚛ Strong acid, HCl, fully dissociates in water. The reacting species is H+(aq) ⚛ The reaction is therefore an Arrhenius neutralisation reaction: H+(aq) + OH-(aq) → H2O(l) ⚛ No bonds need to broken in the strong acid or strong base, no energy is lost in breaking bonds. ⚛ Energy is released when the H-O bonds form in the H2O product. (ii) Weak Acid + Strong Base ⚛ Strong base, NaOH, fully dissociates in water. The reacting species is OH-(aq) ⚛ Weak acid, HCN(aq), only partially dissociates in water. Most of the acid species in solution are undissociated HCN molecules. ⚛ The reaction is therefore a Brønsted-Lowry proton transfer reaction: HCN(aq) + OH-(aq) → CN-(aq) + H2O(l) ⚛ In order for this reaction to occur, H-C bonds in HCN molecules must be broken before H2O molecules can be produced. Breaking covalent bonds is an endothermic reaction, it absorbs energy. ⚛ So, even though we might expect the same amount of energy to be produced in both the HCl(aq) and HCN(aq) reactions because both reactions are producing the same number of moles of H2O, we see that energy will be consumed in breaking bonds in HCN so the amount of energy produced overall will be less than that for the HCl + NaOH reaction. The heat liberated per mole when a weak acid neutralises a strong base is less than the amount of heat liberated per mole when a strong acid neutralises a strong base. Need teaching resources? Join AUS-e-TUTE! Register your group. ↪ Back to top Problem Solving: Molar Enthalpy of Neutralisation The Problem: The heat released when an aqueous solution of acetic acid reacted with aqueous sodium hydroxide was measured in the laboratory using a cup calorimeter. The calorimeter consisted of two styrofoam™ coffee cups, one inside the other. This was covered with a styrofoam™ lid through which a thermometer and a stirring rod were placed. The two reacting aqueous solutions, acetic acid and sodium hydroxide, were placed in the inner cup. The following data were obtained: \| \| \| \| --- \| Volume of 1.0 mol L-1 CH3COOH(aq) used \| = \| 50.0 mL \| \| Volume of 1.0 mol L-1 NaOH(aq) used \| = \| 50.0 mL \| \| Initial temperature of CH3COOH(aq) \| = \| 25.0 °C \| \| Initial temperature of NaOH(aq) \| = \| 25.2 °C \| \| Maximum temperature of mixture \| = \| 31.0 °C \| \| Heat capacity of water and of the mixture \| = \| 4.18 J g-1°C-1 \| Determine the molar enthalpy of neutralisation for this reaction in kJ mol-1. Solving the Problem using the StoPGoPS model for problem solving: \| \| \| \| --- \| STOP! \| State the question. \| What is the question asking you to do? Determine (calculate) the molar enthalpy of neutralisation in kJ mol-1 ΔneutH = ? kJ mol-1 \| \| PAUSE! \| Pause to Plan. \| What information (data) have you been given? c(CH3COOH(aq)) = 1.0 mol L-1 V(CH3COOH(aq)) = 50.0 mL c(NaOH(aq)) = 1.0 mol L-1 V(NaOH(aq)) = 50.0 mL Ti(CH3COOH(aq)) = 25.0 °C Ti(NaOH(aq)) = 25.2 °C Tf = 31.0 °C Cg = 4.18 J g-1°C-1 What is your plan for solving this problem? Step 1: Calculate the heat produced, q, in J : q = m(total) × Cg × ΔT Step 2: Calculate the enthalpy change for the reaction, ΔH, in J : ΔH = −q Step 3: Calculate the heat produced per mole of water produced, ΔneutH, in J mol-1 : ΔneutH = ΔH ÷ n(H2O(l)) Step 4: Convert J mol-1 to kJ mol-1 ΔneutH kJ mol-1 = ΔneutH J mol-1 ÷ 1000 J/kJ \| \| GO! \| Go with the Plan. \| Step 1: Calculate the heat produced, q, in J : q = m(total) × Cg × ΔT V(total) = V(CH3COOH(aq)) + V(NaOH(aq)) = 50.0 + 50.0 = 100.0 mL assume density of mixture = 1.0 g mL-1, then m(total) = 100.0 g Cg = 4.18 J g-1 °C-1 Ti(mixture) = [Ti(CH3COOH(aq)) + Ti(NaOH(aq))] ÷ 2 = [25.0 + 25.2]/2 = 25.1 °C Tf(mixture) = 31.0 °C ΔI(mixture) = 31.0 − 25.1 = 5.9 °C q = 100.0 × 4.18 × 5.9 = 2466.2 J Step 2: Calculate the enthalpy change for the reaction, ΔH, in J : ΔH = −q = −2466.2 J Step 3: Calculate the heat produced per mole of water produced, ΔneutH, in J mol-1 : ΔneutH = ΔH ÷ n(H2O(l)) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) n(H2O(l)) = n(CH3COOH(aq)) = c × V = 1.0 mol L-1 × 50.0 mL/1000 mL/L = 0.050 mol ΔneutH = −2466.2 J ÷ 0.050 mol = −49324 J mol-1 Step 4: Convert J mol-1 to kJ mol-1 ΔneutH kJ mol-1 = ΔneutH J mol-1 ÷ 1000 J/kJ ΔneutH kJ mol-1 = −49324 J mol-1 ÷ 1000 J/kJ = −49 kJ mol-1 (Only 2 significant figures are justified) \| \| PAUSE! \| Ponder Plausability. \| Have you answered the question that was asked? Yes, we have calculated the molar enthalpy of neutralisation in kJ mol-1 Is your solution to the question reasonable? Acetic acid is a weak acid, so the heat released per mole of water formed will be less than that for a strong acid. 49 kJ mol-1 is less than 55.2 kJ mol -1, so our value is plausible. Neutralisation reactions release heat, they are exothermic reactions, so the sign of the molar enthalpy of neutralisation is negative. Our value, ΔneutH = −49 kJ mol -1 is negative so it is plausible. \| \| STOP! \| State the solution. \| What is the molar enthalpy of neutralisation? ΔneutH = −49 kJ mol-1 \| ↪ Back to top Sample Question: Molar Enthalpy of Neutralisation 25.00 mL of 0.255 mol L-1 HCl(aq) is neutralised by exactly 19.73 mL of NaOH(aq). If the initial temperature of the reactants was 18.78 °C, calculate the final temperature of the mixture. ↪ Back to top Footnotes: (1) The Brønsted-Lowry equivalent proton-transfer reaction is : H3O+(aq) + OH-(aq) → 2H2O(l) This reaction will NOT be used in the following discussion because it produces 2 moles of water per mole of hydroxide ion used. (2) see IUPAC Green Book, "Quantities, Units and Symbols in Physical Chemistry" Third Edition 2007 (3) A better method for measuring heat of neutralization is to use an adiabatic calorimeter fitted with an electrical heater. A desciption of this type of calorimeter can be found in the calorimetry tutorial. (4) You will find slightly different values quoted for molar heat of neutralisation mostly because the neutralisation reaction is dependent on the temperature at which the reaction occurs. In general the values you see quoted will be between 55 kJ mol-1 and 58 kJ mol-1 and refer to reactions that take place at ambient temperatures in a laboratory. (5) The temperature is likely to decrease as more acid is added because the heat that was generated by the completed reaction is being dissipated in a greater mass of solution. ↪ Back to top
3275	https://www.tsfx.edu.au/resources/W_-_Dilution__Titration_Problems_Worksheet_with_answers.pdf?srsltid=AfmBOop9FL9YN4A7zu-jy9zSMmhAdaKwdVdPGobDu8RpQtjuIlo0QkLQ	For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing – All Rights Reserved Molarity Review Problems 1) What is the molarity of a solution in which 0.45 grams of sodium nitrate are dissolved in 265 mL of solution. 2) How many grams of beryllium chloride are needed to make 125 mL of a 0.050 M solution? Dilutions Worksheet 1) If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? 2) If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? 3) How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl? 4) I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will the molarity of the solution be? 5) How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution? Titrations Practice Worksheet Find the requested quantities in the following problems: 1) If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl? 2) If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution? 3) Explain the difference between an endpoint and equivalence point in a titration. For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing – All Rights Reserved Molarity Review Problems- Solutions 1) What is the molarity of a solution in which 0.45 grams of sodium nitrate (NaNO 3) are dissolved in 265 mL of solution. Molarity = mass in grams x 1 mole = .45g x 1 mole Volume in liters x molar mass .265 L x 85.00g 0.020 M 2) How many grams of beryllium chloride (BeCl 2) are needed to make 125 mL of a 0.050 M solution? Molarity = mass in grams x 1 mole 0.050M = X x 1 mole Volume in liters x molar mass .125L x 79.91g 0.50 grams Dilutions Worksheet - Solutions 1) If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? M1V1 = M 2V2 (0.15 M)(125 mL) = x (150 mL) x = 0.125 M 2) If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? M1V1 = M 2V2 (0.15 M)(100 mL) = x (150 mL) x = 0.100 M 3) How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl? M1V1 = M 2V2 (10 M)(250 mL) = (0.05 M) x x = 50,000 mL 4) I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will the molarity of the solution be? M1V1 = M 2V2 (1.5 M)(345 mL) = x (250 mL) x = 2.07 M For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing – All Rights Reserved 5) How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution? M1V1 = M 2V2 (2.4 M)(500 mL) = (1.0 M) x x = 1200 mL 1200 mL will be the final volume of the solution. However, since there ’s already 500 mL of solution present, you only need to add 700 mL of water to get 1200 mL as your final volume. The answer: 700 mL. Titrations Practice Worksheet- Solutions 1) If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl? 2) If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution? For questions 1 and 2, the units for your final answer should be “M”, or “molar”, because you’re trying to find the molarity of the acid or base solution. To solve these problems, use M 1V1 = M 2V2. 1) 0.043 M HCl 2) 0.0036 M NaOH 3) Explain the difference between an endpoint and equivalence point in a titration. 3) Endpoint: When you actually stop doing the titration (usually, this is determined by a color change in an indicator or an indication of pH=7.0 on an electronic pH probe) Equivalence point: Whe n the solution is exactly neutralized. It’s important to keep in mind that the equivalence point and the endpoint are not exactly the same because indicators don’t change color at exactly 7.0000 pH and pH probes aren’t infinitely accurate. Generally, you can measure the effectiveness of a titration by the closeness of the endpoint to the equivalence point.
3276	https://www.geeksforgeeks.org/dsa/minimax-algorithm-in-game-theory-set-1-introduction/	Minimax Algorithm in Game Theory \| Set 1 (Introduction) Last Updated : 13 Jun, 2022 Suggest changes 148 Likes Minimax is a kind of backtracking algorithm that is used in decision making and game theory to find the optimal move for a player, assuming that your opponent also plays optimally. It is widely used in two player turn-based games such as Tic-Tac-Toe, Backgammon, Mancala, Chess, etc. In Minimax the two players are called maximizer and minimizer. The maximizer tries to get the highest score possible while the minimizer tries to do the opposite and get the lowest score possible. Every board state has a value associated with it. In a given state if the maximizer has upper hand then, the score of the board will tend to be some positive value. If the minimizer has the upper hand in that board state then it will tend to be some negative value. The values of the board are calculated by some heuristics which are unique for every type of game. Example: Consider a game which has 4 final states and paths to reach final state are from root to 4 leaves of a perfect binary tree as shown below. Assume you are the maximizing player and you get the first chance to move, i.e., you are at the root and your opponent at next level. Which move you would make as a maximizing player considering that your opponent also plays optimally? Since this is a backtracking based algorithm, it tries all possible moves, then backtracks and makes a decision. Maximizer goes LEFT: It is now the minimizers turn. The minimizer now has a choice between 3 and 5. Being the minimizer it will definitely choose the least among both, that is 3 Maximizer goes RIGHT: It is now the minimizers turn. The minimizer now has a choice between 2 and 9. He will choose 2 as it is the least among the two values. Being the maximizer you would choose the larger value that is 3. Hence the optimal move for the maximizer is to go LEFT and the optimal value is 3. Now the game tree looks like below : The above tree shows two possible scores when maximizer makes left and right moves. Note: Even though there is a value of 9 on the right subtree, the minimizer will never pick that. We must always assume that our opponent plays optimally. Below is the implementation for the same. C++ ```` // A simple C++ program to find // maximum score that // maximizing player can get. include using namespace std; // Returns the optimal value a maximizer can obtain. // depth is current depth in game tree. // nodeIndex is index of current node in scores[]. // isMax is true if current move is // of maximizer, else false // scores[] stores leaves of Game tree. // h is maximum height of Game tree int minimax(int depth, int nodeIndex, bool isMax, int scores[], int h) { // Terminating condition. i.e // leaf node is reached if (depth == h) return scores[nodeIndex]; // If current move is maximizer, // find the maximum attainable // value if (isMax) return max(minimax(depth+1, nodeIndex2, false, scores, h), minimax(depth+1, nodeIndex2 + 1, false, scores, h)); // Else (If current move is Minimizer), find the minimum // attainable value else return min(minimax(depth+1, nodeIndex2, true, scores, h), minimax(depth+1, nodeIndex2 + 1, true, scores, h)); } // A utility function to find Log n in base 2 int log2(int n) { return (n==1)? 0 : 1 + log2(n/2); } // Driver code int main() { // The number of elements in scores must be // a power of 2. int scores[] = {3, 5, 2, 9, 12, 5, 23, 23}; int n = sizeof(scores)/sizeof(scores); int h = log2(n); int res = minimax(0, 0, true, scores, h); cout << "The optimal value is : " << res << endl; return 0; } ```` // A simple C++ program to find // A simple C++ program to find // maximum score that // maximum score that // maximizing player can get. // maximizing player can get. ``` include#include ``` using namespace std; using namespace std // Returns the optimal value a maximizer can obtain. // Returns the optimal value a maximizer can obtain. // depth is current depth in game tree. // depth is current depth in game tree. // nodeIndex is index of current node in scores[]. // nodeIndex is index of current node in scores[]. // isMax is true if current move is // isMax is true if current move is // of maximizer, else false // of maximizer, else false // scores[] stores leaves of Game tree. // scores[] stores leaves of Game tree. // h is maximum height of Game tree // h is maximum height of Game tree int minimax(int depth, int nodeIndex, bool isMax, int minimax int depth int nodeIndex bool isMax int scores[], int h) int scores int h { // Terminating condition. i.e // Terminating condition. i.e // leaf node is reached // leaf node is reached if (depth == h) if depth == h return scores[nodeIndex]; return scores nodeIndex // If current move is maximizer, // If current move is maximizer, // find the maximum attainable // find the maximum attainable // value // value if (isMax) if isMax return max(minimax(depth+1, nodeIndex2, false, scores, h), return max minimax depth + 1 nodeIndex 2 false scores h minimax(depth+1, nodeIndex2 + 1, false, scores, h)); minimax depth + 1 nodeIndex 2 + 1 false scores h // Else (If current move is Minimizer), find the minimum // Else (If current move is Minimizer), find the minimum // attainable value // attainable value else else return min(minimax(depth+1, nodeIndex2, true, scores, h), return min minimax depth + 1 nodeIndex 2 true scores h minimax(depth+1, nodeIndex2 + 1, true, scores, h)); minimax depth + 1 nodeIndex 2 + 1 true scores h } // A utility function to find Log n in base 2 // A utility function to find Log n in base 2 int log2(int n) int log2 int n { return (n==1)? 0 : 1 + log2(n/2); return n == 1? 0 1 + log2 n/ 2 } // Driver code // Driver code int main() int main { // The number of elements in scores must be // The number of elements in scores must be // a power of 2. // a power of 2. int scores[] = {3, 5, 2, 9, 12, 5, 23, 23}; int scores = 3 5 2 9 12 5 23 23 int n = sizeof(scores)/sizeof(scores); int n = sizeof scores/ sizeof scores 0 int h = log2(n); int h = log2 n int res = minimax(0, 0, true, scores, h); int res = minimax 0 0 true scores h cout << "The optimal value is : " << res << endl; cout<<"The optimal value is : "<< res<< endl return 0; return 0 } Java ```` // A simple java program to find maximum score that // maximizing player can get. import java.io.; class GFG { // Returns the optimal value a maximizer can obtain. // depth is current depth in game tree. // nodeIndex is index of current node in scores[]. // isMax is true if current move is of maximizer, else false // scores[] stores leaves of Game tree. // h is maximum height of Game tree static int minimax(int depth, int nodeIndex, boolean isMax, int scores[], int h) { // Terminating condition. i.e leaf node is reached if (depth == h) return scores[nodeIndex]; // If current move is maximizer, find the maximum attainable // value if (isMax) return Math.max(minimax(depth+1, nodeIndex2, false, scores, h), minimax(depth+1, nodeIndex2 + 1, false, scores, h)); // Else (If current move is Minimizer), find the minimum // attainable value else return Math.min(minimax(depth+1, nodeIndex2, true, scores, h), minimax(depth+1, nodeIndex2 + 1, true, scores, h)); } // A utility function to find Log n in base 2 static int log2(int n) { return (n==1)? 0 : 1 + log2(n/2); } // Driver code public static void main (String[] args) { // The number of elements in scores must be // a power of 2. int scores[] = {3, 5, 2, 9, 12, 5, 23, 23}; int n = scores.length; int h = log2(n); int res = minimax(0, 0, true, scores, h); System.out.println( "The optimal value is : " +res); } } // This code is contributed by vt_m ```` C# ```` // A simple C# program to find maximum score that // maximizing player can get. using System; public class GFG { // Returns the optimal value a maximizer can obtain. // depth is current depth in game tree. // nodeIndex is index of current node in scores[]. // isMax is true if current move is of maximizer, else false // scores[] stores leaves of Game tree. // h is maximum height of Game tree static int minimax(int depth, int nodeIndex, bool isMax, int []scores, int h) { // Terminating condition. i.e leaf node is reached if (depth == h) return scores[nodeIndex]; // If current move is maximizer, find the maximum attainable // value if (isMax) return Math.Max(minimax(depth+1, nodeIndex2, false, scores, h), minimax(depth+1, nodeIndex2 + 1, false, scores, h)); // Else (If current move is Minimizer), find the minimum // attainable value else return Math.Min(minimax(depth+1, nodeIndex2, true, scores, h), minimax(depth+1, nodeIndex2 + 1, true, scores, h)); } // A utility function to find Log n in base 2 static int log2(int n) { return (n==1)? 0 : 1 + log2(n/2); } // Driver code static public void Main () { // The number of elements in scores must be // a power of 2. int []scores = {3, 5, 2, 9, 12, 5, 23, 23}; int n = scores.Length; int h = log2(n); int res = minimax(0, 0, true, scores, h); Console.WriteLine( "The optimal value is : " +res); } } // This code is contributed by ajit. ```` Python3 ```` A simple Python3 program to find maximum score that maximizing player can get import math def minimax (curDepth, nodeIndex, maxTurn, scores, targetDepth): # base case : targetDepth reached if (curDepth == targetDepth): return scores[nodeIndex] if (maxTurn): return max(minimax(curDepth + 1, nodeIndex 2, False, scores, targetDepth), minimax(curDepth + 1, nodeIndex 2 + 1, False, scores, targetDepth)) else: return min(minimax(curDepth + 1, nodeIndex 2, True, scores, targetDepth), minimax(curDepth + 1, nodeIndex 2 + 1, True, scores, targetDepth)) Driver code scores = [3, 5, 2, 9, 12, 5, 23, 23] treeDepth = math.log(len(scores), 2) print("The optimal value is : ", end = "") print(minimax(0, 0, True, scores, treeDepth)) This code is contributed by rootshadow ```` JavaScript ```` // Javascript program to find maximum score that // maximizing player can get. // Returns the optimal value a maximizer can obtain. // depth is current depth in game tree. // nodeIndex is index of current node in scores[]. // isMax is true if current move is of maximizer, else false // scores[] stores leaves of Game tree. // h is maximum height of Game tree function minimax(depth, nodeIndex, isMax, scores, h) { // Terminating condition. i.e leaf node is reached if (depth == h) return scores[nodeIndex]; // If current move is maximizer, find the maximum attainable // value if (isMax) return Math.max(minimax(depth+1, nodeIndex2, false, scores, h), minimax(depth+1, nodeIndex2 + 1, false, scores, h)); // Else (If current move is Minimizer), find the minimum // attainable value else return Math.min(minimax(depth+1, nodeIndex2, true, scores, h), minimax(depth+1, nodeIndex2 + 1, true, scores, h)); } // A utility function to find Log n in base 2 function log2(n) { return (n==1)? 0 : 1 + log2(n/2); } // Driver Code // The number of elements in scores must be // a power of 2. let scores = [3, 5, 2, 9, 12, 5, 23, 23]; let n = scores.length; let h = log2(n); let res = minimax(0, 0, true, scores, h); document.write( "The optimal value is : " +res); ```` Output: The optimal value is: 12 Time complexity : O(b^d) b is the branching factor and d is count of depth or ply of graph or tree. Space Complexity : O(bd) where b is branching factor into d is maximum depth of tree similar to DFS. The idea of this article is to introduce Minimax with a simple example. In the above example, there are only two choices for a player. In general, there can be more choices. In that case, we need to recur for all possible moves and find the maximum/minimum. For example, in Tic-Tac-Toe, the first player can make 9 possible moves. In the above example, the scores (leaves of Game Tree) are given to us. For a typical game, we need to derive these values We will soon be covering Tic Tac Toe with Minimax algorithm. This article is contributed by Akshay L. Aradhya. 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3277	https://www.storyofmathematics.com/a-projectile-is-shot-from-the-edge-of-a-cliff/	A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal. Determine the following quantities: – The horizontal and vertical components of the velocity vector. – The maximum height reached by the projectile above the launch point. The aim of this question is to understand the different parameters during 2D projectile motion. The most important parameters during the flight of a projectile are its range, time of flight, and maximum height. The range of a projectile is given by the following formula: [ R \ = \ \dfrac{ v_i^2 \ sin ( 2 \theta ) }{ g } ] The time of flight of a projectile is given by the following formula: [ t \ = \ \dfrac{ 2 v_i \ sin \theta }{ g } ] The maximum height of a projectile is given by the following formula: [ h \ = \ \dfrac{ v_i^2 \ sin^2 \theta }{ 2 g } ] The same problem can be solved with the fundamental equations of motion. Which are given below: [ v_{ f } \ = \ v_{ i } + a t ] [ S = v_{i} t + \dfrac{ 1 }{ 2 } a t^2 ] [ v_{ f }^2 \ = \ v_{ i }^2 + 2 a S ] Expert Answer Given that: [ v_i \ =\ 65 \ m/s ] [ \theta \ =\ 37^{ \circ } ] [ h_i \ =\ 125 \ m ] Part (a) – The horizontal and vertical components of the velocity vector. [ v_{i_{x}} \ =\ v_i cos ( \theta ) \ = \ 65 cos( 37^{ \circ } ) \ = \ 52 \ m/s ] [ v_{i_{y}} \ =\ v_i sin ( \theta ) \ = \ 65 sin( 37^{ \circ } ) \ = \ 39 \ m/s ] Part (b) – The maximum height reached by the projectile above the launch point. For upward motion: [ v_i \ =\ 39 \ m/s ] [ v_f \ =\ 0 \ m/s ] [ a \ =\ -9.8 \ m/s^{ 2 } ] Using 3rd equation of motion: [ S \ = \ \dfrac{ v_f^2 – v_i^2 }{ 2a } ] [ S \ = \ \dfrac{ 0^2 – 39^2 }{ 2(-9.8) } ] [ S \ = \ \dfrac{ 1521 }{ 19.6 } ] [ S \ = \ 77.60 \ m ] Numerical Result Part (a) – The horizontal and vertical components of the velocity vector: [ v_{i_{x}} \ = \ 52 \ m/s ] [ v_{i_{y}} \ = \ 39 \ m/s ] Part (b) – The maximum height reached by the projectile above the launch point: [ S \ = \ 77.60 \ m ] Example For the same projectile given in the question above, find thetime elapsed before hitting the ground level. For upward motion: [ v_i \ =\ 39 \ m/s ] [ v_f \ =\ 0 \ m/s ] [ a \ =\ -9.8 \ m/s^{ 2 } ] Using 1st equation of motion: [ t_1 \ = \ \dfrac{ v_f – v_i }{ a } ] [ t_1 \ = \ \dfrac{ 0 – 39 }{ -9.8 } ] [ t_1 \ = \ 3.98 \ s ] For downward motion: [ v_i \ =\ 0 \ m/s ] [ S \ = \ 77.60 + 125 \ = \ 180.6 \ m ] [ a \ =\ 9.8 \ m/s^{ 2 } ] Using 2nd equation of motion: [ S \ = \ v_{i} t_2 + \dfrac{ 1 }{ 2 } a t_2^2 ] [ 180.6 \ = \ (0) t_2 + \dfrac{ 1 }{ 2 } ( 9.8 ) t_2^2 ] [ 180.6 \ = \ \dfrac{ 1 }{ 2 } ( 9.8 ) t_2^2 ] [ t_2^2 \ = \ 36.86 ] [ t_2 \ = \ 6.07 \ s ] So the total time: [ t \ = \ t_1 + t_2 \ = \ 3.98 + 6.07 \ = \ 10.05 \ s ] Previous Question < > Next Question Post navigation Back to Top of Page © 2025 SOM – STORY OF MATHEMATICS
3278	https://blogs.cornell.edu/info2040/2022/09/13/56590/	The Minimax Algorithm in Tic-Tac-Toe: When graphs, game theory and algorithms come together : Networks Course blog for INFO 2040/CS 2850/Econ 2040/SOC 2090 Skip to main content Search Cornell Networks Course blog for INFO 2040/CS 2850/Econ 2040/SOC 2090 The Minimax Algorithm in Tic-Tac-Toe: When graphs, game theory and algorithms come together The graphs are very fun to explore and study! … until they get too big. Consider, for the example, the game of Tic-Tac-Toe. If we make each possible state of the board a node and connect two nodes with an edge if there is a difference of one move between them, the graph we will get is going to consist of 362,880 different nodes! By eliminating symmetry and recognizing when different sequences of moves lead to the same position, Gary Fredericks got this number down to 765 distinct board positions. Here is the graph visualization he came up with: Source of the picture: To truly become unbeatable at the game and always make the move with the greatest payoff, one would have to be able to model every possible outcome for every possible state of the field. This sounds like something nearly impossible to do for a human, and that is when the Minimax algorithm (and the game theory) come for the rescue. The name of the algorithm is a product of two words: minimize and maximize. That is, minimize the payoff for the opponent, maximize the payoff for yourself. If a sequence of moves leads to a victory, its payoff is evaluated at +10 points. If it leads to a loss, the payoff is -10. If a sequence ultimately results in a draw, the payoff it brings to the table is 0. As an example, let’s look at the near end of a game, when it’s turn for Player X to make a move: Source of the picture: Out of the three available nodes the initial state node is connected to, the one on the left yields the greatest payoff (+10 points), so this is the move to make. Similarly, player O is trying to maximize their own payoff. Player X knows this and evaluates their opponent’s moves as well. The following picture serves as a good example of this process: Source of the picture: Both top nodes are states of the board (move sequences) which bring a payoff of -10. Even though there is a two-step path to a node with a payoff of +10, there is an obvious choice Player O will make (the algorithm assumes them to be a perfect minimizer) that will yield a good outcome for them (and a bad one for Player X). The Minimax algorithm evaluates the payoff of each available move by taking turns for the minimizer and the maximizer. The following picture is a perfect example of the evaluation process: Source of the picture: Here, the initial state is node 1, so X has three possible moves. The payoff for node 2 is -10 (I explained why in the previous example). The payoff for node 3 is 0, because Player O is assumed to be the perfect minimizer, and therefore will not allow X to get +10. The payoff for node 4 is +10, so this is the move to make. The Minimax Tic-Tac-Toe algorithm is impossible to beat, and when two Minimaxes play against each other, every move they make is the best response to what the opponent could possibly do (Nash equilibrium), resulting in 100% chance of a draw. Want to try it yourself? Check out this website developed by Jason Fox of Never Stop Building LLC: . Sources: Aradhya, A. L. (2022, April 29). Minimax algorithm in Game theory: Set 3 (tic-tac-toe ai – finding optimal move). GeeksforGeeks. Retrieved September 13, 2022, from Fox, J. (2021, November 10). Tic Tac Toe: Understanding the minimax algorithm. Never Stop Building – Crafting Wood with Japanese Techniques. Retrieved September 13, 2022, from Fredericks, G. (2010, September 18). Visualizing Tic-tac-Toe. gfredericks.com. Retrieved September 13, 2022, from September 13, 2022 \| category: Uncategorized 1 Comment Comments One Response to “ The Minimax Algorithm in Tic-Tac-Toe: When graphs, game theory and algorithms come together ” Какова Временная Сложность Игры «Крестики-Нолики» Для ИИ? - Олдскульные геймерыon September 23rd, 2023 at 6:40 am[…] Запрос на удаление Посмотреть полный ответ на blogs.cornell.edu […] Leave a Reply Name (required) Mail (will not be published) (required) Website Anti-spam To prove you are a person (not a spam script), type the words from the following picture or audio file. Listen to the Anti-spam wordLoad new Captcha refreshed. Blogging Calendar September 2022\| M \| T \| W \| T \| F \| S \| S \| --- --- --- \| \| 1 \| 2 \| 3 \| 4 \| \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| \| 12 \| 13 \| 14 \| 15 \| 16 \| 17 \| 18 \| \| 19 \| 20 \| 21 \| 22 \| 23 \| 24 \| 25 \| \| 26 \| 27 \| 28 \| 29 \| 30 \| \| « AugOct » Google Search Archives Archives EntriesComments ©2025 Cornell University Powered by Edublogs Campus and running on blogs.cornell.edu Skip to toolbar Log In Search
3279	https://bioinformatics.stackexchange.com/questions/2959/duplicate-genes-with-rsem-counts-which-one-to-choose	r - Duplicate genes with RSEM counts: Which one to choose? - Bioinformatics Stack Exchange Join Bioinformatics 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 Bioinformatics helpchat Bioinformatics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Bioinformatics 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 Duplicate genes with RSEM counts: Which one to choose? Ask Question Asked 7 years, 10 months ago Modified7 years, 10 months ago Viewed 2k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. I have Ensembl ids in the first column and samples with RSEM counts data in other columns. I converted Ensembl ids to gene symbols. Now I see there are three genes repeated twice. r Gene S1 S2 S3 COG8 804.07 1475.16 323.80 COG8 249.23 867.94 113.64 SDHD 1464.84 3508.02 75 SDHD 4.16 3.97 0 SOGA3 81.65 30.76 32.69 SOGA3 0.00 10.77 25.65 These are the Ensembl ids: SOGA3: ENSG00000214338, ENSG00000255330 SDHD: ENSG00000204370, ENSG00000255292 COG8: ENSG00000213380, ENSG00000272617. This is not for DE analysis, I want to do subtyping. Should I calculate mean or median on that? How should I go forward with this duplicate genes? r rna-seq bioconductor rsem Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Nov 30, 2017 at 12:18 terdon♦ 11.4k 6 6 gold badges 26 26 silver badges 51 51 bronze badges asked Nov 30, 2017 at 10:09 stack_learnerstack_learner 1,282 16 16 silver badges 28 28 bronze badges 4 I will use this data for subtyping. And for subtyping there should not be any duplicate genes.stack_learner –stack_learner 2017-11-30 10:14:23 +00:00 Commented Nov 30, 2017 at 10:14 Then you can add them up: if each transcript (ENST..) is of a gene and your subtyping method doesn't handle multiple transcripts, assign them to the same gene. BTW how do you do subtyping (I am looking for applying those methods myself)llrs –llrs 2017-11-30 10:16:29 +00:00 Commented Nov 30, 2017 at 10:16 just summing them is fine or not should I calculate median or mean on those duplicates?stack_learner –stack_learner 2017-11-30 10:17:38 +00:00 Commented Nov 30, 2017 at 10:17 Then you need to judge if each of these identifiers are really the same gene or not. I would say they are but that's up to you. Why do you think you need to calculate the median or mean? Aren't you interested in knowing how much each gene is being transcripted?llrs –llrs 2017-11-30 10:33:54 +00:00 Commented Nov 30, 2017 at 10:33 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. Let's look into this a bit more deeply. For instance: HUGO: SOGA3 Ensembl 1: ENSG00000214338 Ensembl 2: ENSG00000255330 The Ensembl pages (linked above) for both ENSG00000214338 and ENSG00000255330 show that these genes have 3 paralogs and each lists the other as one of the paralogs. Your other two examples are similar, they are also paralogs of each other. You can also see that the position of each gene in the genome is slightly different: ENSG00000255330 : Chromosome 6: 127,438,406-127,519,001 reverse strand. ENSG00000214338 : Chromosome 6: 127,472,794-127,519,191 reverse strand. So these two genes refer to two distinct regions of the genome. And, as expected, they have different transcripts. However, two of those transcripts (ENST00000481848.6 from ENSG00000255330 and ENST00000525778.5 from ENSG00000214338) code for the same protein sequence. This means that we have one protein but two different genes for it. To illustrate, here are the two transcripts aligned: ```r ENST00000525778.5 CCGTGTTCGAGCGGTTTTGGGATCCTCTGCCAAGTACGGGATAAGAGAAT ENST00000481848.6 CTC----------------------------------------------- ENST00000525778.5 AGAGCAAGGAGTTTGAGCGCTGCTGCCAATCTTCCATCTCGGGCGTGGCT ENST00000481848.6 -------------------------------------------------- ENST00000525778.5 CTGGCCTTTTTGTCTCTTTATCCCGCCACTCCCCACCCCCGTCCCCCGTC ENST00000481848.6 -------------------------------------------------- ENST00000525778.5 CCCCTCCCCGCCTTCTTCTCTTTCCTCTTGAAAGTAACCTCTCTTTGGTT ENST00000481848.6 -------------------------------------------TTTGGTT ENST00000525778.5 ACTTCTACCCTATCGTCCCTTGCAATCCCGCCTTCTCTCCTGCCCAGTTC ENST00000481848.6 ACTTCTACCCTATCGTCCCTTGCAATCCCGCCTTCTCTCCTGCCCAGTTC ENST00000525778.5 GGTGAGGTCAGCTCATGAACATCTCTGCACTTCCTGGGGAGCGTCTGGCC ENST00000481848.6 GGTGAGGTCAGCTCATGAACATCTCTGCACTTCCTGGGGAGCGTCTGGCC ENST00000525778.5 CAGCTTGGTACCTCGCGTGTTAGCCTGCTGAGAGCTGCAGCTCCAGCCCA ENST00000481848.6 CAGCTTGGTACCTCGCGTGTTAGCCTGCTGAGAGCTGCAGCTCCAGCCCA ENST00000525778.5 GCCGCGGACCGCCGGGAAGTCACCCGCTCTGGGGGCAGCTGAATCTTGAC ENST00000481848.6 GCCGCGGACCGCCGGGAAGTCACCCGCTCTGGGGGCAGCTGAATCTTGAC ENST00000525778.5 GTTACCGCCATTGTTTGGAGAAATCGAGTGGGTGTGGAACTGCGGCGAGC ENST00000481848.6 GTTACCGCCATTGTTTGGAGAAATCGAGTGGGTGTGGAACTGCGGCGAGC ENST00000525778.5 GAGCCCAAAGTCATGCTCAGCCGGAAAGTCGCATACAATAATATGAAGTT ENST00000481848.6 GAGCCCAAAGTCATGCTCAGCCGGAAAGTCGCATACAATAATATGAAGTT ENST00000525778.5 GAACTTTCCTCGGCGCAGAAAGGCACAATAATAAACCCTGAGATCCCCGG ENST00000481848.6 GAACTTTCCTCGGCGCAGAAAGGCACAATAATAAACCCTGAGATCCCCGG ENST00000525778.5 AATATCAGAAGGAAAGTGCTGTTGATTTGTTCAGCTACTTGGAAACCTCG ENST00000481848.6 AATATCAGAAGGAAAGTGCTGTTGATTTGTTCAGCTACTTGGAAACCTCG ENST00000525778.5 CCTCTGCTGAGAAGTGAGCAAGTTCTGGATTTCTAAAGGGTTGGGGATGA ENST00000481848.6 CCTCTGCTGAGAA------------------------------------- ENST00000525778.5 ATGATGGGCTTAGGAAGGAAGGTTGGCGGCGGTGAAGAGGGAAGAGGATG ENST00000481848.6 --------CTTAGGAAGGAAGGTTGGCGGCGGTGAAGAGGGAAGAGGATG ENST00000525778.5 ATCTAGTAATGGGCTCTTCCAAATCGCATCTGGTAGGTTCATAGCCATGA ENST00000481848.6 ATCTAGTAATGGGCTCTTCCAAATCGCATCTGGTAGGTTCATAGCCATGA ENST00000525778.5 GTCAGCCACCAATCGGGGGCGCTGCCCCTGCCACAGCAGCAGCTTCTCCC ENST00000481848.6 GTCAGCCACCAATCGGGGGCGCTGCCCCTGCCACAGCAGCAGCTTCTCCC ENST00000525778.5 GCCGCTGCAGCCACTGAGGCTCGTCTGCACCCGGAGGGCAGCAGCAGAAA ENST00000481848.6 GCCGCTGCAGCCACTGAGGCTCGTCTGCACCCGGAGGGCAGCAGCAGAAA ENST00000525778.5 GCAGCAGCGAGCTCAGTCACCCGCTAGACCGAGGGACAGTTCGCTGCGAC ENST00000481848.6 GCAGCAGCGAGCTCAGTCACCCGCTAGACCGAGGGACAGTTCGCTGCGAC ENST00000525778.5 AGACAATCGCGGCCACCCGGTCCCCAGTGGGGGCCGGCACTAAACTCAAT ENST00000481848.6 AGACAATCGCGGCCACCCGGTCCCCAGTGGGGGCCGGCACTAAACTCAAT ENST00000525778.5 TCTGTTCGGCAGCAGCAGCTGCAGCAGCAGCAACAGCAGGGTAACAAGAC ENST00000481848.6 TCTGTTCGGCAGCAGCAGCTGCAGCAGCAGCAACAGCAGGGTAACAAGAC ENST00000525778.5 CGGGAGCCGCACGGGGCCGCCCGCCAGCATCCGAGGAGGCGGAGGTGGGG ENST00000481848.6 CGGGAGCCGCACGGGGCCGCCCGCCAGCATCCGAGGAGGCGGAGGTGGGG ENST00000525778.5 CCGAGAAGGCGACGCCCCTGGCGCCCAAAGGGGCCGCTCCGGGAGCTGTC ENST00000481848.6 CCGAGAAGGCGACGCCCCTGGCGCCCAAAGGGGCCGCTCCGGGAGCTGTC ENST00000525778.5 CAGCCTGTGGCTGGTGCTGAAGCGGCCCCCGCCGCGACCCTGGCCGCTTT ENST00000481848.6 CAGCCTGTGGCTGGTGCTGAAGCGGCCCCCGCCGCGACCCTGGCCGCTTT ENST00000525778.5 GGGCGGCAGGAGGCCCGGACCGCCCGAGGAGCCGCCTCGGGAGCTAGAGT ENST00000481848.6 GGGCGGCAGGAGGCCCGGACCGCCCGAGGAGCCGCCTCGGGAGCTAGAGT ENST00000525778.5 CCGTGCCCTCGAAACTCGGGGAACCCCCTCCGCTCGGAGAAGGGGGAGGA ENST00000481848.6 CCGTGCCCTCGAAACTCGGGGAACCCCCTCCGCTCGGAGAAGGGGGAGGA ENST00000525778.5 GGGGGCGGGGAAGGAGGAGGAGCCGGCGGCGGCTCCGGGGAAAGGGAGGG ENST00000481848.6 GGGGGCGGGGAAGGAGGAGGAGCCGGCGGCGGCTCCGGGGAAAGGGAGGG ENST00000525778.5 GGGCGCTCCGCAGCCGCCGCCGCCCAGGGGCTGGCGAGGGAAAGGCGTAC ENST00000481848.6 GGGCGCTCCGCAGCCGCCGCCGCCCAGGGGCTGGCGAGGGAAAGGCGTAC ENST00000525778.5 GCGCTCAGCAGAGGGGCGGCAGCGGCGGGGAGGGGGCCTCCCCTTCTCCA ENST00000481848.6 GCGCTCAGCAGAGGGGCGGCAGCGGCGGGGAGGGGGCCTCCCCTTCTCCA ENST00000525778.5 TCCTCCTCTTCTGCGGGCAAAACCCCAGGAACCGGCAGCAGAAACTCCGG ENST00000481848.6 TCCTCCTCTTCTGCGGGCAAAACCCCAGGAACCGGCAGCAGAAACTCCGG ENST00000525778.5 AAGCGGCGTTGCGGGGGGCGGCAGCGGTGGTGGAGGGAGCTACTGGAAAG ENST00000481848.6 AAGCGGCGTTGCGGGGGGCGGCAGCGGTGGTGGAGGGAGCTACTGGAAAG ENST00000525778.5 AAGGATGTCTGCAGTCTGAGCTCATCCAGTTCCATCTCAAGAAGGAGCGG ENST00000481848.6 AAGGATGTCTGCAGTCTGAGCTCATCCAGTTCCATCTCAAGAAGGAGCGG ENST00000525778.5 GCGGCAGCGGCGGCGGCCGCGGCTCAGATGCACGCTAAGAACGGCGGCGG ENST00000481848.6 GCGGCAGCGGCGGCGGCCGCGGCTCAGATGCACGCTAAGAACGGCGGCGG ENST00000525778.5 CAGCAGTAGCCGCAGCTCCCCGGTGTCTGGCCCCCCTGCCGTTTGCGAGA ENST00000481848.6 CAGCAGTAGCCGCAGCTCCCCGGTGTCTGGCCCCCCTGCCGTTTGCGAGA ENST00000525778.5 CCCTGGCCGTCGCCTCCGCCTCCCCAATGGCGGCGGCGGCGGAGGGCCCC ENST00000481848.6 CCCTGGCCGTCGCCTCCGCCTCCCCAATGGCGGCGGCGGCGGAGGGCCCC ENST00000525778.5 CAGCAGAGCGCAGAGGGCAGCGCGAGCGGCGGGGGCATGCAGGCGGCAGC ENST00000481848.6 CAGCAGAGCGCAGAGGGCAGCGCGAGCGGCGGGGGCATGCAGGCGGCAGC ENST00000525778.5 GCCCCCTTCGTCGCAGCCGCACCCGCAGCAGCTCCAAGAGCAGGAAGAAA ENST00000481848.6 GCCCCCTTCGTCGCAGCCGCACCCGCAGCAGCTCCAAGAGCAGGAAGAAA ENST00000525778.5 TGCAAGAGGAGATGGAGAAGCTGCGAGAGGAAAACGAGACTCTCAAGAAC ENST00000481848.6 TGCAAGAGGAGATGGAGAAGCTGCGAGAGGAAAACGAGACTCTCAAGAAC ENST00000525778.5 GAGATCGATGAGCTGAGAACCGAGATGGACGAGATGAGGGACACTTTCTT ENST00000481848.6 GAGATCGATGAGCTGAGAACCGAGATGGACGAGATGAGGGACACTTTCTT ENST00000525778.5 CGAGGAGGATGCCTGTCAACTGCAGGAAATGCGCCACGAGTTGGAGAGAG ENST00000481848.6 CGAGGAGGATGCCTGTCAACTGCAGGAAATGCGCCACGAGTTGGAGAGAG ENST00000525778.5 CCAACAAAAACTGCCGGATCCTGCAGTACCGCCTCCGCAAAGCCGAGCGC ENST00000481848.6 CCAACAAAAACTGCCGGATCCTGCAGTACCGCCTCCGCAAAGCCGAGCGC ENST00000525778.5 AAAAGGCTCCGCTACGCCCAGACCGGGGAAATCGACGGGGAGCTGTTGCG ENST00000481848.6 AAAAGGCTCCGCTACGCCCAGACCGGGGAAATCGACGGGGAGCTGTTGCG ENST00000525778.5 CAGCCTGGAGCAGGACCTCAAGGTTGCAAAGGATGTATCTGTGAGACTTC ENST00000481848.6 CAGCCTGGAGCAGGACCTCAAGGTTGCAAAGGATGTATCTGTGAGACTTC ENST00000525778.5 ACCATGAATTAGAAAATGTGGAAGAAAAGAGAACAACAACAGAAGATGAA ENST00000481848.6 ACCATGAATTAGAAAATGTGGAAGAAAAGAGAACAACAACAGAAGATGAA ENST00000525778.5 AATGAGAAACTGAGGCAACAGCTCATAGAAGTTGAAATTGCAAAGCAAGC ENST00000481848.6 AATGAGAAACTGAGGCAACAGCTCATAGAAGTTGAAATTGCAAAGCAAGC ENST00000525778.5 TTTACAGAATGAACTGGAAAAAATGAAAGAGTTATCCTTAAAAAGAAGAG ENST00000481848.6 TTTACAGAATGAACTGGAAAAAATGAAAGAGTTATCCTTAAAAAGAAGAG ENST00000525778.5 GAAGCAAAGATTTGCCAAAATCTGAAAAAAAGGCTCAACAGACTCCCACA ENST00000481848.6 GAAGCAAAGATTTGCCAAAATCTGAAAAAAAGGCTCAACAGACTCCCACA ENST00000525778.5 GAGGAGGACAATGAAGATCTGAAGTGCCAGCTGCAGTTTGTTAAGGAAGA ENST00000481848.6 GAGGAGGACAATGAAGATCTGAAGTGCCAGCTGCAGTTTGTTAAGGAAGA ENST00000525778.5 AGCCGCTTTGATGAGAAAGAAAATGGCCAAGATTGATAAAGAAAAGGACA ENST00000481848.6 AGCCGCTTTGATGAGAAAGAAAATGGCCAAGATTGATAAAGAAAAGGACA ENST00000525778.5 GATTTGAACACGAGCTCCAGAAGTACAGATCCTTTTATGGGGATCTGGAC ENST00000481848.6 GATTTGAACACGAGCTCCAGAAGTACAGATCCTTTTATGGGGATCTGGAC ENST00000525778.5 AGTCCTTTGCCCAAAGGAGAAGCCGGAGGCCCTCCCAGCACTAGGGAGGC ENST00000481848.6 AGTCCTTTGCCCAAAGGAGAAGCCGGAGGCCCTCCCAGCACTAGGGAGGC ENST00000525778.5 CGAGCTCAAGCTACGGCTAAGGCTGGTGGAGGAAGAAGCCAACATCCTGG ENST00000481848.6 CGAGCTCAAGCTACGGCTAAGGCTGGTGGAGGAAGAAGCCAACATCCTGG ENST00000525778.5 GCAGGAAAATCGTCGAACTGGAGGTGGAGAACAGAGGCCTGAAGGCGGAA ENST00000481848.6 GCAGGAAAATCGTCGAACTGGAGGTGGAGAACAGAGGCCTGAAGGCGGAA ENST00000525778.5 CTGGACGACCTTAGGGGCGATGACTTCAACGGCTCGGCCAACCCGCTCAT ENST00000481848.6 CTGGACGACCTTAGGGGCGATGACTTCAACGGCTCGGCCAACCCGCTCAT ENST00000525778.5 GAGGGAGCAGAGCGAATCCCTGTCGGAGCTGCGGCAGCACCTGCAGCTGG ENST00000481848.6 GAGGGAGCAGAGCGAATCCCTGTCGGAGCTGCGGCAGCACCTGCAGCTGG ENST00000525778.5 TGGAAGACGAGACGGAGCTGCTGCGGAGGAACGTGGCCGACCTGGAGGAG ENST00000481848.6 TGGAAGACGAGACGGAGCTGCTGCGGAGGAACGTGGCCGACCTGGAGGAG ENST00000525778.5 CAGAACAAGCGCATCACGGCGGAGCTCAACAAGTACAAGTACAAGTCCGG ENST00000481848.6 CAGAACAAGCGCATCACGGCGGAGCTCAACAAGTACAAGTACAAGTCCGG ENST00000525778.5 CGGCCACGACAGCGCGCGGCACCACGACAACGCCAAGACCGAGGCCCTGC ENST00000481848.6 CGGCCACGACAGCGCGCGGCACCACGACAACGCCAAGACCGAGGCCCTGC ENST00000525778.5 AGGAGGAGCTGAAGGCGGCGCGCCTGCAGATCAACGAGCTCAGCGGCAAG ENST00000481848.6 AGGAGGAGCTGAAGGCGGCGCGCCTGCAGATCAACGAGCTCAGCGGCAAG ENST00000525778.5 GTCATGCAGCTGCAGTACGAGAACCGCGTGCTTATGTCCAACATGCAGCG ENST00000481848.6 GTCATGCAGCTGCAGTACGAGAACCGCGTGCTTATGTCCAACATGCAGCG ENST00000525778.5 CTACGACCTGGCCTCGCACCTGGGCATCCGCGGCAGCCCCCGCGACAGCG ENST00000481848.6 CTACGACCTGGCCTCGCACCTGGGCATCCGCGGCAGCCCCCGCGACAGCG ENST00000525778.5 ACGCCGAGAGCGACGCGGGCAAGAAGGAGAGCGACGACGACTCGCGGCCT ENST00000481848.6 ACGCCGAGAGCGACGCGGGCAAGAAGGAGAGCGACGACGACTCGCGGCCT ENST00000525778.5 CCGCACCGCAAGCGCGAAGGGCCCATCGGCGGCGAGAGCGACTCGGAGGA ENST00000481848.6 CCGCACCGCAAGCGCGAAGGGCCCATCGGCGGCGAGAGCGACTCGGAGGA ENST00000525778.5 GGTGCGCAACATCCGCTGCCTCACGCCCACTCGCTCCTTCTACCCGGCGC ENST00000481848.6 GGTGCGCAACATCCGCTGCCTCACGCCCACTCGCTCCTTCTACCCGGCGC ENST00000525778.5 CCGGGCCCTGGCCCAAGAGCTTCTCCGATCGGCAGCAGATGAAGGACATC ENST00000481848.6 CCGGGCCCTGGCCCAAGAGCTTCTCCGATCGGCAGCAGATGAAGGACATC ENST00000525778.5 CGCTCGGAGGCCGAGCGCCTGGGCAAGACCATCGACCGGCTCATCGCCGA ENST00000481848.6 CGCTCGGAGGCCGAGCGCCTGGGCAAGACCATCGACCGGCTCATCGCCGA ENST00000525778.5 CACGAGCACCATCATCACCGAGGCGCGCATCTACGTGGCCAACGGGGACC ENST00000481848.6 CACGAGCACCATCATCACCGAGGCGCGCATCTACGTGGCCAACGGGGACC ENST00000525778.5 TGTTCGGACTCATGGACGAGGAGGACGACGGCAGCCGCATCCGGGAGCAC ENST00000481848.6 TGTTCGGACTCATGGACGAGGAGGACGACGGCAGCCGCATCCGGGAGCAC ENST00000525778.5 GAGCTGCTCTACCGCATCAACGCTCAGATGAAGGCCTTCCGCAAGGAGCT ENST00000481848.6 GAGCTGCTCTACCGCATCAACGCTCAGATGAAGGCCTTCCGCAAGGAGCT ENST00000525778.5 GCAGACCTTCATCGACCGCCTCGAGGTGCCCAAGTCTGCGGACGACCGCG ENST00000481848.6 GCAGACCTTCATCGACCGCCTCGAGGTGCCCAAGTCTGCGGACGACCGCG ENST00000525778.5 GCGCCGAGGAGCCCATTTCCGTGAGTCAGATGTTCCAGCCTATCATTTTA ENST00000481848.6 GCGCCGAGGAGCCCATTTCCGTGAGTCAGATGTTCCAGCCTATCATTTTA ENST00000525778.5 CTTATTCTCATTCTTGTATTATTTTCATCACTTTCTTACACAACAATATT ENST00000481848.6 CTTATTCTCATTCTTGTATTATTTTCATCACTTTCTTACACAACAATATT ENST00000525778.5 TAAACTTGTCTTCCTTTTTACACTGTTTTTTGTACTGTAAATCTTTCATC ENST00000481848.6 TAAACTTGTCTTCCTTTTTACACTGTTTTTTGTACTGTAAATCTTTCATC ENST00000525778.5 ATTTACCATTCATTGTAGTATTTTCAGTTTGTTTATTTTGTTCACCCTTC ENST00000481848.6 ATTTACCATTCATTGTAGTATTTTCAGTTTGTTTATTTTGTTCACCCTTC ENST00000525778.5 AAGACAAGAAGTAAAAGAAGTATAATTTCTGTAGTAACCAATGCTATAAA ENST00000481848.6 AAGACAAGAAGTAAAAGAAGTATAATTTCTGTAGTAACCAATGCTATAAA ENST00000525778.5 AACACTGAAGACTGCTTATTTCTTTAAAAAGATACAACTCATCTTACCAA ENST00000481848.6 AACACTGAAGACTGCTTATTTCTTTAAAAAGATACAACTCATCTTACCAA ENST00000525778.5 GACCAAA------------------------------------------- ENST00000481848.6 GACCAAATTCAATAAGAAGCCCAAACACTAAAATATTTCAGGCTTTATTT ENST00000525778.5 -------------------------------------------------- ENST00000481848.6 TAAAGGCAAGTGAGACTGCTTCAAATAAAACAACTTCAAGCTTCCAAGAA [. . .] ``` The ENST00000481848.6 transcript continues quite a bit after this but there is no more similarity. The transcripts are not identical, but are very similar. Now, look at the proteins each transcript encodes: ```r ENST00000481848.6 MSQPPIGGAAPATAAASPAAAATEARLHPEGSSRKQQRAQSPARPRDSSL ENST00000525778.5 MSQPPIGGAAPATAAASPAAAATEARLHPEGSSRKQQRAQSPARPRDSSL ENST00000481848.6 RQTIAATRSPVGAGTKLNSVRQQQLQQQQQQGNKTGSRTGPPASIRGGGG ENST00000525778.5 RQTIAATRSPVGAGTKLNSVRQQQLQQQQQQGNKTGSRTGPPASIRGGGG ENST00000481848.6 GAEKATPLAPKGAAPGAVQPVAGAEAAPAATLAALGGRRPGPPEEPPREL ENST00000525778.5 GAEKATPLAPKGAAPGAVQPVAGAEAAPAATLAALGGRRPGPPEEPPREL ENST00000481848.6 ESVPSKLGEPPPLGEGGGGGGEGGGAGGGSGEREGGAPQPPPPRGWRGKG ENST00000525778.5 ESVPSKLGEPPPLGEGGGGGGEGGGAGGGSGEREGGAPQPPPPRGWRGKG ENST00000481848.6 VRAQQRGGSGGEGASPSPSSSSAGKTPGTGSRNSGSGVAGGGSGGGGSYW ENST00000525778.5 VRAQQRGGSGGEGASPSPSSSSAGKTPGTGSRNSGSGVAGGGSGGGGSYW ENST00000481848.6 KEGCLQSELIQFHLKKERAAAAAAAAQMHAKNGGGSSSRSSPVSGPPAVC ENST00000525778.5 KEGCLQSELIQFHLKKERAAAAAAAAQMHAKNGGGSSSRSSPVSGPPAVC ENST00000481848.6 ETLAVASASPMAAAAEGPQQSAEGSASGGGMQAAAPPSSQPHPQQLQEQE ENST00000525778.5 ETLAVASASPMAAAAEGPQQSAEGSASGGGMQAAAPPSSQPHPQQLQEQE ENST00000481848.6 EMQEEMEKLREENETLKNEIDELRTEMDEMRDTFFEEDACQLQEMRHELE ENST00000525778.5 EMQEEMEKLREENETLKNEIDELRTEMDEMRDTFFEEDACQLQEMRHELE ENST00000481848.6 RANKNCRILQYRLRKAERKRLRYAQTGEIDGELLRSLEQDLKVAKDVSVR ENST00000525778.5 RANKNCRILQYRLRKAERKRLRYAQTGEIDGELLRSLEQDLKVAKDVSVR ENST00000481848.6 LHHELENVEEKRTTTEDENEKLRQQLIEVEIAKQALQNELEKMKELSLKR ENST00000525778.5 LHHELENVEEKRTTTEDENEKLRQQLIEVEIAKQALQNELEKMKELSLKR ENST00000481848.6 RGSKDLPKSEKKAQQTPTEEDNEDLKCQLQFVKEEAALMRKKMAKIDKEK ENST00000525778.5 RGSKDLPKSEKKAQQTPTEEDNEDLKCQLQFVKEEAALMRKKMAKIDKEK ENST00000481848.6 DRFEHELQKYRSFYGDLDSPLPKGEAGGPPSTREAELKLRLRLVEEEANI ENST00000525778.5 DRFEHELQKYRSFYGDLDSPLPKGEAGGPPSTREAELKLRLRLVEEEANI ENST00000481848.6 LGRKIVELEVENRGLKAELDDLRGDDFNGSANPLMREQSESLSELRQHLQ ENST00000525778.5 LGRKIVELEVENRGLKAELDDLRGDDFNGSANPLMREQSESLSELRQHLQ ENST00000481848.6 LVEDETELLRRNVADLEEQNKRITAELNKYKYKSGGHDSARHHDNAKTEA ENST00000525778.5 LVEDETELLRRNVADLEEQNKRITAELNKYKYKSGGHDSARHHDNAKTEA ENST00000481848.6 LQEELKAARLQINELSGKVMQLQYENRVLMSNMQRYDLASHLGIRGSPRD ENST00000525778.5 LQEELKAARLQINELSGKVMQLQYENRVLMSNMQRYDLASHLGIRGSPRD ENST00000481848.6 SDAESDAGKKESDDDSRPPHRKREGPIGGESDSEEVRNIRCLTPTRSFYP ENST00000525778.5 SDAESDAGKKESDDDSRPPHRKREGPIGGESDSEEVRNIRCLTPTRSFYP ENST00000481848.6 APGPWPKSFSDRQQMKDIRSEAERLGKTIDRLIADTSTIITEARIYVANG ENST00000525778.5 APGPWPKSFSDRQQMKDIRSEAERLGKTIDRLIADTSTIITEARIYVANG ENST00000481848.6 DLFGLMDEEDDGSRIREHELLYRINAQMKAFRKELQTFIDRLEVPKSADD ENST00000525778.5 DLFGLMDEEDDGSRIREHELLYRINAQMKAFRKELQTFIDRLEVPKSADD ENST00000481848.6 RGAEEPISVSQMFQPIILLILILVLFSSLSYTTIFKLVFLFTLFFVL ENST00000525778.5 RGAEEPISVSQMFQPIILLILILVLFSSLSYTTIFKLVFLFTLFFVL ``` As you can see, they are 100% identical. So, we have two genes (distinct regions of the DNA) which produce the same protein. Because the gene product is the same, the "function" of the genes is also the same so HUGO treats them as the same gene and gives them the same name. Whether or not you want to treat them as one gene or two will depend on what you are doing. If you are comparing the relative abundance of each transcript, you must treat them separately. If you want to examine the expression level of the protein product and are using the RNASEM data as a proxy for protein information, then you should treat them as one "gene" and simply sum your RNASEM values. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Nov 30, 2017 at 12:14 answered Nov 30, 2017 at 12:08 terdon♦terdon 11.4k 6 6 gold badges 26 26 silver badges 51 51 bronze badges Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. There is no one-to-one mapping of gene ids from one database to the other. Ensembl (who maintain Ensembl ENSG IDs), ncbi (who maintain EntrezGene IDs and RefSeq transcript ids) and HUGO (who maintain gene symbols/names) all have different ideas about what a gene is, which part of the genome belongs to which gene etc. In fact, it's worse than that because HUGO gene IDs refer to the abstract idea of a gene, rather than a particular part of any chromosome. My suggestion would be to leave the IDs as Ensembl IDs (as these actually refer to a physical part of a chromosome, which is what you are quantifying), right up until you have to report the results to someone who really NEEDS gene names. If you still have duplicates, you could sum, or pick the most interesting, but I would suggest just leaving the duplicates in there, and treat it as thought there are two genes with the same name. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Nov 30, 2017 at 11:31 terdon♦ 11.4k 6 6 gold badges 26 26 silver badges 51 51 bronze badges answered Nov 30, 2017 at 10:54 Ian SudberyIan Sudbery 3,351 1 1 gold badge 13 13 silver badges 22 22 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Bioinformatics 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. 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3280	https://www.ncbi.nlm.nih.gov/books/NBK532916/	Vitamin A Toxicity - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Vitamin A Toxicity Jazmine M. Olson; Muhammad Atif Ameer; Amandeep Goyal. Author Information and Affiliations Authors Jazmine M. Olson; Muhammad Atif Ameer 1; Amandeep Goyal 2. Affiliations 1 Suburban Community Hospital 2 University of Kansas Medical Center Last Update: September 2, 2023. Go to: Continuing Education Activity Vitamin A plays a crucial role in regulating various physiological processes of the body and maintaining vision and immune system function to support skin health and cell growth.Although vitamin A is essential for the body, excessive intake can pose various adverse effects, disrupting the body's equilibrium and overall well-being.Vitamin A toxicity, also known as hypervitaminosis A, can result from either the excessive consumption of vitamin A and related compounds or its topical application.This activity describes the causes, evaluation, management, treatment, and prevention of vitamin A toxicity, emphasizing the essential role of the interprofessional healthcare team in enhancing care for affected patients. Objectives: Identify the risk factors associated with excessive vitamin A intake, considering dietary sources, supplements, and medication history. Assess laboratory findings, including serum vitamin A levels, liver function tests, lipid profiles, and hematological parameters, to diagnose and monitor vitamin A toxicity in patients. Select and recommend appropriate treatment modalities, alternative therapies, or supplements to patients based on the severity and presentation of vitamin A toxicity. Coordinate with the interprofessional healthcare team to guide pregnant women about safe vitamin A use during pregnancy and prevent teratogenic malformations. Access free multiple choice questions on this topic. Go to: Introduction Vitamin A is essential for maintaining the body's vision, cell division, reproduction, and immune function. Vitamin A belongs to the category of lipid-soluble compounds called retinoic acids.Beta-carotene is the most well-known form of 2 clinically significant forms of vitamin A: preformed vitamin A and provitamin A carotenoids.Mammals metabolize carotenoids into active vitamin A.Preformed vitamin A encompasses metabolically active compounds such as retinol, retinal, retinoic acid, and retinyl esters.Although essential for overall health, excessive consumption of preformed vitamin A can lead to acute and chronic toxicity.The condition characterized by elevated levels of vitamin A in the body is referred to as hypervitaminosis A.Vitamin A, also known as teratogen, is capable of causing severe malformations. This article comprehensively reviews the prevention, presentation, laboratory findings, and treatment of the 3 recognized syndromes of vitamin A toxicity: acute, chronic, and teratogenic. Go to: Etiology Hypervitaminosis A is relatively uncommon and usually arises from excessive supplementation or medication usage. The primary cause of toxicity is the consumption of substantial quantities of vitamin A through dietary supplements and foods.This usually occurs when individuals consume high doses of vitamin A without proper medical supervision. Most diets contain a combination of preformed vitamin A and provitamin A carotenoids. Preformed vitamin A is derived from animal sources and is found in egg yolks, butter, chicken, beef, organ meats, fish, fish oils, and fortified foods. Preformed vitamin A is readily absorbed in the small intestine and subsequently stored in the liver.Therefore, excessive intake of dietary preformed vitamin A from animal-based sources and supplements can contribute to the risk of toxicity. Provitamin A carotenoids, such as beta-carotene, are plant pigments that the body converts into active vitamin A. These carotenoids are plentiful in leafy greens and vibrantly colored vegetables and fruits, including carrots, sweet potatoes, and papayas. The absorption of provitamin A is variable and subject to feedback regulation, making it unlikely to lead to toxicity with excessive intake. Certain dermatological medications, such as isotretinoin, contain analogs of vitamin A. Prolonged or excessive usage of these medications can lead to an accumulation of vitamin A stores, resulting in hypervitaminosis A, toxicity, and teratogenic effects. The recommended daily allowance (RDA) for vitamin A is measured in retinol activity equivalents (RAE) to account for the different bioactivities of retinol and provitamin A carotenoids. Notably, 1 RAE is equivalent to 1 mcg of retinol or 3 International units (IU). The RDA for vitamin A is as follows: 700 RAE for adult women 900 RAE for adult men 750 to 770 RAE for pregnant women 1200 to 1300 RAE for lactating women The tolerable upper intake levels (UL) for vitamin A have been established to prevent toxicity. The UL of vitamin A for adults is 3000 RAE per day, including preformed vitamin A from foods and supplements and the amount that can be converted from plant-based sources such as beta-carotene. The consumption of preformed vitamin A from dietary supplements in developed countries frequently surpasses the RDA. Go to: Epidemiology Few published reports of vitamin A toxicity have been published, with fewer than 10 cases per year documented from 1976 to 1987. Infants and children are at a higher risk of toxicity due to their smaller body size and reduced tolerance for high doses.Accidental ingestion of vitamin A supplements by children is a common cause of acute toxicity. There is no substantial difference in the occurrence of vitamin A toxicity between males and females. However, pregnant women are at risk if they take high doses of vitamin supplements. Excessive supplement intake typically occurs in developed countries when individuals exceed the RDA.Supplementation programs have been implemented in regions where vitamin deficiency is a significant public health concern. Moreover, toxicity can arise if dosing guidelines are not carefully followed. Populations heavily reliant on liver-based food sources rich in vitamin A, such as polar bear liver and chicken liver, face an increased risk of vitamin A toxicity. However, only rare cases of this phenomenon are reported in the published literature, and it is not considered a significant public health concern. Go to: Pathophysiology The symptoms of systemic vitamin A toxicity vary depending on the severity and duration of excessive intake. The 3 recognized syndromes of vitamin A toxicity are acute, chronic, and teratogenic. Acute systemic vitamin A toxicity typically arises when an individual consumes over 100,000 RAE within a short period, often from supplements or high-dose medications. The toxicity symptoms include nausea, vomiting, headache, dizziness, irritability, blurred vision, and muscular incoordination.Acute toxicity is rare and is more likely to occur after consuming synthetic forms of vitamin A, such as the retinoid medication isotretinoin. Mucocutaneous effects include cheilitis and dryness of lips and oral, ophthalmic, and nasal mucosa. The suggested mechanism involves decreased sebum production, reduced epidermal thickness, and altered barrier function. Additional cutaneous effects include dry skin, pruritus, peeling of the palms and soles, and fissuring of the fingertips. Higher doses of vitamin A may lead to telogen effluvium. Severe cases may manifest with bone pain and increased intracranial pressure. Chronic vitamin A toxicity is associated with prolonged ingestion of excessive vitamin A, typically exceeding 8000 RAE per day. This condition can develop after consuming substantial quantities of animal-based foods rich in preformed vitamin A, such as liver, or through the prolonged use of high-dose vitamin A supplements.The toxicity symptoms include dry, cracked skin, hair loss, brittle nails, fatigue, loss of appetite, bone and joint pain, and hepatomegaly.Chronic retinoid toxicity affects various organ systems. Bone-related effects include bone spurs, calcinosis, and bone resorption, leading to hypercalcemia, osteoporosis, and hip fractures.Central nervous system effects include headache, nausea, and vomiting. Pseudotumor cerebri syndrome has been associated with vitamin A toxicity. Hypothyroidism, reversible upon discontinuation of therapy, was observed in 40% of patients during clinical trials for cutaneous T-cell lymphoma with bexarotene, a derivative of vitamin A.Reversible renal dysfunction with elevated serum creatinine has been observed after etretinate use, but not after isotretinoin use.Etretinate, a psoriasis medication,is no longer prescribed in most countries because of its teratogenic effects. Hypertriglyceridemia is the most common laboratory abnormality associated with oral retinoid use. Triglyceride, low-density lipoprotein, and total cholesterol levels increase in many patients using bexarotene, isotretinoin, etretinate, and acitretin.Elevated serum transaminases may also occur with oral retinoid usage, more often with etretinate or acitretin compared to isotretinoin and bexarotene. These levels typically show an increase within 2 to 8 weeks after the initiation of therapy, followed by normalization over the subsequent 2 to 4 weeks. Prolonged liver damage can include fibrosis and cirrhosis. No definitive causal association has been established between isotretinoin and depression, psychosis, or suicide attempts, although such a link had been previously suggested. The teratogenic effects of vitamin A intake were first discovered in animal studies during the 1950s. Subsequently, extensive research has been conducted to elucidate the underlying causes of the effects.Excessive vitamin A consumption during pregnancy has been associated with various congenital malformations in humans. The disabilities associated with this condition include abnormalities in the central nervous system, such as microcephaly and hydrocephalus; cardiac issues, such as transposition of the great vessels; craniofacial deformities, such as cleft lip and palate; limb abnormalities; and urinary tract disorders. Vitamin A is critical for regulating gene expression and guiding cell differentiation during embryonic development. The embryo is particularly susceptible to the teratogenic effects of excessive amounts of vitamin A during the first trimester, which is a phase marked by rapid organogenesis. Teratogenic effects are primarily associated with high doses of retinoid medications such as isotretinoin rather than dietary sources of vitamin A. A minimum safe dose of oral retinoids during pregnancy has not been established. Isotretinoin is estimated to increase the risk of fetal malformations by 25-fold. The mechanism is believed to be linked to a toxic effect on neural crest cells, possibly affecting the regulation of axial patterning in the embryo via the expression of the homeobox gene Hoxb-1. Pregnant women should refrain from using oral retinoid medications and excessive vitamin A supplements to mitigate the risk of vitamin A-related malformations. They should consult healthcare professionals and prenatal care providers to ensure proper prenatal nutrition and avoid drugs or supplements that are contraindicated during pregnancy. Prescribers should only recommend oral retinoids to women of childbearing age who are not pregnant and use reliable birth control methods to prevent the potential risk of vitamin A-related teratogenic effects. Go to: Histopathology Hypervitaminosis A affects multiple organ systems of the body and can damage the liver, bones, central nervous system, and skin. The specific histopathological findings differ based on the severity and duration of the toxicity, some of which are listed below. Liver: Hepatic steatosis, hepatocellular injury, and fibrosis with collagen deposits Bone: Osteoporosis and thickening of cortical bone Central nervous system: Pseudotumor cerebri and cerebral edema Skin: Exfoliative dermatitis and epidermal hyperplasia Go to: Toxicokinetics The toxicokinetics of vitamin A toxicity involve its absorption, distribution, metabolism, and elimination from the body. Absorption:Dietary vitamin A is sourced from 2 main categories: preformed vitamin A, including retinol and retinyl esters, which are found in animal-derived products and supplemented foods, and provitamin A carotenoids present in fruits and vegetables. High-fat meals enhance the absorption of vitamin A in the small intestine. Distribution:After absorption, vitamin A is transported to the liver, where it is stored as retinyl esters. From there,it is released into the bloodstream bound to retinol-binding protein (RBP) and transthyretin. These carrier proteins help transport vitamin A to target tissues, including the eyes, skin, and other organs. Metabolism:Retinol is converted into its active forms, retinal and retinoic acid, within target tissues.Lecithin retinol acyltransferase (LRAT) is the enzyme responsible for catalyzing retinoid esterification for storage in the liver.Cellular RBP assists LRAT in regulating retinoid uptake and metabolism. These 2 proteins are essential in the metabolism of retinoids, which may be responsible for the toxic effects of vitamin A. Elimination: Fat-soluble vitamin A and its metabolites are primarily excreted through bile and feces, with only a small portion being eliminated in the urine. Go to: History and Physical The medical history will likely reveal excessive consumption of vitamin A-rich foods, supplements, or oral retinoid medications. Symptoms include blurred vision, headaches, dry skin, hair loss, fatigue, and bone pain. Typical physical examination findings may include dryness of the conjunctiva and mucous membranes, scaly skin, alopecia, papilledema, bone tenderness, and hepatomegaly. Patients with a history of using topical vitamin A derivatives will likely exhibit localized skin peeling and erythema, but systemic signs and symptoms are not expected. Go to: Evaluation A patient's history and physical examination findings guide the selection of diagnostic studies. Patients with persistent headaches while taking vitamin A medications should be evaluated for increased intracranial pressure and pseudotumor cerebri syndrome. Laboratory findings obtained during assessing patients with vitamin A toxicity can reveal several abnormalities, a few of which are mentioned below. Serum vitamin A levels:The serum vitamin A levels exceed 80 mcg/dL. Liver function tests:Elevated levels of alanine transaminase (ALT), aspartate transaminase (AST), and alkaline phosphatase (ALP) indicate potential hepatic damage or inflammation. Serum lipids:Increased levels of triglycerides and cholesterol are observed. Hematological abnormalities:The identified abnormalities in the blood include leukocytosis, thrombocytopenia, and anemia. Bone markers: Increased levels of ALP are associated with bone turnover. Renal function: Elevations of blood urea nitrogen and creatinine occur. Electrolyte imbalances: Alterations in electrolyte levels, such as sodium, potassium, and calcium occur. Some patients require laboratory testing to monitor their condition. Individuals taking isotretinoin should undergo regular lipid level checks. Elevations in liver enzymes are usually mild and reversible, although ALT and AST may need to be monitored based on the dosage and any comorbidities the patient may have. Women of childbearing age must have 2 negative pregnancy tests, conducted 30 days apart, before initiating isotretinoin treatment. Additional testing during and immediately after discontinuing the treatment is also necessary. Free T4 should be monitored both before and during treatment with bexarotene.In addition, baseline fasting serum lipid levels should be measured at treatment initiation and then every 1 to 2 weeks during therapy until they stabilize. Patients with a history of kidney disease and undergoing etretinate therapy should regularly monitor their renal function throughout the treatment course. Go to: Treatment / Management The treatment of systemic vitamin A toxicity includes discontinuing vitamin A intake, providing supportive care, and managing patient symptoms. The management of skin irritation caused by topical retinoids involves reducing the volume and frequency of applications while increasing the use of emollients. Patients can be reassured that this adverse effect will likely improve with the continued use of the skincare products. Acute retinoid toxicity is rare, but in documented cases, recovery is typically rapid upon discontinuation of the medication. The sources of excess vitamin A intake, including dietary supplements and medications, must be identified as a primary step. Furthermore, healthcare professionals should recommend avoiding foods rich in preformed vitamin A to patients, such as liver and fortified products, until the toxicity resolves. Supportive care involves close monitoring of the patient's condition,including vital signs and overall health status. Severe cases may necessitate hospital admission. Hypotension can be managed with fluid administration, and hypercalcemia can be treated with calcitonin and corticosteroids.When there is evidence of hepatic damage, liver function tests can assess the extent of the injury and guide further management decisions. Patients with dry or peeling skin are advised to use moisturizers or emollients to alleviate discomfort and facilitate skin healing. For patients experiencing dry eyes, artificial tears and lubricating eye drops, including those containing methylcellulose, can be beneficial. The treatment of medication-induced chronic toxicity depends on the specific drug and its adverse effects. In cases where oral retinoids can lead to an elevation of the fasting triglyceride level to 800 mcg/dL or higher, options such as discontinuation, dose reduction, or the addition of lipid-lowering medication can be considered to prevent pancreatitis. Milder elevations can be monitored or treated similarly. Patients experiencing toxicity from bexarotene can consider using a statin or fibrate concurrently to manage retinoid-induced hyperlipidemia and lower the risk of pancreatitis. Therapy may need to be discontinued if values remain elevated despite intervention and exceed 3 times the upper limit of normal. Patients diagnosed with pseudotumor cerebri syndrome should discontinue using vitamin A medications and might need acetazolamide to lower the intracranial pressure. Teratogenic malformations result in permanent damage, and the necessary supportive or surgical care varies based on the affected organ system. Go to: Differential Diagnosis When assessing the differential diagnosis of hypervitaminosis A, it is crucial to exclude other conditions that might manifest with similar symptoms, some of which are listed below. Acute liver injury: Viral hepatitis, alcohol use, or drug toxicity. Hypercalcemia: Hyperparathyroidism, malignancy, or excessive calcium and vitamin D intake. Hypothyroidism: Primary thyroid disorders or medications. Chronic renal failure: Diabetes, hypertension, glomerulonephritis, or polycystic kidney disease. Central nervous system:Headache, nausea, and vomiting may be symptoms of pseudotumor cerebri and are not always caused by vitamin A toxicity. Dermatological conditions: Exfoliative dermatitis and other skin conditions. Nonspecific symptoms:The nonspecific symptoms, including skin rashes, joint pain, fatigue, and renal and liver abnormalities, can also indicate conditions such as systemic lupus erythematosus and may not solely point to vitamin A toxicity. Go to: Prognosis The prognosis of hypervitaminosis A depends on the severity of the condition and the duration of exposure to high levels of vitamin A. In most patients who discontinue the source of excess vitamin A, toxicity symptoms gradually reverse, and complete recovery is expected. Symptoms such as dry skin, headache, and nausea typically improve within a few weeks or months with no long-term complications. However, severe cases can have serious consequences. Prolonged and excessive vitamin A intake can result in significant organ damage, including the liver, bones, central nervous system, and skin. Liver fibrosis or cirrhosis may be irreversible. Bone abnormalities, such as reduced bone density or fractures, may persist even after discontinuing vitamin A intake. In cases of pseudotumor cerebri, visual impairment may be permanent. The prognosis depends on the extent of organ damage and the promptness of medical intervention. Go to: Complications Complications of vitamin A toxicity include those listed below. Acute toxicity: Nausea, vomiting, headache, dizziness, irritability, blurred vision, and intracranial hypertension Chronic toxicity: Dry, itchy skin, hair loss, bone and joint pain, fatigue, anorexia, and weight loss Teratogenic effects: Craniofacial abnormalities, central nervous system malformations, and cardiovascular malformations Hepatic toxicity:Hepatomegaly, elevated liver enzymes, jaundice, fibrosis, and cirrhosis Skeletal abnormalities: Increased bone resorption, decreased bone formation, osteoporosis, fractures, and bone pain Hematological disturbances:Leukocytosis, anemia, and thrombocytopenia Central nervous system: Pseudotumor cerebri, with symptoms of headache, visual disturbances, and papilledema Go to: Deterrence and Patient Education Patient education regarding vitamin A toxicity is a shared responsibility among healthcare professionals, including physicians, advanced care practitioners, registered dietitians, nurses, and pharmacists. The significance of maintaining a balanced diet that includes essential vitamins and minerals is emphasized during routine medical appointments. Information regarding dietary sources of vitamin A, such as liver, fish, eggs, dairy products, and colorful fruits and vegetables, is provided to patients.Clinicians should advise patients about their recommended daily vitamin A intake according to their age, sex, and overall health. Pharmacists provide information about both over-the-counter and prescription drugs containing vitamin A or its derivatives, including retinoids. They can elucidate the proper usage and potential adverse effects of these medications, underscoring the significance of adhering to the recommended dosage and treatment duration. Obstetricians and other clinicians responsible for caring for women of childbearing age are responsible for educating them about the risks associated with excessive vitamin A intake from supplements or medications during pregnancy and the potential for teratogenic effects. If clinicians suspect toxicity, they should communicate the possible complications to patients and recommend discontinuing sources of vitamin A. This may involve dietary modifications and adjustments to medication and supplement usage. Go to: Pearls and Other Issues Teratogenicity is the most significant adverse effect of vitamin A toxicity. To mitigate this risk, patients should be counseled to avoid consuming more than the recommended amount of supplemental vitamin A during pregnancy. Most other adverse effects, such as skin irritation, dryness, and increased intracranial pressure, typically resolve when vitamin A intake or application is reduced or discontinued. Elevated triglycerides, cholesterol, or transaminases generally show improvement even with the continued use of the medication. Nevertheless, these levels should be closely monitored, and treatment should be discontinued if elevations persist or worsen. Go to: Enhancing Healthcare Team Outcomes Many individuals regularly consume substantial amounts of vitamins and supplements, often assuming they are safe due to their over-the-counter availability.However, poison control centers receive thousands of calls yearly related to vitamin overdoses and potential toxicity. An interprofessional team can collaborate to educate the public about these risks. Healthcare professionals, including physicians, nurses, and pharmacists, must possess a comprehensive skill set to address vitamin A toxicity effectively. Pharmacists are critical in disseminating vitamin safety information to the public, advocating for a balanced, healthy diet over supplement reliance.They also instruct consumers about the vitamin A content in both over-the-counter and prescription medications. The nursing staff encourages patients to consume a healthy diet and refrain from taking vitamin A supplements unless advised by their medical provider. Primary care physicians should collaborate with nurses and pharmacists to ensure that patients maintain safe levels of vitamin A intake from dietary sources, supplements, and medications. Pregnancy is an absolute contraindication for isotretinoin therapy. Despite this, patients may still take other retinoid medications and supplements that have teratogenic effects. The nursing and medical staff, along with obstetricians and clinicians, are responsible for obtaining and coordinating patients' medication history, ensuring thorough documentation and periodic review of all medications, vitamins, and supplements. Patients should be informed about the teratogenic risks associated with excessive vitamin A intake and provided with guidance on the maximum recommended dosages during pregnancy. By combining these skills, strategic approaches, shared responsibilities, effective communication, and seamless care coordination, healthcare professionals can deliver patient-centered care, optimize outcomes, enhance patient safety, and improve overall interprofessional team performance when managing vitamin A toxicity. Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Go to: References 1. Hunt JR. Teratogenicity of high vitamin A intake. N Engl J Med. 1996 May 02;334(18):1197. 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Vitamin A toxicity secondary to excessive intake of yellow-green vegetables, liver and laver. J Hepatol. 1999 Jul;31(1):142-8. [PubMed: 10424294] 32. Wiegand UW, Chou RC. Pharmacokinetics of acitretin and etretinate. J Am Acad Dermatol. 1998 Aug;39(2 Pt 3):S25-33. [PubMed: 9703120] Disclosure:Jazmine Olson declares no relevant financial relationships with ineligible companies. Disclosure:Muhammad Atif Ameer declares no relevant financial relationships with ineligible companies. Disclosure:Amandeep Goyal declares no relevant financial relationships with ineligible companies. Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology Histopathology Toxicokinetics History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Deterrence and Patient Education Pearls and Other Issues Enhancing Healthcare Team Outcomes Review Questions References Copyright © 2025, StatPearls Publishing LLC. 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StatPearls. 2025 Jan High provitamin A carotenoid serum concentrations, elevated retinyl esters, and saturated retinol-binding protein in Zambian preschool children are consistent with the presence of high liver vitamin A stores.[Am J Clin Nutr. 2015]High provitamin A carotenoid serum concentrations, elevated retinyl esters, and saturated retinol-binding protein in Zambian preschool children are consistent with the presence of high liver vitamin A stores.Mondloch S, Gannon BM, Davis CR, Chileshe J, Kaliwile C, Masi C, Rios-Avila L, Gregory JF 3rd, Tanumihardjo SA. Am J Clin Nutr. 2015 Aug; 102(2):497-504. Epub 2015 Jul 15. Review Vitamin A Transporters in Visual Function: A Mini Review on Membrane Receptors for Dietary Vitamin A Uptake, Storage, and Transport to the Eye.[Nutrients. 2021]Review Vitamin A Transporters in Visual Function: A Mini Review on Membrane Receptors for Dietary Vitamin A Uptake, Storage, and Transport to the Eye.Martin Ask N, Leung M, Radhakrishnan R, Lobo GP. Nutrients. 2021 Nov 9; 13(11). Epub 2021 Nov 9. Overview of current knowledge of metabolism of vitamin A and carotenoids.[J Natl Cancer Inst. 1984]Overview of current knowledge of metabolism of vitamin A and carotenoids.Goodman DS. J Natl Cancer Inst. 1984 Dec; 73(6):1375-9. Review The importance of vitamin A in nutrition.[Curr Pharm Des. 2000]Review The importance of vitamin A in nutrition.Dawson MI. Curr Pharm Des. 2000 Feb; 6(3):311-25. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Vitamin A Toxicity - StatPearlsVitamin A Toxicity - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close Olson JM, Ameer MA, Goyal A. 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3281	https://zhuanlan.zhihu.com/p/339804760	一起习题：鸡兔同笼题库大全，附解析过程及答案 - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册一起习题：鸡兔同笼题库大全，附解析过程及答案切换模式一起习题：鸡兔同笼题库大全，附解析过程及答案题霸公众号：一起习题；高效整理错题辅导神器 30 人赞同了该文章鸡兔同笼问题是我国古代著名趣题之一。通过学习解鸡兔同笼问题，可以提高孩子的分析问题、解决问题的能力。例如：现有一笼子，里面有鸡和兔子若干只，数一数，共有头14个，腿38条，聪明的小朋友，你能算出鸡和兔子各有多少只吗？列举法就是将各种情况一一地罗列出来，再针对要求，筛选符合题意的答案。根据上面的表格，我们可以看出，鸡为9只，兔子为5只。我们在列表的时候不要按顺序列，否则做题的速度会很慢；比如说列完鸡为0只，兔子为14只，发现腿的数量56条，和实际38条相差较大，那么下一个你可以跳过鸡的数量为2只这种情况，直接列鸡的数量为3只，这样做速度会快一些！今天为大家收集整理了上百道鸡兔同笼题库，包含有详细的解析方法和答案。希望对孩子的学习有一些帮助。下面就随便列举几道： 1、笼子的鸡和兔，共10个头，34只脚，其中鸡有__只，兔有____只．答案与解析：假设全部是鸡，则兔的只数：（34-10×2）÷（4-2） =（34-20）÷2， =14÷2， =7（只）；鸡的只数：10-7=3（只）；答：其中鸡有3只，兔有7只．故答案为：3，7． 2、鸡与兔共100只，鸡的腿数比兔的腿数少28条，问鸡与兔各有几只？答案与解析：解：4100＝400，400-0＝400假设都是兔子，一共有400只兔子的脚，那么鸡的脚为0只，鸡的脚比兔子的脚少400只。 400-28＝372实际鸡的脚数比兔子的脚数只少28只，相差372只，这是为什么？ 4+2＝6这是因为只要将一只兔子换成一只鸡，兔子的总脚数就会减少4只（从400只变为396只）；鸡的总脚数就会增加2只（从0只到2只），它们的相差数就会少4+2＝6只（也就是原来的相差数是400-0＝400，现在的相差数为396-2＝394，相差数少了400-394＝6）。 372÷6＝62表示鸡的只数，也就是说因为假设中的100只兔子中有62只改为了鸡，所以脚的相差数从400改为28，一共改了372只 100-62＝38表示兔的只数。以上只是题库中随机抽取的2道题。如需获取所有鸡兔同笼题库。鸡兔同笼问题可以让孩子在不同的生活场境中解决实际问题，既可使学生在实践中得到锻炼，又能增强学生数学应用的意识与能力。发布于 2020-12-26 15:07 家禽养殖技术鸡兔同笼问题赞同 301 条评论分享喜欢收藏申请转载写下你的评论... 1 条评论默认最新百川 2023-06-11 回复喜欢关于作者题霸公众号：一起习题；高效整理错题辅导神器回答 147文章 39关注者 5,379 关注发私信推荐阅读鸡兔同笼解题方法之假设法，附详细讲解步骤及分析，易懂！ =========================== 孩子在学习数学的时候，不仅仅要学会用一种方法解决一道题，还要学会用多种方法解决一道题。这样的锻炼对训练孩子思维的灵活性具有重要作用，学会了一题多解，孩子在接下来的学习中才会举… 木子李的李【跟娃学数学】四年级复杂鸡兔同笼、例扣类、多或少类的鸡兔同笼问题，这一篇就够啦 ======================================= 二宝妈跟娃学【视频】5.小学奥数：鸡兔同笼问题 ================= 鸡兔同笼问题是小学奥数中很常见的一个考点。解决鸡兔同笼问题，我们首先要知道下面的常识：每只鸡有1个头，2只脚；每只兔子有1个头，4只脚。其中的一种解决方案如下：假设让所有的鸡和兔… 数学王·许老师小学数学《鸡兔同笼》教案 ============ 一、教学目标【知识与技能】理解掌握并会运用列表法、假设法解决“鸡兔同笼”问题。【过程与方法】经历自主探索解决问题的过程，体验解决问题的策略的多样化;在解决问题的过程中，提高逻… 中公教师考试想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
3282	https://www.umsl.edu/~defreeseca/intalg/ch2extra/triangle.htm	Geometry Problems Involving the Angles of a Triangle Geometry Problems Involving the Angles of a Triangle First, let’s remember some triangle trivia: An isosceles triangle has two equal sides and two equal angles. A right triangle has one 90 o angle. If a triangle is neither isosceles nor right, we will call it a generic triangle. A piece of trivia that is true for all triangles: The sum of the three angles of any triangle is equal to 180 degrees. Now let’s try a problem. The largest angle of a triangle is 5 times as big as the smallest angle. The third angle is 12 degrees more than the smallest angle. Find the measure of all three angles. What are we asked to find in this problem? Look at the last sentence of the problem, and you will see that we want to find the measures of all three angles of the triangle. To define the variables in this problem, we will start by drawing a picture of a generic triangle: Since the problem tells us a lot about the largest angle and the middle angle but not much about the smallest angle, we’ll call the smallest angle x degrees. The largest angle is 5 times the smallest angle, so the largest angle is 5x degrees. The middle angle is 12 degrees more than the smallest angle, so we’ll call the middle angle x +12 degrees. We’ll write all of this information in our picture: Our equation will come from the fact that the sum of the three angles of any triangle must be 180 degrees. This gives us: (5x) + (x + 12) + (x) = 180 Now we’ll solve the equation: 5x + x + 12 + x = 180 7x + 12 = 180 7x = 168 x = 24 So the smallest angle measures 24 degrees. Have we finished the problem? Not quite. We still need to find the measures of the other two angles. We’ll refer back to our triangle to see that the largest angle measures 5x=5(24)=120 o, and the middle angle measures x+12=24+12=36 o. So the three angles measure 24 o, 36 o, and 120 o. Before we leave the problem, let’s check our answer with the words of the problem: wordscheck The largest angle of a triangle is 5 times as big as the smallest angle.Is 120 five times as big as 24? Yes! The third angle is 12 degrees more than the smallest angle.Is 36 equal to 12 more than 24? Yes! This is a triangle: do the angles all add up to 180 degrees?24 + 36 + 120 = 180 Our answer checks. We’re done!
3283	https://www.amboss.com/us/snippet/Coagulation_cascade	Coagulation cascade AMBOSS Register / Log in Menu Coagulation cascade A series of reactions involving a system of three pathways that results in the coagulation of blood. Consists of the extrinsic pathway (tissue factor pathway), intrinsic pathway (contact activation pathway), and final common pathway. © 2025 AMBOSS \| Terms and Conditions \| Privacy \| Legal Notice × ×
3284	https://askfilo.com/user-question-answers-smart-solutions/a-non-conducting-disc-of-radius-having-charge-uniformly-3331333330393031	Question asked by Filo student A non-conducting disc of radius R having charge q uniformly distributed over it is rotated with angular speed ω, then the magnetic moment associated with it will be (1) qR2ω (2) 21qR2ω (3) 41qR2ω (4) 81qR2ω Views: 5,723 students Updated on: Apr 8, 2025 Text SolutionText solutionverified iconVerified Concepts: Magnetic moment, Rotational motion, Charge distribution Explanation: To find the magnetic moment associated with a rotating charged disc, we can use the formula for the magnetic moment (μ) of a rotating charge distribution. The magnetic moment is given by the formula: μ=21qR2ω where q is the total charge, R is the radius of the disc, and ω is the angular speed. This formula arises from considering the current produced by the rotating charge and the area of the disc. Therefore, the correct answer is option (2) 21qR2ω. Step by Step Solution: Step 1 Identify the total charge (q) and the radius (R) of the disc. Step 2 Use the formula for the magnetic moment of a rotating charge distribution: μ=21qR2ω. Step 3 Substitute the values into the formula to find the magnetic moment. Final Answer: The magnetic moment associated with the rotating disc is 21qR2ω. Students who ask this question also asked Views: 5,973 Topic: Smart Solutions View solution Views: 5,325 Topic: Smart Solutions View solution Views: 5,274 Topic: Smart Solutions View solution Views: 5,467 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| A non-conducting disc of radius R having charge q uniformly distributed over it is rotated with angular speed ω, then the magnetic moment associated with it will be (1) qR2ω (2) 21qR2ω (3) 41qR2ω (4) 81qR2ω \| \| Updated On \| Apr 8, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 12 Passed \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
3285	https://www.youtube.com/watch?v=fCelOk2M96M	Absolute Value - Understand In 10 Minutes TabletClass Math 880000 subscribers 765 likes Description 59422 views Posted: 9 Oct 2019 Absolute Value is the distance a number is from zero. In algebra we study many topics about absolute value to include the definition of absolute value, absolute value equations, absolute value inequalities and absolute value graphs. In this video I will go over the basics of absolute value which will help you later in more advance absolute value topics. If you need extra help with math check out my math learning program below: TabletClass Math Academy Math Courses: 38 comments Transcript: okay let's understand absolute value and I'm gonna try to keep this video right around ten minutes I might go a little bit shorter or maybe a little bit longer but if you stick with me you're going to you know I'm really upgrade your knowledge of absolute value or maybe you pretty much know a lot about absolute value just want a review so if you're new to my channel I am a math teacher teach middle school high school even some college so I love teaching hopefully you'll consider subscribing if you like what you're gonna see in this particular video and one other thing too if you really do like my teaching style I'll leave the link in the description below to my full complete learning program various levels of math and I go really above and beyond what I do on YouTube but without being said let's understand absolute value so much confusion here so here is let's just do a real basic thing here let's say absolute value of negative 3 now let's just start from the very very beginning what like how do you even recognize like absolute value well absolute value recognize it has these little bars around a number or an algebraic expression okay so this would be the absolute value negative 3 this is the absolute value of x plus 1 ok these long little bars when you see these bars not something like this not grouping symbols or parentheses or these type of deals specifically these a little straight up and down vertical bars ok that's absolute value in mathematics now just as an aside real quick there is something later on it looks like this if you're studying a little bit more advanced algebra they have long vertical bars so longer this is finding the determinant of a matrix so you don't want to confuse this with absolute value but with that being said let's just stick with the absolute value real basic situation like this now what I hear a lot from students when they're first learning mathematics you know the basic algebra level or middle school level they'll say oh the absolute value of a number is the positive version of the number right so the absolute value negative three most you would would know that that is a positive three so what's the absolute value of positive three most you out there would say oh that's just positive three so you would be correct however your understanding of the definition oftentimes is the absolute value means the positive version of any number whether it's positive or negative and that's not correct okay we really need to understand what the definition of absolute value is because this is going to come into play little and later different topics of absolute value so as you progress in your math education you're gonna be learning about absolute value working with just the basic operation of it and then you're gonna be learning absolute value equations I'm gonna make various videos I probably already have them out my youtube video certainly solve a lot of these problems but you can study absolute value equations absolute value inequalities absolute value graphs and there's even some other things in there as well so it's a big topic in algebra let's get back to the what we want to accomplish in this video is just understanding the essence of absolute value so what is absolute value now these two problems with three I answer correctly okay the absolute value negative 3 is a positive 3 an absolute valuable positive 3 is a positive 3 so what is absolute value well absolute value is actually technically by definition the distance a number is from zero on a number line so this is a number line I'm relying you've probably seen this we really want to refer to this as the real number line but that's kind of an aside topic so here we have zero in the middle we go out one two three and ice we continue to the right numbers get larger right and this way as we go left and we go.we numbers get smaller so we have negative one negative two negative three so what I'm saying here is absolute value negative 3 and the absolute value of 3 or both positive three so what does that mean well what it means is that the distance the distance negative three is from zero is in fact three units away okay now this is a real big point that I'm gonna make distance distance is always measured in positive units okay we kind of there's another word out there a little technical says it's like displacement or distance positive units so if you had like a kind of roar and you say hey you know or a tape measure said go ahead and tell me from zero to negative three how you know how far away is that and you might say well that's three units but not your tape measure or your ruler is not going to be saying negative units now likewise three to zero is also three units away okay so these are positive values again distance is always positive so the definition of absolute value is the distance a number is from zero on the number line okay you really really need to understand that so let's take a look at a couple of basic type of problems let's say I have some like this two times the absolute value negative three minus the absolute value of seven so let's do a couple little arithmetic problems so if you want to pause the video and try this yourself you know go right ahead I'm gonna go ahead and solve it now so what you want to do here is let's just go ahead and take care of these absolute value functions right now so this would be two times the absolute value negative three is what positive three so let's write that like this minus the absolute value of a positive seven is simply seven so we can just write this problem now in this manner so we went from the absolute values into actual just numbers okay so now we just finish this up two times three is what's six six minus seven hopefully you did not say one okay the answer is negative one so if you're studying absolute value at this point you're likely have already studied positive and negative numbers okay I'm the the rules of those very very important a lot of students must you know mess this up so this is a real basic illustration of working with absolute values and and uh you know basic order of operation type of problems now let's take a look at something a little bit different okay I'm gonna write a real basic absolute value equation so the absolute value of x is equal to three okay so what is X what obviously X is a variable right so it's a variable in algebra but what really is a variable okay there's a little pop quiz a variable is what what's some number right so any variable X Y when I just represent some number so I want to take a look at this absolute value equation let's do it kind of like differently let's say the absolute value of some number the absolute value of some number is equal to three yeah now hopefully you can figure this out I said hey Sarah you know what's I got I got some number here and it's absolute value is equal to three what is that number okay some number is its absolute value is equal to three well you would say oh yeah well you're probably plugging in three and you would be correct but also I could find the absolute value of negative three and that would also be correct okay so the main thing that you want to remember about absolute value equations there's always two solutions always two solutions to absolute value of absolute value equations okay oh there's always going to be two solutions now real quick I'm gonna just write something out 2x plus one apps of active play X plus one is equal to nine and absolute value of 2x plus 1 is less than nine okay so I have an absolute value equation and an absolute value inequality I'm not going to get into these but I want to make a point as we wrap up this video okay both of these look pretty similar however the way we solve them is completely different so another very common mistake is students they'll see an absolute value equation and they'll start doing what you do for an inequality they'll confuse this with this and that is what happens when you just try to remember math like in a row two way okay I know if you're familiar with that word wrote that just means like trying to learn math by just a bunch of flashcards not truly learning it but just trying to remember it like okay I'm just gonna remember all this stuff today I hate it gonna remember it by looking at it a million times and just hope I kind of remember for a test that's not the way to approach mathematics because you're gonna end up confusing things now how do I know this because I've been teaching for several years and you know you just see the mistakes students make so everything I'm telling here comes from years of experience so as you get into these topics you may not have started to study this yet but you will you're going to you gonna you know you got to be careful here that you don't confuse these two but I can go on and on we can even talk about the absolute value graph I'll just tell you very briefly the graph of an absolute value function is a V looking thing you can happen on up V a down type of V situation and that's even for another video but absolute value very critical in algebra in mathematics so you definitely need to know it and hopefully this video helped you out again if you like my teaching style please consider subscribing to my channel right now literally I have hundreds of YouTube videos that will help you out for sure and I'm posting like all the time every day so if you'd like my my work and my teaching style then hopefully you consider follow me again if you're looking for something extremely comprehensive like my best you know learning material then you want to check out the link below and I have all different type of courses various math so you can check that off you're like and if you liked the video I would certainly appreciate a thumbs up and leave me some feedback how are things going in your math class are you struggling in a particular area what's working for you do you have any other questions I try to read as many comments as I can I do get a lot of comments on my videos which I'm grateful for but um you know lets me know how I'm doing and also gives me ideas for future videos but with that being said I hope I've kept it pretty close to 10 minutes I wish you all the best in your math class thanks for watching and have a great day
3286	https://www.healthcare.uiowa.edu/familymedicine/fpinfo/OB/OB2020/Severe%20HTN%20Pregnancy%2010.28.2020.pdf	UIHC Guidelines Emergent Therapy for Acute-Onset, Severe Hypertension during Pregnancy and the Postpartum Period Note: this guideline does not address the comprehensive treatments necessary for patients with preeclampsia and eclampsia In these guidelines, Provider is defined as a clinical staff member who is authorized to prescribe medications in compliance with Iowa Code and UIHC Bylaws, Rules and Regulations.  Diagnosis Acute-onset, severe systolic hypertension, severe diastolic hypertension, or both can occur during the prenatal, intrapartum, or postpartum periods. Pregnant women or women in the postpartum period with acute-onset, severe systolic hypertension, severe diastolic hypertension, or both require urgent antihypertensive therapy (within 30–60 minutes). Hypertensive Emergency Definition Severe hypertension occurs when one or both of these occur for 15 minutes or longer: Value Systolic blood pressure (SBP) ≥ 160 mm Hg Diastolic blood pressure (DBP) ≥ 110 mm Hg Blood pressure should be accurately measured using standard techniques (including correct cuff size). • In clinic, patient should be: o Comfortably seated or semi-reclining, legs uncrossed, back and arm supported o Relaxed and not talking o Expose the upper arm fully by removing any constricting clothing. Do not place the BP cuff over clothing. o Middle of blood pressure cuff on upper arm should be level with the heart o NOTE: Patient should NOT be repositioned to reclining or side because it will provide a falsely low reading • In the hospital, patient should be: o As listed above “in clinic” o Acceptable alternative if clinically indicated: Left lateral recumbent position with arm at heart level Once the criteria for severe hypertension is met: 1. Notify the LIP 2. If diagnosed in the clinic setting, transfer patient to the hospital for treatment 3. Antepartum: a. Begin continuous electronic fetal monitoring b. Prepare for emergent delivery c. Must be either in Emergency Department or Labor & Delivery or Mother Baby Care Unit (6JPP or 3JPP) prior to antihypertensive treatment i. Exception: If severe hypertension is identified in clinic nifedipine may be administered while awaiting transport to the hospital if a Provider specializing Obstetrics is present. Postpartum: a. Must be on either in Emergency Department or Labor & Delivery or Mother Baby Care Unit (6JPP or 3JPP) for antihypertensive treatment b. Exception: If severe hypertension is identified in clinic in a patient who is 6 or less weeks post-partum, nifedipine may be administered while awaiting transport to the hospital if a Provider specializing Obstetrics is present.  Treatment Requirements for acute onset, severe hypertension treatment in antepartum and postpartum patients: • Antepartum patients must either be in the ED or L&D or Mother Baby Care Unit (6JPP or 3JPP) o Exception: If severe hypertension is identified in clinic, nifedipine may be administered while awaiting transport to the hospital if a Provider specializing in Obstetrics is present. • Postpartum patients must be either in ED or L&D or Mother Baby (6JPP or 3JPP) o Exception: If severe hypertension is identified in clinic in a patient who is 6 or less weeks post-partum, nifedipine may be administered while awaiting transport to the hospital if a Provider specializing Obstetrics is present. • The Provider must complete a timely beside evaluation and be readily available on the OB units during the administration for all doses. • In order to not delay treatment, the medication may be administered prior to completion of the bedside evaluation. Goals Achieve a range of: Value Systolic blood pressure (SBP) 140 – 150 mm Hg Diastolic blood pressure (DBP) 90 - 100 mm Hg First-Line Regimen: • Labetalol IV (refer to Appendix A) o Dose: 20 mg IV bolus; if not effective in 10-15 min, give 40 mg IV; if not effective in 10-15 min, give 80 mg IV; if there is not an adequate response, try hydralazine IV o Hold for maternal HR < 60 BPM o Adverse events to be aware of: maternal bradycardia, neonatal bradycardia Second-Line Regimens: • Hydralazine IV (refer to Appendix B) o Dose: 5-10 mg IV bolus; if not effective in 20 min, give 10 mg IV o May be considered first-line for patients with asthma, heart disease, or congestive heart failure o Adverse event to be aware of: maternal hypotension • Nifedipine Immediate Release Orally (refer to Appendix C) May be considered first-line if no IV access or contraindications to labetalol or hydralazine o Dose: 10 mg orally; if not effective in 20 minutes, give 20 mg orally; if not effective in 20 minutes, give 20 mg orally o Note: Administer orally and do not puncture or otherwise administer sublingually o May be considered first-line if no IV access or contraindications to labetalol or hydralazine o Adverse event to be aware of: maternal hypotension, maternal tachycardia Alternative Regimens: • Labetalol Orally o Dose: 200 mg orally; if not effective in 30 minutes, give 200 mg orally o Adverse events to be aware of: maternal bradycardia, neonatal bradycardia • If elevations in blood pressure are refractory to the agents listed above (if given in successive, appropriate doses), emergent consultation with a maternal-fetal medicine specialist, anesthesiologist, or intensivist is recommended as patient should be evaluated for treatment with an infusion (with invasive arterial blood pressure monitoring): Nicardipine Esmolol Nitroprusside: reserved for extreme emergencies and use for the shortest amount of time possible because of concerns about cyanide and thiocyanate toxicity in mom and fetus/newborn and increased intracranial pressure with potential worsening of cerebral edema in the mom Nitroglycerin: avoid around the time of delivery due to the effect on uterine tone and risk for postpartum hemorrhage  Monitoring Close maternal and continuous fetal monitoring are advised during the treatment of acute-onset, severe hypertension. Post-drug administration monitoring is dependent on the pharmacokinetics of the drug being used. Drug Onset (min) Peak (min) Duration of effect (hours) Monitoring Labetolol IV 5 30 3 - 8 Once goal BP is met, check BP, HR, RR every 10 min for 1 hour, then every 15 min for 1 hour, then every 30 min for 1 hour, then every hour for 4 hours post-dose Hydralazine IV 5 - 15 10 - 60 3 - 8 Once goal BP is met, check BP, HR, RR every 10 min for 1 hour, then every 15 min for 1 hour, then every 30 min for 1 hour, then every hour for 4 hours post-dose Nifedipine Oral, Immediate Release 10 30 4 - 8 Once goal BP is met, check BP, HR, RR every 10 min for 1 hour, then every 15 min for 1 hour, then every 30 min for 1 hour, then every hour for 4 hours post-dose During/after administration of antihypertensives, notify LIP immediately if: • Systolic blood pressure drops below 110 mm Hg • Diastolic blood pressure drops below 70 mm Hg • Maternal HR > 120 beats per minute or HR < 60 beats per minute • Changes in fetal heart rate including minimal variability, absent variability, variable decelerations, or late decelerations. • Signs/symptoms of hypotension (e.g., flushing, headache, nausea, dizziness, shortness of breath). If hypotension occurs: o Position patient on left side with head of bead flat o Elevate patient’s legs o Apply oxygen at 10 L via mask if there is evidence of fetal compromise o Notify Anesthesia After initial stabilization, the team should monitor blood pressure closely and institute maintenance therapy as needed. Ensure patients are carefully monitored well into the postpartum period. References: American College of Obstetricians and Gynecologists. Emergent therapy for acute-onset, severe hypertension during pregnancy and the postpartum period. Committee opinion no. 692. 2017 April. Retrieved from American College of Obstetricians and Gynecologists. Severe hypertension ion Pregnancy. Algorithms for labetalol, hydralazine, oral nifedipine. 2017 April. Retrieved from American College of Obstetricians and Gynecologists. Emergent Therapy for Acute-Onset, Severe Hypertension During Pregnancy and the Postpartum Period. 2019 February. Retrieved from %20for%20Acute-Onset%20Severe%20Hypertension%20During%20Pregnancy%20and%20the%20Postpartum%20Period Bernstein PS, Martin JN Jr, Barton JR, et al. Consensus bundle on severe hypertension during pregnancy and the postpartum period. J Obstet Gynecol Neonatal Nurs. 2017 Jul 1. pii: S0884-2175(17)30284-8. doi: 10.1016/j.jogn.2017.05.003. [Epub ahead of print]. Podovei M, Bateman BT. The consensus bundle on hypertension in pregnancy and the anesthesiologist: doing all the right things for all patients all of the time. Anesth Analg. 2017;125:383-5. Last revision: October 2020 Approved at P&T: October 27, 2020 Appendix A Labetalol IV for Severe Hypertension during Pregnancy and the Postpartum Period Continuous fetal monitoring is advised during treatment of acute onset severe hypertension Administer: labetalol 20 mg intravenously (IV) over 2 min Repeat BP in 10 minutes Obtain baseline blood pressure (BP), heart rate (HR), and respiratory rate (RR) At 10 - 15 min post-labetalol administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? Administer: labetalol 40 mg IV over 2 min BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Repeat BP in 10 minutes Administer: Labetalol 80 mg IV over 2 min Yes No At 10 - 15 min post-labetalol administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? Yes No BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Repeat BP in 10 minutes At 10 - 15 min post-labetalol administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? No BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Move to hydralazine IV. If contraindicated, move to another antihypertensive. Yes Appendix B Hydralazine IV for Severe Hypertension during Pregnancy and the Postpartum Period Continuous fetal monitoring is advised during treatment of acute onset severe hypertension Administer: hydralazine 5-10 mg IV over 2 min Repeat BP in 20 minutes Obtain baseline blood pressure (BP), heart rate (HR), and respiratory rate (RR) At 20 min post-hydralazine administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? Administer: hydralazine 10 mg IV over 2 min BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Repeat BP in 20 minutes Yes No At 20 min post-hydralazine administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? Yes No BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Move to another antihypertensive Appendix C Immediate-Release Nifedipine for Severe Hypertension during Pregnancy and the Postpartum Period Continuous fetal monitoring is advised during treatment of acute onset severe hypertension Administer: Immediate-release (IR) nifedipine 10 mg orally Repeat BP in 20 minutes Obtain baseline blood pressure (BP) and heart rate (HR) At 20 min post-nifedipine administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? Administer: IR nifedipine 20 mg orally BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Repeat BP in 20 minutes Yes No At 20 min post-nifedipine administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? Yes No BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Administer: IR nifedipine 20 mg orally Repeat BP in 20 minutes At 20 min post-nifedipine administration: Is SBP ≥160 mm Hg and/or DBP ≥ 110 mm Hg? BP is within goal range: BP, HR, RR: every 10 min x 1 hour, then every 15 min x 1 hour, then every 30 min x 1 hour, then every hour x 4 hours, then per LIP Move to another antihypertensive No Yes Approved at P&T: 10/27/2020
3287	https://en.wikipedia.org/wiki/Sum_coloring	Jump to content Sum coloring Magyar Edit links From Wikipedia, the free encyclopedia In graph theory, a sum coloring of a graph is a labeling of its vertices by positive integers, with no two adjacent vertices having equal labels, that minimizes the sum of the labels. The minimum sum that can be achieved is called the chromatic sum of the graph. Chromatic sums and sum coloring were introduced by Supowit in 1987 using non-graph-theoretic terminology, and first studied in graph theoretic terms by Ewa Kubicka (independently of Supowit) in her 1989 doctoral thesis. Obtaining the chromatic sum may require using more distinct labels than the chromatic number of the graph, and even when the chromatic number of a graph is bounded, the number of distinct labels needed to obtain the optimal chromatic sum may be arbitrarily large. Computing the chromatic sum is NP-hard. However it may be computed in linear time for trees and pseudotrees, and in polynomial time for outerplanar graphs. There is a constant-factor approximation algorithm for interval graphs and for bipartite graphs. The interval graph case remains NP-hard. It is the case arising in Supowit's original application in VLSI design, and also has applications in scheduling. References [edit] ^ Małafiejski, Michał (2004), "Sum coloring of graphs", in Kubale, Marek (ed.), Graph Colorings, Contemporary Mathematics, vol. 352, Providence, RI: American Mathematical Society, pp. 55–65, doi:10.1090/conm/352/06372, ISBN 9780821834589, MR 2076989 ^ Supowit, K. J. (1987), "Finding a maximum planar subset of a set of nets in a channel", IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems, 6 (1): 93–94, doi:10.1109/tcad.1987.1270250, S2CID 14949711 ^ Kubicka, Ewa Maria (1989), The chromatic sum and efficient tree algorithms, Ph.D. thesis, Western Michigan University, MR 2637573 ^ Erdős, Paul; Kubicka, Ewa; Schwenk, Allen J. (1990), "Graphs that require many colors to achieve their chromatic sum", Proceedings of the Twentieth Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1989), Congressus Numerantium, 71: 17–28, MR 1041612 ^ Kubicka, Ewa; Schwenk, Allen J. (1989), "An introduction to chromatic sums", Proceedings of the 17th ACM Computer Science Conference (CSC '89), New York, NY, USA: ACM, pp. 39–45, doi:10.1145/75427.75430, ISBN 978-0-89791-299-0, S2CID 28544302 ^ Jump up to: a b Kubicka, Ewa M. (2005), "Polynomial algorithm for finding chromatic sum for unicyclic and outerplanar graphs", Ars Combinatoria, 76: 193–201, MR 2152758 ^ Jump up to: a b Halldórsson, Magnús M.; Kortsarz, Guy; Shachnai, Hadas (2001), "Minimizing average completion of dedicated tasks and interval graphs", Approximation, randomization, and combinatorial optimization (Berkeley, CA, 2001), Lecture Notes in Computer Science, vol. 2129, Berlin: Springer, pp. 114–126, doi:10.1007/3-540-44666-4_15, ISBN 978-3-540-42470-3, MR 1910356 ^ Giaro, Krzysztof; Janczewski, Robert; Kubale, Marek; Małafiejski, Michał (2002), "A 27/26-approximation algorithm for the chromatic sum coloring of bipartite graphs", Approximation algorithms for combinatorial optimization, Lecture Notes in Computer Science, vol. 2462, Berlin: Springer, pp. 135–145, doi:10.1007/3-540-45753-4_13, ISBN 978-3-540-44186-1, MR 2091822 ^ Marx, Dániel (2005), "A short proof of the NP-completeness of minimum sum interval coloring", Operations Research Letters, 33 (4): 382–384, CiteSeerX 10.1.1.5.2707, doi:10.1016/j.orl.2004.07.006, MR 2127409 Retrieved from " Categories: Graph coloring NP-complete problems
3288	https://www.khanacademy.org/math/ka-math-class-11/x0419e5b3b578592a:relations-and-functions-ncert-new/x0419e5b3b578592a:functions/v/greatest-integer-function	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
3289	https://collegedunia.com/exams/preparation-of-benzene-chemistry-articleid-7120	Preparation of Benzene: Aromatic Compounds in Organic Chemistry Select Goal & City Select Goal Search for Colleges, Exams, Courses and More.. Write a Review Get Upto ₹300 Explore Explore More Study Abroad Get upto 50% discount on Visa Fees Top Universities & Colleges Abroad Exams Top Courses Exams Read College Reviews News Admission Alerts 2025 Education Loan Institute (Counselling, Coaching and More) Ask a Question College Predictor Test Series Practice Questions Course Finder Scholarship All Courses B.Tech MBA M.Tech MBBS B.Com B.Sc B.Sc (Nursing) BA BBA BCA Course Finder All Courses B.Tech MBA M.Tech MBBS B.Com B.Sc B.Sc (Nursing) BA BBA BCA B.Arch B.Ed B.Pharm B.Sc (Agriculture) BAMS LLB LLM M.Pharm M.Sc MCA Bachelor of Physiotherapy B.Des M.Planning B.Planning Agriculture Arts Commerce Computer Applications Design Engineering Law Management Medical Paramedical Pharmacy Science Architecture Aviation Dental Education Hotel Management Mass Communications Veterinary Sciences Animation Exams Class 11 Chemistry Hydrocarbons Preparation of Benzene: Aromatic Compounds in Organic Chemistry Namrata DasExams Prep Master Exams Prep Master Benzene is prepared by cyclic polymerisation of ethyne or by decarboxylation of aromatic acids or reduction of phenol. Benzene, an organic compound discovered by Michael Faraday, is an important cyclic hydrocarbon with the chemical formula C 6 H 6. The benzene molecule is composed of six carbon atoms arranged in a ring and each carbon atom is bonded to only one hydrogen atom. It contains only carbon and hydrogen atoms, thus it is categorized as a hydrocarbon. NCERT Solutions For:Class 11 Chemistry Chapter 13 Hydrocarbons Table of Content Properties and Uses of Benzene Preparation of Benzene from Ethyne Preparation of Benzene from Aromatic Acids Preparation of Benzene by Reduction of Phenol Preparation of Benzene from Sulphonic Acids Things to Remember Previous Years’ Questions Sample Questions Key Terms: Benzene, Preparation of Benzene, Aromatic Compounds, Ethyne, Aromatic Acids, Sulphonic Acids Properties and Uses of Benzene [Click Here for Sample Questions] Benzene is a colorless chemical with a distinctive odor and is also categorized as an aromatic compound. It occurs naturally in natural gas, crude oil, or coal. Benzene is highly volatile, flammable, and toxic. It is also a known carcinogenic compound. It is primarily used as an intermediate agent in making and synthesizing other industrial chemicals, such as ethylbenzene, cumene and cyclohexane which are further used to manufacture plastics, resins, nylon and synthetic fibres. Natural sources of benzene are volcanoes, forest fires, crude oil and gasoline. However, it can also be prepared through various laboratory techniques discussed below. Read More: \| Related Topics \| \| Polymerisation \| Lindlar Catayst \| Conformational isomers \| \| Cycloalkanes \| Conformation \| Huckel Rule of Aromaticity \| \| Unsaturated Hydrocarbons \| Hydrocarbons \| Alkanes \| Preparation of Benzene from Ethyne [Click Here for Previous Years Questions] Once ethyne passes through a red-hot iron tube at 873 K, It undergoes cyclic polymerization to form benzene. Three molecules of ethyne polymerize to form benzene. Preparation of Benzene from Aromatic Acids [Click Here for Sample Questions] In this process, benzene is obtained from benzoic Acid (Sodium benzoate). The sodium salt of benzoic acid when heated with Soda-lime (NaOH) causes decarboxylation. Through Decarboxylation, carbon dioxide is removed to produce benzene and sodium carbonate as the by-product. Preparation of Benzene by Reduction of Phenol [Click Here for Previous Years Questions] When vapors of phenol pass over heated zinc dust, phenol reacts with zinc dust to convert to a phenoxide ion and a proton, which accepts an electron from Zn forming an H radical. This results in the formation of ZnO and the phenoxide ion that was formed converts itself into benzene. Preparation of Benzene from Sulphonic Acids [Click Here for Sample Questions] Sulphonic Acids undergo hydrolysis to form benzene. Benzene is obtained when the compound benzene sulphonic acid (C 6 H 5- SO 3 H) passes through a high amount of heated steam. C6H5-SO3H + H2O → C6H6+ H 2 SO 4 Read More: \| Chapter-Related Concepts \| \| Alkenes \| Ozonolysis \| Alkynes \| \| Aromaticity \| Electrophilic substitution reactions \| Nitration \| \| Degree of Unsaturation Formula \| Alliphatic Hydrocarbons \| Conformation \| Things to Remember Benzene is prepared from ethyne by the cyclic polymerization process. Here, Ethyne is passed through a red-hot iron tube at 873 K. The ethyne molecule then undergoes cyclic polymerization to form benzene. The benzene molecule (C 6 H 6) is composed of six carbon atoms arranged in a ring and each carbon atom is bonded to only one hydrogen atom. Benzene occurs naturally in the environment but it can also be prepared by various scientific methods. Alkynes such as ethyne, aromatic acids such as Benzoic Acid, and Sulphonic Acids are used to prepare Benzene. It can also be prepared by the reduction of phenol. Previous Years’ Questions Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by… [NEET 2010] In pyrrole the electron density is maximum on… [NEET 2016] In preparation of alkene from alcohol using Al 2 O 3, which is effective factor?... [NEET 2001] In Friedel-Craft's synthesis of toluene, the reactants in addition to anhydrous… [NEET 2000] Dihedral angle in staggered form of ethane is… [NEET 2000] Consider the nitration of benzene using mixed conc… [NEET 2016] Acetylenic hydrogens are acidic because… [NEET 1989] A compound is treated with NaNH 2 to give sodium salt. Identify the compound… [NEET 1993] Which one of these, is not compatible with arenes?... [NEET 1998] Correct order of stability is… [NEET 2000] Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?... [NEET 1980] Sample Questions Ques. Why is benzene extraordinarily stable though it contains three double bonds? (4 marks) Ans.The six carbon atoms in benzene are sp² hybridized. The two sp² hybrid orbitals of each carbon atom overlap with the sp² hybrid orbitals of adjacent carbon atoms to form six C–C sigma bonds in the hexagonal plane. The remaining sp² hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six C - H sigma bonds. Each carbon atom is now left with one unhybridized p orbital. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three π bonds by the lateral overlap of three π bonds by overlap of p orbitals of C1 –C2, C3 – C4, C5 – C6 or C2 – C3, C4 – C5, C6 – C1 respectively. The six π electrons are thus delocalized and can move freely about the six carbon nuclei. The delocalized π electron cloud is attracted more strongly by the nuclei of the carbon atoms than the electron cloud localized between two carbon atoms. Therefore, the presence of delocalized π electrons in benzene makes it more stable. Ques. Explain the carcinogenic and toxic natures of benzene. (3 marks) Ans.Benzene and polynuclear hydrocarbons consisting of more than two benzene rings combined together are toxic and are said to contain cancer-producing (carcinogenic) properties. Such polynuclear hydrocarbons are formed due to the incomplete combustion of organic materials such as tobacco, coal, and petroleum. They find their way into the human body and go through various biochemical reactions and finally damage DNA and cause cancer. Some of the carcinogenic hydrocarbons are 1,2-Benzanthracene, 3-Methylcholanthrene, 1,2-Benzpyrene, 1,2,5,6-Dibenzanthracene, and 9,10-Dimethyl-1,2-benzanthracene. Ques. State four characteristics of Benzene. (3 marks) Ans.The characteristic of Benzene are as follows: Volatile and flammable nature carcinogenic and toxic colorless characteristic odor Ques. Shed some light on the aromaticity of Benzene. (3 marks) Ans. A compound is considered an aromatic compound if it fulfills the Hückel Rule which states that the compound should have the following characteristics: It is planar The π electrons in the ring are completely delocalized. (4n + 2) π electrons in the ring are present where n is an integer (n= 0, 1, 2, .....) Ques. Who discovered benzene? (2 marks) Ans. The English scientist, Michael Faraday discovered benzene in 1825. However, August Kekulé in 1865 proposed that benzene has a cyclic arrangement of six carbon atoms with alternate single and double bonds and one hydrogen atom attached to each carbon atom. Ques. What are the various methods of preparing benzene in the laboratory? (4 marks) Ans. The various methods of preparing benzene in the laboratory are as follows: Cyclic polymerization of ethyne Decarboxylation of benzoic acid reduction of phenol hydrolysis of sulphonic acid Ques. State three applications of benzene. (3 marks) Ans. Benzene has a wide range of uses. Some of its uses are as follows: It is used in the printing industry as many products used by them contain benzene. Some of the products that have benzene are used for cleaning printing equipment to increase the duration of their life cycle. Benzene is used in the production of dyes, nylon fibres, medicines, pigments, insecticides, adhesives, soaps and cleansers etc. It is used in manufacturing different chemicals such as ethylbenzene, cumene and cyclohexane. Ques. At what temperature, the polymerisation of ethyne occurs? (1 mark) Ans. The polymerisation of ethyne occurs at 873 K. Ques. What are the physical properties of benzene? (2 marks) Ans. Benzene is a non-polar molecule and is colorless with a gasoline-like aroma. It is immiscible with water but is readily miscible with organic solvents. Also, it burns with a sooty flame. Ques. How is phenol reduced to benzene? (2 marks) Ans.When vapors of phenol pass over heated zinc dust, phenol reacts with zinc dust to convert to a phenoxide ion and a proton, which accepts an electron from Zn forming an H radical. This results in the formation of ZnO and the phenoxide ion that was formed, converts itself into Benzene. 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Reason (R): Chlorine has higher oxidation potential than H 2 _2 2O. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). Assertion (A) is true, but Reason (R) is false. Assertion (A) is false, but Reason (R) is true. View Solution 2.Among the following outermost electronic configurations of transition metals, which one shows the highest oxidation state? 3 d 6 4 s 2 3d^6\,4s^2 3 d 6 4 s 2 3 d 5 4 s 1 3d^5\,4s^1 3 d 5 4 s 1 3 d 5 4 s 2 3d^5\,4s^2 3 d 5 4 s 2 3 d 3 4 s 2 3d^3\,4s^2 3 d 3 4 s 2 View Solution 3.Calculate E cell E_{\text{cell}} E cell of a galvanic cell in which the following reaction takes place at 25°C: Zn(s)+Pb 2+(0.02 M)⟶Zn 2+(0.1 M)+Pb(s)[Given:E Zn 2+/Zn 0=−0.76 V,E Pb 2+/Pb 0=−0.13 V,log⁡2=0.3010,log⁡4=0.6021,log⁡5=0.6990] \text{Zn(s)} + \text{Pb}^{2+}(0.02M) \longrightarrow \text{Zn}^{2+}(0.1M) + \text{Pb(s)} \quad \text{[Given: } E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76V, \; E^0_{\text{Pb}^{2+}/\text{Pb}} = -0.13V, \; \log 2 = 0.3010, \; \log 4 = 0.6021, \; \log 5 = 0.6990 \text{]} Zn(s)+Pb 2+(0.02 M)⟶Zn 2+(0.1 M)+Pb(s)[Given:E Zn 2+/Zn 0=−0.76 V,E Pb 2+/Pb 0=−0.13 V,lo g 2=0.3010,lo g 4=0.6021,lo g 5=0.6990] View Solution 4.Reactant ‘A’ underwent a decomposition reaction. The concentration of ‘A’ was measured periodically and recorded in the table given below: \| Time (Hours) \| [A] (M) \| --- \| \| 0 \| 0.40 \| \| 1 \| 0.20 \| \| 2 \| 0.10 \| \| 3 \| 0.05 \| Based on the above data, predict the order of the reaction and write the expression for the rate law. View Solution 5.0.3 g of acetic acid (Molar mass = 60 g mol−1^{-1}−1) dissolved in 30 g of benzene shows a depression in freezing point equal to 0.45°C. Calculate the percentage association of acid if it forms a dimer in the solution. (Given: K f K_f K f for benzene = 5.12 K kg mol−1^{-1}−1) View Solution 6.Among the following, which will have inversion of configuration on reaction with aqueous alkali and why? (i) 1-chloropropane (ii) 2-chloro-2-methylpropane View Solution View All Similar Chemistry Concepts Huckel's Rule: Applications & Exceptions Conformation: Definition, Types, Difference, Representation Aliphatic Hydrocarbons: Alkanes, Alkenes & Alkynes, Examples Benzene: Characteristics, Structure, Properties & Uses Butane (C4H10): Structure, Properties, Uses, Production Hypochlorous Acid: Properties, Preparation, Reactions & Uses Conformational Isomers: Definition, Types, Difference and Representation Markovnikov’s Rule: Statement, Mechanism & Examples Hydrocarbons: Types, Properties & Uses NCERT Solutions for class 11 Chemistry Chapter 13 :Hydrocarbons download guide free counselling CBSE CLASS XII Previous Year Papers Chemistry Chemistry Preparation Guides Physical and Chemical Classification of Matter Law of Multiple Proportions: Definition, Examples & Limitations Discovery of Electron: Characteristics, JJ Thomson experiment, Discovery of cathode rays and Solved Questions Standard Enthalpy of Formation: Explanation and Bond Dissociation Enthalpy Dalton’s Atomic Theory: Definition, Postulates, Limitations and Solved Questions Isomerism: Definition, Examples of Isomer compounds, Types, Sample Questions Atomic Radius: Definition, Types, Variation, and Periodic Trends of Atomic Radii Wave Nature of Electromagnetic Radiation: Relation Between Frequency and Energy Mole Fraction: Definition, Formula, Advantages, Properties Exothermic Reactions: Definition, Enthalpy Change & Examples Uncertainty in Measurement: Scientific Notation Bond Enthalpy: Definition, Average Bond Enthalpies, and Enthalpy of a Reaction Critical Temperature - Volume, Pressure and Point, Graphical Representation and Sample Questions Classification of Elements and Periodicity in Properties: Genesis & Types Dipole Moment: Definition, Formula, Bond Dipole Moment & Examples Precipitation Reaction: Examples, Properties and Uses Electronegativity: Explanation, Factors Affecting, and Table Pauli Exclusion Principle: Explanation, Formulation, Examples, and Applications Covalent Bond: Definition, Types, Polarization, Properties, Important Questions Ionic Radius: Definition, Electrical Charge, Trends and Solved Questions Chemistry NCERT Solutions Degree of Unsaturation Formula: Equation, Calculation, Solved Examples Effects of Acid Rain: Soil Acidification and Ocean Acidification Carbanions: Explanation, Occurrence, Properties and Examples Bond Dissociation Enthalpy: Definition, Effects and Difference Energies of Orbitals: Aufbau, Principle and Principle Quantum Number Valence Electrons: Definition, Characteristics & Formula Atomic Number and Mass number: Isotopes & Properties Bond Order: Nature, Important Equations and Importance Buffer Solutions: Preparation, Mechanism, Types, and Uses Photon Energy: Properties, Formula, and Kinetic Energy of Photons Comments No Comments To Show SUBSCRIBE TO OUR NEWS LETTER COLLEGE NOTIFICATIONS EXAM NOTIFICATIONS NEWS UPDATES SUBMIT TOP COLLEGES m.b.a b.tech/b.e mca bca m.tech ma ba TOP UNIVERSITIES engineering management medical law commerce science arts Top Exam cat gate jee-main neet xat clat mat study abroad canada usa uk uae australia germany sweden ireland new zealand hong kong singapore malaysia netherlands italy other links about collegedunia contact us advertising career terms & conditions Download the Collegedunia app on © 2025 Collegedunia Web Pvt. 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3290	https://www.vedantu.com/neet/difference-between-c3-c4-and-cam-pathway	Published Time: 2020-06-13, 12:26:18,+5:30 Difference Between C₃, C₄, and CAM Pathway: A Comprehensive Guide Sign In All Courses NCERT, book solutions, revision notes, sample papers & more Find courses by class Starting @ ₹1,350 Find courses by target Starting @ ₹1,350 Long Term Courses Full Year Courses Starting @ just Rs 9000 One-to-one LIVE classes Learn one-to-one with a teacher for a personalised experience Courses for Kids Courses for Kids Confidence-building & personalised learning courses for Class LKG-8 students English Superstar Age 4 - 8 Level based holistic English program Summer Camp For Lkg - Grade 10 Limited-time summer learning experience Spoken English Class 3 - 5 See your child speak fluently Learn Maths Class 1 - 5 Turn your child into a Math wizard Coding Classes Class 1 - 8 Learn to build apps and games, be future ready Free study material Get class-wise, author-wise, & board-wise free study material for exam preparation NCERT SolutionsCBSEJEE MainJEE AdvancedNEETQuestion and AnswersPopular Book Solutions Subject wise Concepts ICSE & State Boards Kids Concept Online TuitionCompetative Exams and Others Offline Centres Online Tuition Get class-wise, subject-wise, & location-wise online tuition for exam preparation Online Tuition By Class Online Tuition By Subject Online Tuition By Location More Know about our results, initiatives, resources, events, and much more Our results A celebration of all our success stories Child safety Creating a safe learning environment for every child Help India Learn Helps in learning for Children affected by the Pandemic WAVE Highly-interactive classroom that makes learning fun Vedantu Improvement Promise (VIP) We guarantee improvement in school and competitive exams Master talks Heartfelt and insightful conversations with super achievers Our initiatives Resources About us Know more about our passion to revolutionise online education Careers Check out the roles we're currently hiring for Our Culture Dive into Vedantu's Essence - Living by Values, Guided by Principles Become a teacher Apply now to join the team of passionate teachers Contact us Got questions? Please get in touch with us Vedantu Store NEET Difference Between Biology Difference Between C3 C4 And Cam Pathway C₃ vs C₄ Vs CAM: Key Distinctions in Photosynthesis Download PDF Courses Class 11 NEET Course (2023-25) Class 12 NEET Course (2023-24) NEET Repeater Course (2023-24) Class 8 NEET Foundation Course Class 9 NEET Foundation Course Class 10 NEET Foundation Course Exam Info Important Dates (Exam Dates) Eligibility Criteria Application Form Registration Form Correction Centres (Exam Centres) Admit Card Reservation Criteria Slot Booking Neet Syllabus Neet Physics Syllabus Neet Chemistry Syllabus Neet Biology Syllabus Weightage Pattern Study Material Important Questions Revision Notes Formulas Difference between Chapter specific previous year question and answer Question Papers Previous Year Paper Solutions Question Paper Analysis Mock Test Sample paper Answer Key Cut off Neet Colleges Government College Private College Results Percentage Vs Marks Rank List Counselling Marks and Rank Neet Question Answers Neet Physics Question Answers Neet Chemistry Question Answers Neet Biology Question Answers News Videos FAQs Latest Updates Predict Your Score : NEET Rank and College Predictor Understanding the Difference Between C₃, C₄, and CAM Pathway Photosynthesis is the fundamental process by which plants convert sunlight into energy. However, different plants have adapted unique carbon fixation pathways to optimize this process under varying environmental conditions. These pathways—C₃, C₄, and CAM—help plants minimise photorespiration, enhance water efficiency, and maximise growth. Understanding their differences is crucial for studying plant adaptations, especially in the face of climate change. What are the Differences Between C₃, C₄, and CAM Pathway C₃ Pathway C₄ Pathway CAM Pathway The first stable product is 3-phosphoglycerate (3-PGA), a three-carbon compound.The first stable product is oxaloacetate, a four-carbon compound.The first stable product is malic acid, stored as a four-carbon compound. Photosynthesis occurs only in mesophyll cells.Photosynthesis is divided between mesophyll and bundle sheath cells.Carbon fixation occurs at night, and photosynthesis takes place during the day in mesophyll cells. Photorespiration is high because RuBisCO binds with oxygen.Photorespiration is low due to spatial separation of carbon fixation and the Calvin cycle.Photorespiration is very low due to temporal separation of processes. Water-use efficiency is low.Water-use efficiency is moderate.Water-use efficiency is high as stomata open at night to minimize water loss. The CO₂ compensation point is 50–100 ppm.The CO₂ compensation point is 10–20 ppm.The CO₂ compensation point is very low. Optimal temperature for photosynthesis is 15–25°C.Optimal temperature is 30–40°C.Optimal temperature is 35–45°C. Efficiency in hot, dry conditions is low.Efficiency in hot, dry conditions is high.Efficiency in hot, dry conditions is very high. Stomata remain open during the day.Stomata are partially open during the day.Stomata are open at night and closed during the day. Common in plants like rice, wheat, soybean, and barley.Found in maize, sugarcane, sorghum, and millet.Seen in cactus, pineapple, and agave. Essential Study Materials for NEET UG Success Study Materials for NEET 2025 S. No Study Materials Links 1NEET Revision Notes 2NEET Important Questions 3NEET Question Papers 4NEET Sample Papers 5NEET Syllabus 6NEET Mock Test Series 7NEET Formulas 8NEET Difference Between 9NEET Most Important Chapters 10NEET Question and Answers 11NEET UG Preparation Books 12NEET Chapter-Wise Weightage FAQs on C₃ vs C₄ Vs CAM: Key Distinctions in Photosynthesis What is the main difference between C₃, C₄, and CAM pathways? The primary difference lies in how these pathways fix carbon dioxide. C₃ plants directly fix CO₂ into a three-carbon compound (3-PGA), C₄ plants fix CO₂ into a four-carbon compound (oxaloacetate) to reduce photorespiration, and CAM plants store CO₂ at night and use it during the day to conserve water. Why do C₃ plants experience higher photorespiration? C₃ plants have no adaptations to prevent RuBisCO from binding with oxygen instead of CO₂, leading to higher photorespiration, especially in hot and dry conditions. Which plants use the C₃ pathway? C₃ plants include major crops like wheat, rice, barley, and soybean, as well as most trees and shrubs. How does the C₄ pathway help plants survive in hot climates? C₄ plants, such as maize and sugarcane, separate carbon fixation and the Calvin cycle into different cells (mesophyll and bundle sheath cells), reducing photorespiration and increasing water-use efficiency in high temperatures. What is the distinguishing difference between C₃, C₄, and CAM pathways? The key distinguishing difference is how each pathway fixes carbon. C₃ plants produce a three-carbon compound directly in the Calvin cycle, C₄ plants create a four-carbon intermediate to reduce photorespiration, and CAM plants fix carbon dioxide at night to conserve water in dry environments. How can we distinguish the difference between C₃ and C₄ plants based on photorespiration? C₃ plants have high photorespiration due to RuBisCO binding with oxygen, while C₄ plants have low photorespiration because carbon fixation and the Calvin cycle occur in separate cells, reducing oxygen interference. What is the distinguishing factor that makes CAM plants more water-efficient than C₃ and C₄ plants? CAM plants open their stomata at night to take in CO₂, storing it as malic acid, which is used for photosynthesis during the day. This prevents water loss, making them more water-efficient than C₃ and C₄ plants. How do we distinguish between C₃ and C₄ plants in terms of their first stable product? C₃ plants form a three-carbon compound (3-PGA), while C₄ plants produce a four-carbon compound (oxaloacetate) as the first stable product. What is the distinguishing difference in stomatal activity among C₃, C₄, and CAM plants? C₃ plants: Stomata are open during the day. C₄ plants: Stomata are partially open during the day. CAM plants: Stomata are open at night and closed during the day to prevent water loss. How do the distinguishing differences in temperature preferences affect C₃, C₄, and CAM plants? C₃ plants thrive in cool to moderate temperatures (15–25°C), C₄ plants prefer warm temperatures (30–40°C), and CAM plants excel in hot and arid conditions (35–45°C). 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3291	https://www.nagwa.com/en/videos/254105656138/	Question Video: Determining the Equation of a Straight Line in Two-Intercept Form Mathematics • First Year of Secondary School Which of the following represents the equation of a straight line in two-intercept form? [A] (𝑥/𝑎) + (𝑦/𝑏) = 1 [B] (𝑥/𝑎) + (𝑦/𝑏) = 𝑐 [C] 𝑦 = 𝑚𝑥 + 𝑐 [D] 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 [E] 𝑎𝑥 + 𝑏𝑦 = 1 Video Transcript Which of the following represents the equation of a straight line in two-intercept form? Is it (A) 𝑥 over 𝑎 plus 𝑦 over 𝑏 equals one? (B) 𝑥 over 𝑎 plus 𝑦 over 𝑏 equals 𝑐. (C) 𝑦 equals 𝑚𝑥 plus 𝑐. (D) 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero. Or (E) 𝑎𝑥 plus 𝑏𝑦 equals one. We recall that there are many ways of writing the equation of a straight line. For example, option (D) 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero is known as the general form of the equation of a straight line. As we are looking for two-intercept form, we can rule out this option. Option (C) is written in slope–intercept form. This is also sometimes written as 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope or gradient of the line and 𝑏, or 𝑐 in this example, is the 𝑦-intercept. We can therefore also rule out this option. Let’s consider the straight line that intercepts the 𝑥- and 𝑦-axis as shown. If this line intercepts the 𝑥-axis at 𝑎 and the 𝑦-axis at 𝑏, we know that the points of intersection have coordinates 𝑎, zero and zero, 𝑏. We then define the two-intercept form as follows. The two-intercept form of the equation of the straight line that intercepts the 𝑥-axis at 𝑎, zero and 𝑦-axis at zero, 𝑏 is 𝑥 over 𝑎 plus 𝑦 over 𝑏 equals one. The correct answer is option (A). We can derive this answer as follows. We begin by recalling the slope or gradient of a line is equal to 𝑦 sub two minus 𝑦 sub one over 𝑥 sub two minus 𝑥 sub one, where 𝑥 sub one, 𝑦 sub one and 𝑥 sub two, 𝑦 sub two are two points that lie on the line. This is also sometimes referred to as the change in 𝑦 over the change in 𝑥 or the rise over the run. From our diagram, we see that 𝑚 is equal to 𝑏 minus zero over zero minus 𝑎. This simplifies to 𝑏 over negative 𝑎, which can also be written as negative 𝑏 over 𝑎. Next, we recall the point–slope form for the equation of a straight line. This states that 𝑦 minus 𝑦 sub one is equal to 𝑚 multiplied by 𝑥 minus 𝑥 sub one. Substituting in the values of 𝑚, 𝑥 sub one, and 𝑦 sub one, we have 𝑦 minus zero is equal to negative 𝑏 over 𝑎 multiplied by 𝑥 minus 𝑎. Distributing the parentheses or expanding the brackets on the right-hand side, we have 𝑦 is equal to negative 𝑏 over 𝑎 𝑥 plus 𝑏. We can then divide through by 𝑏, giving us 𝑦 over 𝑏 is equal to negative 𝑥 over 𝑎 plus one. Adding 𝑥 over 𝑎 to both sides gives us the required equation 𝑥 over 𝑎 plus 𝑦 over 𝑏 is equal to one. This is the two-intercept form of the equation of a straight line which intercepts the 𝑥-axis at 𝑎, zero and the 𝑦-axis at zero, 𝑏. Lesson Menu Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company Content Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
3292	https://www.wikihow.com/Convert-Celsius-to-Kelvin	Skip to Content Courses Hot Guides Tech Help Pro Expert Videos About wikiHow Pro Upgrade QUIZZES All Quizzes Hot Love Quizzes Personality Quizzes Trivia Quizzes Taylor Swift Quizzes EDIT Edit this Article EXPLORE Tech Help ProAbout UsRandom ArticleQuizzes Request a New ArticleCommunity DashboardTrendingForums Arts and EntertainmentArtworkBooksMovies Computers and ElectronicsComputersPhone SkillsTechnology Hacks HealthMen's HealthMental HealthWomen's Health RelationshipsDatingLoveRelationship Issues Hobbies and CraftsCraftsDrawingGames Education & CommunicationCommunication SkillsPersonal DevelopmentStudying Personal Care and StyleFashionHair CarePersonal Hygiene Quizzes Love Quizzes Personality Quizzes Fun Games ForumsArts and EntertainmentFinance and BusinessHome and Garden Relationship QuizzesCars & Other VehiclesFood and EntertainingPersonal Care and Style Sports and FitnessComputers and ElectronicsHealthPets and Animals TravelEducation & CommunicationHobbies and CraftsPhilosophy and Religion Work WorldFamily LifeHolidays and TraditionsRelationships LOG IN ### Log in wikiHow Account No account yet? Create an account RANDOM Home Random Browse Articles TrendingNew Quizzes & Games All QuizzesHot Love Quizzes Personality Quizzes Fun Games Dating Simulator Learn Something New Forums Courses Happiness Hub Explore More Support wikiHow About wikiHow Log in / Sign up Terms of Use wikiHow is where trusted research and expert knowledge come together. Learn why people trust wikiHow Categories Education and Communications Studying Mathematics Conversion Aids How to Convert Celsius to Kelvin Download Article Reviewed by Anne Schmidt Last Updated: February 24, 2025 References Download Article Convert Celsius to Kelvin \| Understanding Kelvin \| Convert Fahrenheit to Kelvin (Optional) \| Video \| Expert Interview \| Expert Q&A \| Things You'll Need X This article was reviewed by Anne Schmidt. Anne Schmidt is a Chemistry Instructor in Wisconsin. Anne has been teaching high school chemistry for over 20 years and is passionate about providing accessible and educational chemistry content. She has over 9,000 subscribers to her educational chemistry YouTube channel. She has presented at the American Association of Chemistry Teachers (AATC) and was an Adjunct General Chemistry Instructor at Northeast Wisconsin Technical College. Anne was published in the Journal of Chemical Education as a Co-Author, has an article in ChemEdX, and has presented twice and was published with the AACT. Anne has a BS in Chemistry from the University of Wisconsin, Oshkosh, and an MA in Secondary Education and Teaching from Viterbo University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been viewed 71,413 times. Converting Celsius to Kelvin, luckily, is easy. The Kelvin temperature scale is an absolute thermodynamic scale used commonly in the physical sciences. It uses zero as absolute zero, unlike Fahrenheit or Celsius, where there are negative numbers. In order to read temperatures over a wide variety of disciplines, you must learn to convert Fahrenheit to Celsius and Celsius to Kelvin. With practice, you may be able to do this in your head, rather than using a calculator. Steps Part 1 Part 1 of 3: Convert Celsius to Kelvin Download Article 1 Write down the temperature in degrees Celsius. The conversion to Kelvin is remarkably easy-- all it takes is some simple addition. The following three examples will be used throughout the section: 30°C 0°C 100°C 2. 2 Add 273.15 to your Celsius temperature. For example, 30 plus 273.15 is 303.15. This is all you need to do to make your conversion. Just add 273.15 and you're done. X Expert Source Dr. Vivek PrakashBiomedical Engineer Expert Interview X Research source Advertisement 3. 3 Replace the °C with a simple K. Do not put a degree sign in, as it would be incorrect. Once you've made your calculation, simply add a K and call it a day. X Research source Advertisement Part 2 Part 2 of 3: Understanding Kelvin Download Article 1 Never use the word "degree" when talking about Kelvin. The correct way to say "292 K" is simply "two-hundred and ninety-two Kelvin." Kelvin is known as the "absolute temperature," and does not use degrees. X Research source Each unit is actually called "a Kelvin." It doesn't get two degrees hotter, but two Kelvins hotter. 2. 2 Know that 0 Kelvin is the theoretical point where an ideal gas has no volume. Absolute zero, or 0 K, is the point where molecules theoretically stop moving. This is the "perfect" cold. While absolute zero is not possible to create, scientists have come close. The point of the Kelvin scale is to make this sort of calculation -- absolute zero -- easier to work with. X Research source 3. 3 Use Kelvin for scientific research. Because 0 K is the lowest temperature possible in the universe, Kelvin has no negative numbers. This makes it much, much easier to work with mathematically. You can better compare temperatures, find differences or averages, and note relationships when you don't have to work with positive and negative temperatures. X Research source Kelvin is also used to measure color temperature -- thus the 3000K, 6000K, etc. settings on cameras and professional lighting kits and bulbs. X Research source 4. 4 Dive into the technical definitions of Kelvin for advanced classes. The Kelvin is defined as the thermodynamic temperature of the triple point of water. Accordingly, the number 273.15 is frequently used to convert temperatures to Kelvin. Don't worry if this explanation doesn't make much sense -- it is mostly for high-level chemists and physicists. X Research source Advertisement Part 3 Part 3 of 3: Convert Fahrenheit to Kelvin (Optional) Download Article 1 Convert Fahrenheit to Celsius before converting to Kelvin. You cannot go directly from Fahrenheit to Kelvin without converting to Celsius first. The Celsius to Kelvin conversion is much easier than the Fahrenheit to Celsius conversion. You will almost definitely need a calculator for this. 86°F 2. 2 Subtract 32 from your Fahrenheit reading. For example, 86 minus 32 is 54. As an interesting aside, know that this is because the freezing point of Celsius is 32 less than Fahrenheit. X Research source Multiply the number you just calculated by , or 0.5555.. For example, 54 times 0.5555 is 30. In some formulas, you may also be asked to divide by 1.8, which is the same calculation as multiplying by 0.5555. This finishes the conversion to Celsius. 3 Add 273.15 to finish converting to Kelvin. Once you've subtracted 32 and multiplied by , you're in Celsius. Now just add 273.15 to get to Kelvin and call it a day. X Research source Advertisement Expert Q&A Search Add New Question Question How do I convert Celsius to Fahrenheit? Dr. Vivek Prakash Biomedical Engineer Dr. Vivek Prakash is a Biomedical Engineer based in Oxford, Ohio. He works as a Research Associate at Miami University, working on cell-based hormone replacement therapy. He received his PhD from the Department of Biosciences and Bioengineering at the Indian Institute of Technology. Dr. Vivek Prakash Biomedical Engineer Expert Answer To convert Celsius to Fahrenheit, multiply the Celsius temperature by 9/5 and then add 32 to the result. For example, if you have a Celsius temperature of 20 degrees, the conversion would be 20×95+32=6820×59+32=68. Therefore, 20 degrees Celsius is equivalent to 68 degrees Fahrenheit. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 2 Helpful 1 Question How do I convert Fahrenheit to Celsius? Donagan Top Answerer Subtract 32 from the Fahrenheit number. The multiply by 5, and divide by 9. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 1 Helpful 5 Question How do I convert absolute zero on the Kelvin scale to the temperature on Fahrenheit scale? Donagan Top Answerer °F = (1.8)(K - 273) + 32. When K is zero, the formula becomes (1.8)(-273) + 32 = -491.4 + 32 = -459.4°F. (That answer is an approximation, because -273 is an approximation. The actual number is -273.15, making the actual answer -459.67°F.) Thanks! We're glad this was helpful. Thank you for your feedback. 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Things You'll Need Calculator Pen Paper You Might Also Like How to Convert Between Fahrenheit, Celsius, and KelvinHow to Convert Kelvin to Fahrenheit or Celsius How to Convert Fahrenheit to Kelvin°C to °F Conversion: A Complete GuideHow to Estimate Celsius Temperatures in FahrenheitHow to Convert Units EasilyHow to Determine Relative Humidity: Formulas & MoreConversion Factor for Grams to Kilograms (and Kilograms to Grams)How to Convert Within Metric MeasurementsHow to Convert Units With PrefixesHow to Convert Metric Weight to PoundsHow to Understand the Metric SystemHow to Change Newtons to KilogramsHow to Convert Kilojoules to Calories Advertisement Expert Interview Thanks for reading our article! If you’d like to learn more about converting measures, check out our in-depth interview with Dr. Vivek Prakash. References ↑ Dr. Vivek Prakash. Biomedical Engineer. Expert Interview ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ More References (1) ↑ About This Article Reviewed by: Anne Schmidt Chemistry Instructor This article was reviewed by Anne Schmidt. Anne Schmidt is a Chemistry Instructor in Wisconsin. Anne has been teaching high school chemistry for over 20 years and is passionate about providing accessible and educational chemistry content. She has over 9,000 subscribers to her educational chemistry YouTube channel. She has presented at the American Association of Chemistry Teachers (AATC) and was an Adjunct General Chemistry Instructor at Northeast Wisconsin Technical College. Anne was published in the Journal of Chemical Education as a Co-Author, has an article in ChemEdX, and has presented twice and was published with the AACT. Anne has a BS in Chemistry from the University of Wisconsin, Oshkosh, and an MA in Secondary Education and Teaching from Viterbo University. 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3293	https://www.quora.com/How-can-we-prove-a-right-isosceles-triangle	How can we prove a right isosceles triangle? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Geometry Proofs (mathematics) Isosceles Right Triangle Two Triangles Geometrical Theorems Right Angles Congruent Triangles Geometry Formulas Triangles 5 How can we prove a right isosceles triangle? All related (32) Sort Recommended Vinayak Mandrawadker Works at Retirement · Author has 121 answers and 107.3K answer views ·6y Isosceles triangle means 2 sides will be equal.It is right angle triangle base and altitude are equal. In a triangle ABC , AC is hypotenuse, AB is altitude and BC is base. AB is equal to BC. Angle ABC is 90 degree. Angle BAC and angle BCA are 45 degrees. Hence triangle ABC is a right isosceles triangle. Upvote · Sponsored by Amazon Web Services (AWS) Reliability you can trust with AWS Databases. Your databases should be secure, reliable, and built for performance. Explore how to build powerful apps. Sign Up 99 30 Related questions More answers below How do I identify whether an isosceles triangle is right angled isosceles triangle? Can a right angled triangle be isosceles? How do we prove an isosceles triangle? How can you prove that all right triangles are not isosceles? How do you prove that the base angles of an isosceles triangle is equal? Bikki Chhantyal I like Tan c/Sin c people · Author has 292 answers and 996.9K answer views ·8y Related How do you determine the formula of an isosceles right triangle? Consider an isosceles triangle of two equal sides a and an unequal side b. From Heron’s formula. Derivation of Heron's / Hero's Formula for Area of Triangle A =√s(s−a)(s−b)(s−c)s(s−a)(s−b)(s−c) s = a+a+b 2=2 a+b 2 a+a+b 2=2 a+b 2 For isosceles triangle A =√s(s−a)(s−b)(s−a)s(s−a)(s−b)(s−a) A =(s−a)√s(s−b)(s−a)s(s−b) A =(2 a+b 2−a)√(2 a+b 2)(2 a+b 2−b)(2 a+b 2−a)(2 a+b 2)(2 a+b 2−b) A = b 2√(2 a+b 2)2 a−b 2 b 2(2 a+b 2)2 a−b 2 A = b 4√4 a 2−b 2 b 4 4 a 2−b 2 Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 2 Dave Williamson MathNerd/MS Ed/age 65/99th %ile GRE/seeks free tuition Univ adv.study.math study · Author has 1.7K answers and 12.3M answer views ·8y Related How do you determine the formula of an isosceles right triangle? Are you asking how to figure out the lengths of the sides and the measures of the angles of an isosceles right triangle? A right triangle always has a 90° angle. All three angles must add up to 180°, so the other two angles must add up to 90°. In an isosceles triangle, the other two angles are equal, so 45° + 45° = 90° In order to figure out the lengths of the three sides, you must know one of the lengths. If you know the length of one of the shorter sides, then you know the length of the other short side, because, in an isosceles triangle, the two short sides are always equal to each other. Multipl Continue Reading Are you asking how to figure out the lengths of the sides and the measures of the angles of an isosceles right triangle? A right triangle always has a 90° angle. All three angles must add up to 180°, so the other two angles must add up to 90°. In an isosceles triangle, the other two angles are equal, so 45° + 45° = 90° In order to figure out the lengths of the three sides, you must know one of the lengths. If you know the length of one of the shorter sides, then you know the length of the other short side, because, in an isosceles triangle, the two short sides are always equal to each other. Multiply the length of one side by √2 to get the length of the hypotenuse. If you only know the length of the longest side, the hypotenuse, then divide that length by √2 to get the length of the other two sides. Do you want the formula for the perimeter of that triangle? Add the lengths of the three sides P = a + b + c Do you want the formula for the area of that triangle? Multiply the height by the base, then divide by 2. A = h⋅b 2 h⋅b 2 If you need something else, please add a comment below, or revise the details of your question. The more information that you give when you ask a question, the easier it is for someone to reply. Upvote · David Smith BSc (Hons) in Mathematics&Computer Science, University of Bristol (Graduated 1986) · Author has 3.6K answers and 4M answer views ·6y Related How can you prove that all right triangles are not isosceles? I suspect you have deliberately phrased the question so as to allow ambiguity. It could mean either How can you prove that some right triangles are not isosceles? or How can you prove that no right triangles are isosceles? These two questions leave much less room for ambiguity. The first is easy to prove; just draw a 3,4,5 3,4,5 triangle. The second cannot be proved because it is false; draw a 1,1,√2 1,1,2 triangle. Upvote · 9 2 Promoted by HP HP Tech Takes Tech Enthusiast \| Insights, Tips & Guidance ·Updated Sep 18 What are the pros and cons of getting an expensive laser printer like HP versus a cheap but good quality inkjet/multi-function machine like Canon Pixma MX490? Choosing between an expensive laser printer and a more affordable inkjet or multi-function machine ultimately depends on your printing habits, the type of documents you produce, and your long-term budget. Laser printers and inkjet printers serve different purposes, and understanding their strengths and limitations can help you make a more informed decision. Laser printers are typically designed for speed and efficiency, making them ideal for office environments or users who print frequently. Inkjet and ink tank printers are slower but they offer excellent colour reproduction and are often more Continue Reading Choosing between an expensive laser printer and a more affordable inkjet or multi-function machine ultimately depends on your printing habits, the type of documents you produce, and your long-term budget. Laser printers and inkjet printers serve different purposes, and understanding their strengths and limitations can help you make a more informed decision. Laser printers are typically designed for speed and efficiency, making them ideal for office environments or users who print frequently. Inkjet and ink tank printers are slower but they offer excellent colour reproduction and are often more compact and affordable upfront. If your printing needs are frequent and primarily text-based, a laser printer such as the HP Color Laser 179fnw or the HP LaserJet M234sdw would be a strong choice. These models deliver fast print speeds and sharp text output. Toner cartridges used in these printers last significantly longer than ink cartridges, resulting in a lower cost per page over time. Although the initial investment is higher, the long-term savings and reliability could make them well-suited for your business use or heavy personal workloads. LaserJet Printers - Black & White or Color Document Printers Alternatively, if your printing volume is low to moderate and you value colour accuracy for photos or creative projects, an ink-based solution like the HP Smart Tank 7605 or HP Smart Tank 5105 may be more convenient. These printers offer refillable ink tanks that reduce running costs compared to traditional cartridge-based inkjets. They are also compact and versatile, supporting scanning and copying functions in addition to printing. However, they require occasional maintenance to prevent ink from drying out or clogging, and their print speed is generally slower than laser models. HP Smart Tank Printers – Refillable Ink Tank Printers So to break it down, if you prioritise speed, durability, and cost-efficiency for high-volume printing, a laser printer from HP’s 200 or 3000 series is the better investment. On the other hand, if your needs include occasional printing with a focus on colour and photo output, HP’s Smart Tank series provides a more economical and flexible alternative. The right choice will depend on how often you print and what kind of documents you’re producing. Check out the blog linked below to learn more about the different models and which is best for your printing needs, hope this helps! Inkjet vs LaserJet vs OfficeJet: HP Printer Buying Guide By Lizzie - HP Tech Expert Upvote · 999 113 99 36 9 4 Related questions More answers below Is it possible for a triangle to be right and isosceles? Can an isosceles triangle be a right triangle? Why or why not? Can a right angle be isosceles? Why or why not? In an isosceles triangle, how do you prove that the attitude is equal? How many lines of symmetry does the isosceles right triangle have? Assistant Bot · 1y To prove that a triangle is a right isosceles triangle, you need to demonstrate two key properties: Two sides are equal (isosceles property). One angle is 90 degrees (right triangle property). Here’s a step-by-step approach to proving a triangle is a right isosceles triangle: Given: A triangle A B C A B C. To Prove: A B=A C A B=A C (isosceles) and ∠A=90∘∠A=90∘ (right angle). Proof Steps: Show that two sides are equal: If you have the lengths of the sides, you can directly show that A B=A C A B=A C. Alternatively, if you have coordinates for points A A, B B, and C C, you can use the distance formula: AB=\sqrt AB=\sqrt Continue Reading To prove that a triangle is a right isosceles triangle, you need to demonstrate two key properties: Two sides are equal (isosceles property). One angle is 90 degrees (right triangle property). Here’s a step-by-step approach to proving a triangle is a right isosceles triangle: Given: A triangle A B C A B C. To Prove: A B=A C A B=A C (isosceles) and ∠A=90∘∠A=90∘ (right angle). Proof Steps: Show that two sides are equal: If you have the lengths of the sides, you can directly show that A B=A C A B=A C. Alternatively, if you have coordinates for points A A, B B, and C C, you can use the distance formula: A B=√(x B−x A)2+(y B−y A)2 A B=(x B−x A)2+(y B−y A)2 A C=√(x C−x A)2+(y C−y A)2 A C=(x C−x A)2+(y C−y A)2 If A B=A C A B=A C, then the triangle is isosceles. Show that one angle is 90 degrees: If you know the angles, you can check if ∠A=90∘∠A=90∘. If you have the slopes of the sides, you can show that the two sides are perpendicular. For example, if m 1 m 1 is the slope of A B A B and m 2 m 2 is the slope of A C A C, then A B A B and A C A C are perpendicular if: m 1⋅m 2=−1 m 1⋅m 2=−1 Alternatively, you can use the Pythagorean theorem. If you know the lengths of all three sides a a, b b, and c c (where c c is the length of the hypotenuse), verify: a 2+b 2=c 2 a 2+b 2=c 2 Conclusion: If both conditions are satisfied (i.e., A B=A C A B=A C and ∠A=90∘∠A=90∘), then triangle A B C A B C is a right isosceles triangle. Example: If A(0,0)A(0,0), B(1,0)B(1,0), and C(0,1)C(0,1): Calculate distances: A B=1 A B=1, A C=1 A C=1 (is the same). Angle A A is 90∘90∘ (since A B A B and A C A C are perpendicular). Thus, triangle A B C A B C is a right isosceles triangle. Upvote · Robert Nichols Author has 5K answers and 15.6M answer views ·9y Related Is an isosceles triangle always a right angled triangle? No, an isosceles triangle is not ALWAYS a right triangle. An Isosceles Triangle can be a right triangle, it is a 45–45–90 degree triangle. Continue Reading No, an isosceles triangle is not ALWAYS a right triangle. An Isosceles Triangle can be a right triangle, it is a 45–45–90 degree triangle. Upvote · 9 4 9 3 Steve Baker Senior Software Engineer (2013–present) · Author has 40.2K answers and 295.1M answer views ·8y Related How do you prove that in a right triangle c² = a² + b²? There are dozens and dozens of proofs out there. This is my favorite. Take four copies of your triangle and arrange them like this: The area of the yellow square is c 2 c 2 of course. The area of the entire figure is (a+b)2(a+b)2 The area of each of the triangles is a×b/2 a×b/2 So - we know that the area of the entire figure is (a+b)2(a+b)2 but it’s also the sum of the areas of the yellow square and the four blue triangles: c 2+4×a b/2 c 2+4×a b/2 OK - so c 2=(a+b)2−4 a b/2 c 2=(a+b)2−4 a b/2 Multiply out the (a+b)2(a+b)2 term: c 2=a 2+b 2+2 a b−4 a b/2 c 2=a 2+b 2+2 a b−4 a b/2 …simplify a bit and you get… c 2=a 2+b 2 c 2=a 2+b 2 Easy! Continue Reading There are dozens and dozens of proofs out there. This is my favorite. Take four copies of your triangle and arrange them like this: The area of the yellow square is c 2 c 2 of course. The area of the entire figure is (a+b)2(a+b)2 The area of each of the triangles is a×b/2 a×b/2 So - we know that the area of the entire figure is (a+b)2(a+b)2 but it’s also the sum of the areas of the yellow square and the four blue triangles: c 2+4×a b/2 c 2+4×a b/2 OK - so c 2=(a+b)2−4 a b/2 c 2=(a+b)2−4 a b/2 Multiply out the (a+b)2(a+b)2 term: c 2=a 2+b 2+2 a b−4 a b/2 c 2=a 2+b 2+2 a b−4 a b/2 …simplify a bit and you get… c 2=a 2+b 2 c 2=a 2+b 2 Easy! Upvote · 99 44 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 210 William Mccoy A former high school math teacher (Grades 9, 10, and 11). I have a Bachelor of Science Degree in Math Ed. (1980) from Miami U. in Oxford, O. · Author has 2.6K answers and 27M answer views ·6y Related How do you know the triangle is an isosceles right? I know that a triangle is an isosceles right triangle if: (1.) It has a right angle, i.e., an angle which measures 90 degrees, and ... (2.) The legs, i.e., the two sides of the triangle that are shorter than the hypotenuse, are congruent, i.e, they are equal in length or the two acute angles both measure 45 degrees. Upvote · Bob Hock Euclidean Geometry & Solid & Analytic Geometry. Descriptive Geometry. · Author has 290 answers and 453.3K answer views ·8y Related How do you determine the formula of an isosceles right triangle? There is no formula for an isosceles right triangle, however, the Pythagorean Theorem is used to find sides of a right triangle. The isosceles right triangle obviously contains a 90 Deg angle. The 2 sides that form the 90 Deg angle are equal to each other, so the remaining side, the hypotenuse, can be found using the Pythagorean Theorem. Upvote · Promoted by Unbiased Unbiased UK's leading platform connecting you with financial experts ·Updated Aug 13 Is paying for financial advice really worth it? If you’ve ever wondered whether financial advice is worth the money, you’re not alone. While financial advice does incur a fee, studies show it could make you tens of thousands better off in the long run. A report in 2019 found that people who got professional financial advice built up, on average, £47,000 more wealth over a decade compared to those who didn’t. And this wasn’t just high earners—it helped people from all walks of life. Why? Because advisers help you make better long-term decisions, such as growing your pension, avoiding unnecessary tax, and making the most of your savings and in Continue Reading If you’ve ever wondered whether financial advice is worth the money, you’re not alone. While financial advice does incur a fee, studies show it could make you tens of thousands better off in the long run. A report in 2019 found that people who got professional financial advice built up, on average, £47,000 more wealth over a decade compared to those who didn’t. And this wasn’t just high earners—it helped people from all walks of life. Why? Because advisers help you make better long-term decisions, such as growing your pension, avoiding unnecessary tax, and making the most of your savings and investments. 💡 Top tip: You don’t need to be rich to benefit. Even a one-off chat with an adviser can help you: Understand your money better Avoid costly mistakes Make smarter choices for your future AtUnbiased, we match you with one qualified financial adviser who will meet your unique needs. The first consultation is always free, so you can see if advice is right for you— so there’s no pressure and no cost. 👉 Want to see how much better off you could be? Try our adviser matching tool and take the first step toward a healthier financial future. Find your adviser today. Report by the International Longevity Centre UK and Royal London Upvote · 999 289 99 51 9 6 Haresh Sagar Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views ·May 31 Related How do you prove that an isosceles triangle has maximum area when its equal sides are perpendicular to each other? In the diagram we can see five isoceles triangles, ABC, ABD, ABE, ABF & ABG. The base of these triangles is AB and height related to AB is maximum when equal sides are perpendicular to each other. So if we have equal sides of triangle as r r the trig. formula for area is… K=1 2 r 2 s i n θ K=1 2 r 2 s i n θ With respect to θ θ, K′=1 2 r 2 c o s θ K′=1 2 r 2 c o s θ For maximum and minimum, we have to set it equal to zero… 1 2 r 2 c o s θ=0⟹1 2 r 2 c o s θ=0⟹ c o s θ=0⟹θ=90°c o s θ=0⟹θ=90° Proved!!! Continue Reading In the diagram we can see five isoceles triangles, ABC, ABD, ABE, ABF & ABG. The base of these triangles is AB and height related to AB is maximum when equal sides are perpendicular to each other. So if we have equal sides of triangle as r r the trig. formula for area is… K=1 2 r 2 s i n θ K=1 2 r 2 s i n θ With respect to θ θ, K′=1 2 r 2 c o s θ K′=1 2 r 2 c o s θ For maximum and minimum, we have to set it equal to zero… 1 2 r 2 c o s θ=0⟹1 2 r 2 c o s θ=0⟹ c o s θ=0⟹θ=90°c o s θ=0⟹θ=90° Proved!!! Upvote · 99 13 9 3 Dave Williamson MathNerd/MS Ed/age 65/99th %ile GRE/seeks free tuition Univ adv.study.math study · Author has 1.7K answers and 12.3M answer views ·9y Related Can an isosceles triangle be a right triangle? Why or why not? Ask yourself this question: What is an isosceles triangle? Go ahead. Pause here. Answer the question in your head. then scroll down. Yes, I agree, an isosceles triangle is a triangle that has two sides of equal length (and also has two angles of equal size). How do you create an isosceles right triangle? 1) Draw a triangle, no matter how ugly it is. _ (It does not matter if it looks like the sides are equal or not.) 2) Label the squarest of the angles of the triangle as being 90 degrees. _ (This is the right angle that makes your triangle a right triangle.) 3) How many degrees must the oth Continue Reading Ask yourself this question: What is an isosceles triangle? Go ahead. Pause here. Answer the question in your head. then scroll down. Yes, I agree, an isosceles triangle is a triangle that has two sides of equal length (and also has two angles of equal size). How do you create an isosceles right triangle? 1) Draw a triangle, no matter how ugly it is. _ (It does not matter if it looks like the sides are equal or not.) 2) Label the squarest of the angles of the triangle as being 90 degrees. _ (This is the right angle that makes your triangle a right triangle.) 3) How many degrees must the other two angles add up to? _ (Correct, the other two sum to 90, because the sum of three angles must be 180.) 4) Label each of the other angles as half of that remaining 90 degrees. _ (Yes, the three angles of an isosceles right triangle must be 90, 45 and 45. 5) The two sides adjacent to the right angle must be equal to each other in an isosceles triangle, so pick a length for both of these sides and label it, or use two lines to represent that they are of equal length. 6) Multiply the length of one of those sides by √2 2 to get the length of the hypotenuse. I bet you ended up with a good triangle. I hope your drawings look better than mine. Upvote · 9 7 Jesus Sanchez Software Developer at TrianCal (2015–present) · Author has 342 answers and 1.1M answer views ·9y Related Is an isosceles triangle always a right angled triangle? No, examples: Isosceles right angled triangle ( 90º : 45º : 45º ), link: TrianCal Isosceles non-right angled triangle ( 120º : 30º : 30º ), link: TrianCal Continue Reading No, examples: Isosceles right angled triangle ( 90º : 45º : 45º ), link: TrianCal Isosceles non-right angled triangle ( 120º : 30º : 30º ), link: TrianCal Upvote · 9 4 John K WilliamsSon Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.4M answer views ·6y Related How can you prove that all right triangles are not isosceles? it is very simple to prove that all right triangles are not isosceles. Simply construct one right triangle that is not isosceles. For example, with sides 3, 4, 5. It is a right triangle. It is not isosceles. Next, construct one right triangle that is isosceles. For example, with sides 1, 1, and √2. It is a right triangle. It is isosceles. Thus, SOME right triangles are isosceles. SOME right triangles are not isosceles. Therefore, not ALL right triangles are isosceles because SOME are them are. QED. Upvote · 9 2 9 4 Related questions How do I identify whether an isosceles triangle is right angled isosceles triangle? Can a right angled triangle be isosceles? How do we prove an isosceles triangle? How can you prove that all right triangles are not isosceles? How do you prove that the base angles of an isosceles triangle is equal? Is it possible for a triangle to be right and isosceles? Can an isosceles triangle be a right triangle? Why or why not? Can a right angle be isosceles? Why or why not? In an isosceles triangle, how do you prove that the attitude is equal? How many lines of symmetry does the isosceles right triangle have? How do you determine the formula of an isosceles right triangle? Is an isosceles triangle always a right angled triangle? How can you prove (by contradiction) that all right triangles are not isosceles? How can you prove that a right-angled isosceles triangle has its base as the altitude? What are the reasons and how do you prove an isosceles triangle? Related questions How do I identify whether an isosceles triangle is right angled isosceles triangle? Can a right angled triangle be isosceles? How do we prove an isosceles triangle? How can you prove that all right triangles are not isosceles? How do you prove that the base angles of an isosceles triangle is equal? Is it possible for a triangle to be right and isosceles? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. 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3294	https://www.brookings.edu/articles/its-not-just-cities-suburbs-and-exurbs-need-to-adopt-and-implement-climate-plans-too/	It’s not just cities—suburbs and exurbs need to adopt and implement climate plans too Subscribe to Infrastructure at Metro Commentary It’s not just cities—suburbs and exurbs need to adopt and implement climate plans too Megumi Tamura and Megumi Tamura Megumi Tamura Research Intern - Brookings Metro Joseph W. Kane Joe Kane Joseph W. Kane Fellow - Brookings Metro April 26, 2023 More On Climate Change Climate Policy Brookings Metro Brookings Initiative on Climate Research and Action Promoting equitable and effective climate action in every community As the country increasingly faces devastating floods, historic blizzards, and extreme droughts, there is an urgent need to take climate action. Adapting to these events is becoming a daily reality for many people and places, but avoiding the worst impacts in the long term depends on reducing greenhouse gas (GHG) emissions now. And while striving for GHG reductions globally and nationally matters, the ultimate responsibility rests in the hands of local leaders, who must plan and invest in sustainable infrastructure, land uses, and more. A recent Brookings analysis showed that nearly every major U.S. city has adopted some form of climate action plan—a document pledging GHG reductions across the transportation sector, buildings, and other infrastructure. But cities often struggle to specify detailed strategies to execute on these pledges, including how to pay for them. A lack of fiscal, technical, and programmatic capacity is a huge barrier to action. However, regional planning and action can help overcome these barriers in cities’ climate plans. This requires local leaders go beyond individual cities (or the urban core of a metro area) and coordinate with the suburban and exurban areas surrounding them. Doing so can address some of the country’s largest GHG emissions sources, which stretch across different jurisdictions—car-centric transportation networks, for instance—while also promoting scalable strategies in the process. And there is a huge federal carrot to doing so, including a recent Biden administration initiative that commits $250 million toward regional planning efforts and other innovative climate proposals. The need for suburbs to take climate action is obvious, especially given their role fueling the sprawl that has exacerbated many of the country’s climate challenges over the last several decades. Yet the need for action might be even more urgent in exurbs, which are located at the fringes of major metro areas and have seen dramatic growth in recent years. GHG emissions continue to surge out of these areas, where larger single-family homes consume more energy, there is a greater reliance on personal vehicles, and spread-out communities lead to less walking and biking as well as longer drives. In metropolitan regions, suburbs emit up to four times the household emissions of their urban cores. While households located in more densely populated neighborhoods have a carbon footprint 50% below the national average, those in the suburbs emit up to twice the average. In metro areas such as New York, GHG emissions in these outlying jurisdictions are readily apparent: Emissions in Manhattan average lower than 38 tons per household annually, but in exurban jurisdictions such as Sussex County, N.J., these emissions exceed 66 tons per household annually. It’s also important to note that looking just at households does not fully capture all the emissions spilling out of these areas, such as those from far-flung industrial, commercial, and logistics activities. Fundamentally rethinking existing patterns of development may require widespread zoning changes, extensive building retrofits, and a surge in funding to consider new, more sustainable public infrastructure improvements. Yet political leadership may have an unwillingness to even consider new approaches. Suburbs and exurbs may be adding to the country’s climate challenges, but they are also central to addressing them. After all, higher GHG emissions in these areas are not just the result of individual households and businesses—they are largely determined by the decisions of planners and policymakers. The amount of emissions depends on whether public transportation is available in neighborhoods, prevailing land uses, the design of buildings, and the types of energy sources available. Greater regional coordination on these factors can help advance climate action. City, suburban, and exurban leaders can work together across a variety of fronts, including: No city is an island. The relationship between urban cores and the suburbs and exurbs surrounding them should not be a one-way street in which only the central cities guide climate action. Leaders across entire regions need to look beyond the planning efforts of individual jurisdictions and test out new measurement approaches, investments, and collaborations to jump-start climate action at a greater geographic scale. Doing so not only offers promise in bridging climate planning gaps today, but also promoting more sustainable outcomes for years to come. Authors The Brookings Institution is committed to quality, independence, and impact.We are supported by a diverse array of funders. In line with our values and policies, each Brookings publication represents the sole views of its author(s). Climate Change Climate Policy Brookings Metro Brookings Initiative on Climate Research and Action Promoting equitable and effective climate action in every community Matt Posner, Xavier de Souza Briggs September 15, 2025 Anthony F. Pipa September 9, 2025 Matthew A. Kraft, Sohil Malik, Grace T. Falken August 26, 2025 Get the latest from Brookings The Brookings Institution is a nonprofit organization based in Washington, D.C. Our mission is to conduct in-depth, nonpartisan research to improve policy and governance at local, national, and global levels. Copyright 2025 The Brookings Institution
3295	https://www.msdmanuals.com/professional/dental-disorders/dental-emergencies/postextraction-problems	Postextraction Problems - Dental Disorders - MSD Manual Professional Edition honeypot link skip to main content Professional Consumer Professional edition active ENGLISH MSD Manual Professional Version MEDICAL TOPICSRESOURCESCOMMENTARYPROCEDURESQUIZZESABOUT US MEDICAL TOPICSRESOURCESCOMMENTARYPROCEDURESQUIZZES Professional/ Dental Disorders/ Dental Emergencies/ Postextraction Problems/ OTHER TOPICS IN THIS CHAPTER Overview of Dental Emergencies Fractured and Avulsed Teeth Mandibular Dislocation Postextraction Problems Medication-Related Osteonecrosis of the Jaw (MRONJ) Postextraction Problems ByJohn Safar, DDS, MAGD, ABGD, Texas A&M University College of Dentistry Reviewed ByDavid F. Murchison, DDS, MMS, The University of Texas at Dallas Reviewed/Revised Modified Nov 2024 v950445 View Patient Education Topic Resources 3D Models (0) Audios (0) Calculators (0) Images (1) Tables (0) Videos (0) Osteomyelitis Postextraction problems are a subset of dental emergencies that require immediate treatment. These problems include Swelling and pain Bleeding Alveolitis Osteomyelitis Osteonecrosis of the jaw Swelling and pain Swelling is normal after oral surgery and is proportional to the degree of manipulation and trauma. An ice pack (or a plastic bag of frozen peas or corn, which adapts to facial contours) should be used for the first day. Cold is applied for 25-minute periods every hour or 2. If swelling persists or increases after 3 days, or if the pain becomes severe, then the patient should be referred back to their dentist or surgeon (1). Postoperative pain varies from moderate to severe and is treated with analgesics (see Treatment of Pain). Bleeding Postextraction bleeding usually occurs in the small vessels. Any clots extending out of the socket are removed with gauze, and a 4-inch gauze pad (folded) or a tea bag (which contains tannic acid) is placed over the socket. Then the patient is instructed to apply continuous pressure by biting for 1 hour. The procedure may have to be repeated 2 or 3 times. Patients are told to wait at least 1 hour before checking the site so as not to disrupt clot formation. They also are informed that a few drops of blood diluted in a mouth full of saliva appear to be more blood than is actually present. If bleeding continues, the site may be anesthetized by nerve block or local infiltration with 2% lidocaine containing 1:100,000 epinephrine. The socket is then curetted to remove the existing clot and to freshen the bone and is irrigated with normal saline. Then the area is sutured under gentle tension. Local hemostatic agents, such as oxidized cellulose, topical thrombin on a gelatin sponge, or microfibrillar collagen, may be placed in the socket before suturing.If bleeding continues, the site may be anesthetized by nerve block or local infiltration with 2% lidocaine containing 1:100,000 epinephrine. The socket is then curetted to remove the existing clot and to freshen the bone and is irrigated with normal saline. Then the area is sutured under gentle tension. Local hemostatic agents, such as oxidized cellulose, topical thrombin on a gelatin sponge, or microfibrillar collagen, may be placed in the socket before suturing. In most cases, patients taking anticoagulants (eg, aspirin, clopidogrel, warfarin, direct-acting oral anticoagulants) need not stop therapy before dental surgery (In most cases, patients taking anticoagulants (eg, aspirin, clopidogrel, warfarin, direct-acting oral anticoagulants) need not stop therapy before dental surgery (2). In those who are at increased risk of bleeding due to comorbid disease or in those undergoing more extensive procedures, consulting with their patient's physician about timing of antiplatelet or anticoagulant dosing or a brief 24- to 48-hour interruption in therapy is indicated. Postextraction alveolitis (dry socket) Postextraction alveolitis is pain emanating from bare bone if the socket’s clot lyses. Although this condition is self-limited, it is quite painful and usually requires some type of intervention. It is much more common among people who smoke or use oral contraceptives and occurs mainly after removal of mandibular molars, usually wisdom teeth. Typically, the pain begins on the second or third postoperative day, is referred to the ear, and lasts from a few days to many weeks. The socket should be rinsed with saline (chlorhexidine may be used for debridement only). For pain relief, a topical local anesthetic gel can be applied or a local anesthetic can be injected. Another option for symptom relief has been to place a 1- to 2-inch iodoform gauze strip saturated in eugenol (an analgesic) or coated with an anesthetic ointment, such as lidocaine 2.5% or tetracaine 0.5%, into the socket (The socket should be rinsed with saline (chlorhexidine may be used for debridement only). For pain relief, a topical local anesthetic gel can be applied or a local anesthetic can be injected. Another option for symptom relief has been to place a 1- to 2-inch iodoform gauze strip saturated in eugenol (an analgesic) or coated with an anesthetic ointment, such as lidocaine 2.5% or tetracaine 0.5%, into the socket (3). The gauze is changed every 1 to 3 days until symptoms do not return after the gauze is left out for a few hours. More recently, a commercially available mixture of butamben (an anesthetic), eugenol, and iodoform (antimicrobial) has become more commonly used. Although not resorbable, this mixture washes out of the socket spontaneously after a few days. These procedures typically eliminate the need for systemic analgesics, although nonsteroidal anti-inflammatory drugs (NSAIDs) may be given if additional pain relief is needed. Patients should follow up with a dentist in 24 hours. Osteomyelitis Osteomyelitis, which in rare cases is confused with alveolitis, is differentiated by fever, local tenderness, and swelling. If symptoms last a month, a sequestrum (ie, localized area of necrotic bone), which is diagnostic of osteomyelitis, should be sought by dedicated dental radiograph or CT scan. Osteomyelitis requires long-term treatment with antibiotics effective against both gram-positive and gram-negative organisms and referral to an oral surgeon for monitoring and/or definitive care. Osteomyelitis Image Image courtesy of Byron (Pete) Benson, DDS, MS, Texas A&M University Baylor College of Dentistry. Osteonecrosis of the jaw (ONJ) Osteonecrosis of the jaw is an oral lesion involving persistent exposure of mandibular or maxillary bone, which usually manifests with pain, loosening of teeth, and purulent discharge. ONJ may occur after dental extraction but also may develop after trauma or radiation therapy to the head and neck. Medication-related ONJ (MRONJ) refers to the association discovered between use of antiresorptive agents and ONJ. These agents include antiresorptive medications (eg, bisphosphonates, denosumab) and antiangiogenic medications (eg, sirolimus, bevacizumab). Although similar in clinical course to osteonecrosis of the jaw, MRONJ is not always preceded by a tooth extraction or other oral surgery. (MRONJ) refers to the association discovered between use of antiresorptive agents and ONJ. These agents include antiresorptive medications (eg, bisphosphonates, denosumab) and antiangiogenic medications (eg, sirolimus, bevacizumab). Although similar in clinical course to osteonecrosis of the jaw, MRONJ is not always preceded by a tooth extraction or other oral surgery. Management of osteonecrosis of the jaw is challenging and typically involves palliation, limited debridement, antibiotics, and oral rinses. References Lockhart PB, Tampi MP, Abt E, et al. Evidence-based clinical practice guideline on antibiotic use for the urgent management of pulpal- and periapical-related dental pain and intraoral swelling: A report from the American Dental Association.J Am Dent Assoc. 2019;150(11):906-921.e12. doi:10.1016/j.adaj.2019.08.020 American Dental Association Library & Archives, Research Services and Scientific Information. Oral anticoagulant and antiplatelet medications and dental procedures. Key Points. Accessed November 1, 2024. Daly BJ, Sharif MO, Jones K, Worthington HV, Beattie A. Local interventions for the management of alveolar osteitis (dry socket).Cochrane Database Syst Rev. 2022;9(9):CD006968. Published 2022 Sep 26. doi:10.1002/14651858.CD006968.pub3 Test your Knowledge Take a quiz on this topic Copyright © 2025 Merck & Co., Inc., Rahway, NJ, USA and its affiliates. All rights reserved. About us Disclaimer Cookie Preferences Copyright© 2025 Merck & Co., Inc., Rahway, NJ, USA and its affiliates. All rights reserved. Find In Topic This Site Uses Cookies and Your Privacy Choice Is Important to Us We suggest you choose Customize my Settings to make your individualized choices. Accept Cookies means that you are choosing to accept third-party Cookies and that you understand this choice. See our Privacy Policy Customize my Settings Reject Cookies Accept Cookies
3296	https://www.datacamp.com/tutorial/demand-curve	Skip to main content The Demand Curve: How to Understand the Relationship Between Price and Sales Learn how to use the demand curve to reveal a relationship between price and quantity sold, and use this knowledge to guide smarter decisions in competitive markets. Explore methods to find the optimal price to maximize overall revenue. Updated Apr 10, 2025 · 14 min read The demand curve is one of the most interesting and useful tools for understanding consumer behavior and pricing strategies. It’s commonly used for analyzing the relationship between the price of a good or service and the quantity sold, although in other contexts it can be used to analyze market dynamics or forecast revenue outcomes. In this article, we will learn all about the demand curve. We will explore both the defining characteristics and related terms. Later on, I will also find time to show how the demand curve can be extended to find the optimal price that maximizes profit - a useful bit of applied algebra to create an elementary optimization technique that I found useful in my own work. By following these methods, I guarantee that you will be able to not only make decisions on pricing strategies but you will also be to able to clearly articulate the methods in a way that provides the framework for future projects. While I’ll be using R and SQL in this article, know that you can use these same methods with other major tools out there, like Excel, Python, Tableau, and Power BI. Finally, I want to say, to really become a pricing expert, I strongly recommend our course, Forecasting Product Demand in R. What is a Demand Curve? A demand curve is a fundamental concept in pricing strategy and economics more generally. It is a compelling visual way to represent the relationship between the price of a good or service and the quantity demanded by consumers. The main components of a demand curve include the price on the one axis and the quantity demanded on the other. Here is one example: One basic demand curve shape. Image by Author. Components of a demand curve In the above graph, you will notice something that is somewhat obvious at first glance: that price and quantity sold have an inverse relationship. More formally, we could say that the demand curve relates very much to what is known as the law of demand, which states that, all else being equal, as the price of a product decreases, the quantity demanded increases. Sometimes, this same idea is referred to as a price and quantity relationship, meaning that the price of a product directly influences how much consumers are willing to buy. However, this particular graph also contains something a little less obvious, which is that the quantity sold falls more precipitously when the price is low and flattens out when the price is high. The curvature here relates to other concepts, like changing elasticity and diminishing marginal utility which I’ll cover in the next section. Price elasticity of demand One key concept that emerges from the inverse relationship between price and quantity demanded is the price elasticity of demand, which measures how sensitive the quantity demanded is to a change in price. Formally, elasticity is often calculated as: A product is said to have elastic demand if a relatively small change in price leads to a large change in quantity demanded. High-end electronics or designer handbags might see a sharp drop in demand if their prices rise even a bit, because consumers can find substitutes. A product has inelastic demand if large changes in price lead to relatively small changes in the quantity demanded. Necessities such as gasoline or medicine are classic examples. Even if the price goes up, consumers are still compelled to buy nearly the same amount because there are no alternatives. Types of demand curves Although I showed one example of a demand curve, do know that it can take on many different forms depending on the underlying characteristics of the market, the product, and consumer preferences. While it is always downward-sloping, this relationship does not have to follow any single mathematical shape. Linear demand curves Linear demand curves feature a constant rate of change between price and quantity. These would be common in simple market scenarios. They are represented by a straight line. Non-linear demand curves As we saw, demand curves aren’t always drawn as perfectly straight lines because the relationship between price and quantity is not always well represented by a simple one-to-one linear function. There are several reasons why, which I started to mention earlier: Changing elasticity: Demand elasticity, meaning how sensitive demand is to a change in price, often varies at different points along the curve. For example, when prices are very high, buyers may react strongly to even a slight price drop. Diminishing marginal utility: In many economic models, each additional unit of a good provides a bit less utility to the consumer than the previous one. As a result, consumers might be willing to buy a lot more if prices fall from high to moderate levels, but once you hit a certain range, lowering the price further doesn’t spark the same big increase in quantity wanted or sold. In practice, non-linear demand curves may be more realistic because they are better at depicting a variable rate of change. After all, real-world data is messy, and consumer behavior is complex. That said, there are often ways to linearize a non-linear curve, as I will show later on. Individual demand curve vs. market demand curve We can also compare differences in demand at the consumer and market level. An individual demand curve represents the relationship between the price of a product and the quantity demanded by a single consumer. It reflects personal preferences and willingness to pay. A market demand curve combines the demand from all consumers in a market. It aggregates individual demand curves, showing the total quantity demanded at each price point. Because of how they are constructed, market demand curves are used by businesses to gauge market potential. Demand curve vs. supply curve If you are studying demand curves, you might have also heard of what is called a supply curve. Essentially, a supply curve and demand curve are two sides of the same coin. While the demand curve shows how much of a product consumers are willing to buy at different prices, the supply curve represents how much producers are willing to sell at those same prices. Now, if you plot both a demand curve and a supply curve on the same chart, you will see a place where they cross. The intersection of these curves, known as the equilibrium point, is thought to determine the market price and quantity. So, when demand outpaces supply, prices tend to rise, which incentivizes producers to produce more. But when supply exceeds demand, prices drop, which encourages more buying. Demand curve vs. supply curve. Image by Author. Using a Demand Curve in Data Science There are a lot of different ways to make use of demand curves to make business decisions. Here, I’m going to walk you through one specific way we can make demand curves useful: We will use our data to create both a demand curve and a demand model using a quadratic regression technique. I’ll show how you can use this technique in both R and SQL, where it can be used in interactive Power BI or Tableau dashboards. Before I continue, I want to say that, if you are reading this and think you might want to upskill an entire team at once, such as a team of department of business analysts or data scientists, DataCamp for Business is here to help. We will help your team build skills and techniques that are helpful in your specific business context. A lot of heavy data analysis projects require multiple contributors - so reach out to our DataCamp team to learn more. Advance Your Team's Data Science Skills Unlock the full potential of data science with DataCamp for Business. Access comprehensive courses, projects, and centralized reporting for teams of 2 or more. Request a Demo Today! Using a demand curve to analyze prices in R Let’s create and analyze a demand curve using R. To simulate a dataset, I’ve created an exponential decay-type curve with scattered noise around it. The idea here is that each data point represents a week of sales for a car dealership, where the dealership offers different specials and tries out different pricing levels for the same type of car as an attempt to increase sales. ``` Load necessary library library(tidyverse) Define parameters for the exponential decay a <- 100 # Initial value b <- 0.5 # Decay rate Define a realistic price range price <- seq(10000, 40000, length.out = 200) # Price from $10,000 to $40,000 Calculate the exponential decay y <- a exp(-b (price / 10000)) # Adjust decay rate to match price scaling Add mixed noise (additive + slight proportional) set.seed(42) # For reproducibility additive_noise <- rnorm(length(y), mean = 0, sd = 1) # Constant additive noise proportional_noise <- rnorm(length(y), mean = 0, sd = 0.05 y) # Small proportional noise y_noisy <- y + additive_noise + proportional_noise Ensure no negative values (optional, as sales can't be negative) y_noisy[y_noisy < 0] <- 0.01 Combine price and noisy y into a data frame demand_curve_data <- data.frame( price = price, quantity_sold = y_noisy ) Clean and prepare the data demand_curve_data <- demand_curve_data %>% mutate(quantity_sold = round(quantity_sold, 0)) %>% filter(quantity_sold != 0) Plot the noisy exponential decay ggplot(demand_curve_data, aes(x = price, y = quantity_sold)) + geom_point(color = '#203147', alpha = 0.6) + geom_line( aes(y = a exp(-b (price / 10000))), color = '#01ef63', size = 1.2, linetype = "dashed" ) + labs( title = "Cars Sold", subtitle = "Demand Curve (Cars Sold)", x = "Price ($)", y = "Quantity Sold" ) + scale_x_continuous(labels = scales::dollar_format()) # Format x-axis as dollars Powered By Was this AI assistant helpful? ``` Demand curve with noise. Image by Author Here, I’ve graphed a relationship between price and quantity sold, and I used scatterplot points to represent natural variation, which could be caused by many different reasons. Reading this graph, we see that as price increases, sales decrease, and this relationship is more sensitive in some price ranges than others. But, depending on what your doing with the data, this general insight might not be enough to inform a practical business decision. In my experience, bosses often want to see a normative insight that provides a concrete answer to a more pressing business question. One such question is: What is the car price that actually maximizes overall revenue? We could start by trying to make some guesses just by looking at the graph. On the low end, we see, following the green line, that the quantity sold is 60 where the x value for price is $10,000. This would translate to $600,000 in total revenue. On the high end, we see the green line hit 10 or so where the x values are $40,000, equaling $400,000 in revenue. Right in the middle, we see the green line hit 30 where the price is $25,000. This equals $750,000. You might be starting to get the feeling now that the price that maximizes overall revenue is somewhere in the middle. Conceptually, this makes sense: We expect super low prices to yield a high volume of sales but relatively small revenue, while super high prices would yield much fewer sales and still also relatively small revenue. But somewhere in the middle there must be an optimal price that maximizes revenue. Now, instead of eyeballing the graph, let’s learn a better, more efficient and solid way to maximize overall revenue. We can figure out the optimal price by solving a rudimentary optimization problem. We first create a linear model, then bend that model into a parabola, and find the highest point. Our first step here is to linearize the relationship. Because I constructed all of this with an exponential decay function, I know it will be linearized well by taking the logarithm of quantity sold. For simplicity, let’s ignore any heteroscedasticity. ``` Create a linearized version of the data demand_curve_data$log_quantity_sold <- log(demand_curve_data$quantity_sold) Plot the linearized relationship ggplot(demand_curve_data, aes(x = price, y = log_quantity_sold)) + geom_point(color = '#203147', alpha = 0.6) + geom_smooth(method = "lm", color = '#01ef63', se = FALSE) + labs( title = "Cars Sold", subtitle = "Demand Curve (Log Quantity Sold)", x = "Price ($)", y = "Log of Quantity Sold" ) + scale_x_continuous(labels = scales::dollar_format()) # Format x-axis as dollars Powered By Was this AI assistant helpful? ``` Transformed demand curve. Image by Author Now, we can create a new variable called revenue which is defined as price times the log_quantity_sold. As a note, we create a linear model object using the lm() function because we want access to the numbers associated with the model coefficients. For our next step, we insert our linear model equation into our formula. If we had had no log transform: This would mean that our equation: revenue = price quantity_sold can now be written as revenue = price (b0 + b1 price). Then, we would distribute terms. Just like we would distribute terms in x(1-x) to make x-x2, we would rewrite our equation as, revenue = b0 Price + b1 price2. Notice now that we would have revenue defined by an equation that has price as a squared term. This is really the key part because, by graphing price as a squared term, we have now created a maximizing function. However, since we do have a log transform, we have to do anexponential back-transform of the log-linear model. This means that our equation: revenue = price log_quantity_sold is now going to be written as revenue = price eb0 + b1 price. We can now factor this equation into a new form, if we prefer: revenue``=``price price price eb0 eb0 eb0 b0 b0 b0 b0 b0 b0 b0 0 0 0 0 0 eb1⋅price eb1⋅price e b1⋅price b1⋅price b1⋅price b1⋅price b1⋅price b1⋅price b1 b 1 1 1 1 1 1 ⋅ price price. But this particular equation doesn't distribute in the same way as a linear or polynomial equation because the exponential function does not allow terms to be separated or distributed like the more standard algebraic terms, above. At any rate, let's continue: ``` Fit a linear model to the log-transformed quantity sold model <- lm(log_quantity_sold ~ price, data = demand_curve_data) Create a revenue model based on the linear model coefficients revenue_model_data <- demand_curve_data %>% mutate( revenue_model = price exp(coef(model) + coef(model) price) # Modeled revenue ) Plot the revenue model ggplot(revenue_model_data, aes(x = price, y = revenue_model)) + geom_point(color = '#203147', alpha = 0.6) + # Actual revenue points geom_line(color = '#01ef63', size = 1.2) + # Revenue model line labs( title = "Car Sales", subtitle = "Revenue Model", x = "Price ($)", y = "Revenue ($)" ) + scale_y_continuous(labels = scales::dollar_format()) + # Format y-axis as dollars theme_minimal() + # Apply a minimal theme for aesthetics theme( text = element_text(family = "Arial", size = 12), plot.title = element_text(face = "bold", size = 16), plot.subtitle = element_text(size = 14) ) Powered By ``` Revenue model created from a demand curve. Image by Author. Interestingly, even though the original model used log⁡_quantity_sold, the graph we have created is entirely on the original scales of price and revenue. The log transform only affects the underlying model, and our back-transform involving e put our graph on the original scale so it's still easy to interpret. Finding where the derivative equals zero Now, our final key point is this: The place at the top of the parabola, where the derivative equals zero, is going to be the theoretical price that maximizes revenue. We might remember that we find the derivative of an equation like 3x² + 2x by multiplying with the exponents to get 6x + 2. This same reasoning applies, but again, remember, in this example, we first have to back-transform our logged values before forming the revenue function. By translating back to the original scale before finding the revenue maximum, we know we are finding the price that maximizes actual revenue, not some transformed metric. So this is the correct revenue function to differentiate: revenue = price e(b0 + b1 price). Let’s work it out. Here, I take the linear model coefficients. I then define a revenue function. Finally, I calculate the optimal price and learn about the associated expected total revenue. ``` Model coefficients b0 <- 4.596 # Intercept b1 <- -4.974e-05 # Slope for price Define the revenue function revenue <- function(P, b0, b1) { P exp(b0 + b1 P) } Calculate the optimal price (analytical solution from dR/dP = 0) - shortcut code here (optimal_price <- -1 / b1) Calculate maximum revenue (max_revenue <- revenue(optimal_price, b0, b1)) Powered By ``` ``` 20104.54 732853.5 Powered By Was this AI assistant helpful? ``` You might have noticed that I had one line of code for the optimal price that wasn't well explained in the context. I'm referring to this part of the code: (optimal_price <- -1 / b1). Let me take a minute to show you how I found this out, that the optimal price is -1 minus the slope. First, we have to remember what is called the product rule in calculus, which states that, if the function of x "f of x" is equal to u(x) times v(x), then the the derivative, which we can write as f'(x), is equal to the derivative of the first thing multiplied by the section thing, then added to the first thing times these derivative of the second thing. ``` the product rule in calculus f(x) = u(x) v(x) f'(x) = u'(x) v(x) + u(x) v'(x) Powered By ``` Here is our equation again, along with the derivative: ``` our equation: revenue = price e^(b0 + b1 price) the derivative of our equation derivative = 1 e^(b0 + b1 price) + price b1 e^(b0 + b1 price) Powered By ``` I know that the derivative of price here is 1 because that is the thing we are finding the derivative of. "How much does price change as price changes?" Well, it's a one-to-one relationship. On the other side, the derivative of e(b0 + b1 price) is not neat like a simple function is neat, like 2x2, for example, which has as its derivative 4x. In the expression that we are working with here, the exponent is a polynomial. The trick here is to look closely at all the terms: To start, b0 is just a single value, so the derivative there is zero price is 1, as we already handled. b1 is the scaling factor, so we put that number in front, then, as per rule, we leave the polynomial exponent the same. So the derivative of e(b0 + b1 price) is b1 e(b0 + b1 price). This is known as the chain rule, as I'll show here: ``` the chain rule in calculus f(x) = x^g(x) f'(x) = g'(x) x^g(x) Powered By ``` The last thing is to do some factoring. Algebra tells us that these are the same: ``` A + B A = A(1+B) Powered By ``` And when I look closely at the derivative I'm working with, I realize there's an A and a B part: The "A" is e(b0 + b1 price) and the "B" is price b1. So I can factor this ``` e^(b0 + b1 price) + price b1 e^(b0 + b1 price) Powered By ``` to this: ``` e^(b0 + b1 price)(1 + price b1) Powered By ``` Finally, I have to set this derivative to zero to find the place where the slope stops going up and starts going down, which is the top of the parabola. ``` e^(b0 + b1 price)(1 + price b1) = 0 Powered By ``` As a final trick, I know that, when A B equals zero, either A or B (or both) have to equal zero. But conveniently, I know that e(b0 + b1 price) can never equal zero because it's an exponent. If it were positive, the number would be large; if the exponent evaluated to 0, the result would be 1; if the exponent were negative, the result would be a small fraction. So I can discard half when solving for zero. The only thing I have to do now is move things around and solve for price. ``` (1 + b1 price) = 0 b1 price = -1 price = - 1 / b1 Powered By ``` This is how I found -1 / b1 to put into our R code custom-built function. Now, to pick up where we left off, we can graph the optimal price and see its associated revenue. The optimal price is $20,104.54, which yields an expected $732,853.50 in total revenue, more or less. I'm saying 'more or less' because, if you check the math, we are assuming 36.45 cars sold, but we can't sell half a car. ``` ggplot(revenue_model_data, aes(x = price, y = revenue_model)) + geom_point() + geom_line(color = '#01ef63') + ggtitle("Car Sales") + labs(subtitle = "revenue model") + ylab("Revenue") + scale_y_continuous(labels = scales::dollar_format()) + geom_vline(xintercept = optimal_price, linetype = "dashed", color = '#203147') + geom_point(aes(x = optimal_price, y = max_revenue), color = '#203147', size = 3) Powered By Was this AI assistant helpful? ``` Finding the price that maximizes revenue. Image by Author. In another workflow, we could even extend these ideas to consider profit instead of revenue. I won’t work it all out here, but to do this, know that we might consider profit = revenue - cost, where cost is considered as quantity unit cost. If unit cost is constant and quantity can be replaced by an expression like 1 - price, then we get an upside down parabola again. If you are interested in learning more about how calculus and derivatives feature in machine learning, try our Introduction to Deep Learning with PyTorch course which explores gradient descent in detail in the context of hyperparameter tuning. The context is different, but you can apply some similar thinking. Optimization, in its different forms, as you are seeing, is a key skill in data science, and practice helps. Demand curve vs. Laffer curve If you’ve studied economics, you might recognize this parabola as something similar to the Laffer curve, which is a somewhat controversial concept that illustrates how tax rates influence government tax revenue. While I would consider both the Laffer curve and this transformation on the demand curve to be “maximize vs. rate” type problems, the Laffer curve is specific to taxation: the x-axis would represent the tax rate (0% to 100%), and the y-axis would be the total tax revenue collected. Conceptually, the Laffer curve is also an inverse-U shape because, at the high end, we would consider that, if the tax rate were too high, it would yield no revenue because there would be no incentive for individuals or businesses to engage in taxable activities. Economists might disagree on that 'incentive' part. The Laffer curve in Ferris Bueller's Day Off. Source: YouTube Using a demand curve to analyze prices in SQL Now that we’ve gone through the idea, I want to also share a SQL script that will do the same work. This function will perform a regression on quantity_sold versus price. It will then allow you to find the optimal price to maximize revenue based on the regression coefficients, and determine the corresponding maximum revenue. I think the following script will be particularly useful when creating dynamic dashboards in Power BI or Tableau because you will have real-time data updates without the need to take anything offline. You will see also that I used a manual approach to find the regression coefficients using SQL aggregate functions, since most SQL dialects do not have specialized regression functions that you can use. You might be surprised to know that simple linear regression works well in SQL because the model coefficients for simple linear regression can be expressed in terms of the correlation, standard deviation, and mean values between x and y. ``` -- SQL script to perform linear regression on log(quantity_sold) vs price, -- calculate the optimal price to maximize revenue, and determine the maximum revenue. WITH regression_data AS ( SELECT COUNT() AS N, SUM(price) AS sum_x, SUM(LOG(quantity_sold)) AS sum_y, SUM(price LOG(quantity_sold)) AS sum_xy, SUM(price price) AS sum_x2, SUM(LOG(quantity_sold) LOG(quantity_sold)) AS sum_y2 FROM sales_data ), regression_coefficients AS ( SELECT N, sum_x, sum_y, sum_xy, sum_x2, sum_y2, (N sum_xy - sum_x sum_y) / (N sum_x2 - sum_x sum_x) AS slope, (sum_y - ((N sum_xy - sum_x sum_y) / (N sum_x2 - sum_x sum_x)) sum_x) / N AS intercept FROM regression_data ), optimal_price AS ( SELECT slope, intercept, (-1 / slope) AS optimal_P FROM regression_coefficients ), max_revenue AS ( SELECT optimal_P, slope, intercept, optimal_P EXP(intercept + slope optimal_P) AS max_R FROM optimal_price ) SELECT optimal_P AS Optimal_Price, max_R AS Maximum_Revenue FROM max_revenue; Powered By ``` ``` Optimal_Price \| Maximum_Revenue ------------- \| ---------------- 20104.32 \| 402082.58 Powered By Was this AI assistant helpful? ``` Please know that the SQL script provided might need adjustments depending on the SQL dialect used, as the syntax can vary. Also, for troubleshooting purposes, I would make sure that your sales data table contains valid and positive values for both price and quantity_sold, since negative or zero values can lead to errors. Also, if you wanted to incorporate a log transform, as we did in the R section, you could do that, but, depending on your dialect, the natural logarithm function might be named differently, either LN or LOG. In DuckDB, LN and LOG are actually different: LN is the natural logarithm e, while LOG is base 10. If you are interesting in creating a demand curve to analyze the relationship between price and sales in SQL, tune in to my Analyzing Product Demand in SQL code-along. Finally, I would say that, to enhance your SQL proficiency as an analyst, you should explore the Associate Data Analyst in SQL career track, so you develop a full set of tools. Exceptions to the Demand Curve There are notable exceptions where consumer behavior can defy the standard pattern of the demand curve: Giffen goods Giffen goods are a rare type of inferior good for which the law of demand does not hold. When the price of a Giffen good rises, consumers may paradoxically buy more of it rather than less. This occurs because the good makes up such a large part of a consumer’s budget that when its price goes up, the consumer can no longer afford more desirable substitutes and is forced to purchase even more of the Giffen good. A classic (though debated) historical example is the rise in bread prices during the 19th century in some places (England, maybe), where people ended up consuming more bread because they could not afford other, more expensive foods. Veblen goods Veblen goods are luxury or status-symbol products—such as designer handbags, fine wines, or high-end watches—whose demand increases as price increases. Part of their appeal comes from the perception that a higher price signals exclusivity or higher social status, making the good more desirable as it becomes more expensive. You might think of this as “conspicuous consumption.” Other special cases Bandwagon or network effects: Some goods become more valuable to each consumer as more people use them (e.g., social media platforms, certain software ecosystems). Demand can rise not just because of price but because of the good’s growing popularity or network size. Snob effect: Some consumers purchase goods precisely because few other people have them. In these cases, lower prices could make the good seem “less exclusive,” decreasing the appeal. Speculative or investment assets: For assets like certain cryptocurrencies or commodities, if the market believes prices will continue rising, demand can jump in response to price increases, rather than fall. In all of these examples, the usual price–quantity relationship flips or becomes more complex due to psychological, cultural, or situational factors that override the classic assumption that consumers simply want “more at lower prices.” Demand Curve Common Terms We considered a few different terms in this article. Let’s create a table to help keep this all straight. \| Concept \| Description \| --- \| \| Demand Curve \| A graphical representation showing the relationship between the price of a good and the quantity demanded by consumers. \| \| Law of Demand \| States that, all else being equal, as the price of a product decreases, the quantity demanded increases. \| \| Price-Quantity Relationship \| Indicates that the price of a product directly influences the quantity consumers are willing to buy. \| \| Elasticity \| Measures how sensitive the quantity demanded is to a change in price. \| \| Diminishing Marginal Utility \| Refers to the decrease in satisfaction or usefulness from consuming additional units of a product. \| Final Thoughts We covered a lot of ground: We practiced everything from demand curves and demand models, to log and exponential transforms, to quadratic regression techniques and optimization. Keep learning business analysis with DataCamp. Consider enrolling in courses like the ones I mentioned earlier: Forecasting Product Demand in R and Data-Driven Decision Making for Business, both of which will teach you how to use data to enhance your decisions in business and finance. There’s no substitute for having some technical skills and knowing the right techniques. Also, keep practicing the core things of data science. In addition to the core regression courses (Introduction to Regression with statsmodels in Python and Introduction to Regression in R) learn Nonlinear Modeling with Generalized Additive Models (GAMs) in R or Statistics Fundamentals in Python. I also recommend undertaking our full Machine Learning Scientist in Python career track. Choose the thing you are weakest in, so you learn the most. Author Josef Waples I'm a data science writer and editor with contributions to research articles in scientific journals. I'm especially interested in linear algebra, statistics, R, and the like. I also play a fair amount of chess! Become an ML Scientist Upskill in Python to become a machine learning scientist. Start Learning for Free Topics Data Analysis Data Science Learn Data Analysis and Data Science with DataCamp Course Forecasting in R 5 hr 50.6K Learn how to make predictions about the future using time series forecasting in R including ARIMA models and exponential smoothing methods. See Details Start Course Course Forecasting Product Demand in R 4 hr 9.3K Learn how to identify important drivers of demand, look at seasonal effects, and predict demand for a hierarchy of products from a real world example. See Details Start Course Course Data-Driven Decision Making for Business 2 hr 25.3K Discover how to make better business decisions by applying practical data frameworks—no coding required. See Details Start Course See More Related blog Data Science in Sales: Customer Sentiment Analysis Learn how data science can be used to analyze customer emotions and deliver valuable insights for sales optimization. Elena Kosourova 9 min Tutorial Simple Linear Regression: Everything You Need to Know Learn simple linear regression. Master the model equation, understand key assumptions and diagnostics, and learn how to interpret the results effectively. Josef Waples Tutorial Learning Curves Tutorial: What Are Learning Curves? Learn about how learning curves can help you evaluate your data and identify optimal solutions. Kurtis Pykes code-along Analyzing Product Demand in SQL Find out how to unlock a new dimension of your analytical power by bringing simple linear regression into SQL! Josef Waples code-along Using Joins to Analyze Book Sales in SQL Learn how to join data together to get it into a state for analysis. Tim Sangster code-along Visualizing Sales Data with PivotCharts in Excel In this code along, we delve into the world of data analysis and visualization using PivotCharts and PivotTables on a sales dataset. Agata Bak-Geerinck See More See More
3297	https://bowelinterestgroup.co.uk/resources/red-flag-symptoms-in-gi-conditions/	Red flag symptoms in GI conditions - Bowel Interest Group This site is intended for UK healthcare professionals only HOME ABOUT MEET THE BOARD CONTACT WEBINARS NEWS RESOURCES SUBSCRIBE Select Page HOME ABOUT MEET THE BOARD CONTACT WEBINARS NEWS RESOURCES SUBSCRIBE Red flag symptoms in GI conditions This was originally published by AllergyAI Education. The original and references can be readhere. What are the red flag symptoms you should look out for? Red flag symptoms that should prompt referral to secondary care: Abdominal masses A family history of ovarian cancer Rectal masses Anaemia Rectal bleeding Unintentional and unexplained weight loss A family history of bowel cancer Aged >60 years change in bowel habit lasting >6 weeks JOIN OUR MAILING LIST SUBSCRIBE Latest Resources Making Sense of NICE Guidance for Bowel Care January 16, 2023 The Cost of Opioid-Induced Constipation – Strategies for Change Webinar October 5, 2022 The Cost of Opioid-Induced Constipation Report July 4, 2022 Effective bowel management beyond laxatives December 15, 2021 Watch BIG’s experts review the Importance of Adherence Report August 31, 2021 Sponsors Sponsors fund meetings of the expert group and pay for the services of E4H to provide secretariat support to the group. Sponsors have no editorial control over the content of expert group materials except for reviewing the materials to ensure compliance with industry codes of ethical conduct. Website owned and developed by E4H Privacy Policy \| Terms & Conditions
3298	https://math.libretexts.org/Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/07%3A_Probability/7.03%3A_Combinations	7.3: Combinations - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7: Probability Contemporary 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Home 2. Bookshelves 3. Applied Mathematics 4. Contemporary Mathematics (OpenStax) 5. 7: Probability 6. 7.3: Combinations Expand/collapse global location Contemporary Mathematics (OpenStax) Front Matter 1: Sets 2: Logic 3: Real Number Systems and Number Theory 4: Number Representation and Calculation 5: Algebra 6: Money Management 7: Probability 8: Statistics 9: Metric Measurement 10: Geometry 11: Voting and Apportionment 12: Graph Theory 13: Math and... 14: Appendix 15: Answer Key Back Matter 7.3: Combinations Last updated Jan 2, 2025 Save as PDF 7.2.0: Exercises 7.3.0: Exercises Page ID 129595 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Combinations: When Order Doesn’t Matter 1. Example 7.3.1: Distinguishing Between Permutations and Combinations 2. Your Turn 7.3.1 Counting Combinations FORMULA Example 7.3.2: Using the Combination Formula Your Turn 7.3.2 Example 7.3.3: Applying the Combination Formula Your Turn 7.3.3 Checkpoint People in Mathematics: Early Eastern Mathematicians Example 7.3.4: Combining Combinations with the Multiplication Rule for Counting Your Turn 7.3.4 Check Your Understanding Figure 7.3.1: Combinations help us count things like the number of possible card hands, when the order in which the cards were drawn doesn’t matter. (credit: “IMG_3177” by Zanaca/Flickr, CC BY 2.0) Learning Objectives Distinguish between permutation and combination uses. Compute combinations. Apply combinations to solve applications. In Permutations, we studied permutations, which we use to count the number of ways to generate an ordered list of a given length from a group of objects. An important property of permutations is that the order of the list matters: The results of a race and the selection of club officers are examples of lists where the order is important. In other situations, the order is not important. For example, in most card games where a player receives a hand of cards, the order in which the cards are received is irrelevant; in fact, players often rearrange the cards in a way that helps them keep the cards organized. Combinations: When Order Doesn’t Matter In situations in which the order of a list of objects doesn’t matter, the lists are no longer permutations. Instead, we call them combinations. Example 7.3.1: Distinguishing Between Permutations and Combinations For each of the following situations, decide whether the chosen subset is a permutation or a combination. A social club selects 3 members to form a committee. Each of the members has an equal share of responsibility. You are prompted to reset your email password; you select a password consisting of 10 characters without repeats. At a dog show, the judge must choose first-, second-, and third-place finishers from a group of 16 dogs. At a restaurant, the special of the day comes with the customer’s choice of 3 sides taken from a list of 6 possibilities. Answer 1. Since there is no distinction among the responsibilities of the 3 committee members, the order isn’t important. So, this is a combination. 2. The order of the characters in a password matter, so this is a permutation. 3. The order of finish matters in a dog show, so this is a permutation. 4. A plate with mashed potatoes, peas, and broccoli is functionally the same as a plate with peas, broccoli, and mashed potatoes, so this is a combination. Your Turn 7.3.1 Decide whether the following represent permutations or combinations: On Halloween, you give each kid who comes to your door 3 pieces of candy, taken randomly from a candy dish. Your class is going on a field trip, but there are too many people for one vehicle. Your instructor chooses half the class to take the first vehicle. (\frac{12!}{{10!}) (\frac{8!}{{4!4!}) Counting Combinations Permutations and combinations are certainly related, because they both involve choosing a subset of a large group. Let’s explore that connection, so that we can figure out how to use what we know about permutations to help us count combinations. We’ll take a basic example. How many ways can we select 3 letters from the group A, B, C, D, and E? If order matters, that number is msub msub P 3=60 5 P 3=60. That’s small enough that we can list them all out in the table below. ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED BAC BAD BAE BCA BCD BCE BDA BDC BDE BEA BEC BED CAB CAD CAE CBA CBD CBE CDA CDB CDE CEA CEB CED DCA DAC DAE DBA DBC DBE DCA DCB DCE DEA DEB DEC EAB EAC EAD EBA EBC EBD ECA ECB ECD EDA EDB EDC Now, let’s look back at that list and color-code it so that groupings of the same 3 letters get the same color, as shown in Figure 7.9: Figure 7.3.2 After color-coding, we see that the 60 cells can be seen as 10 groups (colors) of 6. That’s no coincidence! We’ve already seen how to compute the number of permutations using the formula To compute the number of combinations, let’s count them another way using the Multiplication Rule for Counting. We’ll do this in two steps: Step 1: Choose 3 letters (paying no attention to order). Step 2: Put those letters in order. The number of ways to choose 3 letters from this group of 5 (A, B, C, D, E) is the number of combinations we’re looking for; let’s call that number msub msub C 3 5 C 3 (read “the number of combinations of 5 objects taken 3 at a time”). We can see from our chart that this is ten (the number of colors used). We can generalize our findings this way: remember that the number of permutations of n n things taken r r at a time is msub msub P r=n!(n−r)!n P r=n!(n−r)!. That number is also equal to msub msub C r×r!n C r×r!, and so it must be the case that n!(n−r)!msub msub⁢C r×r!n!(n−r)!=n C r×r!. Dividing both sides of that equation by r!r! gives us the formula below. FORMULA msub msub n⁢C r=n!r!(n−r)! n C r=n!r!(n−r)! Example 7.3.2: Using the Combination Formula Compute the following: 8⁢C 3 12⁢C 5 15⁢C 9 Answer 1. 8⁢C 3=8!3!(8−3)!=8×7×6×5!3×2×1×5!=8×7=56 12⁢C 5=12!5!(12−5)!=12×11×10×9×8×7!5×4×3×2×1×7!=11×9×8=792 15⁢C 9=15!9!(15−9)!=1⁢5×14×13×1⁢2×11×10×9!9!×6×5×4×3×2×1=14×13×11×10 4=5,005 Your Turn 7.3.2 Compute the following: 6⁢C 4 10⁢C 8 14⁢C 5 Example 7.3.3: Applying the Combination Formula In the card game Texas Hold’em (a variation of poker), players are dealt 2 cards from a standard deck to form their hands. How many different hands are possible? The board game Clue uses a deck of 21 cards. If 3 people are playing, each person gets 6 cards for their hand. How many different 6-card Clue hands are possible? Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. How many different sets of winning numbers are possible? Answer 1. A standard deck has 52 cards, and a hand has 2 cards. Since the order doesn’t matter, we use the formula for counting combinations: msub msub C 2=52!2!(52−2)!=52×51×50!2×1×50!=52×51 2=1,326.52 C 2=52!2!(52−2)!=52×51×50!2×1×50!=52×51 2=1,326. Again, the order doesn’t matter, so the number of combinations is: msub msub C 6=21!6!(21−6)!=21×20×19×18×17×16×15!6×5×4×3×2×1×15!=21×19×17×16 2=54,264.21 C 6=21!6!(21−6)!=21×20×19×18×17×16×15!6×5×4×3×2×1×15!=21×19×17×16 2=54,264. There are 38 numbers to choose from, and we must pick 5. Since order doesn’t matter, the number of combinations is: msub msub C 5=38!5!(38−5)!=38×37×36×35×34×33!5×4×3×2×1×33!=501,492.38 C 5=38!5!(38−5)!=38×37×36×35×34×33!5×4×3×2×1×33!=501,492. Your Turn 7.3.3 At a charity event with 58 people in attendance, 3 raffle winners are chosen. All receive the same prize, so order doesn’t matter. How many different groups of 3 winners can be chosen? A sorority with 42 members needs to choose a committee with 4 members, each with equal responsibility. How many committees are possible? Checkpoint The notation and nomenclature used for the number of combinations is not standard across all sources. You’ll sometimes see(n r)(n r)instead of msub msub C r n C r . Sometimes you’ll hear that expression read as “n n choose r r” as shorthand for “the number of combinations of n n objects taken r r at a time.” People in Mathematics: Early Eastern Mathematicians Although combinations weren’t really studied in Europe until around the 13th century, mathematicians of the Middle and Far East had already been working on them for hundreds of years. The Indian mathematician known as Pingala had described them by the second century BCE; Varāhamihira (fl. sixth century) and Halayudha (fl. 10th century) extended Pingala’s work. In the ninth century, a Jain mathematician named Mahāvīra gave the formula for combinations that we use today. In 10th-century Baghdad, a mathematician named Al-Karaji also knew formulas for combinations; though his work is now lost, it was known to (and repeated by) Persian mathematician Omar Khayyam, whose work survives. Khayyam is probably best remembered as a poet, with his Rubaiyat being his most famous work. Meanwhile, in 11th-century China, Jia Xian also was working with combinations, as was his 13th-century successor Yang Hui. It is not known whether the discoveries of any of these men were known in the other regions, or if the Indians, Persians, and Chinese all came to their discoveries independently. We do know that mathematical knowledge and sometimes texts did get passed along trade routes, so it can’t be ruled out. Example 7.3.4: Combining Combinations with the Multiplication Rule for Counting The student government at a university consists of 10 seniors, 8 juniors, 6 sophomores, and 4 first-years. How many ways are there to choose a committee of 8 people from this group? How many ways are to choose a committee of 8 people if the committee must consist of 2 people from each class? Answer 1. There are 28 people to choose from, and we need 8. So, the number of possible committees is msub msub C 8=3,108,105 28 C 8=3,108,105. 2. Break the selection of the committee members down into a 4-step process: Choose the seniors, then choose the juniors, then the sophomores, and then the first-years, as shown in the table below: \| Class \| Number of Ways to Choose Committee Representatives \| --- \| \| senior \| msub msub C 2=45 10 C 2=45 \| \| junior \| msub msub C 2=28 8 C 2=28 \| \| sophomore \| msub msub C 2=15 6 C 2=15 \| \| first-year \| msub msub C 2=6 4 C 2=6 \| The Multiplication Rule for Counting tells us that we can get the total number of ways to complete this task by multiplying together the number of ways to do each of the four subtasks. So, there are 45×28×15×6=113,400 45×28×15×6=113,400 possible committees with these restrictions. Your Turn 7.3.4 How many ways are there to choose a hand of 6 cards from a standard deck with the constraint that 3 are ♠, 2 are ♡, and 1 is ♣? Check Your Understanding Suppose you want to count the number of ways that you can arrange the apps on the home screen on your phone. Should you use permutations or combinations? Your little brother is packing up for a family vacation, but there’s only room for 3 of his toys. If you want to know how many possible groups of toys he can bring, should you use permutations or combinations? Compute 12⁢C 10. Compute 16⁢C 3. You’re planning a road trip with some friends. Though you have 6 friends you’d consider bringing along, you only have room for 3 other people in the car. How many different possibilities are there for your road trip squad? You’re packing for a trip, for which you need 3 shirts and 3 skirts. If you have 8 shirts and 5 skirts that would work for the trip, how many different ways are there to pack for the trip? This page titled 7.3: Combinations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 7.2.0: Exercises 7.3.0: Exercises Was this article helpful? Yes No Recommended articles 7.3.0: Exercises 7.1: The Multiplication Rule for Counting 7.2: Permutations 7.4: Tree Diagrams, Tables, and Outcomes 7.5: Basic Concepts of Probability Article typeSection or PageAuthorOpenStaxLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow Page TOCno Tags source@ © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 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3299	https://mathoverflow.net/questions/143664/about-the-ratio-of-the-areas-of-a-convex-pentagon-and-the-inner-pentagon-made-by	Skip to main content About the ratio of the areas of a convex pentagon and the inner pentagon made by the five diagonals Ask Question Asked Modified 6 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 34 Save this question. Show activity on this post. Question : Letting S′ be the area of the inner pentagon made by the five diagonals of a convex pentagon whose area is S, then find the max of S′S. I've been interested in this simple question. It seems that a regular pentagon and its affine images would give the max, but I'm facing difficulty. Remark : This question has been asked previously on math.SE without receiving any answers. mg.metric-geometry euclidean-geometry plane-geometry polyhedra Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Apr 13, 2017 at 12:19 CommunityBot 122 silver badges33 bronze badges asked Oct 1, 2013 at 8:29 mathlovemathlove 4,79722 gold badges2121 silver badges4949 bronze badges 7 1 You might want to look up "pentagram map." – user25199 Commented Oct 1, 2013 at 9:00 2 @Carl: Thanks, but I'm not sure this is a clue. – mathlove Commented Oct 2, 2013 at 10:56 @mathlove: just out of curiosity, have you experimented with a dynamic geometry program to test the conjecture? – alvarezpaiva Commented Dec 4, 2013 at 12:56 @alvarezpaiva: Well, sort of. To be honest, I would like anyone who knows well about this kind of experimentation to test this conjecture. – mathlove Commented Dec 5, 2013 at 6:26 Thank you everyone. I'm going to give my +50 reputation to Vít Tuček. By the way, if we consider the same question for a convex n-gon, only the n=5 (pentagon) case might be meaningful since in the n≥6 cases S′/S do not seem to have the max. – mathlove Commented Dec 15, 2013 at 7:07 \| Show 2 more comments 4 Answers 4 Reset to default This answer is useful 16 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by mathlove Show activity on this post. Not an answer, just a long comment. This comment slowly evolved into a partial numerical solution: the regular pentagon is the local maximum. I've used Maple to produce the formula for S′S and the result is a rational function on R10 with both numerator and denominator of degree 12. The numerator has 11495 monomial terms while the denominator is factored into product of six degree two polynomials. This means that the symmetry of the problem is somehow lost during the computation and so one must be a bit careful when using such programs. Nevertheless, restricting the pentagon to have the first side on the x-axes I get numerator that factorizes as product of x2 and a polynomial with 182 monomials. While analytical solution is still most probably out of the question, it may be possible to use this for numerical experiments. For example, working on R10, I was able to verify that the regular pentagon is a stationary point. To my surprise, Maple computed the eigenvalues of the Hessian at this point analytically. Four of the eigenvalues are negative, but unfortunately there are six eigenvalues that are numerically zero (≈10−9). I think it could be possible to "prove" this way that the regular pentagon is a local maximum. I'll try to look at it again tomorrow. Edit: Let me summarize the comments so far. As David Speyer mentioned, all pentagons are affinely congruent to those whose vertices lies on a conic. If one uses the rational parametrization of the circle t→(t1+t2,1−t21+t2), then the degree of the resulting polynomials in the Jacobian rise to more than 30. Nevertheless, one can factor out terms like t or t2+1. Of course, we can still fix one vertex and surprisingly, it matters (at least in Maple) which one. Fixing (0,1) as a vertex leads to better results with degrees (29,29,29,29) and number of monomials (2348,2754,2754,2348). This is still unsolvable (at least in Maple on my computer). But weep not for we can at least compute the Hessian at the regular pentagon and get it's eigenvalues (in this case only numerically, analytic computation timed out): −0.174,−0.075,−0.001,−0.003. We can conclude (if we believe Maple) that the regular pentagon is the local maximum among "circle pentagons". Dag Oskar Madsen reminded me that I didn't use all the freedom we have in the problem. By fixing (0,0),(1,0) and (0,1) I obtained rational function in four variables with numerator of degree 8 with 53 terms and denominator of degree 10 factored into a product of 6 terms. The numerators of the Jacobian have the following structure: "Degrees:", [16, 16, 15, 15], "Terms:", [637, 633, 424, 510] from which we see that this resulted in smallest system so far. On the other hand, any symmetry in the equations was most probably destroyed and thus it may be harder to solve it. Fixing vertices 1,2,3 of the regular pentagon with edge 12 on the x-axis I verified that the eigenvalues of the Hessian are indeed negative, namely −0.027,−0.012,−0.148,−0.136. This establishes numerically that the regular pentagon is a local maximum in the set of all pentagons. I've tried to play with reformulation by Noam D. Elkies and treat it like a constrained maximization problem. That is S=12(a+a2−4b−−−−−−√)S′=S−a+∑i mod 5Ai−2Ai+2S−Ai,wherea=∑i mod 5Ai,b=∑i mod 5AiAi+1 and Ai is the area of triangle formed by three consecutive vertices Pi−1PiPi+1. The problem is to maximize S′ relative to S=1. The system for Lagrange multipliers reduces to 5 rational equations with numerators of degree 12 with 1544 terms ∂L∂ai=fi+(λ+∑j mod 5Aj−2Aj+2(1−Aj)2)gi=0, where gi=12(1+Ai−4Ai−1−4Ai+12−a)fi=gi−1+∑j mod 5(Aj−2δi,j+2+Aj+2δi,j−2)(1−Aj)−Aj−2Aj+2(gi−δi,j)(1−Aj)2; plus the constraint a−b=1, which is quadratic with eleven terms. I encourage everybody to try this out in their CAS of choice. I got curious about the current state of the art in Groebner basis and asked another question. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Apr 13, 2017 at 12:58 CommunityBot 122 silver badges33 bronze badges answered Dec 9, 2013 at 5:15 Vít TučekVít Tuček 9,10722 gold badges3131 silver badges6565 bronze badges 13 One idea I had for making the computation simpler was to WLOG that the conic through the vertices of the pentagon is the unit circle, and the first vertex is at (1,0). We can then parametrize the other 4 points as (2ti1+t2i,1−t2i1+t2i), getting a rational function of 4 variables. This still wasn't simple enough to do by hand, but does it turn out nicely when plugged into your function? – David E Speyer Commented Dec 9, 2013 at 5:30 There are two things I don't understand. First, why is it sufficient to constraint on the circle? And second, after plugin it it the numerator has degree 2 less than the denominator. The numerator has degree 20 and the denominator has degree 22. Both are factorized into 6 terms. – Vít Tuček Commented Dec 9, 2013 at 5:56 1 Technically the circle isn't enough: We need to try a circle, a parabola and a hyperbola. Any 5 points are on a conic, and a conic up to affine transformation is either x+y2=1, y=x2 or y2−x1=1. I figured I'd do the circle first because it includes the case we believe to be optimal. Your result, however, suggests that this is not a useful suggestion. – David E Speyer Commented Dec 9, 2013 at 14:06 2 An alternative is to fix three neighbouring vertices at (0,1), (0,0) and (1,0). (We can do this without loss of generality since we only need to consider pentagons up to affine transformation.) Does this make the formula simpler? – Dag Oskar Madsen Commented Dec 9, 2013 at 15:37 2 Fixing the three vertices at (0,0), (1,0) and (0,1) to be in positions 1, 3 and 4 seems a little better: denominator is degree 7, a product of 6 terms; numerator is degree 8 with 49 terms. – David E Speyer Commented Dec 9, 2013 at 22:01 \| Show 8 more comments This answer is useful 14 Save this answer. Show activity on this post. [Corrected (two typos noted by V.T.) and expanded (alternative coordinates)] Also not an answer, but this time a simpler algebraic formulation. All indices are cyclic mod 5. Let the vertices be Pi (imod5), and let Ai be the area of triangle Pi−1PiPi+1. Then S is the larger root of the quadratic equation S2−S∑imod5Ai+∑imod5AiAi+1, and S′=S−∑imod5Ai+∑imod5Ai−2Ai+2S−Ai. So we are to prove that as Ai vary over all positive numbers the ratio S′/S is maximized when the Ai are all equal. The formula for S was obtained by choosing coordinates so that the Pi are at (0,0), (0,1), (1,0), (x2,y2), (x3,y3) (as suggested by Dag Oskar Madsen) and seeking a simple relation between S and the Ai that has the appropriate symmetry in the Ai. In retrospect this amounts to the quadratic relation on the six Plücker coordinates for a 2-dimensional subspace of R4. To prove the formula for S′ we can then argue as follows. Let Qi be the vertex of the inner pentagon opposite Pi. Then S−∑imod5Ai almost gives the area S′ of the inner pentagon except we must add the areas of the five triangles Pi−2QiPi+2 which were subtracted twice. We evaluate the area of this triangle by using the following observation. Let WXYZ be a convex quadrilateral whose diagonals intersect at O. Then Area(XOY)=Area(WXY)Area(XYZ)Area(WXYZ). (Proof: use the area formula 12absinC for the areas of WOX, XOY, YOZ, ZOW.) In our setting WXYZ is Pi+1Pi+2Pi−2Pi−1 and its area is S−Ai. The formula above for S′ is less pleasant to work with than it might look because each denominator S−Ai contains a square root. It might be better to use as coordinates the quadrilateral areas Bi:=S−Ai=12(Pi+1−Pi−2)×(Pi−1−Pi+2) instead of Ai. Since the relation between S and the Ai is homogeneous quadratic, the relation between S and the Bi is also homogeneous quadratic, and indeed it turns out to be given by the same polynomial S2−S∑imod5Bi+∑imod5BiBi+1=0, though this time S is the smaller root. So now we are to prove that the ratio between S′=S−∑imod5(S−Bi)+∑imod5(S−Bi−2)(S−Bi+2)Bi. and S is maximal when all the Bi are equal. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Dec 11, 2013 at 22:49 answered Dec 9, 2013 at 22:43 Noam D. ElkiesNoam D. Elkies 81.1k1515 gold badges287287 silver badges381381 bronze badges 12 By C you mean S or is it something different? Is it clear that if the areas Ai are equal, then the pentagon must be regular? – Vít Tuček Commented Dec 10, 2013 at 17:14 Good questions. 1) Sorry, you're right: "formula for C" should read "formula for S". I'll fix it in the next edit. 2) Yes, if all the Ai are equal then the pentagon must be an affine image of a regular pentagon. More generally, the Ai determine the pentagon up to area-preserving affine transformation. That's because once we've solved for S you know the areas of the other five triangles spanned by the vertices (e.g. P0P2P3 has area S−A1−A4), [cont'd] – Noam D. Elkies Commented Dec 10, 2013 at 17:30 ... and then once you've placed three of the points, say P0 and P±1, you can locate each of the others by intersecting lines parallel to P0P1 and P0P−1. – Noam D. Elkies Commented Dec 10, 2013 at 17:31 Huh? Lines parallel to P0P1 and P0P−1? Anyway, I am inclined to believe the result since we need to place two more points and we have at least three conditions. By the way, there is another C in your post in 12absinC. – Vít Tuček Commented Dec 10, 2013 at 18:44 Yes, parallel. Once you've placed P0 and P1, the area of triangle P0P1Pi is proportional to the distance from Pi to the line P0P1, so knowing the area puts Pi on a line parallel to P0P1. If you also know the area of P0P−1Pi then Pi is the intersection of that line with one parallel to P0P−1. And yes, I know about the other C, but that's just the usual notation for the angle of triangle ABC at vertex C. – Noam D. Elkies Commented Dec 10, 2013 at 19:53 \| Show 7 more comments This answer is useful 5 Save this answer. Show activity on this post. Here's a possible strategy: Consider deformations of a vertex of the convex (outer) pentagon which preserve area. The vertex is constrained to move along a line parallel to the segment connecting the adjacent vertices, and clearly must remain in a compact interval if the pentagon is to remain convex. The area of the inner pentagon may be written as a sum and difference of areas of triangles, each of which has at most one vertex move along a line segment during the deformation, so has area proportional to the length of an edge. The vertices of the inner pentagon are determined by projective transformations of the moving vertex, so have coordinates along the fixed segments given by linear fractional transformations of the displacement. Thus, the area of the pentagon is a sum of rational functions, and it turns out each is concave down on the interval, so the area is a concave rational function with a unique maximum. One should therefore get 5 constraints on the pentagon vertices. Up to affine transformations, there are a 4-parameter family of pentagons (e.g. by fixing 3 vertices, as David Speyer suggested in a comment). So one obtains an overdetermined system, which hopefully has a unique (or finitely many) solutions, one of which is the (affine) regular pentagon. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Dec 11, 2013 at 15:35 Ian AgolIan Agol 70.7k33 gold badges198198 silver badges366366 bronze badges 4 If there is a unique maximum, then I think we are more or less done. I've already computed in Maple that the regular pentagon is a local maximum in the set of "circular pentagons" and right now I am trying to extend this result to all pentagons. What bothers me though is your claim that "The area of the inner pentagon may be written as a sum and difference of areas of triangles, each of which has at most one vertex move along a line segment during the deformation." I can't imagine not using e.g. area of P1P2Q4 (with notation as in the Noam Elkies's answer). – Vít Tuček Commented Dec 11, 2013 at 16:07 @VítTuček: the area formula I'm using is P0P2P3−P0P1P4+P2P3Q0+P0P4Q2+P0P1Q3−P2P3Q1−P3P2Q4. The first 3 areas are constant, and the others have a single vertex moving along a diagonal. If I computed the linear fractional transformations correctly, then the signed areas of these last 4 triangles are concave down as a function of the displacement of P0 along the line parallel to P1P4. – Ian Agol Commented Dec 11, 2013 at 20:12 Also, I'm only claiming concavity in certain directions in the space of pentagons - this does not prove concavity globally, but it might give enough constraints to characterize the maximum. – Ian Agol Commented Dec 11, 2013 at 20:21 Nice! Your formula actually explains why it's better to fix P0,P2,P3 instead of three consecutive vertices. – Vít Tuček Commented Dec 12, 2013 at 3:15 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. A complete solution is available at On the polygon determined by the short diagonals of a convex polygon, Jacqueline Cho, Dan Ismailescu, Yiwon Kim, Andrew Woojong Lee Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Dec 21, 2018 at 2:32 user84909user84909 23111 silver badge33 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 mg.metric-geometry euclidean-geometry plane-geometry polyhedra See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Linked 14 What is the state of art in Groebner bases Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.

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